Advanced Engineering Mathematics Solutions Manual.pdf

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1. -4-2 2 4
t
-4
-2
2
4
X
1
Introduction to Differential Equations
Exercises 1.1
1.
Second-order; linear.
2.
Third-order; nonlinear because of (
dy/dx
)
4
.
3.
The differential equation is first-order. Writing it in the form
x
(
dy/dx
)+
y
2
= 1, we see that it is nonlinear in
y
because of
y
2
. However, writing it in the form (
y
2
−
1)(
dx/dy
)+
x
= 0, we see that it is linear in
x
.
4.
The differential equation is first-order. Writing it in the form
u
(
dv/du
) + (1+
u
)
v
=
ue
u
we see that it is linear
in
v
. However, writing it in the form (
v
+
uv
−
ue
u
)(
du/dv
)+
u
= 0, we see that it is nonlinear in
u
.
5.
Fourth-order; linear
6.
Second-order; nonlinear because of cos(
r
+
u
)
7.
Second-order; nonlinear because of
x
1+(
dy/dx
)
2
8.
Second-order; nonlinear because of 1
/R
2
9.
Third-order; linear
10.
Second-order; nonlinear because of ˙
x
2
11.
From
y
=
e
−
x/
2
we obtain
y
x
=
−
1
2
e
−
x/
2
. Then 2
y
x
+
y
=
−
e
−
x/
2
+
e
−
x/
2
=0.
12.
From
y
=
6
5
−
6
5
e
−
20
t
we obtain
dy/dt
=24
e
−
20
t
, so that
dy
dt
+20
y
=24
e
−
20
t
+20
/
6
5
−
6
5
e
−
20
t
t
=24
.
13.
From
y
=
e
3
x
cos 2
x
we obtain
y
x
=3
e
3
x
cos 2
x
−
2
e
3
x
sin 2
x
and
y
xx
=5
e
3
x
cos 2
x
−
12
e
3
x
sin 2
x
, so that
y
xx
−
6
y
x
+13
y
=0.
14.
From
y
=
−
cos
x
ln(sec
x
+ tan
x
) we obtain
y
x
=
−
1+ sin
x
ln(sec
x
+ tan
x
) and
y
xx
= tan
x
+ cos
x
ln(sec
x
+ tan
x
). Then
y
xx
+
y
= tan
x
.
15.
Writing ln(2
X
−
1)
−
ln(
X
−
1)=
t
and differentiating implicitly we obtain
2
2
X
−
1
dX
dt
−
1
X
−
1
dX
dt
=1
/
2
2
X
−
1
−
1
X
−
1
t
dX
dt
=1
2
X
−
2
−
2
X
+1
(2
X
−
1)(
X
−
1)
dX
dt
=1
dX
dt
=
−
(2
X
−
1)(
X
−
1)=(
X
−
1)(1
−
2
X
)
.
Exponentiating both sides of the implicit solution we obtain
2
X
−
1
X
−
1
=
e
t
=
⇒
2
X
−
1=
Xe
t
−
e
t
=
⇒
(
e
t
−
1)=(
e
t
−
2)
X
=
⇒
X
=
e
t
−
1
e
t
−
2
.
Solving
e
t
−
2 = 0 we get
t
= ln 2. Thus, the solution is defined on (
−∞
,
ln 2) or on (ln 2
,
∞
). The graph of the
solution defined on (
−∞
,
ln 2) is dashed, and the graph of the solution defined on (ln 2
,
∞
) is solid.
1

2. -4-2 2 4
x
-4
-2
2
4
y
Exercises 1.1
16.
Implicitly differentiating the solution we obtain
−
2
x
2
dy
dx
−
4
xy
+2
y
dy
dx
=0 =
⇒−
x
2
dy
−
2
xy dx
+
ydy
=0
=
⇒
2
xy dx
+(
x
2
−
y
)
dy
=0
.
Using the quadratic formula to solve
y
2
−
2
x
2
y
−
1= 0 for
y
, we get
y
=
k
2
x
2
±
√
4
x
4
+4
m
/
2=
x
2
±
√
x
4
+ 1. Thus, two explicit solutions are
y
1
=
x
2
+
√
x
4
+ 1and
y
2
=
x
2
−
√
x
4
+ 1. Both solutions are defined on
(
−∞
,
∞
). The graph of
y
1
(
x
) is solid and the graph of
y
2
is dashed.
17.
Differentiating
P
=
c
1
e
t
/
(1+
c
1
e
t
) we obtain
dP
dt
=
(1+
c
1
e
t
)
c
1
e
t
−
c
1
e
t
·
c
1
e
t
(1+
c
1
e
t
)
2
=
c
1
e
t
1+
c
1
e
t
[(1+
c
1
e
t
)
−
c
1
e
t
]
1+
c
1
e
t
=
P
(1
−
P
)
.
18.
Differentiating
y
=
e
−
x
2
w
x
0
e
t
2
dt
+
c
1
e
−
x
2
we obtain
y
x
=
e
−
x
2
e
x
2
−
2
xe
−
x
2
w
x
0
e
t
2
dt
−
2
c
1
xe
−
x
2
=1
−
2
xe
−
x
2
w
x
0
e
t
2
dt
−
2
c
1
xe
−
x
2
.
Substituting into the differential equation, we have
y
x
+2
xy
=1
−
2
xe
−
x
2
w
x
0
e
t
2
dt
−
2
c
1
xe
−
x
2
+2
xe
−
x
2
w
x
0
e
t
2
dt
+2
c
1
xe
−
x
2
=1
.
19.
From
y
=
c
1
e
2
x
+
c
2
xe
2
x
we obtain
dy
dx
=(2
c
1
+
c
2
)
e
2
x
+2
c
2
xe
2
x
and
d
2
y
dx
2
=(4
c
1
+4
c
2
)
e
2
x
+4
c
2
xe
2
x
, so that
d
2
y
dx
2
−
4
dy
dx
+4
y
=(4
c
1
+4
c
2
−
8
c
1
−
4
c
2
+4
c
1
)
e
2
x
+(4
c
2
−
8
c
2
+4
c
2
)
xe
2
x
=0
.
20.
From
y
=
c
1
x
−
1
+
c
2
x
+
c
3
x
ln
x
+4
x
2
we obtain
dy
dx
=
−
c
1
x
−
2
+
c
2
+
c
3
+
c
3
ln
x
+8
x,
d
2
y
dx
2
=2
c
1
x
−
3
+
c
3
x
−
1
+8
,
and
d
3
y
dx
3
=
−
6
c
1
x
−
4
−
c
3
x
−
2
,
so that
x
3
d
3
y
dx
3
+2
x
2
d
2
y
dx
2
−
x
dy
dx
+
y
=(
−
6
c
1
+4
c
1
+
c
1
+
c
1
)
x
−
1
+(
−
c
3
+2
c
3
−
c
2
−
c
3
+
c
2
)
x
+(
−
c
3
+
c
3
)
x
ln
x
+(16
−
8+4)
x
2
=12
x
2
.
21.
From
y
=
h
−
x
2
,x<
0
x
2
,x
≥
0
we obtain
y
x
=
h
−
2
x, x <
0
2
x, x
≥
0
so that
xy
x
−
2
y
=0.
2

3. Exercises 1.1
22.
The function
y
(
x
) is not continuous at
x
= 0 since lim
x
→
0
−
y
(
x
) = 5 and lim
x
→
0
+
y
(
x
)=
−
5. Thus,
y
x
(
x
)doesnot
exist at
x
=0.
23.
From
x
=
e
−
2
t
+3
e
6
t
and
y
=
−
e
−
2
t
+5
e
6
t
we obtain
dx
dt
=
−
2
e
−
2
t
+18
e
6
t
and
dy
dt
=2
e
−
2
t
+30
e
6
t
.
Then
x
+3
y
=(
e
−
2
t
+3
e
6
t
)+3(
−
e
−
2
t
+5
e
6
t
)
=
−
2
e
−
2
t
+18
e
6
t
=
dx
dt
and
5
x
+3
y
=5(
e
−
2
t
+3
e
6
t
)+3(
−
e
−
2
t
+5
e
6
t
)
=2
e
−
2
t
+30
e
6
t
=
dy
dt
.
24.
From
x
= cos 2
t
+ sin 2
t
+
1
5
e
t
and
y
=
−
cos 2
t
−
sin 2
t
−
1
5
e
t
we obtain
dx
dt
=
−
2 sin 2
t
+ 2 cos 2
t
+
1
5
e
t
and
dy
dt
= 2 sin 2
t
−
2 cos 2
t
−
1
5
e
t
and
d
2
x
dt
2
=
−
4 cos 2
t
−
4 sin 2
t
+
1
5
e
t
and
d
2
y
dt
2
= 4 cos 2
t
+ 4 sin 2
t
−
1
5
e
t
.
Then
4
y
+
e
t
=4(
−
cos 2
t
−
sin 2
t
−
1
5
e
t
)+
e
t
=
−
4 cos 2
t
−
4 sin 2
t
+
1
5
e
t
=
d
2
x
dt
2
and
4
x
−
e
t
= 4(cos 2
t
+ sin 2
t
+
1
5
e
t
)
−
e
t
= 4 cos 2
t
+ 4 sin 2
t
−
1
5
e
t
=
d
2
y
dt
2
.
25.
An interval on which tan 5
t
is continuous is
−
π/
2
<
5
t<π/
2, so 5 tan 5
t
will be a solution on (
−
π/
10
,π/
10).
26.
For (1
−
sin
t
)
−
1
/
2
to be continuous we must have 1
−
sin
t>
0orsin
t<
1. Thus, (1
−
sin
t
)
−
1
/
2
will be a
solution on (
π/
2
,
5
π/
2).
27.
(
y
x
)
2
+ 1= 0 has no real solution.
28.
The only solution of (
y
x
)
2
+
y
2
=0is
y
= 0, since if
y
h
=0,
y
2
>
0 and (
y
x
)
2
+
y
2
≥
y
2
>
0.
29.
The first derivative of
f
(
t
)=
e
t
is
e
t
. The first derivative of
f
(
t
)=
e
kt
is
ke
kt
. The differential equations are
y
x
=
y
and
y
x
=
ky
, respectively.
30.
Any function of the form
y
=
ce
t
or
y
=
ce
−
t
is its own second derivative. The corresponding differential
equation is
y
xx
−
y
= 0. Functions of the form
y
=
c
sin
t
or
y
=
c
cos
t
have second derivatives that are the
negatives of themselves. The differential equation is
y
xx
+
y
=0.
31.
Since the
n
th derivative of
φ
(
x
) must exist if
φ
(
x
) is a solution of the
n
th order differential equation, all lower-
order derivatives of
φ
(
x
) must exist and be continuous. [Recall that a differentiable function is continuous.]
32.
Solving the system
c
1
y
1
(0) +
c
2
y
2
(0) = 2
c
1
y
x
1
(0) +
c
2
y
x
2
(0) = 0
3

4. Exercises 1.1
for
c
1
and
c
2
we get
c
1
=
2
y
x
2
(0)
y
1
(0)
y
x
2
(0)
−
y
x
1
(0)
y
2
(0)
and
c
2
=
−
2
y
x
1
(0)
y
1
(0)
y
x
2
(0)
−
y
x
1
(0)
y
2
(0)
.
Thus, a particular solution is
y
=
2
y
x
2
(0)
y
1
(0)
y
x
2
(0)
−
y
x
1
(0)
y
2
(0)
y
1
−
2
y
x
1
(0)
y
1
(0)
y
x
2
(0)
−
y
x
1
(0)
y
2
(0)
y
2
,
where we assume that
y
1
(0)
y
x
2
(0)
−
y
x
1
(0)
y
2
(0)
h
=0.
33.
For the first-order differential equation integrate
f
(
x
). For the second-order differential equation integrate twice.
In the latter case we get
y
=

(

f
(
t
)
dt
)
dt
+
c
1
t
+
c
2
.
34.
Solving for
y
x
using the quadratic formula we obtain the two differential equations
y
x
=
1
t

2+2
x
1+3
t
6

and
y
x
=
1
t

2
−
2
x
1+3
t
6

,
so the differential equation cannot be put in the form
dy/dt
=
f
(
t, y
).
35.
The differential equation
yy
x
−
ty
= 0 has normal form
dy/dt
=
t
. These are not equivalent because
y
= 0 is a
solution of the first differential equation but not a solution of the second.
36.
Differentiating we get
y
x
=
c
1
+3
c
2
t
2
and
y
xx
=6
c
2
t
. Then
c
2
=
y
xx
/
6
t
and
c
1
=
y
x
−
ty
xx
/
2, so
y
=
/
y
x
−
ty
xx
2
t
t
+
/
y
xx
6
t
t
t
3
=
ty
x
−
1
3
t
2
y
xx
and the differential equation is
t
2
y
xx
−
3
ty
x
+3
y
=0.
37. (a)
From
y
=
e
mt
we obtain
y
x
=
me
mt
. Then
y
x
+2
y
= 0 implies
me
mt
+2
e
mt
=(
m
+2)
e
mt
=0
.
Since
e
mt
>
0 for all
t
,
m
=
−
2. Thus
y
=
e
−
2
t
is a solution.
(b)
From
y
=
e
mt
we obtain
y
x
=
me
mt
and
y
xx
=
m
2
e
mt
. Then
y
xx
−
5
y
x
+6
y
= 0 implies
m
2
e
mt
−
5
me
mt
+6
e
mt
=(
m
−
2)(
m
−
3)
e
mt
=0
.
Since
e
mt
>
0 for all
t
,
m
= 2 and
m
=3. Thus
y
=
e
2
t
and
y
=
e
3
t
are solutions.
(c)
From
y
=
t
m
we obtain
y
x
=
mt
m
−
1
and
y
xx
=
m
(
m
−
1)
t
m
−
2
. Then
ty
xx
+2
y
x
= 0 implies
tm
(
m
−
1)
t
m
−
2
+2
mt
m
−
1
=[
m
(
m
−
1)+2
m
]
t
m
−
1
=(
m
2
+
m
)
t
m
−
1
=
m
(
m
+1)
t
m
−
1
=0
.
Since
t
m
−
1
>
0 for
t>
0,
m
= 0 and
m
=
−
1.Thus
y
= 1and
y
=
t
−
1
are solutions.
(d)
From
y
=
t
m
we obtain
y
x
=
mt
m
−
1
and
y
xx
=
m
(
m
−
1)
t
m
−
2
. Then
t
2
y
xx
−
7
ty
x
+15
y
= 0 implies
t
2
m
(
m
−
1)
t
m
−
2
−
7
tmt
m
−
1
+15
t
m
=[
m
(
m
−
1)
−
7
m
+15]
t
m
=(
m
2
−
8
m
+15)
t
m
=(
m
−
3)(
m
−
5)
t
m
=0
.
Since
t
m
>
0 for
t>
0,
m
= 3 and
m
=5. Thus
y
=
t
3
and
y
=
t
5
are solutions.
38.
When
g
(
t
)=0,
y
= 0 is a solution of a linear equation.
39. (a)
Solving (10
−
5
y
)
/
3
x
= 0 we see that
y
= 2 is a constant solution.
(b)
Solving
y
2
+2
y
−
3=(
y
+ 3)(
y
−
1) = 0 we see that
y
=
−
3 and
y
= 1are constant solutions.
(c)
Since 1
/
(
y
−
1) = 0 has no solutions, the differential equation has no constant solutions.
4

5. t
y
−5 5
5
10
Exercises 1.1
(d)
Setting
y
x
=0wehave
y
xx
= 0 and 6
y
= 10. Thus
y
=5
/
3 is a constant solution.
40.
From
y
x
=(1
−
y
)
/
(
x
−
2) we see that a tangent line to the graph of
y
(
x
) is possibly vertical at
x
= 2. Intervals
of existence could be (
−∞
,
2) and (2
,
∞
).
41.
One solution is given by the upper portion of the graph with domain approximately (0
,
2
.
6). The other solution
is given by the lower portion of the graph, also with domain approximately (0
,
2
.
6).
42.
One solution, with domain approximately (
−∞
,
1
.
6) is the portion of the graph in the second quadrant together
with the lower part of the graph in the first quadrant. A second solution, with domain approximately (0
,
1
.
6)
is the upper part of the graph in the first quadrant. The third solution, with domain (0
,
∞
), is the part of the
graph in the fourth quadrant.
43.
Differentiating (
x
3
+
y
3
)
/xy
=3
c
we obtain
xy
(3
x
2
+3
y
2
y
x
)
−
(
x
3
+
y
3
)(
xy
x
+
y
)
x
2
y
2
=0
3
x
3
+3
xy
3
y
x
−
x
4
y
x
−
x
3
y
−
xy
3
y
x
−
y
4
=0
(3
xy
3
−
x
4
−
xy
3
)
y
x
=
−
3
x
3
y
+
x
3
y
+
y
4
y
x
=
y
4
−
2
x
3
y
2
xy
3
−
x
4
=
y
(
y
3
−
2
x
3
)
x
(2
y
3
−
x
3
)
.
44.
A tangent line will be vertical where
y
x
is undefined, or in this case, where
x
(2
y
3
−
x
3
) = 0. This gives
x
=0
and 2
y
3
=
x
3
. Substituting
y
3
=
x
3
/
2into
x
3
+
y
3
=3
xy
we get
x
3
+
1
2
x
3
=3
x
/
1
2
1
/
3
x
t
3
2
x
3
=
3
2
1
/
3
x
2
x
3
=2
2
/
3
x
2
x
2
(
x
−
2
2
/
3
)=0
.
Thus, there are vertical tangent lines at
x
= 0 and
x
=2
2
/
3
,orat(0
,
0) and (2
2
/
3
,
2
1
/
3
). Since 2
2
/
3
≈
1
.
59, the
estimates of the domains in Problem 42 were close.
45.
Since
φ
x
(
x
)
>
0 for all
x
in
I
,
φ
(
x
) is an increasing function on
I
. Hence, it can have no relative extrema on
I
.
46. (a)
When
y
=5,
y
x
=0,so
y
= 5 is a solution of
y
x
=5
−
y
.
(b)
When
y>
5,
y
x
<
0, and the solution must be decreasing. When
y<
5,
y
x
>
0, and the solution must be
increasing. Thus, none of the curves in color can be solutions.
(c)
47. (a)
y
= 0 and
y
=
a/b
.
5

6. y=a b
y=0
x
y
Exercises 1.1
(b)
Since
dy/dx
=
y
(
a
−
by
)
>
0 for 0
<y<a/b
,
y
=
φ
(
x
) is increasing on this interval. Since
dy/dx <
0 for
y<
0or
y > a/b
,
y
=
φ
(
x
) is decreasing on these intervals.
(c)
Using implicit differentiation we compute
d
2
y
dx
2
=
y
(
−
by
1
)+
y
1
(
a
−
by
)=
y
1
(
a
−
2
by
)
.
Solving
d
2
y/dx
2
= 0 we obtain
y
=
a/
2
b
. Since
d
2
y/dx
2
>
0 for 0
<y<a/
2
b
and
d
2
y/dx
2
<
0 for
a/
2
b < y < a/b
, the graph of
y
=
φ
(
x
) has a point of inflection at
y
=
a/
2
b
.
(d)
48. (a)
In
Mathematica
use
Clear[y]
y[x
]:= x Exp[5x] Cos[2x]
y[x]
y
''''
[x]
−
20 y
'''
[x] + 158 y
''
[x]
−
580 y
'
[x] + 841 y[x] // Simplify
(b)
In
Mathematica
use
Clear[y]
y[x
]:= 20 Cos[5 Log[x]]/x
−
3 Sin[5 Log[x]]/x
y[x]
xˆ3 y
'''
[x]+2xˆ2y
''
[x]+20xy
'
[x]
−
78 y[x] // Simplify
Exercises 1.2
1.
Solving
−
1
3
=
1
1+
c
1
we get
c
1
=
−
4. The solution is
y
=
1
1
−
4
e
−
t
.
2.
Solving 2 =
1
1+
c
1
e
we get
c
1
=
−
1
2
e
−
1
. The solution is
y
=
2
2
−
e
−
(
t
+1)
.
3.
Using
x
1
=
−
c
1
sin
t
+
c
2
cos
t
we obtain
c
1
=
−
1and
c
2
= 8. The solution is
x
=
−
cos
t
+ 8 sin
t
.
4.
Using
x
1
=
−
c
1
sin
t
+
c
2
cos
t
we obtain
c
2
= 0 and
−
c
1
= 1. The solution is
x
=
−
cos
t
.
5.
Using
x
1
=
−
c
1
sin
t
+
c
2
cos
t
we obtain
√
3
2
c
1
+
1
2
c
2
=
1
2
−
1
2
c
1
+
√
3
2
c
2
=0
.
Solving we find
c
1
=
√
3
4
and
c
2
=
1
4
. The solution is
x
=
√
3
4
cos
t
+
1
4
sin
t
.
6

7. Exercises 1.2
6.
Using
x
x
=
−
c
1
sin
t
+
c
2
cos
t
we obtain
√
2
2
c
1
+
√
2
2
c
2
=
√
2
−
√
2
2
c
1
+
√
2
2
c
2
=2
√
2
.
Solving we find
c
1
=
−
1and
c
2
= 3. The solution is
x
=
−
cos
t
+ 3 sin
t
.
7.
From the initial conditions we obtain the system
c
1
+
c
2
=1
c
1
−
c
2
=2
.
Solving we get
c
1
=
3
2
and
c
2
=
−
1
2
. A solution of the initial-value problem is
y
=
3
2
e
x
−
1
2
e
−
x
.
8.
From the initial conditions we obtain the system
c
1
e
+
c
2
e
−
1
=0
c
1
e
−
c
2
e
−
1
=
e.
Solving we get
c
1
=
1
2
and
c
2
=
−
1
2
e
2
. A solution of the initial-value problem is
y
=
1
2
e
x
−
1
2
e
2
−
x
.
9.
From the initial conditions we obtain
c
1
e
−
1
+
c
2
e
=5
c
1
e
−
1
−
c
2
e
=
−
5
.
Solving we get
c
1
= 0 and
c
2
=5
e
−
1
. A solution of the initial-value problem is
y
=5
e
−
x
−
1
.
10.
From the initial conditions we obtain
c
1
+
c
2
=0
c
1
−
c
2
=0
.
Solving we get
c
1
=
c
2
= 0. A solution of the initial-value problem is
y
=0.
11.
Two solutions are
y
= 0 and
y
=
x
3
.
12.
Two solutions are
y
= 0 and
y
=
x
2
. (Also, any constant multiple of
x
2
is a solution.)
13.
For
f
(
x, y
)=
y
2
/
3
we have
∂f
∂y
=
2
3
y
−
1
/
3
. Thus the differential equation will have a unique solution in any
rectangular region of the plane where
y
h
=0.
14.
For
f
(
x, y
)=
√
xy
we have
∂f
∂y
=
1
2

x
y
. Thus the differential equation will have a unique solution in any
region where
x>
0 and
y>
0 or where
x<
0 and
y<
0.
15.
For
f
(
x, y
)=
y
x
we have
∂f
∂y
=
1
x
. Thus the differential equation will have a unique solution in any region
where
x
h
=0.
16.
For
f
(
x, y
)=
x
+
y
we have
∂f
∂y
= 1. Thus the differential equation will have a unique solution in the entire
plane.
17.
For
f
(
x, y
)=
x
2
4
−
y
2
we have
∂f
∂y
=
2
x
2
y
(4
−
y
2
)
2
. Thus the differential equation will have a unique solution in
any region where
y<
−
2,
−
2
<y<
2, or
y>
2.
7

8. Exercises 1.2
18.
For
f
(
x, y
)=
x
2
1+
y
3
we have
∂f
∂y
=
−
3
x
2
y
2
(1+
y
3
)
2
. Thus the differential equation will have a unique solution in
any region where
y
6
=
−
1.
19.
For
f
(
x, y
)=
y
2
x
2
+
y
2
we have
∂f
∂y
=
2
x
2
y
(
x
2
+
y
2
)
2
. Thus the differential equation will have a unique solution in
any region not containing (0
,
0).
20.
For
f
(
x, y
)=
y
+
x
y
−
x
we have
∂f
∂y
=
−
2
x
(
y
−
x
)
2
. Thus the differential equation will have a unique solution in any
region where
y<x
or where
y>x
.
21.
The differential equation has a unique solution at (1
,
4).
22.
The differential equation is not guaranteed to have a unique solution at (5
,
3).
23.
The differential equation is not guaranteed to have a unique solution at (2
,
−
3).
24.
The differential equation is not guaranteed to have a unique solution at (
−
1
,
1).
25. (a)
A one-parameter family of solutions is
y
=
cx
. Since
y
⇒
=
c
,
xy
⇒
=
xc
=
y
and
y
(0) =
c
·
0=0.
(b)
Writing the equation in the form
y
⇒
=
y/x
we see that
R
cannot contain any point on the
y
-axis. Thus,
any rectangular region disjoint from the
y
-axis and containing (
x
0
,y
0
) will determine an interval around
x
0
and a unique solution through (
x
0
,y
0
). Since
x
0
= 0 in part (a) we are not guaranteed a unique solution
through (0
,
0).
(c)
The piecewise-defined function which satisfies
y
(0) = 0 is not a solution since it is not differentiable at
x
=0.
26. (a)
Since
d
dx
tan(
x
+
c
) = sec
2
(
x
+
c
)=1+tan
2
(
x
+
c
), we see that
y
= tan(
x
+
c
) satisfies the differential
equation.
(b)
Solving
y
(0) = tan
c
= 0 we obtain
c
= 0 and
y
= tan
x
. Since tan
x
is discontinuous at
x
=
±
π/
2, the
solution is not defined on (
−
2
,
2) because it contains
±
π/
2.
(c)
The largest interval on which the solution can exist is (
−
π/
2
,π/
2).
27. (a)
Since
d
dt

−
1
t
+
c

=
1
(
t
+
c
)
2
=
y
2
, we see that
y
=
−
1
t
+
c
is a solution of the differential equation.
(b)
Solving
y
(0) =
−
1
/c
= 1we obtain
c
=
−
1and
y
=1
/
(1
−
t
). Solving
y
(0) =
−
1
/c
=
−
1we obtain
c
=1
and
y
=
−
1
/
(1+
t
). Being sure to include
t
= 0, we see that the interval of existence of
y
=1
/
(1
−
t
)is
(
−∞
,
1), while the interval of existence of
y
=
−
1
/
(1+
t
)is(
−
1
,
∞
).
(c)
Solving
y
(0) =
−
1
/c
=
y
0
we obtain
c
=
−
1
/y
0
and
y
=
−
1
−
1
/y
0
+
t
=
y
0
1
−
y
0
t
,y
0
6
=0
.
Since we must have
−
1
/y
0
+
t
6
= 0, the largest interval of existence (which must contain 0) is either
(
−∞
,
1
/y
0
) when
y
0
>
0or(1
/y
0
,
∞
) when
y
0
<
0.
(d)
By inspection we see that
y
= 0 is a solution on (
−∞
,
∞
).
28. (a)
Differentiating 3
x
2
−
y
2
=
c
we get 6
x
−
2
yy
⇒
=0or
yy
⇒
=3
x
.
8

9. -4-2 2 4
x
-4
-2
2
4
y
-4-2 2 4
x
-4
-2
2
4
y
-4-2 24
x
1
2
3
4
5
y
-4-2 24
x
1
2
3
4
5
y
Exercises 1.2
(b)
Solving 3
x
2
−
y
2
= 3 for
y
we get
y
=
φ
1
(
x
)=
1
3(
x
2
−
1)
,
1
<x<
∞
,
y
=
φ
2
(
x
)=
−
1
3(
x
2
−
1)
,
1
<x<
∞
,
y
=
φ
3
(
x
)=
1
3(
x
2
−
1)
,
−∞
<x<
−
1
,
y
=
φ
4
(
x
)=
−
1
3(
x
2
−
1)
,
−∞
<x<
−
1
.
Only
y
=
φ
3
(
x
) satisfies
y
(
−
2) = 3.
(c)
Setting
x
= 2 and
y
=
−
4in3
x
2
−
y
2
=
c
we get 12
−
16=
−
4=
c
, so the
explicit solution is
y
=
−
1
3
x
2
+4
,
−∞
<x<
∞
.
(d)
Setting
c
= 0 we have
y
=
√
3
x
and
y
=
−
√
3
x
, both defined on (
−∞
,
∞
).
29.
When
x
= 0 and
y
=
1
2
,
y
1
=
−
1, so the only plausible solution curve is the one with negative slope at (0
,
1
2
),
or the black curve.
30.
If the solution is tangent to the
x
-axis at (
x
0
,
0), then
y
1
= 0 when
x
=
x
0
and
y
= 0. Substituting these values
into
y
1
+2
y
=3
x
−
6weget0+0=3
x
0
−
6or
x
0
=2.
31.
The theorem guarantees a unique (meaning single) solution through any point. Thus, there cannot be two
distinct solutions through any point.
32.
Differentiating
y
=
x
4
/
16 we obtain
y
1
=
x
3
/
4=
x
(
x
4
/
16)
1
/
2
, so the first
function is a solution of the differential equation. Since
y
= 0 satisfies the
differential equation and lim
x
→
0
+
x
3
/
4 = 0, the second function is also a solution
of the differential equation. Both functions satisfy the condition
y
(2)=1.
Theorem 1.1 simply guarantees the existence of a unique solution on
some
interval containing (2
,
1). In this, such an interval could be (0
,
4), where the two
functions are identical.
33.
The antiderivative of
y
1
=8
e
2
x
+6
x
is
y
=4
e
2
x
+3
x
2
+
c
. Setting
x
= 0 and
y
=9weget9=4+
c
,so
c
=5
and
y
=4
e
2
x
+3
x
2
+5.
34.
The antiderivative of
y
11
=12
x
−
2is
y
1
=6
x
2
−
2
x
+
c
1
. From the equation of the tangent line we see that
when
x
=1,
y
= 4 and
y
1
=
−
1(the slope of the tangent line). Solving
−
1= 6(1)
2
−
2(1) +
c
1
we get
c
1
=
−
5.
The antiderivative of
y
1
=6
x
2
−
2
x
−
5is
y
=2
x
3
−
x
2
−
5
x
+
c
2
. Setting
x
= 1and
y
= 4 we get 4 =
−
4+
c
2
,
so
c
2
= 8 and
y
=2
x
3
−
x
2
−
5
x
+8.
9

10. Exercises 1.3
Exercises 1.3
1.
dP
dt
=
kP
+
r
;
dP
dt
=
kP
−
r
2.
Let
b
be the rate of births and
d
the rate of deaths. Then
b
=
k
1
P
and
d
=
k
2
P
. Since
dP/dt
=
b
−
d
, the
differential equation is
dP/dt
=
k
1
P
−
k
2
P
.
3.
Let
b
be the rate of births and
d
the rate of deaths. Then
b
=
k
1
P
and
d
=
k
2
P
2
. Since
dP/dt
=
b
−
d
, the
differential equation is
dP/dt
=
k
1
P
−
k
2
P
2
.
4.
Let
P
(
t
) be the number of owls present at time
t
. Then
dP/dt
=
k
(
P
−
200+10
t
).
5.
From the graph we estimate
T
0
=180
◦
and
T
m
=75
◦
. We observe that when
T
= 85,
dT/dt
≈−
1. From the
differential equation we then have
k
=
dT/dt
T
−
T
m
=
−
1
85
−
75
=
−
0
.
1
.
6.
By inspecting the graph we take
T
m
to be
T
m
(
t
)=80
−
30 cos
πt/
12. Then the temperature of the body at time
t
is determined by the differential equation
dT
dt
=
k

T
−

80
−
30 cos
π
12
t
9a
,t>
0
.
7.
The number of students with the flu is
x
and the number not infected is 1000
−
x
,so
dx/dt
=
kx
(1000
−
x
).
8.
By analogy with differential equation modeling the spread of a disease we assume that the rate at which the
technological innovation is adopted is proportional to the number of people who have adopted the innovation
and also to the number of people,
y
(
t
), who have not yet adopted it. If one person who has adopted the
innovation is introduced into the population then
x
+
y
=
n
+ 1and
dx
dt
=
kx
(
n
+1
−
x
)
,x
(0)=1
.
9.
The rate at which salt is leaving the tank is
(3 gal/min)
·
/
A
300
lb/gal
t
=
A
100
lb/min.
Thus
dA/dt
=
A/
100.
10.
The rate at which salt is entering the tank is
R
1
= (3 gal/min)
·
(2 lb/gal) = 6 lb/min.
Since the solution is pumped out at a slower rate, it is accumulating at the rate of (3
−
2)gal/min = 1gal/min.
After
t
minutes there are 300 +
t
gallons of brine in the tank. The rate at which salt is leaving is
R
2
= (2 gal/min)
·
/
A
300 +
t
lb/gal
t
=
2
A
300 +
t
lb/min
.
The differential equation is
dA
dt
=6
−
2
A
300 +
t
.
11.
The volume of water in the tank at time
t
is
V
=
A
w
h
. The differential equation is then
dh
dt
=
1
A
w
dV
dt
=
1
A
w

−
cA
0
x
2
gh

=
−
cA
0
A
w
x
2
gh.
10

11. Exercises 1.3
Using
A
0
=
π
/
2
12
t
2
=
π
36
,
A
w
=10
2
= 100, and
g
= 32, this becomes
dh
dt
=
−
cπ/
36
100
√
64
h
=
−
cπ
450
√
h.
12.
The volume of water in the tank at time
t
is
V
=
1
3
πr
2
h
=
1
3
A
w
h
. Using the formula from Problem 11 for the
volume of water leaving the tank we see that the differential equation is
dh
dt
=
3
A
w
dV
dt
=
3
A
w
(
−
cA
h
x
2
gh
)=
−
3
cA
h
A
w
x
2
gh.
Using
A
h
=
π
(2
/
12)
2
=
π/
36,
g
= 32, and
c
=0
.
6, this becomes
dh
dt
=
−
3(0
.
6)
π/
36
A
w
√
64
h
=
−
0
.
4
π
A
w
h
1
/
2
.
To find
A
w
we let
r
be the radius of the top of the water. Then
r/h
=8
/
20, so
r
=2
h/
5 and
A
w
=
π
(2
h/
5)
2
=
4
πh
2
/
25. Thus
dh
dt
=
−
0
.
4
π
4
πh
2
/
25
h
1
/
2
=
−
2
.
5
h
−
3
/
2
.
13.
Since
i
=
dq
dt
and
L
d
2
q
dt
2
+
R
dq
dt
=
E
(
t
) we obtain
L
di
dt
+
Ri
=
E
(
t
).
14.
By Kirchoff’s second law we obtain
R
dq
dt
+
1
C
q
=
E
(
t
).
15.
From Newton’s second law we obtain
m
dv
dt
=
−
kv
2
+
mg
.
16.
We have from Archimedes’ principle
upward force of water on barrel = weight of water displaced
= (62
.
4)
×
(volume of water displaced)
= (62
.
4)
π
(
s/
2)
2
y
=15
.
6
πs
2
y.
It then follows from Newton’s second law that
w
g
d
2
y
dt
2
=
−
15
.
6
πs
2
y
or
d
2
y
dt
2
+
15
.
6
πs
2
g
w
y
= 0, where
g
=32
and
w
is the weight of the barrel in pounds.
17.
The net force acting on the mass is
F
=
ma
=
m
d
2
x
dt
2
=
−
k
(
s
+
x
)+
mg
=
−
kx
+
mg
−
ks.
Since the condition of equilibrium is
mg
=
ks
, the differential equation is
m
d
2
x
dt
2
=
−
ks.
18.
From Problem 17, without a damping force, the differential equation is
md
2
x/dt
2
=
−
kx
. With a damping
force proportional to velocity the differential equation becomes
m
d
2
x
dt
2
=
−
kx
−
β
dx
dt
or
m
d
2
x
dt
2
+
β
dx
dt
+
kx
=0
.
19.
Let
x
(
t
) denote the height of the top of the chain at time
t
with the positive direction upward. The weight of
the portion of chain off the ground is
W
=(
x
ft)
·
(1lb/ft) =
x
. The mass of the chain is
m
=
W/g
=
x/
32.
The net force is
F
=5
−
W
=5
−
x
. By Newton’s second law,
d
dt

x
32
v

=5
−
x
or
x
dv
dt
+
v
dx
dt
=160
−
32
x.
11

12. (x,y)
x
y
α
α
θ
θ
θ
φ
x
y
Exercises 1.3
Thus, the differential equation is
x
d
2
x
dt
+

dx
dt

2
+32
x
=160
.
20.
The force is the weight of the chain, 2
L
, so by Newton’s second law,
d
dt
[
mv
]=2
L
. Since the mass of the portion
of chain off the ground is
m
=2(
L
−
x
)
/g
, we have
d
dt

2(
L
−
x
)
g
v
a
=2
L
or (
L
−
x
)
dv
dt
+
v

−
dx
dt

=
Lg.
Thus, the differential equation is
(
L
−
x
)
d
2
x
dt
2
−

dx
dt

2
=
Lg.
21.
From
g
=
k/R
2
we find
k
=
gR
2
. Using
a
=
d
2
r/dt
2
and the fact that the positive direction is upward we get
d
2
r
dt
2
=
−
a
=
−
k
r
2
=
−
gR
2
r
2
or
d
2
r
dt
2
+
gR
2
r
2
=0
.
22.
The gravitational force on
m
is
F
=
−
kM
r
m/r
2
. Since
M
r
=4
πδr
3
/
3 and
M
=4
πδR
3
/
3wehave
M
r
=
r
3
M/R
3
and
F
=
−
k
M
r
m
r
2
=
−
k
r
3
Mm/R
3
r
2
=
−
k
mM
R
3
r.
Now from
F
=
ma
=
d
2
r/dt
2
we have
m
d
2
r
dt
2
=
−
k
mM
R
3
r
or
d
2
r
dt
2
=
−
kM
R
3
r.
23.
The differential equation is
dA
dt
=
k
(
M
−
A
).
24.
The differential equation is
dA
dt
=
k
1
(
M
−
A
)
−
k
2
A
.
25.
The differential equation is
x
1
(
t
)=
r
−
kx
(
t
) where
k>
0.
26.
By the Pythagorean Theorem the slope of the tangent line is
y
1
=
−
y
1
s
2
−
y
2
.
27.
We see from the figure that 2
θ
+
α
=
π
.Thus
y
−
x
= tan
α
= tan(
π
−
2
θ
)=
−
tan 2
θ
=
−
2 tan
θ
1
−
tan
2
θ
.
Since the slope of the tangent line is
y
1
= tan
θ
we have
y/x
=2
y
1
[1
−
(
y
1
)
2
]or
y
−
y
(
y
1
)
2
=2
xy
1
, which is the quadratic equation
y
(
y
1
)
2
+2
xy
1
−
y
=0in
y
1
.
Using the quadratic formula we get
y
1
=
−
2
x
±
1
4
x
2
+4
y
2
2
y
=
−
x
±
1
x
2
+
y
2
y
.
Since
dy/dx >
0, the differential equation is
dy
dx
=
−
x
+
1
x
2
+
y
2
y
or
y
dy
dx
−
1
x
2
+
y
2
+
x
=0
.
28.
The differential equation is
dP/dt
=
kP
, so from Problem 29 in Exercises 1
.
1,
P
=
e
kt
, and a one-parameter
family of solutions is
P
=
ce
kt
.
12

13. Exercises 1.3
29.
The differential equation in (3) is
dT/dt
=
k
(
T
−
T
m
). When the body is cooling,
T>T
m
,so
T
−
T
m
>
0.
Since
T
is decreasing,
dT/dt <
0 and
k<
0. When the body is warming,
T<T
m
,so
T
−
T
m
<
0. Since
T
is
increasing,
dT/dt >
0 and
k<
0.
30.
The differential equation in (8) is
dA/dt
=6
−
A/
100. If
A
(
t
) attains a maximum, then
dA/dt
= 0 at this time
and
A
= 600. If
A
(
t
) continues to increase without reaching a maximum then
A
x
(
t
)
>
0 for
t>
0 and
A
cannot
exceed 600. In this case, if
A
x
(
t
) approaches 0 as
t
increases to infinity, we see that
A
(
t
) approaches 600 as
t
increases to infinity.
31.
This differential equation could describe a population that undergoes periodic fluctuations.
32.
(1):
dP
dt
=
kP
is linear (2):
dA
dt
=
kA
is linear
(3):
dT
dt
=
k
(
T
−
T
m
) is linear (5):
dx
dt
=
kx
(
n
+1
−
x
) is nonlinear
(6):
dX
dt
=
k
(
α
−
X
)(
β
−
X
) is nonlinear (8):
dA
dt
=6
−
A
100
is linear
(10):
dh
dt
=
−
A
h
A
w
x
2
gh
is nonlinear (11):
L
d
2
q
dt
2
+
R
dq
dt
+
1
C
q
=
E
(
t
) is linear
(12):
d
2
s
dt
2
=
−
g
is linear (14):
m
dv
dt
=
mg
−
kv
is linear
(15):
m
d
2
s
dt
2
+
k
ds
dt
=
mg
is linear (16):
d
2
x
dt
2
−
64
L
x
= 0 is linear
33.
From Problem 21,
d
2
r/dt
2
=
−
gR
2
/r
2
. Since
R
is a constant, if
r
=
R
+
s
, then
d
2
r/dt
2
=
d
2
s/dt
2
and, using
a Taylor series, we get
d
2
s
dt
2
=
−
g
R
2
(
R
+
s
)
2
=
−
gR
2
(
R
+
s
)
−
2
≈−
gR
2
[
R
−
2
−
2
sR
−
3
+
···
]=
−
g
+
2
gs
R
3
+
···
.
Thus, for
R
much larger than
s
, the differential equation is approximated by
d
2
s/dt
2
=
−
g
.
34.
If
ρ
is the mass density of the raindrop, then
m
=
ρV
and
dm
dt
=
ρ
dV
dt
=
ρ
d
dt

4
3
πr
3
a
=
ρ

4
πr
2
dr
dt

=
ρS
dr
dt
.
If
dr/dt
is a constant, then
dm/dt
=
kS
where
ρ dr/dt
=
k
or
dr/dt
=
k/ρ
. Since the radius is decreasing,
k<
0. Solving
dr/dt
=
k/ρ
we get
r
=(
k/ρ
)
t
+
c
0
. Since
r
(0) =
r
0
,
c
0
=
r
0
and
r
=
kt/ρ
+
r
0
.
From Newton’s second law,
d
dt
[
mv
]=
mg
, where
v
is the velocity of the raindrop. Then
m
dv
dt
+
v
dm
dt
=
mg
or
ρ

4
3
πr
3

dv
dt
+
v
(
k
4
πr
2
)=
ρ

4
3
πr
3

g.
Dividing by 4
ρπr
3
/
3weget
dv
dt
+
3
k
ρr
v
=
g
or
dv
dt
+
3
k/ρ
kt/ρ
+
r
0
v
=
g, k <
0
.
35.
We assume that the plow clears snow at a constant rate of
k
cubic miles per hour. Let
t
be the time in hours
after noon,
x
(
t
) the depth in miles of the snow at time
t
, and
y
(
t
) the distance the plow has moved in
t
hours.
Then
dy/dt
is the velocity of the plow and the assumption gives
wx
dy
dt
=
k
13

14. Exercises 1.3
where
w
is the width of the plow. Each side of this equation simply represents the volume of snow plowed in
one hour. Now let
t
0
be the number of hours before noon when it started snowing and let
s
be the constant rate
in miles per hour at which
x
increases. Then for
t>
−
t
0
,
x
=
s
(
t
+
t
0
). The differential equation then becomes
dy
dt
=
k
ws
1
t
+
t
0
.
Integrating we obtain
y
=
k
ws
[ln(
t
+
t
0
)+
c
]
where
c
is a constant. Now when
t
=0,
y
=0so
c
=
−
ln
t
0
and
y
=
k
ws
ln
/
1+
t
t
0
t
.
Finally, from the fact that when
t
=1,
y
= 2 and when
t
=2,
y
= 3, we obtain
/
1+
2
t
0
t
2
=
/
1+
1
t
0
t
3
.
Expanding and simplifying gives
t
2
0
+
t
0
−
1= 0. Since
t
0
>
0, we find
t
0
≈
0
.
618 hours
≈
37 minutes. Thus it
started snowing at about 11:23 in the morning.
Chapter 1 Review Exercises
1.
d
dx
c
1
x
=
−
c
1
x
2
=
−
c
1
/x
x
;
dy
dx
=
−
y
x
2.
d
dx
(5 +
c
1
e
−
2
x
)=
−
2
c
1
e
−
2
x
=
−
2(5 +
c
1
e
−
2
x
−
5);
dy
dx
=
−
2(
y
−
5) or
dy
dx
=
−
2
y
+10
3.
d
dx
(
c
1
cos
kx
+
c
2
sin
kx
)=
−
kc
1
sin
kx
+
kc
2
cos
kx
;
d
2
dx
2
(
c
1
cos
kx
+
c
2
sin
kx
)=
−
k
2
c
1
cos
kx
−
k
2
c
2
sin
kx
=
−
k
2
(
c
1
cos
kx
+
c
2
sin
kx
);
d
2
y
dx
2
=
−
k
2
y
or
d
2
y
dx
2
+
k
2
y
=0
4.
d
dx
(
c
1
cosh
kx
+
c
2
sinh
kx
)=
kc
1
sinh
kx
+
kc
2
cosh
kx
;
d
2
dx
2
(
c
1
cosh
kx
+
c
2
sinh
kx
)=
k
2
c
1
cosh
kx
+
k
2
c
2
sinh
kx
=
k
2
(
c
1
cosh
kx
+
c
2
sinh
kx
);
d
2
y
dx
2
=
k
2
y
or
d
2
y
dx
2
−
k
2
y
=0
5.
y
x
=
c
1
e
x
+
c
2
xe
x
+
c
2
e
x
;
y
xx
=
c
1
e
x
+
c
2
xe
x
+2
c
2
e
x
;
y
xx
+
y
=2(
c
1
e
x
+
c
2
xe
x
)+2
c
2
e
x
=2(
c
1
e
x
+
c
2
xe
x
+
c
2
e
x
)=2
y
x
;
y
xx
−
2
y
x
+
y
=0
6.
y
x
=
−
c
1
e
x
sin
x
+
c
1
e
x
cos
x
+
c
2
e
x
cos
x
+
c
2
e
x
sin
x
;
y
xx
=
−
c
1
e
x
cos
x
−
c
1
e
x
sin
x
−
c
1
e
x
sin
x
+
c
1
e
x
cos
x
−
c
2
e
x
sin
x
+
c
2
e
x
cos
x
+
c
2
e
x
cos
x
+
c
2
e
x
sin
x
=
−
2
c
1
e
x
sin
x
+2
c
2
e
x
cos
x
;
y
xx
−
2
y
x
=
−
2
c
1
e
x
cos
x
−
2
c
2
e
x
sin
x
=
−
2
y
;
y
xx
−
2
y
x
+2
y
=0
7.
a,d
8.
c
9.
b
10.
a,c
11.
b
12.
a,b,d
13.
A few solutions are
y
=0,
y
=
c
, and
y
=
e
x
.
14

15. -3-2-1 123
x
-3
-2
-1
1
2
3
y
-3-2-1 123
x
-3
-2
-1
1
2
3
y
Chapter 1 Review Exercises
14.
Easy solutions to see are
y
= 0 and
y
=3.
15.
The slope of the tangent line at (
x, y
)is
y
x
, so the differential equation is
y
x
=
x
2
+
y
2
.
16.
The rate at which the slope changes is
dy
x
/dx
=
y
xx
, so the differential equation is
y
xx
=
−
y
x
or
y
xx
+
y
x
=0.
17. (a)
The domain is all real numbers.
(b)
Since
y
x
=2
/
3
x
1
/
3
, the solution
y
=
x
2
/
3
is undefined at
x
= 0. This function is a solution of the differential
equation on (
−∞
,
0) and also on (0
,
∞
).
18. (a)
Differentiating
y
2
−
2
y
=
x
2
−
x
+
c
we obtain 2
yy
x
−
2
y
x
=2
x
−
1or (2
y
−
2)
y
x
=2
x
−
1.
(b)
Setting
x
= 0 and
y
= 1in the solution we have 1
−
2=0
−
0+
c
or
c
=
−
1. Thus, a solution of the
initial-value problem is
y
2
−
2
y
=
x
2
−
x
−
1.
(c)
Using the quadratic formula to solve
y
2
−
2
y
−
(
x
2
−
x
−
1) = 0 we get
y
=(2
±
x
4+4(
x
2
−
x
−
1))
/
2
=1
±
√
x
2
−
x
=1
±
x
x
(
x
−
1) . Since
x
(
x
−
1)
≥
0 for
x
≤
0or
x
≥
1, we see that neither
y
=1+
x
x
(
x
−
1)
nor
y
=1
−
x
x
(
x
−
1) is differentiable at
x
= 0. Thus, both functions are solutions of the differential
equation, but neither is a solution of the initial-value problem.
19. (a)
y
=
x
2
+
c
1
y
=
−
x
2
+
c
2
(b)
When
y
=
x
2
+
c
1
,
y
x
=2
x
and (
y
x
)
2
=4
x
2
. When
y
=
−
x
2
+
c
2
,
y
x
=
−
2
x
and (
y
x
)
2
=4
x
2
.
(c)
Pasting together
x
2
,
x
≥
0, and
−
x
2
,
x
≤
0, we get
y
=
h
−
x
2
,x
≤
0
x
2
,x>
0
.
20.
The slope of the tangent line is
y
x
)
)
(
−
1
,
4)
=6
√
4+5(
−
1)
3
=7.
21.
Differentiating
y
= sin(ln
x
) we obtain
y
x
= cos(ln
x
)
/x
and
y
xx
=
−
[sin(ln
x
) + cos(ln
x
)]
/x
2
. Then
x
2
y
xx
+
xy
x
+
y
=
x
2

−
sin(ln
x
) + cos(ln
x
)
x
2

+
x
cos(ln
x
)
x
+ sin(ln
x
)=0
.
22.
Differentiating
y
= cos(ln
x
) ln(cos(ln
x
)) + (ln
x
) sin(ln
x
) we obtain
y
x
= cos(ln
x
)
1
cos(ln
x
)

−
sin(ln
x
)
x

+ ln(cos(ln
x
))

−
sin(ln
x
)
x

+ln
x
cos(ln
x
)
x
+
sin(ln
x
)
x
=
−
ln(cos(ln
x
)) sin(ln
x
)
x
+
(ln
x
) cos(ln
x
)
x
and
y
xx
=
−
x
b
ln(cos(ln
x
))
cos(ln
x
)
x
+ sin(ln
x
)
1
cos(ln
x
)

−
sin(ln
x
)
x

c
1
x
2
+ ln(cos(ln
x
)) sin(ln
x
)
1
x
2
+
x
b
(ln
x
)

−
sin(ln
x
)
x

+
cos(ln
x
)
x
c
1
x
2
−
(ln
x
) cos(ln
x
)
1
x
2
=
1
x
2
b
−
ln(cos(ln
x
) cos(ln
x
)+
sin
2
(ln
x
)
cos(ln
x
)
+ ln(cos(ln
x
) sin(ln
x
)
−
(ln
x
) sin(ln
x
) + cos(ln
x
)
−
(ln
x
) cos(ln
x
)
c
.
15

16. Chapter 1 Review Exercises
Then
x
2
y
xx
+
xy
x
+
y
=
−
ln(cos(ln
x
)) cos(ln
x
)+
sin
2
(ln
x
)
cos(ln
x
)
+ ln(cos(ln
x
)) sin(ln
x
)
−
(ln
x
) sin(ln
x
)
+ cos(ln
x
)
−
(ln
x
) cos(ln
x
)
−
ln(cos(ln
x
)) sin(ln
x
)
+ (ln
x
) cos(ln
x
) + cos(ln
x
) ln(cos(ln
x
)) + (ln
x
) sin(ln
x
)
=
sin
2
(ln
x
)
cos(ln
x
)
+ cos(ln
x
)=
sin
2
(ln
x
) + cos
2
(ln
x
)
cos(ln
x
)
=
1
cos(ln
x
)
= sec(ln
x
)
.
(This problem is easily done using
Mathematica
.)
23.
From the graph we see that estimates for
y
0
and
y
1
are
y
0
=
−
3 and
y
1
=0.
24.
The differential equation is
dh
dt
=
−
cA
0
A
w
x
2
gh.
Using
A
0
=
π
(1
/
24)
2
=
π/
576,
A
w
=
π
(2)
2
=4
π
, and
g
= 32, this becomes
dh
dt
=
−
cπ/
576
4
π
√
64
h
=
c
288
√
h.
25.
From Newton’s second law we obtain
m
dv
dt
=
1
2
mg
−
µ
√
3
2
mg
or
dv
dt
=16

1
−
√
3
µ

.
16

17. y
tt
y
x
y
tt
y
x
x
y
y
tt
y
x
x
y y
x
y
x x
y
2
First-Order Differential Equations
Exercises 2.1
1. 2.
3. 4.
5. 6.
7. 8.
17

18. y
x x
y
y
x x
y
y
x x
y
-1 0 1
1 2
x
5
4
3
2
1
y
1 2-1-2
x
1
y
1 2-1-2
x
-1
y
1 2
x
-5
-4
-3
-2
-1
y
Exercises 2.1
9. 10.
11. 12.
13. 14.
15.
Writing the differential equation in the form
dy/dx
=
y
(1
−
y
)(1 +
y
) we see that critical points are located at
y
=
−
1,
y
= 0, and
y
= 1. The phase portrait is shown below.
(a) (b)
(c) (d)
18

19. -1 0 1
1 2
x
5
4
3
2
1
y
1 2-1-2
x
1
y
1 2-1-2
x
-1
y
-1-2
x
-5
-4
-3
-2
-1
y
03
1
0
2
-2 5
-2 0 2
024
Exercises 2.1
16.
Writing the differential equation in the form
dy/dx
=
y
2
(1
−
y
)(1 +
y
) we see that critical points are located at
y
=
−
1,
y
= 0, and
y
= 1. The phase portrait is shown below.
(a) (b)
(c) (d)
17.
Solving
y
2
−
3
y
=
y
(
y
−
3) = 0 we obtain the critical points 0 and 3.
From the phase portrait we see that 0 is asymptotically stable and 3 is unstable.
18.
Solving
y
2
−
y
3
=
y
2
(1
−
y
) = 0 we obtain the critical points 0 and 1.
From the phase portrait we see that 1 is asymptotically stable and 0 is semi-stable.
19.
Solving (
y
−
2)
2
= 0 we obtain the critical point 2.
From the phase portrait we see that 2 is semi-stable.
20.
Solving 10 + 3
y
−
y
2
=(5
−
y
)(2 +
y
) = 0 we obtain the critical points
−
2 and 5.
From the phase portrait we see that 5 is asymptotically stable and
−
2 is unstable.
21.
Solving
y
2
(4
−
y
2
)=
y
2
(2
−
y
)(2 +
y
) = 0 we obtain the critical points
−
2, 0, and 2.
From the phase portrait we see that 2 is asymptotically stable, 0 is semi-stable, and
−
2 is unstable.
22.
Solving
y
(2
−
y
)(4
−
y
) = 0 we obtain the critical points 0, 2, and 4.
From the phase portrait we see that 2 is asymptotically stable and 0 and 4 are unstable.
19

20. -2 -1 0
0 ln 9
mg/k
mg/k
√
α
β
α
Exercises 2.1
23.
Solving
y
ln(
y
+ 2) = 0 we obtain the critical points
−
1 and 0.
From the phase portrait we see that
−
1 is asymptotically stable and 0 is unstable.
24.
Solving
ye
y
−
9
y
=
y
(
e
y
−
9) = 0 we obtain the critical points 0 and ln 9.
From the phase portrait we see that 0 is asymptotically stable and ln 9 is unstable.
25. (a)
Writing the differential equation in the form
dv
dt
=
k
m
y
mg
k
−
v
/
we see that a critical point is
mg/k
.
From the phase portrait we see that
mg/k
is an asymptotically stable critical point. Thus,
lim
t
→∞
v
=
mg/k
.
(b)
Writing the differential equation in the form
dv
dt
=
k
m
y
mg
k
−
v
2
/
=
k
m
x
e
mg
k
−
v
vx
e
mg
k
+
v
v
we see that a critical point is
t
mg/k
.
From the phase portrait we see that
t
mg/k
is an asymptotically stable critical point. Thus,
lim
t
→∞
v
=
t
mg/k
.
26. (a)
From the phase portrait we see that critical points are
α
and
β
. Let
X
(0) =
X
0
.
If
X
0
<α
, we see that
X
→
α
as
t
→∞
.If
α<X
0
<β
, we see that
X
→
α
as
t
→∞
.
If
X
0
>β
, we see that
X
(
t
) increases in an unbounded manner, but more specific behavior of
X
(
t
)as
t
→∞
is not known.
(b)
When
α
=
β
the phase portrait is as shown.
If
X
0
<α
, then
X
(
t
)
→
α
as
t
→∞
.If
X
0
>α
, then
X
(
t
) increases in an unbounded manner. This
could happen in a finite amount of time. That is, the phase portrait does not indicate that
X
becomes
unbounded as
t
→∞
.
20

21. t
X
−2/α
α
α/2
t
X
1/α
α
2α
x
t
y
x
c0c
><>
3-3
3
-3
y
x
-2.2 0.5 1.7
><><
Exercises 2.1
(c)
When
k
= 1 and
α
=
β
the differential equation is
dX/dt
=(
α
−
X
)
2
. Separating variables and integrating
we have
dX
(
α
−
X
)
2
=
dt
1
α
−
X
=
t
+
c
α
−
X
=
1
t
+
c
X
=
α
−
1
t
+
c
.
For
X
(0) =
α/
2 we obtain
X
(
t
)=
α
−
1
t
+2
/α
.
For
X
(0) = 2
α
we obtain
X
(
t
)=
α
−
1
t
−
1
/α
.
For
X
0
>α
,
X
(
t
) increases without bound up to
t
=1
/α
.For
t>
1
/α
,
X
(
t
) increases but
X
→
α
as
t
→∞
.
27.
Critical points are
y
= 0 and
y
=
c
.
28.
Critical points are
y
=
−
2
.
2,
y
=0
.
5, and
y
=1
.
7.
29.
At each point on the circle of radius
c
the lineal element has slope
c
2
.
21

22. -3-2-10 1 2 3
-3
-2
-1
0
1
2
3
x
y
x
y
-
55
-
5
5
Exercises 2.1
30.
When
y<
1
2
x
2
,
y
/
=
x
2
−
2
y
is positive and the portions of solution curves
“inside” the nullcline parabola are increasing. When
y>
1
2
x
2
,
y
/
=
x
2
−
2
y
is
negative and the portions of the solution curves “outside” the nullcline parabola
are decreasing.
31.
Recall that for
dy/dx
=
f
(
y
) we are assuming that
f
and
f
/
are continuous functions of
y
on some interval
I
. Now suppose that the graph of a nonconstant solution of the differential equation crosses the line
y
=
c
.
If the point of intersection is taken as an initial condition we have two distinct solutions of the initial-value
problem. This violates uniqueness, so the graph of any nonconstant solution must lie entirely on one side of any
equilibrium solution. Since
f
is continuous it can only change signs around a point where it is 0. But this is a
critical point. Thus,
f
(
y
) is completely positive or completely negative in each region
R
i
.If
y
(
x
) is oscillatory
or has a relative extremum, then it must have a horizontal tangent line at some point (
x
0
,y
0
). In this case
y
0
would be a critical point of the differential equation, but we saw above that the graph of a nonconstant solution
cannot intersect the graph of the equilibrium solution
y
=
y
0
.
32.
By Problem 31, a solution
y
(
x
)of
dy/dx
=
f
(
y
) cannot have relative extrema and hence must be monotone.
Since
y
/
(
x
)=
f
(
y
)
>
0,
y
(
x
) is monotone increasing, and since
y
(
x
) is bounded above by
c
2
, lim
x
→∞
y
(
x
)=
L
,
where
L
≤
c
2
. We want to show that
L
=
c
2
. Since
L
is a horizontal asymptote of
y
(
x
), lim
x
→∞
y
/
(
x
)=0.
Using the fact that
f
(
y
) is continuous we have
f
(
L
)=
f
( lim
x
→∞
y
(
x
)) = lim
x
→∞
f
(
y
(
x
)) = lim
x
→∞
y
/
(
x
)=0
.
But then
L
is a critical point of
f
. Since
c
1
<L
≤
c
2
, and
f
has no critical points between
c
1
and
c
2
,
L
=
c
2
.
33. (a)
Assuming the existence of the second derivative, points of inflection of
y
(
x
) occur where
y
//
(
x
) = 0. From
dy/dx
=
g
(
y
)wehave
d
2
y/dx
2
=
g
/
(
y
)
dy/dx
. Thus, the
y
-coordinate of a point of
inflection can be located by solving
g
/
(
y
) = 0. (Points where
dy/dx
= 0 correspond to constant solutions
of the differential equation.)
(b)
Solving
y
2
−
y
−
6=(
y
−
3)(
y
+ 2) = 0 we see that 3 and
−
2 are critical
points. Now
d
2
y/dx
2
=(2
y
−
1)
dy/dx
=(2
y
−
1)(
y
−
3)(
y
+ 2), so the only
possible point of inflection is at
y
=
1
2
, although the concavity of solutions
can be different on either side of
y
=
−
2 and
y
= 3. Since
y
//
(
x
)
<
0 for
y<
−
2 and
1
2
<y<
3, and
y
//
(
x
)
>
0 for
−
2
<y<
1
2
and
y>
3, we
see that solution curves are concave down for
y<
−
2 and
1
2
<y<
3 and
concave up for
−
2
<y<
1
2
and
y>
3. Points of inflection of solutions of
autonomous differential equations will have the same
y
-coordinates because
between critical points they are horizontal translates of each other.
22

23. Exercises 2.2
Exercises 2.2
In many of the following problems we will encounter an expression of the form ln
|
g
(
y
)
|
=
f
(
x
)+
c
. To solve for
g
(
y
)
we exponentiate both sides of the equation. This yields
|
g
(
y
)
|
=
e
f
(
x
)+
c
=
e
c
e
f
(
x
)
which implies
g
(
y
)=
±
e
c
e
f
(
x
)
.
Letting
c
1
=
±
e
c
we obtain
g
(
y
)=
c
1
e
f
(
x
)
.
1.
From
dy
= sin 5
xdx
we obtain
y
=
−
1
5
cos 5
x
+
c
.
2.
From
dy
=(
x
+1)
2
dx
we obtain
y
=
1
3
(
x
+1)
3
+
c
.
3.
From
dy
=
−
e
−
3
x
dx
we obtain
y
=
1
3
e
−
3
x
+
c
.
4.
From
1
(
y
−
1)
2
dy
=
dx
we obtain
−
1
y
−
1
=
x
+
c
or
y
=1
−
1
x
+
c
.
5.
From
1
y
dy
=
4
x
dx
we obtain ln
|
y
|
=4ln
|
x
|
+
c
or
y
=
c
1
x
4
.
6.
From
1
y
dy
=
−
2
xdx
we obtain ln
|
y
|
=
−
x
2
+
c
or
y
=
c
1
e
−
x
2
.
7.
From
e
−
2
y
dy
=
e
3
x
dx
we obtain 3
e
−
2
y
+2
e
3
x
=
c
.
8.
From
ye
y
dy
=
k
e
−
x
+
e
−
3
x

dx
we obtain
ye
y
−
e
y
+
e
−
x
+
1
3
e
−
3
x
=
c
.
9.
From
x
y
+2+
1
y
v
dy
=
x
2
ln
xdx
we obtain
y
2
2
+2
y
+ln
|
y
|
=
x
3
3
ln
|
x
|−
1
9
x
3
+
c
.
10.
From
1
(2
y
+3)
2
dy
=
1
(4
x
+5)
2
dx
we obtain
2
2
y
+3
=
1
4
x
+5
+
c
.
11.
From
1
csc
y
dy
=
−
1
sec
2
x
dx
or sin
ydy
=
−
cos
2
xdx
=
−
1
2
(1 + cos 2
x
)
dx
we obtain
−
cos
y
=
−
1
2
x
−
1
4
sin 2
x
+
c
or 4 cos
y
=2
x
+ sin 2
x
+
c
1
.
12.
From 2
ydy
=
−
sin 3
x
cos
3
3
x
dx
=
−
tan 3
x
sec
2
3
xdx
we obtain
y
2
=
−
1
6
sec
2
3
x
+
c
.
13.
From
e
y
(
e
y
+1)
2
dy
=
−
e
x
(
e
x
+1)
3
dx
we obtain
−
(
e
y
+1)
−
1
=
1
2
(
e
x
+1)
−
2
+
c
.
14.
From
y
(1 +
y
2
)
1
/
2
dy
=
x
(1 +
x
2
)
1
/
2
dx
we obtain
k
1+
y
2

1
/
2
=
k
1+
x
2

1
/
2
+
c
.
15.
From
1
S
dS
=
kdr
we obtain
S
=
ce
kr
.
16.
From
1
Q
−
70
dQ
=
kdt
we obtain ln
|
Q
−
70
|
=
kt
+
c
or
Q
−
70 =
c
1
e
kt
.
17.
From
1
P
−
P
2
dP
=
x
1
P
+
1
1
−
P
v
dP
=
dt
we obtain ln
|
P
|−
ln
|
1
−
P
|
=
t
+
c
so that ln
P
1
−
P
=
t
+
c
or
P
1
−
P
=
c
1
e
t
. Solving for
P
we have
P
=
c
1
e
t
1+
c
1
e
t
.
18.
From
1
N
dN
=
k
te
t
+2
−
1

dt
we obtain ln
|
N
|
=
te
t
+2
−
e
t
+2
−
t
+
c
.
19.
From
y
−
2
y
+3
dy
=
x
−
1
x
+4
dx
or
x
1
−
5
y
+3
v
dy
=
x
1
−
5
x
+4
v
dx
we obtain
y
−
5ln
|
y
+3
|
=
x
−
5ln
|
x
+4
|
+
c
or
x
x
+4
y
+3
v
5
=
c
1
e
x
−
y
.
23

24. Exercises 2.2
20.
From
y
+1
y
−
1
dy
=
x
+2
x
−
3
dx
or
x
1+
2
y
−
1
v
dy
=
x
1+
5
x
−
3
v
dx
we obtain
y
+2ln
|
y
−
1
|
=
x
+5ln
|
x
−
3
|
+
c
or
(
y
−
1)
2
(
x
−
3)
5
=
c
1
e
x
−
y
.
21.
From
xdx
=
1
t
1
−
y
2
dy
we obtain
1
2
x
2
= sin
−
1
y
+
c
or
y
= sin
x
x
2
2
+
c
1
v
.
22.
From
1
y
2
dy
=
1
e
x
+
e
−
x
dx
=
e
x
(
e
x
)
2
+1
dx
we obtain
−
1
y
= tan
−
1
e
x
+
c
or
y
=
−
1
tan
−
1
e
x
+
c
.
23.
From
1
x
2
+1
dx
=4
dt
we obtain tan
−
1
x
=4
t
+
c
. Using
x
(
π/
4) = 1 we find
c
=
−
3
π/
4. The solution of the
initial-value problem is tan
−
1
x
=4
t
−
3
π
4
or
x
= tan
x
4
t
−
3
π
4
v
.
24.
From
1
y
2
−
1
dy
=
1
x
2
−
1
dx
or
1
2
x
1
y
−
1
−
1
y
+1
v
dy
=
1
2
x
1
x
−
1
−
1
x
+1
v
dx
we obtain
ln
|
y
−
1
|−
ln
|
y
+1
|
=ln
|
x
−
1
|−
ln
|
x
+1
|
+ln
c
or
y
−
1
y
+1
=
c
(
x
−
1)
x
+1
. Using
y
(2) = 2 we find
c
= 1. The solution of the initial-value problem is
y
−
1
y
+1
=
x
−
1
x
+1
or
y
=
x
.
25.
From
1
y
dy
=
1
−
x
x
2
dx
=
x
1
x
2
−
1
x
v
dx
we obtain ln
|
y
|
=
−
1
x
−
ln
|
x
|
=
c
or
xy
=
c
1
e
−
1
/x
. Using
y
(
−
1) =
−
1
we find
c
1
=
e
−
1
. The solution of the initial-value problem is
xy
=
e
−
1
−
1
/x
.
26.
From
1
1
−
2
y
dy
=
dt
we obtain
−
1
2
ln
|
1
−
2
y
|
=
t
+
c
or 1
−
2
y
=
c
1
e
−
2
t
. Using
y
(0)=5
/
2wefind
c
1
=
−
4.
The solution of the initial-value problem is 1
−
2
y
=
−
4
e
−
2
t
or
y
=2
e
−
2
t
+
1
2
.
27.
Separating variables and integrating we obtain
dx
√
1
−
x
2
−
dy
t
1
−
y
2
= 0 and sin
−
1
x
−
sin
−
1
y
=
c.
Setting
x
= 0 and
y
=
√
3
/
2 we obtain
c
=
−
π/
3. Thus, an implicit solution of the initial-value problem is
sin
−
1
x
−
sin
−
1
y
=
π/
3. Solving for
y
and using a trigonometric identity we get
y
= sin
y
sin
−
1
x
+
π
3
/
=
x
cos
π
3
+
t
1
−
x
2
sin
π
3
=
x
2
+
√
3
√
1
−
x
2
2
.
28.
From
1
1+(2
y
)
2
dy
=
−
x
1+(
x
2
)
2
dx
we obtain
1
2
tan
−
1
2
y
=
−
1
2
tan
−
1
x
2
+
c
or tan
−
1
2
y
+ tan
−
1
x
2
=
c
1
.
Using
y
(1) = 0 we find
c
1
=
π/
4. The solution of the initial-value problem is
tan
−
1
2
y
+ tan
−
1
x
2
=
π
4
.
29. (a)
The equilibrium solutions
y
(
x
)=2and
y
(
x
)=
−
2 satisfy the initial conditions
y
(0) = 2 and
y
(0) =
−
2,
respectively. Setting
x
=
1
4
and
y
=1in
y
= 2(1 +
ce
4
x
)
/
(1
−
ce
4
x
) we obtain
1=2
1+
ce
1
−
ce
,
1
−
ce
=2+2
ce,
−
1=3
ce,
and
c
=
−
1
3
e
.
24

25. Exercises 2.2
The solution of the corresponding initial-value problem is
y
=2
1
−
1
3
e
4
x
−
1
1+
1
3
e
4
x
−
1
=2
3
−
e
4
x
−
1
3+
e
4
x
−
1
.
(b)
Separating variables and integrating yields
1
4
ln
|
y
−
2
|−
1
4
ln
|
y
+2
|
+ln
c
1
=
x
ln
|
y
−
2
|−
ln
|
y
+2
|
+ln
c
=4
x
ln



c
(
y
−
2)
y
+2



=4
x
c
y
−
2
y
+2
=
e
4
x
.
Solving for
y
we get
y
=2(
c
+
e
4
x
)
/
(
c
−
e
4
x
). The initial condition
y
(0) =
−
2 can be solved for, yielding
c
= 0 and
y
(
x
)=
−
2. The initial condition
y
(0) = 2 does not correspond to a value of
c
, and it must
simply be recognized that
y
(
x
) = 2 is a solution of the initial-value problem. Setting
x
=
1
4
and
y
=1in
y
=2(
c
+
e
4
x
)
/
(
c
−
e
4
x
) leads to
c
=
−
3
e
. Thus, a solution of the initial-value problem is
y
=2
−
3
e
+
e
4
x
−
3
e
−
e
4
x
=2
3
−
e
4
x
−
1
3+
e
4
x
−
1
.
30.
From
x
1
y
−
1
+
−
1
y
v
dy
=
1
x
dx
we obtain ln
|
y
−
1
|−
ln
|
y
|
=ln
|
x
|
+
c
or
y
=
1
1
−
c
1
x
. Another solution is
y
=0.
(a)
If
y
(0) = 1 then
y
=1.
(b)
If
y
(0) = 0 then
y
=0.
(c)
If
y
(1
/
2) = 1
/
2 then
y
=
1
1+2
x
.
(d)
Setting
x
= 2 and
y
=
1
4
we obtain
1
4
=
1
1
−
c
1
(2)
,
1
−
2
c
1
=4
,
and
c
1
=
−
3
2
.
Thus,
y
=
1
1+
3
2
x
=
2
2+3
x
.
31.
Singular solutions of
dy/dx
=
x
t
1
−
y
2
are
y
=
−
1 and
y
= 1. A singular solution of (
e
x
+
e
−
x
)
dy/dx
=
y
2
is
y
=0.
32.
Differentiating ln(
x
2
+ 10) + csc
y
=
c
we get
2
x
x
2
+10
−
cot
y
csc
y
dy
dx
=0
,
2
x
x
2
+10
−
cos
y
sin
y
1
sin
y
dy
dx
=0
,
or
2
x
sin
2
ydx
−
(
x
2
+ 10) cos
ydy
=0
.
Writing the differential equation in the form
dy
dx
=
2
x
sin
2
y
(
x
2
+ 10) cos
y
we see that singular solutions occur when sin
2
y
=0,or
y
=
kπ
, where
k
is an integer.
25

26. -0.004-0.002 0.0020.004
x
0.97
0.98
1
1.01
y
-0.004-0.002 0.0020.004
x
0.98
0.99
1.01
1.02
y
-0.004-0.002 0.0020.004
x
0.9996
0.9998
1.0002
1.0004
y
-0.004-0.002 0.0020.004
x
0.9996
0.9998
1.0002
1.0004
y
Exercises 2.2
33.
The singular solution
y
= 1 satisfies the initial-value problem.
34.
Separating variables we obtain
dy
(
y
−
1)
2
=
dx
. Then
−
1
y
−
1
=
x
+
c
and
y
=
x
+
c
−
1
x
+
c
.
Setting
x
= 0 and
y
=1
.
01 we obtain
c
=
−
100. The solution is
y
=
x
−
101
x
−
100
.
35.
Separating variables we obtain
dy
(
y
−
1)
2
+0
.
01
=
dx
. Then
10 tan
−
1
10(
y
−
1) =
x
+
c
and
y
=1+
1
10
tan
x
+
c
10
.
Setting
x
= 0 and
y
= 1 we obtain
c
= 0. The solution is
y
=1+
1
10
tan
x
10
.
36.
Separating variables we obtain
dy
(
y
−
1)
2
−
0
.
01
=
dx
. Then
5ln




10
y
−
11
10
y
−
9




=
x
+
c.
Setting
x
= 0 and
y
= 1 we obtain
c
= 5 ln 1 = 0. The solution is
5ln




10
y
−
11
10
y
−
9




=
x.
37.
Separating variables, we have
dy
y
−
y
3
=
dy
y
(1
−
y
)(1 +
y
)
=
y
1
y
+
1
/
2
1
−
y
−
1
/
2
1+
y
/
dy
=
dx.
Integrating, we get
ln
|
y
|−
1
2
ln
|
1
−
y
|−
1
2
ln
|
1+
y
|
=
x
+
c.
26

27. 12345
x
-4
-2
2
4
y
-4-2 24
x
-4
-2
2
4
y
-4-2 24
x
-4
-2
2
4
y
12345
x
-4
-2
2
4
y
-4-2 2 4
x
-2
2
4
6
8
y
3
<>
Exercises 2.2
When
y>
1, this becomes
ln
y
−
1
2
ln(
y
−
1)
−
1
2
ln(
y
+1)=ln
y
t
y
2
−
1
=
x
+
c.
Letting
x
= 0 and
y
= 2 we find
c
=ln(2
/
√
3 ). Solving for
y
we get
y
1
(
x
)=2
e
x
/
√
4
e
2
x
−
3 , where
x>
ln(
√
3
/
2).
When 0
<y<
1wehave
ln
y
−
1
2
ln(1
−
y
)
−
1
2
ln(1 +
y
)=ln
y
t
1
−
y
2
=
x
+
c.
Letting
x
= 0 and
y
=
1
2
we find
c
= ln(1
/
√
3 ). Solving for
y
we get
y
2
(
x
)=
e
x
/
√
e
2
x
+ 3 , where
−∞
<x<
∞
.
When
−
1
<y<
0 we have
ln(
−
y
)
−
1
2
ln(1
−
y
)
−
1
2
ln(1 +
y
)=ln
−
y
t
1
−
y
2
=
x
+
c.
Letting
x
= 0 and
y
=
−
1
2
we find
c
= ln(1
/
√
3 ). Solving for
y
we get
y
3
(
x
)=
−
e
x
/
√
e
2
x
+ 3 , where
−∞
<x<
∞
.
When
y<
−
1wehave
ln(
−
y
)
−
1
2
ln(1
−
y
)
−
1
2
ln(
−
1
−
y
)=ln
−
y
t
y
2
−
1
=
x
+
c.
Letting
x
= 0 and
y
=
−
2wefind
c
= ln(2
/
√
3 ). Solving for
y
we get
y
4
(
x
)=
−
2
e
x
/
√
4
e
2
x
−
3 , where
x>
ln(
√
3
/
2).
38. (a)
The second derivative of
y
is
d
2
y
dx
2
=
−
dy/dx
(
y
−
1)
2
=
−
1
/
(
y
−
3)
(
y
−
3)
2
=
−
1
(
y
−
3)
3
.
The solution curve is concave up when
d
2
y/dx
2
>
0or
y>
3, and concave
down when
d
2
y/dx
2
<
0or
y<
3. From the phase portrait we see that the
solution curve is decreasing when
y<
3 and increasing when
y>
3.
27

28. -1 12345
x
-2
2
4
6
8
y
-5-4-3-2-1 12
x
-5
-4
-3
-2
-1
1
2
y
-2-1.5-1-0.5
x
-2
-1
1
2
y
Exercises 2.2
(b)
Separating variables and integrating we obtain
(
y
−
3)
dy
=
dx
1
2
y
2
−
3
y
=
x
+
c
y
2
−
6
y
+9=2
x
+
c
1
(
y
−
3)
2
=2
x
+
c
1
y
=3
±
√
2
x
+
c
1
.
The initial condition dictates whether to use the plus or minus sign.
When
y
1
(0) = 4 we have
c
1
= 1 and
y
1
(
x
)=3+
√
2
x
+1.
When
y
2
(0) = 2 we have
c
1
= 1 and
y
2
(
x
)=3
−
√
2
x
+1.
When
y
3
(1) = 2 we have
c
1
=
−
1 and
y
3
(
x
)=3
−
√
2
x
−
1.
When
y
4
(
−
1) = 4 we have
c
1
= 3 and
y
4
(
x
)=3+
√
2
x
+3.
39. (a)
Separating variables we have 2
ydy
=(2
x
+1)
dx
. Integrating gives
y
2
=
x
2
+
x
+
c
. When
y
(
−
2) =
−
1we
find
c
=
−
1, so
y
2
=
x
2
+
x
−
1 and
y
=
−
√
x
2
+
x
−
1 . The negative square root is chosen because of the
initial condition.
(b)
The interval of definition appears to be approximately (
−∞
,
−
1
.
65).
(c)
Solving
x
2
+
x
−
1 = 0 we get
x
=
−
1
2
±
1
2
√
5 , so the exact interval of definition is (
−∞
,
−
1
2
−
1
2
√
5).
40. (a)
From Problem 7 the general solution is 3
e
−
2
y
+2
e
3
x
=
c
. When
y
(0) = 0 we find
c
=5,so3
e
−
2
y
+2
e
3
x
=5.
Solving for
y
we get
y
=
−
1
2
ln
1
3
(5
−
2
e
3
x
).
(b)
The interval of definition appears to be approximately (
−∞
,
0
.
3).
(c)
Solving
1
3
(5
−
2
e
3
x
)=0weget
x
=
1
3
ln(
5
2
), so the exact interval of definition is (
−∞
,
1
3
ln(
5
2
)).
41. (a)
While
y
2
(
x
)=
−
√
25
−
x
2
is defined at
x
=
−
5 and
x
=5,
y
/
1
(
x
) is not defined at these values, and so the
interval of definition is the open interval (
−
5
,
5).
(b)
At any point on the
x
-axis the derivative of
y
(
x
) is undefined, so no solution curve can cross the
x
-axis.
Since
−
x/y
is not defined when
y
= 0, the initial-value problem has no solution.
42. (a)
Separating variables and integrating we obtain
x
2
−
y
2
=
c
.For
c
k
= 0 the graph is a square hyperbola
centered at the origin. All four initial conditions imply
c
= 0 and
y
=
±
x
. Since the differential equation is
28

29. -6-4-2 2 4 6 8
x
0.5
1
1.5
2
2.5
3
3.5
y
x
y
−3 3
−3
3
Exercises 2.2
not defined for
y
= 0, solutions are
y
=
±
x
,
x<
0 and
y
=
±
x
,
x>
0. The solution for
y
(
a
)=
a
is
y
=
x
,
x>
0; for
y
(
a
)=
−
a
is
y
=
−
x
; for
y
(
−
a
)=
a
is
y
=
−
x
,
x<
0; and for
y
(
−
a
)=
−
a
is
y
=
x
,
x<
0.
(b)
Since
x/y
is not defined when
y
= 0, the initial-value problem has no solution.
(c)
Setting
x
= 1 and
y
=2in
x
2
−
y
2
=
c
we get
c
=
−
3, so
y
2
=
x
2
+ 3 and
y
(
x
)=
√
x
2
+ 3 , where
the positive square root is chosen because of the initial condition. The domain is all real numbers since
x
2
+3
>
0 for all
x
.
43.
Separating variables we have
dy/
k
t
1+
y
2
sin
2
y

=
dx
which is
not readily integrated (even by a CAS). We note that
dy/dx
≥
0
for all values of
x
and
y
and that
dy/dx
= 0 when
y
= 0 and
y
=
π
,
which are equilibrium solutions.
44.
Separating variables we have
dy/
(
√
y
+
y
)=
dx/
(
√
x
+
x
). To integrate

dt/
(
√
t
+
t
) we substitute
u
2
=
t
and
get

2
u
u
+
u
2
du
=

2
1+
u
du
=2ln
|
1+
u
|
+
c
= 2 ln(1 +
√
x
)+
c.
Integrating the separated differential equation we have
2ln(1+
√
y
) = 2 ln(1 +
√
x
)+
c
or ln(1 +
√
y
) = ln(1 +
√
x
)+ln
c
1
.
Solving for
y
we get
y
=[
c
1
(1 +
√
x
)
−
1]
2
.
45.
We are looking for a function
y
(
x
) such that
y
2
+
x
dy
dx
v
2
=1
.
Using the positive square root gives
dy
dx
=
t
1
−
y
2
=
⇒
dy
t
1
−
y
2
=
dx
=
⇒
sin
−
1
y
=
x
+
c.
Thus a solution is
y
= sin(
x
+
c
). If we use the negative square root we obtain
y
= sin(
c
−
x
)=
−
sin(
x
−
c
)=
−
sin(
x
+
c
1
)
.
Note also that
y
= 1 and
y
=
−
1 are solutions.
46. (a)
(b)
For
|
x
|
>
1 and
|
y
|
>
1 the differential equation is
dy/dx
=
t
y
2
−
1
/
√
x
2
−
1 . Separating variables and
integrating, we obtain
dy
t
y
2
−
1
=
dx
√
x
2
−
1
and cosh
−
1
y
= cosh
−
1
x
+
c.
29

30. -4-2024
-4
-2
0
2
4
x
y
-4 -2 0 2 4
-4
-2
0
2
4
x
y
c=-2
c=10
c=67
c=-31
-6-4-2 0 2 4 6
-4
-2
0
2
4
x
y
-2 0 2 4 6
-4
-2
0
2
4
x
y
-4-2 0 2 4 6 8 10
-8
-6
-4
-2
0
2
4
x
y
Exercises 2.2
Setting
x
= 2 and
y
= 2 we find
c
= cosh
−
1
2
−
cosh
−
1
2 = 0 and cosh
−
1
y
= cosh
−
1
x
. An explicit solution
is
y
=
x
.
47. (a)
Separating variables and integrating, we have
(3
y
2
+1)
dy
=
−
(8
x
+5)
dx
and
y
3
+
y
=
−
4
x
2
−
5
x
+
c.
Using a CAS we show various contours of
f
(
x, y
)=
y
3
+
y
+4
x
2
+5
x
. The plots
shown on [
−
5
,
5]
×
[
−
5
,
5] correspond to
c
-values of 0,
±
5,
±
20,
±
40,
±
80, and
±
125.
(b)
The value of
c
corresponding to
y
(0) =
−
1is
f
(0
,
−
1) =
−
2; to
y
(0) = 2 is
f
(0
,
2) = 10; to
y
(
−
1)=4is
f
(
−
1
,
4) = 67; and to
y
(
−
1) =
−
3is
−
31.
48. (a)
Separating variables and integrating, we have
(
−
2
y
+
y
2
)
dy
=(
x
−
x
2
)
dx
and
−
y
2
+
1
3
y
3
=
1
2
x
2
−
1
3
x
3
+
c.
Using a CAS we show some contours of
f
(
x, y
)=2
y
3
−
6
y
2
+2
x
3
−
3
x
2
.
The plots shown on [
−
7
,
7]
×
[
−
5
,
5] correspond to
c
-values of
−
450,
−
300,
−
200,
−
120,
−
60,
−
20,
−
10,
−
8
.
1,
−
5,
−
0
.
8, 20, 60, and 120.
(b)
The value of
c
corresponding to
y
(0) =
3
2
is
f
k
0
,
3
2

=
−
27
4
. The
portion of the graph between the dots corresponds to the solution
curve satisfying the intial condition. To determine the interval of
definition we find
dy/dx
for 2
y
3
−
6
y
2
+2
x
3
−
3
x
2
=
−
27
4
. Using
implicit differentiation we get
y
/
=(
x
−
x
2
)
/
(
y
2
−
2
y
), which is infinite
when
y
= 0 and
y
= 2. Letting
y
= 0 in 2
y
3
−
6
y
2
+2
x
3
−
3
x
2
=
−
27
4
and using a CAS to solve for
x
we get
x
=
−
1
.
13232. Similarly,
letting
y
= 2, we find
x
=1
.
71299. The largest interval of definition
is approximately (
−
1
.
13232
,
1
.
71299).
(c)
The value of
c
corresponding to
y
(0) =
−
2is
f
(0
,
−
2) =
−
40. The
portion of the graph to the right of the dot corresponds to the solu-
tion curve satisfying the initial condition. To determine the interval
of definition we find
dy/dx
for 2
y
3
−
6
y
2
+2
x
3
−
3
x
2
=
−
40. Using
implicit differentiation we get
y
/
=(
x
−
x
2
)
/
(
y
2
−
2
y
), which is infinite
when
y
= 0 and
y
= 2. Letting
y
= 0 in 2
y
3
−
6
y
2
+2
x
3
−
3
x
2
=
−
40
and using a CAS to solve for
x
we get
x
=
−
2
.
29551. The largest
interval of definition is approximately (
−
2
.
29551
,
∞
).
30

31. Exercises 2.3
Exercises 2.3
1.
For
y
/
−
5
y
= 0 an integrating factor is
e
−

5
dx
=
e
−
5
x
so that
d
dx

e
−
5
x
y
<
= 0 and
y
=
ce
5
x
for
−∞
<x<
∞
.
2.
For
y
/
+2
y
= 0 an integrating factor is
e

2
dx
=
e
2
x
so that
d
dx

e
2
x
y
<
= 0 and
y
=
ce
−
2
x
for
−∞
<x<
∞
.
The transient term is
ce
−
2
x
.
3.
For
y
/
+
y
=
e
3
x
an integrating factor is
e

dx
=
e
x
so that
d
dx
[
e
x
y
]=
e
4
x
and
y
=
1
4
e
3
x
+
ce
−
x
for
−∞
<x<
∞
.
The transient term is
ce
−
x
.
4.
For
y
/
+4
y
=
4
3
an integrating factor is
e

4
dx
=
e
4
x
so that
d
dx

e
4
x
y
<
=
4
3
e
4
x
and
y
=
1
3
+
ce
−
4
x
for
−∞
<x<
∞
. The transient term is
ce
−
4
x
.
5.
For
y
/
+3
x
2
y
=
x
2
an integrating factor is
e

3
x
2
dx
=
e
x
3
so that
d
dx
>
e
x
3
y
c
=
x
2
e
x
3
and
y
=
1
3
+
ce
−
x
3
for
−∞
<x<
∞
. The transient term is
ce
−
x
3
.
6.
For
y
/
+2
xy
=
x
3
an integrating factor is
e

2
xdx
=
e
x
2
so that
d
dx
>
e
x
2
y
c
=
x
3
e
x
2
and
y
=
1
2
x
2
−
1
2
+
ce
−
x
2
for
−∞
<x<
∞
. The transient term is
ce
−
x
2
.
7.
For
y
/
+
1
x
y
=
1
x
2
an integrating factor is
e

(1
/x
)
dx
=
x
so that
d
dx
[
xy
]=
1
x
and
y
=
1
x
ln
x
+
c
x
for 0
<x<
∞
.
8.
For
y
/
−
2
y
=
x
2
+ 5 an integrating factor is
e
−

2
dx
=
e
−
2
x
so that
d
dx

e
−
2
x
y
<
=
x
2
e
−
2
x
+5
e
−
2
x
and
y
=
−
1
2
x
2
−
1
2
x
−
11
4
+
ce
2
x
for
−∞
<x<
∞
.
9.
For
y
/
−
1
x
y
=
x
sin
x
an integrating factor is
e
−

(1
/x
)
dx
=
1
x
so that
d
dx
.
1
x
y
f
= sin
x
and
y
=
cx
−
x
cos
x
for
0
<x<
∞
.
10.
For
y
/
+
2
x
y
=
3
x
an integrating factor is
e

(2
/x
)
dx
=
x
2
so that
d
dx

x
2
y
<
=3
x
and
y
=
3
2
+
cx
−
2
for 0
<x<
∞
.
11.
For
y
/
+
4
x
y
=
x
2
−
1 an integrating factor is
e

(4
/x
)
dx
=
x
4
so that
d
dx

x
4
y
<
=
x
6
−
x
4
and
y
=
1
7
x
3
−
1
5
x
+
cx
−
4
for 0
<x<
∞
.
12.
For
y
/
−
x
(1 +
x
)
y
=
x
an integrating factor is
e
−

[
x/
(1+
x
)]
dx
=(
x
+1)
e
−
x
so that
d
dx

(
x
+1)
e
−
x
y
<
=
x
(
x
+1)
e
−
x
and
y
=
−
x
−
2
x
+3
x
+1
+
ce
x
x
+1
for
−
1
<x<
∞
.
13.
For
y
/
+
x
1+
2
x
v
y
=
e
x
x
2
an integrating factor is
e

[1+(2
/x
)]
dx
=
x
2
e
x
so that
d
dx

x
2
e
x
y
<
=
e
2
x
and
y
=
1
2
e
x
x
2
+
ce
−
x
x
2
for 0
<x<
∞
. The transient term is
ce
−
x
x
2
.
14.
For
y
/
+
x
1+
1
x
v
y
=
1
x
e
−
x
sin 2
x
an integrating factor is
e

[1+(1
/x
)]
dx
=
xe
x
so that
d
dx
[
xe
x
y
] = sin 2
x
and
y
=
−
1
2
x
e
−
x
cos 2
x
+
ce
−
x
x
for 0
<x<
∞
. The entire solution is transient.
15.
For
dx
dy
−
4
y
x
=4
y
5
an integrating factor is
e
−

(4
/y
)
dy
=
y
−
4
so that
d
dy

y
−
4
x
<
=4
y
and
x
=2
y
6
+
cy
4
for
0
<y<
∞
.
31

32. Exercises 2.3
16.
For
dx
dy
+
2
y
x
=
e
y
an integrating factor is
e

(2
/y
)
dy
=
y
2
so that
d
dy

y
2
x
<
=
y
2
e
y
and
x
=
e
y
−
2
y
e
y
+
2
y
2
+
c
y
2
for 0
<y<
∞
. The transient term is
2+
c
y
2
.
17.
For
y
/
+ (tan
x
)
y
= sec
x
an integrating factor is
e

tan
xdx
= sec
x
so that
d
dx
[(sec
x
)
y
] = sec
2
x
and
y
= sin
x
+
c
cos
x
for
−
π/
2
<x<π/
2.
18.
For
y
/
+ (cot
x
)
y
= sec
2
x
csc
x
an integrating factor is
e

cot
xdx
= sin
x
so that
d
dx
[(sin
x
)
y
] = sec
2
x
and
y
= sec
x
+
c
csc
x
for 0
<x<π/
2.
19.
For
y
/
+
x
+2
x
+1
y
=
2
xe
−
x
x
+1
an integrating factor is
e

[(
x
+2)
/
(
x
+1)]
dx
=(
x
+1)
e
x
so that
d
dx
[(
x
+1)
e
x
y
]=2
x
and
y
=
x
2
x
+1
e
−
x
+
c
x
+1
e
−
x
for
−
1
<x<
∞
. The entire solution is transient.
20.
For
y
/
+
4
x
+2
y
=
5
(
x
+2)
2
an integrating factor is
e

[4
/
(
x
+2)]
dx
=(
x
+2)
4
so that
d
dx

(
x
+2)
4
y
<
=5(
x
+2)
2
and
y
=
5
3
(
x
+2)
−
1
+
c
(
x
+2)
−
4
for
−
2
<x<
∞
. The entire solution is transient.
21.
For
dr
dθ
+
r
sec
θ
= cos
θ
an integrating factor is
e

sec
θdθ
= sec
θ
+ tan
θ
so that
d
dθ
[
r
(sec
θ
+ tan
θ
)] = 1 + sin
θ
and
r
(sec
θ
+ tan
θ
)=
θ
−
cos
θ
+
c
for
−
π/
2
<θ<π/
2.
22.
For
dP
dt
+(2
t
−
1)
P
=4
t
−
2 an integrating factor is
e

(2
t
−
1)
dt
=
e
t
2
−
t
so that
d
dt
>
Pe
t
2
−
t
c
=(4
t
−
2)
e
t
2
−
t
and
P
=2+
ce
t
−
t
2
for
−∞
<t<
∞
. The transient term is
ce
t
−
t
2
.
23.
For
y
/
+
x
3+
1
x
v
y
=
e
−
3
x
x
an integrating factor is
e

[3+(1
/x
)]
dx
=
xe
3
x
so that
d
dx

xe
3
x
y
<
= 1 and
y
=
e
−
3
x
+
ce
−
3
x
x
for 0
<x<
∞
. The transient term is
ce
−
3
x
/x
.
24.
For
y
/
+
2
x
2
−
1
y
=
x
+1
x
−
1
an integrating factor is
e

[2
/
(
x
2
−
1)]
dx
=
x
−
1
x
+1
so that
d
dx
.
x
−
1
x
+1
y
f
= 1 and
(
x
−
1)
y
=
x
(
x
+1)+
c
(
x
+ 1) for
−
1
<x<
1.
25.
For
y
/
+
1
x
y
=
1
x
e
x
an integrating factor is
e

(1
/x
)
dx
=
x
so that
d
dx
[
xy
]=
e
x
and
y
=
1
x
e
x
+
c
x
for 0
<x<
∞
.
If
y
(1) = 2 then
c
=2
−
e
and
y
=
1
x
e
x
+
2
−
e
x
.
26.
For
dx
dy
−
1
y
x
=2
y
an integrating factor is
e
−

(1
/y
)
dy
=
1
y
so that
d
dy
.
1
y
x
f
= 2 and
x
=2
y
2
+
cy
for
−∞
<y<
∞
.If
y
(1) = 5 then
c
=
−
49
/
5 and
x
=2
y
2
−
49
5
y
.
27.
For
di
dt
+
R
L
i
=
E
L
an integrating factor is
e

(
R/L
)
dt
=
e
Rt/L
so that
d
dt
>
ie
Rt/L
c
=
E
L
e
Rt/L
and
i
=
E
R
+
ce
−
Rt/L
for
−∞
<t<
∞
.If
i
(0) =
i
0
then
c
=
i
0
−
E/R
and
i
=
E
R
+
x
i
0
−
E
R
v
e
−
Rt/L
.
28.
For
dT
dt
−
kT
=
−
T
m
k
an integrating factor is
e

(
−
k
)
dt
=
e
−
kt
so that
d
dt
[
Te
−
kt
]=
−
T
m
ke
−
kt
and
T
=
T
m
+
ce
kt
for
−∞
<t<
∞
.If
T
(0) =
T
0
then
c
=
T
0
−
T
m
and
T
=
T
m
+(
T
0
−
T
m
)
e
kt
.
29.
For
y
/
+
1
x
+1
y
=
ln
x
x
+1
an integrating factor is
e

[1
/
(
x
+1)]
dx
=
x
+ 1 so that
d
dx
[(
x
+1)
y
]=ln
x
and
y
=
x
x
+1
ln
x
−
x
x
+1
+
c
x
+1
for 0
<x<
∞
.If
y
(1) = 10 then
c
= 21 and
y
=
x
x
+1
ln
x
−
x
x
+1
+
21
x
+1
.
32

33. x
y
5
1
x
y
5
1
-
1
x
y
3
2
x
y
5
-
1
1
Exercises 2.3
30.
For
y
β
+ (tan
x
)
y
= cos
2
x
an integrating factor is
e

tan
xdx
= sec
x
so that
d
dx
[(sec
x
)
y
] = cos
x
and
y
=
sin
x
cos
x
+
c
cos
x
for
−
π/
2
<x<π/
2. If
y
(0) =
−
1 then
c
=
−
1 and
y
= sin
x
cos
x
−
cos
x
.
31.
For
y
β
+2
y
=
f
(
x
) an integrating factor is
e
2
x
so that
ye
2
x
=
I
1
2
e
2
x
+
c
1
,
0
≤
x
≤
3;
c
2
,x>
3.
If
y
(0) = 0 then
c
1
=
−
1
/
2 and for continuity we must have
c
2
=
1
2
e
6
−
1
2
so that
y
=
R
1
2
(1
−
e
−
2
x
)
,
0
≤
x
≤
3;
1
2
(
e
6
−
1)
e
−
2
x
,x>
3.
32.
For
y
β
+
y
=
f
(
x
) an integrating factor is
e
x
so that
ye
x
=
I
e
x
+
c
1
,
0
≤
x
≤
1;
−
e
x
+
c
2
,x>
1.
If
y
(0) = 1 then
c
1
= 0 and for continuity we must have
c
2
=2
e
so that
y
=
I
1
,
0
≤
x
≤
1;
2
e
1
−
x
−
1
,x>
1.
33.
For
y
β
+2
xy
=
f
(
x
) an integrating factor is
e
x
2
so that
ye
x
2
=
R
1
2
e
x
2
+
c
1
,
0
≤
x
≤
1;
c
2
,x>
1.
If
y
(0) = 2 then
c
1
=3
/
2 and for continuity we must have
c
2
=
1
2
e
+
3
2
so that
y
=
R
1
2
+
3
2
e
−
x
2
,
0
≤
x
≤
1;

1
2
e
+
3
2

e
−
x
2
,x>
1.
34.
For
y
β
+
2
x
1+
x
2
y
=





x
1+
x
2
,
0
≤
x
≤
1;
−
x
1+
x
2
,x>
1
an integrating factor is 1 +
x
2
so that

1+
x
2

y
=
R
1
2
x
2
+
c
1
,
0
≤
x
≤
1;
−
1
2
x
2
+
c
2
,x>
1.
If
y
(0) = 0 then
c
1
= 0 and for continuity we must have
c
2
= 1 so that
y
=







1
2
−
1
2(1+
x
2
)
,
0
≤
x
≤
1;
3
2(1+
x
2
)
−
1
2
,x>
1.
33

34. x
y
3
5
10
15
20
0 1 2 3 4 5
x
0.2
0.4
0.6
0.8
1
y
Exercises 2.3
35.
We need

P
(
x
)
dx
=
I
2
x,
0
≤
x
≤
1
−
2ln
x, x >
1
.
An integrating factor is
e

P
(
x
)
dx
=
I
e
2
x
,
0
≤
x
≤
1
1
/x
2
,x>
1
and
d
dx
.I
ye
2
x
,
0
≤
x
≤
1
y/x
2
,x>
1
f
=
I
4
xe
2
x
,
0
≤
x
≤
1
4
/x, x >
1
.
Integrating we get
I
ye
2
x
,
0
≤
x
≤
1
y/x
2
,x>
1
=
I
2
xe
2
x
−
e
2
x
+
c
1
,
0
≤
x
≤
1
4ln
x
+
c
2
,x>
1
.
Using
y
(0) = 3 we find
c
1
= 4. For continuity we must have
c
2
=2
−
1+4
e
−
2
=1+4
e
−
2
. Then
y
=
I
2
x
−
1+4
e
−
2
x
,
0
≤
x
≤
1
4
x
2
ln
x
+(1+4
e
−
2
)
x
2
,x>
1
.
36. (a)
An integrating factor for
y
β
−
2
xy
=
−
1is
e
−
x
2
.Thus
d
dx
[
e
−
x
2
y
]=
−
e
−
x
2
e
−
x
2
y
=
−

x
0
e
−
t
2
dt
=
−
√
π
2
erf(
x
)+
c.
From
y
(0) =
√
π/
2, and noting that erf(0) = 0, we get
c
=
√
π/
2. Thus
y
=
e
x
2
−
−
√
π
2
erf(
x
)+
√
π
2
/
=
√
π
2
e
x
2
(1
−
erf(
x
)) =
√
π
2
e
x
2
erfc(
x
)
.
(b)
Using
Mathematica
we find
y
(2)
≈
0
.
226339.
37.
An integrating factor for
y
β
−
2
xy
=1is
e
−
x
2
.Thus
d
dx
[
e
−
x
2
y
]=
e
−
x
2
e
−
x
2
y
=

x
0
e
−
t
2
dt
= erf(
x
)+
c
and
y
=
e
x
2
erf(
x
)+
ce
x
2
.
From
y
(1) = 1 we get 1 =
e
erf(1) +
ce
, so that
c
=
e
−
1
−
erf(1). Thus
y
=
e
x
2
erf(
x
)+(
e
−
1
−
erf(1))
e
x
2
=
e
x
2
−
1
+
e
x
2
(erf(
x
)
−
erf(1))
.
38. (a)
An integrating factor for
y
β
+
2
x
y
=
10 sin
x
x
3
34

35. 12345
x
-5
-4
-3
-2
-1
1
2
y
2 4 6 8
x
2
4
6
8
10
12
14
y
Exercises 2.3
is
x
2
.Thus
d
dx
[
x
2
y
]=10
sin
x
x
x
2
y
=10

x
0
sin
t
t
dt
+
c
y
=10
x
−
2
Si(
x
)+
cx
−
2
.
From
y
(1) = 0 we get
c
=
−
10Si(1). Thus
y
=10
x
−
2
Si(
x
)
−
10
x
−
2
Si(1) = 10
x
−
2
(Si(
x
)
−
Si(1))
.
(b)
Using
Mathematica
we find
y
(2)
≈
1
.
64832.
39. (a)
Separating variables and integrating, we have
dy
y
= sin
x
2
dx
and ln
|
y
|
=

x
0
sin
t
2
dt
+
c.
Now, letting
t
=
t
π/
2
u
we have

x
0
sin
t
2
dt
=
e
π
2

√
2
/π x
0
sin
y
π
2
u
2
/
du,
so
y
=
c
1
e

x
0
sin
t
2
dt
=
c
1
e
√
π/
2

√
2
/π x
0
sin(
πu
2
/
2)
du
=
c
1
e
√
π/
2
S
(
√
2
/π x
)
.
Using
S
(0) = 0 and
y
(0) = 5 we see that
c
1
= 5 and
y
=5
e
√
π/
2
S
(
√
2
/π x
)
.
(b)
Using
Mathematica
we find
y
(2)
≈
11
.
181. From the graph we
see that as
x
→∞
,
y
(
x
) oscillates with decreasing amplitude
approaching 9
.
35672.
40.
For
y
/
+
e
x
y
= 1 an integrating factor is
e
e
x
.Thus
d
dx

e
e
x
y
<
=
e
e
x
and
e
e
x
y
=

x
0
e
e
t
dt
+
c.
From
y
(0) = 1 we get
c
=
e
,so
y
=
e
−
e
x

x
0
e
e
t
dt
+
e
1
−
e
x
.
When
y
/
+
e
x
y
= 0 we can separate variables and integrate:
dy
y
=
−
e
x
dx
and ln
|
y
|
=
−
e
x
+
c.
Thus
y
=
c
1
e
−
e
x
. From
y
(0) = 1 we get
c
1
=
e
,so
y
=
e
1
−
e
x
.
When
y
/
+
e
x
y
=
e
x
we can see by inspection that
y
= 1 is a solution.
35

36. Exercises 2.3
41.
On the interval (
−
3
,
3) the integrating factor is
e

x dx/
(
x
2
−
9)
=
e
−

x dx/
(9
−
x
2
)
=
e
1
2
ln(9
−
x
2
)
=
t
9
−
x
2
and so
d
dx
>
t
9
−
x
2
y
c
= 0 and
y
=
c
√
9
−
x
2
.
42.
We want 4 to be a critical point, so use
y
/
=4
−
y
.
43. (a)
All solutions of the form
y
=
x
5
e
x
−
x
4
e
x
+
cx
4
satisfy the initial condition. In this case, since 4
/x
is
discontinuous at
x
= 0, the hypotheses of Theorem 1
.
1 are not satisfied and the initial-value problem does
not have a unique solution.
(b)
The differential equation has no solution satisfying
y
(0) =
y
0
,
y
0
k
=0.
(c)
In this case, since
x
0
k
= 0, Theorem 1
.
1 applies and the initial-value problem has a unique solution given by
y
=
x
5
e
x
−
x
4
e
x
+
cx
4
where
c
=
y
0
/x
4
0
−
x
0
e
x
0
+
e
x
0
.
44.
We want the general solution to be
y
=3
x
−
5+
ce
−
x
. (Rather than
e
−
x
, any function that approaches 0 as
x
→∞
could be used.) Differentiating we get
y
/
=3
−
ce
−
x
=3
−
(
y
−
3
x
+5)=
−
y
+3
x
−
2
,
so the differential equation
y
/
+
y
=3
x
−
2 has solutions asymptotic to the line
y
=3
x
−
5.
45.
The left-hand derivative of the function at
x
= 1 is 1
/e
and the right-hand derivative at
x
= 1 is 1
−
1
/e
. Thus,
y
is not differentiable at
x
=1.
46.
Differentiating
y
c
=
c/x
3
we get
y
/
c
=
−
3
c
x
4
=
−
3
x
c
x
3
=
−
3
x
y
c
so a differential equation with general solution
y
c
=
c/x
3
is
xy
/
+3
y
=0. Now
xy
/
p
+3
y
p
=
x
(3
x
2
)+3(
x
3
)=6
x
3
so a differential equation with general solution
y
=
c/x
3
+
x
3
is
xy
/
+3
y
=6
x
3
.
47.
Since
e

P
(
x
)
dx
+
c
=
e
c
e

P
(
x
)
dx
=
c
1
e

P
(
x
)
dx
, we would have
c
1
e

P
(
x
)
dx
y
=
c
2
+

c
1
e

P
(
x
)
dx
f
(
x
)
dx
and
e

P
(
x
)
dx
y
=
c
3
+

e

P
(
x
)
dx
f
(
x
)
dx,
which is the same as (6) in the text.
48.
We see by inspection that
y
= 0 is a solution.
36

37. Exercises 2.4
Exercises 2.4
1.
Let
M
=2
x
−
1 and
N
=3
y
+ 7 so that
M
y
=0=
Nx
. From
f
x
=2
x
−
1 we obtain
f
=
x
2
−
x
+
h
(
y
),
h
/
(
y
)=3
y
+ 7, and
h
(
y
)=
3
2
y
2
+7
y
. The solution is
x
2
−
x
+
3
2
y
2
+7
y
=
c
.
2.
Let
M
=2
x
+
y
and
N
=
−
x
−
6
y
. Then
M
y
= 1 and
N
x
=
−
1, so the equation is not exact.
3.
Let
M
=5
x
+4
y
and
N
=4
x
−
8
y
3
so that
M
y
=4=
N
x
. From
f
x
=5
x
+4
y
we obtain
f
=
5
2
x
2
+4
xy
+
h
(
y
),
h
/
(
y
)=
−
8
y
3
, and
h
(
y
)=
−
2
y
4
. The solution is
5
2
x
2
+4
xy
−
2
y
4
=
c
.
4.
Let
M
= sin
y
−
y
sin
x
and
N
= cos
x
+
x
cos
y
−
y
so that
M
y
= cos
y
−
sin
x
=
N
x
. From
f
x
= sin
y
−
y
sin
x
we obtain
f
=
x
sin
y
+
y
cos
x
+
h
(
y
),
h
/
(
y
)=
−
y
, and
h
(
y
)=
1
2
y
2
. The solution is
x
sin
y
+
y
cos
x
−
1
2
y
2
=
c
.
5.
Let
M
=2
y
2
x
−
3 and
N
=2
yx
2
+4 so that
M
y
=4
xy
=
N
x
. From
f
x
=2
y
2
x
−
3 we obtain
f
=
x
2
y
2
−
3
x
+
h
(
y
),
h
/
(
y
) = 4, and
h
(
y
)=4
y
. The solution is
x
2
y
2
−
3
x
+4
y
=
c
.
6.
Let
M
=4
x
3
−
3
y
sin 3
x
−
y/x
2
and
N
=2
y
−
1
/x
+cos 3
x
so that
M
y
=
−
3 sin 3
x
−
1
/x
2
and
N
x
=1
/x
2
−
3 sin 3
x
.
The equation is not exact.
7.
Let
M
=
x
2
−
y
2
and
N
=
x
2
−
2
xy
so that
M
y
=
−
2
y
and
N
x
=2
x
−
2
y
. The equation is not exact.
8.
Let
M
=1+ln
x
+
y/x
and
N
=
−
1+ln
x
so that
M
y
=1
/x
=
N
x
. From
f
y
=
−
1+ln
x
we obtain
f
=
−
y
+
y
ln
x
+
h
(
y
),
h
/
(
x
)=1+ln
x
, and
h
(
y
)=
x
ln
x
. The solution is
−
y
+
y
ln
x
+
x
ln
x
=
c
.
9.
Let
M
=
y
3
−
y
2
sin
x
−
x
and
N
=3
xy
2
+2
y
cos
x
so that
M
y
=3
y
2
−
2
y
sin
x
=
N
x
. From
f
x
=
y
3
−
y
2
sin
x
−
x
we obtain
f
=
xy
3
+
y
2
cos
x
−
1
2
x
2
+
h
(
y
),
h
/
(
y
) = 0, and
h
(
y
) = 0. The solution is
xy
3
+
y
2
cos
x
−
1
2
x
2
=
c
.
10.
Let
M
=
x
3
+
y
3
and
N
=3
xy
2
so that
M
y
=3
y
2
=
N
x
. From
f
x
=
x
3
+
y
3
we obtain
f
=
1
4
x
4
+
xy
3
+
h
(
y
),
h
/
(
y
) = 0, and
h
(
y
) = 0. The solution is
1
4
x
4
+
xy
3
=
c
.
11.
Let
M
=
y
ln
y
−
e
−
xy
and
N
=1
/y
+
x
ln
y
so that
M
y
=1+ln
y
+
ye
−
xy
and
N
x
=ln
y
. The equation is not
exact.
12.
Let
M
=3
x
2
y
+
e
y
and
N
=
x
3
+
xe
y
−
2
y
so that
M
y
=3
x
2
+
e
y
=
N
x
. From
f
x
=3
x
2
y
+
e
y
we obtain
f
=
x
3
y
+
xe
y
+
h
(
y
),
h
/
(
y
)=
−
2
y
, and
h
(
y
)=
−
y
2
. The solution is
x
3
y
+
xe
y
−
y
2
=
c
.
13.
Let
M
=
y
−
6
x
2
−
2
xe
x
and
N
=
x
so that
M
y
=1=
N
x
. From
f
x
=
y
−
6
x
2
−
2
xe
x
we obtain
f
=
xy
−
2
x
3
−
2
xe
x
+2
e
x
+
h
(
y
),
h
/
(
y
) = 0, and
h
(
y
) = 0. The solution is
xy
−
2
x
3
−
2
xe
x
+2
e
x
=
c
.
14.
Let
M
=1
−
3
/x
+
y
and
N
=1
−
3
/y
+
x
so that
M
y
=1=
N
x
. From
f
x
=1
−
3
/x
+
y
we obtain
f
=
x
−
3ln
|
x
|
+
xy
+
h
(
y
),
h
/
(
y
)=1
−
3
y
, and
h
(
y
)=
y
−
3ln
|
y
|
. The solution is
x
+
y
+
xy
−
3ln
|
xy
|
=
c
.
15.
Let
M
=
x
2
y
3
−
1
/
k
1+9
x
2

and
N
=
x
3
y
2
so that
M
y
=3
x
2
y
2
=
N
x
. From
f
x
=
x
2
y
3
−
1
/
k
1+9
x
2

we
obtain
f
=
1
3
x
3
y
3
−
1
3
arctan(3
x
)+
h
(
y
),
h
/
(
y
) = 0, and
h
(
y
) = 0. The solution is
x
3
y
3
−
arctan(3
x
)=
c
.
16.
Let
M
=
−
2
y
and
N
=5
y
−
2
x
so that
M
y
=
−
2=
N
x
. From
f
x
=
−
2
y
we obtain
f
=
−
2
xy
+
h
(
y
),
h
/
(
y
)=5
y
,
and
h
(
y
)=
5
2
y
2
. The solution is
−
2
xy
+
5
2
y
2
=
c
.
17.
Let
M
= tan
x
−
sin
x
sin
y
and
N
= cos
x
cos
y
so that
M
y
=
−
sin
x
cos
y
=
N
x
. From
f
x
= tan
x
−
sin
x
sin
y
we obtain
f
=ln
|
sec
x
|
+ cos
x
sin
y
+
h
(
y
),
h
/
(
y
) = 0, and
h
(
y
) = 0. The solution is ln
|
sec
x
|
+ cos
x
sin
y
=
c
.
18.
Let
M
=2
y
sin
x
cos
x
−
y
+2
y
2
e
xy
2
and
N
=
−
x
+ sin
2
x
+4
xye
xy
2
so that
M
y
= 2 sin
x
cos
x
−
1+4
xy
3
e
xy
2
+4
ye
xy
2
=
N
x
.
From
f
x
=2
y
sin
x
cos
x
−
y
+2
y
2
e
xy
2
we obtain
f
=
y
sin
2
x
−
xy
+2
e
xy
2
+
h
(
y
),
h
/
(
y
)=0,and
h
(
y
) = 0. The
solution is
y
sin
2
x
−
xy
+2
e
xy
2
=
c
.
37

38. Exercises 2.4
19.
Let
M
=4
t
3
y
−
15
t
2
−
y
and
N
=
t
4
+3
y
2
−
t
so that
M
y
=4
t
3
−
1=
N
t
. From
f
t
=4
t
3
y
−
15
t
2
−
y
we obtain
f
=
t
4
y
−
5
t
3
−
ty
+
h
(
y
),
h
/
(
y
)=3
y
2
, and
h
(
y
)=
y
3
. The solution is
t
4
y
−
5
t
3
−
ty
+
y
3
=
c
.
20.
Let
M
=1
/t
+1
/t
2
−
y/
k
t
2
+
y
2

and
N
=
ye
y
+
t/
k
t
2
+
y
2

so that
M
y
=
k
y
2
−
t
2

/
k
t
2
+
y
2

2
=
N
t
. From
f
t
=1
/t
+1
/t
2
−
y/
k
t
2
+
y
2

we obtain
f
=ln
|
t
|−
1
t
−
arctan
x
t
y
v
+
h
(
y
),
h
/
(
y
)=
ye
y
, and
h
(
y
)=
ye
y
−
e
y
.
The solution is
ln
|
t
|−
1
t
−
arctan
x
t
y
v
+
ye
y
−
e
y
=
c.
21.
Let
M
=
x
2
+2
xy
+
y
2
and
N
=2
xy
+
x
2
−
1 so that
M
y
=2(
x
+
y
)=
N
x
. From
f
x
=
x
2
+2
xy
+
y
2
we obtain
f
=
1
3
x
3
+
x
2
y
+
xy
2
+
h
(
y
),
h
/
(
y
)=
−
1, and
h
(
y
)=
−
y
. The general solution is
1
3
x
3
+
x
2
y
+
xy
2
−
y
=
c
.If
y
(1) = 1 then
c
=4
/
3 and the solution of the initial-value problem is
1
3
x
3
+
x
2
y
+
xy
2
−
y
=
4
3
.
22.
Let
M
=
e
x
+
y
and
N
=2+
x
+
ye
y
so that
M
y
=1=
N
x
. From
f
x
=
e
x
+
y
we obtain
f
=
e
x
+
xy
+
h
(
y
),
h
/
(
y
)=2+
ye
y
, and
h
(
y
)=2
y
+
ye
y
−
y
. The general solution is
e
x
+
xy
+2
y
+
ye
y
−
e
y
=
c
.If
y
(0) = 1 then
c
= 3 and the solution of the initial-value problem is
e
x
+
xy
+2
y
+
ye
y
−
e
y
=3.
23.
Let
M
=4
y
+2
t
−
5 and
N
=6
y
+4
t
−
1 so that
M
y
=4=
N
t
. From
f
t
=4
y
+2
t
−
5 we obtain
f
=4
ty
+
t
2
−
5
t
+
h
(
y
),
h
/
(
y
)=6
y
−
1, and
h
(
y
)=3
y
2
−
y
. The general solution is 4
ty
+
t
2
−
5
t
+3
y
2
−
y
=
c
.
If
y
(
−
1) = 2 then
c
= 8 and the solution of the initial-value problem is 4
ty
+
t
2
−
5
t
+3
y
2
−
y
=8.
24.
Let
M
=
t/
2
y
4
and
N
=
k
3
y
2
−
t
2

/y
5
so that
M
y
=
−
2
t/y
5
=
N
t
. From
f
t
=
t/
2
y
4
we obtain
f
=
t
2
4
y
4
+
h
(
y
),
h
/
(
y
)=
3
y
3
, and
h
(
y
)=
−
3
2
y
2
. The general solution is
t
2
4
y
4
−
3
2
y
2
=
c
.If
y
(1) = 1 then
c
=
−
5
/
4 and the
solution of the initial-value problem is
t
2
4
y
4
−
3
2
y
2
=
−
5
4
.
25.
Let
M
=
y
2
cos
x
−
3
x
2
y
−
2
x
and
N
=2
y
sin
x
−
x
3
+ln
y
so that
M
y
=2
y
cos
x
−
3
x
2
=
N
x
. From
f
x
=
y
2
cos
x
−
3
x
2
y
−
2
x
we obtain
f
=
y
2
sin
x
−
x
3
y
−
x
2
+
h
(
y
),
h
/
(
y
)=ln
y
, and
h
(
y
)=
y
ln
y
−
y
. The
general solution is
y
2
sin
x
−
x
3
y
−
x
2
+
y
ln
y
−
y
=
c
.If
y
(0) =
e
then
c
= 0 and the solution of the initial-value
problem is
y
2
sin
x
−
x
3
y
−
x
2
+
y
ln
y
−
y
=0.
26.
Let
M
=
y
2
+
y
sin
x
and
N
=2
xy
−
cos
x
−
1
/
k
1+
y
2

so that
M
y
=2
y
+ sin
x
=
N
x
. From
f
x
=
y
2
+
y
sin
x
we obtain
f
=
xy
2
−
y
cos
x
+
h
(
y
),
h
/
(
y
)=
−
1
1+
y
2
, and
h
(
y
)=
−
tan
−
1
y
. The general solution
is
xy
2
−
y
cos
x
−
tan
−
1
y
=
c
.If
y
(0) = 1 then
c
=
−
1
−
π/
4 and the solution of the initial-value problem is
xy
2
−
y
cos
x
−
tan
−
1
y
=
−
1
−
π
4
.
27.
Equating
M
y
=3
y
2
+4
kxy
3
and
N
x
=3
y
2
+40
xy
3
we obtain
k
= 10.
28.
Equating
M
y
=18
xy
2
−
sin
y
and
N
x
=4
kxy
2
−
sin
y
we obtain
k
=9
/
2.
29.
Let
M
=
−
x
2
y
2
sin
x
+2
xy
2
cos
x
and
N
=2
x
2
y
cos
x
so that
M
y
=
−
2
x
2
y
sin
x
+4
xy
cos
x
=
N
x
. From
f
y
=2
x
2
y
cos
x
we obtain
f
=
x
2
y
2
cos
x
+
h
(
y
),
h
/
(
y
) = 0, and
h
(
y
) = 0. The solution of the differential
equation is
x
2
y
2
cos
x
=
c
.
30.
Let
M
=
k
x
2
+2
xy
−
y
2

/
k
x
2
+2
xy
+
y
2

and
N
=
k
y
2
+2
xy
−
x
2

/
k
y
2
+2
xy
+
x
2

so that
M
y
=
−
4
xy/
(
x
+
y
)
3
=
N
x
. From
f
x
=
k
x
2
+2
xy
+
y
2
−
2
y
2

/
(
x
+
y
)
2
we obtain
f
=
x
+
2
y
2
x
+
y
+
h
(
y
),
h
/
(
y
)=
−
1,
and
h
(
y
)=
−
y
. The solution of the differential equation is
x
2
+
y
2
=
c
(
x
+
y
).
38

39. 1234567
x
0.5
1
1.5
2
2.5
y
-4-2 24
x
-6
-4
-2
2
4
y
y1
y2
Exercises 2.4
31.
We note that (
M
y
−
N
x
)
/N
=1
/x
, so an integrating factor is
e

dx/x
=
x
. Let
M
=2
xy
2
+3
x
2
and
N
=2
x
2
y
so that
M
y
=4
xy
=
N
x
. From
f
x
=2
xy
2
+3
x
2
we obtain
f
=
x
2
y
2
+
x
3
+
h
(
y
),
h
/
(
y
)=0,and
h
(
y
)=0. The
solution of the differential equation is
x
2
y
2
+
x
3
=
c
.
32.
We note that (
M
y
−
N
x
)
/N
= 1, so an integrating factor is
e

dx
=
e
x
. Let
M
=
xye
x
+
y
2
e
x
+
ye
x
and
N
=
xe
x
+2
ye
x
so that
M
y
=
xe
x
+2
ye
x
+
e
x
=
N
x
. From
f
y
=
xe
x
+2
ye
x
we obtain
f
=
xye
x
+
y
2
e
x
+
h
(
x
),
h
/
(
y
) = 0, and
h
(
y
) = 0. The solution of the differential equation is
xye
x
+
y
2
e
x
=
c
.
33.
We note that (
N
x
−
M
y
)
/M
=2
/y
, so an integrating factor is
e

2
dy/y
=
y
2
. Let
M
=6
xy
3
and
N
=4
y
3
+9
x
2
y
2
so that
M
y
=18
xy
2
=
N
x
. From
f
x
=6
xy
3
we obtain
f
=3
x
2
y
3
+
h
(
y
),
h
/
(
y
)=4
y
3
, and
h
(
y
)=
y
4
. The
solution of the differential equation is 3
x
2
y
3
+
y
4
=
c
.
34.
We note that (
M
y
−
N
x
)
/N
=
−
cot
x
, so an integrating factor is
e
−

cot
xdx
= csc
x
. Let
M
= cos
x
csc
x
= cot
x
and
N
=(1+2
/y
) sin
x
csc
x
=1+2
/y
, so that
M
y
=0=
N
x
. From
f
x
= cot
x
we obtain
f
= ln(sin
x
)+
h
(
y
),
h
/
(
y
)=1+2
/y
, and
h
(
y
)=
y
+ln
y
2
. The solution of the differential equation is ln(sin
x
)+
y
+ln
y
2
=
c
.
35. (a)
Write the separable equation as
−
g
(
x
)
dx
+
dy/h
(
y
). Identifying
M
=
−
g
(
x
) and
N
=1
/h
(
y
), we see that
M
y
=0=
N
x
, so the differential equation is exact.
(b)
Separating variables and integrating we have
−
(sin
y
)
dy
= (cos
x
)
dx
and cos
y
= sin
x
+
c.
Using
y
(7
π/
6) =
π/
2 we get
c
=1
/
2, so the solution of the initial-value problem is cos
y
= sin
x
+
1
2
.
(c)
Solving for
y
we have
y
= cos
−
1
(sin
x
+
1
2
). Since the domain of cos
−
1
t
is [
−
1
,
1] we see that
−
1
≤
sin
x
+
1
2
≤
1or
−
3
2
≤
sin
x
≤
1
2
, which is
equivalent to sin
x
≤
1
2
. We also want the interval to contain 7
π/
6, so
the interval of definition is (5
π/
6
,
13
π/
6).
36. (a)
Implicitly differentiating
x
3
+2
x
2
y
+
y
2
=
c
and solving for
dy/dx
we obtain
3
x
2
+2
x
2
dy
dx
+4
xy
+2
y
dy
dx
= 0 and
dy
dx
=
−
3
x
2
+4
xy
2
x
2
+2
y
.
Separating variables we get (4
xy
+3
x
2
)
dx
+(2
y
+2
x
2
)
dy
=0.
(b)
Setting
x
= 0 and
y
=
−
2in
x
3
+2
x
2
y
+
y
2
=
c
we find
c
= 4, and setting
x
=
y
= 1 we also find
c
=4.
Thus, both initial conditions determine the same implicit solution.
(c)
Solving
x
3
+2
x
2
y
+
y
2
= 4 for
y
we get
y
1
(
x
)=
−
x
2
−
t
4
−
x
3
+
x
4
and
y
2
(
x
)=
−
x
2
+
t
4
−
x
3
+
x
4
.
37.
To see that the equations are not equivalent consider
dx
=(
x/y
)
dy
= 0. An integrating factor is
µ
(
x, y
)=
y
resulting in
ydx
+
xdy
= 0. A solution of the latter equation is
y
= 0, but this is not a solution of the original
equation.
38.
The explicit solution is
y
=
t
(3 + cos
2
x
)
/
(1
−
x
2
) . Since 3 + cos
2
x>
0 for all
x
we must have 1
−
x
2
>
0or
−
1
<x<
1. Thus, the interval of definition is (
−
1
,
1).
39

40. Exercises 2.4
39. (a)
Since
f
y
=
N
(
x, y
)=
xe
xy
+2
xy
+1
/x
we obtain
f
=
e
xy
+
xy
2
+
y
x
+
h
(
x
) so that
f
x
=
ye
xy
+
y
2
−
y
x
2
+
h
/
(
x
).
Let
M
(
x, y
)=
ye
xy
+
y
2
−
y
x
2
.
(b)
Since
f
x
=
M
(
x, y
)=
y
1
/
2
x
−
1
/
2
+
x
k
x
2
+
y

−
1
we obtain
f
=2
y
1
/
2
x
1
/
2
+
1
2
ln


x
2
+
y


+
g
(
y
) so that
f
y
=
y
−
1
/
2
x
1
/
2
+
1
2
k
x
2
+
y

−
1
+
g
/
(
x
). Let
N
(
x, y
)=
y
−
1
/
2
x
1
/
2
+
1
2
k
x
2
+
y

−
1
.
40.
First note that
d
k
t
x
2
+
y
2

=
x
t
x
2
+
y
2
dx
+
y
t
x
2
+
y
2
dy.
Then
xdx
+
ydy
=
t
x
2
+
y
2
dx
becomes
x
t
x
2
+
y
2
dx
+
y
t
x
2
+
y
2
dy
=
d
k
t
x
2
+
y
2

=
dx.
The left side is the total differential of
t
x
2
+
y
2
and the right side is the total differential of
x
+
c
.Thus
t
x
2
+
y
2
=
x
+
c
is a solution of the differential equation.
Exercises 2.5
1.
Letting
y
=
ux
we have
(
x
−
ux
)
dx
+
x
(
udx
+
xdu
)=0
dx
+
xdu
=0
dx
x
+
du
=0
ln
|
x
|
+
u
=
c
x
ln
|
x
|
+
y
=
cx.
2.
Letting
y
=
ux
we have
(
x
+
ux
)
dx
+
x
(
udx
+
xdu
)=0
(1+2
u
)
dx
+
xdu
=0
dx
x
+
du
1+2
u
=0
ln
|
x
|
+
1
2
ln
|
1+2
u
|
=
c
x
2
y
1+2
y
x
/
=
c
1
x
2
+2
xy
=
c
1
.
3.
Letting
x
=
vy
we have
vy
(
vdy
+
ydv
)+(
y
−
2
vy
)
dy
=0
vy dv
+
k
v
2
−
2
v
+1

dy
=0
vdv
(
v
−
1)
2
+
dy
y
=0
ln
|
v
−
1
|−
1
v
−
1
+ln
|
y
|
=
c
ln




x
y
−
1




−
1
x/y
−
1
+ln
y
=
c
(
x
−
y
)ln
|
x
−
y
|−
y
=
c
(
x
−
y
)
.
40

41. Exercises 2.5
4.
Letting
x
=
vy
we have
y
(
vdy
+
ydv
)
−
2(
vy
+
y
)
dy
=0
ydv
−
(
v
+2)
dy
=0
dv
v
+2
−
dy
y
=0
ln
|
v
+2
|−
ln
|
y
|
=
c
ln




x
y
+2




−
ln
|
y
|
=
c
x
+2
y
=
c
1
y
2
.
5.
Letting
y
=
ux
we have
k
u
2
x
2
+
ux
2

dx
−
x
2
(
udx
+
xdu
)=0
u
2
dx
−
xdu
=0
dx
x
−
du
u
2
=0
ln
|
x
|
+
1
u
=
c
ln
|
x
|
+
x
y
=
c
y
ln
|
x
|
+
x
=
cy.
6.
Letting
y
=
ux
we have
k
u
2
x
2
+
ux
2

dx
+
x
2
(
udx
+
xdu
)=0
k
u
2
+2
u

dx
+
xdu
=0
dx
x
+
du
u
(
u
+2)
=0
ln
|
x
|
+
1
2
ln
|
u
|−
1
2
ln
|
u
+2
|
=
c
x
2
u
u
+2
=
c
1
x
2
y
x
=
c
1
y
y
x
+2
/
x
2
y
=
c
1
(
y
+2
x
)
.
7.
Letting
y
=
ux
we have
(
ux
−
x
)
dx
−
(
ux
+
x
)(
udx
+
xdu
)=0
k
u
2
+1

dx
+
x
(
u
+1)
du
=0
dx
x
+
u
+1
u
2
+1
du
=0
ln
|
x
|
+
1
2
ln
k
u
2
+1

+ tan
−
1
u
=
c
ln
x
2
x
y
2
x
2
+1
v
+ 2 tan
−
1
y
x
=
c
1
ln
k
x
2
+
y
2

+ 2 tan
−
1
y
x
=
c
1
.
41

42. Exercises 2.5
8.
Letting
y
=
ux
we have
(
x
+3
ux
)
dx
−
(3
x
+
ux
)(
udx
+
xdu
)=0
k
u
2
−
1

dx
+
x
(
u
+3)
du
=0
dx
x
+
u
+3
(
u
−
1)(
u
+1)
du
=0
ln
|
x
|
+2ln
|
u
−
1
|−
ln
|
u
+1
|
=
c
x
(
u
−
1)
2
u
+1
=
c
1
x
y
y
x
−
1
/
2
=
c
1
y
y
x
+1
/
(
y
−
x
)
2
=
c
1
(
y
+
x
)
.
9.
Letting
y
=
ux
we have
−
ux dx
+(
x
+
√
ux
)(
udx
+
xdu
)=0
(
x
+
x
√
u
)
du
+
u
3
/
2
dx
=0
x
u
−
3
/
2
+
1
u
v
du
+
dx
x
=0
−
2
u
−
1
/
2
+ln
|
u
|
+ln
|
x
|
=
c
ln
|
y/x
|
+ln
|
x
|
=2
t
x/y
+
c
y
(ln
|
y
|−
c
)
2
=4
x.
10.
Letting
y
=
ux
we have
y
ux
+
t
x
2
+
u
2
x
2
/
dx
−
x
(
udx
+
xdu
)=0
x
t
1+
u
2
dx
−
x
2
du
=0
dx
x
−
du
√
1+
u
2
=0
ln
|
x
|−
ln



u
+
t
1+
u
2



=
c
u
+
t
1+
u
2
=
c
1
x
y
+
t
y
2
+
x
2
=
c
1
x
2
.
11.
Letting
y
=
ux
we have
k
x
3
−
u
3
x
3

dx
+
u
2
x
3
(
udx
+
xdu
)=0
dx
+
u
2
xdu
=0
dx
x
+
u
2
du
=0
ln
|
x
|
+
1
3
u
3
=
c
3
x
3
ln
|
x
|
+
y
3
=
c
1
x
3
.
Using
y
(1) = 2 we find
c
1
= 8. The solution of the initial-value problem is 3
x
3
ln
|
x
|
+
y
3
=8
x
3
.
42

43. Exercises 2.5
12.
Letting
y
=
ux
we have
k
x
2
+2
u
2
x
2

dx
−
ux
2
(
udx
+
xdu
)=0
k
1+
u
2

dx
−
ux du
=0
dx
x
−
udu
1+
u
2
=0
ln
|
x
|−
1
2
ln
k
1+
u
2

=
c
x
2
1+
u
2
=
c
1
x
4
=
c
1
k
y
2
+
x
2

.
Using
y
(
−
1) = 1 we find
c
1
=1
/
2. The solution of the initial-value problem is 2
x
4
=
y
2
+
x
2
.
13.
Letting
y
=
ux
we have
(
x
+
uxe
u
)
dx
−
xe
u
(
udx
+
xdu
)=0
dx
−
xe
u
du
=0
dx
x
−
e
u
du
=0
ln
|
x
|−
e
u
=
c
ln
|
x
|−
e
y/x
=
c.
Using
y
(1) = 0 we find
c
=
−
1. The solution of the initial-value problem is ln
|
x
|
=
e
y/x
−
1.
14.
Letting
x
=
vy
we have
y
(
vdy
+
ydv
)+
vy
(ln
vy
−
ln
y
−
1)
dy
=0
ydv
+
v
ln
vdy
=0
dv
v
ln
v
+
dy
y
=0
ln
|
ln
|
v
||
+ln
|
y
|
=
c
y
ln




x
y




=
c
1
.
Using
y
(1) =
e
we find
c
1
=
−
e
. The solution of the initial-value problem is
y
ln




x
y




=
−
e
.
15.
From
y
/
+
1
x
y
=
1
x
y
−
2
and
w
=
y
3
we obtain
dw
dx
+
3
x
w
=
3
x
. An integrating factor is
x
3
so that
x
3
w
=
x
3
+
c
or
y
3
=1+
cx
−
3
.
16.
From
y
/
−
y
=
e
x
y
2
and
w
=
y
−
1
we obtain
dw
dx
+
w
=
−
e
x
. An integrating factor is
e
x
so that
e
x
w
=
−
1
2
e
2
x
+
c
or
y
−
1
=
−
1
2
e
x
+
ce
−
x
.
17.
From
y
/
+
y
=
xy
4
and
w
=
y
−
3
we obtain
dw
dx
−
3
w
=
−
3
x
. An integrating factor is
e
−
3
x
so that
e
−
3
x
=
xe
−
3
x
+
1
3
e
−
3
x
+
c
or
y
−
3
=
x
+
1
3
+
ce
3
x
.
18.
From
y
/
−
x
1+
1
x
v
y
=
y
2
and
w
=
y
−
1
we obtain
dw
dx
+
x
1+
1
x
v
w
=
−
1. An integrating factor is
xe
x
so that
xe
x
w
=
−
xe
x
+
e
x
+
c
or
y
−
1
=
−
1+
1
x
+
c
x
e
−
x
.
43

44. Exercises 2.5
19.
From
y
/
−
1
t
y
=
−
1
t
2
y
2
and
w
=
y
−
1
we obtain
dw
dt
+
1
t
w
=
1
t
2
. An integrating factor is
t
so that
tw
=ln
t
+
c
or
y
−
1
=
1
t
ln
t
+
c
t
. Writing this in the form
t
y
=ln
x
+
c
, we see that the solution can also be expressed in the
form
e
t/y
=
c
1
x
.
20.
From
y
/
+
2
3(1+
t
2
)
y
=
2
t
3(1+
t
2
)
y
4
and
w
=
y
−
3
we obtain
dw
dt
−
2
t
1+
t
2
w
=
−
2
t
1+
t
2
. An integrating factor is
1
1+
t
2
so that
w
1+
t
2
=
1
1+
t
2
+
c
or
y
−
3
=1+
c
k
1+
t
2

.
21.
From
y
/
−
2
x
y
=
3
x
2
y
4
and
w
=
y
−
3
we obtain
dw
dx
+
6
x
w
=
−
9
x
2
. An integrating factor is
x
6
so that
x
6
w
=
−
9
5
x
5
+
c
or
y
−
3
=
−
9
5
x
−
1
+
cx
−
6
.If
y
(1) =
1
2
then
c
=
49
5
and
y
−
3
=
−
9
5
x
−
1
+
49
5
x
−
6
.
22.
From
y
/
+
y
=
y
−
1
/
2
and
w
=
y
3
/
2
we obtain
dw
dx
+
3
2
w
=
3
2
. An integrating factor is
e
3
x/
2
so that
e
3
x/
2
w
=
e
3
x/
2
+
c
or
y
3
/
2
=1+
ce
−
3
x/
2
.If
y
(0) = 4 then
c
= 7 and
y
3
/
2
=1+7
e
−
3
x/
2
.
23.
Let
u
=
x
+
y
+ 1 so that
du/dx
=1+
dy/dx
. Then
du
dx
−
1=
u
2
or
1
1+
u
2
du
=
dx
. Thus tan
−
1
u
=
x
+
c
or
u
= tan(
x
+
c
), and
x
+
y
+ 1 = tan(
x
+
c
)or
y
= tan(
x
+
c
)
−
x
−
1.
24.
Let
u
=
x
+
y
so that
du/dx
=1+
dy/dx
. Then
du
dx
−
1=
1
−
u
u
or
udu
=
dx
.Thus
1
2
u
2
=
x
+
c
or
u
2
=2
x
+
c
1
,
and (
x
+
y
)
2
=2
x
+
c
1
.
25.
Let
u
=
x
+
y
so that
du/dx
=1+
dy/dx
. Then
du
dx
−
1 = tan
2
u
or cos
2
udu
=
dx
.Thus
1
2
u
+
1
4
sin 2
u
=
x
+
c
or 2
u
+ sin 2
u
=4
x
+
c
1
, and 2(
x
+
y
) + sin 2(
x
+
y
)=4
x
+
c
1
or 2
y
+ sin 2(
x
+
y
)=2
x
+
c
1
.
26.
Let
u
=
x
+
y
so that
du/dx
=1+
dy/dx
. Then
du
dx
−
1 = sin
u
or
1
1 + sin
u
du
=
dx
. Multiplying by
(1
−
sin
u
)
/
(1
−
sin
u
)wehave
1
−
sin
u
cos
2
u
du
=
dx
or
k
sec
2
u
−
tan
u
sec
u

du
=
dx
. Thus tan
u
−
sec
u
=
x
+
c
or tan(
x
+
y
)
−
sec(
x
+
y
)=
x
+
c
.
27.
Let
u
=
y
−
2
x
+ 3 so that
du/dx
=
dy/dx
−
2. Then
du
dx
+2=2+
√
u
or
1
√
u
du
=
dx
.Thus2
√
u
=
x
+
c
and
2
√
y
−
2
x
+3=
x
+
c
.
28.
Let
u
=
y
−
x
+ 5 so that
du/dx
=
dy/dx
−
1. Then
du
dx
+1=1+
e
u
or
e
−
u
du
=
dx
.Thus
−
e
−
u
=
x
+
c
and
−
e
y
−
x
+5
=
x
+
c
.
29.
Let
u
=
x
+
y
so that
du/dx
=1+
dy/dx
. Then
du
dx
−
1 = cos
u
and
1
1 + cos
u
du
=
dx
.Now
1
1 + cos
u
=
1
−
cos
u
1
−
cos
2
u
=
1
−
cos
u
sin
2
u
= csc
2
u
−
csc
u
cot
u
so we have

(csc
2
u
−
csc
u
cot
u
)
du
=

dx
and
−
cot
u
+ csc
u
=
x
+
c
.Thus
−
cot(
x
+
y
) + csc(
x
+
y
)=
x
+
c
.
Setting
x
= 0 and
y
=
π/
4 we obtain
c
=
√
2
−
1. The solution is
csc(
x
+
y
)
−
cot(
x
+
y
)=
x
+
√
2
−
1
.
30.
Let
u
=3
x
+2
y
so that
du/dx
=3+2
dy/dx
. Then
du
dx
=3+
2
u
u
+2
=
5
u
+6
u
+2
and
u
+2
5
u
+6
du
=
dx
.Now
u
+2
5
u
+6
=
1
5
+
4
25
u
+30
44

45. Exercises 2.5
so we have

x
1
5
+
4
25
u
+30
v
du
=
dx
and
1
5
u
+
4
25
ln
|
25
u
+30
|
=
x
+
c
.Thus
1
5
(3
x
+2
y
)+
4
25
ln
|
75
x
+50
y
+30
|
=
x
+
c.
Setting
x
=
−
1 and
y
=
−
1 we obtain
c
=
4
5
ln 95. The solution is
1
5
(3
x
+2
y
)+
4
25
ln
|
75
x
+50
y
+30
|
=
x
+
4
5
ln 95
or
5
y
−
5
x
+2ln
|
75
x
+50
y
+30
|
=10ln95
.
31.
We write the differential equation
M
(
x, y
)
dx
+
N
(
x, y
)
dy
=0as
dy/dx
=
f
(
x, y
) where
f
(
x, y
)=
−
M
(
x, y
)
N
(
x, y
)
.
The function
f
(
x, y
) must necessarily be homogeneous of degree 0 when
M
and
N
are homogeneous of degree
α
. Since
M
is homogeneous of degree
α
,
M
(
tx, ty
)=
t
α
M
(
x, y
), and letting
t
=1
/x
we have
M
(1
,y/x
)=
1
x
α
M
(
x, y
)or
M
(
x, y
)=
x
α
M
(1
,y/x
)
.
Thus
dy
dx
=
f
(
x, y
)=
−
x
α
M
(1
,y/x
)
x
α
N
(1
,y/x
)
=
−
M
(1
,y/x
)
N
(1
,y/x
)
=
F
y
y
x
/
.
To show that the differential equation also has the form
dy
dx
=
G
x
x
y
v
we use the fact that
M
(
x, y
)=
y
α
M
(
x/y,
1). The forms
F
(
y/x
) and
G
(
x/y
) suggest, respectively, the substitu-
tions
u
=
y/x
or
y
=
ux
and
v
=
x/y
or
x
=
vy
.
32.
As
x
→−∞
,
e
6
x
→
0 and
y
→
2
x
+ 3. Now write (1 +
ce
6
x
)
/
(1
−
ce
6
x
)as(
e
−
6
x
+
c
)
/
(
e
−
6
x
−
c
). Then, as
x
→∞
,
e
−
6
x
→
0 and
y
→
2
x
−
3.
33. (a)
The substitutions
y
=
y
1
+
u
and
dy
dx
=
dy
1
dx
+
du
dx
lead to
dy
1
dx
+
du
dx
=
P
+
Q
(
y
1
+
u
)+
R
(
y
1
+
u
)
2
=
P
+
Qy
1
+
Ry
2
1
+
Qu
+2
y
1
Ru
+
Ru
2
or
du
dx
−
(
Q
+2
y
1
R
)
u
=
Ru
2
.
This is a Bernoulli equation with
n
= 2 which can be reduced to the linear equation
dw
dx
+(
Q
+2
y
1
R
)
w
=
−
R
by the substitution
w
=
u
−
1
.
45

46. Exercises 2.5
(b)
Identify
P
(
x
)=
−
4
/x
2
,
Q
(
x
)=
−
1
/x
, and
R
(
x
) = 1. Then
dw
dx
+
x
−
1
x
+
4
x
v
w
=
−
1. An integrating
factor is
x
3
so that
x
3
w
=
−
1
4
x
4
+
c
or
u
=
.
−
1
4
x
+
cx
−
3
f
−
1
. Thus,
y
=
2
x
+
u
.
34.
Write the differential equation in the form
x
(
y
/
/y
)=ln
x
+ln
y
and let
u
=ln
y
. Then
du/dx
=
y
/
/y
and the
differential equation becomes
x
(
du/dx
)=ln
x
+
u
or
du/dx
−
u/x
= (ln
x
)
/x
, which is first-order, linear. An
integrating factor is
e
−

dx/x
=1
/x
, so that (using integration by parts)
d
dx
>
1
x
u
c
=
ln
x
x
2
and
u
x
=
−
1
x
−
ln
x
x
+
c.
The solution is
ln
y
=
−
1
−
ln
x
+
cx
or
y
=
e
−
cx
−
1
x
.
Exercises 2.6
1.
We identify
f
(
x, y
)=2
x
−
3
y
+ 1. Then, for
h
=0
.
1,
y
n
+1
=
y
n
+0
.
1(2
x
n
−
3
y
n
+1)=0
.
2
x
n
+0
.
7
y
n
+0
.
1
,
and
y
(1
.
1)
≈
y
1
=0
.
2(1) + 0
.
7(5) + 0
.
1=3
.
8
y
(1
.
2)
≈
y
2
=0
.
2(1
.
1)+0
.
7(3
.
8)+0
.
1=2
.
98
.
For
h
=0
.
05
y
n
+1
=
y
n
+0
.
05(2
x
n
−
3
y
n
+1)=0
.
1
x
n
+0
.
85
y
n
+0
.
1
,
and
y
(1
.
05)
≈
y
1
=0
.
1(1) + 0
.
85(5) + 0
.
1=4
.
4
y
(1
.
1)
≈
y
2
=0
.
1(1
.
05)+0
.
85(4
.
4)+0
.
1=3
.
895
y
(1
.
15)
≈
y
3
=0
.
1(1
.
1)+0
.
85(3
.
895) + 0
.
1=3
.
47075
y
(1
.
2)
≈
y
4
=0
.
1(1
.
15)+0
.
85(3
.
47075) + 0
.
1=3
.
11514
2.
We identify
f
(
x, y
)=
x
+
y
2
. Then, for
h
=0
.
1,
y
n
+1
=
y
n
+0
.
1(
x
n
+
y
2
n
)=0
.
1
x
n
+
y
n
+0
.
1
y
2
n
,
and
y
(0
.
1)
≈
y
1
=0
.
1(0)+0+0
.
1(0)
2
=0
y
(0
.
2)
≈
y
2
=0
.
1(0
.
1)+0+0
.
1(0)
2
=0
.
01
.
For
h
=0
.
05
y
n
+1
=
y
n
+0
.
05(
x
n
+
y
2
n
)=0
.
05
x
n
+
y
n
+0
.
05
y
2
n
,
and
y
(0
.
05)
≈
y
1
=0
.
05(0) + 0 + 0
.
05(0)
2
=0
y
(0
.
1)
≈
y
2
=0
.
05(0
.
05)+0+0
.
05(0)
2
=0
.
0025
y
(0
.
15)
≈
y
3
=0
.
05(0
.
1)+0
.
0025 + 0
.
05(0
.
0025)
2
=0
.
0075
y
(0
.
2)
≈
y
4
=0
.
05(0
.
15)+0
.
0075 + 0
.
05(0
.
0075)
2
=0
.
0150
.
46

47. h=0.1
h=0.05
x
n
y
n
True
Value
Abs.
Error
% Rel.
Error
x
n
y
n
True
Value
Abs.
Error
% Rel.
Error
0.00
1.0000 1.0000 0.0000
0.00 0.00
1.0000 1.0000 0.0000
0.00
0.10
1.1000 1.1052 0.0052
0.47 0.05
1.0500 1.0513 0.0013
0.12
0.20
1.2100 1.2214 0.0114
0.93 0.10
1.1025 1.1052 0.0027
0.24
0.30
1.3310 1.3499 0.0189
1.40 0.15
1.1576 1.1618 0.0042
0.36
0.40
1.4641 1.4918 0.0277
1.86 0.20
1.2155 1.2214 0.0059
0.48
0.50
1.6105 1.6487 0.0382
2.32 0.25
1.2763 1.2840 0.0077
0.60
0.60
1.7716 1.8221 0.0506
2.77 0.30
1.3401 1.3499 0.0098
0.72
0.70
1.9487 2.0138 0.0650
3.23 0.35
1.4071 1.4191 0.0120
0.84
0.80
2.1436 2.2255 0.0820
3.68 0.40
1.4775 1.4918 0.0144
0.96
0.90
2.3579 2.4596 0.1017
4.13 0.45
1.5513 1.5683 0.0170
1.08
1.00
2.5937 2.7183 0.1245
4.58 0.50
1.6289 1.6487 0.0198
1.20
0.55
1.7103 1.7333 0.0229
1.32
0.60
1.7959 1.8221 0.0263
1.44
0.65
1.8856 1.9155 0.0299
1.56
0.70
1.9799 2.0138 0.0338
1.68
0.75
2.0789 2.1170 0.0381
1.80
0.80
2.1829 2.2255 0.0427
1.92
0.85
2.2920 2.3396 0.0476
2.04
0.90
2.4066 2.4596 0.0530
2.15
0.95
2.5270 2.5857 0.0588
2.27
1.00
2.6533 2.7183 0.0650
2.39
h=0.1
h=0.05
x
n
y
n
True
Value
Abs.
Error
% Rel.
Error
x
n
y
n
True
Value
Abs.
Error
% Rel.
Error
1.00
1.0000 1.0000 0.0000
0.00 1.00
1.0000 1.0000 0.0000
0.00
1.10
1.2000 1.2337 0.0337
2.73 1.05
1.1000 1.1079 0.0079
0.72
1.20
1.4640 1.5527 0.0887
5.71 1.10
1.2155 1.2337 0.0182
1.47
1.30
1.8154 1.9937 0.1784
8.95 1.15
1.3492 1.3806 0.0314
2.27
1.40
2.2874 2.6117 0.3243
12.42 1.20
1.5044 1.5527 0.0483
3.11
1.50
2.9278 3.4903 0.5625
16.12 1.25
1.6849 1.7551 0.0702
4.00
1.30
1.8955 1.9937 0.0982
4.93
1.35
2.1419 2.2762 0.1343
5.90
1.40
2.4311 2.6117 0.1806
6.92
1.45
2.7714 3.0117 0.2403
7.98
1.50
3.1733 3.4903 0.3171
9.08
h=0.1
h=0.05
x
n
y
n
x
n
y
n
0.00
0.0000
0.00
0.0000
0.10
0.1000
0.05
0.0500
0.20
0.1905
0.10
0.0976
0.30
0.2731
0.15
0.1429
0.40
0.3492
0.20
0.1863
0.50
0.4198
0.25
0.2278
0.30
0.2676
0.35
0.3058
0.40
0.3427
0.45
0.3782
0.50
0.4124
h=0.1
h=0.05
x
n
y
n
x
n
y
n
0.00
1.0000
0.00
1.0000
0.10
1.1000
0.05
1.0500
0.20
1.2220
0.10
1.1053
0.30
1.3753
0.15
1.1668
0.40
1.5735
0.20
1.2360
0.50
1.8371
0.25
1.3144
0.30
1.4039
0.35
1.5070
0.40
1.6267
0.45
1.7670
0.50
1.9332
Exercises 2.6
3.
Separating variables and integrating, we have
dy
y
=
dx
and ln
|
y
|
=
x
+
c.
Thus
y
=
c
1
e
x
and, using
y
(0) = 1, we find
c
=1,so
y
=
e
x
is the solution of the initial-value problem.
4.
Separating variables and integrating, we have
dy
y
=2
xdx
and ln
|
y
|
=
x
2
+
c.
Thus
y
=
c
1
e
x
2
and, using
y
(1) = 1, we find
c
=
e
−
1
,so
y
=
e
x
2
−
1
is the solution of the initial-value problem.
5. 6.
47

48. h=0.1
h=0.05
x
n
y
n
x
n
y
n
0.00
0.5000
0.00
0.5000
0.10
0.5250
0.05
0.5125
0.20
0.5431
0.10
0.5232
0.30
0.5548
0.15
0.5322
0.40
0.5613
0.20
0.5395
0.50
0.5639
0.25
0.5452
0.30
0.5496
0.35
0.5527
0.40
0.5547
0.45
0.5559
0.50
0.5565
h=0.1
h=0.05
x
n
y
n
x
n
y
n
0.00
1.0000
0.00
1.0000
0.10
1.1000
0.05
1.0500
0.20
1.2159
0.10
1.1039
0.30
1.3505
0.15
1.1619
0.40
1.5072
0.20
1.2245
0.50
1.6902
0.25
1.2921
0.30
1.3651
0.35
1.4440
0.40
1.5293
0.45
1.6217
0.50
1.7219
h=0.1
h=0.05
x
n
y
n
x
n
y
n
1.00
1.0000
1.00
1.0000
1.10
1.0000
1.05
1.0000
1.20
1.0191
1.10
1.0049
1.30
1.0588
1.15
1.0147
1.40
1.1231
1.20
1.0298
1.50
1.2194
1.25
1.0506
1.30
1.0775
1.35
1.1115
1.40
1.1538
1.45
1.2057
1.50
1.2696
h=0.1
h=0.05
x
n
y
n
x
n
y
n
0.00
0.5000
0.00
0.5000
0.10
0.5250
0.05
0.5125
0.20
0.5499
0.10
0.5250
0.30
0.5747
0.15
0.5375
0.40
0.5991
0.20
0.5499
0.50
0.6231
0.25
0.5623
0.30
0.5746
0.35
0.5868
0.40
0.5989
0.45
0.6109
0.50
0.6228
5 10-5-10
5
x
y
x
Euler
Runge-Kutta
h=0.25
5 10-5-10
5
h=0.1
x
y
Runge-Kutta
Euler
5 10-5-10
5
Euler
Runge-Kutta
x
y
h=0.05
2 4-2
5
h=0.25
Runge-Kutta
Euler
x
y
2 4-2
5
y
x
Euler
Runge-Kutta
h=0.1
2 4-2
5
h=0.05
Euler
Runge-Kutta
x
y
Exercises 2.6
7. 8.
9. 10.
11.
12.
48

49. h=0.1 Euler
h=0.05
Euler h=0.1 R-K
h=0.05 R-K
x
n
y
n
x
n
y
n
x
n
y
n
x
n
y
n
0.00
1.0000
0.00
1.0000
0.00
1.0000
0.00
1.0000
0.10
1.0000
0.05
1.0000
0.10
1.0101
0.05
1.0025
0.20
1.0200
0.10
1.0050
0.20
1.0417
0.10
1.0101
0.30
1.0616
0.15
1.0151
0.30
1.0989
0.15
1.0230
0.40
1.1292
0.20
1.0306
0.40
1.1905
0.20
1.0417
0.50
1.2313
0.25
1.0518
0.50
1.3333
0.25
1.0667
0.60
1.3829
0.30
1.0795
0.60
1.5625
0.30
1.0989
0.70
1.6123
0.35
1.1144
0.70
1.9607
0.35
1.1396
0.80
1.9763
0.40
1.1579
0.80
2.7771
0.40
1.1905
0.90
2.6012
0.45
1.2115
0.90
5.2388
0.45
1.2539
1.00
3.8191
0.50
1.2776
1.00
42.9931
0.50
1.3333
0.55
1.3592
0.55
1.4337
0.60
1.4608
0.60
1.5625
0.65
1.5888
0.65
1.7316
0.70
1.7529
0.70
1.9608
0.75
1.9679
0.75
2.2857
0.80
2.2584
0.80
2.7777
0.85
2.6664
0.85
3.6034
0.90
3.2708
0.90
5.2609
0.95
4.2336
0.95
10.1973
1.00
5.9363
1.00
84.0132
Exercises 2.7
13.
Using separation of variables we find that the solution of the differential equation is
y
=1
/
(1
−
x
2
), which is
undefined at
x
= 1, where the graph has a vertical asymptote.
Exercises 2.7
1.
Let
P
=
P
(
t
) be the population at time
t
, and
P
0
the initial population. From
dP/dt
=
kP
we obtain
P
=
P
0
e
kt
.
Using
P
(5) = 2
P
0
we find
k
=
1
5
ln 2 and
P
=
P
0
e
(ln 2)
t/
5
. Setting
P
(
t
)=3
P
0
we have
3=
e
(ln 2)
t/
5
=
⇒
ln 3 =
(ln 2)
t
5
=
⇒
t
=
5ln3
ln 2
≈
7
.
9 years.
Setting
P
(
t
)=4
P
0
we have
4=
e
(ln 2)
t/
5
=
⇒
ln 4 =
(ln 2)
t
5
=
⇒
t
= 10 years.
2.
Setting
P
=10,000 and
t
= 3 in Problem 1 we obtain
10,000 =
P
0
e
(ln 2)3
/
5
=
⇒
P
0
= 10,000
e
−
0
.
6ln2
≈
6597
.
5
.
Then
P
(10) =
P
0
e
2ln2
=4
P
0
≈
26,390.
3.
Let
P
=
P
(
t
) be the population at time
t
. From
dP/dt
=
kt
and
P
(0) =
P
0
= 500 we obtain
P
= 500
e
kt
. Using
P
(10) = 575 we find
k
=
1
10
ln 1
.
15. Then
P
(30) = 500
e
3ln1
.
15
≈
760 years.
4.
Let
N
=
N
(
t
) be the number of bacteria at time
t
and
N
0
the initial number. From
dN/dt
=
kN
we obtain
N
=
N
0
e
kt
. Using
N
(3) = 400 and
N
(10) = 2000 we find 400 =
N
0
e
3
k
or
e
k
= (400
/N
0
)
1
/
3
. From
N
(10) = 2000
we then have
2000 =
N
0
e
10
k
=
N
0
x
400
N
0
v
10
/
3
=
⇒
2000
400
10
/
3
=
N
−
7
/
3
0
=
⇒
N
0
=
x
2000
400
10
/
3
v
−
3
/
7
≈
201
.
5.
Let
I
=
I
(
t
) be the intensity,
t
the thickness, and
I
(0) =
I
0
.If
dI/dt
=
kI
and
I
(3)=0
.
25
I
0
then
I
=
I
0
e
kt
,
k
=
1
3
ln 0
.
25, and
I
(15) = 0
.
00098
I
0
.
6.
From
dS/dt
=
rS
we obtain
S
=
S
0
e
rt
where
S
(0) =
S
0
.
49

50. Exercises 2.7
(a)
If
S
0
= $5000 and
r
=5
.
75% then
S
(5) = $6665
.
45.
(b)
If
S
(
t
) =$10,000 then
t
= 12 years.
(c)
S
≈
$6651
.
82
7.
Let
N
=
N
(
t
) be the amount of lead at time
t
. From
dN/dt
=
kN
and
N
(0) = 1 we obtain
N
=
e
kt
. Using
N
(3
.
3) = 1
/
2wefind
k
=
1
3
.
3
ln 1
/
2. When 90% of the lead has decayed, 0
.
1 grams will remain. Setting
N
(
t
)=0
.
1wehave
e
t
(1
/
3
.
3) ln(1
/
2)
=0
.
1=
⇒
t
3
.
3
ln
1
2
=ln0
.
1=
⇒
t
=
3
.
3ln0
.
1
ln 1
/
2
≈
10
.
96 hours.
8.
Let
N
=
N
(
t
) be the amount at time
t
. From
dN/dt
=
kt
and
N
(0) = 100 we obtain
N
= 100
e
kt
. Using
N
(6) = 97 we find
k
=
1
6
ln 0
.
97. Then
N
(24) = 100
e
(1
/
6)(ln 0
.
97)24
= 100(0
.
97)
4
≈
88
.
5 mg.
9.
Setting
N
(
t
) = 50 in Problem 8 we obtain
50 = 100
e
kt
=
⇒
kt
=ln
1
2
=
⇒
t
=
ln 1
/
2
(1
/
6) ln 0
.
97
≈
136
.
5 hours
.
10. (a)
The solution of
dA/dt
=
kA
is
A
(
t
)=
A
0
e
kt
. Letting
A
=
1
2
A
0
and solving for
t
we obtain the half-life
T
=
−
(ln 2)
/k
.
(b)
Since
k
=
−
(ln 2)
/T
we have
A
(
t
)=
A
0
e
−
(ln 2)
t/T
=
A
0
2
−
t/T
.
11.
Assume that
A
=
A
0
e
kt
and
k
=
−
0
.
00012378. If
A
(
t
)=0
.
145
A
0
then
t
≈
15,600 years.
12.
From Example 3, the amount of carbon present at time
t
is
A
(
t
)=
A
0
e
−
0
.
00012378
t
. Letting
t
= 660 and solving
for
A
0
we have
A
(660) =
A
0
e
−
0
.
0001237(660)
=0
.
921553
A
0
. Thus, approximately 92% of the original amount of
C-14 remained in the cloth as of 1988.
13.
Assume that
dT/dt
=
k
(
T
−
10) so that
T
=10+
ce
kt
.If
T
(0)=70
◦
and
T
(1
/
2)=50
◦
then
c
= 60 and
k
= 2 ln(2
/
3) so that
T
(1) = 36
.
67
◦
.If
T
(
t
)=15
◦
then
t
=3
.
06 minutes.
14.
Assume that
dT/dt
=
k
(
T
−
5) so that
T
=5+
ce
kt
.If
T
(1)=55
◦
and
T
(5)=30
◦
then
k
=
−
1
4
ln 2 and
c
=59
.
4611 so that
T
(0) = 64
.
4611
◦
.
15.
Assume that
dT/dt
=
k
(
T
−
100) so that
T
= 100 +
ce
kt
.If
T
(0) = 20
◦
and
T
(1)=22
◦
then
c
=
−
80 and
k
= ln(39
/
40) so that
T
(
t
)=90
◦
implies
t
=82
.
1 seconds. If
T
(
t
)=98
◦
then
t
= 145
.
7 seconds.
16.
Using separation of variables to solve
dT/dt
=
k
(
T
−
T
m
)weget
T
(
t
)=
T
m
+
ce
kt
. Using
T
(0) = 70 we find
c
=70
−
T
m
,so
T
(
t
)=
T
m
+ (70
−
T
m
)
e
kt
. Using the given observations, we obtain
T
y
1
2
/
=
T
m
+ (70
−
T
m
)
e
k/
2
= 110
T
(1) =
T
m
+ (70
−
T
m
)
e
k
= 145
.
Then
e
k/
2
= (110
−
T
m
)
/
(70
−
T
m
) and
e
k
=(
e
k/
2
)
2
=
x
110
−
T
m
70
−
T
m
v
2
=
145
−
T
m
70
−
T
m
(110
−
T
m
)
2
70
−
T
m
= 145
−
T
m
12100
−
220
T
m
+
T
2
m
= 10150
−
250
T
m
+
T
2
m
T
m
= 390
.
50

51. Exercises 2.7
The temperature in the oven is 390
◦
.
17.
We are given
T
m
= 65. By Newton’s law of cooling we then have
dT
dt
=
k
(
T
−
65)
,T
(0) = 82
.
Using linearity or separation of variables we obtain
T
=65+
ce
kt
. From
T
(0) = 82 we find
c
= 17, so
that
T
= 65+17
e
kt
. To find
k
we use the fact that the body temperature at
t
= 2 hours was 75
◦
F. Then
75 = 65 + 17
e
2
k
and
k
=
−
0
.
2653 and
T
(
t
) = 65 + 17
e
−
0
.
2653
t
.
At the time of death,
t
0
,
T
(
t
0
)=98
.
6
◦
F, so 98
.
6=65+17
e
−
0
.
2653
t
, which gives
t
=
−
2
.
568. Thus, the murder
took place about 2
.
568 hours prior to the discovery of the body.
18.
We will assume that the temperature of both the room and the cream is 72
◦
F, and that the temperature of the
coffee when it is first put on the table is 175
◦
F. If we let
T
1
(
t
) represent the temperature of the coffee in Mr.
Jones’ cup at time
t
, then
dT
1
dt
=
k
(
T
1
−
72)
,
which implies
T
1
=72+
c
1
e
kt
. At time
t
= 0 Mr. Jones adds cream to his coffee which immediately reduces
its temperature by an amount
α
, so that
T
1
(0) = 175
−
α
. Thus 175
−
α
=
T
1
(0)=72+
c
1
, which implies
c
1
= 103
−
α
, so that
T
1
(
t
) = 72 + (103
−
α
)
e
kt
.At
t
=5,
T
1
(5) = 72 + (103
−
α
)
e
5
k
. Now we let
T
2
(
t
) represent
the temperature of the coffee in Mrs. Jones’ cup. From
T
2
=72+
c
2
e
kt
and
T
2
(0) = 175 we obtain
c
2
= 103,
so that
T
2
(
t
)=72+103
e
kt
.At
t
=5,
T
2
(5) = 72 + 103
e
5
k
. When cream is added to Mrs. Jones’ coffee the
temperature is reduced by an amount
α
. Using the fact that
k<
0wehave
T
2
(5)
−
α
= 72 + 103
e
5
k
−
α<
72 + 103
e
5
k
−
αe
5
k
= 72 + (103
−
α
)
e
5
k
=
T
1
(5)
.
Thus, the temperature of the coffee in Mr. Jones’ cup is hotter.
19.
From
dA/dt
=4
−
A/
50 we obtain
A
= 200 +
ce
−
t/
50
.If
A
(0) = 30 then
c
=
−
170 and
A
= 200
−
170
e
−
t/
50
.
20.
From
dA/dt
=0
−
A/
50 we obtain
A
=
ce
−
t/
50
.If
A
(0) = 30 then
c
= 30 and
A
=30
e
−
t/
50
.
21.
From
dA/dt
=10
−
A/
100 we obtain
A
= 1000 +
ce
−
t/
100
.If
A
(0) = 0 then
c
=
−
1000 and
A
= 1000
−
1000
e
−
t/
100
.At
t
=5,
A
(5)
≈
48
.
77 pounds.
22.
From
dA
dt
=10
−
10
A
500
−
(10
−
5)
t
=10
−
2
A
100
−
t
we obtain
A
= 1000
−
10
t
+
c
(100
−
t
)
2
.If
A
(0) = 0 then
c
=
−
1
10
. The tank is empty in 100 minutes.
23.
From
dA
dt
=3
−
4
A
100+(6
−
4)
t
=3
−
2
A
50 +
t
we obtain
A
=50+
t
+
c
(50+
t
)
−
2
.If
A
(0) = 10 then
c
=
−
100,000
and
A
(30) = 64
.
38 pounds.
24. (a)
Initially the tank contains 300 gallons of solution. Since brine is pumped in at a rate of
3 gal/min and the solution is pumped out at a rate of 2 gal/min, the net change is an increase of 1
gal/min. Thus, in 100 minutes the tank will contain its capacity of 400 gallons.
(b)
The differential equation describing the amount of salt in the tank is
A
/
(
t
)=6
−
2
A/
(300 +
t
) with solution
A
(
t
) = 600 + 2
t
−
(4
.
95
×
10
7
)(300 +
t
)
−
2
,
0
≤
t
≤
100
,
51

52. 200 400 600
t
200
400
600
800
A
Exercises 2.7
as noted in the discussion following Example 5 in the text. Thus, the amount of salt in the tank when it
overflows is
A
(100) = 800
−
(4
.
95
×
10
7
)(400)
−
2
= 490
.
625 lbs.
(c)
When the tank is overflowing the amount of salt in the tank is governed by the differential equation
dA
dt
= (3 gal/min)(2 lb/gal)
−
(
A
400
lb/gal)(3 gal/min)
=6
−
3
A
400
,A
(100) = 490
.
625
.
Solving the equation we obtain
A
(
t
) = 800 +
ce
−
3
t/
400
. The initial condition yields
c
=
−
654
.
947, so that
A
(
t
) = 800
−
654
.
947
e
−
3
t/
400
.
When
t
= 150,
A
(150) = 587
.
37 lbs.
(d)
As
t
→∞
, the amount of salt is 800 lbs, which is to be expected since
(400 gal)(2 lbs/gal)= 800 lbs.
(e)
25.
Assume
L di/dt
+
Ri
=
E
(
t
),
L
=0
.
1,
R
= 50, and
E
(
t
) = 50 so that
i
=
3
5
+
ce
−
500
t
.If
i
(0) = 0 then
c
=
−
3
/
5
and lim
t
→∞
i
(
t
)=3
/
5.
26.
Assume
L di/dt
+
Ri
=
E
(
t
),
E
(
t
)=
E
0
sin
ωt
, and
i
(0) =
i
0
so that
i
=
E
0
R
L
2
ω
2
+
R
2
sin
ωt
−
E
0
Lω
L
2
ω
2
+
R
2
cos
ωt
+
ce
−
Rt/L
.
Since
i
(0) =
i
0
we obtain
c
=
i
0
+
E
0
Lω
L
2
ω
2
+
R
2
.
27.
Assume
R dq/dt
+(1
/c
)
q
=
E
(
t
),
R
= 200,
C
=10
−
4
, and
E
(
t
) = 100 so that
q
=1
/
100 +
ce
−
50
t
.If
q
(0)=0
then
c
=
−
1
/
100 and
i
=
1
2
e
−
50
t
.
28.
Assume
R dq/dt
+(1
/c
)
q
=
E
(
t
),
R
= 1000,
C
=5
×
10
−
6
, and
E
(
t
) = 200. Then
q
=
1
1000
+
ce
−
200
t
and
i
=
−
200
ce
−
200
t
.If
i
(0)=0
.
4 then
c
=
−
1
500
,
q
(0
.
005) = 0
.
003 coulombs, and
i
(0
.
005) = 0
.
1472 amps. As
t
→∞
we have
q
→
1
1000
.
29.
For 0
≤
t
≤
20 the differential equation is 20
di/dt
+2
i
= 120. An integrating factor is
e
t/
10
,so
d
dt
>
e
t/
10
i
c
=6
e
t/
10
and
i
=60+
c
1
e
−
t/
10
.If
i
(0) = 0 then
c
1
=
−
60 and
i
=60
−
60
e
−
t/
10
.
For
t>
20 the differential equation is 20
di/dt
+2
i
= 0 and
i
=
c
2
e
−
t/
10
.
At
t
=20wewant
c
2
e
−
2
=60
−
60
e
−
2
so that
c
2
=60
k
e
2
−
1

.Thus
i
(
t
)=
R
60
−
60
e
−
t/
10
,
0
≤
t
≤
20;
60
k
e
2
−
1

e
−
t/
10
,t>
20.
52

53. Exercises 2.7
30.
Separating variables we obtain
dq
E
0
−
q/C
=
dt
k
1
+
k
2
t
=
⇒−
C
ln



E
0
−
q
C



=
1
k
2
ln
|
k
1
+
k
2
t
|
+
c
1
=
⇒
(
E
0
−
q/C
)
−
C
(
k
1
+
k
2
t
)
1
/k
2
=
c
2
.
Setting
q
(0) =
q
0
we find
c
2
=
(
E
0
−
q
0
/C
)
−
C
k
1
/k
2
1
,so
(
E
0
−
q/C
)
−
C
(
k
1
+
k
2
t
)
1
/k
2
=
(
E
0
−
q
0
/C
)
−
C
k
1
/k
2
1
=
⇒
y
E
0
−
q
C
/
−
C
=
y
E
0
−
q
0
C
/
−
C
x
k
1
k
+
k
2
t
v
−
1
/k
2
=
⇒
E
0
−
q
C
=
y
E
0
−
q
0
C
/
x
k
1
k
+
k
2
t
v
1
/Ck
2
=
⇒
q
=
E
0
C
+(
q
0
−
E
0
C
)
x
k
1
k
+
k
2
t
v
1
/Ck
2
.
31. (a)
From
m dv/dt
=
mg
−
kv
we obtain
v
=
gm/k
+
ce
−
kt/m
.If
v
(0) =
v
0
then
c
=
v
0
−
gm/k
and the solution
of the initial-value problem is
v
=
gm
k
+
y
v
0
−
gm
k
/
e
−
kt/m
.
(b)
As
t
→∞
the limiting velocity is
gm/k
.
(c)
From
ds/dt
=
v
and
s
(0) = 0 we obtain
s
=
gm
k
t
−
m
k
y
v
0
−
gm
k
/
e
−
kt/m
+
m
k
y
v
0
−
gm
k
/
.
32. (a)
Integrating
d
2
s/dt
2
=
−
g
we get
v
(
t
)=
ds/dt
=
−
gt
+
c
. From
v
(0) = 300 we find
c
= 300, so the velocity
is
v
(
t
)=
−
32
t
+ 300. Integrating again and using
s
(0) = 0 we get
s
(
t
)=
−
16
t
2
+ 300
t
. The maximum
height is attained when
v
= 0, that is, at
t
a
=9
.
375. The maximum height will be
s
(9
.
375) = 1406
.
25 ft.
(b)
When air resistance is proportional to velocity the model for the velocity is
m dv/dt
=
−
mg
−
kv
(using the
fact that the positive direction is upward.) Solving the differential equation using separation of variables
we obtain
v
(
t
)=
−
mg/k
+
ce
−
kt/m
. From
v
(0) = 300 we get
v
(
t
)=
−
mg
k
+
y
300 +
mg
k
/
e
−
kt/m
.
Integrating and using
s
(0) = 0 we find
s
(
t
)=
−
mg
k
t
+
m
k
y
300 +
mg
k
/
(1
−
e
−
kt/m
)
.
Setting
k
=0
.
0025,
m
=16
/
32 = 0
.
5, and
g
=32wehave
s
(
t
) = 1,340,000
−
6,400
t
−
1,340,000
e
−
0
.
005
t
and
v
(
t
)=
−
6,400 + 6,700
e
−
0
.
005
t
.
The maximum height is attained when
v
= 0, that is, at
t
a
=9
.
162. The maximum height will be
s
(9
.
162) = 1363
.
79 ft, which is less than the maximum height in part (a).
33.
We saw in part (a) of Problem 31 that the ascent time is
t
a
=9
.
375. To find when the cannonball hits the
ground we solve
s
(
t
)=
−
16
t
2
+ 300
t
= 0, getting a total time in flight of
t
=18
.
75. Thus, the time of descent
is
t
d
=18
.
75
−
9
.
375 = 9
.
375. The impact velocity is
v
i
=
v
(18
.
75) =
−
300, which has the same magnitude as
the initial velocity.
53

54. Exercises 2.7
We saw in part (b) of Problem 31 that the ascent time in the case of air resistance is
t
a
=9
.
162. Solving
s
(
t
) = 1,340,000
−
6,400
t
−
1,340,000
e
−
0
.
005
t
= 0 we see that the total time of flight is 18
.
466. Thus, the descent
time is
t
d
=18
.
466
−
9
.
162=9
.
304. The impact velocity is
v
i
=
v
(18
.
466) =
−
290
.
91, compared to an initial
velocity of
v
0
= 300.
34.
Assuming that air resistance is proportional to velocity and the positive direction is downward, the model for
the velocity is
m dv/dt
=
mg
−
kv
. Using separation of variables to solve this differential equation we obtain
v
(
t
)=
mg/k
+
ce
−
kt/m
. From
v
(0) = 0 we get
v
(
t
)=(
mg/k
)(1
−
e
−
kt/m
). Letting
k
=0
.
5,
m
= 160
/
32=5,
and
g
=32wehave
v
(
t
) = 320(1
−
e
−
0
.
1
t
). Integrating, we find
s
(
t
) = 320
t
+ 3200
e
−
0
.
1
t
.At
t
= 15, when the
parachute opens,
v
(15) = 248
.
598 and
s
(15) = 5514
.
02. At this point the value of
k
changes to
k
= 10 and the
new initial velocity is
v
0
= 248
.
598. Her velocity with the parachute open (with time measured from the instant
of opening) is
v
p
(
t
) = 16 + 232
.
598
e
−
2
t
. Integrating, we find
s
p
(
t
)=16
t
−
116
.
299
e
−
2
t
. Twenty seconds after
leaving the plane is five seconds after the parachute opens. Her velocity at this time is
v
p
(5) = 16
.
0106 ft/sec and
she has fallen
s
(15) +
s
p
(5) = 5514
.
02 + 79
.
9947 = 5594
.
01 ft. Her terminal velocity is lim
t
→∞
v
p
(
t
) = 16, so she
has very nearly reached her terminal velocity five seconds after the parachute opens. When the parachute opens,
the distance to the ground is 15,000
−
5514
.
02 = 9485
.
98 ft. Solving
s
p
(
t
) = 9485
.
98 we get
t
= 592
.
874 s = 9
.
88
min. Thus, it will take her approximately 9
.
88 minutes to reach the ground after her parachute has opened and
a total of (592
.
874 + 15)
/
60 = 10
.
13 minutes after she exits the plane.
35. (a)
The differential equation is first-order, linear. Letting
b
=
k/ρ
, the integrating factor is
e

3
b dt/
(
bt
+
r
0
)
=
(
r
0
+
bt
)
3
. Then
d
dt
[(
r
0
+
bt
)
3
v
]=
g
(
r
0
+
bt
)
3
and (
r
0
+
bt
)
3
v
=
g
4
b
(
r
0
+
bt
)
4
+
c.
The solution of the differential equation is
v
(
t
)=(
g/
4
b
)(
r
0
+
bt
)+
c
(
r
0
+
bt
)
−
3
. Using
v
(0) = 0 we find
c
=
−
gr
4
0
/
4
b
, so that
v
(
t
)=
g
4
b
(
r
0
+
bt
)
−
gr
4
0
4
b
(
r
0
+
bt
)
3
=
gρ
4
k
y
r
0
+
k
ρ
t
/
−
gρr
4
0
4
k
(
r
0
+
kt/ρ
)
3
.
(b)
Integrating
dr/dt
=
k/ρ
we get
r
=
kt/ρ
+
c
. Using
r
(0) =
r
0
we have
c
=
r
0
,so
r
(
t
)=
kt/ρ
+
r
0
.
(c)
If
r
=0
.
007 ft when
t
= 10 s, then solving
r
(10) = 0
.
007 for
k/ρ
, we obtain
k/ρ
=
−
0
.
0003 and
r
(
t
)=
0
.
01
−
0
.
0003
t
. Solving
r
(
t
)=0weget
t
=33
.
3, so the raindrop will have evaporated completely at 33
.
3
seconds.
36.
Separating variables we obtain
dP
P
=
k
cos
tdt
=
⇒
ln
|
P
|
=
k
sin
t
+
c
=
⇒
P
=
c
1
e
k
sin
t
.
If
P
(0) =
P
0
then
c
1
=
P
0
and
P
=
P
0
e
k
sin
t
.
37. (a)
From
dP/dt
=(
k
1
−
k
2
)
P
we obtain
P
=
P
0
e
(
k
1
−
k
2
)
t
where
P
0
=
P
(0).
(b)
If
k
1
>k
2
then
P
→∞
as
t
→∞
.If
k
1
=
k
2
then
P
=
P
0
for every
t
.If
k
1
<k
2
then
P
→
0as
t
→∞
.
38.
The first equation can be solved by separation of variables. We obtain
x
=
c
1
e
−
λ
1
t
. From
x
(0) =
x
0
we obtain
c
1
=
x
0
and so
x
=
x
0
e
−
λ
1
t
. The second equation then becomes
dy
dt
=
x
0
λ
1
e
−
λ
1
t
−
λ
2
y
or
dy
dt
+
λ
2
y
=
x
0
λ
1
e
−
λ
1
t
54

55. t
rk
x
Exercises 2.7
which is linear. An integrating factor is
e
λ
2
t
.Thus
d
dt
[
e
λ
2
t
y
]=
x
0
λ
1
e
−
λ
1
t
e
λ
2
t
=
x
0
λ
1
e
(
λ
2
−
λ
1
)
t
e
λ
2
t
y
=
x
0
λ
1
λ
2
−
λ
1
e
(
λ
2
−
λ
1
)
t
+
c
2
y
=
x
0
λ
1
λ
2
−
λ
1
e
−
λ
1
t
+
c
2
e
−
λ
2
t
.
From
y
(0) =
y
0
we obtain
c
2
=(
y
0
λ
2
−
y
0
λ
1
−
x
0
λ
1
)
/
(
λ
2
−
λ
1
). The solution is
y
=
x
0
λ
1
λ
2
−
λ
1
e
−
λ
1
t
+
y
0
λ
2
−
y
0
λ
1
−
x
0
λ
1
λ
2
−
λ
1
e
−
λ
2
t
.
39. (a)
Solving
k
1
(
M
−
A
)
−
k
2
A
= 0 for
A
we find the equilibrium solution
A
=
k
1
M/
(
k
1
+
k
2
). From the phase
portrait we see that lim
t
→∞
A
(
t
)=
k
1
M/
(
k
1
+
k
2
).
Since
k
2
>
0, the material will never be completely memorized and the larger
k
2
is, the less the amount of
material will be memorized over time.
(b)
Write the differential equation in the form
dA/dt
+(
k
1
+
k
2
)
A
=
k
1
M
.
Then an integrating factor is
e
(
k
1
+
k
2
)
t
, and
d
dt
>
e
(
k
1
+
k
2
)
t
A
c
=
k
1
Me
(
k
1
+
k
2
)
t
=
⇒
e
(
k
1
+
k
2
)
t
A
=
k
1
M
k
1
+
k
2
e
(
k
1
+
k
2
)
t
+
c
=
⇒
A
=
k
1
M
k
1
+
k
2
+
ce
−
(
k
1
+
k
2
)
t
.
Using
A
(0) = 0 we find
c
=
−
k
1
M
k
1
+
k
2
and
A
=
k
1
M
k
1
+
k
2
y
1
−
e
−
(
k
1
+
k
2
)
t
/
.As
t
→∞
,
A
→
k
1
M
k
1
+
k
2
.
40. (a)
Solving
r
−
kx
= 0 for
x
we find the equilibrium solution
x
=
r/k
. When
x<r/k
,
dx/dt >
0 and when
x>r/k
,
dx/dt <
0. From the phase portrait we see that lim
t
→∞
x
(
t
)=
r/k
.
(b)
From
dx/dt
=
r
−
kx
and
x
(0) = 0 we obtain
x
=
r/k
−
(
r/k
)
e
−
kt
so that
x
→
r/k
as
t
→∞
.If
x
(
T
)=
r/
2
k
then
T
= (ln 2)
/k
.
55

56. 5 10 15 20
t
500
1000
1500
2000
N
Exercises 2.8
Exercises 2.8
1. (a)
Solving
N
(1
−
0
.
0005
N
) = 0 for
N
we find the equilibrium solutions
N
= 0 and
N
= 2000. When
0
<N<
2000,
dN/dt >
0. From the phase portrait we see that lim
t
→∞
N
(
t
) = 2000.
A graph of the solution is shown in part (b).
(b)
Separating variables and integrating we have
dN
N
(1
−
0
.
0005
N
)
=
y
1
N
−
1
N
−
2000
/
dN
=
dt
and
ln
N
−
ln(
N
−
2000) =
t
+
c.
Solving for
N
we get
N
(
t
) = 2000
e
c
+
t
/
(1 +
e
c
+
t
) = 2000
e
c
e
t
/
(1 +
e
c
e
t
). Using
N
(0) = 1 and solving for
e
c
we find
e
c
=1
/
1999 and so
N
(
t
) = 2000
e
t
/
(1999 +
e
t
). Then
N
(10) = 1833
.
59, so 1834 companies are
expected to adopt the new technology when
t
= 10.
2.
From
dN
dt
=
N
(
a
−
bN
) and
N
(0) = 500 we obtain
N
=
500
a
500
b
+(
a
−
500
b
)
e
−
at
. Since lim
t
→∞
N
=
a
b
=50,000
and
N
(1) = 1000 we have
a
=0
.
7033,
b
=0
.
00014, and
N
=
50,000
1+99
e
−
0
.
7033
t
.
3.
From
dP
dt
=
P
k
10
−
1
−
10
−
7
P

and
P
(0) = 5000 we obtain
P
=
500
0
.
0005 + 0
.
0995
e
−
0
.
1
t
so that
P
→
1,000,000
as
t
→∞
.If
P
(
t
) = 500,000 then
t
=52
.
9 months.
4. (a)
We have
dP/dt
=
P
(
a
−
bP
) with
P
(0) = 3
.
929 million. Using separation of variables we obtain
P
(
t
)=
3
.
929
a
3
.
929
b
+(
a
−
3
.
929
b
)
e
−
at
=
a/b
1+(
a/
3
.
929
b
−
1)
e
−
at
=
c
1+(
c/
3
.
929
−
1)
e
−
at
.
At
t
= 60(1850) the population is 23.192 million, so
23
.
192 =
c
1+(
c/
3
.
929
−
1)
e
−
60
a
or
c
=23
.
192+23
.
192(
c/
3
.
929
−
1)
e
−
60
a
.At
t
= 120(1910)
91
.
972 =
c
1+(
c/
3
.
929
−
1)
e
−
120
a
or
c
=91
.
972+91
.
972(
c/
3
.
929
−
1)(
e
−
60
a
)
2
. Combining the two equations for
c
we get
x
(
c
−
23
.
192)
/
23
.
192
c/
3
.
929
−
1
v
2
y
c
3
.
929
−
1
/
=
c
−
91
.
972
91
.
972
or
91
.
972(3
.
929)(
c
−
23
.
192)
2
= (23
.
192)
2
(
c
−
91
.
972)(
c
−
3
.
929)
.
The solution of this quadratic equation is
c
= 197
.
274. This in turn gives
a
=0
.
0313. Therefore
P
(
t
)=
197
.
274
1+49
.
21
e
−
0
.
0313
t
.
56

57. Census Predicted %
Year Population
Population
Error Error
1790
3.929 3.929
0.000
0.00
1800
5.308 5.334
-0.026
-0.49
1810
7.240 7.222
0.018
0.24
1820
9.638 9.746
-0.108
-1.12
1830
12.866 13.090
-0.224
-1.74
1840
17.069 17.475
-0.406
-2.38
1850
23.192 23.143
0.049
0.21
1860
31.433 30.341
1.092
3.47
1870
38.558 39.272
-0.714
-1.85
1880
50.156 50.044
0.112
0.22
1890
62.948 62.600
0.348
0.55
1900
75.996 76.666
-0.670
-0.88
1910
91.972 91.739
0.233
0.25
1920
105.711 107.143
-1.432
-1.35
1930
122.775 122.140
0.635
0.52
1940
131.669 136.068
-4.399
-3.34
1950
150.697 148.445
2.252
1.49
Exercises 2.8
(b)
(c)
The model predicts a population of 159.0 million for 1960 and 167.8 million for 1970. The census populations
for these years were 179.3 and 203.3, respectively. The percentage errors are 12.8 and 21.2, respectively.
5. (a)
The differential equation is
dP/dt
=
P
(5
−
P
)
−
4. Solving
P
(5
−
P
)
−
4 = 0 for
P
we obtain equilibrium
solutions
P
= 1 and
P
= 4. The phase portrait is shown below and solution curves are shown in part (b).
We see that for
P
0
>
4 and 1
<P
0
<
4 the population approaches 4 as
t
increases. For 0
<P
0
<
1 the
population decreases to 0 in finite time.
(b)
The differential equation is
dP
dt
=
P
(5
−
P
)
−
4=
−
(
P
2
−
5
P
+4)=
−
(
P
−
4)(
P
−
1)
.
Separating variables and integrating, we obtain
dP
(
P
−
4)(
P
−
1)
=
−
dt
x
1
/
3
P
−
4
−
1
/
3
P
−
1
v
dP
=
−
dt
1
3
ln




P
−
4
P
−
1




=
−
t
+
c
P
−
4
P
−
1
=
c
1
e
−
3
t
.
Setting
t
= 0 and
P
=
P
0
we find
c
1
=(
P
0
−
4)
/
(
P
0
−
1). Solving for
P
we obtain
P
(
t
)=
4(
P
0
−
1)
−
(
P
0
−
4)
e
−
3
t
(
P
0
−
1)
−
(
P
0
−
4)
e
−
3
t
.
(c)
To find when the population becomes extinct in the case 0
<P
0
<
1 we set
P
=0in
P
−
4
P
−
1
=
P
0
−
4
P
0
−
1
e
−
3
t
57

58. t
e
P
t
1e
P
0.511.522.53
p
-2
-1
1
2
f
Exercises 2.8
from part (a) and solve for
t
. This gives the time of extinction
t
=
−
1
3
ln
4(
P
0
−
1)
P
0
−
4
.
6. (a)
Solving
P
(5
−
P
)
−
25
4
= 0 for
P
we obtain the equilibrium solution
P
=
5
2
.For
P
k
=
5
2
,
dP/dt <
0. Thus,
if
P
0
<
5
2
, the population becomes extinct (otherwise there would be another equilibrium solution.) Using
separation of variables to solve the initial-value problem we get
P
(
t
)=[4
P
0
+(10
P
0
−
25)
t
]
/
[4+(4
P
0
−
10)
t
].
To find when the population becomes extinct for
P
0
<
5
2
we solve
P
(
t
) = 0 for
t
. We see that the time of
extinction is
t
=4
P
0
/
5(5
−
2
P
0
).
(b)
Solving
P
(5
−
P
)
−
7 = 0 for
P
we obtain complex roots, so there are no equilibrium solutions. Since
dP/dt <
0 for all values of
P
, the population becomes extinct for any initial condition. Using separation of
variables to solve the initial-value problem we get
P
(
t
)=
5
2
+
√
3
2
tan
.
tan
−
1
x
2
P
0
−
5
√
3
v
−
√
3
2
t
f
.
Solving
P
(
t
) = 0 for
t
we see that the time of extinction is
t
=
2
3
y
√
3 tan
−
1
(5
/
√
3)+
√
3 tan
−
1

(2
P
0
−
5)
/
√
3
<
/
.
7. (a)
The differential equation is
dP/dt
=
P
(1
−
ln
P
), which has equilibrium solution
P
=
e
. When
P
0
>e
,
dP/dt <
0, and when
P
0
<e
,
dP/dt >
0.
(b)
The differential equation is
dP/dt
=
P
(1 + ln
P
), which has equilibrium solution
P
=1
/e
. When
P
0
>
1
/e
,
dP/dt >
0, and when
P
o
<
1
/e
,
dP/dt <
0.
(c)
From
dP
dt
=
P
(
a
−
b
ln
P
) we obtain
−
1
b
ln
|
a
−
b
ln
P
|
=
t
+
c
1
so that
P
=
e
a/b
e
−
ce
−
bt
.If
P
(0) =
P
0
then
c
=
a
b
−
ln
P
0
.
8. (a)
Using a CAS to solve
P
(1
−
P
)+0
.
3
e
−
P
= 0 for
P
we see that
P
=1
.
09216 is an equilibrium solution.
(b)
Since
f
(
P
)
>
0 for 0
<P <
1
.
09216, the solution
P
(
t
)of
dP/dt
=
P
(1
−
P
)+0
.
3
e
−
P
,P
(0) =
P
0
,
is increasing for
P
0
<
1
.
09216. Since
f
(
P
)
<
0 for
P>
1
.
09216, the solution
P
(
t
) is decreasing for
P
0
>
1
.
09216. Thus
P
=1
.
09216 is an attractor.
58

59. 246810
t
0.5
1
1.5
2
p
50010001500
t
2
4
6
8
10
h
Exercises 2.8
(c)
The curves for the second initial-value problem are thicker. The equilibrium
solution for the logic model is
P
= 1. Comparing 1
.
09216 and 1, we see that the
percentage increase is 9
.
216%.
9.
Let
X
=
X
(
t
) be the amount of
C
at time
t
and
dX
dt
=
k
(120
−
2
X
)(150
−
X
). If
X
(0) = 0 and
X
(5)=10
then
X
=
150
−
150
e
180
kt
1
−
2
.
5
e
180
kt
where
k
=
.
0001259, and
X
(20) = 29
.
3 grams. Now
X
→
60 as
t
→∞
, so that the
amount of
A
→
0 and the amount of
B
→
30 as
t
→∞
.
10.
From
dX
dt
=
k
(150
−
X
)
2
,
X
(0) = 0, and
X
(5) = 10 we obtain
X
= 150
−
150
150
kt
+1
where
k
=
.
000095238.
Then
X
(20) = 33
.
3 grams and
X
→
150 as
t
→∞
so that the amount of
A
→
0 and the amount of
B
→
0as
t
→∞
.If
X
(
t
) = 75 then
t
= 70 minutes.
11. (a)
The initial-value problem is
dh/dt
=
−
8
A
h
√
h/A
w
,
h
(0) =
H
. Sepa-
rating variables and integrating we have
dh
√
h
=
−
8
A
h
A
w
dt
and 2
√
h
=
−
8
A
h
A
w
t
+
c.
Using
h
(0) =
H
we find
c
=2
√
H
, so the solution of the initial-value problem is
t
h
(
t
)=(
A
w
√
H
−
4
A
h
t
)
/A
w
, where
A
w
√
H
−
4
A
h
t
≥
0. Thus,
h
(
t
)=(
A
w
√
H
−
4
A
h
t
)
2
/A
2
w
for 0
≤
t
≤
A
w
H/
4
A
h
.
(b)
Identifying
H
= 10,
A
w
=4
π
, and
A
h
=
π/
576 we have
h
(
t
)=
t
2
/
331,776
−
(
t
5
/
2
/
144)
t
+ 10. Solving
h
(
t
) = 0 we see that the tank empties in 576
√
10 seconds or 30
.
36 minutes.
12.
To obtain the solution of this differential equation we use
h
(
t
) from part (a) in Problem 11 with
A
h
replaced
by
cA
h
. Then
h
(
t
)=(
A
w
√
H
−
4
cA
h
t
)
2
/A
2
w
. Solving
h
(
t
) = 0 with
c
=0
.
6 and the values from Problem 11 we
see that the tank empties in 3035
.
79 seconds or 50
.
6 minutes.
13. (a)
Separating variables and integrating we have
6
h
3
/
2
dh
=
−
5
t
and
12
5
h
5
/
2
=
−
5
t
+
c.
Using
h
(0) = 20 we find
c
= 1920
√
5 , so the solution of the initial-value problem is
h
(
t
)=
k
800
√
5
−
25
12
t

2
/
5
.
Solving
h
(
t
) = 0 we see that the tank empties in 384
√
5 seconds or 14
.
31 minutes.
(b)
When the height of the water is
h
, the radius of the top of the water is
r
=
h
tan 30
◦
=
h/
√
3 and
A
w
=
πh
2
/
3. The differential equation is
dh
dt
=
−
c
A
h
A
w
t
2
gh
=
−
0
.
6
π
(2
/
12)
2
πh
2
/
3
√
64
h
=
−
2
5
h
3
/
2
.
Separating variables and integrating we have
5
h
3
/
2
dh
=
−
2
dt
and 2
h
5
/
2
=
−
2
t
+
c.
Using
h
(0) = 9 we find
c
= 486, so the solution of the initial-value problem is
h
(
t
) = (243
−
t
)
2
/
5
. Solving
h
(
t
) = 0 we see that the tank empties in 24
.
3 seconds or 4
.
05 minutes.
59

60. Exercises 2.8
14.
When the height of the water is
h
, the radius of the top of the water is
2
5
(20
−
h
) and
A
w
=4
π
(20
−
h
)
2
/
25.
The differential equation is
dh
dt
=
−
c
A
h
A
w
t
2
gh
=
−
0
.
6
π
(2
/
12)
2
4
π
(20
−
h
)
2
/
25
√
64
h
=
−
5
6
√
h
(20
−
h
)
2
.
Separating variables and integrating we have
(20
−
h
)
2
√
h
dh
=
−
5
6
dt
and 800
√
h
−
80
3
h
3
/
2
+
2
5
h
5
/
2
=
−
5
6
t
+
c.
Using
h
(0) = 20 we find
c
= 2560
√
5
/
3, so an implicit solution of the initial-value problem is
800
√
h
−
80
3
h
3
/
2
+
2
5
h
5
/
2
=
−
5
6
t
+
2560
√
5
3
.
To find the time it takes the tank to empty we set
h
= 0 and solve for
t
. The tank empties in 1024
√
5 seconds
or 38
.
16 minutes. Thus, the tank empties more slowly when the base of the cone is on the bottom.
15. (a)
Separating variables we obtain
mdv
mg
−
kv
2
=
dt
1
g
dv
1
−
(
kv/mg
)
2
=
dt
√
mg
√
kg
t
k/mg dv
1
−
(
√
kv/
√
mg
)
2
=
dt
e
m
kg
tanh
−
1
√
kv
√
mg
=
t
+
c
tanh
−
1
√
kv
√
mg
=
e
kg
m
t
+
c
1
.
Thus the velocity at time
t
is
v
(
t
)=
e
mg
k
tanh
Q
e
kg
m
t
+
c
1
P
.
Setting
t
= 0 and
v
=
v
0
we find
c
1
= tanh
−
1
(
√
kv
0
/
√
mg
).
(b)
Since tanh
t
→
1as
t
→∞
,wehave
v
→
t
mg/k
as
t
→∞
.
(c)
Integrating the expression for
v
(
t
) in part (a) we obtain
s
(
t
)=
e
mg
k

tanh
Q
e
kg
m
t
+
c
1
P
dt
=
m
k
ln
N
cosh
Q
e
kg
m
t
+
c
1
Pπ
+
c
2
.
Setting
t
= 0 and
s
=
s
0
we find
c
2
=
s
0
−
m
k
ln(cosh
c
1
).
16.
We solve
m
dv
dt
=
−
mg
−
kv
2
,v
(0) = 300
using separation of variables. This gives
v
(
t
)=
e
mg
k
tan
Q
tan
−
1
300
a
k
mg
−
e
kg
m
t
P
.
60

61. Exercises 2.8
Integrating and using
s
(0) = 0 we find
s
(
t
)=
m
k
ln




cos
e
kg
m
t
−
tan
−
1
300
a
k
mg




+
m
2
k
ln
x
1+
90000
k
mg
v
.
Solving
v
(
t
) = 0 we see that
t
a
=6
.
60159. The maximum height is
s
(
t
a
) = 823
.
84 ft.
17.
To find
t
d
we solve
m
dv
dt
=
mg
−
kv
2
,v
(0) = 0
using separation of variables. This gives
v
(
t
)=
e
mg
k
tanh
e
kg
m
t.
Integrating and using
s
(0) = 0 gives
s
(
t
)=
m
k
ln
Q
cosh
e
kg
m
t
P
.
To find the time of descent we solve
s
(
t
) = 823
.
84 and find
t
d
=7
.
77882. The impact velocity is
v
(
t
d
) = 182
.
998,
which is positive because the positive direction is downward.
18. (a)
Solving
v
t
=
t
mg/k
for
k
we obtain
k
=
mg/v
2
t
. The differential equation then becomes
m
dv
dt
=
mg
−
mg
v
2
t
v
2
or
dv
dt
=
g
x
1
−
1
v
2
t
v
2
v
.
Separating variables and integrating gives
v
t
tanh
−
1
v
v
t
=
gt
+
c
1
.
The initial condition
v
(0) = 0 implies
c
1
=0,so
v
(
t
)=
v
t
tanh
gt
v
t
.
We find the distance by integrating:
s
(
t
)=

v
t
tanh
gt
v
t
dt
=
v
2
t
g
ln
x
cosh
gt
v
t
v
+
c
2
.
The initial condition
s
(0) = 0 implies
c
2
=0,so
s
(
t
)=
v
2
t
g
ln
x
cosh
gt
v
t
v
.
In 25 seconds she has fallen 20,000
−
14,800 = 5,200 feet. Solving 5200 = (
v
2
t
/
32) ln
y
cosh
32(25)
v
t
/
for
v
t
gives
v
t
≈
271
.
711 ft/s. Then
s
(
t
)=
v
2
t
g
ln
y
cosh
gt
v
t
/
= 2307
.
08 ln(cosh 0
.
117772
t
)
.
(b)
At
t
= 15,
s
(15) = 2,542.94 ft and
v
(15) =
s
/
(15) = 256
.
287 ft/sec.
19.
While the object is in the air its velocity is modeled by the linear differential equation
m dv/dt
=
mg
−
kv
. Using
m
= 160,
k
=
1
4
, and
g
= 32, the differential equation becomes
dv/dt
+(1
/
640)
v
= 32. The integrating factor
is
e

dt/
640
=
e
t/
640
and the solution of the differential equation is
e
t/
640
v
=

32
e
t/
640
dt
= 20,480
e
t/
640
+
c
.
Using
v
(0) = 0 we see that
c
=
−
20,480 and
v
(
t
) = 20,480
−
20,480
e
−
t/
640
. Integrating we get
s
(
t
) = 20,480
t
+
13,107,200
e
−
t/
640
+
c
. Since
s
(0) = 0,
c
=
−
13,107,200 and
s
(
t
)=
−
13,107,200 + 20,480
t
+ 13,107,200
e
−
t/
640
.
61

62. Exercises 2.8
To find when the object hits the liquid we solve
s
(
t
) = 500
−
75 = 425, obtaining
t
a
=5
.
16018. The velocity
at the time of impact with the liquid is
v
a
=
v
(
t
a
) = 164
.
482. When the object is in the liquid its velocity is
modeled by
m dv/dt
=
mg
−
kv
2
. Using
m
= 160,
g
= 32, and
k
=0
.
1 this becomes
dv/dt
= (51,200
−
v
2
)
/
1600.
Separating variables and integrating we have
dv
51,200
−
v
2
=
dt
1600
and
√
2
640
ln




v
−
160
√
2
v
+ 160
√
2




=
1
1600
t
+
c.
Solving
v
(0) =
v
a
= 164
.
482 we obtain
c
=
−
0
.
00407537. Then, for
v<
160
√
2 = 226
.
274,




v
−
160
√
2
v
+ 160
√
2




=
e
√
2
t/
5
−
1
.
8443
or
−
v
−
160
√
2
v
+ 160
√
2
=
e
√
2
t/
5
−
1
.
8443
.
Solving for
v
we get
v
(
t
)=
13964
.
6
−
2208
.
29
e
√
2
t/
5
61
.
7153 + 9
.
75937
e
√
2
t/
5
.
Integrating we find
s
(
t
) = 226
.
275
t
−
1600 ln(6
.
3237 +
e
√
2
t/
5
)+
c.
Solving
s
(0) = 0 we see that
c
= 3185
.
78, so
s
(
t
) = 3185
.
78 + 226
.
275
t
−
1600 ln(6
.
3237 +
e
√
2
t/
5
)
.
To find when the object hits the bottom of the tank we solve
s
(
t
) = 75, obtaining
t
b
=0
.
466273. The time
from when the object is dropped from the helicopter to when it hits the bottom of the tank is
t
a
+
t
b
=5
.
62708
seconds.
20. (a)
Let
ρ
be the weight density of the water and V the volume of the object. Archimedes’ principle states that
the upward buoyant force has magnitude equal to the weight of the water displaced. Taking the positive
direction to be down, the differential equation is
m
dv
dt
=
mg
−
kv
2
−
ρV.
(b)
Using separation of variables we have
mdv
(
mg
−
ρV
)
−
kv
2
=
dt
m
√
k
√
kdv
(
√
mg
−
ρV
)
2
−
(
√
kv
)
2
=
dt
m
√
k
1
√
mg
−
ρV
tanh
−
1
√
kv
√
mg
−
ρV
=
t
+
c.
Thus
v
(
t
)=
e
mg
−
ρV
k
tanh
x
√
kmg
−
kρV
m
t
+
c
1
v
.
(c)
Since tanh
t
→
1as
t
→∞
, the terminal velocity is
t
(
mg
−
ρV
)
/k
.
21. (a)
Writing the equation in the form (
x
−
t
x
2
+
y
2
)
dx
+
ydy
we identify
M
=
x
−
t
x
2
+
y
2
and
N
=
y
. Since
62

63. Exercises 2.8
M
and
N
are both homogeneous of degree 1 we use the substitution
y
=
ux
. It follows that
y
x
−
t
x
2
+
u
2
x
2
/
dx
+
ux
(
udx
+
xdu
)=0
x
>y
1
−
t
1+
u
2
/
+
u
2
c
dx
+
x
2
udu
=0
−
udu
1+
u
2
−
√
1+
u
2
=
dx
x
udu
√
1+
u
2
(1
−
√
1+
u
2
)
=
dx
x
.
Letting
w
=1
−
√
1+
u
2
we have
dw
=
−
u du/
√
1+
u
2
so that
−
ln
y
1
−
t
1+
u
2
/
=ln
x
+
c
1
1
−
√
1+
u
2
=
c
1
x
1
−
t
1+
u
2
=
−
c
2
x
(
−
c
2
=1
/c
1
)
1+
c
2
x
=
e
1+
y
2
x
2
1+
2
c
2
x
+
c
2
2
x
2
=1+
y
2
x
2
.
Solving for
y
2
we have
y
2
=2
c
2
x
+
c
2
2
=4
y
c
2
2
/y
x
+
c
2
2
/
which is a family of parabolas symmetric with respect to the
x
-axis with vertex at (
−
c
2
/
2
,
0) and focus at
the origin.
(b)
Let
u
=
x
2
+
y
2
so that
du
dx
=2
x
+2
y
dy
dx
.
Then
y
dy
dx
=
1
2
du
dx
−
x
and the differential equation can be written in the form
1
2
du
dx
−
x
=
−
x
+
√
u
or
1
2
du
dx
=
√
u.
Separating variables and integrating we have
du
2
√
u
=
dx
√
u
=
x
+
c
u
=
x
2
+2
cx
+
c
2
x
2
+
y
2
=
x
2
+2
cx
+
c
2
y
2
=2
cx
+
c
2
.
22. (a)
From 2
W
2
−
W
3
=
W
2
(2
−
W
) = 0 we see that
W
= 0 and
W
= 2 are constant solutions.
(b)
Separating variables and using a CAS to integrate we get
dW
W
√
4
−
2
W
=
dx
and
−
tanh
−
1
y
1
2
√
4
−
2
W
/
=
x
+
c.
63

64. x
W
−3 3
2
200040006000800010000
t
2
4
6
8
10
h
Exercises 2.8
Using the facts that the hyperbolic tangent is an odd function and 1
−
tanh
2
x
= sech
2
x
we have
1
2
√
4
−
2
W
= tanh(
−
x
−
c
)=
−
tanh(
x
+
c
)
1
4
(4
−
2
W
) = tanh
2
(
x
+
c
)
1
−
1
2
W
= tanh
2
(
x
+
c
)
1
2
W
=1
−
tanh
2
(
x
+
c
) = sech
2
(
x
+
c
)
.
Thus,
W
(
x
) = 2 sech
2
(
x
+
c
).
(c)
Letting
x
= 0 and
W
= 2 we find that sech
2
(
c
) = 1 and
c
=0.
23. (a)
Solving
r
2
+ (10
−
h
)
2
=10
2
for
r
2
we see that
r
2
=20
h
−
h
2
. Combining the rate of input of water,
π
,
with the rate of output due to evaporation,
kπr
2
=
kπ
(20
h
−
h
2
), we have
dV/dt
=
π
−
kπ
(20
h
−
h
2
). Using
V
=10
πh
2
−
1
3
πh
3
, we see also that
dV/dt
= (20
πh
−
πh
2
)
dh/dt
. Thus,
(20
πh
−
πh
2
)
dh
dt
=
π
−
kπ
(20
h
−
h
2
) and
dh
dt
=
1
−
20
kh
+
kh
2
20
h
−
h
2
.
(b)
Letting
k
=1
/
100, separating variables and integrating (with the help of a
CAS), we get
100
h
(
h
−
20)
(
h
−
10)
2
dh
=
dt
and
100(
h
2
−
10
h
+ 100)
10
−
h
=
t
+
c.
Using
h
(0) = 0 we find
c
= 1000, and solving for
h
we get
h
(
t
)=0
.
005

√
t
2
+ 4000
t
−
t

, where the positive square root is
chosen because
h
≥
0.
(c)
The volume of the tank is
V
=
2
3
π
(10)
3
feet, so at a rate of
π
cubic feet per minute, the tank will fill in
2
3
(10)
3
≈
666
.
67 minutes
≈
11
.
11 hours.
(d)
At 666
.
67 minutes, the depth of the water is
h
(666
.
67) = 5
.
486 feet. From the graph in (b) we suspect that
lim
t
→∞
h
(
t
) = 10, in which case the tank will never completely fill. To prove this we compute the limit of
h
(
t
):
lim
t
→∞
h
(
t
)=0
.
005 lim
t
→∞
α
1
t
2
+ 4000
t
−
t
β
=0
.
005 lim
t
→∞
t
2
+ 4000
t
−
t
2
√
t
2
+ 4000
t
+
t
=0
.
005 lim
t
→∞
4000
t
t
1
1 + 4000
/t
+
t
=0
.
005 lim
t
→∞
4000
1+1
=0
.
005(2000) = 10
.
64

65. t P(t)
Q(t)
0 3.929
0.035
10 5.308
0.036
20 7.240
0.033
30 9.638
0.033
40 12.866
0.033
50 17.069
0.036
60 23.192
0.036
70 31.433
0.023
80 38.558
0.030
90 50.156
0.026
100 62.948
0.021
110 75.996
0.021
120 91.972
0.015
130 105.711
0.016
140 122.775
0.007
150 131.669
0.014
160 150.697
0.019
170 179.300
20406080100120140
P
0.005
0.01
0.015
0.02
0.025
0.03
0.035
Q
255075100125150
t
25
50
75
100
125
150
175
P
Exercises 2.8
24. (a)
(b)
The regression line is
Q
=0
.
0348391
−
0
.
000168222
P
.
(c)
The solution of the logistic equation is given in equation (5) in the text. Identifying
a
=0
.
0348391 and
b
=0
.
000168222 we have
P
(
t
)=
aP
0
bP
0
+(
a
−
bP
0
)
e
−
at
.
(d)
With
P
0
=3
.
929 the solution becomes
P
(
t
)=
0
.
136883
0
.
000660944 + 0
.
0341781
e
−
0
.
0348391
t
.
(e)
(f)
We identify
t
= 180 with 1970,
t
= 190 with 1980, and
t
= 200 with 1990. The model predicts
P
(180) =
188
.
661,
P
(190) = 193
.
735, and
P
(200) = 197
.
485. The actual population figures for these years are
203
.
303, 226
.
542, and 248
.
765 millions. As
t
→∞
,
P
(
t
)
→
a/b
= 207
.
102.
65

66. 255075100125150
t
5
10
15
20
x, y, z
x(t)
y(t)
z(t)
Exercises 2.9
Exercises 2.9
1.
The equation
dx/dt
=
−
λ
1
x
can be solved by separation of variables. Integrating both sides of
dx/x
=
−
λ
1
dt
we obtain ln
|
x
|
=
−
λ
1
t
+
c
from which we get
x
=
c
1
e
−
λ
1
t
. Using
x
(0) =
x
0
we find
c
1
x
0
so that
x
=
x
0
e
−
λ
1
t
.
Substituting this result into the second differential equation we have
dy
dt
+
λ
2
y
=
λ
1
x
0
e
−
λ
1
t
which is linear. An integrating factor is
e
λ
2
t
so that
d
dt

e
λ
2
t
y
<
=
λ
1
x
0
e
(
λ
2
−
λ
1
)
t
+
c
2
y
=
λ
1
x
0
λ
2
−
λ
1
e
(
λ
2
−
λ
1
)
t
e
−
λ
2
t
+
c
2
e
−
λ
2
t
=
λ
1
x
0
λ
2
−
λ
1
e
−
λ
1
t
+
c
2
e
−
λ
2
t
.
Using
y
(0) = 0 we find
c
2
=
−
λ
1
x
0
/
(
λ
2
−
λ
1
). Thus
y
=
λ
1
x
0
λ
2
−
λ
1
k
e
−
λ
1
t
−
e
−
λ
2
t

.
Substituting this result into the third differential equation we have
dz
dt
=
λ
1
λ
2
x
0
λ
2
−
λ
1
k
e
−
λ
1
t
−
e
−
λ
2
t

.
Integrating we find
z
=
−
λ
2
x
0
λ
2
−
λ
1
e
−
λ
1
t
+
λ
1
x
0
λ
2
−
λ
1
e
−
λ
2
t
+
c
3
.
Using
z
(0) = 0 we find
c
3
=
x
0
.Thus
z
=
x
x
1
−
λ
2
λ
2
−
λ
1
e
−
λ
1
t
+
λ
1
λ
2
−
λ
1
e
−
λ
2
t
v
.
2.
We see from the graph that the half-life of
A
is approximately 4
.
7 days.
To determine the half-life of
B
we use
t
= 50 as a base, since at this
time the amount of substance
A
is so small that it contributes very little
to substance
B
. Now we see from the graph that
y
(50)
≈
16
.
2 and
y
(191)
≈
8
.
1. Thus, the half-life of
B
is approximately 141 days.
3.
The amounts of
x
and
y
are the same at about
t
= 5 days. The amounts of
x
and
z
are the same at about
t
= 20 days. The amounts of
y
and
z
are the same at about
t
= 147 days. The time when
y
and
z
are the same
makes sense because most of
A
and half of
B
are gone, so half of
C
should have been formed.
4.
Suppose that the series is described schematically by
W
=
⇒−
λ
1
X
=
⇒−
λ
2
Y
=
⇒−
λ
3
Z
where
−
λ
1
,
−
λ
2
, and
−
λ
3
are the decay constants for
W
,
X
and
Y
, respectively, and
Z
is a stable element. Let
w
(
t
),
x
(
t
),
y
(
t
), and
z
(
t
) denote the amounts of substances
W
,
X
,
Y
, and
Z
, respectively. A model for the radioactive series is
dw
dt
=
−
λ
1
w
dx
dt
=
λ
1
w
−
λ
2
x
dy
dt
=
λ
2
x
−
λ
3
y
dz
dt
=
λ
3
y.
66

67. Exercises 2.9
5.
The system is
x
/
1
=2
·
3+
1
50
x
2
−
1
50
x
1
·
4=
−
2
25
x
1
+
1
50
x
2
+6
x
/
2
=
1
50
x
1
·
4
−
1
50
x
2
−
1
50
x
2
·
3=
2
25
x
1
−
2
25
x
2
.
6.
Let
x
1
,
x
2
, and
x
3
be the amounts of salt in tanks A, B, and C, respectively, so that
x
/
1
=
1
100
x
2
·
2
−
1
100
x
1
·
6=
1
50
x
2
−
3
50
x
1
x
/
2
=
1
100
x
1
·
6+
1
100
x
3
−
1
100
x
2
·
2
−
1
100
x
2
·
5=
3
50
x
1
−
7
100
x
2
+
1
100
x
3
x
/
3
=
1
100
x
2
·
5
−
1
100
x
3
−
1
100
x
3
·
4=
1
20
x
2
−
1
20
x
3
.
7. (a)
A model is
dx
1
dt
=3
·
x
2
100
−
t
−
2
·
x
1
100 +
t
,x
1
(0) = 100
dx
2
dt
=2
·
x
1
100 +
t
−
3
·
x
2
100
−
t
,x
2
(0) = 50
.
(b)
Since the system is closed, no salt enters or leaves the system and
x
1
(
t
)+
x
2
(
t
) = 100 + 50 = 150 for all
time. Thus
x
1
= 150
−
x
2
and the second equation in part (a) becomes
dx
2
dt
=
2(150
−
x
2
)
100 +
t
−
3
x
2
100
−
t
=
300
100 +
t
−
2
x
2
100 +
t
−
3
x
2
100
−
t
or
dx
2
dt
+
x
2
100 +
t
+
3
100
−
t
v
x
2
=
300
100 +
t
,
which is linear in
x
2
. An integrating factor is
e
2 ln(100+
t
)
−
3 ln(100
−
t
)
= (100 +
t
)
2
(100
−
t
)
−
3
so
d
dt
[(100 +
t
)
2
(100
−
t
)
−
3
x
2
] = 300(100 +
t
)(100
−
t
)
−
3
.
Using integration by parts, we obtain
(100 +
t
)
2
(100
−
t
)
−
3
x
2
= 300
.
1
2
(100 +
t
)(100
−
t
)
−
2
−
1
2
(100
−
t
)
−
1
+
c
f
.
Thus
x
2
=
300
(100 +
t
)
2
.
c
(100
−
t
)
3
−
1
2
(100
−
t
)
2
+
1
2
(100 +
t
)(100
−
t
)
f
=
300
(100 +
t
)
2
[
c
(100
−
t
)
3
+
t
(100
−
t
)]
.
Using
x
2
(0) = 50 we find
c
=5
/
3000. At
t
= 30,
x
2
= (300
/
130
2
)(70
3
c
+30
·
70)
≈
47
.
4lbs.
8.
A model is
dx
1
dt
= (4 gal/min)(0 lb/gal)
−
(4 gal/min)
x
1
200
x
1
lb/gal
v
dx
2
dt
= (4 gal/min)
x
1
200
x
1
lb/gal
v
−
(4 gal/min)
x
1
150
x
2
lb/gal
v
dx
3
dt
= (4 gal/min)
x
1
150
x
2
lb/gal
v
−
(4 gal/min)
x
1
100
x
3
lb/gal
v
67

68. x
y
t
x,
y
50 100
5
10
x
y
t
x,y
10 20
10
5
x
y
t
x,
y
10 20
10
5
x
y
t
x,y
10 20
10
5
x
y
t
x,y
10 20
10
5
x
y
t
x,y
20 40
10
5
x
y
t
x,y
20 40
10
5
x
y
t
x,y
20 40
10
5
x
y
t
x,y
20 40
10
5
Exercises 2.9
or
dx
1
dt
=
−
1
50
x
1
dx
2
dt
=
1
50
x
1
−
2
75
x
2
dx
3
dt
=
2
75
x
2
−
1
25
x
3
.
Over a long period of time we would expect
x
1
,
x
2
, and
x
3
to approach 0 because the entering pure water should
flush the salt out of all three tanks.
9.
From the graph we see that the populations are first equal at about
t
=5
.
6. The approximate periods of
x
and
y
are both 45.
10. (a)
The population
y
(
t
) approaches 10,000, while the population
x
(
t
)
approaches extinction.
(b)
The population
x
(
t
) approaches 5,000, while the population
y
(
t
)
approaches extinction.
(c)
The population
y
(
t
) approaches 10,000, while the population
x
(
t
)
approaches extinction.
(d)
The population
x
(
t
) approaches 5,000, while the population
y
(
t
)
approaches extinction.
11. (a) (b)
(c) (d)
68

69. Exercises 2.9
In each case the population
x
(
t
) approaches 6,000, while the population
y
(
t
) approaches 8,000.
12.
By Kirchoff’s first law we have
i
1
=
i
2
+
i
3
. By Kirchoff’s second law, on each loop we have
E
(
t
)=
Li
/
1
+
R
1
i
2
and
E
(
t
)=
Li
/
1
+
R
2
i
3
+
1
C
q
so that
q
=
CR
1
i
2
−
CR
2
i
3
. Then
i
3
=
q
/
=
CR
1
i
/
2
−
CR
2
i
3
so that the system is
Li
/
2
+
Li
/
3
+
R
1
i
2
=
E
(
t
)
−
R
1
i
/
2
+
R
2
i
/
3
+
1
C
i
3
=0
.
13.
By Kirchoff’s first law we have
i
1
=
i
2
+
i
3
. Applying Kirchoff’s second law to each loop we obtain
E
(
t
)=
i
1
R
1
+
L
1
di
2
dt
+
i
2
R
2
and
E
(
t
)=
i
1
R
1
+
L
2
di
3
dt
+
i
3
R
3
.
Combining the three equations, we obtain the system
L
1
di
2
dt
+(
R
1
+
R
2
)
i
2
+
R
1
i
3
=
E
L
2
di
3
dt
+
R
1
i
2
+(
R
1
+
R
3
)
i
3
=
E.
14.
By Kirchoff’s first law we have
i
1
=
i
2
+
i
3
. By Kirchoff’s second law, on each loop we have
E
(
t
)=
Li
/
1
+
Ri
2
and
E
(
t
)=
Li
/
1
+
1
C
q
so that
q
=
CRi
2
. Then
i
3
=
q
/
=
CRi
/
2
so that system is
Li
/
+
Ri
2
=
E
(
t
)
CRi
/
2
+
i
2
−
i
1
=0
.
15.
We first note that
s
(
t
)+
i
(
t
)+
r
(
t
)=
n
. Now the rate of change of the number of susceptible persons,
s
(
t
),
is proportional to the number of contacts between the number of people infected and the number who are
susceptible; that is,
ds/dt
=
−
k
1
s
i
. We use
−
k
1
because
s
(
t
) is decreasing. Next, the rate of change of the
number of persons who have recovered is proportional to the number infected; that is,
dr/dt
=
k
2
i
where
k
2
is
positive since
r
is increasing. Finally, to obtain
di/dt
we use
d
dt
(
s
+
i
+
r
)=
d
dt
n
=0
.
This gives
di
dt
=
−
dr
dt
−
ds
dt
=
−
k
2
i
+
k
1
si.
The system of equations is then
ds
dt
=
−
k
1
si
di
dt
=
−
k
2
i
+
k
1
si
dr
dt
=
k
2
i.
A reasonable set of initial conditions is
i
(0) =
i
0
, the number of infected people at time 0,
s
(0) =
n
−
i
0
, and
r
(0) = 0.
16. (a)
If we know
s
(
t
) and
i
(
t
) then we can determine
r
(
t
) from
s
+
i
+
r
=
n
.
69

70. s
i
t
s,
i
510
5
10
s
i
t
s,
i
510
5
10
s
i
t
s,
i
510
5
10
s
i
t
s,
i
510
5
10
Exercises 2.9
(b)
In this case the system is
ds
dt
=
−
0
.
2
si
di
dt
=
−
0
.
7
i
+0
.
2
si.
We also note that when
i
(0) =
i
0
,
s
(0)=10
−
i
0
since
r
(0) = 0 and
i
(
t
)+
s
(
t
)+
r
(
t
) = 0 for all values of
t
.Now
k
2
/k
1
=0
.
7
/
0
.
2=3
.
5, so we consider initial conditions
s
(0) = 2,
i
(0)=8;
s
(0) = 3
.
4,
i
(0)=6
.
6;
s
(0) = 7,
i
(0) = 3; and
s
(0) = 9,
i
(0) = 1.
We see that an initial susceptible population greater than
k
2
/k
1
results in an epidemic in the sense that
the number of infected persons increases to a maximum before decreasing to 0. On the other hand, when
s
(0)
<k
2
/k
1
, the number of infected persons decreases from the start and there is no epidemic.
Chapter 2 Review Exercises
1.
Writing the differential equation in the form
y
/
=
k
(
y
+
A/k
) we see that the critical point
−
A/k
is a repeller
for
k>
0 and an attractor for
k<
0.
2.
Separating variables and integrating we have
dy
y
=
4
x
dx
ln
y
=4ln
x
+
c
=ln
x
4
+
c
y
=
c
1
x
4
.
We see that when
x
=0,
y
= 0, so the initial-value problem has an infinite number of solutions for
k
= 0 and
no solutions for
k
k
=0.
3.
dy
dx
=(
y
−
1)
2
(
y
−
3)
2
4.
dy
dx
=
y
(
y
−
2)
2
(
y
−
4)
70

71. t
x
t
x
t
x
t
x
0
>>
0
<>
0
<<
0
><
x
y
Chapter 2 Review Exercises
5.
x
n
x
even
x
n
x
odd
−
x
n
x
even
−
x
n
x
odd
For
dx/dt
=
x
n
,when
n
is even, 0 is semi-stable; when
n
is odd, 0 is unstable. For
dx/dt
=
−
x
n
,when
n
is even,
0 is semi-stable; when
n
is odd, 0 is stable.
6.
The zero of
f
occurs at approximately 1
.
3. Since
P
/
(
t
)=
f
(
P
)
>
0 for
P<
1
.
3 and
P
/
(
t
)=
f
(
P
)
>
0 for
P>
1
.
3, lim
t
→∞
P
(
t
)=1
.
3.
7.
8. (a)
linear in
y
, homogeneous, exact
(b)
linear in
x
(c)
separable, exact, linear in
x
and
y
(d)
Bernoulli in
x
(e)
separable
(f)
separable, linear in
x
, Bernoulli
(g)
linear in
x
(h)
homogeneous
(i)
Bernoulli
(j)
homogeneous, exact, Bernoulli
(k)
linear in
x
and
y
, exact, separable, homogeneous
(l)
exact, linear in
y
(m)
homogeneous
(n)
separable
9.
Separating variables we obtain
cos
2
xdx
=
y
y
2
+1
dy
=
⇒
1
2
x
+
1
4
sin 2
x
=
1
2
ln
k
y
2
+1

+
c
=
⇒
2
x
+ sin 2
x
=2ln
k
y
2
+1

+
c.
71

72. Chapter 2 Review Exercises
10.
Write the differential equation in the form
y
ln
x
y
dx
=
x
x
ln
x
y
−
y
v
dy
. This is a homogeneous equation, so let
x
=
uy
. Then
dx
=
udy
+
ydu
and the differential equation becomes
y
ln
u
(
udy
+
ydu
)=(
uy
ln
u
−
y
)
dy
or
y
ln
udu
=
−
dy.
Separating variables we obtain
ln
udu
=
−
dy
y
=
⇒
u
ln
|
u
|−
u
=
−
ln
|
y
|
+
c
=
⇒
x
y
ln




x
y




−
x
y
=
−
ln
|
y
|
+
c
=
⇒
x
(ln
x
−
ln
y
)
−
x
=
−
y
ln
|
y
|
+
cy.
11.
The differential equation
dy
dx
+
2
6
x
+1
y
=
−
3
x
2
6
x
+1
y
−
2
is Bernoulli. Using
w
=
y
3
we obtain
dw
dx
+
6
6
x
+1
w
=
−
9
x
2
6
x
+1
. An integrating factor is 6
x
+1,so
d
dx
[(6
x
+1)
w
]=
−
9
x
2
=
⇒
w
=
−
3
x
3
6
x
+1
+
c
6
x
+1
=
⇒
(6
x
+1)
y
3
=
−
3
x
3
+
c.
(Note: The differential equation is also exact.)
12.
Write the differential equation in the form (3
y
2
+2
x
)
dx
+(4
y
2
+6
xy
)
dy
= 0. Letting
M
=3
y
2
+2
x
and
N
=4
y
2
+6
xy
we see that
M
y
=6
y
=
N
x
so the differential equation is exact. From
f
x
=3
y
2
+2
x
we obtain
f
=3
xy
2
+
x
2
+
h
(
y
). Then
f
y
=6
xy
+
h
/
(
y
)=4
y
2
+6
xy
and
h
/
(
y
)=4
y
2
so
h
(
y
)=
4
3
y
3
. The general solution
is
3
xy
2
+
x
2
+
4
3
y
3
=
c.
13.
Write the equation in the form
dQ
dt
+
1
t
Q
=
t
3
ln
t.
An integrating factor is
e
ln
t
=
t
,so
d
dt
[
tQ
]=
t
4
ln
t
=
⇒
tQ
=
−
1
25
t
5
+
1
5
t
5
ln
t
+
c
=
⇒
Q
=
−
1
25
t
4
+
1
5
t
4
ln
t
+
c
t
.
14.
Letting
u
=2
x
+
y
+1wehave
du
dx
=2+
dy
dx
,
and so the given differential equation is transformed into
u
x
du
dx
−
2
v
=1 or
du
dx
=
2
u
+1
u
.
Separating variables and integrating we get
u
2
u
+1
du
=
dx
x
1
2
−
1
2
1
2
u
+1
v
du
=
dx
1
2
u
−
1
4
ln
|
2
u
+1
|
=
x
+
c
2
u
−
ln
|
2
u
+1
|
=2
x
+
c
1
.
72

73. t
y
Chapter 2 Review Exercises
Resubstituting for
u
gives the solution
4
x
+2
y
+2
−
ln
|
4
x
+2
y
+3
|
=2
x
+
c
1
or
2
x
+2
y
+2
−
ln
|
4
x
+2
y
+3
|
=
c
1
.
15.
Write the equation in the form
dy
dx
+
8
x
x
2
+4
y
=
2
x
x
2
+4
. An integrating factor is
k
x
2
+4

4
,so
d
dx
>
k
x
2
+4

4
y
c
=2
x
k
x
2
+4

3
=
⇒
k
x
2
+4

4
y
=
1
4
k
x
2
+4

4
+
c
=
⇒
y
=
1
4
+
c
k
x
2
+4

−
4
.
16.
Letting
M
=2
r
2
cos
θ
sin
θ
+
r
cos
θ
and
N
=4
r
+ sin
θ
−
2
r
cos
2
θ
we see that
M
r
=4
r
cos
θ
sin
θ
+ cos
θ
=
N
θ
so the differential equation is exact. From
f
θ
=2
r
2
cos
θ
sin
θ
+
r
cos
θ
we obtain
f
=
−
r
2
cos
2
θ
+
r
sin
θ
+
h
(
r
).
Then
f
r
=
−
2
r
cos
2
θ
+ sin
θ
+
h
/
(
r
)=4
r
+ sin
θ
−
2
r
cos
2
θ
and
h
/
(
r
)=4
r
so
h
(
r
)=2
r
2
. The general solution
is
−
r
2
cos
2
θ
+
r
sin
θ
+2
r
2
=
c.
17.
The differential equation has the form
d
dx
[(sin
x
)
y
] = 0. Integrating we have (sin
x
)
y
=
c
or
y
=
c/
sin
x
. The initial condition implies
c
=
−
2 sin(7
π/
6) = 1. Thus,
y
=1
/
sin
x
, where
π<x<
2
π
is
chosen to include
x
=7
π/
6.
18.
For
y>t
the slopes are negative and for
y<t
the slopes are positive, so the
differential equation is (a).
19. (a)
For
y<
0,
√
y
is not a real number.
(b)
Separating variables and integrating we have
dy
√
y
=
dx
and 2
√
y
=
x
+
c.
Letting
y
(
x
0
)=
y
0
we get
c
=2
√
y
0
−
x
0
, so that
2
√
y
=
x
+2
√
y
0
−
x
0
and
y
=
1
4
(
x
+2
√
y
0
−
x
0
)
2
.
Since
√
y>
0 for
y
k
= 0, we see that
dy/dx
=
1
2
(
x
+2
√
y
0
−
x
0
) must be positive. Thus, the interval on
which the solution is defined is (
x
0
−
2
√
y
0
,
∞
).
73

74. -2-1 12
x
-2
-1
1
2
y
Chapter 2 Review Exercises
20. (a)
The differential equation is homogeneous and we let
y
=
ux
. Then
(
x
2
−
y
2
)
dx
+
xy dy
=0
(
x
2
−
u
2
x
2
)
dx
+
ux
2
(
udx
+
xdu
)=0
dx
+
ux du
=0
udu
=
−
dx
x
1
2
u
2
=
−
ln
|
x
|
+
c
y
2
x
2
=
−
2ln
|
x
|
+
c
1
.
The initial condition gives
c
1
= 2, so an implicit solution is
y
2
=
x
2
(2
−
2ln
|
x
|
).
(b)
Solving for
y
in part (a) and being sure that the initial condition is still satisfied, we have
y
=
−
√
2
|
x
|
(1
−
ln
|
x
|
)
1
/
2
, where
−
e
≤
x
≤
e
so that 1
−
ln
|
x
|≥
0. A graph of this function shown be-
low indicates that the derivative is not defined at
x
= 0 and
x
=
e
. Thus, the solution of the initial-value
problem is
y
=
−
√
2
x
(1
−
ln
x
)
1
/
2
, for 0
<x<e
.
21.
The graph of
y
1
(
x
) is the portion of the closed black curve lying in the fourth quadrant. Its interval of definition
is approximately (0
.
7
,
4
.
3). The graph of
y
2
(
x
) is the portion of the left-hand black curve lying in the third
quadrant. Its interval of definition is (
−∞
,
0).
22.
The first step of Euler’s method gives
y
(1
.
1)
≈
9+0
.
1(1 + 3) = 9
.
4. Applying Euler’s method one more time
gives
y
(1
.
2)
≈
9
.
4+0
.
1(1+1
.
1
√
9
.
4)
≈
9
.
8373.
23.
From
dP
dt
=0
.
018
P
and
P
(0) = 4 billion we obtain
P
=4
e
0
.
018
t
so that
P
(45) = 8
.
99 billion.
24.
Let
A
=
A
(
t
) be the volume of CO
2
at time
t
. From
dA
dt
=1
.
2
−
A
4
and
A
(0) = 16 ft
3
we obtain
A
=
4
.
8+11
.
2
e
−
t/
4
. Since
A
(10) = 5
.
7ft
3
, the concentration is 0
.
017%. As
t
→∞
we have
A
→
4
.
8ft
3
or 0
.
06%.
25.
From
dE/dt
=
−
E/RC
and
E
(
t
1
)=
E
0
we obtain
E
=
E
0
e
(
t
1
−
t
)
/RC
.
26.
From
V dC/dt
=
kA
(
C
s
−
C
) and
C
(0) =
C
0
we obtain
C
=
C
s
+(
C
0
−
C
s
)
e
−
kAt/V
.
27. (a)
The differential equation is
dT
dt
=
k
[
T
−
T
2
−
B
(
T
1
−
T
)] =
k
[(1 +
B
)
T
−
(
BT
1
+
T
2
)]
.
Separating variables we obtain
dT
(1 +
B
)
T
−
(
BT
1
+
T
2
)
=
kdt
. Then
1
1+
B
ln
|
(1 +
B
)
T
−
(
BT
1
+
T
2
)
|
=
kt
+
c
and
T
(
t
)=
BT
1
+
T
2
1+
B
+
c
3
e
k
(1+
B
)
t
.
Since
T
(0) =
T
1
we must have
c
3
=
T
1
−
T
2
1+
B
and so
T
(
t
)=
BT
1
+
T
2
1+
B
+
T
1
−
T
2
1+
B
e
k
(1+
B
)
t
.
74

75. 1020
10
20
x
y
-
55
-
5
5
Chapter 2 Review Exercises
(b)
Since
k<
0, lim
t
→∞
e
k
(1+
B
)
t
= 0 and lim
t
→∞
T
(
t
)=
BT
1
+
T
2
1+
B
.
(c)
Since
T
s
=
T
2
+
B
(
T
1
−
T
), lim
t
→∞
T
s
=
T
2
+
BT
1
−
B
−
BT
1
+
T
2
1+
B
/
=
BT
1
+
T
2
1+
B
.
28.
We first solve
−
1
−
t
10
/
di
dt
+0
.
2
i
= 4. Separating variables we obtain
di
40
−
2
i
=
dt
10
−
t
.
Then
−
1
2
ln
|
40
−
2
i
|
=
−
ln
|
10
−
t
|
+
c
or
√
40
−
2
i
=
c
1
(10
−
t
)
.
Since
i
(0) = 0 we must have
c
1
=2
/
√
10 . Solving for
i
we get
i
(
t
)=4
t
−
1
5
t
2
,
0
≤
t<
10. For
t
≥
10 the equation for the current becomes 0
.
2
i
=4or
i
= 20. Thus
i
(
t
)=
I
4
t
−
1
5
t
2
,
0
≤
t<
10
20
,t
≥
10
.
29.
From
y

1+(
y
β
)
2
<
=
k
we obtain
dx
=
√
y
√
k
−
y
dy
.If
y
=
k
sin
2
θ
then
dy
=2
k
sin
θ
cos
θdθ, dx
=2
k
−
1
2
−
1
2
cos 2
θ
/
dθ,
and
x
=
kθ
−
k
2
sin 2
θ
+
c.
If
x
= 0 when
θ
= 0 then
c
=0.
30. (a)
From
y
=
−
x
−
1+
c
1
e
x
we obtain
y
β
=
y
+
x
so that the differential equation of the orthogonal family is
dy
dx
=
−
1
y
+
x
or
dx
dy
+
x
=
−
y
. An integrating factor is
e
y
,so
d
dy
[
e
y
x
]=
−
ye
y
=
⇒
e
y
x
=
−
ye
y
+
e
y
+
c
=
⇒
x
=
−
y
+1+
ce
−
y
.
(b)
Differentiating the family of curves, we have
y
β
=
−
1
(
x
+
c
1
)
2
=
−
1
y
2
.
The differential equation for the family of orthogonal trajectories is then
y
β
=
y
2
. Separating variables and integrating we get
dy
y
2
=
dx
−
1
y
=
x
+
c
1
y
=
−
1
x
+
c
1
.
31.
From
dx
dt
=
k
1
x
(
α
−
x
) we obtain
−
1
/α
x
+
1
/α
α
−
x
/
dx
=
k
1
dt
so that
x
=
αc
1
e
αk
1
t
1+
c
1
e
αk
1
t
. From
dy
dt
=
k
2
xy
we
obtain
ln
|
y
|
=
k
2
k
1
ln


1+
c
1
e
αk
1
t


+
c
or
y
=
c
2

1+
c
1
e
αk
1
t

k
2
/k
1
.
75

76. Chapter 2 Review Exercises
32.
In tank A the salt input is
x
7
gal
min
vx
2
lb
gal
v
+
x
1
gal
min
vx
x
2
100
lb
gal
v
=
x
14 +
1
100
x
2
v
lb
min
.
The salt output is
x
3
gal
min
vx
x
1
100
lb
gal
v
+
x
5
gal
min
vx
x
1
100
lb
gal
v
=
2
25
x
1
lb
min
.
In tank B the salt input is
x
5
gal
min
vx
x
1
100
lb
gal
v
=
1
20
x
1
lb
min
.
The salt output is
x
1
gal
min
vx
x
2
100
lb
gal
v
+
x
4
gal
min
vx
x
2
100
lb
gal
v
=
1
20
x
2
lb
min
.
The system of differential equations is then
dx
1
dt
=14+
1
100
x
2
−
2
25
x
1
dx
2
dt
=
1
20
x
1
−
1
20
x
2
.
76

77. 3
Higher-Order Differential Equations
Exercises 3.1
1.
From
y
=
c
1
e
x
+
c
2
e
−
x
we find
y
1
=
c
1
e
x
−
c
2
e
−
x
. Then
y
(0) =
c
1
+
c
2
=0,
y
1
(0) =
c
1
−
c
2
= 1 so that
c
1
=1
/
2
and
c
2
=
−
1
/
2. The solution is
y
=
1
2
e
x
−
1
2
e
−
x
.
2.
From
y
=
c
1
e
4
x
+
c
2
e
−
x
we find
y
1
=4
c
1
e
4
x
−
c
2
e
−
x
. Then
y
(0) =
c
1
+
c
2
=1,
y
1
(0)=4
c
1
−
c
2
= 2 so that
c
1
=3
/
5 and
c
2
=2
/
5. The solution is
y
=
3
5
e
4
x
+
2
5
e
−
x
.
3.
From
y
=
c
1
x
+
c
2
x
ln
x
we find
y
1
=
c
1
+
c
2
(1+ln
x
). Then
y
(1) =
c
1
=3,
y
1
(1) =
c
1
+
c
2
=
−
1 so that
c
1
=3
and
c
2
=
−
4. The solution is
y
=3
x
−
4
x
ln
x
.
4.
From
y
=
c
1
+
c
2
cos
x
+
c
3
sin
x
we find
y
1
=
−
c
2
sin
x
+
c
3
cos
x
and
y
11
=
−
c
2
cos
x
−
c
3
sin
x
. Then
y
(
π
)=
c
1
−
c
2
=0,
y
1
(
π
)=
−
c
3
=2,
y
11
(
π
)=
c
2
=
−
1 so that
c
1
=
−
1,
c
2
=
−
1, and
c
3
=
−
2. The solution is
y
=
−
1
−
cos
x
−
2 sin
x
.
5.
From
y
=
c
1
+
c
2
x
2
we find
y
1
=2
c
2
x
. Then
y
(0) =
c
1
=0,
y
1
(0)=2
c
2
·
0 = 0 and
y
1
(0) = 1 is not possible.
Since
a
2
(
x
)=
x
is 0 at
x
= 0, Theorem 3
.
1 is not violated.
6.
In this case we have
y
(0) =
c
1
=0,
y
1
(0)=2
c
2
·
0=0so
c
1
= 0 and
c
2
is arbitrary. Two solutions are
y
=
x
2
and
y
=2
x
2
.
7.
From
x
(0) =
x
0
=
c
1
we see that
x
(
t
)=
x
0
cos
ωt
+
c
2
sin
ωt
and
x
1
(
t
)=
−
x
0
sin
ωt
+
c
2
ω
cos
ωt
. Then
x
1
(0) =
x
1
=
c
2
ω
implies
c
2
=
x
1
/ω
.Thus
x
(
t
)=
x
0
cos
ωt
+
x
1
ω
sin
ωt.
8.
Solving the system
x
(
t
0
)=
c
1
cos
ωt
0
+
c
2
sin
ωt
0
=
x
x
1
(
t
0
)=
−
c
1
ω
sin
ωt
0
+
c
2
ω
cos
ωt
0
=
x
1
for
c
1
and
c
2
gives
c
1
=
ωx
0
cos
ωt
0
−
x
1
sin
ωt
0
ω
and
c
2
=
x
1
cos
ωt
0
+
ωx
0
sin
ωt
0
ω
.
Thus
x
(
t
)=
ωx
0
cos
ωt
0
−
x
1
sin
ωt
0
ω
cos
ωt
+
x
1
cos
ωt
0
+
ωx
0
sin
ωt
0
ω
sin
ωt
=
x
0
(cos
ωt
cos
ωt
0
+ sin
ωt
sin
ωt
0
)+
x
1
ω
(sin
ωt
cos
ωt
0
−
cos
ωt
sin
ωt
0
)
=
x
0
cos
ω
(
t
−
t
0
)+
x
1
ω
sin
ω
(
t
−
t
0
)
.
9.
Since
a
2
(
x
)=
x
−
2 and
x
0
= 0 the problem has a unique solution for
−∞
<x<
2.
10.
Since
a
0
(
x
) = tan
x
and
x
0
= 0 the problem has a unique solution for
−
π/
2
<x<π/
2.
11.
We have
y
(0) =
c
1
+
c
2
=0,
y
1
(1) =
c
1
e
+
c
2
e
−
1
= 1 so that
c
1
=
e/
1
e
2
−
1
.
and
c
2
=
−
e/
1
e
2
−
1
.
. The
solution is
y
=
e
(
e
x
−
e
−
x
)
/
1
e
2
−
1
.
.
12.
In this case we have
y
(0) =
c
1
=1,
y
1
(1) = 2
c
2
= 6 so that
c
1
= 1 and
c
2
= 3. The solution is
y
=1+3
x
2
.
77

78. Exercises 3.1
13.
From
y
=
c
1
e
x
cos
x
+
c
2
e
x
sin
x
we find
y
1
=
c
1
e
x
(
−
sin
x
+ cos
x
)+
c
2
e
x
(cos
x
+ sin
x
).
(a)
We have
y
(0) =
c
1
=1,
y
1
(0) =
c
1
+
c
2
= 0 so that
c
1
= 1 and
c
2
=
−
1. The solution is
y
=
e
x
cos
x
−
e
x
sin
x
.
(b)
We have
y
(0) =
c
1
=1,
y
(
π
)=
−
c
1
e
π
=
−
1, which is not possible.
(c)
We have
y
(0) =
c
1
=1,
y
(
π/
2) =
c
2
e
π/
2
= 1 so that
c
1
= 1 and
c
2
=
e
−
π/
2
. The solution is
y
=
e
x
cos
x
+
e
−
π/
2
e
x
sin
x
.
(d)
We have
y
(0) =
c
1
=0,
y
(
π
)=
−
c
1
e
π
= 0 so that
c
1
= 0 and
c
2
is arbitrary. Solutions are
y
=
c
2
e
x
sin
x
,
for any real numbers
c
2
.
14. (a)
We have
y
(
−
1) =
c
1
+
c
2
+3=0,
y
(1) =
c
1
+
c
2
+ 3 = 4, which is not possible.
(b)
We have
y
(0) =
c
1
·
0+
c
2
·
0 + 3 = 1, which is not possible.
(c)
We have
y
(0) =
c
1
·
0+
c
2
·
0+3=3,
y
(1) =
c
1
+
c
2
+ 3 = 0 so that
c
1
is arbitrary and
c
2
=
−
3
−
c
1
.
Solutions are
y
=
c
1
x
2
−
(
c
1
+3)
x
4
+3.
(d)
We have
y
(1) =
c
1
+
c
2
+3=3,
y
(2)=4
c
1
+16
c
2
+ 3 = 15 so that
c
1
=
−
1 and
c
2
= 1. The solution is
y
=
−
x
2
+
x
4
+3.
15.
Since (
−
4)
x
+ (3)
x
2
+ (1)(4
x
−
3
x
2
) = 0 the functions are linearly dependent.
16.
Since (1)0 + (0)
x
+ (0)
e
x
= 0 the functions are linearly dependent. A similar argument shows that any set of
functions containing
f
(
x
) = 0 will be linearly dependent.
17.
Since (
−
1
/
5)5 + (1) cos
2
x
+ (1) sin
2
x
= 0 the functions are linearly dependent.
18.
Since (1) cos 2
x
+(1)1+(
−
2) cos
2
x
= 0 the functions are linearly dependent.
19.
Since (
−
4)
x
+ (3)(
x
−
1) + (1)(
x
+ 3) = 0 the functions are linearly dependent.
20.
From the graphs of
f
1
(
x
)=2+
x
and
f
2
(
x
)=2+
|
x
|
we
see that the functions are linearly independent since they
cannot be multiples of each other.
21.
The functions are linearly independent since
W
1
1+
x, x, x
2
.
=
2
2
2
2
2
2
2
1+
xxx
2
112
x
002
2
2
2
2
2
2
2
=2
3
=0.
22.
Since (
−
1
/
2)
e
x
+(1
/
2)
e
−
x
+ (1) sinh
x
= 0 the functions are linearly dependent.
23.
The functions satisfy the differential equation and are linearly independent since
W
1
e
−
3
x
,e
4
x
.
=7
e
x
3
=0
for
−∞
<x<
∞
. The general solution is
y
=
c
1
e
−
3
x
+
c
2
e
4
x
.
24.
The functions satisfy the differential equation and are linearly independent since
W
(cosh 2
x,
sinh 2
x
)=2
for
−∞
<x<
∞
. The general solution is
y
=
c
1
cosh 2
x
+
c
2
sinh 2
x.
78

79. Exercises 3.1
25.
The functions satisfy the differential equation and are linearly independent since
W
(
e
x
cos 2
x, e
x
sin 2
x
)=2
e
2
x
3
=0
for
−∞
<x<
∞
. The general solution is
y
=
c
1
e
x
cos 2
x
+
c
2
e
x
sin 2
x
.
26.
The functions satisfy the differential equation and are linearly independent since
W
3
e
x/
2
,xe
x/
2
4
=
e
x
3
=0
for
−∞
<x<
∞
. The general solution is
y
=
c
1
e
x/
2
+
c
2
xe
x/
2
.
27.
The functions satisfy the differential equation and are linearly independent since
W
1
x
3
,x
4
.
=
x
6
3
=0
for 0
<x<
∞
. The general solution is
y
=
c
1
x
3
+
c
2
x
4
.
28.
The functions satisfy the differential equation and are linearly independent since
W
(cos(ln
x
)
,
sin(ln
x
)) = 1
/x
3
=0
for 0
<x<
∞
. The general solution is
y
=
c
1
cos(ln
x
)+
c
2
sin(ln
x
)
.
29.
The functions satisfy the differential equation and are linearly independent since
W
1
x, x
−
2
,x
−
2
ln
x
.
=9
x
−
6
3
=0
for 0
<x<
∞
. The general solution is
y
=
c
1
x
+
c
2
x
−
2
+
c
3
x
−
2
ln
x.
30.
The functions satisfy the differential equation and are linearly independent since
W
(1
,x,
cos
x,
sin
x
)=1
for
−∞
<x<
∞
. The general solution is
y
=
c
1
+
c
2
x
+
c
3
cos
x
+
c
4
sin
x.
31. (a)
From the graphs of
y
1
=
x
3
and
y
2
=
|
x
|
3
we see that
the functions are linearly independent since they cannot be
multiples of each other. It is easily shown that
y
1
=
x
3
solves
x
2
y
11
−
4
xy
1
+6
y
= 0. To show that
y
2
=
|
x
|
3
is a
solution let
y
2
=
x
3
for
x
≥
0 and let
y
2
=
−
x
3
for
x<
0.
(b)
If
x
≥
0 then
y
2
=
x
3
and
W
(
y
1
,y
2
)=
2
2
2
2
x
3
x
3
3
x
2
3
x
2
2
2
2
2
=0. If
x<
0 then
y
2
=
−
x
3
and
W
(
y
1
,y
2
)=
2
2
2
2
x
3
−
x
3
3
x
2
−
3
x
2
2
2
2
2
=0.
This does not violate Theorem 3
.
3 since
a
2
(
x
)=
x
2
is zero at
x
=0.
79

80. Exercises 3.1
(c)
The functions
Y
1
=
x
3
and
Y
2
=
x
2
are solutions of
x
2
y
11
−
4
xy
1
+6
y
= 0. They are linearly independent
since
W
1
x
3
,x
2
.
=
x
4
3
= 0 for
−∞
<x<
∞
.
(d)
The function
y
=
x
3
satisfies
y
(0) = 0 and
y
1
(0) = 0.
(e)
Neither is the general solution since we form a general solution on an interval for which
a
2
(
x
)
3
= 0 for every
x
in the interval.
32.
By the superposition principle, if
y
1
=
e
x
and
y
2
=
e
−
x
are both solutions of a homogeneous linear differential
equation, then so are
1
2
(
y
1
+
y
2
)=
e
x
+
e
−
x
2
= cosh
x
and
1
2
(
y
1
−
y
2
)=
e
x
−
e
−
x
2
= sinh
x.
33.
Since
e
x
−
3
=
e
−
3
e
x
=(
e
−
5
e
2
)
e
x
=
e
−
5
e
x
+2
, we see that
e
x
−
3
is a constant multiple of
e
x
+2
and the functions
are linearly dependent.
34.
Since 0
y
1
+0
y
2
+
···
+0
y
k
+1
y
k
+1
= 0, the set of solutions is linearly dependent.
35.
The functions
y
1
=
e
2
x
and
y
2
=
e
5
x
form a fundamental set of solutions of the homogeneous equation, and
y
p
=6
e
x
is a particular solution of the nonhomogeneous equation.
36.
The functions
y
1
= cos
x
and
y
2
= sin
x
form a fundamental set of solutions of the homogeneous equation, and
y
p
=
x
sin
x
+ (cos
x
) ln(cos
x
) is a particular solution of the nonhomogeneous equation.
37.
The functions
y
1
=
e
2
x
and
y
2
=
xe
2
x
form a fundamental set of solutions of the homogeneous equation, and
y
p
=
x
2
e
2
x
+
x
−
2 is a particular solution of the nonhomogeneous equation.
38.
The functions
y
1
=
x
−
1
/
2
and
y
2
=
x
−
1
form a fundamental set of solutions of the homogeneous equation, and
y
p
=
1
15
x
2
−
1
6
x
is a particular solution of the nonhomogeneous equation.
39. (a)
We have
y
1
p
1
=6
e
2
x
and
y
11
p
1
=12
e
2
x
,so
y
11
p
1
−
6
y
1
p
1
+5
y
p
1
=12
e
2
x
−
36
e
2
x
+15
e
2
x
=
−
9
e
2
x
.
Also,
y
1
p
2
=2
x
+ 3 and
y
11
p
2
=2,so
y
11
p
2
−
6
y
1
p
2
+5
y
p
2
=2
−
6(2
x
+3)+5(
x
2
+3
x
)=5
x
2
+3
x
−
16
.
(b)
By the superposition principle for nonhomogeneous equations a particular solution of
y
11
−
6
y
1
+5
y
=
5
x
2
+3
x
−
16
−
9
e
2
x
is
y
p
=
x
2
+3
x
+3
e
2
x
. A particular solution of the second equation is
y
p
=
−
2
y
p
2
−
1
9
y
p
1
=
−
2
x
2
−
6
x
−
1
3
e
2
x
.
40. (a)
y
p
1
=5
(b)
y
p
2
=
−
2
x
(c)
y
p
=
y
p
1
+
y
p
2
=5
−
2
x
(d)
y
p
=
1
2
y
p
1
−
2
y
p
2
=
5
2
+4
x
80

81. Exercises 3.2
Exercises 3.2
In Problems 1-8 we use reduction of order to find a second solution. In Problems 9-16 we use formula (5) from the
text.
1.
Define
y
=
u
(
x
)
e
2
x
so
y
1
=2
ue
2
x
+
u
1
e
2
x
,y
11
=
e
2
x
u
11
+4
e
2
x
u
1
+4
e
2
x
u,
and
y
11
−
4
y
1
+4
y
=4
e
2
x
u
11
=0
.
Therefore
u
11
= 0 and
u
=
c
1
x
+
c
2
. Taking
c
1
= 1 and
c
2
= 0 we see that a second solution is
y
2
=
xe
2
x
.
2.
Define
y
=
u
(
x
)
xe
−
x
so
y
1
=(1
−
x
)
e
−
x
u
+
xe
−
x
u
1
,y
11
=
xe
−
x
u
11
+ 2(1
−
x
)
e
−
x
u
1
−
(2
−
x
)
e
−
x
u,
and
y
11
+2
y
1
+
y
=
e
−
x
(
xu
11
+2
u
1
)=0 or
u
11
+
2
x
u
1
=0
.
If
w
=
u
1
we obtain the first-order equation
w
1
+
2
x
w
= 0 which has the integrating factor
e
2
5
dx/x
=
x
2
.Now
d
dx
[
x
2
w
] = 0 gives
x
2
w
=
c.
Therefore
w
=
u
1
=
c/x
2
and
u
=
c
1
/x
. A second solution is
y
2
=
1
x
xe
−
x
=
e
−
x
.
3.
Define
y
=
u
(
x
) cos 4
x
so
y
1
=
−
4
u
sin 4
x
+
u
1
cos 4
x, y
11
=
u
11
cos 4
x
−
8
u
1
sin 4
x
−
16
u
cos 4
x
and
y
11
+16
y
= (cos 4
x
)
u
11
−
8(sin 4
x
)
u
1
=0 or
u
11
−
8(tan 4
x
)
u
1
=0
.
If
w
=
u
1
we obtain the first-order equation
w
1
−
8(tan 4
x
)
w
= 0 which has the integrating factor
e
−
8
5
tan 4
xdx
=
cos
2
4
x
.Now
d
dx
[(cos
2
4
x
)
w
] = 0 gives (cos
2
4
x
)
w
=
c.
Therefore
w
=
u
1
=
c
sec
2
4
x
and
u
=
c
1
tan 4
x
. A second solution is
y
2
= tan 4
x
cos 4
x
= sin 4
x
.
4.
Define
y
=
u
(
x
) sin 3
x
so
y
1
=3
u
cos 3
x
+
u
1
sin 3
x, y
11
=
u
11
sin 3
x
+6
u
1
cos 3
x
−
9
u
sin 3
x,
and
y
11
+9
y
= (sin 3
x
)
u
11
+ 6(cos 3
x
)
u
1
=0 or
u
11
+ 6(cot 3
x
)
u
1
=0
.
If
w
=
u
1
we obtain the first-order equation
w
1
+ 6(cot 3
x
)
w
= 0 which has the integrating factor
e
6
5
cot 3
xdx
=
sin
2
3
x
.Now
d
dx
[(sin
2
3
x
)
w
] = 0 gives (sin
2
3
x
)
w
=
c.
Therefore
w
=
u
1
=
c
csc
2
3
x
and
u
=
c
1
cot 3
x
. A second solution is
y
2
= cot 3
x
sin 3
x
= cos 3
x
.
5.
Define
y
=
u
(
x
) cosh
x
so
y
1
=
u
sinh
x
+
u
1
cosh
x, y
11
=
u
11
cosh
x
+2
u
1
sinh
x
+
u
cosh
x
and
y
11
−
y
= (cosh
x
)
u
11
+ 2(sinh
x
)
u
1
=0 or
u
11
+ 2(tanh
x
)
u
1
=0
.
81

82. Exercises 3.2
If
w
=
u
1
we obtain the first-order equation
w
1
+ 2(tanh
x
)
w
= 0 which has the integrating factor
e
2
5
tanh
xdx
=
cosh
2
x
.Now
d
dx
[(cosh
2
x
)
w
] = 0 gives (cosh
2
x
)
w
=
c.
Therefore
w
=
u
1
=
c
sech
2
x
and
u
=
c
1
tanh
x
. A second solution is
y
2
= tanh
x
cosh
x
= sinh
x
.
6.
Define
y
=
u
(
x
)
e
5
x
so
y
1
=5
e
5
x
u
+
e
5
x
u
1
,y
11
=
e
5
x
u
11
+10
e
5
x
u
1
+25
e
5
x
u
and
y
11
−
25
y
=
e
5
x
(
u
11
+10
u
1
)=0 or
u
11
+10
u
1
=0
.
If
w
=
u
1
we obtain the first-order equation
w
1
+10
w
= 0 which has the integrating factor
e
10
5
dx
=
e
10
x
.Now
d
dx
[
e
10
x
w
] = 0 gives
e
10
x
w
=
c.
Therefore
w
=
u
1
=
ce
−
10
x
and
u
=
c
1
e
−
10
x
. A second solution is
y
2
=
e
−
10
x
e
5
x
=
e
−
5
x
.
7.
Define
y
=
u
(
x
)
e
2
x/
3
so
y
1
=
2
3
e
2
x/
3
u
+
e
2
x/
3
u
1
,y
11
=
e
2
x/
3
u
11
+
4
3
e
2
x/
3
u
1
+
4
9
e
2
x/
3
u
and
9
y
11
−
12
y
1
+4
y
=9
e
2
x/
3
u
11
=0
.
Therefore
u
11
= 0 and
u
=
c
1
x
+
c
2
. Taking
c
1
= 1 and
c
2
= 0 we see that a second solution is
y
2
=
xe
2
x/
3
.
8.
Define
y
=
u
(
x
)
e
x/
3
so
y
1
=
1
3
e
x/
3
u
+
e
x/
3
u
1
,y
11
=
e
x/
3
u
11
+
2
3
e
x/
3
u
1
+
1
9
e
x/
3
u
and
6
y
11
+
y
1
−
y
=
e
x/
3
(6
u
11
+5
u
1
)=0 or
u
11
+
5
6
u
1
=0
.
If
w
=
u
1
we obtain the first-order equation
w
1
+
5
6
w
= 0 which has the integrating factor
e
(5
/
6)
5
dx
=
e
5
x/
6
.
Now
d
dx
[
e
5
x/
6
w
] = 0 gives
e
5
x/
6
w
=
c.
Therefore
w
=
u
1
=
ce
−
5
x/
6
and
u
=
c
1
e
−
5
x/
6
. A second solution is
y
2
=
e
−
5
x/
6
e
x/
3
=
e
−
x/
2
.
9.
Identifying
P
(
x
)=
−
7
/x
we have
y
2
=
x
4
6
e
−
5
−
(7
/x
)
dx
x
8
dx
=
x
4
6
1
x
dx
=
x
4
ln
|
x
|
.
A second solution is
y
2
=
x
4
ln
|
x
|
.
10.
Identifying
P
(
x
)=2
/x
we have
y
2
=
x
2
6
e
−
5
(2
/x
)
dx
x
4
dx
=
x
2
6
x
−
6
dx
=
−
1
5
x
−
3
.
A second solution is
y
2
=
x
−
3
.
11.
Identifying
P
(
x
)=1
/x
we have
y
2
=ln
x
6
e
−
5
dx/x
(ln
x
)
2
dx
=ln
x
6
dx
x
(ln
x
)
2
=ln
x

−
1
ln
x

=
−
1
.
82

83. Exercises 3.2
A second solution is
y
2
=1.
12.
Identifying
P
(
x
)=0wehave
y
2
=
x
1
/
2
ln
x
6
e
−
5
0
dx
x
(ln
x
)
2
=
x
1
/
2
ln
x

−
1
ln
x

=
−
x
1
/
2
.
A second solution is
y
2
=
x
1
/
2
.
13.
Identifying
P
(
x
)=
−
1
/x
we have
y
2
=
x
sin(ln
x
)
6
e
−
5
−
dx/x
x
2
sin
2
(ln
x
)
dx
=
x
sin(ln
x
)
6
x
x
2
sin
2
(ln
x
)
dx
=[
x
sin(ln
x
)] [
−
cot(ln
x
)] =
−
x
cos(ln
x
)
.
A second solution is
y
2
=
x
cos(ln
x
).
14.
Identifying
P
(
x
)=
−
3
/x
we have
y
2
=
x
2
cos(ln
x
)
6
e
−
5
−
3
dx/x
x
4
cos
2
(ln
x
)
dx
=
x
2
cos(ln
x
)
6
x
3
x
4
cos
2
(ln
x
)
dx
=
x
2
cos(ln
x
) tan(ln
x
)=
x
2
sin(ln
x
)
.
A second solution is
y
2
=
x
2
sin(ln
x
).
15.
Identifying
P
(
x
) = 2(1 +
x
)
/
1
1
−
2
x
−
x
2
.
we have
y
2
=(
x
+1)
6
e
−
5
2(1+
x
)
dx/
(
1
−
2
x
−
x
2
)
(
x
+1)
2
dx
=(
x
+1)
6
e
ln
(
1
−
2
x
−
x
2
)
(
x
+1)
2
dx
=(
x
+1)
6
1
−
2
x
−
x
2
(
x
+1)
2
dx
=(
x
+1)
6

2
(
x
+1)
2
−
1

dx
=(
x
+1)

−
2
x
+1
−
x

=
−
2
−
x
2
−
x.
A second solution is
y
2
=
x
2
+
x
+2.
16.
Identifying
P
(
x
)=
−
2
x/
1
1
−
x
2
.
we have
y
2
=
6
e
−
5
−
2
x dx/
(
1
−
x
2
)
dx
=
6
e
−
ln
(
1
−
x
2
)
dx
=
6
1
1
−
x
2
dx
=
1
2
ln
2
2
2
2
1+
x
1
−
x
2
2
2
2
.
A second solution is
y
2
=ln
|
(1 +
x
)
/
(1
−
x
)
|
.
17.
Define
y
=
u
(
x
)
e
−
2
x
so
y
1
=
−
2
ue
−
2
x
+
u
1
e
−
2
x
,y
11
=
u
11
e
−
2
x
−
4
u
1
e
−
2
x
+4
ue
−
2
x
and
y
11
−
4
y
=
e
−
2
x
u
11
−
4
e
−
2
x
u
1
=0 or
u
11
−
4
u
1
=0
.
If
w
=
u
1
we obtain the first order equation
w
1
−
4
w
= 0 which has the integrating factor
e
−
4
5
dx
=
e
−
4
x
.Now
d
dx
[
e
−
4
x
w
] = 0 gives
e
−
4
x
w
=
c.
Therefore
w
=
u
1
=
ce
4
x
and
u
=
c
1
e
4
x
. A second solution is
y
2
=
e
−
2
x
e
4
x
=
e
2
x
. We see by observation that a
particular solution is
y
p
=
−
1
/
2. The general solution is
y
=
c
1
e
−
2
x
+
c
2
e
2
x
−
1
2
.
83

84. Exercises 3.2
18.
Define
y
=
u
(
x
)
·
1so
y
1
=
u
1
,y
11
=
u
11
and
y
11
+
y
1
=
u
11
+
u
1
=0
.
If
w
=
u
1
we obtain the first order equation
w
1
+
w
= 0 which has the integrating factor
e
5
dx
=
e
x
.Now
d
dx
[
e
x
w
] = 0 gives
e
x
w
=
c.
Therefore
w
=
u
1
=
ce
−
x
and
u
=
c
1
e
−
x
. A second solution is
y
2
=1
·
e
−
x
=
e
−
x
. We see by observation that
a particular solution is
y
p
=
x
. The general solution is
y
=
c
1
+
c
2
e
−
x
+
x.
19.
Define
y
=
u
(
x
)
e
x
so
y
1
=
ue
x
+
u
1
e
x
,y
11
=
u
11
e
x
+2
u
1
e
x
+
ue
x
and
y
11
−
3
y
1
+2
y
=
e
x
u
11
−
e
x
u
1
=0 or
u
11
−
u
1
=0
.
If
w
=
u
1
we obtain the first order equation
w
1
−
w
= 0 which has the integrating factor
e
−
5
dx
=
e
−
x
.Now
d
dx
[
e
−
x
w
] = 0 gives
e
−
x
w
=
c.
Therefore
w
=
u
1
=
ce
x
and
u
=
ce
x
. A second solution is
y
2
=
e
x
e
x
=
e
2
x
. To find a particular solution we
try
y
p
=
Ae
3
x
. Then
y
1
=3
Ae
3
x
,
y
11
=9
Ae
3
x
, and 9
Ae
3
x
−
3
1
3
Ae
3
x
.
+2
Ae
3
x
=5
e
3
x
.Thus
A
=5
/
2 and
y
p
=
5
2
e
3
x
. The general solution is
y
=
c
1
e
x
+
c
2
e
2
x
+
5
2
e
3
x
.
20.
Define
y
=
u
(
x
)
e
x
so
y
1
=
ue
x
+
u
1
e
x
,y
11
=
u
11
e
x
+2
u
1
e
x
+
ue
x
and
y
11
−
4
y
1
+3
y
=
e
x
u
11
−
2
e
x
u
1
=0 or
u
11
−
2
u
1
=0
.
If
w
=
u
1
we obtain the first order equation
w
1
−
2
w
= 0 which has the integrating factor
e
−
2
5
dx
=
e
−
2
x
.Now
d
dx
[
e
−
2
x
w
] = 0 gives
e
−
2
x
w
=
c.
Therefore
w
=
u
1
=
ce
2
x
and
u
=
c
1
e
2
x
. A second solution is
y
2
=
e
x
e
2
x
=
e
3
x
. To find a particular solution
we try
y
p
=
ax
+
b
. Then
y
1
p
=
a
,
y
11
p
= 0, and 0
−
4
a
+3(
ax
+
b
)=3
ax
−
4
a
+3
b
=
x
. Then 3
a
= 1 and
−
4
a
+3
b
=0so
a
=1
/
3 and
b
=4
/
9. A particular solution is
y
p
=
1
3
x
+
4
9
and the general solution is
y
=
c
1
e
x
+
c
2
e
3
x
+
1
3
x
+
4
9
.
21. (a)
For
m
1
constant, let
y
1
=
e
m
1
x
. Then
y
1
1
=
m
1
e
m
1
x
and
y
11
1
=
m
2
1
e
m
1
x
. Substituting into the differential
equation we obtain
ay
11
1
+
by
1
1
+
cy
1
=
am
2
1
e
m
1
x
+
bm
1
e
m
1
x
+
ce
m
1
x
=
e
m
1
x
(
am
2
1
+
bm
1
+
c
)=0
.
Thus,
y
1
=
e
m
1
x
will be a solution of the differential equation whenever
am
2
1
+
bm
1
+
c
= 0. Since a quadratic
equation always has at least one real or complex root, the differential equation must have a solution of the
form
y
1
=
e
m
1
x
.
(b)
Write the differential equation in the form
y
11
+
b
a
y
1
+
c
a
y
=0
,
84

85. Exercises 3.2
and let
y
1
=
e
m
1
x
be a solution. Then a second solution is given by
y
2
=
e
m
1
x
6
e
−
bx/a
e
2
m
1
x
dx
=
e
m
1
x
6
e
−
(
b/a
+2
m
1
)
x
dx
=
−
1
b/a
+2
m
1
e
m
1
x
e
−
(
b/a
+2
m
1
)
(
m
1
3
=
−
b/
2
a
)
=
−
1
b/a
+2
m
1
e
−
(
b/a
+
m
1
)
.
Thus, when
m
1
3
=
−
b/
2
a
, a second solution is given by
y
2
=
e
m
2
x
where
m
2
=
−
b/a
−
m
1
. When
m
1
=
−
b/
2
a
a second solution is given by
y
2
=
e
m
1
x
6
dx
=
xe
m
1
x
.
(c)
The functions
sin
x
=
1
2
i
(
e
ix
−
e
−
ix
)
sinh
x
=
1
2
(
e
x
−
e
−
x
)
cos
x
=
1
2
(
e
ix
+
e
−
ix
)
cosh
x
=
1
2
(
e
x
+
e
−
x
)
are all expressible in terms of exponential functions.
22.
We have
y
1
1
= 1 and
y
11
1
=0,so
xy
11
1
−
xy
1
1
+
y
1
=0
−
x
+
x
= 0 and
y
1
(
x
)=
x
is a solution of the differential
equation. Letting
y
=
u
(
x
)
y
1
(
x
)=
xu
(
x
) we get
y
1
=
xu
1
(
x
)+
u
(
x
) and
y
11
=
xu
11
(
x
)+2
u
1
(
x
)
.
Then
xy
11
−
xy
1
+
y
=
x
2
u
11
+2
xu
1
−
x
2
u
1
−
xu
+
xu
=
x
2
u
11
−
(
x
2
−
2
x
)
u
1
= 0. If we make the substitution
w
=
u
1
, the second-order linear differential equation becomes
x
2
w
1
−
(
x
2
−
x
)
w
= 0, which is separable:
dw
dx
=
3
1
−
1
x
4
w
dw
w
=
3
1
−
1
x
4
dx
ln
w
=
x
−
ln
x
+
c
w
=
c
1
e
x
x
.
Then
u
1
=
c
1
e
x
/x
and
u
=
c
1
5
e
x
dx/x
. To integrate
e
x
/x
we use the series representation for
e
x
. Thus, a
second solution is
y
2
=
xu
(
x
)=
c
1
x
6
e
x
x
dx
=
c
1
x
6
1
x

1+
x
+
1
2!
x
2
+
1
3!
x
3
+
···

dx
=
c
1
x
6

1
x
+1+
1
2!
x
+
1
3!
x
2
+
···

dx
=
c
1
x

ln
x
+
x
+
1
2(2!)
x
2
+
1
3(3!)
x
3
+
···

=
c
1

x
ln
x
+
x
2
+
1
2(2!)
x
3
+
1
3(3!)
x
4
+
···

.
An interval of definition is probably (0
,
∞
) because of the ln
x
term.
85

86. Exercises 3.3
Exercises 3.3
1.
From 4
m
2
+
m
= 0 we obtain
m
= 0 and
m
=
−
1
/
4 so that
y
=
c
1
+
c
2
e
−
x/
4
.
2.
From
m
2
−
36 = 0 we obtain
m
= 6 and
m
=
−
6 so that
y
=
c
1
e
6
x
+
c
2
e
−
6
x
.
3.
From
m
2
−
m
−
6 = 0 we obtain
m
= 3 and
m
=
−
2 so that
y
=
c
1
e
3
x
+
c
2
e
−
2
x
.
4.
From
m
2
−
3
m
+ 2 = 0 we obtain
m
= 1 and
m
= 2 so that
y
=
c
1
e
x
+
c
2
e
2
x
.
5.
From
m
2
+8
m
+ 16 = 0 we obtain
m
=
−
4 and
m
=
−
4 so that
y
=
c
1
e
−
4
x
+
c
2
xe
−
4
x
.
6.
From
m
2
−
10
m
+ 25 = 0 we obtain
m
= 5 and
m
= 5 so that
y
=
c
1
e
5
x
+
c
2
xe
5
x
.
7.
From 12
m
2
−
5
m
−
2 = 0 we obtain
m
=
−
1
/
4 and
m
=2
/
3 so that
y
=
c
1
e
−
x/
4
+
c
2
e
2
x/
3
.
8.
From
m
2
+4
m
−
1 = 0 we obtain
m
=
−
2
±
√
5 so that
y
=
c
1
e
(
−
2+
√
5
)
x
+
c
2
e
(
−
2
−
√
5
)
x
.
9.
From
m
2
+ 9 = 0 we obtain
m
=3
i
and
m
=
−
3
i
so that
y
=
c
1
cos 3
x
+
c
2
sin 3
x
.
10.
From 3
m
2
+ 1 = 0 we obtain
m
=
i/
√
3 and
m
=
−
i/
√
3 so that
y
=
c
1
cos
x/
√
3+
c
2
sin
x/
√
3.
11.
From
m
2
−
4
m
+ 5 = 0 we obtain
m
=2
±
i
so that
y
=
e
2
x
(
c
1
cos
x
+
c
2
sin
x
).
12.
From 2
m
2
+2
m
+ 1 = 0 we obtain
m
=
−
1
/
2
±
i/
2 so that
y
=
e
−
x/
2
(
c
1
cos
x/
2+
c
2
sin
x/
2)
.
13.
From 3
m
2
+2
m
+ 1 = 0 we obtain
m
=
−
1
/
3
±
√
2
i/
3 so that
y
=
e
−
x/
3
3
c
1
cos
√
2
x/
3+
c
2
sin
√
2
x/
3
4
.
14.
From 2
m
2
−
3
m
+ 4 = 0 we obtain
m
=3
/
4
±
√
23
i/
4 so that
y
=
e
3
x/
4
3
c
1
cos
√
23
x/
4+
c
2
sin
√
23
x/
4
4
.
15.
From
m
3
−
4
m
2
−
5
m
= 0 we obtain
m
=0,
m
= 5, and
m
=
−
1 so that
y
=
c
1
+
c
2
e
5
x
+
c
3
e
−
x
.
16.
From
m
3
−
1 = 0 we obtain
m
= 1 and
m
=
−
1
/
2
±
√
3
i/
2 so that
y
=
c
1
e
x
+
e
−
x/
2
3
c
2
cos
√
3
x/
2+
c
3
sin
√
3
x/
2
4
.
17.
From
m
3
−
5
m
2
+3
m
+ 9 = 0 we obtain
m
=
−
1,
m
= 3, and
m
= 3 so that
y
=
c
1
e
−
x
+
c
2
e
3
x
+
c
3
xe
3
x
.
18.
From
m
3
+3
m
2
−
4
m
−
12 = 0 we obtain
m
=
−
2,
m
= 2, and
m
=
−
3 so that
y
=
c
1
e
−
2
x
+
c
2
e
2
x
+
c
3
e
−
3
x
.
19.
From
m
3
+
m
2
−
2 = 0 we obtain
m
= 1 and
m
=
−
1
±
i
so that
u
=
c
1
e
t
+
e
−
t
(
c
2
cos
t
+
c
3
sin
t
)
.
20.
From
m
3
−
m
2
−
4 = 0 we obtain
m
= 2 and
m
=
−
1
/
2
±
√
7
i/
2 so that
x
=
c
1
e
2
t
+
e
−
t/
2
3
c
2
cos
√
7
t/
2+
c
3
sin
√
7
t/
2
4
.
86

87. Exercises 3.3
21.
From
m
3
+3
m
2
+3
m
+ 1 = 0 we obtain
m
=
−
1,
m
=
−
1, and
m
=
−
1 so that
y
=
c
1
e
−
x
+
c
2
xe
−
x
+
c
3
x
2
e
−
x
.
22.
From
m
3
−
6
m
2
+12
m
−
8 = 0 we obtain
m
=2,
m
= 2, and
m
= 2 so that
y
=
c
1
e
2
x
+
c
2
xe
2
x
+
c
3
x
2
e
2
x
.
23.
From
m
4
+
m
3
+
m
2
= 0 we obtain
m
=0,
m
= 0, and
m
=
−
1
/
2
±
√
3
i/
2 so that
y
=
c
1
+
c
2
x
+
e
−
x/
2
3
c
3
cos
√
3
x/
2+
c
4
sin
√
3
x/
2
4
.
24.
From
m
4
−
2
m
2
+ 1 = 0 we obtain
m
=1,
m
=1,
m
=
−
1, and
m
=
−
1 so that
y
=
c
1
e
x
+
c
2
xe
x
+
c
3
e
−
x
+
c
4
xe
−
x
.
25.
From 16
m
4
+24
m
2
+ 9 = 0 we obtain
m
=
±
√
3
i/
2 and
m
=
±
√
3
i/
2 so that
y
=
c
1
cos
√
3
x/
2+
c
2
sin
√
3
x/
2+
c
3
x
cos
√
3
x/
2+
c
4
x
sin
√
3
x/
2
.
26.
From
m
4
−
7
m
2
−
18 = 0 we obtain
m
=3,
m
=
−
3, and
m
=
±
√
2
i
so that
y
=
c
1
e
3
x
+
c
2
e
−
3
x
+
c
3
cos
√
2
x
+
c
4
sin
√
2
x.
27.
From
m
5
+5
m
4
−
2
m
3
−
10
m
2
+
m
+ 5 = 0 we obtain
m
=
−
1,
m
=
−
1,
m
= 1, and
m
= 1, and
m
=
−
5so
that
u
=
c
1
e
−
r
+
c
2
re
−
r
+
c
3
e
r
+
c
4
re
r
+
c
5
e
−
5
r
.
28.
From 2
m
5
−
7
m
4
+12
m
3
+8
m
2
= 0 we obtain
m
=0,
m
=0,
m
=
−
1
/
2, and
m
=2
±
2
i
so that
x
=
c
1
+
c
2
s
+
c
3
e
−
s/
2
+
e
2
s
(
c
4
cos 2
s
+
c
5
sin 2
s
)
.
29.
From
m
2
+ 16 = 0 we obtain
m
=
±
4
i
so that
y
=
c
1
cos 4
x
+
c
2
sin 4
x
.If
y
(0) = 2 and
y
1
(0) =
−
2 then
c
1
=2,
c
2
=
−
1
/
2, and
y
= 2 cos 4
x
−
1
2
sin 4
x
.
30.
From
m
2
+ 1 = 0 we obtain
m
=
±
i
so that
y
=
c
1
cos
θ
+
c
2
sin
θ
.If
y
(
π/
3) = 0 and
y
1
(
π/
3) = 2 then
1
2
c
1
+
√
3
2
c
2
=0,
−
√
3
2
c
1
+
1
2
c
2
=2,so
c
1
=
−
√
3,
c
2
= 1, and
y
=
−
√
3 cos
θ
+ sin
θ
.
31.
From
m
2
−
4
m
−
5 = 0 we obtain
m
=
−
1 and
m
= 5, so that
y
=
c
1
e
−
x
+
c
2
e
5
x
.If
y
(1) = 0 and
y
1
(1)=2,
then
c
1
e
−
1
+
c
2
e
5
=0,
−
c
1
e
−
1
+5
c
2
e
5
=2,so
c
1
=
−
e/
3,
c
2
=
e
−
5
/
3, and
y
=
−
1
3
e
1
−
x
+
1
3
e
5
x
−
5
.
32.
From 4
m
2
−
4
m
−
3 = 0 we obtain
m
=
−
1
/
2 and
m
=3
/
2 so that
y
=
c
1
e
−
x/
2
+
c
2
e
3
x/
2
.If
y
(0) = 1 and
y
1
(0) = 5 then
c
1
+
c
2
=1,
−
1
2
c
1
+
3
2
c
2
=5,so
c
1
=
−
7
/
4,
c
2
=11
/
4, and
y
=
−
7
4
e
−
x/
2
+
11
4
e
3
x/
2
.
33.
From
m
2
+
m
+ 2 = 0 we obtain
m
=
−
1
/
2
±
√
7
i/
2 so that
y
=
e
−
x/
2
1
c
1
cos
√
7
x/
2+
c
2
sin
√
7
x/
2
.
.If
y
(0) = 0 and
y
1
(0) = 0 then
c
1
= 0 and
c
2
= 0 so that
y
=0.
34.
From
m
2
−
2
m
+ 1 = 0 we obtain
m
= 1 and
m
= 1 so that
y
=
c
1
e
x
+
c
2
xe
x
.If
y
(0) = 5 and
y
1
(0) = 10 then
c
1
=5,
c
1
+
c
2
=10so
c
1
=5,
c
2
= 5, and
y
=5
e
x
+5
xe
x
.
35.
From
m
3
+12
m
2
+36
m
= 0 we obtain
m
=0,
m
=
−
6, and
m
=
−
6 so that
y
=
c
1
+
c
2
e
−
6
x
+
c
3
xe
−
6
x
.If
y
(0) = 0,
y
1
(0) = 1, and
y
11
(0) =
−
7 then
c
1
+
c
2
=0
,
−
6
c
2
+
c
3
=1
,
36
c
2
−
12
c
3
=
−
7
,
so
c
1
=5
/
36,
c
2
=
−
5
/
36,
c
3
=1
/
6, and
y
=
5
36
−
5
36
e
−
6
x
+
1
6
xe
−
6
x
.
87

88. Exercises 3.3
36.
From
m
3
+2
m
2
−
5
m
−
6 = 0 we obtain
m
=
−
1,
m
= 2, and
m
=
−
3 so that
y
=
c
1
e
−
x
+
c
2
e
2
x
+
c
3
e
−
3
x
.
If
y
(0) = 0,
y
1
(0) = 0, and
y
11
(0) = 1 then
c
1
+
c
2
+
c
3
=0
,
−
c
1
+2
c
2
−
3
c
3
=0
,c
1
+4
c
2
+9
c
3
=1
,
so
c
1
=
−
1
/
6,
c
2
=1
/
15,
c
3
=1
/
10, and
y
=
−
1
6
e
−
x
+
1
15
e
2
x
+
1
10
e
−
3
x
.
37.
From
m
4
−
3
m
3
+3
m
2
−
m
= 0 we obtain
m
=0,
m
=1,
m
= 1, and
m
= 1 so that
y
=
c
1
+
c
2
e
x
+
c
3
xe
x
+
c
4
x
2
e
x
.If
y
(0) = 0,
y
1
(0) = 0,
y
11
(0) = 1, and
y
111
(0) = 1 then
c
1
+
c
2
=0
,c
2
+
c
3
=0
,c
2
+2
c
3
+2
c
4
=1
,c
2
+3
c
3
+6
c
4
=1
,
so
c
1
=2,
c
2
=
−
2,
c
3
=2,
c
4
=
−
1
/
2, and
y
=2
−
2
e
x
+2
xe
x
−
1
2
x
2
e
x
.
38.
From
m
4
−
1 = 0 we obtain
m
=1,
m
=
−
1, and
m
=
±
i
so that
y
=
c
1
e
x
+
c
2
e
−
x
+
c
3
cos
x
+
c
4
sin
x
.If
y
(0) = 0,
y
1
(0) = 0,
y
11
(0) = 0, and
y
111
(0) = 1 then
c
1
+
c
2
+
c
3
=0
,c
1
−
c
2
+
c
4
=0
,c
1
+
c
2
−
c
3
=0
,c
1
−
c
2
−
c
4
=1
,
so
c
1
=1
/
4,
c
2
=
−
1
/
4,
c
3
=0,
c
4
=
−
1
/
2, and
y
=
1
4
e
x
−
1
4
e
−
x
−
1
2
sin
x.
39.
From
m
2
−
10
m
+ 25 = 0 we obtain
m
= 5 and
m
= 5 so that
y
=
c
1
e
5
x
+
c
2
xe
5
x
.If
y
(0) = 1 and
y
(1) = 0
then
c
1
=1,
c
1
e
5
+
c
2
e
5
=0,so
c
1
=1,
c
2
=
−
1, and
y
=
e
5
x
−
xe
5
x
.
40.
From
m
2
+ 4 = 0 we obtain
m
=
±
2
i
so that
y
=
c
1
cos 2
x
+
c
2
sin 2
x
.If
y
(0) = 0 and
y
(
π
) = 0 then
c
1
= 0 and
y
=
c
2
sin 2
x
.
41.
From
m
2
+ 1 = 0 we obtain
m
=
±
i
so that
y
=
c
1
cos
x
+
c
2
sin
x
.If
y
1
(0) = 0 and
y
1
(
π/
2) = 2 then
c
1
=
−
2,
c
2
= 0, and
y
=
−
2 cos
x
.
42.
From
m
2
−
2
m
+ 2 = 0 we obtain
m
=1
±
i
so that
y
=
e
x
(
c
1
cos
x
+
c
2
sin
x
). If
y
(0) = 1 and
y
(
π
) = 1 then
c
1
= 1 and
y
(
π
)=
e
π
cos
π
=
−
e
π
. Since
−
e
π
3
= 1, the boundary-value problem has no solution.
43.
The auxiliary equation is
m
2
−
3 = 0 which has roots
−
√
3 and
√
3 . By (10) the general solution is
y
=
c
1
e
√
3
x
+
c
2
e
−
√
3
x
. By (11) the general solution is
y
=
c
1
cosh
√
3
x
+
c
2
sinh
√
3
x
.For
y
=
c
1
e
√
3
x
+
c
2
e
−
√
3
x
the initial conditions imply
c
1
+
c
2
=1,
√
3
c
1
−
√
3
c
2
= 5. Solving for
c
1
+
c
2
we find
c
1
=
1
6
(3+5
√
3 ) and
c
2
=
1
6
(3
−
5
√
3)so
y
=
1
6
(3+5
√
3)
e
√
3
x
+
1
6
(3
−
5
√
3)
e
√
3
x
.For
y
=
c
1
cosh
√
3
x
+
c
2
sinh
√
3
x
the initial conditions
imply
c
1
=1,
√
3
c
2
= 5. Solving for
c
1
and
c
2
we find
c
1
= 1 and
c
2
=
5
3
√
3so
y
= cosh
√
3
x
+
5
3
√
3 sinh
√
3
x
.
44.
The auxiliary equation is
m
2
−
1 = 0 which has roots
−
1 and 1. By (10) the general solution is
y
=
c
1
e
x
+
c
2
e
−
x
.
By (11) the general solution is
y
=
c
1
cosh
x
+
c
2
sinh
x
.For
y
=
c
1
e
x
+
c
2
e
−
x
the boundary conditions
imply
c
1
+
c
2
=1,
c
1
e
−
c
2
e
−
1
= 0. Solving for
c
1
and
c
2
we find
c
1
=1
/
(1 +
e
2
) and
c
2
=
e
2
/
(1 +
e
2
)
so
y
=
e
x
/
(1 +
e
2
)+
e
2
e
−
x
/
(1 +
e
2
). For
y
=
c
1
cosh
x
+
c
2
sinh
x
the boundary conditions imply
c
1
=1,
c
2
=
−
tanh 1, so
y
= cosh
x
−
(tanh 1) sinh
x
.
45.
Using a CAS to solve the auxiliary equation
m
3
−
6
m
2
+2
m
+ 1 we find
m
1
=
−
0
.
270534,
88

89. Exercises 3.3
m
2
=0
.
658675, and
m
3
=5
.
61186. The general solution is
y
=
c
1
e
−
0
.
270534
x
+
c
2
e
0
.
658675
x
+
c
3
e
5
.
61186
x
.
46.
Using a CAS to solve the auxiliary equation 6
.
11
m
3
+8
.
59
m
2
+7
.
93
m
+0
.
778 = 0 we find
m
1
=
−
0
.
110241,
m
2
=
−
0
.
647826 + 0
.
857532
i
, and
m
3
=
−
0
.
647826
−
0
.
857532
i
. The general solution is
y
=
c
1
e
−
0
.
110241
x
+
e
−
0
.
647826
x
(
c
2
cos 0
.
857532
x
+
c
3
sin 0
.
857532
x
)
.
47.
Using a CAS to solve the auxiliary equation 3
.
15
m
4
−
5
.
34
m
2
+6
.
33
m
−
2
.
03 = 0 we find
m
1
=
−
1
.
74806,
m
2
=0
.
501219,
m
3
=0
.
62342+0
.
588965
i
, and
m
4
=0
.
62342
−
0
.
588965
i
. The general solution
is
y
=
c
1
e
−
1
.
74806
x
+
c
2
e
0
.
501219
x
+
e
0
.
62342
x
(
c
3
cos 0
.
588965
x
+
c
4
sin 0
.
588965
x
)
.
48.
Using a CAS to solve the auxiliary equation
m
4
+2
m
2
−
m
+2 = 0 we find
m
1
=1
/
2+
√
3
i/
2,
m
2
=1
/
2
−
√
3
i/
2,
m
3
=
−
1
/
2+
√
7
i/
2, and
m
4
=
−
1
/
2
−
√
7
i/
2. The general solution is
y
=
e
x/
2

c
1
cos
√
3
2
x
+
c
2
sin
√
3
2
x
a
+
e
−
x/
2

c
3
cos
√
7
2
x
+
c
4
sin
√
7
2
x
a
.
49.
The auxiliary equation should have two positive roots, so that the solution has the form
y
=
c
1
e
k
1
x
+
c
2
e
k
2
x
.
Thus, the differential equation is (f).
50.
The auxiliary equation should have one positive and one negative root, so that the solution has the form
y
=
c
1
e
k
1
x
+
c
2
e
−
k
2
x
. Thus, the differential equation is (a).
51.
The auxiliary equation should have a pair of complex roots
a
±
bi
where
a<
0, so that the solution has the
form
e
ax
(
c
1
cos
bx
+
c
2
sin
bx
). Thus, the differential equation is (e).
52.
The auxiliary equation should have a repeated negative root, so that the solution has the form
y
=
c
1
e
−
x
+
c
2
xe
−
x
. Thus, the differential equation is (c).
53.
The differential equation should have the form
y
11
+
k
2
y
= 0 where
k
= 1 so that the period of the solution is
2
π
. Thus, the differential equation is (d).
54.
The differential equation should have the form
y
11
+
k
2
y
= 0 where
k
= 2 so that the period of the solution is
π
. Thus, the differential equation is (b).
55. (a)
The auxiliary equation is
m
2
−
64
/L
= 0 which has roots
±
8
/
√
L
. Thus, the general solution of the
differential equation is
x
=
c
1
cosh(8
t/
√
L
)+
c
2
sinh(8
t/
√
L
).
(b)
Setting
x
(0) =
x
0
and
x
1
(0)=0wehave
c
1
=
x
0
,8
c
2
/
√
L
= 0. Solving for
c
1
and
c
2
we get
c
1
=
x
0
and
c
2
=0,so
x
(
t
)=
x
0
cosh(8
t/
√
L
).
(c)
When
L
= 20 and
x
0
=1,
x
(
t
) = cosh(4
t
√
5 ). The chain will last touch the peg when
x
(
t
) = 10.
Solving
x
(
t
)=10for
t
we get
t
1
=
1
4
√
5 cosh
−
1
10
≈
1
.
67326. The velocity of the chain at this instant is
x
1
(
t
1
)=12
)
11
/
5
≈
17
.
7989 ft/s.
56.
Both
−
C
[1] and
c
1
represent arbitrary constants, and each may take on any real value.
57.
Since (
m
−
4)(
m
+5)
2
=
m
3
+6
m
2
−
15
m
−
100 the differential equation is
y
111
+6
y
11
−
15
y
1
−
100
y
= 0. The
differential equation is not unique since any constant multiple of the left-hand side of the differential equation
would lead to the auxiliary roots.
58.
A third root must be
m
3
=3
−
i
and the auxiliary equation is
3
m
+
1
2
4
[
m
−
(3 +
i
)][
m
−
(3
−
i
)] =
3
m
+
1
2
4
(
m
2
−
6
x
+ 10) =
m
3
−
11
2
m
2
+7
m
+5
.
89

90. Exercises 3.3
The differential equation is
y
111
−
11
2
y
11
+7
y
1
+5
y
=0
.
59.
From the solution
y
1
=
e
−
4
x
cos
x
we conclude that
m
1
=
−
4+
i
and
m
2
=
−
4
−
i
are roots of the auxiliary
equation. Hence another solution must be
y
2
=
e
−
4
x
sin
x
. Now dividing the polynomial
m
3
+6
m
2
+
m
−
34 by
b
m
−
(
−
4+
i
)
cb
m
−
(
−
4
−
i
)
c
=
m
2
+8
m
+ 17 gives
m
−
2. Therefore
m
3
= 2 is the third root of the auxiliary
equation, and the general solution of the differential equation is
y
=
c
1
e
−
4
x
cos
x
+
c
2
e
−
4
x
sin
x
+
c
3
e
2
x
.
60.
Factoring the difference of two squares we obtain
m
4
+1=(
m
2
+1)
2
−
2
m
2
=(
m
2
+1
−
√
2
m
)(
m
2
+1+
√
2
m
)=0
.
Using the quadratic formula on each factor we get
m
=
±
√
2
/
2
±
√
2
i/
2. The solution of the differential equation
is
y
(
x
)=
e
√
2
x/
2

c
1
cos
√
2
2
x
+
c
2
sin
√
2
2
x
a
+
e
−
√
2
x/
2

c
3
cos
√
2
2
x
+
c
4
sin
√
2
2
x
a
.
61.
The auxiliary equation is
m
2
+
λ
= 0 and we consider three cases.
Case I
When
λ
= 0 the general solution of the differential equation is
y
=
c
1
+
c
2
x
. The boundary conditions
imply 0 =
y
(0) =
c
1
and 0 =
y
(
π/
2) =
c
2
π/
2, so that
c
1
=
c
2
= 0 and the problem possesses only the trivial
solution.
Case II
When
λ<
0 the general solution of the differential equation is
y
=
c
1
e
√
−
λx
+
c
2
e
−
√
−
λx
, or alterna-
tively,
y
=
c
1
cosh
√
−
λx
+
c
2
sinh
√
−
λx
. Again,
y
(0) = 0 implies
c
1
=0so
y
=
c
2
sinh
√
−
λx
. The second
boundary condition implies 0 =
y
(
π/
2) =
c
2
sinh
√
−
λπ/
2or
c
2
= 0. In this case also, the problem possesses
only the trivial solution.
Case III
When
λ>
0 the general solution of the differential equation is
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx
.
In this case also,
y
(0) = 0 yields
c
1
= 0, so that
y
=
c
2
sin
√
λx
. The second boundary condition implies
0=
c
2
sin
√
λπ/
2. When
√
λπ/
2 is an integer multiple of
π
, that is, when
√
λ
=2
k
for
k
a nonzero integer,
the problem will have nontrivial solutions. Thus, for
λ
=4
k
2
the boundary-value problem will have nontrivial
solutions
y
=
c
2
sin 2
kx
, where
k
is a nonzero integer. On the other hand, when
√
λ
is not an even integer, the
boundary-value problem will have only the trivial solution.
62.
Applying integration by parts twice we have
6
e
ax
f
(
x
)
dx
=
1
a
e
ax
f
(
x
)
−
1
a
6
e
ax
f
1
(
x
)
dx
=
1
a
e
ax
f
(
x
)
−
1
a

1
a
e
ax
f
1
(
x
)
−
1
a
6
e
ax
f
11
(
x
)
dx

=
1
a
e
ax
f
(
x
)
−
1
a
2
e
ax
f
1
(
x
)+
1
a
2
6
e
ax
f
11
(
x
)
dx.
Collecting the integrals we get
6
e
ax

f
(
x
)
−
1
a
2
f
11
(
x
)

dx
=
1
a
e
ax
f
(
x
)
−
1
a
2
e
ax
f
1
(
x
)
.
In order for the technique to work we need to have
6
e
ax

f
(
x
)
−
1
a
2
f
11
(
x
)

dx
=
k
6
e
ax
f
(
x
)
dx
90

91. Exercises 3.4
or
f
(
x
)
−
1
a
2
f
11
(
x
)=
kf
(
x
)
,
where
k
3
= 0. This is the second-order differential equation
f
11
(
x
)+
a
2
(
k
−
1)
f
(
x
)=0
.
If
k<
1,
k
3
= 0, the solution of the differential equation is a pair of exponential functions, in which case
the original integrand is an exponential function and does not require integration by parts for its evaluation.
Similarly, if
k
=1,
f
11
(
x
) = 0 and
f
(
x
) has the form
f
(
x
)=
ax
+
b
. In this case a single application of integration
by parts will suffice. Finally, if
k>
1, the solution of the differential equation is
f
(
x
)=
c
1
cos
a
√
k
−
1
x
+
c
2
sin
a
√
k
−
1
x,
and we see that the technique will work for linear combinations of cos
αx
and sin
αx
.
Exercises 3.4
1.
From
m
2
+3
m
+ 2 = 0 we find
m
1
=
−
1 and
m
2
=
−
2. Then
y
c
=
c
1
e
−
x
+
c
2
e
−
2
x
and we assume
y
p
=
A
.
Substituting into the differential equation we obtain 2
A
= 6. Then
A
=3,
y
p
= 3 and
y
=
c
1
e
−
x
+
c
2
e
−
2
x
+3
.
2.
From 4
m
2
+9=0wefind
m
1
=
−
3
2
i
and
m
2
=
3
2
i
. Then
y
c
=
c
1
cos
3
2
x
+
c
2
sin
3
2
x
and we assume
y
p
=
A
.
Substituting into the differential equation we obtain 9
A
= 15. Then
A
=
5
3
,
y
p
=
5
3
and
y
=
c
1
cos
3
2
x
+
c
2
sin
3
2
x
+
5
3
.
3.
From
m
2
−
10
m
+25=0wefind
m
1
=
m
2
= 5. Then
y
c
=
c
1
e
5
x
+
c
2
xe
5
x
and we assume
y
p
=
Ax
+
B
.
Substituting into the differential equation we obtain 25
A
= 30 and
−
10
A
+25
B
= 3. Then
A
=
6
5
,
B
=
6
5
,
y
p
=
6
5
x
+
6
5
, and
y
=
c
1
e
5
x
+
c
2
xe
5
x
+
6
5
x
+
6
5
.
4.
From
m
2
+
m
−
6=0wefind
m
1
=
−
3 and
m
2
= 2. Then
y
c
=
c
1
e
−
3
x
+
c
2
e
2
x
and we assume
y
p
=
Ax
+
B
.
Substituting into the differential equation we obtain
−
6
A
= 2 and
A
−
6
B
= 0. Then
A
=
−
1
3
,
B
=
−
1
18
,
y
p
=
−
1
3
x
−
1
18
, and
y
=
c
1
e
−
3
x
+
c
2
e
2
x
−
1
3
x
−
1
18
.
5.
From
1
4
m
2
+
m
+ 1 = 0 we find
m
1
=
m
2
= 0. Then
y
c
=
c
1
e
−
2
x
+
c
2
xe
−
2
x
and we assume
y
p
=
Ax
2
+
Bx
+
C
.
Substituting into the differential equation we obtain
A
=1,2
A
+
B
=
−
2, and
1
2
A
+
B
+
C
= 0. Then
A
=1,
B
=
−
4,
C
=
7
2
,
y
p
=
x
2
−
4
x
+
7
2
, and
y
=
c
1
e
−
2
x
+
c
2
xe
−
2
x
+
x
2
−
4
x
+
7
2
.
91

92. Exercises 3.4
6.
From
m
2
−
8
m
+20=0wefind
m
1
=2+4
i
and
m
2
=2
−
4
i
. Then
y
c
=
e
2
x
(
c
1
cos 4
x
+
c
2
sin 4
x
) and we
assume
y
p
=
Ax
2
+
Bx
+
C
+(
Dx
+
E
)
e
x
. Substituting into the differential equation we obtain
2
A
−
8
B
+20
C
=0
−
6
D
+13
E
=0
−
16
A
+20
B
=0
13
D
=
−
26
20
A
= 100
.
Then
A
=5,
B
=4,
C
=
11
10
,
D
=
−
2,
E
=
−
12
13
,
y
p
=5
x
2
+4
x
+
11
10
+
1
−
2
x
−
12
13
.
e
x
and
y
=
e
2
x
(
c
1
cos 4
x
+
c
2
sin 4
x
)+5
x
2
+4
x
+
11
10
+

−
2
x
−
12
13

e
x
.
7.
From
m
2
+ 3 = 0 we find
m
1
=
√
3
i
and
m
2
=
−
√
3
i
. Then
y
c
=
c
1
cos
√
3
x
+
c
2
sin
√
3
x
and we assume
y
p
=(
Ax
2
+
Bx
+
C
)
e
3
x
. Substituting into the differential equation we obtain 2
A
+6
B
+12
C
=0,12
A
+12
B
=0,
and 12
A
=
−
48. Then
A
=
−
4,
B
=4,
C
=
−
4
3
,
y
p
=
1
−
4
x
2
+4
x
−
4
3
.
e
3
x
and
y
=
c
1
cos
√
3
x
+
c
2
sin
√
3
x
+

−
4
x
2
+4
x
−
4
3

e
3
x
.
8.
From 4
m
2
−
4
m
−
3=0wefind
m
1
=
3
2
and
m
2
=
−
1
2
. Then
y
c
=
c
1
e
3
x/
2
+
c
2
e
−
x/
2
and we assume
y
p
=
A
cos 2
x
+
B
sin 2
x
. Substituting into the differential equation we obtain
−
19
−
8
B
= 1 and 8
A
−
19
B
=0.
Then
A
=
−
19
425
,
B
=
−
8
425
,
y
p
=
−
19
425
cos 2
x
−
8
425
sin 2
x
, and
y
=
c
1
e
3
x/
2
+
c
2
e
−
x/
2
−
19
425
cos 2
x
−
8
425
sin 2
x.
9.
From
m
2
−
m
= 0 we find
m
1
= 1 and
m
2
= 0. Then
y
c
=
c
1
e
x
+
c
2
and we assume
y
p
=
Ax
. Substituting into
the differential equation we obtain
−
A
=
−
3. Then
A
=3,
y
p
=3
x
and
y
=
c
1
e
x
+
c
2
+3
x
.
10.
From
m
2
+2
m
= 0 we find
m
1
=
−
2 and
m
2
= 0. Then
y
c
=
c
1
e
−
2
x
+
c
2
and we assume
y
p
=
Ax
2
+
Bx
+
Cxe
−
2
x
.
Substituting into the differential equation we obtain 2
A
+2
B
=5,4
A
= 2, and
−
2
C
=
−
1. Then
A
=
1
2
,
B
=2,
C
=
1
2
,
y
p
=
1
2
x
2
+2
x
+
1
2
xe
−
2
x
, and
y
=
c
1
e
−
2
x
+
c
2
+
1
2
x
2
+2
x
+
1
2
xe
−
2
x
.
11.
From
m
2
−
m
+
1
4
= 0 we find
m
1
=
m
2
=
1
2
. Then
y
c
=
c
1
e
x/
2
+
c
2
xe
x/
2
and we assume
y
p
=
A
+
Bx
2
e
x/
2
.
Substituting into the differential equation we obtain
1
4
A
= 3 and 2
B
= 1. Then
A
= 12,
B
=
1
2
,
y
p
=
12 +
1
2
x
2
e
x/
2
, and
y
=
c
1
e
x/
2
+
c
2
xe
x/
2
+12+
1
2
x
2
e
x/
2
.
12.
From
m
2
−
16 = 0 we find
m
1
= 4 and
m
2
=
−
4. Then
y
c
=
c
1
e
4
x
+
c
2
e
−
4
x
and we assume
y
p
=
Axe
4
x
.
Substituting into the differential equation we obtain 8
A
= 2. Then
A
=
1
4
,
y
p
=
1
4
xe
4
x
and
y
=
c
1
e
4
x
+
c
2
e
−
4
x
+
1
4
xe
4
x
.
13.
From
m
2
+4=0wefind
m
1
=2
i
and
m
2
=
−
2
i
. Then
y
c
=
c
1
cos 2
x
+
c
2
sin 2
x
and we assume
y
p
=
Ax
cos 2
x
+
Bx
sin 2
x
. Substituting into the differential equation we obtain 4
B
= 0 and
−
4
A
= 3. Then
A
=
−
3
4
,
B
=0,
y
p
=
−
3
4
x
cos 2
x
, and
y
=
c
1
cos 2
x
+
c
2
sin 2
x
−
3
4
x
cos 2
x.
92

93. Exercises 3.4
14.
From
m
2
+ 4 = 0 we find
m
1
=2
i
and
m
2
=
−
2
i
. Then
y
c
=
c
1
cos 2
x
+
c
2
sin 2
x
and we assume
y
p
=(
Ax
3
+
Bx
2
+
Cx
) cos 2
x
+(
Dx
3
+
Ex
2
+
Fx
) sin 2
x
. Substituting into the differential equation we
obtain
2
B
+4
F
=0
6
A
+8
E
=0
12
D
=0
−
4
C
+2
E
=
−
3
−
8
B
+6
D
=0
−
12
A
=1
.
Then
A
=
−
1
12
,
B
=0,
C
=
25
32
,
D
=0,
E
=
1
16
,
F
=0,
y
p
=
1
−
1
12
x
3
+
25
32
x
.
cos 2
x
+
1
16
x
2
sin 2
x
, and
y
=
c
1
cos 2
x
+
c
2
sin 2
x
+

−
1
12
x
3
+
25
32
x

cos 2
x
+
1
16
x
2
sin 2
x.
15.
From
m
2
+ 1 = 0 we find
m
1
=
i
and
m
2
=
−
i
. Then
y
c
=
c
1
cos
x
+
c
2
sin
x
and we assume
y
p
=
(
Ax
2
+
Bx
) cos
x
+(
Cx
2
+
Dx
) sin
x
. Substituting into the differential equation we obtain 4
C
=0,2
A
+2
D
=0,
−
4
A
= 2, and
−
2
B
+2
C
= 0. Then
A
=
−
1
2
,
B
=0,
C
=0,
D
=
1
2
,
y
p
=
−
1
2
x
2
cos
x
+
1
2
x
sin
x
, and
y
=
c
1
cos
x
+
c
2
sin
x
−
1
2
x
2
cos
x
+
1
2
x
sin
x.
16.
From
m
2
−
5
m
= 0 we find
m
1
= 5 and
m
2
= 0. Then
y
c
=
c
1
e
5
x
+
c
2
and we assume
y
p
=
Ax
4
+
Bx
3
+
Cx
2
+
Dx
.
Substituting into the differential equation we obtain
−
20
A
=2,12
A
−
15
B
=
−
4, 6
B
−
10
C
=
−
1, and
2
C
−
5
D
= 6. Then
A
=
−
1
10
,
B
=
14
75
,
C
=
53
250
,
D
=
−
697
625
,
y
p
=
−
1
10
x
4
+
14
75
x
3
+
53
250
x
2
−
697
625
x
, and
y
=
c
1
e
5
x
+
c
2
−
1
10
x
4
+
14
75
x
3
+
53
250
x
2
−
697
625
x.
17.
From
m
2
−
2
m
+ 5 = 0 we find
m
1
=1+2
i
and
m
2
=1
−
2
i
. Then
y
c
=
e
x
(
c
1
cos 2
x
+
c
2
sin 2
x
) and we assume
y
p
=
Axe
x
cos 2
x
+
Bxe
x
sin 2
x
. Substituting into the differential equation we obtain 4
B
= 1 and
−
4
A
=0.
Then
A
=0,
B
=
1
4
,
y
p
=
1
4
xe
x
sin 2
x
, and
y
=
e
x
(
c
1
cos 2
x
+
c
2
sin 2
x
)+
1
4
xe
x
sin 2
x.
18.
From
m
2
−
2
m
+ 2 = 0 we find
m
1
=1+
i
and
m
2
=1
−
i
. Then
y
c
=
e
x
(
c
1
cos
x
+
c
2
sin
x
) and we assume
y
p
=
Ae
2
x
cos
x
+
Be
2
x
sin
x
. Substituting into the differential equation we obtain
A
+2
B
= 1 and
−
2
A
+
B
=
−
3.
Then
A
=
7
5
,
B
=
−
1
5
,
y
p
=
7
5
e
2
x
cos
x
−
1
5
e
2
x
sin
x
and
y
=
e
x
(
c
1
cos
x
+
c
2
sin
x
)+
7
5
e
2
x
cos
x
−
1
5
e
2
x
sin
x.
19.
From
m
2
+2
m
+1 = 0 we find
m
1
=
m
2
=
−
1. Then
y
c
=
c
1
e
−
x
+
c
2
xe
−
x
and we assume
y
p
=
A
cos
x
+
B
sin
x
+
C
cos 2
x
+
D
sin 2
x
. Substituting into the differential equation we obtain 2
B
=0,
−
2
A
=1,
−
3
C
+4
D
= 3, and
−
4
C
−
3
D
= 0. Then
A
=
−
1
2
,
B
=0,
C
=
−
9
25
,
D
=
12
25
,
y
p
=
−
1
2
cos
x
−
9
25
cos 2
x
+
12
25
sin 2
x
, and
y
=
c
1
e
−
x
+
c
2
xe
−
x
−
1
2
cos
x
−
9
25
cos 2
x
+
12
25
sin 2
x.
20.
From
m
2
+2
m
−
24 = 0 we find
m
1
=
−
6 and
m
2
= 4. Then
y
c
=
c
1
e
−
6
x
+
c
2
e
4
x
and we assume
y
p
=
A
+(
Bx
2
+
Cx
)
e
4
x
. Substituting into the differential equation we obtain
−
24
A
= 16, 2
B
+10
C
=
−
2, and
20
B
=
−
1. Then
A
=
−
2
3
,
B
=
−
1
20
,
C
=
−
19
100
,
y
p
=
−
2
3
−
1
1
20
x
2
+
19
100
x
.
e
4
x
, and
y
=
c
1
e
−
6
x
+
c
2
e
4
x
−
2
3
−

1
20
x
2
+
19
100
x

e
4
x
.
93

94. Exercises 3.4
21.
From
m
3
−
6
m
2
= 0 we find
m
1
=
m
2
= 0 and
m
3
= 6. Then
y
c
=
c
1
+
c
2
x
+
c
3
e
6
x
and we assume
y
p
=
Ax
2
+
B
cos
x
+
C
sin
x
. Substituting into the differential equation we obtain
−
12
A
=3,6
B
−
C
=
−
1,
and
B
+6
C
= 0. Then
A
=
−
1
4
,
B
=
−
6
37
,
C
=
1
37
,
y
p
=
−
1
4
x
2
−
6
37
cos
x
+
1
37
sin
x
, and
y
=
c
1
+
c
2
x
+
c
3
e
6
x
−
1
4
x
2
−
6
37
cos
x
+
1
37
sin
x.
22.
From
m
3
−
2
m
2
−
4
m
+8=0wefind
m
1
=
m
2
= 2 and
m
3
=
−
2. Then
y
c
=
c
1
e
2
x
+
c
2
xe
2
x
+
c
3
e
−
2
x
and we
assume
y
p
=(
Ax
3
+
Bx
2
)
e
2
x
. Substituting into the differential equation we obtain 24
A
= 6 and 6
A
+8
B
=0.
Then
A
=
1
4
,
B
=
−
3
16
,
y
p
=
1
1
4
x
3
−
3
16
x
2
.
e
2
x
, and
y
=
c
1
e
2
x
+
c
2
xe
2
x
+
c
3
e
−
2
x
+

1
4
x
3
−
3
16
x
2

e
2
x
.
23.
From
m
3
−
3
m
2
+3
m
−
1=0wefind
m
1
=
m
2
=
m
3
= 1. Then
y
c
=
c
1
e
x
+
c
2
xe
x
+
c
3
x
2
e
x
and we assume
y
p
=
Ax
+
B
+
Cx
3
e
x
. Substituting into the differential equation we obtain
−
A
=1,3
A
−
B
= 0, and 6
C
=
−
4.
Then
A
=
−
1,
B
=
−
3,
C
=
−
2
3
,
y
p
=
−
x
−
3
−
2
3
x
3
e
x
, and
y
=
c
1
e
x
+
c
2
xe
x
+
c
3
x
2
e
x
−
x
−
3
−
2
3
x
3
e
x
.
24.
From
m
3
−
m
2
−
4
m
+4=0wefind
m
1
=1,
m
2
= 2, and
m
3
=
−
2. Then
y
c
=
c
1
e
x
+
c
2
e
2
x
+
c
3
e
−
2
x
and we
assume
y
p
=
A
+
Bxe
x
+
Cxe
2
x
. Substituting into the differential equation we obtain 4
A
=5,
−
3
B
=
−
1, and
4
C
= 1. Then
A
=
5
4
,
B
=
1
3
,
C
=
1
4
,
y
p
=
5
4
+
1
3
xe
x
+
1
4
xe
2
x
, and
y
=
c
1
e
x
+
c
2
e
2
x
+
c
3
e
−
2
x
+
5
4
+
1
3
xe
x
+
1
4
xe
2
x
.
25.
From
m
4
+2
m
2
+1 = 0 we find
m
1
=
m
3
=
i
and
m
2
=
m
4
=
−
i
. Then
y
c
=
c
1
cos
x
+
c
2
sin
x
+
c
3
x
cos
x
+
c
4
x
sin
x
and we assume
y
p
=
Ax
2
+
Bx
+
C
. Substituting into the differential equation we obtain
A
=1,
B
=
−
2, and
4
A
+
C
= 1. Then
A
=1,
B
=
−
2,
C
=
−
3,
y
p
=
x
2
−
2
x
−
3, and
y
=
c
1
cos
x
+
c
2
sin
x
+
c
3
x
cos
x
+
c
4
x
sin
x
+
x
2
−
2
x
−
3
.
26.
From
m
4
−
m
2
= 0 we find
m
1
=
m
2
=0,
m
3
= 1, and
m
4
=
−
1. Then
y
c
=
c
1
+
c
2
x
+
c
3
e
x
+
c
4
e
−
x
and
we assume
y
p
=
Ax
3
+
Bx
2
+(
Cx
2
+
Dx
)
e
−
x
. Substituting into the differential equation we obtain
−
6
A
=4,
−
2
B
=0,10
C
−
2
D
= 0, and
−
4
C
= 2. Then
A
=
−
2
3
,
B
=0,
C
=
−
1
2
,
D
=
−
5
2
,
y
p
=
−
2
3
x
3
−
1
1
2
x
2
+
5
2
x
.
e
−
x
,
and
y
=
c
1
+
c
2
x
+
c
3
e
x
+
c
4
e
−
x
−
2
3
x
3
−

1
2
x
2
+
5
2
x

e
−
x
.
27.
We have
y
c
=
c
1
cos 2
x
+
c
2
sin 2
x
and we assume
y
p
=
A
. Substituting into the differential equation we find
A
=
−
1
2
.Thus
y
=
c
1
cos 2
x
+
c
2
sin 2
x
−
1
2
. From the initial conditions we obtain
c
1
= 0 and
c
2
=
√
2, so
y
=
√
2 sin 2
x
−
1
2
.
28.
We have
y
c
=
c
1
e
−
2
x
+
c
2
e
x/
2
and we assume
y
p
=
Ax
2
+
Bx
+
C
. Substituting into the differential equation we
find
A
=
−
7,
B
=
−
19, and
C
=
−
37. Thus
y
=
c
1
e
−
2
x
+
c
2
e
x/
2
−
7
x
2
−
19
x
−
37. From the initial conditions
we obtain
c
1
=
−
1
5
and
c
2
=
186
5
,so
y
=
−
1
5
e
−
2
x
+
186
5
e
x/
2
−
7
x
2
−
19
x
−
37
.
29.
We have
y
c
=
c
1
e
−
x/
5
+
c
2
and we assume
y
p
=
Ax
2
+
Bx
. Substituting into the differential equation we find
A
=
−
3 and
B
= 30. Thus
y
=
c
1
e
−
x/
5
+
c
2
−
3
x
2
+30
x
. From the initial conditions we obtain
c
1
= 200 and
c
2
=
−
200, so
y
= 200
−
x/
5
−
200
−
3
x
2
+30
x.
94

95. Exercises 3.4
30.
We have
y
c
=
c
1
e
−
2
x
+
c
2
xe
−
2
x
and we assume
y
p
=(
Ax
3
+
Bx
2
)
e
−
2
x
. Substituting into the differential equation
we find
A
=
1
6
and
B
=
3
2
.Thus
y
=
c
1
e
−
2
x
+
c
2
xe
−
2
x
+
1
1
6
x
3
+
3
2
x
2
.
e
−
2
x
. From the initial conditions we
obtain
c
1
= 2 and
c
2
=9,so
y
=2
e
−
2
x
+9
xe
−
2
x
+

1
6
x
3
+
3
2
x
2

e
−
2
x
.
31.
We have
y
c
=
e
−
2
x
(
c
1
cos
x
+
c
2
sin
x
) and we assume
y
p
=
Ae
−
4
x
. Substituting into the differential equation
we find
A
=5. Thus
y
=
e
−
2
x
(
c
1
cos
x
+
c
2
sin
x
)+7
e
−
4
x
. From the initial conditions we obtain
c
1
=
−
10 and
c
2
=9,so
y
=
e
−
2
x
(
−
10 cos
x
+ 9 sin
x
+7
e
−
4
x
)
.
32.
We have
y
c
=
c
1
cosh
x
+
c
2
sinh
x
and we assume
y
p
=
Ax
cosh
x
+
Bx
sinh
x
. Substituting into the differential
equation we find
A
= 0 and
B
=
1
2
.Thus
y
=
c
1
cosh
x
+
c
2
sinh
x
+
1
2
x
sinh
x.
From the initial conditions we obtain
c
1
= 2 and
c
2
= 12, so
y
= 2 cosh
x
+ 12 sinh
x
+
1
2
x
sinh
x.
33.
We have
x
c
=
c
1
cos
ωt
+
c
2
sin
ωt
and we assume
x
p
=
At
cos
ωt
+
Bt
sin
ωt
. Substituting into the differential
equation we find
A
=
−
F
0
/
2
ω
and
B
=0. Thus
x
=
c
1
cos
ωt
+
c
2
sin
ωt
−
(
F
0
/
2
ω
)
t
cos
ωt
. From the initial
conditions we obtain
c
1
= 0 and
c
2
=
F
0
/
2
ω
2
,so
x
=(
F
0
/
2
ω
2
) sin
ωt
−
(
F
0
/
2
ω
)
t
cos
ωt.
34.
We have
x
c
=
c
1
cos
ωt
+
c
2
sin
ωt
and we assume
x
p
=
A
cos
γt
+
B
sin
γt
, where
γ
3
=
ω
. Substituting into the
differential equation we find
A
=
F
0
/
(
ω
2
−
γ
2
) and
B
=0. Thus
x
=
c
1
cos
ωt
+
c
2
sin
ωt
+
F
0
(
ω
2
−
γ
2
)
cos
γt.
From the initial conditions we obtain
c
1
=
F
0
/
(
ω
2
−
γ
2
) and
c
2
=0,so
x
=
F
0
(
ω
2
−
γ
2
)
cos
ωt
+
F
0
(
ω
2
−
γ
2
)
cos
γt.
35.
We have
y
c
=
c
1
+
c
2
e
x
+
c
3
xe
x
and we assume
y
p
=
Ax
+
Bx
2
e
x
+
Ce
5
x
. Substituting into the differential
equation we find
A
=2,
B
=
−
12, and
C
=
1
2
.Thus
y
=
c
1
+
c
2
e
x
+
c
3
xe
x
+2
x
−
12
x
2
e
x
+
1
2
e
5
x
.
From the initial conditions we obtain
c
1
= 11,
c
2
=
−
11, and
c
3
=9,so
y
=11
−
11
e
x
+9
xe
x
+2
x
−
12
x
2
e
x
+
1
2
e
5
x
.
36.
We have
y
c
=
c
1
e
−
2
x
+
e
x
(
c
2
cos
√
3
x
+
c
3
sin
√
3
x
) and we assume
y
p
=
Ax
+
B
+
Cxe
−
2
x
. Substituting into
the differential equation we find
A
=
1
4
,
B
=
−
5
8
, and
C
=
2
3
.Thus
y
=
c
1
e
−
2
x
+
e
x
(
c
2
cos
√
3
x
+
c
3
sin
√
3
x
)+
1
4
x
−
5
8
+
2
3
xe
−
2
x
.
From the initial conditions we obtain
c
1
=
−
23
12
,
c
2
=
−
59
24
, and
c
3
=
17
72
√
3, so
y
=
−
23
12
e
−
2
x
+
e
x

−
59
24
cos
√
3
x
+
17
72
√
3 sin
√
3
x

+
1
4
x
−
5
8
+
2
3
xe
−
2
x
.
95

96. Exercises 3.4
37.
We have
y
c
=
c
1
cos
x
+
c
2
sin
x
and we assume
y
p
=
A
2
+
Bx
+
C
. Substituting into the differential equation
we find
A
=1,
B
= 0, and
C
=
−
1. Thus
y
=
c
1
cos
x
+
c
2
sin
x
+
x
2
−
1. From
y
(0) = 5 and
y
(1) = 0 we obtain
c
1
−
1=5
(cos 1)
c
1
+ sin(1)
c
2
=0
.
Solving this system we find
c
1
= 6 and
c
2
=
−
6 cot 1. The solution of the boundary-value problem is
y
= 6 cos
x
−
6(cot 1) sin
x
+
x
2
−
1
.
38.
We have
y
c
=
e
x
(
c
1
cos
x
+
c
2
sin
x
) and we assume
y
p
=
Ax
+
B
. Substituting into the differential equation we
find
A
= 1 and
B
=0. Thus
y
=
e
x
(
c
1
cos
x
+
c
2
sin
x
)+
x
. From
y
(0) = 0 and
y
(
π
)=
π
we obtain
c
1
=0
π
−
e
π
c
1
=
π.
Solving this system we find
c
1
= 0 and
c
2
is any real number. The solution of the boundary-value problem is
y
=
c
2
e
x
sin
x
+
x.
39.
We have
y
c
=
c
1
cos 2
x
+
c
2
sin 2
x
and we assume
y
p
=
A
cos
x
+
B
sin
x
on [0
,π/
2]. Substituting into the
differential equation we find
A
= 0 and
B
=
1
3
.Thus
y
=
c
1
cos 2
x
+
c
2
sin 2
x
+
1
3
sin
x
on [0
,π/
2]. On (
π/
2
,
∞
)
we have
y
=
c
3
cos 2
x
+
c
4
sin 2
x
. From
y
(0) = 1 and
y
1
(0) = 2 we obtain
c
1
=1
1
3
+2
c
2
=2
.
Solving this system we find
c
1
= 1 and
c
2
=
5
6
.Thus
y
= cos 2
x
+
5
6
sin 2
x
+
1
3
sin
x
on [0
,π/
2]. Now continuity
of
y
at
x
=
π/
2 implies
cos
π
+
5
6
sin
π
+
1
3
sin
π
2
=
c
3
cos
π
+
c
4
sin
π
or
−
1+
1
3
=
−
c
3
. Hence
c
3
=
2
3
. Continuity of
y
1
at
x
=
π/
2 implies
−
2 sin
π
+
5
3
cos
π
+
1
3
cos
π
2
=
−
2
c
3
sin
π
+2
c
4
cos
π
or
−
5
3
=
−
2
c
4
. Then
c
4
=
5
6
and the solution of the boundary-value problem is
y
(
x
)=
d
cos 2
x
+
5
6
sin 2
x
+
1
3
sin
x,
0
≤
x
≤
π/
2
2
3
cos 2
x
+
5
6
sin 2
x, x > π/
2.
40.
The complementary function is
y
c
=
e
2
x
(
c
1
cos 2
x
+
c
2
sin 2
x
). We assume a particular solution of the form
y
p
=(
Ax
3
+
Bx
2
+
Cx
)
e
2
x
cos 2
x
+(
Dx
3
+
Ex
2
+
F
)
e
2
x
sin 2
x
. Substituting into the differential equation and
using a CAS to simplify yields
[12
Dx
2
+(6
A
+8
E
)
x
+(2
B
+4
F
)]
e
2
x
cos 2
x
+[
−
12
Ax
2
+(
−
8
B
+6
D
)
x
+(
−
4
C
+2
E
)]
e
2
x
sin 2
x
=(2
x
2
−
3
x
)
e
2
x
cos 2
x
+ (10
x
2
−
x
−
1)
e
2
x
sin 2
x.
This gives the system of equations
12
D
=2
,
−
12
A
=10
,
6
A
+8
E
=
−
3
,
−
8
B
+6
D
=
−
1
,
2
B
+4
F
=0
,
−
4
C
+2
E
=
−
1
,
96

97. Exercises 3.5
from which we find
A
=
−
5
6
,
B
=
1
4
,
C
=
3
8
,
D
=
1
6
,
E
=
1
4
, and
F
=
−
1
8
. Thus, a particular solution of the
differential equation is
y
p
=
3
−
5
6
x
3
+
1
4
x
2
+
3
8
x
4
e
2
x
cos 2
x
+
3
1
6
x
3
+
1
4
x
2
−
1
8
x
4
e
2
x
sin 2
x.
41.
f
(
t
)=
e
t
sin
t
. We see that
y
p
→∞
as
t
→∞
and
y
p
→
0as
t
→−∞
.
42.
f
(
t
)=
e
−
t
. We see that
y
p
→∞
as
t
→∞
and
y
p
→∞
as
t
→−∞
.
43.
f
(
t
) = sin 2
t
. We see that
y
p
is sinusoidal.
44.
f
(
t
) = 1. We see that
y
p
is constant and simply translates
y
c
vertically.
Exercises 3.5
The particular solution,
y
p
=
u
1
y
1
+
u
2
y
2
, in the following problems can take on a variety of forms, especially where
trigonometric functions are involved. The validity of a particular form can best be checked by substituting it back
into the differential equation.
1.
The auxiliary equation is
m
2
+1=0,so
y
c
=
c
1
cos
x
+
c
2
sin
x
and
W
=
2
2
2
2
cos
x
sin
x
−
sin
x
cos
x
2
2
2
2
=1
.
Identifying
f
(
x
) = sec
x
we obtain
u
1
1
=
−
sin
x
sec
x
1
=
−
tan
x
u
1
2
=
cos
x
sec
x
1
=1
.
Then
u
1
=ln
|
cos
x
|
,
u
2
=
x
, and
y
=
c
1
cos
x
+
c
2
sin
x
+ cos
x
ln
|
cos
x
|
+
x
sin
x.
2.
The auxiliary equation is
m
2
+1=0,so
y
c
=
c
1
cos
x
+
c
2
sin
x
and
W
=
2
2
2
2
cos
x
sin
x
−
sin
x
cos
x
2
2
2
2
=1
.
Identifying
f
(
x
) = tan
x
we obtain
u
1
1
=
−
sin
x
tan
x
=
cos
2
x
−
1
cos
x
= cos
x
−
sec
x
u
1
2
= sin
x.
Then
u
1
= sin
x
−
ln
|
sec
x
+ tan
x
|
,
u
2
=
−
cos
x
, and
y
=
c
1
cos
x
+
c
2
sin
x
+ cos
x
(sin
x
−
ln
|
sec
x
+ tan
x
|
)
−
cos
x
sin
x.
3.
The auxiliary equation is
m
2
+1=0,so
y
c
=
c
1
cos
x
+
c
2
sin
x
and
W
=
2
2
2
2
cos
x
sin
x
−
sin
x
cos
x
2
2
2
2
=1
.
Identifying
f
(
x
) = sin
x
we obtain
u
1
1
=
−
sin
2
x
u
1
2
= cos
x
sin
x.
97

98. Exercises 3.5
Then
u
1
=
1
4
sin 2
x
−
1
2
x
=
1
2
sin
x
cos
x
−
1
2
x
u
2
=
−
1
2
cos
2
x.
and
y
=
c
1
cos
x
+
c
2
sin
x
+
1
2
sin
x
cos
2
x
−
1
2
x
cos
x
−
1
2
cos
2
x
sin
x
=
c
1
cos
x
+
c
2
sin
x
−
1
2
x
cos
x.
4.
The auxiliary equation is
m
2
+1=0,so
y
c
=
c
1
cos
x
+
c
2
sin
x
and
W
=
2
2
2
2
cos
x
sin
x
−
sin
x
cos
x
2
2
2
2
=1
.
Identifying
f
(
x
) = sec
x
tan
x
we obtain
u
1
1
=
−
sin
x
(sec
x
tan
x
)=
−
tan
2
x
=1
−
sec
2
x
u
1
2
= cos
x
(sec
x
tan
x
) = tan
x.
Then
u
1
=
x
−
tan
x
,
u
2
=
−
ln
|
cos
x
|
, and
y
=
c
1
cos
x
+
c
2
sin
x
+
x
cos
x
−
sin
x
−
sin
x
ln
|
cos
x
|
=
c
1
cos
x
+
c
3
sin
x
+
x
cos
x
−
sin
x
ln
|
cos
x
|
.
5.
The auxiliary equation is
m
2
+1=0,so
y
c
=
c
1
cos
x
+
c
2
sin
x
and
W
=
2
2
2
2
cos
x
sin
x
−
sin
x
cos
x
2
2
2
2
=1
.
Identifying
f
(
x
) = cos
2
x
we obtain
u
1
1
=
−
sin
x
cos
2
x
u
1
2
= cos
3
x
= cos
x
1
1
−
sin
2
x
.
.
Then
u
1
=
1
3
cos
3
x
,
u
2
= sin
x
−
1
3
sin
3
x
, and
y
=
c
1
cos
x
+
c
2
sin
x
+
1
3
cos
4
x
+ sin
2
x
−
1
3
sin
4
x
=
c
1
cos
x
+
c
2
sin
x
+
1
3
1
cos
2
x
+ sin
2
x
.1
cos
2
x
−
sin
2
x
.
+ sin
2
x
=
c
1
cos
x
+
c
2
sin
x
+
1
3
cos
2
x
+
2
3
sin
2
x
=
c
1
cos
x
+
c
2
sin
x
+
1
3
+
1
3
sin
2
x.
6.
The auxiliary equation is
m
2
+1=0,so
y
c
=
c
1
cos
x
+
c
2
sin
x
and
W
=
2
2
2
2
cos
x
sin
x
−
sin
x
cos
x
2
2
2
2
=1
.
Identifying
f
(
x
) = sec
2
x
we obtain
u
1
1
=
−
sin
x
cos
2
x
u
1
2
= sec
x.
98

99. Exercises 3.5
Then
u
1
=
−
1
cos
x
=
−
sec
x
u
2
=ln
|
sec
x
+ tan
x
|
and
y
=
c
1
cos
x
+
c
2
sin
x
−
cos
x
sec
x
+ sin
x
ln
|
sec
x
+ tan
x
|
=
c
1
cos
x
+
c
2
sin
x
−
1 + sin
x
ln
|
sec
x
+ tan
x
|
.
7.
The auxiliary equation is
m
2
−
1 = 0, so
y
c
=
c
1
e
x
+
c
2
e
−
x
and
W
=
2
2
2
2
e
x
e
−
x
e
x
−
e
−
x
2
2
2
2
=
−
2
.
Identifying
f
(
x
) = cosh
x
=
1
2
(
e
−
x
+
e
x
) we obtain
u
1
1
=
1
4
e
2
x
+
1
4
u
1
2
=
−
1
4
−
1
4
e
2
x
.
Then
u
1
=
−
1
8
e
−
2
x
+
1
4
x
u
2
=
−
1
8
e
2
x
−
1
4
x
and
y
=
c
1
e
x
+
c
2
e
−
x
−
1
8
e
−
x
+
1
4
xe
x
−
1
8
e
x
−
1
4
xe
−
x
=
c
3
e
x
+
c
4
e
−
x
+
1
4
x
(
e
x
−
e
−
x
)
=
c
3
e
x
+
c
4
e
−
x
+
1
2
x
sinh
x.
8.
The auxiliary equation is
m
2
−
1 = 0, so
y
c
=
c
1
e
x
+
c
2
e
−
x
and
W
=
2
2
2
2
e
x
e
−
x
e
x
−
e
−
x
2
2
2
2
=
−
2
.
Identifying
f
(
x
) = sinh 2
x
we obtain
u
1
1
=
−
1
4
e
−
3
x
+
1
4
e
x
u
1
2
=
1
4
e
−
x
−
1
4
e
3
x
.
Then
u
1
=
1
12
e
−
3
x
+
1
4
e
x
u
2
=
−
1
4
e
−
x
−
1
12
e
3
x
.
and
y
=
c
1
e
x
+
c
2
e
−
x
+
1
12
e
−
2
x
+
1
4
e
2
x
−
1
4
e
−
2
x
−
1
12
e
2
x
=
c
1
e
x
+
c
2
e
−
x
+
1
6
1
e
2
x
−
e
−
2
x
.
=
c
1
e
x
+
c
2
e
−
x
+
1
3
sinh 2
x.
99

100. Exercises 3.5
9.
The auxiliary equation is
m
2
−
4 = 0, so
y
c
=
c
1
e
2
x
+
c
2
e
−
2
x
and
W
=
2
2
2
2
e
2
x
e
−
2
x
2
e
2
x
−
2
e
−
2
x
2
2
2
2
=
−
4
.
Identifying
f
(
x
)=
e
2
x
/x
we obtain
u
1
1
=1
/
4
x
and
u
1
2
=
−
e
4
x
/
4
x.
Then
u
1
=
1
4
ln
|
x
|
,
u
2
=
−
1
4
6
x
x
0
e
4
t
t
dt
and
y
=
c
1
e
2
x
+
c
2
e
−
2
x
+
1
4

e
2
x
ln
|
x
|−
e
−
2
x
6
x
x
0
e
4
t
t
dt

,x
0
>
0
.
10.
The auxiliary equation is
m
2
−
9 = 0, so
y
c
=
c
1
e
3
x
+
c
2
e
−
3
x
and
W
=
2
2
2
2
e
3
x
e
−
3
x
3
e
3
x
−
3
e
−
3
x
2
2
2
2
=
−
6
.
Identifying
f
(
x
)=9
x/e
3
x
we obtain
u
1
1
=
3
2
xe
−
6
x
and
u
1
2
=
−
3
2
x
. Then
u
1
=
−
1
24
e
−
6
x
−
1
4
xe
−
6
x
,
u
2
=
−
3
4
x
2
and
y
=
c
1
e
3
x
+
c
2
e
−
3
x
−
1
24
e
−
3
x
−
1
4
xe
−
3
x
−
3
4
x
2
e
−
3
x
=
c
1
e
3
x
+
c
3
e
−
3
x
−
1
4
xe
−
3
x
(1
−
3
x
)
.
11.
The auxiliary equation is
m
2
+3
m
+2=(
m
+ 1)(
m
+2)=0,so
y
c
=
c
1
e
−
x
+
c
2
e
−
2
x
and
W
=
2
2
2
2
e
−
x
e
−
2
x
−
e
−
x
−
2
e
−
2
x
2
2
2
2
=
−
e
−
3
x
.
Identifying
f
(
x
)=1
/
(1 +
e
x
) we obtain
u
1
1
=
e
x
1+
e
x
u
1
2
=
−
e
2
x
1+
e
x
=
e
x
1+
e
x
−
e
x
.
Then
u
1
= ln(1 +
e
x
),
u
2
=ln(1+
e
x
)
−
e
x
, and
y
=
c
1
e
−
x
+
c
2
e
−
2
x
+
e
−
x
ln(1 +
e
x
)+
e
−
2
x
ln(1 +
e
x
)
−
e
−
x
=
c
3
e
−
x
+
c
2
e
−
2
x
+(1+
e
−
x
)
e
−
x
ln(1 +
e
x
)
.
12.
The auxiliary equation is
m
2
−
2
m
+1=(
m
−
1)
2
=0,so
y
c
=
c
1
e
x
+
c
2
xe
x
and
W
=
2
2
2
2
e
x
xe
x
e
x
xe
x
+
e
x
2
2
2
2
=
e
2
x
.
Identifying
f
(
x
)=
e
x
/
1
1+
x
2
.
we obtain
u
1
1
=
−
xe
x
e
x
e
2
x
(1 +
x
2
)
=
−
x
1+
x
2
u
1
2
=
e
x
e
x
e
2
x
(1 +
x
2
)
=
1
1+
x
2
.
100

101. Exercises 3.5
Then
u
1
=
−
1
2
ln
1
1+
x
2
.
,
u
2
= tan
−
1
x
, and
y
=
c
1
e
x
+
c
2
xe
x
−
1
2
e
x
ln
1
1+
x
2
.
+
xe
x
tan
−
1
x.
13.
The auxiliary equation is
m
2
+3
m
+2=(
m
+ 1)(
m
+2)=0,so
y
c
=
c
1
e
−
x
+
c
2
e
−
2
x
and
W
=
2
2
2
2
e
−
x
e
−
2
x
−
e
−
x
−
2
e
−
2
x
2
2
2
2
=
−
e
−
3
x
.
Identifying
f
(
x
) = sin
e
x
we obtain
u
1
1
=
e
−
2
x
sin
e
x
e
−
3
x
=
e
x
sin
e
x
u
1
2
=
e
−
x
sin
e
x
−
e
−
3
x
=
−
e
2
x
sin
e
x
.
Then
u
1
=
−
cos
e
x
,
u
2
=
e
x
cos
x
−
sin
e
x
, and
y
=
c
1
e
−
x
+
c
2
e
−
2
x
−
e
−
x
cos
e
x
+
e
−
x
cos
e
x
−
e
−
2
x
sin
e
x
=
c
1
e
−
x
+
c
2
e
−
2
x
−
e
−
2
x
sin
e
x
.
14.
The auxiliary equation is
m
2
−
2
m
+1=(
m
−
1)
2
=0,so
y
c
=
c
1
e
t
+
c
2
te
t
and
W
=
2
2
2
2
e
t
te
t
e
t
te
t
+
e
t
2
2
2
2
=
e
2
t
.
Identifying
f
(
t
)=
e
t
tan
−
1
t
we obtain
u
1
1
=
−
te
t
e
t
tan
−
1
t
e
2
t
=
−
t
tan
−
1
t
u
1
2
=
e
t
e
t
tan
−
1
t
e
2
t
= tan
−
1
t.
Then
u
1
=
−
1+
t
2
2
tan
−
1
t
+
t
2
u
2
=
t
tan
−
1
t
−
1
2
ln
1
1+
t
2
.
and
y
=
c
1
e
t
+
c
2
te
t
+

−
1+
t
2
2
tan
−
1
t
+
t
2

e
t
+

t
tan
−
1
t
−
1
2
ln
1
1+
t
2
.

te
t
=
c
1
e
t
+
c
3
te
t
+
1
2
e
t
b1
t
2
−
1
.
tan
−
1
t
−
ln
1
1+
t
2
.c
.
15.
The auxiliary equation is
m
2
+2
m
+1=(
m
+1)
2
=0,so
y
c
=
c
1
e
−
t
+
c
2
te
−
t
and
W
=
2
2
2
2
e
−
t
te
−
t
−
e
−
t
−
te
−
t
+
e
−
t
2
2
2
2
=
e
−
2
t
.
Identifying
f
(
t
)=
e
−
t
ln
t
we obtain
u
1
1
=
−
te
−
t
e
−
t
ln
t
e
−
2
t
=
−
t
ln
t
u
1
2
=
e
−
t
e
−
t
ln
t
e
−
2
t
=ln
t.
101

102. Exercises 3.5
Then
u
1
=
−
1
2
t
2
ln
t
+
1
4
t
2
u
2
=
t
ln
t
−
t
and
y
=
c
1
e
−
t
+
c
2
te
−
t
−
1
2
t
2
e
−
t
ln
t
+
1
4
t
2
e
−
t
+
t
2
e
−
t
ln
t
−
t
2
e
−
t
=
c
1
e
−
t
+
c
2
te
−
t
+
1
2
t
2
e
−
t
ln
t
−
3
4
t
2
e
−
t
.
16.
The auxiliary equation is 2
m
2
+2
m
+1=0,so
y
c
=
e
−
x/
2
(
c
1
cos
x/
2+
c
2
sin
x/
2) and
W
=
2
2
2
2
2
e
−
x/
2
cos
x
2
e
−
x/
2
sin
x
2
−
1
2
e
−
x/
2
cos
x
2
−
1
2
e
−
x/
2
sin
x
2
1
2
e
−
x/
2
cos
x
2
−
1
2
e
x/
2
sin
x
2
2
2
2
2
2
=
1
2
e
−
x
.
Identifying
f
(
x
)=2
√
x
we obtain
u
1
1
=
−
e
−
x/
2
sin(
x/
2)2
√
x
e
−
x/
2
=
−
4
e
x/
2
√
x
sin
x
2
u
1
2
=
−
e
−
x/
2
cos(
x/
2)2
√
x
e
−
x/
2
=4
e
x/
2
√
x
cos
x
2
.
Then
u
1
=
−
4
6
x
x
0
e
t/
2
√
t
sin
t
2
dt
u
2
=4
6
x
x
0
e
t/
2
√
t
cos
t
2
dt
and
y
=
e
−
x/
2
3
c
1
cos
x
2
+
c
2
sin
x
2
4
−
4
e
−
x/
2
cos
x
2
6
x
x
0
e
t/
2
√
t
sin
t
2
dt
+4
e
−
x/
2
sin
x
2
6
x
x
0
e
t/
2
√
t
cos
t
2
dt.
17.
The auxiliary equation is 3
m
2
−
6
m
+6=0,so
y
c
=
e
x
(
c
1
cos
x
+
c
2
sin
x
) and
W
=
2
2
2
2
e
x
cos
xe
x
sin
x
e
x
cos
x
−
e
x
sin
xe
x
cos
x
+
e
x
sin
x
2
2
2
2
=
e
2
x
.
Identifying
f
(
x
)=
1
3
e
x
sec
x
we obtain
u
1
1
=
−
(
e
x
sin
x
)(
e
x
sec
x
)
/
3
e
2
x
=
−
1
3
tan
x
u
1
2
=
(
e
x
cos
x
)(
e
x
sec
x
)
/
3
e
2
x
=
1
3
.
Then
u
1
=
1
3
ln(cos
x
),
u
2
=
1
3
x
, and
y
=
c
1
e
x
cos
x
+
c
2
e
x
cos
x
+
1
3
ln(cos
x
)
e
x
cos
x
+
1
3
xe
x
sin
x.
18.
The auxiliary equation is 4
m
2
−
4
m
+1=(2
m
−
1)
2
=0,so
y
c
=
c
1
e
x/
2
+
c
2
xe
x/
2
and
W
=
2
2
2
2
e
x/
2
xe
x/
2
1
2
e
x/
2
1
2
xe
x/
2
+
e
x/
2
2
2
2
2
=
e
x
.
102

103. Exercises 3.5
Identifying
f
(
x
)=
1
4
e
x/
2
√
1
−
x
2
we obtain
u
1
1
=
−
xe
x/
2
e
x/
2
√
1
−
x
2
4
e
x
=
−
1
4
x
)
1
−
x
2
u
1
2
=
e
x/
2
e
x/
2
√
1
−
x
2
4
e
x
=
1
4
)
1
−
x
2
.
Then
u
1
=
1
12
1
1
−
x
2
.
3
/
2
u
2
=
x
8
)
1
−
x
2
+
1
8
sin
−
1
x
and
y
=
c
1
e
x/
2
+
c
2
xe
x/
2
+
1
12
e
x/
2
1
1
−
x
2
.
3
/
2
+
1
8
x
2
e
x/
2
)
1
−
x
2
+
1
8
xe
x/
2
sin
−
1
x.
19.
The auxiliary equation is 4
m
2
−
1=(2
m
−
1)(2
m
+ 1) = 0, so
y
c
=
c
1
e
x/
2
+
c
2
e
−
x/
2
and
W
=
2
2
2
2
e
x/
2
e
−
x/
2
1
2
e
x/
2
−
1
2
e
−
x/
2
2
2
2
2
=
−
1
.
Identifying
f
(
x
)=
xe
x/
2
/
4 we obtain
u
1
1
=
x/
4 and
u
1
2
=
−
xe
x
/
4. Then
u
1
=
x
2
/
8 and
u
2
=
−
xe
x
/
4+
e
x
/
4.
Thus
y
=
c
1
e
x/
2
+
c
2
e
−
x/
2
+
1
8
x
2
e
x/
2
−
1
4
xe
x/
2
+
1
4
e
x/
2
=
c
3
e
x/
2
+
c
2
e
−
x/
2
+
1
8
x
2
e
x/
2
−
1
4
xe
x/
2
and
y
1
=
1
2
c
3
e
x/
2
−
1
2
c
2
e
−
x/
2
+
1
16
x
2
e
x/
2
+
1
8
xe
x/
2
−
1
4
e
x/
2
.
The initial conditions imply
c
3
+
c
2
=1
1
2
c
3
−
1
2
c
2
−
1
4
=0
.
Thus
c
3
=3
/
4 and
c
2
=1
/
4, and
y
=
3
4
e
x/
2
+
1
4
e
−
x/
2
+
1
8
x
2
e
x/
2
−
1
4
xe
x/
2
.
20.
The auxiliary equation is 2
m
2
+
m
−
1=(2
m
−
1)(
m
+1)=0,so
y
c
=
c
1
e
x/
2
+
c
2
e
−
x
and
W
=
2
2
2
2
e
x/
2
e
−
x
1
2
e
x/
2
−
e
−
x
2
2
2
2
=
−
3
2
e
−
x/
2
.
Identifying
f
(
x
)=(
x
+1)
/
2 we obtain
u
1
1
=
1
3
e
−
x/
2
(
x
+1)
u
1
2
=
−
1
3
e
x
(
x
+1)
.
Then
u
1
=
−
e
−
x/
2

2
3
x
−
2

u
2
=
−
1
3
xe
x
.
103

104. Exercises 3.5
Thus
y
=
c
1
e
x/
2
+
c
2
e
−
x
−
x
−
2
and
y
1
=
1
2
c
1
e
x/
2
−
c
2
e
−
x
−
1
.
The initial conditions imply
c
1
−
c
2
−
2=1
1
2
c
1
−
c
2
−
1=0
.
Thus
c
1
=8
/
3 and
c
2
=1
/
3, and
y
=
8
3
e
x/
2
+
1
3
e
−
x
−
x
−
2
.
21.
The auxiliary equation is
m
2
+2
m
−
8=(
m
−
2)(
m
+4)=0,so
y
c
=
c
1
e
2
x
+
c
2
e
−
4
x
and
W
=
2
2
2
2
e
2
x
e
−
4
x
2
e
2
x
−
4
e
−
4
x
2
2
2
2
=
−
6
e
−
2
x
.
Identifying
f
(
x
)=2
e
−
2
x
−
e
−
x
we obtain
u
1
1
=
1
3
e
−
4
x
−
1
6
e
−
3
x
u
1
2
=
−
1
6
e
3
x
−
1
3
e
2
x
.
Then
u
1
=
−
1
12
e
−
4
x
+
1
18
e
−
3
x
u
2
=
1
18
e
3
x
−
1
6
e
2
x
.
Thus
y
=
c
1
e
2
x
+
c
2
e
−
4
x
−
1
12
e
−
2
x
+
1
18
e
−
x
+
1
18
e
−
x
−
1
6
e
−
2
x
=
c
1
e
2
x
+
c
2
e
−
4
x
−
1
4
e
−
2
x
+
1
9
e
−
x
and
y
1
=2
c
1
e
2
x
−
4
c
2
e
−
4
x
+
1
2
e
−
2
x
−
1
9
e
−
x
.
The initial conditions imply
c
1
+
c
2
−
5
36
=1
2
c
1
−
4
c
2
+
7
18
=0
.
Thus
c
1
=25
/
36 and
c
2
=4
/
9, and
y
=
25
36
e
2
x
+
4
9
e
−
4
x
−
1
4
e
−
2
x
+
1
9
e
−
x
.
22.
The auxiliary equation is
m
2
−
4
m
+4=(
m
−
2)
2
=0,so
y
c
=
c
1
e
2
x
+
c
2
xe
2
x
and
W
=
2
2
2
2
e
2
x
xe
2
x
2
e
2
x
2
xe
2
x
+
e
2
x
2
2
2
2
=
e
4
x
.
104

105. Exercises 3.5
Identifying
f
(
x
)=
1
12
x
2
−
6
x
.
e
2
x
we obtain
u
1
1
=6
x
2
−
12
x
3
u
1
2
=12
x
2
−
6
x.
Then
u
1
=2
x
3
−
3
x
4
u
2
=4
x
3
−
3
x
2
.
Thus
y
=
c
1
e
2
x
+
c
2
xe
2
x
+
1
2
x
3
−
3
x
4
.
e
2
x
+
1
4
x
3
−
3
x
2
.
xe
2
x
=
c
1
e
2
x
+
c
2
xe
2
x
+
e
2
x
1
x
4
−
x
3
.
and
y
1
=2
c
1
e
2
x
+
c
2
1
2
xe
2
x
+
e
2
x
.
+
e
2
x
1
4
x
3
−
3
x
2
.
+2
e
2
x
1
x
4
−
x
3
.
.
The initial conditions imply
c
1
=1
2
c
1
+
c
2
=0
.
Thus
c
1
= 1 and
c
2
=
−
2, and
y
=
e
2
x
−
2
xe
2
x
+
e
2
x
1
x
4
−
x
3
.
=
e
2
x
1
x
4
−
x
3
−
2
x
+1
.
.
23.
Write the equation in the form
y
11
+
1
x
y
1
+

1
−
1
4
x
2

y
=
x
−
1
/
2
and identify
f
(
x
)=
x
−
1
/
2
. From
y
1
=
x
−
1
/
2
cos
x
and
y
2
=
x
−
1
/
2
sin
x
we compute
W
(
y
1
,y
2
)=
2
2
2
2
x
−
1
/
2
cos
xx
−
1
/
2
sin
x
−
x
−
1
/
2
sin
x
−
1
2
x
−
3
/
2
cos
xx
−
1
/
2
cos
x
−
1
2
x
−
3
/
2
sin
x
2
2
2
2
=
1
x
.
Now
u
1
1
= sin
x
so
u
1
= cos
x,
and
u
1
2
= cos
x
so
u
2
= sin
x.
Thus
y
=
c
1
x
−
1
/
2
cos
x
+
c
2
x
−
1
/
2
sin
x
+
x
−
1
/
2
cos
2
x
+
x
−
1
/
2
sin
2
x
=
c
1
x
−
1
/
2
cos
x
+
c
2
x
−
1
/
2
sin
x
+
x
−
1
/
2
.
24.
Write the equation in the form
y
11
+
1
x
y
1
+
1
x
2
y
=
sec(ln
x
)
x
2
and identify
f
(
x
) = sec(ln
x
)
/x
2
. From
y
1
= cos(ln
x
) and
y
2
= sin(ln
x
) we compute
W
=
2
2
2
2
2
2
cos(ln
x
) sin(ln
x
)
−
sin(ln
x
)
x
cos(ln
x
)
x
2
2
2
2
2
2
=
1
x
.
Now
u
1
1
=
−
tan(ln
x
)
x
so
u
1
=ln
|
cos(ln
x
)
|
,
105

106. Exercises 3.5
and
u
1
2
=
1
x
so
u
2
=ln
x.
Thus, a particular solution is
y
p
= cos(ln
x
)ln
|
cos(ln
x
)
|
+ (ln
x
) sin(ln
x
)
.
25.
The auxiliary equation is
m
3
+
m
=
m
(
m
2
+1)=0,so
y
c
=
c
1
+
c
2
cos
x
+
c
3
sin
x
and
W
=
2
2
2
2
2
2
2
1 cos
x
sin
x
0
−
sin
x
cos
x
0
−
cos
x
−
sin
x
2
2
2
2
2
2
2
=1
.
Identifying
f
(
x
) = tan
x
we obtain
u
1
1
=
W
1
=
2
2
2
2
2
2
2
0 cos
x
sin
x
0
−
sin
x
cos
x
tan
x
−
cos
x
−
sin
x
2
2
2
2
2
2
2
= tan
x
u
1
2
=
W
2
=
2
2
2
2
2
2
2
1 0 sin
x
0 0 cos
x
0 tan
x
−
sin
x
2
2
2
2
2
2
2
=
−
sin
x
u
1
3
=
W
3
=
2
2
2
2
2
2
2
1 cos
x
0
0
−
sin
x
0
0
−
cos
x
tan
x
2
2
2
2
2
2
2
=
−
sin
x
tan
x
=
cos
2
x
−
1
cos
x
= cos
x
−
sec
x.
Then
u
1
=
−
ln
|
cos
x
|
u
2
= cos
x
u
3
= sin
x
−
ln
|
sec
x
+ tan
x
|
and
y
=
c
1
+
c
2
cos
x
+
c
3
sin
x
−
ln
|
cos
x
|
+ cos
2
x
+ sin
2
x
−
sin
x
ln
|
sec
x
+ tan
x
|
=
c
4
+
c
2
cos
x
+
c
3
sin
x
−
ln
|
cos
x
|−
sin
x
ln
|
sec
x
+ tan
x
|
for
−∞
<x<
∞
.
26.
The auxiliary equation is
m
3
+4
m
=
m
1
m
2
+4
.
=0,so
y
c
=
c
1
+
c
2
cos 2
x
+
c
3
sin 2
x
and
W
=
2
2
2
2
2
2
2
1 cos 2
x
sin 2
x
0
−
2 sin 2
x
2 cos 2
x
0
−
4 cos 2
x
−
4 sin 2
x
2
2
2
2
2
2
2
=8
.
106

107. Exercises 3.5
Identifying
f
(
x
) = sec 2
x
we obtain
u
1
1
=
1
8
W
1
=
1
8
2
2
2
2
2
2
2
0 cos 2
x
sin 2
x
0
−
2 sin 2
x
2 cos 2
x
sec 2
x
−
4 cos 2
x
−
4 sin 2
x
2
2
2
2
2
2
2
=
1
4
sec 2
x
u
1
2
=
1
8
W
2
=
1
8
2
2
2
2
2
2
2
1 0 sin 2
x
0 0 2 cos 2
x
0 sec 2
x
−
4 sin 2
x
2
2
2
2
2
2
2
=
−
1
4
u
1
3
=
1
8
W
3
=
1
8
2
2
2
2
2
2
2
1 cos 2
x
0
0
−
2 sin 2
x
0
0
−
4 cos 2
x
sec 2
x
2
2
2
2
2
2
2
=
−
1
4
tan 2
x.
Then
u
1
=
1
8
ln
|
sec 2
x
+ tan 2
x
|
u
2
=
−
1
4
x
u
3
=
1
8
ln
|
cos 2
x
|
and
y
=
c
1
+
c
2
cos 2
x
+
c
3
sin 2
x
+
1
8
ln
|
sec 2
x
+ tan 2
x
|−
1
4
x
cos 2
x
+
1
8
sin 2
x
ln
|
cos 2
x
|
for
−
π/
4
<x<π/
4.
27.
The auxiliary equation is 3
m
2
−
6
m
+30 = 0, which has roots 1+3
i
,so
y
c
=
e
x
(
c
1
cos 3
x
+
c
2
sin 3
x
). We consider
first the differential equation 3
y
11
−
6
y
1
+30
y
= 15 sin
x
, which can be solved using undetermined coefficients.
Letting
y
p
1
=
A
cos
x
+
B
sin
x
and substituting into the differential equation we get
(27
A
−
6
B
) cos
x
+(6
a
+27
b
) sin
x
= 15 sin
x.
Then
27
A
−
6
B
= 0 and 6
a
+27
b
=15
,
so
A
=
2
17
and
B
=
9
17
. Thus,
y
p
1
=
2
17
cos
x
+
9
17
sin
x
. Next, we consider the differential equation 3
y
11
−
6
y
1
+30
y
,
for which a particular solution
y
p
2
can be found using variation of parameters. The Wronskian is
W
=
2
2
2
2
e
x
cos 3
xe
x
sin 3
x
e
x
cos 3
x
−
3
e
x
sin 3
x
3
e
x
cos 3
x
+
e
x
sin 3
x
2
2
2
2
=3
e
2
x
.
Identifying
f
(
x
)=
1
3
e
x
tan
x
we obtain
u
1
1
=
−
1
9
sin 3
x
tan 3
x
and
u
1
2
=
1
9
sin 3
x.
Then
u
1
=
1
27
sin 3
x
+
1
27

ln

cos
3
x
2
−
sin
3
x
2

−
ln

cos
3
x
2
+ sin
3
x
2

u
2
=
−
1
27
cos 3
x.
Thus
y
p
2
=
1
27
e
x
cos 3
x

ln

cos
3
x
2
−
sin
3
x
2

−
ln

cos
3
x
2
+ sin
3
x
2

107

108. Exercises 3.5
and the general solution of the original differential equation is
y
=
e
x
(
c
1
cos 3
x
+
c
2
sin 3
x
)+
y
p
1
(
x
)+
y
p
2
(
x
)
.
28.
The auxiliary equation is
m
2
−
2
m
+1=(
m
−
1)
2
= 0, which has repeated root 1, so
y
c
=
c
1
e
x
+
c
2
xe
x
.
We consider first the differential equation
y
11
−
2
y
1
+
y
=4
x
2
−
3, which can be solved using undetermined
coefficients. Letting
y
p
1
=
Ax
2
+
Bx
+
C
and substituting into the differential equation we get
Ax
2
+(
−
4
A
+
B
)
x
+(2
A
−
2
B
+
C
)=4
x
2
−
3
.
Then
A
=4
,
−
4
A
+
B
=0
,
and 2
A
−
2
B
+
C
=
−
3
,
so
A
=4,
B
= 16, and
C
= 21. Thus,
y
p
1
=4
x
2
+16
x
+ 21. Next we consider the differential equation
y
11
−
2
y
1
+
y
=
x
−
1
e
x
, for which a particular solution
y
p
2
can be found using variation of parameters. The
Wronskian is
W
=
2
2
2
2
e
x
xe
x
e
x
xe
x
+
e
x
2
2
2
2
=
e
2
x
.
Identifying
f
(
x
)=
e
x
/x
we obtain
u
1
1
=
−
1 and
u
1
2
=1
/x
. Then
u
1
=
−
x
and
u
2
=ln
x
, so that
y
p
2
=
−
xe
x
+
xe
x
ln
x,
and the general solution of the original differential equation is
y
=
y
c
+
y
p
1
+
y
p
2
=
c
1
e
x
+
c
2
xe
x
+4
x
2
+16
x
+21
−
xe
x
+
xe
x
ln
x.
29.
The interval of definition for Problem 1 is (
−
π/
2
,π/
2), for Problem 7 is (
−∞
,
∞
), for Problem 9 is (0
,
∞
), and
for Problem 18 is (
−
1
,
1). In Problem 24 the general solution is
y
=
c
1
cos(ln
x
)+
c
2
sin(ln
x
) + cos(ln
x
)ln
|
cos(ln
x
)
|
+ (ln
x
) sin(ln
x
)
for
−
π/
2
<
ln
x<π/
2or
e
−
π/
2
<x<e
π/
2
. The bounds on ln
x
are due to the presence of sec(ln
x
) in the
differential equation.
30.
We are given that
y
1
=
x
2
is a solution of
x
4
y
11
+
x
3
y
1
−
4
x
2
y
= 0. To find a second solution we use reduction
of order. Let
y
=
x
2
u
(
x
). Then the product rule gives
y
1
=
x
2
u
1
+2
xu
and
y
11
=
x
2
u
11
+4
xu
1
+2
u,
so
x
4
y
11
+
x
3
y
1
−
4
x
2
y
=
x
5
(
xu
11
+5
u
1
)=0
.
Letting
w
=
u
1
, this becomes
xw
1
+5
w
= 0. Separating variables and integrating we have
dw
w
=
−
5
x
dx
and ln
|
w
|
=
−
5ln
x
+
c.
Thus,
w
=
x
−
5
and
u
=
−
1
4
x
−
4
. A second solution is then
y
2
=
x
2
x
−
4
=1
/x
2
, and the general solution of the
homogeneous differential equation is
y
c
=
c
1
x
2
+
c
2
/x
2
. To find a particular solution,
y
p
, we use variation of
parameters. The Wronskian is
W
=
2
2
2
2
x
2
1
/x
2
2
x
−
2
/x
3
2
2
2
2
=
−
4
x
.
Identifying
f
(
x
)=1
/x
4
we obtain
u
1
1
=
1
4
x
−
5
and
u
1
2
=
−
1
4
x
−
1
. Then
u
1
=
−
1
16
x
−
4
and
u
2
=
−
1
4
ln
x
,so
y
p
=
−
1
16
x
−
4
x
2
−
1
4
(ln
x
)
x
−
2
=
−
1
16
x
−
2
−
1
4
x
−
2
ln
x.
108

109. Exercises 3.6
The general solution is
y
=
c
1
x
2
+
c
2
x
2
−
1
16
x
2
−
1
4
x
2
ln
x.
Exercises 3.6
1.
The auxiliary equation is
m
2
−
m
−
2=(
m
+ 1)(
m
−
2) = 0 so that
y
=
c
1
x
−
1
+
c
2
x
2
.
2.
The auxiliary equation is 4
m
2
−
4
m
+1=(2
m
−
1)
2
= 0 so that
y
=
c
1
x
1
/
2
+
c
2
x
1
/
2
ln
x
.
3.
The auxiliary equation is
m
2
= 0 so that
y
=
c
1
+
c
2
ln
x
.
4.
The auxiliary equation is
m
2
−
4
m
=
m
(
m
−
4) = 0 so that
y
=
c
1
+
c
2
x
4
.
5.
The auxiliary equation is
m
2
+ 4 = 0 so that
y
=
c
1
cos(2 ln
x
)+
c
2
sin(2 ln
x
).
6.
The auxiliary equation is
m
2
+4
m
+3=(
m
+ 1)(
m
+ 3) = 0 so that
y
=
c
1
x
−
1
+
c
2
x
−
3
.
7.
The auxiliary equation is
m
2
−
4
m
−
2 = 0 so that
y
=
c
1
x
2
−
√
6
+
c
2
x
2+
√
6
.
8.
The auxiliary equation is
m
2
+2
m
−
4 = 0 so that
y
=
c
1
x
−
1+
√
5
+
c
2
x
−
1
−
√
5
.
9.
The auxiliary equation is 25
m
2
+ 1 = 0 so that
y
=
c
1
cos
1
1
5
ln
x
.
+
c
2
1
1
5
ln
x
.
.
10.
The auxiliary equation is 4
m
2
−
1=(2
m
−
1)(2
m
+ 1) = 0 so that
y
=
c
1
x
1
/
2
+
c
2
x
−
1
/
2
.
11.
The auxiliary equation is
m
2
+4
m
+4=(
m
+2)
2
= 0 so that
y
=
c
1
x
−
2
+
c
2
x
−
2
ln
x
.
12.
The auxiliary equation is
m
2
+7
m
+6=(
m
+ 1)(
m
+ 6) = 0 so that
y
=
c
1
x
−
1
+
c
2
x
−
6
.
13.
The auxiliary equation is 3
m
2
+3
m
+ 1 = 0 so that
y
=
x
−
1
/
2
e
c
1
cos
3
√
3
6
ln
x
4
+
c
2
sin
3
√
3
6
ln
x
4C
.
14.
The auxiliary equation is
m
2
−
8
m
+ 41 = 0 so that
y
=
x
4
[
c
1
cos(5 ln
x
)+
c
2
sin(5 ln
x
)].
15.
Assuming that
y
=
x
m
and substituting into the differential equation we obtain
m
(
m
−
1)(
m
−
2)
−
6=
m
3
−
3
m
2
+2
m
−
6=(
m
−
3)(
m
2
+2)=0
.
Thus
y
=
c
1
x
3
+
c
2
cos
3
√
2ln
x
4
+
c
3
sin
3
√
2ln
x
4
.
16.
Assuming that
y
=
x
m
and substituting into the differential equation we obtain
m
(
m
−
1)(
m
−
2) +
m
−
1=
m
3
−
3
m
2
+3
m
−
1=(
m
−
1)
3
=0
.
Thus
y
=
c
1
x
+
c
2
x
ln
x
+
c
3
x
(ln
x
)
2
.
17.
Assuming that
y
=
x
m
and substituting into the differential equation we obtain
m
(
m
−
1)(
m
−
2)(
m
−
3)+6
m
(
m
−
1)(
m
−
2) =
m
4
−
7
m
2
+6
m
=
m
(
m
−
1)(
m
−
2)(
m
+3)=0
.
Thus
y
=
c
1
+
c
2
x
+
c
3
x
2
+
c
4
x
−
3
.
18.
Assuming that
y
=
x
m
and substituting into the differential equation we obtain
m
(
m
−
1)(
m
−
2)(
m
−
3)+6
m
(
m
−
1)(
m
−
2)+9
m
(
m
−
1)+3
m
+1=
m
4
+2
m
2
+1=(
m
2
+1)
2
=0
.
Thus
y
=
c
1
cos(ln
x
)+
c
2
sin(ln
x
)+
c
3
ln
x
cos(ln
x
)+
c
4
ln
x
sin(ln
x
)
.
109

110. Exercises 3.6
19.
The auxiliary equation is
m
2
−
5
m
=
m
(
m
−
5) = 0 so that
y
c
=
c
1
+
c
2
x
5
and
W
(1
,x
5
)=
2
2
2
2
1
x
5
05
x
4
2
2
2
2
=5
x
4
.
Identifying
f
(
x
)=
x
3
we obtain
u
1
1
=
−
1
5
x
4
and
u
1
2
=1
/
5
x
. Then
u
1
=
−
1
25
x
5
,
u
2
=
1
5
ln
x
, and
y
=
c
1
+
c
2
x
5
−
1
25
x
5
+
1
5
x
5
ln
x
=
c
1
+
c
3
x
5
+
1
5
x
5
ln
x.
20.
The auxiliary equation is 2
m
2
+3
m
+1=(2
m
+ 1)(
m
+ 1) = 0 so that
y
c
=
c
1
x
−
1
+
c
2
x
−
1
/
2
and
W
(
x
−
1
,x
−
1
/
2
)=
2
2
2
2
x
−
1
x
−
1
/
2
−
x
−
2
−
1
2
x
−
3
/
2
2
2
2
2
=
1
2
x
−
5
/
2
.
Identifying
f
(
x
)=
1
2
−
1
2
x
we obtain
u
1
1
=
x
−
x
2
and
u
1
2
=
x
3
/
2
−
x
1
/
2
. Then
u
1
=
1
2
x
2
−
1
3
x
3
,
u
2
=
2
5
x
5
/
2
−
2
3
x
3
/
2
, and
y
=
c
1
x
−
1
+
c
2
x
−
1
/
2
+
1
2
x
−
1
3
x
2
+
2
5
x
2
−
2
3
x
=
c
1
x
−
1
+
c
2
x
−
1
/
2
−
1
6
x
+
1
15
x
2
.
21.
The auxiliary equation is
m
2
−
2
m
+1=(
m
−
1)
2
= 0 so that
y
c
=
c
1
x
+
c
2
x
ln
x
and
W
(
x, x
ln
x
)=
2
2
2
2
xx
ln
x
1 1+ln
x
2
2
2
2
=
x.
Identifying
f
(
x
)=2
/x
we obtain
u
1
1
=
−
2ln
x/x
and
u
1
2
=2
/x
. Then
u
1
=
−
(ln
x
)
2
,
u
2
=2ln
x
, and
y
=
c
1
x
+
c
2
x
ln
x
−
x
(ln
x
)
2
+2
x
(ln
x
)
2
=
c
1
x
+
c
2
x
ln
x
+
x
(ln
x
)
2
.
22.
The auxiliary equation is
m
2
−
3
m
+2=(
m
−
1)(
m
−
2) = 0 so that
y
c
=
c
1
x
+
c
2
x
2
and
W
(
x, x
2
)=
2
2
2
2
xx
2
12
x
2
2
2
2
=
x
2
.
Identifying
f
(
x
)=
x
2
e
x
we obtain
u
1
1
=
−
x
2
e
x
and
u
1
2
=
xe
x
. Then
u
1
=
−
x
2
e
x
+2
xe
x
−
2
e
x
,
u
2
=
xe
x
−
e
x
,
and
y
=
c
1
x
+
c
2
x
2
−
x
3
e
x
+2
x
2
e
x
−
2
xe
x
+
x
3
e
x
−
x
2
e
x
=
c
1
x
+
c
2
x
2
+
x
2
e
x
−
2
xe
x
.
23.
The auxiliary equation is
m
2
+2
m
=
m
(
m
+ 2) = 0, so that
y
=
c
1
+
c
2
x
−
2
and
y
1
=
−
2
c
2
x
−
3
.
The initial conditions imply
c
1
+
c
2
=0
−
2
c
2
=4
.
Thus,
c
1
=2,
c
2
=
−
2, and
y
=2
−
2
x
−
2
.
24.
The auxiliary equation is
m
2
−
6
m
+8=(
m
−
2)(
m
−
4) = 0, so that
y
=
c
1
x
2
+
c
2
x
4
and
y
1
=2
c
1
x
+4
c
2
x
3
.
The initial conditions imply
4
c
1
+16
c
2
=32
4
c
1
+32
c
2
=0
.
110

111. Exercises 3.6
Thus,
c
1
= 16,
c
2
=
−
2, and
y
=16
x
2
−
2
x
4
.
25.
The auxiliary equation is
m
2
+ 1 = 0, so that
y
=
c
1
cos(ln
x
)+
c
2
sin(ln
x
) and
y
1
=
−
c
1
1
x
sin(ln
x
)+
c
2
1
x
cos(ln
x
)
.
The initial conditions imply
c
1
= 1 and
c
2
=2. Thus
y
= cos(ln
x
) + 2 sin(ln
x
).
26.
The auxiliary equation is
m
2
−
4
m
+4=(
m
−
2)
2
= 0, so that
y
=
c
1
x
2
+
c
2
x
2
ln
x
and
y
1
=2
c
1
x
+
c
2
(
x
+2
x
ln
x
)
.
The initial conditions imply
c
1
= 5 and
c
2
+ 10 = 3. Thus
y
=5
x
2
−
7
x
2
ln
x
.
27.
The auxiliary equation is
m
2
= 0 so that
y
c
=
c
1
+
c
2
ln
x
and
W
(1
,
ln
x
)=
2
2
2
2
1ln
x
01
/x
2
2
2
2
=
1
x
.
Identifying
f
(
x
) = 1 we obtain
u
1
1
=
−
x
ln
x
and
u
1
2
=
x
. Then
u
1
=
1
4
x
2
−
1
2
x
2
ln
x
,
u
2
=
1
2
x
2
, and
y
=
c
1
+
c
2
ln
x
+
1
4
x
2
−
1
2
x
2
ln
x
+
1
2
x
2
ln
x
=
c
1
+
c
2
ln
x
+
1
4
x
2
.
The initial conditions imply
c
1
+
1
4
= 1 and
c
2
+
1
2
=
−
1
2
. Thus,
c
1
=
3
4
,
c
2
=
−
1, and
y
=
3
4
−
ln
x
+
1
4
x
2
.
28.
The auxiliary equation is
m
2
−
6
m
+8=(
m
−
2)(
m
−
4) = 0, so that
y
c
=
c
1
x
2
+
c
2
x
4
and
W
=
2
2
2
2
x
2
x
4
2
x
4
x
3

=2
x
5
.
Identifying
f
(
x
)=8
x
4
we obtain
u
1
1
=
−
4
x
3
and
u
1
2
=4
x
. Then
u
1
=
−
x
4
,
u
2
=2
x
2
, and
y
=
c
1
x
2
+
c
2
x
4
+
x
6
.
The initial conditions imply
1
4
c
1
+
1
16
c
2
=
−
1
64
c
1
+
1
2
c
2
=
−
3
16
.
Thus
c
1
=
1
16
,
c
2
=
−
1
2
, and
y
=
1
16
x
2
−
1
2
x
4
+
x
6
.
29.
Substituting into the differential equation we obtain
d
2
y
dt
2
+9
dy
dt
+8
y
=
e
2
t
.
The auxiliary equation is
m
2
+9+8=(
m
+ 1)(
m
+ 8) = 0 so that
y
c
=
c
1
e
−
t
+
c
2
e
−
8
t
. Using undetermined
coefficients we try
y
p
=
Ae
2
t
. This leads to 30
Ae
2
t
=
e
2
t
, so that
A
=1
/
30 and
y
=
c
1
e
−
t
+
c
2
e
−
8
t
+
1
30
e
2
t
=
c
1
x
−
1
+
c
2
x
−
8
+
1
30
x
2
.
30.
Substituting into the differential equation we obtain
d
2
y
dt
2
−
5
dy
dt
+6
y
=2
t.
The auxiliary equation is
m
2
−
5
m
+6=(
m
−
2)(
m
−
3) = 0 so that
y
c
=
c
1
e
2
t
+
c
2
e
3
t
. Using undetermined
coefficients we try
y
p
=
At
+
B
. This leads to (
−
5
A
+6
B
)+6
At
=2
t
, so that
A
=1
/
3,
B
=5
/
18, and
y
=
c
1
e
2
t
+
c
2
e
3
t
+
1
3
t
+
5
18
=
c
1
x
2
+
c
2
x
3
+
1
3
ln
x
+
5
18
.
31.
Substituting into the differential equation we obtain
d
2
y
dt
2
−
4
dy
dt
+13
y
=4+3
e
t
.
111

112. Exercises 3.6
The auxiliary equation is
m
2
−
4
m
+13 = 0 so that
y
c
=
e
2
t
(
c
1
cos 3
t
+
c
2
sin 3
t
). Using undetermined coefficients
we try
y
p
=
A
+
Be
t
. This leads to 13
A
+10
Be
t
=4+3
e
t
, so that
A
=4
/
13,
B
=3
/
10, and
y
=
e
2
t
(
c
1
cos 3
t
+
c
2
sin 3
t
)+
4
13
+
3
10
e
t
=
x
2
[
c
1
cos(3 ln
x
)+
c
2
sin(3 ln
x
)] +
4
13
+
3
10
x.
32.
From
d
2
y
dx
2
=
1
x
2

d
2
y
dt
2
−
dy
dt

it follows that
d
3
y
dx
3
=
1
x
2
d
dx

d
2
y
dt
2
−
dy
dt

−
2
x
3

d
2
y
dt
2
−
dy
dt

=
1
x
2
d
dx

d
2
y
dt
2

−
1
x
2
d
dx

dy
dt

−
2
x
3
d
2
y
dt
2
+
2
x
3
dy
dt
=
1
x
2
d
3
y
dt
3

1
x

−
1
x
2
d
2
y
dt
2

1
x

−
2
x
3
d
2
y
dt
2
+
2
x
3
dy
dt
=
1
x
3

d
3
y
dt
3
−
3
d
2
y
dt
2
+2
dy
dt

.
Substituting into the differential equation we obtain
d
3
y
dt
3
−
3
d
2
y
dt
2
+2
dy
dt
−
3

d
2
y
dt
2
−
dy
dt

+6
dy
dt
−
6
y
=3+3
t
or
d
3
y
dt
3
−
6
d
2
y
dt
2
+11
dy
dt
−
6
y
=3+3
t.
The auxiliary equation is
m
3
−
6
m
2
+11
m
−
6=(
m
−
1)(
m
−
2)(
m
−
3) = 0 so that
y
c
=
c
1
e
t
+
c
2
e
2
t
+
c
3
e
3
t
. Using
undetermined coefficients we try
y
p
=
A
+
Bt
. This leads to (11
B
−
6
A
)
−
6
Bt
=3+3
t
, so that
A
=
−
17
/
12,
B
=
−
1
/
2, and
y
=
c
1
e
t
+
c
2
e
2
t
+
c
3
e
3
t
−
17
12
−
1
2
t
=
c
1
x
+
c
2
x
2
+
c
3
x
3
−
17
12
−
1
2
ln
x.
33.
The auxiliary equation is 2
m
(
m
−
1)(
m
−
2)
−
10
.
98
m
(
m
−
1) + 8
.
5
m
+1
.
3 = 0, so that
m
1
=
−
0
.
053299,
m
2
=1
.
81164,
m
3
=6
.
73166, and
y
=
c
1
x
−
0
.
053299
+
c
2
x
1
.
81164
+
c
3
x
6
.
73166
.
34.
The auxiliary equation is
m
(
m
−
1)(
m
−
2) + 4
m
(
m
−
1)+5
m
−
9 = 0, so that
m
1
=1
.
40819 and the two
complex roots are
−
1
.
20409
±
2
.
22291
i
. The general solution of the differential equation is
y
=
c
1
x
1
.
40819
+
x
−
1
.
20409
[
c
2
cos(2
.
22291 ln
x
)+
c
3
sin(2
.
22291 ln
x
)]
.
35.
The auxiliary equation is
m
(
m
−
1)(
m
−
2)(
m
−
3)+6
m
(
m
−
1)(
m
−
2)+3
m
(
m
−
1)
−
3
m
+ 4 = 0, so that
m
1
=
m
2
=
√
2 and
m
3
=
m
4
=
−
√
2 . The general solution of the differential equation is
y
=
c
1
x
√
2
+
c
2
x
√
2
ln
x
+
c
3
x
−
√
2
+
c
4
x
−
√
2
ln
x.
36.
The auxiliary equation is
m
(
m
−
1)(
m
−
2)(
m
−
3)
−
6
m
(
m
−
1)(
m
−
2) + 33
m
(
m
−
1)
−
105
m
+ 169 = 0, so
that
m
1
=
m
2
=3+2
i
and
m
3
=
m
4
=3
−
2
i
. The general solution of the differential equation is
y
=
x
3
[
c
1
cos(2 ln
x
)+
c
2
sin(2 ln
x
)] +
x
3
ln
x
[
c
3
cos(2 ln
x
)+
c
4
sin(2 ln
x
)]
.
112

113. Exercises 3.6
In the next two problems we use the substitution
t
=
−
x
since the initial conditions are on the interval (
−∞
,
0). In
this case
dy
dt
=
dy
dx
dx
dt
=
−
dy
dx
and
d
2
y
dt
2
=
d
dt

dy
dt

=
d
dt

−
dy
dx

=
−
d
dt
(
y
1
)=
−
dy
1
dx
dx
dt
=
−
d
2
y
dx
2
dx
dt
=
d
2
y
dx
2
.
37.
The differential equation and initial conditions become
4
t
2
d
2
y
dt
2
+
y
=0;
y
(
t
)
2
2
2
2
t
=1
=2
,y
1
(
t
)
2
2
2
2
t
=1
=
−
4
.
The auxiliary equation is 4
m
2
−
4
m
+1=(2
m
−
1)
2
= 0, so that
y
=
c
1
t
1
/
2
+
c
2
t
1
/
2
ln
t
and
y
1
=
1
2
c
1
t
−
1
/
2
+
c
2

t
−
1
/
2
+
1
2
t
−
1
/
2
ln
t

.
The initial conditions imply
c
1
= 2 and 1 +
c
2
=
−
4. Thus
y
=2
t
1
/
2
−
5
t
1
/
2
ln
t
=2(
−
x
)
1
/
2
−
5(
−
x
)
1
/
2
ln(
−
x
)
,x<
0
.
38.
The differential equation and initial conditions become
t
2
d
2
y
dt
2
−
4
t
dy
dt
+6
y
=0;
y
(
t
)
2
2
2
2
t
=2
=8
,y
1
(
t
)
2
2
2
2
t
=2
=0
.
The auxiliary equation is
m
2
−
5
m
+6=(
m
−
2)(
m
−
3) = 0, so that
y
=
c
1
t
2
+
c
2
t
3
and
y
1
=2
c
1
t
+3
c
2
t
2
.
The initial conditions imply
4
c
1
+8
c
2
=8
4
c
1
+12
c
2
=0
from which we find
c
1
= 6 and
c
2
=
−
2. Thus
y
=6
t
2
−
2
t
3
=6
x
2
+2
x
3
,x<
0
.
39.
Letting
u
=
x
+ 2 we obtain
dy
dx
=
dy
du
and, using the chain rule,
d
2
y
dx
2
=
d
dx

dy
du

=
d
2
y
du
2
du
dx
=
d
2
y
du
2
(1) =
d
2
y
du
2
.
Substituting into the differential equation we obtain
u
2
d
2
y
du
2
+
u
dy
du
+
y
=0
.
The auxiliary equation is
m
2
+ 1 = 0 so that
y
=
c
1
cos(ln
u
)+
c
2
sin(ln
u
)=
c
1
cos[ ln(
x
+ 2)] +
c
2
sin[ ln(
x
+ 2)]
.
40.
If 1
−
i
is a root of the auxiliary equation then so is 1 +
i
, and the auxiliary equation is
(
m
−
2)[
m
−
(1 +
i
)][
m
−
(1
−
i
)] =
m
3
−
4
m
2
+6
m
−
4=0
.
113

114. Exercises 3.6
We need
m
3
−
4
m
2
+6
m
−
4 to have the form
m
(
m
−
1)(
m
−
2) +
bm
(
m
−
1) +
cm
+
d
. Expanding this last
expression and equating coefficients we get
b
=
−
1,
c
= 3, and
d
=
−
4. Thus, the differential equation is
x
3
y
111
−
x
2
y
11
+3
xy
1
−
4
y
=0
.
41.
For
x
2
y
11
= 0 the auxiliary equation is
m
(
m
−
1) = 0 and the general solution is
y
=
c
1
+
c
2
x
. The initial
conditions imply
c
1
=
y
0
and
c
2
=
y
1
,so
y
=
y
0
+
y
1
x
. The initial conditions are satisfied for all real values of
y
0
and
y
1
.
For
x
2
y
11
−
2
xy
1
+2
y
= 0 the auxiliary equation is
m
2
−
3
m
+2=(
m
−
1)(
m
−
2) = 0 and the general solution
is
y
=
c
1
x
+
c
2
x
2
. The initial condition
y
(0) =
y
0
implies 0 =
y
0
and the condition
y
1
(0) =
y
1
implies
c
1
=
y
1
.
Thus, the initial conditions are satisfied for
y
0
= 0 and for all real values of
y
1
.
For
x
2
y
11
−
4
xy
1
+6
y
= 0 the auxiliary equation is
m
2
−
5
m
+6=(
m
−
2)(
m
−
3) = 0 and the general solution
is
y
=
c
1
x
2
+
c
2
x
3
. The initial conditions imply
y
(0) = 0 =
y
0
and
y
1
(0) = 0. Thus, the initial conditions are
satisfied only for
y
0
=
y
1
=0.
42.
The function
y
(
x
)=
−
√
x
cos(ln
x
) is defined for
x>
0 and has
x
-intercepts where ln
x
=
π/
2+
kπ
for
k
an
integer or where
x
=
e
π/
2+
kπ
. Solving
π/
2+
kπ
=0
.
5weget
k
≈−
0
.
34, so
e
π/
2+
kπ
<
0
.
5 for all negative
integers and the graph has infinitely many
x
-intercepts in (0
,
0
.
5).
Exercises 3.7
1.
We have
y
1
1
=
y
11
1
=
e
x
,so
(
y
11
1
)
2
=(
e
x
)
2
=
e
2
x
=
y
2
1
.
Also,
y
1
2
=
−
sin
x
and
y
11
2
=
−
cos
x
,so
(
y
11
2
)
2
=(
−
cos
x
)
2
= cos
2
x
=
y
2
2
.
However, if
y
=
c
1
y
1
+
c
2
y
2
,wehave(
y
11
)
2
=(
c
1
e
x
−
c
2
cos
x
)
2
and
y
2
=(
c
1
e
x
+
c
2
cos
x
)
2
.Thus(
y
11
)
2
3
=
y
2
.
2.
We have
y
1
1
=
y
11
1
=0,so
y
1
y
11
1
=1
·
0=0=
1
2
(0)
2
=
1
2
(
y
1
1
)
2
.
Also,
y
1
2
=2
x
and
y
11
2
=2,so
y
2
y
11
2
=
x
2
(2) = 2
x
2
=
1
2
(2
x
)
2
=
1
2
(
y
1
2
)
2
.
However, if
y
=
c
1
y
1
+
c
2
y
2
, we have
yy
11
=(
c
1
·
1+
c
2
x
2
)(
c
1
·
0+2
c
2
)=2
c
2
(
c
1
+
c
2
x
2
) and
1
2
(
y
1
)
2
=
1
2
[
c
1
·
0+
c
2
(2
x
)]
2
=2
c
2
2
x
2
.Thus
yy
11
3
=
1
2
(
y
1
)
2
.
3.
Let
u
=
y
1
so that
u
1
=
y
11
. The equation becomes
u
1
=
−
u
−
1 which is separable. Thus
du
u
2
+1
=
−
dx
=
⇒
tan
−
1
u
=
−
x
+
c
1
=
⇒
y
1
= tan(
c
1
−
x
)=
⇒
y
=ln
|
cos(
c
1
−
x
)
|
+
c
2
.
4.
Let
u
=
y
1
so that
u
1
=
y
11
. The equation becomes
u
1
=1+
u
2
. Separating variables we obtain
du
1+
u
2
=
dx
=
⇒
tan
−
1
u
=
x
+
c
1
=
⇒
u
= tan(
x
+
c
1
)=
⇒
y
=
−
ln
|
cos(
x
+
c
1
)
|
+
c
2
.
114

115. x
y
−10
10
−π/2 3π/2
Exercises 3.7
5.
Let
u
=
y
1
so that
u
1
=
y
11
. The equation becomes
x
2
u
1
+
u
2
= 0. Separating variables we obtain
du
u
2
=
−
dx
x
2
=
⇒−
1
u
=
1
x
+
c
1
=
c
1
x
+1
x
=
⇒
u
=
−
1
c
1

x
x
+1
/c
1

=
1
c
1

1
c
1
x
+1
−
1

=
⇒
y
=
1
c
2
1
ln
|
c
1
x
+1
|−
1
c
1
x
+
c
2
.
6.
Let
u
=
y
1
so that
y
11
=
u
du
dy
. The equation becomes (
y
+1)
u
du
dy
=
u
2
. Separating variables we obtain
du
u
=
dy
y
+1
=
⇒
ln
|
u
|
=ln
|
y
+1
|
+ln
c
1
=
⇒
u
=
c
1
(
y
+1)
=
⇒
dy
dx
=
c
1
(
y
+1) =
⇒
dy
y
+1
=
c
1
dx
=
⇒
ln
|
y
+1
|
=
c
1
x
+
c
2
=
⇒
y
+1=
c
3
e
c
1
x
.
7.
Let
u
=
y
1
so that
y
11
=
u
du
dy
. The equation becomes
u
du
dy
+2
yu
3
= 0. Separating variables we obtain
du
u
2
+2
ydy
=0 =
⇒−
1
u
+
y
2
=
c
=
⇒
u
=
1
y
2
+
c
1
=
⇒
y
1
=
1
y
2
+
c
1
=
⇒
1
y
2
+
c
1
0
dy
=
dx
=
⇒
1
3
y
3
+
c
1
y
=
x
+
c
2
.
8.
Let
u
=
y
1
so that
y
11
=
u
du
dy
. The equation becomes
y
2
u
du
dy
=
u
. Separating variables we obtain
du
=
dy
y
2
=
⇒
u
=
−
1
y
+
c
1
=
⇒
y
1
=
c
1
y
−
1
y
=
⇒
y
c
1
y
−
1
dy
=
dx
=
⇒
1
c
1

1+
1
c
1
y
−
1

dy
=
dx
(for
c
1
/
=0) =
⇒
1
c
1
y
+
1
c
2
1
ln
|
y
−
1
|
=
x
+
c
2
.
If
c
1
= 0, then
ydy
=
−
dx
and another solution is
1
2
y
2
=
−
x
+
c
2
.
9.
Let
u
=
y
1
so that
y
11
=
u
du
dy
. The equation becomes
u
du
dy
+
yu
= 0. Separating variables
we obtain
du
=
−
ydy
=
⇒
u
=
−
1
2
y
2
+
c
1
=
⇒
y
1
=
−
1
2
y
2
+
c
1
.
When
x
=0,
y
= 1 and
y
1
=
−
1so
−
1=
−
1
2
+
c
1
and
c
1
=
−
1
2
. Then
dy
dx
=
−
1
2
y
2
−
1
2
=
⇒
dy
y
2
+1
=
−
1
2
dx
=
⇒
tan
−
1
y
=
−
1
2
x
+
c
2
=
⇒
y
= tan

−
1
2
x
+
c
2

.
When
x
=0,
y
= 1 so 1 = tan
c
2
and
c
2
=
π/
4. The solution of the initial-value problem
is
y
= tan

π
4
−
1
2
x

,
−
π
2
<x<
3
π
2
.
115

116. x
y
2
−2π 2π
x
y
−1
1
−2π 2π
Exercises 3.7
10.
Let
u
=
y
1
so that
u
1
=
y
11
. The equation becomes (
u
1
)
2
+
u
2
=1
which results in
u
1
=
±
√
1
−
u
2
. To solve
u
1
=
√
1
−
u
2
we separate
variables:
du
√
1
−
u
2
=
dx
=
⇒
sin
−
1
u
=
x
+
c
1
=
⇒
u
= sin(
x
+
c
1
)
=
⇒
y
1
= sin(
x
+
c
1
)
.
When
x
=
π
2
,
y
1
=
√
3
2
,so
√
3
2
= sin
3
π
2
+
c
1
4
and
c
1
=
−
π
6
.Thus
y
1
= sin
3
x
−
π
6
4
=
⇒
y
=
−
cos
3
x
−
π
6
4
+
c
2
.
When
x
=
π
2
,
y
=
1
2
,so
1
2
=
−
cos
3
π
2
−
π
6
4
+
c
2
=
−
1
2
+
c
2
and
c
2
= 1. The solution of the initial-value
problem is
y
=1
−
cos
3
x
−
π
6
4
.
To solve
u
1
=
−
√
1
−
u
2
we separate variables:
du
√
1
−
u
2
=
−
dx
=
⇒
cos
−
1
u
=
x
+
c
1
=
⇒
u
= cos(
x
+
c
1
)=
⇒
y
1
= cos(
x
+
c
1
)
.
When
x
=
π
2
,
y
1
=
√
3
2
,so
√
3
2
= cos
3
π
2
+
c
1
4
and
c
1
=
−
π
3
.Thus
y
1
= cos
3
x
−
π
3
4
=
⇒
y
= sin
3
x
−
π
3
4
+
c
2
.
When
x
=
π
2
,
y
=
1
2
,so
1
2
= sin
3
π
2
−
π
3
4
+
c
2
=
1
2
+
c
2
and
c
2
= 0. The solution of the initial-value
problem is
y
= sin
3
x
−
π
3
4
.
11.
Let
u
=
y
1
so that
u
1
=
y
11
. The equation becomes
u
1
−
1
x
u
=
1
x
u
3
, which is Bernoulli. Using
w
=
u
−
2
we obtain
dw
dx
+
2
x
w
=
−
2
x
. An integrating factor is
x
2
,so
d
dx
[
x
2
w
]=
−
2
x
=
⇒
x
2
w
=
−
x
2
+
c
1
=
⇒
w
=
−
1+
c
1
x
2
=
⇒
u
−
2
=
−
1+
c
1
x
2
=
⇒
u
=
x
√
c
1
−
x
2
=
⇒
dy
dx
=
x
√
c
1
−
x
2
=
⇒
y
=
−
)
c
1
−
x
2
+
c
2
=
⇒
c
1
−
x
2
=(
c
2
−
y
)
2
=
⇒
x
2
+(
c
2
−
y
)
2
=
c
1
.
12.
Let
u
=
y
1
so that
u
1
=
y
11
. The equation becomes
u
1
−
1
x
u
=
u
2
, which is Bernoulli. Using the substitution
w
=
u
−
1
we obtain
dw
dx
+
1
x
w
=
−
1. An integrating factor is
x
,so
d
dx
[
xw
]=
−
x
=
⇒
w
=
−
1
2
x
+
1
x
c
=
⇒
1
u
=
c
1
−
x
2
2
x
=
⇒
u
=
2
x
c
1
−
x
2
=
⇒
y
=
−
ln
2
2
c
1
−
x
2
2
2
+
c
2
.
116

117. 0.511.522.53
x
10
20
30
40
y
0.511.522.53
x
-10
-5
5
10
y
0.511.522.533.5
x
10
20
30
40
y
Exercises 3.7
In Problems 13-16 the thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the
Taylor polynomial.
13.
We look for a solution of the form
y
(
x
)=
y
(0) +
y
1
(0) +
1
2
y
11
(0) +
1
3!
y
111
(0) +
1
4!
y
(4)
(
x
)+
1
5!
y
(5)
(
x
)
.
From
y
11
(
x
)=
x
+
y
2
we compute
y
111
(
x
)=1+2
yy
1
y
(4)
(
x
)=2
yy
11
+2(
y
1
)
2
y
(5)
(
x
)=2
yy
111
+6
y
1
y
11
.
Using
y
(0) = 1 and
y
1
(0) = 1 we find
y
11
(0) = 1
,y
111
(0) = 3
,y
(4)
(0) = 4
,y
(5)
(0)=12
.
An approximate solution is
y
(
x
)=1+
x
+
1
2
x
2
+
1
2
x
3
+
1
6
x
4
+
1
10
x
5
.
14.
We look for a solution of the form
y
(
x
)=
y
(0) +
y
1
(0) +
1
2
y
11
(0) +
1
3!
y
111
(0) +
1
4!
y
(4)
(
x
)+
1
5!
y
(5)
(
x
)
.
From
y
11
(
x
)=1
−
y
2
we compute
y
111
(
x
)=
−
2
yy
1
y
(4)
(
x
)=
−
2
yy
11
−
2(
y
1
)
2
y
(5)
(
x
)=
−
2
yy
111
−
6
y
1
y
11
.
Using
y
(0) = 2 and
y
1
(0) = 3 we find
y
11
(0) =
−
3
,y
111
(0) =
−
12
,y
(4)
(0) =
−
6
,y
(5)
(0) = 102
.
An approximate solution is
y
(
x
)=2+3
x
−
3
2
x
2
−
2
x
3
−
1
4
x
4
+
17
20
x
5
.
15.
We look for a solution of the form
y
(
x
)=
y
(0) +
y
1
(0) +
1
2
y
11
(0) +
1
3!
y
111
(0) +
1
4!
y
(4)
(
x
)+
1
5!
y
(5)
(
x
)
.
From
y
11
(
x
)=
x
2
+
y
2
−
2
y
1
we compute
y
111
(
x
)=2
x
+2
yy
1
−
2
y
11
y
(4)
(
x
)=2+2(
y
1
)
2
+2
yy
11
−
2
y
111
y
(5)
(
x
)=6
y
1
y
11
+2
yy
111
−
2
y
(4)
.
Using
y
(0) = 1 and
y
1
(0) = 1 we find
y
11
(0) =
−
1
,y
111
(0) = 4
,y
(4)
(0) =
−
6
,y
(5)
(0) = 14
.
117

118. 12345
x
-4
-2
2
4
6
8
10
y
Exercises 3.7
An approximate solution is
y
(
x
)=1+
x
−
1
2
x
2
+
2
3
x
3
−
1
4
x
4
+
7
60
x
5
.
16.
We look for a solution of the form
y
(
x
)=
y
(0) +
y
1
(0) +
1
2
y
11
(0) +
1
3!
y
111
(0) +
1
4!
y
(4)
(
x
)+
1
5!
y
(5)
(
x
)+
1
6!
y
(6)
(
x
)
.
From
y
11
(
x
)=
e
y
we compute
y
111
(
x
)=
e
y
y
1
y
(4)
(
x
)=
e
y
(
y
1
)
2
+
e
y
y
11
y
(5)
(
x
)=
e
y
(
y
1
)
3
+3
e
y
y
1
y
11
+
e
y
y
111
y
(6)
(
x
)=
e
y
(
y
1
)
4
+6
e
y
(
y
1
)
2
y
11
+3
e
y
(
y
11
)
2
+4
e
y
y
1
y
111
+
e
y
y
(4)
.
Using
y
(0) = 0 and
y
1
(0) =
−
1 we find
y
11
(0) = 1
,y
111
(0) =
−
1
,y
(4)
(0) = 2
,y
(5)
(0) =
−
5
,y
(6)
(0) = 16
.
An approximate solution is
y
(
x
)=
−
x
+
1
2
x
2
−
1
6
x
3
+
1
12
x
4
+
1
24
x
5
+
1
45
x
6
.
17.
We need to solve
b
1+(
y
1
)
2
c
3
/
2
=
y
11
. Let
u
=
y
1
so that
u
1
=
y
11
. The equation becomes
1
1+
u
2
.
3
/
2
=
u
1
or
1
1+
u
2
.
3
/
2
=
du
dx
. Separating variables and using the substitution
u
= tan
θ
we have
du
(1 +
u
2
)
3
/
2
=
dx
=
⇒
6
sec
2
θ
1
1 + tan
2
θ
.
3
/
2
dθ
=
x
=
⇒
6
sec
2
θ
sec
3
θ
dθ
=
x
=
⇒
6
cos
θdθ
=
x
=
⇒
sin
θ
=
x
=
⇒
u
√
1+
u
2
=
x
=
⇒
y
1
)
1+(
y
1
)
2
=
x
=
⇒
(
y
1
)
2
=
x
2
b
1+(
y
1
)
2
c
=
x
2
1
−
x
2
=
⇒
y
1
=
x
√
1
−
x
2
(for
x>
0) =
⇒
y
=
−
)
1
−
x
2
.
18.
Let
u
=
dx
dt
so that
d
2
x
dt
2
=
u
du
dx
. The equation becomes
u
du
dx
=
−
k
2
x
2
. Separating variables we obtain
udu
=
−
k
2
x
2
dx
=
⇒
1
2
u
2
=
k
2
x
+
c
=
⇒
1
2
v
2
=
k
2
x
+
c.
When
t
=0,
x
=
x
0
and
v
=0so0=
k
2
x
0
+
c
and
c
=
−
k
2
x
0
. Then
1
2
v
2
=
k
2

1
x
−
1
x
0

and
dx
dt
=
−
k
√
2
s
x
0
−
x
xx
0
.
Separating variables we have
−
s
xx
0
x
0
−
x
dx
=
k
√
2
dt
=
⇒
t
=
−
1
k
s
x
0
2
6
s
x
x
0
−
x
dx.
118

119. x
1
=
0
t
x
10 20
-
2
2
x
1
=
1
t
x
10 20
-
2
2
x
1
=-
1.5
t
x
10 20
-
2
2
x
1
=
0
t
x
10
-
1
1
x
1
=
1
t
x
10
-
1
1
x
1
=-
2.5
t
x
10
-
1
1
Exercises 3.8
Using
Mathematica
to integrate we obtain
t
=
−
1
k
s
x
0
2

−
7
x
(
x
0
−
x
)
−
x
0
2
tan
−
1
(
x
0
−
2
x
)
2
x
s
x
x
0
−
x

=
1
k
s
x
0
2
I
7
x
(
x
0
−
x
)+
x
0
2
tan
−
1
x
0
−
2
x
2
7
x
(
x
0
−
x
)
f
.
19.
For
d
2
x
dt
2
+ sin
x
= 0 the motion appears to be periodic with amplitude 1 when
x
1
= 0. The amplitude and
period are larger for larger magnitudes of
x
1
.
For
d
2
x
dt
2
+
dx
dt
+ sin
x
= 0 the motion appears to be periodic with decreasing amplitude. The
dx/dt
term could
be said to have a damping effect.
20.
From (
y
11
)
2
−
y
2
= 0 we have
y
11
=
±
y
, which can be treated as two linear equations. Since linear combinations
of solutions of linear homogeneous differential equations are also solutions, we see that
y
=
c
1
e
x
+
c
2
e
−
x
and
y
=
c
3
cos
x
+
c
4
sin
x
must satisfy the differential equation.
21.
Letting
u
=
y
11
, separating variables, and integrating we have
du
dx
=
7
1+
u
2
,
du
√
1+
u
2
=
dx,
and sinh
−
1
u
=
x
+
c
1
.
Then
u
=
y
11
= sinh(
x
+
c
1
)
,y
1
= cosh(
x
+
c
1
)+
c
2
,
and
y
= sinh(
x
+
c
1
)+
c
2
x
+
c
3
.
Exercises 3.8
1.
From
1
8
x
1
+16
x
= 0 we obtain
x
=
c
1
cos 8
√
2
t
+
c
2
sin 8
√
2
t
so that the period of motion is 2
π/
8
√
2=
√
2
π/
8 seconds.
2.
From 20
x
11
+
kx
= 0 we obtain
x
=
c
1
cos
1
2
s
k
5
t
+
c
2
sin
1
2
s
k
5
t
119

120. Exercises 3.8
so that the frequency 2
/π
=
1
4
)
k/
5
π
and
k
= 320 N/m. If 80
x
11
+ 320
x
= 0 then
x
=
c
1
cos 2
t
+
c
2
sin 2
t
so
that the frequency is 2
/
2
π
=1
/π
vibrations/second.
3.
From
3
4
x
11
+72
x
=0,
x
(0) =
−
1
/
4, and
x
1
(0) = 0 we obtain
x
=
−
1
4
cos 4
√
6
t
.
4.
From
3
4
x
11
+72
x
=0,
x
(0) = 0, and
x
1
(0) = 2 we obtain
x
=
√
6
12
sin 4
√
6
t
.
5.
From
5
8
x
11
+40
x
=0,
x
(0)=1
/
2, and
x
1
(0) = 0 we obtain
x
=
1
2
cos 8
t
.
(a)
x
(
π/
12) =
−
1
/
4,
x
(
π/
8) =
−
1
/
2,
x
(
π/
6) =
−
1
/
4,
x
(
π/
8) = 1
/
2,
x
(9
π/
32) =
√
2
/
4.
(b)
x
1
=
−
4 sin 8
t
so that
x
1
(3
π/
16) = 4 ft/s directed downward.
(c)
If
x
=
1
2
cos 8
t
= 0 then
t
=(2
n
+1)
π/
16 for
n
=0,1,2,
...
.
6.
From 50
x
11
+ 200
x
=0,
x
(0) = 0, and
x
1
(0) =
−
10 we obtain
x
=
−
5 sin 2
t
and
x
1
=
−
10 cos 2
t
.
7.
From 20
x
11
+20
x
=0,
x
(0) = 0, and
x
1
(0) =
−
10 we obtain
x
=
−
10 sin
t
and
x
1
=
−
10 cos
t
.
(a)
The 20 kg mass has the larger amplitude.
(b)
20 kg:
x
1
(
π/
4) =
−
5
√
2 m/s,
x
1
(
π/
2) = 0 m/s; 50 kg:
x
1
(
π/
4) = 0 m/s,
x
1
(
π/
2) = 10 m/s
(c)
If
−
5 sin 2
t
=
−
10 sin
t
then 2 sin
t
(cos
t
−
1) = 0 so that
t
=
nπ
for
n
=0,1,2,
...
, placing both
masses at the equilibrium position. The 50 kg mass is moving upward; the 20 kg mass is moving
upward when
n
is even and downward when
n
is odd.
8.
From
x
11
+16
x
=0,
x
(0) =
−
1, and
x
1
(0) =
−
2 we obtain
x
=
−
cos 4
t
−
1
2
sin 4
t
=
√
5
2
cos(4
t
−
3
.
6)
.
The period is
π/
2 seconds and the amplitude is
√
5
/
2 feet. In 4
π
seconds it will make 8 complete vibrations.
9.
From
1
4
x
11
+
x
=0,
x
(0) = 1
/
2, and
x
1
(0) = 3
/
2 we obtain
x
=
1
2
cos 2
t
+
3
4
sin 2
t
=
√
13
4
sin(2
t
+0
.
588)
.
10.
From 1
.
6
x
11
+40
x
=0,
x
(0) =
−
1
/
3, and
x
1
(0)=5
/
4 we obtain
x
=
−
1
3
cos 5
t
+
1
4
sin 5
t
=
5
12
sin(5
t
+0
.
927)
.
If
x
=5
/
24 then
t
=
1
5
1
π
6
+0
.
927+2
nπ
.
and
t
=
1
5
1
5
π
6
+0
.
927 + 2
nπ
.
for
n
=0,1,2,
...
.
11.
From 2
x
11
+ 200
x
=0,
x
(0) =
−
2
/
3, and
x
1
(0) = 5 we obtain
(a)
x
=
−
2
3
cos 10
t
+
1
2
sin 10
t
=
5
6
sin(10
t
−
0
.
927).
(b)
The amplitude is 5
/
6 ft and the period is 2
π/
10 =
π/
5
(c)
3
π
=
πk/
5 and
k
= 15 cycles.
(d)
If
x
= 0 and the weight is moving downward for the second time, then 10
t
−
0
.
927 = 2
π
or
t
=0
.
721 s.
120

121. Exercises 3.8
(e)
If
x
1
=
25
3
cos(10
t
−
0
.
927) = 0 then 10
t
−
0
.
927 =
π/
2+
nπ
or
t
=(2
n
+1)
π/
20 + 0
.
0927 for
n
=0,1,2,
...
.
(f)
x
(3) =
−
0
.
597 ft
(g)
x
1
(3) =
−
5
.
814 ft/s
(h)
x
11
(3) = 59
.
702 ft/s
2
(i)
If
x
= 0 then
t
=
1
10
(0
.
927 +
nπ
) for
n
=0,1,2,
...
and
x
1
(
t
)=
±
25
3
ft/s.
(j)
If
x
=5
/
12 then
t
=
1
10
(
π/
6+0
.
927+2
nπ
) and
t
=
1
10
(5
π/
6+0
.
927+2
nπ
) for
n
=0,1,2,
...
.
(k)
If
x
=5
/
12 and
x
1
<
0 then
t
=
1
10
(5
π/
6+0
.
927+2
nπ
) for
n
=0,1,2,
...
.
12.
From
x
1
+9
x
=0,
x
(0) =
−
1, and
x
1
(0) =
−
√
3 we obtain
x
=
−
cos 3
t
−
√
3
3
sin 3
t
=
2
√
3
sin

37 +
4
π
3

and
x
1
=2
√
3 cos(3
t
+4
π/
3). If
x
1
= 3 then
t
=
−
7
π/
18 + 2
nπ/
3 and
t
=
−
π/
2+2
nπ/
3 for
n
=1,2,3,
...
.
13.
From
k
1
= 40 and
k
2
= 120 we compute the effective spring constant
k
= 4(40)(120)
/
160 = 120. Now,
m
=20
/
32 so
k/m
= 120(32)
/
20 = 192 and
x
11
+ 192
x
= 0. Using
x
(0) = 0 and
x
1
(0) = 2 we obtain
x
(
t
)=
√
3
12
sin 8
√
3
t
.
14.
Let
m
denote the mass in slugs of the first weight. Let
k
1
and
k
2
be the spring constants and
k
=4
k
1
k
2
/
(
k
1
+
k
2
)
the effective spring constant of the system. Now, the numerical value of the first weight is
W
=
mg
=32
m
,so
32
m
=
k
1

1
3

and 32
m
=
k
2

1
2

.
From these equations we find 2
k
1
=3
k
2
. The given period of the combined system is 2
π/w
=
π/
15, so
w
= 30.
Since the mass of an 8-pound weight is 1
/
4 slug, we have from
w
2
=
k/m
30
2
=
k
1
/
4
=4
k
or
k
= 225
.
We now have the system of equations
4
k
1
k
2
k
1
+
k
2
= 225
2
k
1
=3
k
2
.
Solving the second equation for
k
1
and substituting in the first equation, we obtain
4(3
k
2
/
2)
k
2
3
k
2
/
2+
k
2
=
12
k
2
2
5
k
2
=
12
k
2
5
= 225
.
Thus,
k
2
= 375
/
4 and
k
1
= 1125
/
8. Finally, the value of the first weight is
W
=32
m
=
k
1
3
=
1125
/
8
3
=
375
8
≈
46
.
88 lb
.
15.
For large values of
t
the differential equation is approximated by
x
11
= 0. The solution of this equation is the
linear function
x
=
c
1
t
+
c
2
. Thus, for large time, the restoring force will have decayed to the point where the
spring is incapable of returning the mass, and the spring will simply keep on stretching.
16.
As
t
becomes larger the spring constant increases; that is, the spring is stiffening. It would seem that the
oscillations would become periodic and the spring would oscillate more rapidly. It is likely that the amplitudes
of the oscillations would decrease as
t
increases.
17. (a)
above
(b)
heading upward
121

122. .5 2
t
-1
-.5
.5
1
x
Exercises 3.8
18. (a)
below
(b)
from rest
19. (a)
below
(b)
heading upward
20. (a)
above
(b)
heading downward
21.
From
1
8
x
11
+
x
1
+2
x
=0,
x
(0) =
−
1, and
x
1
(0) = 8 we obtain
x
=4
te
−
4
t
−
e
−
4
t
and
x
1
=8
e
−
4
t
−
16
te
−
4
t
.If
x
= 0 then
t
=1
/
4 second. If
x
1
= 0 then
t
=1
/
2 second and the extreme displacement is
x
=
e
−
2
feet.
22.
From
1
4
x
11
+
√
2
x
1
+2
x
=0,
x
(0) = 0, and
x
1
(0) = 5 we obtain
x
=5
te
−
2
√
2
t
and
x
1
=5
e
−
2
√
2
t
1
1
−
2
√
2
t
.
.If
x
1
= 0 then
t
=
√
2
/
4 second and the extreme displacement is
x
=5
√
2
e
−
1
/
4 feet.
23. (a)
From
x
11
+10
x
1
+16
x
=0,
x
(0) = 1, and
x
1
(0) = 0 we obtain
x
=
4
3
e
−
2
t
−
1
3
e
−
8
t
.
(b)
From
x
11
+
x
1
+16
x
=0,
x
(0) = 1, and
x
1
(0) =
−
12 then
x
=
−
2
3
e
−
2
t
+
5
3
e
−
8
t
.
24. (a)
x
=
1
3
e
−
8
t
1
4
e
6
t
−
1
.
is never zero; the extreme displacement is
x
(0) = 1 meter.
(b)
x
=
1
3
e
−
8
t
1
5
−
2
e
6
t
.
= 0 when
t
=
1
6
ln
5
2
≈
0
.
153 second; if
x
1
=
4
3
e
−
8
t
1
e
6
t
−
10
.
= 0 then
t
=
1
6
ln 10
≈
0
.
384 second and the extreme displacement is
x
=
−
0
.
232 meter.
25. (a)
From 0
.
1
x
11
+0
.
4
x
1
+2
x
=0,
x
(0) =
−
1, and
x
1
(0) = 0 we obtain =
e
−
2
t
b
−
cos 4
t
−
1
2
sin 4
t
c
.
(b)
x
=
e
−
2
t
√
5
2

−
2
√
5
cos 4
t
−
1
√
5
sin 4
t

=
√
5
2
e
−
2
t
sin(4
t
+4
.
25).
(c)
If
x
= 0 then 4
t
+4
.
25 = 2
π
,3
π
,4
π
,
...
so that the first time heading upward is
t
=1
.
294 seconds.
26. (a)
From
1
4
x
11
+
x
1
+5
x
=0,
x
(0) = 1
/
2, and
x
1
(0) = 1 we obtain
x
=
e
−
2
t
1
1
2
cos 4
t
+
1
2
sin 4
t
.
.
(b)
x
=
e
−
2
t
1
√
2

√
2
2
cos 4
t
+
√
2
2
sin 4
t
a
=
1
√
2
e
−
2
t
sin
3
4
t
+
π
4
4
.
(c)
If
x
= 0 then 4
t
+
π/
4=
π
,2
π
,3
π
,
...
so that the times heading downward are
t
=(7+8
n
)
π/
16 for
n
=0,
1, 2,
...
.
(d)
27.
From
5
16
x
11
+
βx
1
+5
x
= 0 we find that the roots of the auxiliary equation are
m
=
−
8
5
β
±
4
5
)
4
β
2
−
25 .
(a)
If 4
β
2
−
25
>
0 then
β>
5
/
2.
(b)
If 4
β
2
−
25 = 0 then
β
=5
/
2.
(c)
If 4
β
2
−
25
<
0 then 0
<β<
5
/
2.
28.
From 0
.
75
x
11
+
βx
1
+6
x
= 0 and
β>
3
√
2 we find that the roots of the auxiliary equation are
m
=
−
2
3
β
±
2
3
)
β
2
−
18 and
x
=
e
−
2
βt/
3

c
1
cosh
2
3
)
β
2
−
18
t
+
c
2
sinh
2
3
)
β
2
−
18
t

.
122

123. 2 4 6
t
-3
3
x
steady-state
transient
2 4 6
t
-3
3
x
x=x +x
c p
Exercises 3.8
If
x
(0) = 0 and
x
1
(0) =
−
2 then
c
1
= 0 and
c
2
=
−
3
/
)
β
2
−
18.
29.
If
1
2
x
11
+
1
2
x
1
+6
x
= 10 cos 3
t
,
x
(0) =
−
2, and
x
1
(0) = 0 then
x
c
=
e
−
t/
2

c
1
cos
√
47
2
t
+
c
2
sin
√
47
2
t
a
and
x
p
=
10
3
(cos 3
t
+ sin 3
t
) so that the equation of motion is
x
=
e
−
t/
2

−
4
3
cos
√
47
2
t
−
64
3
√
47
sin
√
47
2
t
a
+
10
3
(cos 3
t
+ sin 3
t
)
.
30. (a)
If
x
11
+2
x
1
+5
x
= 12 cos 2
t
+ 3 sin 2
t
,
x
(0) =
−
1, and
x
1
(0) = 5 then
x
c
=
e
−
t
(
c
1
cos 2
t
+
c
2
sin 2
t
) and
x
p
= 3 sin 2
t
so that the equation of motion is
x
=
e
−
t
cos 2
t
+ 3 sin 2
t.
(b) (c)
31.
From
x
11
+8
x
1
+16
x
= 8 sin 4
t
,
x
(0) = 0, and
x
1
(0) = 0 we obtain
x
c
=
c
1
e
−
4
t
+
c
2
te
−
4
t
and
x
p
=
−
1
4
cos 4
t
so
that the equation of motion is
x
=
1
4
e
−
4
t
+
te
−
4
t
−
1
4
cos 4
t.
32.
From
x
11
+8
x
1
+16
x
=
e
−
t
sin 4
t
,
x
(0) = 0, and
x
1
(0) = 0 we obtain
x
c
=
c
1
e
−
4
t
+
c
2
te
−
4
t
and
x
p
=
−
24
625
e
−
t
cos 4
t
−
7
625
e
−
t
sin 4
t
so that
x
=
1
625
e
−
4
t
(24 + 100
t
)
−
1
625
e
−
t
(24 cos 4
t
+ 7 sin 4
t
)
.
As
t
→∞
the displacement
x
→
0.
33.
From 2
x
11
+32
x
=68
e
−
2
t
cos 4
t
,
x
(0) = 0, and
x
1
(0) = 0 we obtain
x
c
=
c
1
cos 4
t
+
c
2
sin 4
t
and
x
p
=
1
2
e
−
2
t
cos 4
t
−
2
e
−
2
t
sin 4
t
so that
x
=
−
1
2
cos 4
t
+
9
4
sin 4
t
+
1
2
e
−
2
t
cos 4
t
−
2
e
−
2
t
sin 4
t.
34.
Since
x
=
√
85
4
sin(4
t
−
0
.
219)
−
√
17
2
e
−
2
t
sin(4
t
−
2
.
897), the amplitude approaches
√
85
/
4as
t
→∞
.
35. (a)
By Hooke’s law the external force is
F
(
t
)=
kh
(
t
) so that
mx
11
+
βx
1
+
kx
=
kh
(
t
).
(b)
From
1
2
x
11
+2
x
1
+4
x
= 20 cos
t
,
x
(0) = 0, and
x
1
(0) = 0 we obtain
x
c
=
e
−
2
t
(
c
1
cos 2
t
+
c
2
sin 2
t
) and
x
p
=
56
13
cos
t
+
32
13
sin
t
so that
x
=
e
−
2
t

−
56
13
cos 2
t
−
72
13
sin 2
t

+
56
13
cos
t
+
32
13
sin
t.
36. (a)
From 100
x
11
+ 1600
x
= 1600 sin 8
t
,
x
(0) = 0, and
x
1
(0) = 0 we obtain
x
c
=
c
1
cos 4
t
+
c
2
sin 4
t
and
x
p
=
−
1
3
sin 8
t
so that
x
=
2
3
sin 4
t
−
1
3
sin 8
t.
123

124. 1 2 3
t
-1
1
x
t
x
−1
1
π9π
Exercises 3.8
(b)
If
x
=
1
3
sin 4
t
(2
−
2 cos 4
t
) = 0 then
t
=
nπ/
4 for
n
=0,1,2,
...
.
(c)
If
x
1
=
8
3
cos 4
t
−
8
3
cos 8
t
=
8
3
(1
−
cos 4
t
)(1 + 2 cos 4
t
) = 0 then
t
=
π/
3+
nπ/
2 and
t
=
π/
6+
nπ/
2 for
n
=0,1,2,
...
at the extreme values.
Note
: There are many other values of
t
for which
x
1
=0.
(d)
x
(
π/
6+
nπ/
2) =
√
3
/
2 cm. and
x
(
π/
3+
nπ/
2) =
−
√
3
/
2 cm.
(e)
37.
From
x
11
+4
x
=
−
5 sin 2
t
+ 3 cos 2
t
,
x
(0) =
−
1, and
x
1
(0) = 1 we obtain
x
c
=
c
1
cos 2
t
+
c
2
sin 2
t
,
x
p
=
3
4
t
sin 2
t
+
5
4
t
cos 2
t
, and
x
=
−
cos 2
t
−
1
8
sin 2
t
+
3
4
t
sin 2
t
+
5
4
t
cos 2
t.
38.
From
x
11
+9
x
= 5 sin 3
t
,
x
(0) = 2, and
x
1
(0) = 0 we obtain
x
c
=
c
1
cos 3
t
+
c
2
sin 3
t
,
x
p
=
−
5
6
t
cos 3
t
, and
x
= 2 cos 3
t
+
5
18
sin 3
t
−
5
6
t
cos 3
t.
39. (a)
From
x
11
+
ω
2
x
=
F
0
cos
γt
,
x
(0) = 0, and
x
1
(0) = 0 we obtain
x
c
=
c
1
cos
ωt
+
c
2
sin
ωt
and
x
p
=(
F
0
cos
γt
)
/
1
ω
2
−
γ
2
0
so that
x
=
−
F
0
ω
2
−
γ
2
cos
ωt
+
F
0
ω
2
−
γ
2
cos
γt.
(b)
lim
γ
→
ω
F
0
ω
2
−
γ
2
(cos
γt
−
cos
ωt
) = lim
γ
→
ω
−
F
0
t
sin
γt
−
2
γ
=
F
0
2
ω
t
sin
ωt
.
40.
From
x
11
+
ω
2
x
=
F
0
cos
ωt
,
x
(0) = 0, and
x
1
(0) = 0 we obtain
x
c
=
c
1
cos
ωt
+
c
2
sin
ωt
and
x
p
=(
F
0
t/
2
ω
) sin
ωt
so that
x
=(
F
0
t/
2
ω
) sin
ωt
and lim
γ
→
ω
F
0
2
ω
t
sin
ωt
=
F
0
2
ω
t
sin
ωt
.
41. (a)
From cos(
u
−
v
) = cos
u
cos
v
+ sin
u
sin
v
and cos(
u
+
v
) = cos
u
cos
v
−
sin
u
sin
v
we obtain sin
u
sin
v
=
1
2
[cos(
u
−
v
)
−
cos(
u
+
v
)]. Letting
u
=
1
2
(
γ
−
ω
)
t
and
v
=
1
2
(
γ
+
ω
)
t
, the result follows.
(b)
If
+
=
1
2
(
γ
−
ω
) then
γ
≈
ω
so that
x
=(
F
0
/
2
+γ
) sin
+t
sin
γt
.
42.
See the article ”Distinguished Oscillations of a Forced Harmonic Oscillator” by T.G. Procter in
The College
Mathematics Journal
, March, 1995. In this article the author illustrates that for
F
0
=1,
λ
=0
.
01,
γ
=22
/
9,
and
ω
= 2 the system exhibits beats oscillations on the interval [0
,
9
π
], but that this phenomenon is transient
as
t
→∞
.
124

125. βγ1
g
2.00 1.41 0.58
1.00 1.87 1.03
0.75 1.93 1.36
0.50 1.97 2.02
0.25 1.99 4.01
1 2 3 4
γ
1
2
3
4
g
β=0.25
β=0.5
β=0.75
β=1
β=2
Exercises 3.8
43. (a)
The general solution of the homogeneous equation is
x
c
(
t
)=
c
1
e
−
λt
cos(
7
ω
2
−
λ
2
t
)+
c
2
e
−
λt
sin(
7
ω
2
−
λ
2
t
)
=
Ae
−
λt
sin[
7
ω
2
−
λ
2
t
+
φ
]
,
where
A
=
7
c
2
1
+
c
2
2
, sin
φ
=
c
1
/A
, and cos
φ
=
c
2
/A
.Now
x
p
(
t
)=
F
0
(
ω
2
−
γ
2
)
(
ω
2
−
γ
2
)
2
+4
λ
2
γ
2
sin
γt
+
F
0
(
−
2
λγ
)
(
ω
2
−
γ
2
)
2
+4
λ
2
γ
2
cos
γt
=
A
sin(
γt
+
θ
)
,
where
sin
θ
=
F
0
(
−
2
λγ
)
(
ω
2
−
γ
2
)
2
+4
λ
2
γ
2
F
0
7
ω
2
−
γ
2
+4
λ
2
γ
2
=
−
2
λγ
7
(
ω
2
−
γ
2
)
2
+4
λ
2
γ
2
and
cos
θ
=
F
0
(
ω
2
−
γ
2
)
(
ω
2
−
γ
2
)
2
+4
λ
2
γ
2
F
0
7
(
ω
2
−
γ
2
)
2
+4
λ
2
γ
2
=
ω
2
−
γ
2
7
(
ω
2
−
γ
2
)
2
+4
λ
2
γ
2
.
(b)
If
g
1
(
γ
) = 0 then
γ
1
γ
2
+2
λ
2
−
ω
2
0
= 0 so that
γ
=0or
γ
=
√
ω
2
−
2
λ
2
. The first derivative test shows
that
g
has a maximum value at
γ
=
√
ω
2
−
2
λ
2
. The maximum value of
g
is
g
/
7
ω
2
−
2
λ
2
2
=
F
0
/
2
λ
7
ω
2
−
λ
2
.
(c)
We identify
ω
2
=
k/m
=4,
λ
=
β/
2, and
γ
1
=
√
ω
2
−
2
λ
2
=
7
4
−
β
2
/
2. As
β
→
0,
γ
1
→
2 and the
resonance curve grows without bound at
γ
1
= 2. That is, the system approaches pure resonance.
44. (a)
For
n
= 2, sin
2
γt
=
1
2
(1
−
cos 2
γt
). The system is in pure resonance when 2
γ
1
/
2
π
=
ω/
2
π
, or when
γ
1
=
ω/
2.
(b)
Note that
sin
3
γt
= sin
γt
sin
2
γt
=
1
2
[sin
γt
−
sin
γt
cos 2
γt
]
.
Now
sin(
A
+
B
) + sin(
A
−
B
) = 2 sin
A
cos
B
125

126. 102030
t
γ1=1/2
-10
-5
5
10
x
n=2
10 20 30
t
γ1=1
-10
-5
5
10
x
n=3
20 40
t
γ2=1/3
-10
-5
5
10
x
n=3
Exercises 3.8
so
sin
γt
cos 2
γt
=
1
2
[sin 3
γt
−
sin
γt
]
and
sin
3
γt
=
3
4
sin
γt
−
1
4
sin 3
γt.
Thus
x
11
+
ω
2
x
=
3
4
sin
γt
−
1
4
sin 3
γt.
The frequency of free vibration is
ω/
2
π
. Thus, when
γ
1
/
2
π
=
ω/
2
π
or
γ
1
=
ω
, and when 3
γ
2
/
2
π
=
ω/
2
π
or 3
γ
2
=
ω
or
γ
3
=
ω/
3, the system will be in pure resonance.
(c)
45.
Solving
1
20
q
11
+2
q
1
+ 100
q
= 0 we obtain
q
(
t
)=
e
−
20
t
(
c
1
cos 40
t
+
c
2
sin 40
t
). The initial conditions
q
(0) = 5 and
q
1
(0) = 0 imply
c
1
= 5 and
c
2
=5
/
2. Thus
q
(
t
)=
e
−
20
t

5 cos 40
t
+
5
2
sin 40
t

≈
7
25+25
/
4
e
−
20
t
sin(40
t
+1
.
1071)
and
q
(0
.
01)
≈
4
.
5676 coulombs. The charge is zero for the first time when 40
t
+0
.
4636 =
π
or
t
≈
0
.
0509
second.
46.
Solving
1
4
q
11
+20
q
1
+ 300
q
= 0 we obtain
q
(
t
)=
c
1
e
−
20
t
+
c
2
e
−
60
t
. The initial conditions
q
(0) = 4 and
q
1
(0) = 0
imply
c
1
= 6 and
c
2
=
−
2. Thus
q
(
t
)=6
e
−
20
t
−
2
e
−
60
t
.
Setting
q
= 0 we find
e
40
t
=1
/
3 which implies
t<
0. Therefore the charge is never 0.
47.
Solving
5
3
q
11
+10
q
1
+30
q
= 300 we obtain
q
(
t
)=
e
−
3
t
(
c
1
cos 3
t
+
c
2
sin 3
t
) + 10. The initial conditions
q
(0) =
q
1
(0) = 0 imply
c
1
=
c
2
=
−
10. Thus
q
(
t
)=10
−
10
e
−
3
t
(cos 3
t
+ sin 3
t
) and
i
(
t
)=60
e
3
t
sin 3
t.
Solving
i
(
t
) = 0 we see that the maximum charge occurs when
t
=
π/
3 and
q
(
π/
3)
≈
10
.
432 coulombs.
48.
Solving
q
11
+ 100
q
1
+ 2500
q
= 30 we obtain
q
(
t
)=
c
1
e
−
50
t
+
c
2
te
−
50
t
+0
.
012. The initial conditions
q
(0)=0
and
q
1
(0) = 2 imply
c
1
=
−
0
.
012 and
c
2
=1
.
4. Thus
q
(
t
)=
−
0
.
012
e
−
50
t
+1
.
4
te
−
50
t
+0
.
012 and
i
(
t
)=2
e
−
50
t
−
70
te
−
50
t
.
Solving
i
(
t
) = 0 we see that the maximum charge occurs when
t
=1
/
35 and
q
(1
/
35)
≈
0
.
01871.
49.
Solving
q
11
+2
q
1
+4
q
= 0 we obtain
y
c
=
e
−
t
1
cos
√
3
t
+ sin
√
3
t
0
. The steady-state charge has the form
y
p
=
A
cos
t
+
B
sin
t
. Substituting into the differential equation we find
(3
A
+2
B
) cos
t
+(3
B
−
2
A
) sin
t
= 50 cos
t.
Thus,
A
= 150
/
13 and
B
= 100
/
13. The steady-state charge is
q
p
(
t
)=
150
13
cos
t
+
100
13
sin
t
126

127. Exercises 3.8
and the steady-state current is
i
p
(
t
)=
−
150
13
sin
t
+
100
13
cos
t.
50.
From
i
p
(
t
)=
E
0
Z

R
Z
sin
γt
−
X
Z
cos
γt

and
Z
=
√
X
2
+
R
2
we see that the amplitude of
i
p
(
t
)is
A
=
s
E
2
0
R
2
Z
4
+
E
2
0
X
2
Z
4
=
E
0
Z
2
)
R
2
+
X
2
=
E
0
Z
.
51.
The differential equation is
1
2
q
11
+20
q
1
+ 1000
q
= 100 sin
t
. To use Example 11 in the text we identify
E
0
= 100
and
γ
= 60. Then
X
=
Lγ
−
1
cγ
=
1
2
(60)
−
1
0
.
001(60)
≈
13
.
3333
,
Z
=
)
X
2
+
R
2
=
)
X
2
+ 400
≈
24
.
0370
,
and
E
0
Z
=
100
Z
≈
4
.
1603
.
From Problem 50, then
i
p
(
t
)
≈
4
.
1603(60
t
+
φ
)
where sin
φ
=
−
X/Z
and cos
φ
=
R/Z
. Thus tan
φ
=
−
X/R
≈−
0
.
6667 and
φ
is a fourth quadrant angle. Now
φ
≈−
0
.
5880 and
i
p
(
t
)
≈
4
.
1603(60
t
−
0
.
5880)
.
52.
Solving
1
2
q
11
+20
q
1
+ 1000
q
= 0 we obtain
q
c
(
t
)=(
c
1
cos 40
t
+
c
2
sin 40
t
). The steady-state charge has the form
q
p
(
t
)=
A
sin 60
t
+
B
cos 60
t
+
C
sin 40
t
+
D
cos 40
t
. Substituting into the differential equation we find
(
−
1600
A
−
2400
B
) sin 60
t
+ (2400
A
−
1600
B
) cos 60
t
+ (400
C
−
1600
D
) sin 40
t
+ (1600
C
+ 400
D
) cos 40
t
= 200 sin 60
t
+ 400 cos 40
t.
Equating coefficients we obtain
A
=
−
1
/
26,
B
=
−
3
/
52,
C
=4
/
17, and
D
=1
/
17. The steady-state charge is
q
p
(
t
)=
−
1
26
sin 60
t
−
3
52
cos 60
t
+
4
17
sin 40
t
+
1
17
cos 40
t
and the steady-state current is
i
p
(
t
)=
−
30
13
cos 60
t
+
45
13
sin 60
t
+
160
17
cos 40
t
−
40
17
sin 40
t.
53.
Solving
1
2
q
11
+10
q
1
+ 100
q
= 150 we obtain
q
(
t
)=
e
−
10
t
(
c
1
cos 10
t
+
c
2
sin 10
t
)+3
/
2. The initial conditions
q
(0) = 1 and
q
1
(0) = 0 imply
c
1
=
c
2
=
−
1
/
2. Thus
q
(
t
)=
−
1
2
e
−
10
t
(cos 10
t
+ sin 10
t
)+
3
2
.
As
t
→∞
,
q
(
t
)
→
3
/
2.
54.
By Problem 50 the amplitude of the steady-state current is
E
0
/Z
, where
Z
=
√
X
2
+
R
2
and
X
=
Lγ
−
1
/Cγ
.
Since
E
0
is constant the amplitude will be a maximum when
Z
is a minimum. Since
R
is constant,
Z
will be a
127

128. Exercises 3.8
minimum when
X
= 0. Solving
Lγ
−
1
/Cγ
= 0 for
γ
we obtain
γ
=1
/
√
LC
. The maximum amplitude will be
E
0
/R
.
55.
By Problem 50 the amplitude of the steady-state current is
E
0
/Z
, where
Z
=
√
X
2
+
R
2
and
X
=
Lγ
−
1
/Cγ
.
Since
E
0
is constant the amplitude will be a maximum when
Z
is a minimum. Since
R
is constant,
Z
will be a
minimum when
X
= 0. Solving
Lγ
−
1
/Cγ
= 0 for
C
we obtain
C
=1
/Lγ
2
.
56.
Solving 0
.
1
q
11
+10
q
= 100 sin
γt
we obtain
q
(
t
)=
c
1
cos 10
t
+
c
2
sin 10
t
+
q
p
(
t
) where
q
p
(
t
)=
A
sin
γt
+
B
cos
γt
.
Substituting
q
p
(
t
) into the differential equation we find
(100
−
γ
2
)
A
sin
γt
+ (100
−
γ
2
)
B
cos
γt
= 100 sin
γt.
Equating coefficients we obtain
A
= 100
/
(100
−
γ
2
) and
B
= 0. Thus,
q
p
(
t
)=
100
100
−
γ
2
sin
γt
. The initial
conditions
q
(0) =
q
1
(0) = 0 imply
c
1
= 0 and
c
2
=
−
10
γ/
(100
−
γ
2
). The charge is
q
(
t
)=
10
100
−
γ
2
(10 sin
γt
−
γ
sin 10
t
)
and the current is
i
(
t
)=
100
γ
100
−
γ
2
(cos
γt
−
cos 10
t
)
.
57.
In an
L
-
C
series circuit there is no resistor, so the differential equation is
L
d
2
q
dt
2
+
1
C
q
=
E
(
t
)
.
Then
q
(
t
)=
c
1
cos
3
t/
√
LC
4
+
c
2
sin
3
t/
√
LC
4
+
q
p
(
t
) where
q
p
(
t
)=
A
sin
γt
+
B
cos
γt
. Substituting
q
p
(
t
)into
the differential equation we find

1
C
−
Lγ
2

A
sin
γt
+

1
C
−
Lγ
2

B
cos
γt
=
E
0
cos
γt.
Equating coefficients we obtain
A
= 0 and
B
=
E
0
C/
(1
−
LCγ
2
). Thus, the charge is
q
(
t
)=
c
1
cos
1
√
LC
t
+
c
2
sin
1
√
LC
t
+
E
0
C
1
−
LCγ
2
cos
γt.
The initial conditions
q
(0) =
q
0
and
q
1
(0) =
i
0
imply
c
1
=
q
o
−
E
0
C/
(1
−
LCγ
2
) and
c
2
=
i
0
√
LC
. The current
is
i
(
t
)=
−
c
1
√
LC
sin
1
√
LC
t
+
c
2
√
LC
cos
1
√
LC
t
−
E
0
Cγ
1
−
LCγ
2
sin
γt
=
i
0
cos
1
√
LC
t
−
1
√
LC

q
0
−
E
0
C
1
−
LCγ
2

sin
1
√
LC
t
−
E
0
Cγ
1
−
LCγ
2
sin
γt.
58.
When the circuit is in resonance the form of
q
p
(
t
)is
q
p
(
t
)=
At
cos
kt
+
Bt
sin
kt
where
k
=1
/
√
LC
. Substituting
q
p
(
t
) into the differential equation we find
q
11
p
+
k
2
q
=
−
2
kA
sin
kt
+2
kB
cos
kt
=
E
0
L
cos
kt.
Equating coefficients we obtain
A
= 0 and
B
=
E
0
/
2
kL
. The charge is
q
(
t
)=
c
1
cos
kt
+
c
2
sin
kt
+
E
0
2
kL
t
sin
kt.
128

129. 0.20.40.60.81
x
1
2
3y
Exercises 3.9
The initial conditions
q
(0) =
q
0
and
q
1
(0) =
i
0
imply
c
1
=
q
0
and
c
2
=
i
0
/k
. The current is
i
(
t
)=
−
c
1
k
sin
kt
+
c
2
k
cos
kt
+
E
0
2
kL
(
kt
cos
kt
+ sin
kt
)
=

E
0
2
kL
−
q
0
k

sin
kt
+
i
0
cos
kt
+
E
0
2
L
t
cos
kt.
Exercises 3.9
1. (a)
The general solution is
y
(
x
)=
c
1
+
c
2
x
+
c
3
x
2
+
c
4
x
3
+
w
0
24
EI
x
4
.
The boundary conditions are
y
(0) = 0,
y
1
(0) = 0,
y
11
(
L
)=0,
y
111
(
L
) = 0. The first two conditions give
c
1
= 0 and
c
2
= 0. The conditions at
x
=
L
give the system
2
c
3
+6
c
4
L
+
w
0
2
EI
L
2
=0
6
c
4
+
w
0
EI
L
=0
.
Solving, we obtain
c
3
=
w
0
L
2
/
4
EI
and
c
4
=
−
w
0
L/
6
EI
. The deflection is
y
(
x
)=
w
0
24
EI
(6
L
2
x
2
−
4
Lx
3
+
x
4
)
.
(b)
2. (a)
The general solution is
y
(
x
)=
c
1
+
c
2
x
+
c
3
x
2
+
c
4
x
3
+
w
0
24
EI
x
4
.
The boundary conditions are
y
(0) = 0,
y
11
(0) = 0,
y
(
L
)=0,
y
11
(
L
) = 0. The first two conditions give
c
1
=0
and
c
3
= 0. The conditions at
x
=
L
give the system
c
2
L
+
c
4
L
3
+
w
0
24
EI
L
4
=0
6
c
4
L
+
w
0
2
EI
L
2
=0
.
Solving, we obtain
c
2
=
w
0
L
3
/
24
EI
and
c
4
=
−
w
0
L/
12
EI
. The deflection is
y
(
x
)=
w
0
24
EI
(
L
3
x
−
2
Lx
3
+
x
4
)
.
129

130. 0.20.40.60.81
x
1y
0.20.40.60.81
x
1
y
Exercises 3.9
(b)
3. (a)
The general solution is
y
(
x
)=
c
1
+
c
2
x
+
c
3
x
2
+
c
4
x
3
+
w
0
24
EI
x
4
.
The boundary conditions are
y
(0) = 0,
y
1
(0) = 0,
y
(
L
)=0,
y
11
(
L
) = 0. The first two conditions give
c
1
=0
and
c
2
= 0. The conditions at
x
=
L
give the system
c
3
L
2
+
c
4
L
3
+
w
0
24
EI
L
4
=0
2
c
3
+6
c
4
L
+
w
0
2
EI
L
2
=0
.
Solving, we obtain
c
3
=
w
0
L
2
/
16
EI
and
c
4
=
−
5
w
0
L/
48
EI
. The deflection is
y
(
x
)=
w
0
48
EI
(3
L
2
x
2
−
5
Lx
3
+2
x
4
)
.
(b)
4. (a)
The general solution is
y
(
x
)=
c
1
+
c
2
x
+
c
3
x
2
+
c
4
x
3
+
w
0
L
4
EIπ
4
sin
π
L
x.
The boundary conditions are
y
(0) = 0,
y
1
(0) = 0,
y
(
L
)=0,
y
11
(
L
) = 0. The first two conditions give
c
1
=0
and
c
2
=
−
w
0
L
3
/EIπ
3
. The conditions at
x
=
L
give the system
c
3
L
2
+
c
4
L
3
+
w
0
EIπ
3
L
4
=0
2
c
3
+6
c
4
L
=0
.
Solving, we obtain
c
3
=3
w
0
L
2
/
2
EIπ
3
and
c
4
=
−
w
0
L/
2
EIπ
3
. The deflection is
y
(
x
)=
w
0
L
2
EIπ
3

−
2
L
2
x
+3
Lx
2
−
x
3
+
2
L
3
π
sin
π
L
x

.
130

131. 0.20.40.60.81
x
1y
Exercises 3.9
(b)
5. (a)
y
max
=
y
(
L
)=
w
0
L
4
8
EI
(b)
Replacing both
L
and
x
by
L/
2in
y
(
x
) we obtain
w
0
L
4
/
128
EI
, which is 1
/
16 of the maximum deflection
when the length of the beam is
L
.
6. (a)
y
max
=
y
(
L/
2) =
5
w
0
L
4
384
EI
(b)
The maximum deflection of the beam in Example 1 is
y
(
L/
2) = (
w
0
/
24
EI
)
L
4
/
16 =
w
0
L
4
/
384
EI
, which is
1
/
5 of the maximum displacement of the beam in Problem 2.
7.
The general solution of the differential equation is
y
=
c
1
cosh
s
P
EI
x
+
c
2
sinh
s
P
EI
x
+
w
0
2
P
x
2
+
w
0
EI
P
2
.
Setting
y
(0) = 0 we obtain
c
1
=
−
w
0
EI/P
2
, so that
y
=
−
w
0
EI
P
2
cosh
s
P
EI
x
+
c
2
sinh
s
P
EI
x
+
w
0
2
P
x
2
+
w
0
EI
P
2
.
Setting
y
1
(
L
)=0wefind
c
2
=

s
P
EI
w
0
EI
P
2
sinh
s
P
EI
L
−
w
0
L
P
ag
s
P
EI
cosh
s
P
EI
L.
8.
The general solution of the differential equation is
y
=
c
1
cos
s
P
EI
x
+
c
2
sin
s
P
EI
x
+
w
0
2
P
x
2
+
w
0
EI
P
2
.
Setting
y
(0) = 0 we obtain
c
1
=
−
w
0
EI/P
2
, so that
y
=
−
w
0
EI
P
2
cos
s
P
EI
x
+
c
2
sin
s
P
EI
x
+
w
0
2
P
x
2
+
w
0
EI
P
2
.
Setting
y
1
(
L
)=0wefind
c
2
=

−
s
P
EI
w
0
EI
P
2
sin
s
P
EI
L
−
w
0
L
P
ag
s
P
EI
cos
s
P
EI
L.
9.
For
λ
≤
0 the only solution of the boundary-value problem is
y
=0. For
λ>
0 we have
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx.
Now
y
(0) = 0 implies
c
1
=0,so
y
(
π
)=
c
2
sin
√
λπ
=0
gives
√
λπ
=
nπ
or
λ
=
n
2
,n
=1
,
2
,
3
,... .
131

132. Exercises 3.9
The eigenvalues
n
2
correspond to the eigenfunctions sin
nx
for
n
=1,2,3,
...
.
10.
For
λ
≤
0 the only solution of the boundary-value problem is
y
=0. For
λ>
0 we have
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx.
Now
y
(0) = 0 implies
c
1
=0,so
y
3
π
4
4
=
c
2
sin
√
λ
π
4
=0
gives
√
λ
π
4
=
nπ
or
λ
=16
n
2
,n
=1
,
2
,
3
,... .
The eigenvalues 16
n
2
correspond to the eigenfunctions sin 4
nx
for
n
=1,2,3,
...
.
11.
For
λ
≤
0 the only solution of the boundary-value problem is
y
=0. For
λ>
0 we have
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx.
Now
y
1
(
x
)=
−
c
1
√
λ
sin
√
λx
+
c
2
√
λ
cos
√
λx
and
y
1
(0) = 0 implies
c
2
=0,so
y
(
L
)=
c
1
cos
√
λL
=0
gives
√
λL
=
(2
n
−
1)
π
2
or
λ
=
(2
n
−
1)
2
π
2
4
L
2
,n
=1
,
2
,
3
,... .
The eigenvalues (2
n
−
1)
2
π
2
/
4
L
2
correspond to the eigenfunctions cos
(2
n
−
1)
π
2
L
x
for
n
=1,2,3,
...
.
12.
For
λ
≤
0 the only solution of the boundary-value problem is
y
=0. For
λ>
0 we have
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx.
Now
y
(0) = 0 implies
c
1
=0,so
y
1
3
π
2
4
=
c
2
√
λ
cos
√
λ
π
2
=0
gives
√
λ
π
2
=
(2
n
−
1)
π
2
or
λ
=(2
n
−
1)
2
,n
=1
,
2
,
3
,... .
The eigenvalues (2
n
−
1)
2
correspond to the eigenfunctions sin(2
n
−
1)
x
.
13.
For
λ<
0 the only solution of the boundary-value problem is
y
=0. For
λ
= 0 we have
y
=
c
1
x
+
c
2
.Now
y
1
=
c
1
and
y
1
(0) = 0 implies
c
1
= 0. Then
y
=
c
2
and
y
1
(
π
) = 0. Thus,
λ
= 0 is an eigenvalue with corresponding
eigenfunction
y
=1.
For
λ>
0wehave
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx.
Now
y
1
(
x
)=
−
c
1
√
λ
sin
√
λx
+
c
2
√
λ
cos
√
λx
and
y
1
(0) = 0 implies
c
2
=0,so
y
1
(
π
)=
−
c
1
√
λ
sin
√
λπ
=0
gives
√
λπ
=
nπ
or
λ
=
n
2
,n
=1
,
2
,
3
,... .
The eigenvalues
n
2
correspond to the eigenfunctions cos
nx
for
n
=0,1,2,
...
.
132

133. Exercises 3.9
14.
For
λ
≤
0 the only solution of the boundary-value problem is
y
=0. For
λ>
0 we have
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx.
Now
y
(
−
π
)=
y
(
π
) = 0 implies
c
1
cos
√
λπ
−
c
2
sin
√
λπ
=0
c
1
cos
√
λπ
+
c
2
sin
√
λπ
=0
.
(
1
)
This homogeneous system will have a nontrivial solution when
2
2
2
2
cos
√
λπ
−
sin
√
λπ
cos
√
λπ
sin
√
λπ
2
2
2
2
= 2 sin
√
λπ
cos
√
λπ
= sin 2
√
λπ
=0
.
Then
2
√
λπ
=
nπ
or
λ
=
n
2
4
;
n
=1
,
2
,
3
,....
When
n
=2
k
−
1 is odd, the eigenvalues are (2
k
−
1)
2
/
4. Since cos(2
k
−
1)
π/
2 = 0 and sin(2
k
−
1)
π/
2
3
=0,
we see from either equation in (1) that
c
2
= 0. Thus, the eigenfunctions corresponding to the eigenvalues
(2
k
−
1)
2
/
4 are
y
= cos(2
k
−
1)
x/
2 for
k
=1,2,3,
...
. Similarly, when
n
=2
k
is even, the eigenvalues are
k
2
with corresponding eigenfunctions
y
= sin
kx
for
k
=1,2,3,
...
.
15.
The auxiliary equation has solutions
m
=
1
2
3
−
2
±
)
4
−
4(
λ
+1)
4
=
−
1
±
√
−
λ.
For
λ<
0wehave
y
=
e
−
x
3
c
1
cosh
√
−
λx
+
c
2
sinh
√
−
λx
4
.
The boundary conditions imply
y
(0) =
c
1
=0
y
(5) =
c
2
e
−
5
sinh 5
√
−
λ
=0
so
c
1
=
c
2
= 0 and the only solution of the boundary-value problem is
y
=0.
For
λ
=0wehave
y
=
c
1
e
−
x
+
c
2
xe
−
x
and the only solution of the boundary-value problem is
y
=0.
For
λ>
0wehave
y
=
e
−
x
3
c
1
cos
√
λx
+
c
2
sin
√
λx
4
.
Now
y
(0) = 0 implies
c
1
=0,so
y
(5) =
c
2
e
−
5
sin 5
√
λ
=0
gives
5
√
λ
=
nπ
or
λ
=
n
2
π
2
25
,n
=1
,
2
,
3
,... .
The eigenvalues
n
2
π
2
/
25 correspond to the eigenfunctions
e
−
x
sin
nπ
5
x
for
n
=1,2,3,
...
.
16.
For
λ<
−
1 the only solution of the boundary-value problem is
y
=0. For
λ
=
−
1wehave
y
=
c
1
x
+
c
2
.
Now
y
1
=
c
1
and
y
1
(0) = 0 implies
c
1
= 0. Then
y
=
c
2
and
y
1
(1) = 0. Thus,
λ
=
−
1 is an eigenvalue with
corresponding eigenfunction
y
=1.
For
λ>
−
1wehave
y
=
c
1
cos
√
λ
+1
x
+
c
2
sin
√
λ
+1
x.
133

134. Exercises 3.9
Now
y
1
=
−
c
1
√
λ
+ 1 sin
√
λ
+1
x
+
c
2
√
λ
+ 1 cos
√
λ
+1
x
and
y
1
(0) = 0 implies
c
2
=0,so
y
1
(1) =
−
c
1
√
λ
+ 1 sin
√
λ
+1=0
gives
√
λ
+1=
nπ
or
λ
=
n
2
π
2
−
1
,n
=1
,
2
,
3
,... .
The eigenvalues
n
2
π
2
−
1 correspond to the eigenfunctions cos
nπx
for
n
=0,1,2,
...
.
17.
For
λ
= 0 the only solution of the boundary-value problem is
y
=0. For
λ
3
= 0 we have
y
=
c
1
cos
λx
+
c
2
sin
λx.
Now
y
(0) = 0 implies
c
1
=0,so
y
(
L
)=
c
2
sin
λL
=0
gives
λL
=
nπ
or
λ
=
nπ
L
,n
=1
,
2
,
3
,... .
The eigenvalues
nπ/L
correspond to the eigenfunctions sin
nπ
L
x
for
n
=1,2,3,
...
.
18.
For
λ
= 0 the only solution of the boundary-value problem is
y
=0. For
λ
3
= 0 we have
y
=
c
1
cos
λx
+
c
2
sin
λx.
Now
y
(0) = 0 implies
c
1
=0,so
y
1
(3
π
)=
c
2
λ
cos 3
πλ
=0
gives
3
πλ
=
(2
n
−
1)
π
2
or
λ
=
2
n
−
1
6
,n
=1
,
2
,
3
,... .
The eigenvalues (2
n
−
1)
/
6 correspond to the eigenfunctions sin
2
n
−
1
6
x
for
n
=1,2,3,
...
.
19.
For
λ>
0 a general solution of the given differential equation is
y
=
c
1
cos(
√
λ
ln
x
)+
c
2
sin(
√
λ
ln
x
)
.
Since ln 1 = 0, the boundary condition
y
(1) = 0 implies
c
1
= 0. Therefore
y
=
c
2
sin(
√
λ
ln
x
)
.
Using ln
e
π
=
π
we find that
y
(
e
π
) = 0 implies
c
2
sin
√
λπ
=0
or
√
λπ
=
nπ
,
n
=1,2,3,
...
. The eigenvalues and eigenfunctions are, in turn,
λ
=
n
2
,n
=1
,
2
,
3
,...
and
y
= sin(
n
ln
x
)
.
For
λ
≤
0 the only solution of the boundary-value problem is
y
=0.
To obtain the self-adjoint form we note that the integrating factor is (1
/x
2
)
e
5
dx/x
=1
/x
. That is, the self-
adjoint form is
d
dx
[
xy
1
]+
λ
x
y
=0
.
134

135. Exercises 3.9
Identifying the weight function
p
(
x
)=1
/x
we can then write the orthogonality relation
6
e
π
1
1
x
sin(
n
ln
x
) sin(
m
ln
x
)
dx
=0
,m
3
=
n.
20.
For
λ
= 0 the general solution is
y
=
c
1
+
c
2
ln
x
.Now
y
1
=
c
2
/x
,so
y
1
(
e
−
1
)=
c
2
e
= 0 implies
c
2
= 0. Then
y
=
c
1
and
y
(1) = 0 gives
c
1
=0. Thus
y
(
x
)=0.
For
λ<
0,
y
=
c
1
x
−
√
−
λ
+
c
2
x
√
−
λ
. The initial conditions give
c
2
=
c
1
e
2
√
−
λ
and
c
1
= 0, so that
c
2
= 0 and
y
(
x
)=0.
For
λ>
0,
y
=
c
1
cos(
√
λ
ln
x
)+
c
2
sin(
√
λ
ln
x
). From
y
(1) = 0 we obtain
c
1
= 0 and
y
=
c
2
sin(
√
λ
ln
x
). Now
y
1
=
c
2
(
√
λ/x
) cos(
√
λ
ln
x
), so
y
1
(
e
−
1
)=
c
2
e
√
λ
cos
√
λ
= 0 implies cos
√
λ
=0or
λ
=(2
n
−
1)
2
π
2
/
4 for
n
=1,2,3,
...
. The corresponding eigenfunctions are
y
= sin

2
n
−
1
2
π
ln
x

.
21.
For
λ
= 0 the general solution is
y
=
c
1
+
c
2
ln
x
.Now
y
1
=
c
2
/x
,so
y
1
(1) =
c
2
= 0 and
y
=
c
1
. Since
y
1
(
e
2
)=0
for any
c
1
we see that
y
(
x
) = 1 is an eigenfunction corresponding to the eigenvalue
λ
=0.
For
λ<
0,
y
=
c
1
x
−
√
−
λ
+
c
2
x
√
−
λ
. The initial conditions imply
c
1
=
c
2
=0,so
y
(
x
)=0.
For
λ>
0,
y
=
c
1
cos(
√
λ
ln
x
)+
c
2
sin(
√
λ
ln
x
). Now
y
1
=
−
c
1
√
λ
x
sin(
√
λ
ln
x
)+
c
2
√
λ
x
cos(
√
λ
ln
x
)
,
and
y
1
(1) =
c
2
√
λ
= 0 implies
c
2
= 0. Finally,
y
1
(
e
2
)=
−
(
c
1
√
λ/e
2
) sin(2
√
λ
) = 0 implies
λ
=
n
2
π
2
/
4 for
n
=1,
2, 3,
...
. The corresponding eigenfunctions are
y
= cos
3
nπ
2
ln
x
4
.
22.
For
λ>
1
/
4 a general solution of the given differential equation is
y
=
c
1
x
−
1
/
2
cos

√
4
λ
−
1
2
ln
x

+
c
2
x
−
1
/
2
sin

√
4
λ
−
1
2
ln
x

.
Since ln 1 = 0, the boundary condition
y
(1) = 0 implies
c
1
= 0. Therefore
y
=
c
2
x
−
1
/
2
sin

√
4
λ
−
1
2
ln
x

.
Using ln
e
2
= 2 we find that
y
(
e
2
) = 0 implies
c
2
e
−
1
sin
3
√
4
λ
−
1
4
=0
or
√
4
λ
−
1=
nπ
,
n
=1,2,3,
...
. The eigenvalues and eigenfunctions are, in turn,
λ
=
1
n
2
π
2
+1
.
/
4
,n
=1
,
2
,
3
,...
and
y
=
x
−
1
/
2
sin
3
nπ
2
ln
x
4
.
For
λ<
0 the only solution of the boundary-value problem is
y
=0.
For
λ
=1
/
4 a general solution of the differential equation is
y
=
c
1
x
−
1
/
2
+
c
2
x
−
1
/
2
ln
x.
From
y
(1) = 0 we obtain
c
1
=0,so
y
=
c
2
x
−
1
/
2
ln
x
. From
y
(
e
2
) = 0 we obtain 2
c
2
e
−
1
=0or
c
2
= 0. Thus,
there are no eigenvalues and eigenfunctions in this case.
135

136. x
L
y
Exercises 3.9
To obtain the self-adjoint form we note that the integrating factor is (1
/x
2
)
e
5
(2
/x
)
dx
=(1
/x
2
)
·
x
2
= 1. That
is, the self-adjoint form is
d
dx
b
x
2
y
1
c
+
λy
=0
.
Identifying the weight function
p
(
x
) = 1 we can then write the orthogonality relation
6
e
2
1
1
·
x
−
1
/
2
sin
3
mπ
2
ln
x
4
x
−
1
/
2
sin
3
nπ
2
ln
x
4
dx
=0
,m
3
=
n,
or
6
e
2
1
x
−
1
sin
3
mπ
2
ln
x
4
sin
3
nπ
2
ln
x
4
dx
=0
,m
3
=
n.
23.
For
λ>
0 the general solution is
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx
. Setting
y
(0) = 0 we find
c
1
= 0, so that
y
=
c
2
sin
√
λx
. The boundary condition
y
(1) +
y
1
(1) = 0 implies
c
2
sin
√
λ
+
c
2
√
λ
cos
√
λ
=0
.
Taking
c
2
3
= 0, this equation is equivalent to tan
√
λ
=
−
√
λ
. Thus, the eigenvalues are
λ
n
=
x
2
n
,
n
=1,2,
3,
...
, where the
x
n
are the consecutive positive roots of tan
√
λ
=
−
√
λ
.
24. (a)
Since
λ
n
=
x
2
n
, there are no new eigenvalues when
x
n
<
0. For
λ
= 0, the differential equation
y
11
= 0 has
general solution
y
=
c
1
x
+
c
2
. The boundary conditions imply
c
1
=
c
2
=0,so
y
=0.
(b)
λ
1
=4
.
1159,
λ
2
=24
.
1393,
λ
3
=63
.
6591,
λ
4
= 122
.
8892.
25.
If restraints are put on the column at
x
=
L/
4,
x
=
L/
2, and
x
=3
L/
4, then the critical
load will be
P
4
.
26. (a)
The general solution of the differential equation is
y
=
c
1
cos
s
P
EI
x
+
c
2
sin
s
P
EI
x
+
δ.
Since the column is embedded at
x
= 0, the initial conditions are
y
(0) =
y
1
(0) = 0. If
δ
= 0 this implies
that
c
1
=
c
2
= 0 and
y
(
x
) = 0. That is, there is no deflection.
(b)
If
δ
3
= 0, the initial conditions give, in turn,
c
1
=
−
δ
and
c
2
= 0. Then
y
=
δ

1
−
cos
s
P
EI
x
a
.
In order to satisfy the condition
y
(
L
)=
δ
we must have
δ
=
δ

1
−
cos
s
P
EI
L
a
or cos
s
P
EI
L
=0
.
136

137. 1 e
x
-1
1
y
n=1
1 e
x
-1
1
y
n=2
1 e
x
-1
1
y
n=3
Exercises 3.9
This gives
)
P/EI L
=
nπ/
2 for
n
=1,2,3,
...
. The smallest value of
P
n
, the Euler load, is then
s
P
1
EI
L
=
π
2
or
P
1
=
1
4

π
2
EI
L
2

.
27.
The general solution is
y
=
c
1
cos
s
ρω
2
T
x
+
c
2
sin
s
ρω
2
T
x.
From
y
(0) = 0 we obtain
c
1
= 0. Setting
y
(
L
)=0wefind
)
ρω
2
/T L
=
nπ
,
n
=1,2,3,
...
. Thus, critical
speeds are
ω
n
=
nπ
√
T/L
√
ρ
,
n
=1,2,3,
...
. The corresponding deflection curves are
y
(
x
)=
c
2
sin
nπ
L
x, n
=1
,
2
,
3
, ... ,
where
c
2
3
=0.
28. (a)
When
T
(
x
)=
x
2
the given differential equation is the Cauchy-Euler equation
x
2
y
11
+2
xy
1
+
ρω
2
y
=0
.
The solutions of the auxiliary equation
m
(
m
−
1)+2
m
+
ρω
2
=
m
2
+
m
+
ρω
2
=0
are
m
1
=
−
1
2
−
1
2
)
4
ρω
2
−
1
i, m
2
=
−
1
2
+
1
2
)
4
ρω
2
−
1
i
when
ρω
2
>
0
.
25. Thus
y
=
c
1
x
−
1
/
2
cos(
λ
ln
x
)+
c
2
x
−
1
/
2
sin(
λ
ln
x
)
where
λ
=
)
4
ρω
2
−
1
/
2. Applying
y
(1) = 0 gives
c
1
= 0 and consequently
y
=
c
2
x
−
1
/
2
sin(
λ
ln
x
)
.
The condition
y
(
e
) = 0 requires
c
2
e
−
1
/
2
sin
λ
= 0. We obtain a nontrivial solution when
λ
n
=
nπ
,
n
=1,2,
3,
...
. But
λ
n
=
)
4
ρω
2
n
−
1
/
2=
nπ.
Solving for
ω
n
gives
ω
n
=
1
2
)
(4
n
2
π
2
+1)
/ρ .
The corresponding solutions are
y
n
(
x
)=
c
2
x
−
1
/
2
sin(
nπ
ln
x
)
.
(b)
29.
The auxiliary equation is
m
2
+
m
=
m
(
m
+1) = 0 so that
u
(
r
)=
c
1
r
−
1
+
c
2
. The boundary conditions
u
(
a
)=
u
0
and
u
(
b
)=
u
1
yield the system
c
1
a
−
1
+
c
2
=
u
0
,
c
1
b
−
1
+
c
2
=
u
1
. Solving gives
c
1
=

u
0
−
u
1
b
−
a

ab
and
c
2
=
u
1
b
−
u
0
a
b
−
a
.
137

138. Exercises 3.9
Thus
u
(
r
)=

u
0
−
u
1
b
−
a

ab
r
+
u
1
b
−
u
0
a
b
−
a
.
30.
The auxiliary equation is
m
2
= 0 so that
u
(
r
)=
c
1
+
c
2
ln
r
. The boundary conditions
u
(
a
)=
u
0
and
u
(
b
)=
u
1
yield the system
c
1
+
c
2
ln
a
=
u
0
,
c
1
+
c
2
ln
b
=
u
1
. Solving gives
c
1
=
u
1
ln
a
−
u
0
ln
b
ln(
a/b
)
and
c
2
=
u
0
−
u
1
ln(
a/b
)
.
Thus
u
(
r
)=
u
1
ln
a
−
u
0
ln
b
ln(
a/b
)
+
u
0
−
u
1
ln(
a/b
)
ln
r
=
u
0
ln(
r/b
)
−
u
1
ln(
r/a
)
ln(
a/b
)
.
31. (a)
The general solution of the differential equation is
y
=
c
1
cos 4
x
+
c
2
sin 4
x
. From
y
0
=
y
(0) =
c
1
we see
that
y
=
y
0
cos 4
x
+
c
2
sin 4
x
. From
y
1
=
y
(
π/
2) =
y
0
we see that any solution must satisfy
y
0
=
y
1
.We
also see that when
y
0
=
y
1
,
y
=
y
0
cos 4
x
+
c
2
sin 4
x
is a solution of the boundary-value problem for any
choice of
c
2
. Thus, the boundary-value problem does not have a unique solution for any choice of
y
0
and
y
1
.
(b)
Whenever
y
0
=
y
1
there are infinitely many solutions.
(c)
When
y
0
3
=
y
1
there will be no solutions.
(d)
The boundary-value problem will have the trivial solution when
y
0
=
y
1
= 0. This solution will not be
unique.
32. (a)
The general solution of the differential equation is
y
=
c
1
cos 4
x
+
c
2
sin 4
x
. From 1 =
y
(0) =
c
1
we see that
y
= cos 4
x
+
c
2
sin 4
x
. From 1 =
y
(
L
) = cos 4
L
+
c
2
sin 4
L
we see that
c
2
=(1
−
cos 4
L
)
/
sin 4
L
. Thus,
y
= cos 4
x
+

1
−
cos 4
L
sin 4
L

sin 4
x
will be a unique solution when sin 4
L
3
= 0; that is, when
L
3
=
kπ/
4 where
k
=1,2,3,
...
.
(b)
There will be infinitely many solutions when sin 4
L
= 0 and 1
−
cos 4
L
= 0; that is, when
L
=
kπ/
2 where
k
=1,2,3,
...
.
(c)
There will be no solution when sin 4
L
3
= 0 and 1
−
cos 4
L
3
= 0; that is, when
L
=
kπ/
4 where
k
=1,3,5,
...
.
(d)
There can be no trivial solution since it would fail to satisfy the boundary conditions.
33. (a)
A solution curve has the same
y
-coordinate at both ends of the interval [
−
π, π
] and the tangent lines at the
endpoints of the interval are parallel.
(b)
For
λ
= 0 the solution of
y
11
=0is
y
=
c
1
x
+
c
2
. From the first boundary condition we have
y
(
−
π
)=
−
c
1
π
+
c
2
=
y
(
π
)=
c
1
π
+
c
2
or 2
c
1
π
= 0. Thus,
c
1
= 0 and
y
=
c
2
. This constant solution is seen to satisfy the boundary-value problem.
For
λ<
0wehave
y
=
c
1
cosh
λx
+
c
2
sinh
λx
. In this case the first boundary condition gives
y
(
−
π
)=
c
1
cosh(
−
λπ
)+
c
2
sinh(
−
λπ
)
=
c
1
cosh
λπ
−
c
2
sinh
λπ
=
y
(
π
)=
c
1
cosh
λπ
+
c
2
sinh
λπ
or 2
c
2
sinh
λπ
=0. Thus
c
2
= 0 and
y
=
c
1
cosh
λx
. The second boundary condition implies in a similar
fashion that
c
1
= 0. Thus, for
λ<
0, the only solution of the boundary-value problem is
y
=0.
138

139. -„
„
x
y
−3
3
-„
„
x
y
−3
3
2 4 6 8
t
1
2
-1
-2
x
t
2
4
6
8
10
-2
x
Exercises 3.10
For
λ>
0wehave
y
=
c
1
cos
λx
+
c
2
sin
λx
. The first boundary condition implies
y
(
−
π
)=
c
1
cos(
−
λπ
)+
c
2
sin(
−
λπ
)
=
c
1
cos
λπ
−
c
2
sin
λπ
=
y
(
π
)=
c
1
cos
λπ
+
c
2
sin
λπ
or 2
c
2
sin
λπ
= 0. Similarly, the second boundary condition implies 2
c
1
λ
sin
λπ
=0. If
c
1
=
c
2
= 0 the
solution is
y
= 0. However, if
c
1
3
=0or
c
2
3
= 0, then sin
λπ
= 0, which implies that
λ
must be an integer,
n
.
Therefore, for
c
1
and
c
2
not both 0,
y
=
c
1
cos
nx
+
c
2
sin
nx
is a nontrivial solution of the boundary-value
problem. Since cos(
−
nx
) = cos
nx
and sin(
−
nx
)=
−
sin
nx
, we may assume without loss of generality that
the eigenvalues are
λ
n
=
n
, for
n
a positive integer. The corresponding eigenfunctions are
y
n
= cos
nx
and
y
n
= sin
nx
.
(c)
y
= 2 sin 3
xy
= sin 4
x
−
2 cos 3
x
Exercises 3.10
1.
The period corresponding to
x
(0) = 1,
x
1
(0) = 1 is approximately 5
.
6. The
period corresponding to
x
(0) = 1
/
2,
x
1
(0) =
−
1 is approximately 6
.
2.
2.
The solutions are not periodic.
139

140. 2 4 6 810
t
2
4
6
8
10
-2
x
2 4 6 8 10
t
1
2
3
-1
-2
-3
x
x1=1.1
x1=1.2
t
x
510
−1
1
2
3
4
t
x
510
−1
1
2
3
Exercises 3.10
3.
The period corresponding to
x
(0) = 1,
x
1
(0) = 1 is approximately 5
.
8.
The second initial-value problem does not have a periodic solution.
4.
Both solutions have periods of approximately 6
.
3.
5.
From the graph we see that
|
x
1
|≈
1
.
2.
6.
From the graphs we see that the interval is approximately (
−
0
.
8
,
1
.
1).
7.
Since
xe
0
.
01
x
=
x
[1+0
.
01
x
+
1
2!
(0
.
01
x
)
2
+
···
]
≈
x
for small values of
x
, a linearization is
d
2
x
dt
2
+
x
=0.
140

141. t
x
51015
−3
3
2 4 6 8
t
2
-2
x
24
t
-2
-1
1
2
3
4
5
x
Exercises 3.10
8.
For
x
(0) = 1 and
x
1
(0) = 1 the oscillations are symmetric about the line
x
= 0 with amplitude slightly greater
than 1.
For
x
(0) =
−
2 and
x
1
(0) = 0
.
5 the oscillations are symmetric about the line
x
=
−
2 with small amplitude.
For
x
(0) =
√
2 and
x
1
(0) = 1 the oscillations are symmetric about the line
x
= 0 with amplitude a little greater
than 2.
For
x
(0) = 2 and
x
1
(0) = 0
.
5 the oscillations are symmetric about the line
x
= 2 with small amplitude.
For
x
(0) =
−
2 and
x
1
(0) = 0 there is no oscillation; the solution is constant.
For
x
(0) =
−
√
2 and
x
1
(0) =
−
1 the oscillations are symmetric about the line
x
= 0 with amplitude a little
greater than 2.
9.
This is a damped hard spring, so all solutions should be oscillatory with
x
→
0as
t
→∞
.
10.
This is a damped soft spring, so so we expect no oscillatory solutions.
141

142. 105
t,
k1 = 20
-2
-3
-1
1
2
3
x
1 2 3
t,
k1 = 100
-2
-3
-1
1
2
3
x
10 20 30
t,
k1 = 0.01
-5
-10
5
10
15
-15
x
10 20
t,
k1 = 1
-2
-3
-1
1
2
3
x
Exercises 3.10
11.
When
k
1
is very small the effect of the nonlinearity is greatly diminished, and the system is close to pure
resonance.
12. (a)
The system appears to be oscillatory for
−
0
.
000471
≤
k
1
<
0 and nonoscillatory for
k
1
≤−
0
.
000472.
(b)
The system appears to be oscillatory for
−
0
.
077
≤
k
1
<
0 and nonoscillatory for
k
1
≤
0
.
078.
13.
Since (
dx/dt
)
2
is always positive, it is necessary to use
|
dx/dt
|
(
dx/dt
) in order to account for the fact that the
motion is oscillatory and the velocity (or its square) should be negative when the spring is contracting.
142

143. 0.20.40.60.811.21.4
x
0.2
0.4
0.6
0.8
1
y
0.20.4
t
0.1
0.2
0.3
0.4
0.5
q
0.20.4
t
0.1
0.2
0.3
0.4
0.5
q
0.20.4
t
0.2
0.4
0.6
0.8
1
q
0.20.4
t
0.2
0.4
0.6
0.8
1
q
earth
moon
t
θ
5
-
2
2
earth
moon
t
θ
5
-
2
2
Exercises 3.10
14. (a)
The approximation is accurate to two decimal places for
θ
1
=0
.
3,
and accurate to one decimal place for
θ
1
=0
.
6.
(b)
The thinner curves are solutions of the nonlinear differential equation, while the thicker curves are solutions
of the linear differential equation.
15. (a)
Write the differential equation as
d
2
θ
dt
2
+
ω
2
sin
θ
=0
,
where
ω
2
=
g/6
. To test for differences between the earth
and the moon we take
6
=3,
θ
(0) = 1, and
θ
1
(0)=2.
Using
g
= 32 on the earth and
g
=5
.
5 on the moon we
obtain the graphs shown in the figure. Comparing the
apparent periods of the graphs, we see that the pendulum
oscillates faster on the earth than on the moon.
(b)
The amplitude is greater on the moon than on the earth.
(c)
The linear model is
d
2
θ
dt
2
+
ω
2
θ
=0
,
where
ω
2
=
g/6
. When
g
= 32,
6
=3,
θ
(0) = 1, and
θ
1
(0) = 2, the general solution is
θ
(
t
) = cos 3
.
266
t
+0
.
612 sin 3
.
266
t.
When
g
=5
.
5 the general solution is
θ
(
t
) = cos 1
.
354
t
+1
.
477 sin 1
.
354
t.
As in the nonlinear case, the pendulum oscillates faster on the earth than on the moon and still has greater
amplitude on the moon.
143

144. 12345
-0.4
-0.2
0.2
0.4
246810
-0.4
-0.2
0.2
0.4
Exercises 3.10
16. (a)
The general solution of
d
2
θ
dt
2
+
θ
=0
is
θ
(
t
)=
c
1
cos
t
+
c
2
sin
t
. From
θ
(0) =
π/
12 and
θ
1
(0) =
−
1
/
3wefind
θ
(
t
)=(
π/
12) cos
t
−
(1
/
3) sin
t
.
Setting
θ
(
t
)=0wehavetan
t
=
π/
4 which implies
t
1
= tan
−
1
(
π/
4)
≈
0
.
66577.
(b)
We set
θ
(
t
)=
θ
(0) +
θ
1
(0)
t
+
1
2
θ
11
(0)
t
+
1
6
θ
111
(0)
t
+
···
and use
θ
11
(
t
)=
−
sin
θ
(
t
) together with
θ
(0) =
π/
12 and
θ
1
(0) =
−
1
/
3. Then
θ
11
(0) =
−
sin(
π/
12) =
−
√
2(
√
3
−
1)
/
4 and
θ
111
(0) =
−
cos
θ
(0)
·
θ
1
(0) =
−
cos(
π/
12)(
−
1
/
3) =
√
2(
√
3+1)
/
12. Thus
θ
(
t
)=
π
12
−
1
3
t
−
√
2(
√
3
−
1)
8
t
2
+
√
2(
√
3+1)
72
t
3
+
···
.
(c)
Setting
π/
12
−
t/
3 = 0 we obtain
t
1
=
π/
4
≈
0
.
785398.
(d)
Setting
π
12
−
1
3
t
−
√
2(
√
3
−
1)
8
t
2
=0
and using the positive root we obtain
t
1
≈
0
.
63088.
(e)
Setting
π
12
−
1
3
t
−
√
2(
√
3
−
1)
8
t
2
+
√
2(
√
3+1)
72
t
3
=0
we find
t
1
≈
0
.
661973 to be the first positive root.
(f)
From the output we see that
y
(
t
) is an interpolating function on the interval
0
≤
t
≤
5, whose graph is shown. The positive root of
y
(
t
) = 0 near
t
=1
is
t
1
=0
.
666404.
(g)
To find the next two positive roots we change the interval used in
NDSolve
and
Plot
from
{
t,0,5
}
to
{
t,0,10
}
. We see from the graph that the second
and third positive roots are near 4 and 7, respectively. Replacing
{
t,1
}
in
FindRoot
with
{
t,4
}
and then
{
t,7
}
we obtain
t
2
=3
.
84411 and
t
3
=7
.
0218.
17.
From the table below we see that the pendulum first passes the vertical position between 1.7 and 1.8 seconds.
To refine our estimate of
t
1
we estimate the solution of the differential equation on [1
.
7
,
1
.
8] using a stepsize of
h
=0
.
01. From the resulting table we see that
t
1
is between 1
.
76 and 1
.
77 seconds. Repeating the process with
h
=0
.
001 we conclude that
t
1
≈
1
.
767. Then the period of the pendulum is approximately 4
t
1
=7
.
068. The
error when using
t
1
=2
π
is 7
.
068
−
6
.
283 = 0
.
785 and the percentage relative error is (0
.
785
/
7
.
068)100 = 11
.
1.
144

145. h=0.1
h=0.01
t
n
θ
n
t
n
θ
n
0.00
0.78540
1.70
0.07706
0.10
0.78523
1.71
0.06572
0.20
0.78407
1.72
0.05428
0.30
0.78092
1.73
0.04275
0.40
0.77482
1.74
0.03111
0.50
0.76482
1.75
0.01938
0.60
0.75004
1.76
0.00755
0.70
0.72962
1.77
-0.00438
0.80
0.70275
1.78
-0.01641
0.90
0.66872
1.79
-0.02854
1.00
0.62687
1.80
-0.04076
1.10
0.57660
1.20
0.51744 h=0.001
1.30
0.44895
1.763
0.00398
1.40
0.37085
1.764
0.00279
1.50
0.28289
1.765
0.00160
1.60
0.18497
1.766
0.00040
1.70
0.07706
1.767
-0.00079
1.80
-0.04076
1.768
-0.00199
1.90
-0.16831
1.769
-0.00318
2.00
-0.30531
1.770
-0.00438
0.511.522.5
t
2
4
6
8
10
x
Exercises 3.10
18. (a)
Setting
dy/dt
=
v
, the differential equation in (13) becomes
dv/dt
=
−
gR
2
/y
2
. But, by the chain rule,
dv/dt
=(
dv/dy
)(
dy/dt
)=
v dv/dt
,so
v dv/dy
=
−
gR
2
/y
2
. Separating variables and integrating we obtain
vdv
=
−
gR
2
dy
y
2
and
1
2
v
2
=
gR
2
y
+
c.
Setting
v
=
v
0
and
y
=
R
we find
c
=
−
gR
+
1
2
v
2
0
and
v
2
=2
g
R
2
y
−
2
gR
+
v
2
0
.
(b)
As
y
→∞
we assume that
v
→
0
+
. Then
v
2
0
=2
gR
and
v
0
=
√
2
gR
.
(c)
Using
g
= 32 ft/s and
R
= 4000(5280) ft we find
v
0
=
7
2(32)(4000)(5280)
≈
36765
.
2 ft/s
≈
25067 mi/hr
.
(d)
v
0
=
7
2(0
.
165)(32)(1080)
≈
7760 ft/s
≈
5291 mi/hr
19. (a)
Intuitively, one might expect that only half of a 10-pound chain could be lifted by a 5-pound force.
(b)
Since
x
= 0 when
t
= 0, and
v
=
dx/dt
=
7
160
−
64
x/
3 , we have
v
(0) =
√
160
≈
12
.
65 ft/s.
(c)
Since
x
should always be positive, we solve
x
(
t
) = 0, getting
t
=0
and
t
=
3
2
7
5
/
2
≈
2
.
3717. Since the graph of
x
(
t
) is a parabola,
the maximum value occurs at
t
m
=
3
4
7
5
/
2 . (This can also be
obtained by solving
x
1
(
t
) = 0.) At this time the height of the
chain is
x
(
t
m
)
≈
7
.
5 ft. This is higher than predicted because of
the momentum generated by the force. When the chain is 5 feet
high it still has a positive velocity of about 7
.
3 ft/s, which keeps
it going higher for a while.
20. (a)
Setting
dx/dt
=
v
, the differential equation becomes (
L
−
x
)
dv/dt
−
v
2
=
Lg
. But, by the chain rule,
dv/dt
=(
dv/dx
)(
dx/dt
)=
v dv/dx
,so(
L
−
x
)
v dv/dx
−
v
2
=
Lg
. Separating variables and integrating we
145

146. Exercises 3.10
obtain
v
v
2
+
Lg
dv
=
1
L
−
x
dx
and
1
2
ln(
v
2
+
Lg
)=
−
ln(
L
−
x
)+ln
c,
so
)
v
2
+
Lg
=
c/
(
L
−
x
). When
x
=0,
v
= 0, and
c
=
L
√
Lg
. Solving for
v
and simplifying we get
dx
dt
=
v
(
x
)=
)
Lg
(2
Lx
−
x
2
)
L
−
x
.
Again, separating variables and integrating we obtain
L
−
x
)
Lg
(2
Lx
−
x
2
)
dx
=
dt
and
√
2
Lx
−
x
2
√
Lg
=
t
+
c
1
.
Since
x
(0)=0,wehave
c
1
= 0 and
√
2
Lx
−
x
2
/
√
Lg
=
t
. Solving for
x
we get
x
(
t
)=
L
−
)
L
2
−
Lgt
2
and
v
(
t
)=
dx
dt
=
√
Lgt
)
L
−
gt
2
.
(b)
The chain will be completely on the ground when
x
(
t
)=
L
or
t
=
)
L/g
.
(c)
The predicted velocity of the upper end of the chain when it hits the ground is infinity.
21. (a)
The weight of
x
feet of the chain is 2
x
, so the corresponding mass is
m
=2
x/
32 =
x/
16. The only force
acting on the chain is the weight of the portion of the chain hanging over the edge of the platform. Thus,
by Newton’s second law,
d
dt
(
mv
)=
d
dt

x
16
v

=
1
16

x
dv
dt
+
v
dx
dt

=
1
16

x
dv
dt
+
v
2

=2
x
and
x dv/dt
+
v
2
=32
x
. Now, by the chain rule,
dv/dt
=(
dv/dx
)(
dx/dt
)=
v dv/dx
,so
xv dv/dx
+
v
2
=32
x
.
(b)
We separate variables and write the differential equation as (
v
2
−
32
x
)
dx
+
xv dv
= 0. This is not an exact
form, but
µ
(
x
)=
x
is an integrating factor. Multiplying by
x
we get (
xv
2
−
32
x
2
)
dx
+
x
2
vdv
= 0. This
form is the total differential of
u
=
1
2
x
2
v
2
−
32
3
x
3
, so an implicit solution is
1
2
x
2
v
2
−
32
3
x
3
=
c
. Letting
x
=3
and
v
= 0 we find
c
=
−
288. Solving for
v
we get
dx
dt
=
v
=
8
√
x
3
−
27
√
3
x
,
3
≤
x
≤
8
.
(c)
Separating variables and integrating we obtain
x
√
x
3
−
27
dx
=
8
√
3
dt
and
6
x
3
s
√
s
3
−
27
ds
=
8
√
3
t
+
c.
Since
x
= 3 when
t
= 0, we see that
c
= 0 and
t
=
√
3
8
6
x
3
s
√
s
3
−
27
ds.
We want to find
t
when
x
= 7. Using a CAS we find
t
(7) = 0
.
576 seconds.
22. (a)
There are two forces acting on the chain as it falls from the platform. One is the force due to gravity on
the portion of the chain hanging over the edge of the platform. This is
F
1
=2
x
. The second is due to the
motion of the portion of the chain stretched out on the platform. By Newton’s second law this is
F
2
=
d
dt
[
mv
]=
d
dt

(8
−
x
)2
32
v

=
d
dt

8
−
x
16
v

=
8
−
x
16
dv
dt
−
1
16
v
dx
dt
=
1
16

(8
−
x
)
dv
dt
−
v
2

.
146

147. Exercises 3.10
From
d
dt
[
mv
]=
F
1
−
F
2
we have
d
dt

2
x
32
v

=2
x
−
1
16

(8
−
x
)
dv
dt
−
v
2

x
16
dv
dt
+
1
16
v
dx
dt
=2
x
−
1
16

(8
−
x
)
dv
dt
−
v
2

x
dv
dt
+
v
2
=32
x
−
(8
−
x
)
dv
dt
+
v
2
x
dv
dt
=32
x
−
8
dv
dt
+
x
dv
dt
8
dv
dt
=32
x.
By the chain rule,
dv/dt
=(
dv/dx
)(
dx/dt
)=
v dv/dx
,so
8
dv
dt
=8
v
dv
dx
=32
x
and
v
dv
dx
=4
x.
(b)
Integrating
vdv
=4
xdx
we get
1
2
v
2
=2
x
2
+
c
. Since
v
= 0 when
x
= 3, we have
c
=
−
18. Then
v
2
=4
x
2
−
36 and
v
=
√
4
x
2
−
36 . Using
v
=
dx/dt
, separating variables, and integrating we obtain
dx
√
x
2
−
9
=2
dt
and cosh
−
1
x
3
=2
t
+
c
1
.
Solving for
x
we get
x
(
t
) = 3 cosh(2
t
+
c
1
). Since
x
= 3 when
t
= 0, we have cosh
c
1
= 1 and
c
1
= 0. Thus,
x
(
t
) = 3 cosh 2
t
. Differentiating, we find
v
(
t
)=
dx/dt
= 6 sinh 2
t
.
(c)
To find when the back end of the chain will leave the platform we solve
x
(
t
) = 3 cosh 2
t
= 8. This gives
t
1
=
1
2
cosh
−
1
8
3
≈
0
.
8184 seconds. The velocity at this instant is
v
(
t
1
) = 6 sinh
1
cosh
−
1
8
3
.
=2
√
55
≈
14
.
83 ft/s.
(d)
Replacing 8 with L and 32 with g in part (a) we have
L dv/dt
=
gx
. Then
L
dv
dt
=
Lv
dv
dx
=
gx
and
v
dv
dx
=
g
L
x.
Integrating we get
1
2
v
2
=
g
2
L
x
2
+
c
. Setting
x
=
x
0
and
v
= 0, we find
c
=
−
g
2
L
x
2
0
. Solving for
v
we find
v
(
x
)=
s
g
L
x
2
−
g
L
x
2
0
.
Then the velocity at which the end of the chain leaves the edge of the platform is
v
(
L
)=
s
g
L
(
L
2
−
x
2
0
)
.
23. (a)
Let (
x, y
) be the coordinates of
S
2
on the curve
C
. The slope at (
x, y
) is then
dy/dx
=(
v
1
t
−
y
)
/
(0
−
x
)=(
y
−
v
1
t
)
/x
or
xy
1
−
y
=
−
v
1
t.
(b)
Differentiating with respect to
x
gives
xy
11
+
y
1
−
y
1
=
−
v
1
dt
dx
xy
11
=
−
v
1
dt
ds
ds
dx
xy
11
=
−
v
1
1
v
2
(
−
)
1+(
y
1
)
2
)
xy
11
=
r
)
1+(
y
1
)
2
.
147

148. Exercises 3.10
Letting
u
=
y
1
and separating variables, we obtain
x
du
dx
=
r
)
1+
u
2
du
√
1+
u
2
=
r
x
dx
sinh
−
1
u
=
r
ln
x
+ln
c
= ln(
cx
r
)
u
= sinh(ln
cx
r
)
dy
dx
=
1
2

cx
r
−
1
cx
r

.
At
t
=0,
dy/dx
= 0 and
x
=
a
,so0=
ca
r
−
1
/ca
r
.Thus
c
=1
/a
r
and
dy
dx
=
1
2
e3
x
a
4
r
−
3
a
x
4
r
C
=
1
2

3
x
a
4
r
−
3
x
a
4
−
r

.
If
r>
1or
r<
1, integrating gives
y
=
a
2

1
1+
r
3
x
a
4
1+
r
−
1
1
−
r
3
x
a
4
1
−
r

+
c
1
.
When
t
=0,
y
= 0 and
x
=
a
,so0=(
a/
2)[1
/
(1 +
r
)
−
1
/
(1
−
r
)] +
c
1
.Thus
c
1
=
ar/
(1
−
r
2
) and
y
=
a
2

1
1+
r
3
x
a
4
1+
r
−
1
1
−
r
3
x
a
4
1
−
r

+
ar
1
−
r
2
.
(c)
If
r>
1,
v
1
>v
2
and
y
→∞
as
x
→
0
+
. In other words,
S
2
always lags behind
S
1
.If
r<
1,
v
1
<v
2
and
y
=
ar/
(1
−
r
2
) when
x
= 0. In other words, when the submarine’s speed is greater than the ship’s, their
paths will intersect at the point (0
,ar/
(1
−
r
2
)).
If
r
= 1, integration gives
y
=
1
2

x
2
2
a
−
1
a
ln
x

+
c
2
.
When
t
=0,
y
= 0 and
x
=
a
,so0=(1
/
2)[
a/
2
−
(1
/a
)ln
a
]+
c
2
.Thus
c
2
=
−
(1
/
2)[
a/
2
−
(1
/a
)ln
a
] and
y
=
1
2

x
2
2
a
−
1
a
ln
x

−
1
2

a
2
−
1
a
ln
a

=
1
2

1
2
a
(
x
2
−
a
2
)+
1
a
ln
a
x

.
Since
y
→∞
as
x
→
0
+
,
S
2
will never catch upwith
S
1
.
24. (a)
Let (
r, θ
) denote the polar coordinates of the destroyer
S
1
. When
S
1
travels the 6 miles from (9
,
0) to (3
,
0)
it stands to reason, since
S
2
travels half as fast as
S
1
, that the polar coordinates of
S
2
are (3
,θ
2
), where
θ
2
is unknown. In other words, the distances of the ships from (0
,
0) are the same and
r
(
t
)=15
t
then gives
the radial distance of both ships. This is necessary if
S
1
is to intercept
S
2
.
(b)
The differential of arc length in polar coordinates is (
ds
)
2
=(
rdθ
)
2
+(
dr
)
2
, so that

ds
dt

2
=
r
2

dθ
dt

2
+

dr
dt

2
.
Using
ds/dt
= 30 and
dr/dt
= 15 then gives
900 = 225
t
2

dθ
dt

2
+ 225
675 = 225
t
2

dθ
dt

2
dθ
dt
=
√
3
t
θ
(
t
)=
√
3ln
t
+
c
=
√
3ln
r
15
+
c.
148

149. λ=2, ω=1
λ=1/3, ω=1
t
θ
51015
−3
3
Exercises 3.11
When
r
=3,
θ
=0,so
c
=
−
√
3 ln(1
/
5) and
θ
(
t
)=
√
3

ln
r
15
−
ln
1
5

=
√
3ln
r
3
.
Thus
r
=3
e
θ/
√
3
, whose graph is a logarithmic spiral.
(c)
The time for
S
1
to go from (9
,
0) to (3
,
0) =
1
5
hour. Now
S
1
must intercept the path of
S
2
for some angle
β
, where 0
<β<
2
π
. At the time of interception
t
2
we have 15
t
2
=3
e
β/
√
3
or
t
=
e
β/
√
3
/
5. The total time
is then
t
=
1
5
+
1
5
e
β/
√
3
<
1
5
(1 +
e
2
π/
√
3
)
.
25.
For
λ
2
−
ω
2
>
0 we choose
λ
= 2 and
ω
= 1 with
x
(0)=1
and
x
1
(0) = 2. For
λ
2
−
ω
2
<
0 we choose
λ
=1
/
3 and
ω
=1
with
x
(0) =
−
2 and
x
1
(0) = 4. In both cases the motion
corresponds to the overdamped and underdamped cases for
spring/mass systems.
Exercises 3.11
1.
From
Dx
=2
x
−
y
and
Dy
=
x
we obtain
y
=2
x
−
Dx
,
Dy
=2
Dx
−
D
2
x
, and (
D
2
−
2
D
+1)
x
= 0. Then
x
=
c
1
e
t
+
c
2
te
t
and
y
=(
c
1
−
c
2
)
e
t
+
c
2
te
t
.
2.
From
Dx
=4
x
+7
y
and
Dy
=
x
−
2
y
we obtain
y
=
1
7
Dx
−
4
7
x
,
Dy
=
1
7
D
2
x
−
4
7
Dx
, and (
D
2
−
2
D
−
15)
x
=0.
Then
x
=
c
1
e
5
t
+
c
2
e
−
3
t
and
y
=
1
7
c
1
e
5
t
−
c
2
e
−
3
t
.
3.
From
Dx
=
−
y
+
t
and
Dy
=
x
−
t
we obtain
y
=
t
−
Dx
,
Dy
=1
−
D
2
x
, and (
D
2
+1)
x
=1+
t
. Then
x
=
c
1
cos
t
+
c
2
sin
t
+1+
t
and
y
=
c
1
sin
t
−
c
2
cos
t
+
t
−
1
.
4.
From
Dx
−
4
y
= 1 and
x
+
Dy
= 2 we obtain
y
=
1
4
Dx
−
1
4
,
Dy
=
1
4
D
2
x
, and (
D
2
+1)
x
= 2. Then
x
=
c
1
cos
t
+
c
2
sin
t
+2
and
y
=
1
4
c
2
cos
t
−
1
4
c
1
sin
t
−
1
4
c
1
sin
t
−
1
4
.
5.
From (
D
2
+5)
x
−
2
y
= 0 and
−
2
x
+(
D
2
+2)
y
= 0 we obtain
y
=
1
2
(
D
2
+5)
x
,
D
2
y
=
1
2
(
D
4
+5
D
2
)
x
, and
(
D
2
+ 1)(
D
2
+6)
x
= 0. Then
x
=
c
1
cos
t
+
c
2
sin
t
+
c
3
cos
√
6
t
+
c
4
sin
√
6
t
and
y
=2
c
1
cos
t
+2
c
2
sin
t
−
1
2
c
3
cos
√
6
t
−
1
2
c
4
sin
√
6
t.
149

150. Exercises 3.11
6.
From (
D
+1)
x
+(
D
−
1)
y
= 2 and 3
x
+(
D
+2)
y
=
−
1 we obtain
x
=
−
1
3
−
1
3
(
D
+2)
y
,
Dx
=
−
1
3
(
D
2
+2
D
)
y
,
and (
D
2
+5)
y
=
−
7. Then
y
=
c
1
cos
√
5
t
+
c
2
sin
√
5
t
−
7
5
and
x
=

−
2
3
c
1
−
√
5
3
c
2
a
cos
√
5
t
+

√
5
3
c
1
−
2
3
c
2
a
sin
√
5
t
+
3
5
.
7.
From
D
2
x
=4
y
+
e
t
and
D
2
y
=4
x
−
e
t
we obtain
y
=
1
4
D
2
x
−
1
4
e
t
,
D
2
y
=
1
4
D
4
x
−
1
4
e
t
, and
(
D
2
+ 4)(
D
−
2)(
D
+2)
x
=
−
3
e
t
. Then
x
=
c
1
cos 2
t
+
c
2
sin 2
t
+
c
3
e
2
t
+
c
4
e
−
2
t
+
1
5
e
t
and
y
=
−
c
1
cos 2
t
−
c
2
sin 2
t
+
c
3
e
2
t
+
c
4
e
−
2
t
−
1
5
e
t
.
8.
From (
D
2
+5)
x
+
Dy
= 0 and (
D
+1)
x
+(
D
−
4)
y
= 0 we obtain (
D
−
5)(
D
2
+4)
x
= 0 and (
D
−
5)(
D
2
+4)
y
=0.
Then
x
=
c
1
e
5
t
+
c
2
cos 2
t
+
c
3
sin 2
t
and
y
=
c
4
e
5
t
+
c
5
cos 2
t
+
c
6
sin 2
t.
Substituting into (
D
+1)
x
+(
D
−
4)
y
= 0 gives
(6
c
1
+
c
4
)
e
5
t
+(
c
2
+2
c
3
−
4
c
5
+2
c
6
) cos 2
t
+(
−
2
c
2
+
c
3
−
2
c
5
−
4
c
6
) sin 2
t
=0
so that
c
4
=
−
6
c
1
,
c
5
=
1
2
c
3
,
c
6
=
−
1
2
c
2
, and
y
=
−
6
c
1
e
5
t
+
1
2
c
3
cos 2
t
−
1
2
c
2
sin 2
t.
9.
From
Dx
+
D
2
y
=
e
3
t
and (
D
+1)
x
+(
D
−
1)
y
=4
e
3
t
we obtain
D
(
D
2
+1)
x
=34
e
3
t
and
D
(
D
2
+1)
y
=
−
8
e
3
t
.
Then
y
=
c
1
+
c
2
sin
t
+
c
3
cos
t
−
4
15
e
3
t
and
x
=
c
4
+
c
5
sin
t
+
c
6
cos
t
+
17
15
e
3
t
.
Substituting into (
D
+1)
x
+(
D
−
1)
y
=4
e
3
t
gives
(
c
4
−
c
1
)+(
c
5
−
c
6
−
c
3
−
c
2
) sin
t
+(
c
6
+
c
5
+
c
2
−
c
3
) cos
t
=0
so that
c
4
=
c
1
,
c
5
=
c
3
,
c
6
=
−
c
2
, and
x
=
c
1
−
c
2
cos
t
+
c
3
sin
t
+
17
15
e
3
t
.
10.
From
D
2
x
−
Dy
=
t
and (
D
+3)
x
+(
D
+3)
y
= 2 we obtain
D
(
D
+ 1)(
D
+3)
x
= 1+3
t
and
D
(
D
+ 1)(
D
+3)
y
=
−
1
−
3
t
. Then
x
=
c
1
+
c
2
e
−
t
+
c
3
e
−
3
t
−
t
+
1
2
t
2
and
y
=
c
4
+
c
5
e
−
t
+
c
6
e
−
3
t
+
t
−
1
2
t
2
.
150

151. Exercises 3.11
Substituting into (
D
+3)
x
+(
D
+3)
y
= 2 and
D
2
x
−
Dy
=
t
gives
3(
c
1
+
c
4
)+2(
c
2
+
c
5
)
e
−
t
=2
and
(
c
2
+
c
5
)
e
−
t
+ 3(3
c
3
+
c
6
)
e
−
3
t
=0
so that
c
4
=
−
c
1
,
c
5
=
−
c
2
,
c
6
=
−
3
c
3
, and
y
=
−
c
1
−
c
2
e
−
t
−
3
c
3
e
−
3
t
+
t
−
1
2
t
2
.
11.
From (
D
2
−
1)
x
−
y
= 0 and (
D
−
1)
x
+
Dy
= 0 we obtain
y
=(
D
2
−
1)
x
,
Dy
=(
D
3
−
D
)
x
, and
(
D
−
1)(
D
2
+
D
+1)
x
= 0. Then
x
=
c
1
e
t
+
e
−
t/
2
I
c
2
cos
√
3
2
t
+
c
3
sin
√
3
2
t
f
and
y
=

−
3
2
c
2
−
√
3
2
c
3
a
e
−
t/
2
cos
√
3
2
t
+

√
3
2
c
2
−
3
2
c
3
a
e
−
t/
2
sin
√
3
2
t.
12.
From (2
D
2
−
D
−
1)
x
−
(2
D
+1)
y
= 1 and (
D
−
1)
x
+
Dy
=
−
1 we obtain (2
D
+ 1)(
D
−
1)(
D
+1)
x
=
−
1 and
(2
D
+ 1)(
D
+1)
y
=
−
2. Then
x
=
c
1
e
−
t/
2
+
c
2
e
−
t
+
c
3
e
t
+1
and
y
=
c
4
e
−
t/
2
+
c
5
e
−
t
−
2
.
Substituting into (
D
−
1)
x
+
Dy
=
−
1 gives

−
3
2
c
1
−
1
2
c
4

e
−
t/
2
+(
−
2
c
2
−
c
5
)
e
−
t
=0
so that
c
4
=
−
3
c
1
,
c
5
=
−
2
c
2
, and
y
=
−
3
c
1
e
−
t/
2
−
2
c
2
e
−
t
−
2
.
13.
From (2
D
−
5)
x
+
Dy
=
e
t
and (
D
−
1)
x
+
Dy
=5
e
t
we obtain
Dy
=(5
−
2
D
)
x
+
e
t
and (4
−
D
)
x
=4
e
t
. Then
x
=
c
1
e
4
t
+
4
3
e
t
and
Dy
=
−
3
c
1
e
4
t
+5
e
t
so that
y
=
−
3
4
c
1
e
4
t
+
c
2
+5
e
t
.
14.
From
Dx
+
Dy
=
e
t
and (
−
D
2
+
D
+1)
x
+
y
= 0 we obtain
y
=(
D
2
−
D
−
1)
x
,
Dy
=(
D
3
−
D
2
−
D
)
x
, and
D
2
(
D
−
1)
x
=
e
t
. Then
x
=
c
1
+
c
2
t
+
c
3
e
t
+
te
t
and
y
=
−
c
1
−
c
2
−
c
2
t
−
c
3
e
t
−
te
t
+
e
t
.
15.
Multiplying the first equation by
D
+ 1 and the second equation by
D
2
+ 1 and subtracting we obtain
(
D
4
−
D
2
)
x
= 1. Then
x
=
c
1
+
c
2
t
+
c
3
e
t
+
c
4
e
−
t
−
1
2
t
2
.
151

152. Exercises 3.11
Multiplying the first equation by
D
+ 1 and subtracting we obtain
D
2
(
D
+1)
y
= 1. Then
y
=
c
5
+
c
6
t
+
c
7
e
−
t
−
1
2
t
2
.
Substituting into (
D
−
1)
x
+(
D
2
+1)
y
= 1 gives
(
−
c
1
+
c
2
+
c
5
−
1)+(
−
2
c
4
+2
c
7
)
e
−
t
+(
−
1
−
c
2
+
c
6
)
t
=1
so that
c
5
=
c
1
−
c
2
+2,
c
6
=
c
2
+ 1, and
c
7
=
c
4
. The solution of the system is
x
=
c
1
+
c
2
t
+
c
3
e
t
+
c
4
e
−
t
−
1
2
t
2
y
=(
c
1
−
c
2
+2)+(
c
2
+1)
t
+
c
4
e
−
t
−
1
2
t
2
.
16.
From
D
2
x
−
2(
D
2
+
D
)
y
= sin
t
and
x
+
Dy
= 0 we obtain
x
=
−
Dy
,
D
2
x
=
−
D
3
y
, and
D
(
D
2
+2
D
+2)
y
=
−
sin
t
.
Then
y
=
c
1
+
c
2
e
−
t
cos
t
+
c
3
e
−
t
sin
t
+
1
5
cos
t
+
2
5
sin
t
and
x
=(
c
2
+
c
3
)
e
−
t
sin
t
+(
c
2
−
c
3
)
e
−
t
cos
t
+
1
5
sin
t
−
2
5
cos
t.
17.
From
Dx
=
y
,
Dy
=
z
. and
Dz
=
x
we obtain
x
=
D
2
y
=
D
3
x
so that (
D
−
1)(
D
2
+
D
+1)
x
=0,
x
=
c
1
e
t
+
e
−
t/
2
I
c
2
sin
√
3
2
t
+
c
3
cos
√
3
2
t
f
,
y
=
c
1
e
t
+

−
1
2
c
2
−
√
3
2
c
3
a
e
−
t/
2
sin
√
3
2
t
+

√
3
2
c
2
−
1
2
c
3
a
e
−
t/
2
cos
√
3
2
t,
and
z
=
c
1
e
t
+

−
1
2
c
2
+
√
3
2
c
3
a
e
−
t/
2
sin
√
3
2
t
+

−
√
3
2
c
2
−
1
2
c
3
a
e
−
t/
2
cos
√
3
2
t.
18.
From
Dx
+
z
=
e
t
,(
D
−
1)
x
+
Dy
+
Dz
= 0, and
x
+2
y
+
Dz
=
e
t
we obtain
z
=
−
Dx
+
e
t
,
Dz
=
−
D
2
x
+
e
t
, and
the system (
−
D
2
+
D
−
1)
x
+
Dy
=
−
e
t
and (
−
D
2
+1)
x
+2
y
= 0. Then
y
=
1
2
(
D
2
−
1)
x
,
Dy
=
1
2
D
(
D
2
−
1)
x
,
and (
D
−
2)(
D
2
+1)
x
=
−
2
e
t
so that
x
=
c
1
e
2
t
+
c
2
cos
t
+
c
3
sin
t
+
e
t
,
y
=
3
2
c
1
e
2
t
−
c
2
cos
t
−
c
3
sin
t,
and
z
=
−
2
c
1
e
2
t
−
c
3
cos
t
+
c
2
sin
t.
19.
Write the system in the form
Dx
−
6
y
=0
x
−
Dy
+
z
=0
x
+
y
−
Dz
=0
.
Multiplying the second equation by
D
and adding to the third equation we obtain (
D
+1)
x
−
(
D
2
−
1)
y
=0.
Eliminating
y
between this equation and
Dx
−
6
y
= 0 we find
(
D
3
−
D
−
6
D
−
6)
x
=(
D
+ 1)(
D
+ 2)(
D
−
3)
x
=0
.
152

153. Exercises 3.11
Thus
x
=
c
1
e
−
t
+
c
2
e
−
2
t
+
c
3
e
3
t
,
and, successively substituting into the first and second equations, we get
y
=
−
1
6
c
1
e
−
t
−
1
3
c
2
e
−
2
t
+
1
2
c
3
e
3
t
z
=
−
5
6
c
1
e
−
t
−
1
3
c
2
e
−
2
t
+
1
2
c
3
e
3
t
.
20.
Write the system in the form
(
D
+1)
x
−
z
=0
(
D
+1)
y
−
z
=0
x
−
y
+
Dz
=0
.
Multiplying the third equation by
D
+1 and adding to the second equation we obtain (
D
+1)
x
+(
D
2
+
D
−
1)
z
=0.
Eliminating
z
between this equation and (
D
+1)
x
−
z
= 0 we find
D
(
D
+1)
2
x
=0. Thus
x
=
c
1
+
c
2
e
−
t
+
c
3
te
−
t
,
and, successively substituting into the first and third equations, we get
y
=
c
1
+(
c
2
−
c
3
)
e
−
t
+
c
3
te
−
t
z
=
c
1
+
c
3
e
−
t
.
21.
From (
D
+5)
x
+
y
= 0 and 4
x
−
(
D
+1)
y
= 0 we obtain
y
=
−
(
D
+5)
x
so that
Dy
=
−
(
D
2
+5
D
)
x
. Then
4
x
+(
D
2
+5
D
)
x
+(
D
+5)
x
= 0 and (
D
+3)
2
x
=0. Thus
x
=
c
1
e
−
3
t
+
c
2
te
−
3
t
and
y
=
−
(2
c
1
+
c
2
)
e
−
3
t
−
2
c
2
te
−
3
t
.
Using
x
(1) = 0 and
y
(1) = 1 we obtain
c
1
e
−
3
+
c
2
e
−
3
=0
−
(2
c
1
+
c
2
)
e
−
3
−
2
c
2
e
−
3
=1
or
c
1
+
c
2
=0
2
c
1
+3
c
2
=
−
e
3
.
Thus
c
1
=
e
3
and
c
2
=
−
e
3
. The solution of the initial value problem is
x
=
e
−
3
t
+3
−
te
−
3
t
+3
y
=
−
e
−
3
t
+3
+2
te
−
3
t
+3
.
22.
From
Dx
−
y
=
−
1 and 3
x
+(
D
−
2)
y
= 0 we obtain
x
=
−
1
3
(
D
−
2)
y
so that
Dx
=
−
1
3
(
D
2
−
2
D
)
y
. Then
−
1
3
(
D
2
−
2
D
)
y
=
y
−
1 and (
D
2
−
2
D
+3)
y
=3. Thus
y
=
e
t
3
c
1
cos
√
2
t
+
c
2
sin
√
2
t
4
+1
and
x
=
1
3
e
t
e3
c
1
−
√
2
c
2
4
cos
√
2
t
+
3
√
2
c
1
+
c
2
4
sin
√
2
t
C
+
2
3
.
153

154. 5 10 15 20
t
-3
-2
-1
1
2
3
x1
5 10 15 20
t
-3
-2
-1
1
2
3
x2
Exercises 3.11
Using
x
(0) =
y
(0) = 0 we obtain
c
1
+1=0
1
3
3
c
1
−
√
2
c
2
4
+
2
3
=0
.
Thus
c
1
=
−
1 and
c
2
=
√
2
/
2. The solution of the initial value problem is
x
=
e
t

−
2
3
cos
√
2
t
−
√
2
6
sin
√
2
t
a
+
2
3
y
=
e
t

−
cos
√
2
t
+
√
2
2
sin
√
2
t
a
+1
.
23. (a)
Write the system as
(
D
2
+3)
x
1
−
x
2
=0
−
2
x
1
+(
D
2
+2)
x
2
=0
.
Then
(
D
2
+ 2)(
D
2
+3)
x
1
−
2
x
1
=(
D
2
+ 1)(
D
2
+4)
x
1
=0
,
and
x
1
(
t
)=
c
1
cos
t
+
c
2
sin
t
+
c
3
cos 2
t
+
c
4
sin 2
t.
Since
x
2
=(
D
2
+3)
x
1
,wehave
x
2
(
t
)=2
c
1
cos
t
+2
c
2
sin
t
−
c
3
cos 2
t
−
c
4
sin 2
t.
The initial conditions imply
x
1
(0) =
c
1
+
c
3
=2
x
1
1
(0) =
c
1
+2
c
4
=1
x
2
(0) = 2
c
1
−
c
3
=
−
1
x
1
2
(0) = 2
c
2
−
2
c
4
=1
,
so
c
1
=
1
3
,
c
2
=
2
3
,
c
3
=
5
3
, and
c
4
=
1
6
.Thus
x
1
(
t
)=
1
3
cos
t
+
2
3
sin
t
+
5
3
cos 2
t
+
1
6
sin 2
t
x
2
(
t
)=
2
3
cos
t
+
4
3
sin
t
−
5
3
cos 2
t
−
1
6
sin 2
t.
(b)
In this problem the motion appears to be periodic with period 2
π
. In Example 4 the motion does not
appear to be periodic.
154

155. -2-1 1 2
x1
-2
-1
1
2
3
x2
Exercises 3.11
(c)
24.
Write the system as
0
.
04
i
1
2
+30
i
2
+10
i
3
=50
0
.
02
i
1
3
+10
i
2
+10
i
3
=50
or
i
1
2
+ 750
i
2
+ 250
i
3
= 1250
i
1
3
+ 500
i
2
+ 500
i
3
= 2500
.
In operator notation this becomes
(
D
+ 750)
i
2
+ 250
i
3
= 1250
500
i
2
+(
D
+ 500)
i
3
= 2500
.
Eliminating
i
3
we get
e
(
D
+ 500)(
D
+ 750)
−
250(500)
C
i
2
=(
D
+ 250)(
D
+ 1000)
i
2
=0
.
Thus
i
2
(
t
)=
c
1
e
−
250
t
+
c
2
e
−
1000
t
.
Since
i
3
=
1
250
[1250
−
(
D
+ 750)
i
2
], we have
i
3
(
t
)=
−
2
c
1
e
−
250
t
+
c
2
e
−
1000
t
+5
.
The initial conditions imply
i
2
(0) =
c
1
+
c
2
=0
i
3
(0) =
−
2
c
1
+
c
2
+5=0
,
so
c
1
=
5
3
and
c
2
=
−
5
3
.Thus
i
2
(
t
)=
5
3
e
−
250
t
−
5
3
e
−
1000
t
i
3
(
t
)=
−
10
3
e
−
250
t
−
5
3
e
−
1000
t
+5
.
25.
Equating Newton’s law with the net forces in the
x
- and
y
-directions gives
m
d
2
x
dt
2
= 0 and
m
d
2
y
dt
2
=
−
mg
,
respectively. From
mD
2
x
= 0 we obtain
x
(
t
)=
c
1
t
+
c
2
, and from
mD
2
y
=
−
mg
or
D
2
y
=
−
g
we obtain
y
(
t
)=
−
1
2
gt
2
+
c
3
t
+
c
4
.
26.
From Newton’s second law in the
x
-direction we have
m
d
2
x
dt
2
=
−
k
cos
θ
=
−
k
1
v
dx
dt
=
−|
c
|
dx
dt
.
In the y-direction we have
m
d
2
y
dt
2
=
−
mg
−
k
sin
θ
=
−
mg
−
k
1
v
dy
dt
=
−
mg
−|
c
|
dy
dt
.
155

156. Exercises 3.11
From
mD
2
x
+
|
c
|
Dx
= 0 we have
D
(
mD
+
|
c
|
)
x
= 0 so that (
mD
+
|
c
|
)
x
=
c
1
. This is a first-order linear
equation. An integrating factor is
e
5
|
c
|
dt/m
e
|
c
|
t/m
so that
d
dt
[
e
|
c
|
t/m
x
]=
c
1
e
|
c
|
t/m
and
e
|
c
|
t
x
=(
c
1
m/
|
c
|
)
e
|
c
|
t/m
+
c
2
. The general solution of this equation is
x
(
t
)=
c
3
+
c
2
e
|
c
|
t/m
. From
(
mD
2
+
|
c
|
D
)
y
=
−
mg
we have
D
(
mD
+
|
c
|
)
y
=
−
mg
so that (
mD
+
|
c
|
)
y
=
−
mgt
+
c
1
. This is a first-order
linear equation with integrating factor
e
|
c
|
t/m
.Thus
d
dt
[
e
|
c
|
t/m
y
]=(
−
mgt
+
c
1
)
e
|
c
|
t/m
e
|
c
|
t/m
y
=
−
m
2
g
|
c
|
te
|
c
|
t/m
+
m
3
g
c
2
e
|
c
|
t/m
+
c
1
m
|
c
|
e
|
c
|
t/m
+
c
2
and
y
(
t
)=
−
m
2
g
|
c
|
t
+
m
3
g
c
2
+
c
3
+
c
2
e
−|
c
|
t/m
.
Chapter 3 Review Exercises
1.
y
=0
2.
Since
y
c
=
c
1
e
x
+
c
2
e
−
x
, a particular solution for
y
11
−
y
=1+
e
x
is
y
p
=
A
+
Bxe
x
.
3.
True
4.
True
5.
8 ft., since
k
=4.
6.
2
π/
5, since
1
4
x
11
+6
.
25
x
=0.
7.
5
/
4 m, since
x
=
−
cos 4
t
+
3
4
sin 4
t
.
8.
From
x
(0) = (
√
2
/
2) sin
φ
=
−
1
/
2 we see that sin
φ
=
−
1
/
√
2, so
φ
is an angle in the third or fourth quadrant.
Since
x
1
(
t
)=
√
2 cos(2
t
+
φ
),
x
1
(0) =
√
2 cos
φ
= 1 and cos
φ>
0. Thus
φ
is in the fourth quadrant and
φ
=
−
π/
4.
9.
They are linearly independent over (
−∞
,
∞
) and linearly dependent over (0
,
∞
).
10. (a)
Since
f
2
(
x
)=2ln
x
=2
f
1
(
x
), the functions are linearly dependent.
(b)
Since
x
n
+1
is not a constant multiple of
x
n
, the functions are linearly independent.
(c)
Since
x
+ 1 is not a constant multiple of
x
, the functions are linearly independent.
(d)
Since
f
1
(
x
) = cos
x
cos(
π/
2)
−
sin
x
sin(
π/
2) =
−
sin
x
=
−
f
2
(
x
), the functions are linearly dependent.
(e)
Since
f
1
(
x
)=0
·
f
2
(
x
), the functions are linearly dependent.
(f)
Since 2
x
is not a constant multiple of 2, the functions are linearly independent.
(g)
Since 3(
x
2
) + 2(1
−
x
2
)
−
(2 +
x
2
) = 0, the functions are linearly dependent.
(h)
since
xe
x
+1
+ 0(4
x
−
5)
e
x
−
exe
x
= 0, the functions are linearly dependent.
11. (a)
The auxiliary eqution is (
m
−
3)(
m
+ 5)(
m
−
1) =
m
3
+
m
2
−
17
m
+ 15 = 0, so the differential equation is
y
111
+
y
11
−
17
y
1
+15
y
=0.
(b)
The form of the auxiliary equation is
m
(
m
−
1)(
m
−
2) +
bm
(
m
−
1) +
cm
+
d
=
m
3
+(
b
−
3)
m
2
+(
c
−
b
+2)
m
+
d
=0
.
156

157. Chapter 3 Review Exercises
Since (
m
−
3)(
m
+ 5)(
m
−
1) =
m
3
+
m
2
−
17
m
+15=0,wehave
b
−
3=1,
c
−
b
+2=
−
17, and
d
= 15.
Thus,
b
= 4 and
c
=
−
15, so the differential equation is
y
111
+4
y
11
−
15
y
1
+15
y
=0.
12. (a)
The auxiliary equation is
am
(
m
−
1) +
bm
+
c
=
am
2
+(
b
−
a
)
m
+
c
= 0. If the roots are 3 and
−
1, then
we want (
m
−
3)(
m
+1)=
m
2
−
2
m
−
3 = 0. Thus, let
a
=1,
b
=
−
1, and
c
=
−
3, so that the differential
equation is
x
2
y
11
−
xy
1
−
3
y
=0.
(b)
In this case we want the auxiliary equation to be
m
2
+ 1 = 0, so let
a
=1,
b
= 1, and
c
= 1. Then the
differential equation is
x
2
y
11
+
xy
1
+
y
=0.
13.
From
m
2
−
2
m
−
2 = 0 we obtain
m
=1
±
√
3 so that
y
=
c
1
e
(1+
√
3)
x
+
c
2
e
(1
−
√
3)
x
.
14.
From 2
m
2
+2
m
+ 3 = 0 we obtain
m
=
−
1
/
2
±
√
5
/
2 so that
y
=
e
−
x/
2

c
1
cos
√
5
2
x
+
c
2
sin
√
5
2
x
a
.
15.
From
m
3
+10
m
2
+25
m
= 0 we obtain
m
=0,
m
=
−
5, and
m
=
−
5 so that
y
=
c
1
+
c
2
e
−
5
x
+
c
3
xe
−
5
x
.
16.
From 2
m
3
+9
m
2
+12
m
+ 5 = 0 we obtain
m
=
−
1,
m
=
−
1, and
m
=
−
5
/
2 so that
y
=
c
1
e
−
5
x/
2
+
c
2
e
−
x
+
c
3
xe
−
x
.
17.
From 3
m
3
+10
m
2
+15
m
+ 4 = 0 we obtain
m
=
−
1
/
3 and
m
=
−
3
/
2
±
√
7
/
2 so that
y
=
c
1
e
−
x/
3
+
e
−
3
x/
2

c
2
cos
√
7
2
x
+
c
3
sin
√
7
2
x
a
.
18.
From 2
m
4
+3
m
3
+2
m
2
+6
m
−
4 = 0 we obtain
m
=1
/
2,
m
=
−
2, and
m
=
±
√
2
i
so that
y
=
c
1
e
x/
2
+
c
2
e
−
2
x
+
c
3
cos
√
2
x
+
c
4
sin
√
2
x.
19.
Applying
D
4
to the differential equation we obtain
D
4
(
D
2
−
3
D
+ 5) = 0. Then
y
=
e
3
x/
2

c
1
cos
√
11
2
x
+
c
2
sin
√
11
2
x
a
h
ij k
y
c
+
c
3
+
c
4
x
+
c
5
x
2
+
c
6
x
3
and
y
p
=
A
+
Bx
+
Cx
2
+
Dx
3
. Substituting
y
p
into the differential equation yields
(5
A
−
3
B
+2
C
)+(5
B
−
6
C
+6
D
)
x
+(5
C
−
9
D
)
x
2
+5
Dx
3
=
−
2
x
+4
x
3
.
Equating coefficients gives
A
=
−
222
/
625,
B
=46
/
125,
C
=36
/
25, and
D
=4
/
5. The general solution is
y
=
e
3
x/
2

c
1
cos
√
11
2
x
+
c
2
sin
√
11
2
x
a
−
222
625
+
46
125
x
+
36
25
x
2
+
4
5
x
3
.
20.
Applying (
D
−
1)
3
to the differential equation we obtain (
D
−
1)
3
(
D
−
2
D
+1)=(
D
−
1)
5
= 0. Then
y
=
c
1
e
x
+
c
2
xe
x
h
ijk
y
c
+
c
3
x
2
e
x
+
c
4
x
3
e
x
+
c
5
x
4
e
x
157

158. Chapter 3 Review Exercises
and
y
p
=
Ax
2
e
x
+
Bx
3
e
x
+
Cx
4
e
x
. Substituting
y
p
into the differential equation yields
12
Cx
2
e
x
+6
Bxe
x
+2
Ae
x
=
x
2
e
x
.
Equating coefficients gives
A
=0,
B
= 0, and
C
=1
/
12. The general solution is
y
=
c
1
e
x
+
c
2
xe
x
+
1
12
x
4
e
x
.
21.
Applying
D
(
D
2
+ 1) to the differential equation we obtain
D
(
D
2
+ 1)(
D
3
−
5
D
2
+6
D
)=
D
2
(
D
2
+ 1)(
D
−
2)(
D
−
3) = 0
.
Then
y
=
c
1
+
c
2
e
2
x
+
c
3
e
3
x
h
ij k
y
c
+
c
4
x
+
c
5
cos
x
+
c
6
sin
x
and
y
p
=
Ax
+
B
cos
x
+
C
sin
x
. Substituting
y
p
into the differential equation yields
6
A
+(5
B
+5
C
) cos
x
+(
−
5
B
+5
C
) sin
x
=8+2sin
x.
Equating coefficients gives
A
=4
/
3,
B
=
−
1
/
5, and
C
=1
/
5. The general solution is
y
=
c
1
+
c
2
e
2
x
+
c
3
e
3
x
+
4
3
x
−
1
5
cos
x
+
1
5
sin
x.
22.
Applying
D
to the differential equation we obtain
D
(
D
3
−
D
2
)=
D
3
(
D
−
1) = 0. Then
y
=
c
1
+
c
2
x
+
c
3
e
x
h
ij k
y
c
+
c
4
x
2
and
y
p
=
Ax
2
. Substituting
y
p
into the differential equation yields
−
2
A
= 6. Equating coefficients gives
A
=
−
3.
The general solution is
y
=
c
1
+
c
2
x
+
c
3
e
x
−
3
x
2
.
23.
The auxiliary equation is
m
2
−
2
m
+2=[
m
−
(1 +
i
)][
m
−
(1
−
i
)] = 0, so
y
c
=
c
1
e
x
sin
x
+
c
2
e
x
cos
x
and
W
=
2
2
2
2
e
x
sin
xe
x
cos
x
e
x
cos
x
+
e
x
sin
x
−
e
x
sin
x
+
e
x
cos
x
2
2
2
2
=
−
e
2
x
.
Identifying
f
(
x
)=
e
x
tan
x
we obtain
u
1
1
=
−
(
e
x
cos
x
)(
e
x
tan
x
)
−
e
2
x
= sin
x
u
1
2
=
(
e
x
sin
x
)(
e
x
tan
x
)
−
e
2
x
=
−
sin
2
x
cos
x
= cos
x
−
sec
x.
Then
u
1
=
−
cos
x
,
u
2
= sin
x
−
ln
|
sec
x
+ tan
x
|
, and
y
=
c
1
e
x
sin
x
+
c
2
e
x
cos
x
−
e
x
sin
x
cos
x
+
e
x
sin
x
cos
x
−
e
x
cos
x
ln
|
sec
x
+ tan
x
|
=
c
1
e
x
sin
x
+
c
2
e
x
cos
x
−
e
x
cos
x
ln
|
sec
x
+ tan
x
|
.
24.
The auxiliary equation is
m
2
−
1 = 0, so
y
c
=
c
1
e
x
+
c
2
e
−
x
and
W
=
2
2
2
2
e
x
e
−
x
e
x
−
e
−
x
2
2
2
2
=
−
2
.
158

159. Chapter 3 Review Exercises
Identifying
f
(
x
)=2
e
x
/
(
e
x
+
e
−
x
) we obtain
u
1
1
=
1
e
x
+
e
−
x
=
e
x
1+
e
2
x
u
1
2
=
−
e
2
x
e
x
+
e
−
x
=
−
e
3
x
1+
e
2
x
=
−
e
x
+
e
x
1+
e
2
x
.
Then
u
1
= tan
−
1
e
x
,
u
2
=
−
e
x
+ tan
−
1
e
x
, and
y
=
c
1
e
x
+
c
2
e
−
x
+
e
x
tan
−
1
e
x
−
1+
e
−
x
tan
−
1
e
x
.
25.
The auxiliary equation is 6
m
2
−
m
−
1 = 0 so that
y
=
c
1
x
1
/
2
+
c
2
x
−
1
/
3
.
26.
The auxiliary equation is 2
m
3
+13
m
2
+24
m
+9=(
m
+3)
2
(
m
+1
/
2) = 0 so that
y
=
c
1
x
−
3
+
c
2
x
−
3
ln
x
+
1
4
x
3
.
27.
The auxiliary equation is
m
2
−
5
m
+6=(
m
−
2)(
m
−
3) = 0 and a particular solution is
y
p
=
x
4
−
x
2
ln
x
so
that
y
=
c
1
x
2
+
c
2
x
3
+
x
4
−
x
2
ln
x.
28.
The auxiliary equation is
m
2
−
2
m
+1=(
m
−
1)
2
= 0 and a particular solution is
y
p
=
1
4
x
3
so that
y
=
c
1
x
+
c
2
x
ln
x
+
1
4
x
3
.
29. (a)
The auxiliary equation is
m
2
+
ω
2
=0,so
y
c
=
c
1
cos
ωt
+
c
2
sin
ωt
. When
ω
3
=
α
,
y
p
=
A
cos
αt
+
B
sin
αt
and
y
=
c
1
cos
ωt
+
c
2
sin
ωt
+
A
cos
αt
+
B
sin
αt.
When
ω
=
α
,
y
p
=
At
cos
αt
+
Bt
sin
αt
and
y
=
c
1
cos
ωt
+
c
2
sin
ωt
+
At
cos
αt
+
Bt
sin
αt.
(b)
The auxiliary equation is
m
2
−
ω
2
=0,so
y
c
=
c
1
e
ωt
+
c
2
e
−
ωt
. When
ω
3
=
α
,
y
p
=
Ae
αt
and
y
=
c
1
e
ωt
+
c
2
e
−
ωt
+
Ae
αt
.
When
ω
=
α
,
y
p
=
Ate
αt
and
y
=
c
1
e
ωt
+
c
2
e
−
ωt
+
Ate
αt
.
30. (a)
If
y
= sin
x
is a solution then so is
y
= cos
x
and
m
2
+ 1 is a factor of the auxiliary equation
m
4
+2
m
3
+
11
m
2
+2
m
+ 10 = 0. Dividing by
m
2
+ 1 we get
m
2
+2
m
+ 10, which has roots
−
1
±
3
i
. The general
solution of the differential equation is
y
=
c
1
cos
x
+
c
2
sin
x
+
e
−
x
(
c
3
cos 3
x
+
c
4
sin 3
x
)
.
(b)
The auxiliary equation is
m
(
m
+1)=
m
2
+
m
= 0, so the associated homogeneous differential equation
is
y
11
+
y
1
= 0. Letting
y
=
c
1
+
c
2
e
−
x
+
1
2
x
2
−
x
and computing
y
11
+
y
1
we get
x
. Thus, the differential
equation is
y
11
+
y
1
=
x
.
31. (a)
The auxiliary equation is
m
4
−
2
m
2
+1=(
m
2
−
1)
2
= 0, so the general solution of the differential equation
is
y
=
c
1
sinh
x
+
c
2
cosh
x
+
c
3
x
sinh
x
+
c
4
x
cosh
x.
159

160. Chapter 3 Review Exercises
(b)
Since both sinh
x
and
x
sinh
x
are solutions of the associated homogeneous differential equation, a particular
solution of
y
(4)
−
2
y
11
+
y
= sinh
x
has the form
y
p
=
Ax
2
sinh
x
+
Bx
2
cosh
x
.
32.
Since
y
1
1
= 1 and
y
11
1
=0,
x
2
y
11
1
−
(
x
2
+2
x
)
y
1
1
+(
x
+2)
y
1
=
−
x
2
−
2
x
+
x
2
+2
x
= 0, and
y
1
=
x
is a solution of the
associated homogeneous equation. Using the method of reduction of order, we let
y
=
ux
. Then
y
1
=
xu
1
+
u
and
y
11
=
xu
11
+2
u
1
,so
x
2
y
11
−
(
x
2
+2
x
)
y
1
+(
x
+2)
y
=
x
3
u
11
+2
x
2
u
1
−
x
3
u
1
−
2
x
2
u
1
−
x
2
u
−
2
xu
+
x
2
u
+2
xu
=
x
3
u
11
−
x
3
u
1
=
x
3
(
u
11
−
u
1
)
.
To find a second solution of the homogeneous equation we note that
u
=
e
x
is a solution of
u
11
−
u
1
= 0. Thus,
y
c
=
c
1
x
+
c
2
xe
x
. To find a particular solution we set
x
3
(
u
11
−
u
1
)=
x
3
so that
u
11
−
u
1
= 1. This differential
equation has a particular solution of the form
Ax
. Substituting, we find
A
=
−
1, so a particular solution of the
original differential equation is
y
p
=
−
x
2
and the general solution is
y
=
c
1
x
+
c
2
xe
x
−
x
2
.
33.
The auxiliary equation is
m
2
−
2
m
+ 2 = 0 so that
m
=1
±
i
and
y
=
e
x
(
c
1
cos
x
+
c
2
sin
x
). Setting
y
(
π/
2) = 0
and
y
(
π
)=
−
1 we obtain
c
1
=
e
−
π
and
c
2
= 0. Thus,
y
=
e
x
−
π
cos
x
.
34.
The auxiliary equation is
m
2
+2
m
+1=(
m
+1)
2
= 0, so that
y
=
c
1
e
−
x
+
c
2
xe
−
x
. Setting
y
(
−
1) = 0 and
y
1
(0) = 0 we get
c
1
e
−
c
2
e
= 0 and
−
c
1
+
c
2
=0. Thus
c
1
=
c
2
and
y
=
ce
−
x
+
cxe
−
x
is a solution of the
boundary-value problem for any real number
c
.
35.
The auxiliary equation is
m
2
−
1=(
m
−
1)(
m
+ 1) = 0 so that
m
=
±
1 and
y
=
c
1
e
x
+
c
2
e
−
x
. Assuming
y
p
=
Ax
+
B
+
C
sin
x
and substituting into the differential equation we find
A
=
−
1,
B
= 0, and
C
=
−
1
2
.
Thus
y
p
=
−
x
−
1
2
sin
x
and
y
=
c
1
e
x
+
c
2
e
−
x
−
x
−
1
2
sin
x.
Setting
y
(0) = 2 and
y
1
(0) = 3 we obtain
c
1
+
c
2
=2
c
1
−
c
2
−
3
2
=3
.
Solving this system we find
c
1
=
13
4
and
c
2
=
−
5
4
. The solution of the initial-value problem is
y
=
13
4
e
x
−
5
4
e
−
x
−
x
−
1
2
sin
x.
36.
The auxiliary equation is
m
2
+1=0,so
y
c
=
c
1
cos
x
+
c
2
sin
x
and
W
=
2
2
2
2
cos
x
sin
x
−
sin
x
cos
x
2
2
2
2
=1
.
Identifying
f
(
x
) = sec
3
x
we obtain
u
1
1
=
−
sin
x
sec
3
x
=
−
sin
x
cos
3
x
u
1
2
= cos
x
sec
3
x
= sec
2
x.
Then
u
1
=
−
1
2
1
cos
2
x
=
−
1
2
sec
2
x
u
2
= tan
x.
Thus
y
=
c
1
cos
x
+
c
2
sin
x
−
1
2
cos
x
sec
2
x
+ sin
x
tan
x
160

161. Chapter 3 Review Exercises
=
c
1
cos
x
+
c
2
sin
x
−
1
2
sec
x
+
1
−
cos
2
x
cos
x
=
c
3
cos
x
+
c
2
sin
x
+
1
2
sec
x.
and
y
1
p
=
−
c
3
sin
x
+
c
2
cos
x
+
1
2
sec
x
tan
x.
The initial conditions imply
c
3
+
1
2
=1
c
2
=
1
2
.
Thus
c
3
=
c
2
=1
/
2 and
y
=
1
2
cos
x
+
1
2
sin
x
+
1
2
sec
x.
37.
Let
u
=
y
1
so that
u
1
=
y
11
. The equation becomes
u
du
dx
=4
x
. Separating variables we obtain
udu
=4
xdx
=
⇒
1
2
u
2
=2
x
2
+
c
1
=
⇒
u
2
=4
x
2
+
c
2
.
When
x
=1,
y
1
=
u
=2,so4=4+
c
2
and
c
2
= 0. Then
u
2
=4
x
2
=
⇒
dy
dx
=2
x
or
dy
dx
=
−
2
x
=
⇒
y
=
x
2
+
c
3
or
y
=
−
x
2
+
c
4
.
When
x
=1,
y
=5,so5=1+
c
3
and 5 =
−
1+
c
4
.Thus
c
3
= 4 and
c
4
= 6. We have
y
=
x
2
+ 4 and
y
=
−
x
2
+ 6. Note however that when
y
=
−
x
2
+6,
y
1
=
−
2
x
and
y
1
(1) =
−
2
3
= 2. Thus, the solution of the
initial-value problem is
y
=
x
2
+4.
38.
Let
u
=
y
1
so that
y
11
=
u
du
dy
. The equation becomes 2
u
du
dy
=3
y
2
. Separating variables we obtain
2
udy
=3
y
2
dy
=
⇒
u
2
=
y
3
+
c
1
.
When
x
=0,
y
= 1 and
y
1
=
u
=1so1=1+
c
1
and
c
1
= 0. Then
u
2
=
y
3
=
⇒

dy
dx

2
=
y
3
=
⇒
dy
dx
=
y
3
/
2
=
⇒
y
−
3
/
2
dy
=
dx
=
⇒−
2
y
−
1
/
2
=
x
+
c
2
=
⇒
y
=
4
(
x
+
c
2
)
2
.
When
x
=0,
y
=1,so1=
4
c
2
2
=
⇒
c
2
=
±
2. Thus,
y
=
4
(
x
+2)
2
and
y
=
4
(
x
−
2)
2
. Note however that
when
y
=
4
(
x
+2)
2
,
y
1
=
−
8
(
x
+2)
3
and
y
1
(0) =
−
1
3
= 1. Thus, the solution of the initial-value problem is
y
=
4
(
x
−
2)
2
.
39. (a)
The auxiliary equation is 12
m
4
+64
m
3
+59
m
2
−
23
m
−
12 = 0 and has roots
−
4,
−
3
/
2,
−
1
/
3, and 1
/
2.
The general solution is
y
=
c
1
e
−
4
x
+
c
2
e
−
3
x/
2
+
c
3
e
−
x/
3
+
c
4
e
x/
2
.
161

162. Chapter 3 Review Exercises
(b)
The system of equations is
c
1
+
c
2
+
c
3
+
c
4
=
−
1
−
4
c
1
−
3
2
c
2
−
1
3
c
3
+
1
2
c
4
=2
16
c
1
+
9
4
c
2
+
1
9
c
3
+
1
4
c
4
=5
−
64
c
1
−
27
8
c
2
−
1
27
c
3
+
1
8
c
4
=0
.
Using a CAS we find
c
1
=
−
73
495
,
c
2
=
109
35
,
c
3
=
−
3726
385
, and
c
4
=
257
45
. The solution of the initial-value
problem is
y
=
−
73
495
e
−
4
x
+
109
35
e
−
3
x/
2
−
3726
385
e
−
x/
3
+
257
45
e
x/
2
.
40.
Consider
xy
11
+
y
1
= 0 and look for a solution of the form
y
=
x
m
. Substituting
into the differential equation we have
xy
11
+
y
1
=
m
(
m
−
1)
x
m
−
1
+
mx
m
−
1
=
m
2
x.
Thus, the general solution of
xy
11
+
y
1
=0is
y
c
=
c
1
+
c
2
ln
x
. To find a particular
solution of
xy
11
+
y
1
=
−
√
x
we use variation of parameters. The Wronskian is
W
=
2
2
2
2
1ln
x
01
/x
2
2
2
2
=
1
x
.
Identifying
f
(
x
)=
−
x
−
1
/
2
we obtain
u
1
1
=
x
−
1
/
2
ln
x
1
/x
=
√
x
ln
x
and
u
1
2
=
−
x
−
1
/
2
1
/x
=
−
√
x,
so that
u
1
=
x
3
/
2
3
2
3
ln
x
−
4
9
4
and
u
2
=
−
2
3
x
3
/
2
.
Then
y
p
=
x
3
/
2
1
2
3
ln
x
−
4
9
.
−
2
3
x
3
/
2
ln
x
=
−
4
9
x
3
/
2
and the general solution of the differential equation is
y
=
c
1
+
c
2
ln
x
−
4
9
x
3
/
2
. The initial conditions are
y
(1) = 0 and
y
1
(1) = 0. These imply that
c
1
=
4
9
and
c
2
=
2
3
. The solution of the initial-value problem is
y
=
4
9
+
2
3
ln
x
−
4
9
x
3
/
2
.
41.
From (
D
−
2)
x
+(
D
−
2)
y
= 1 and
Dx
+(2
D
−
1)
y
= 3 we obtain (
D
−
1)(
D
−
2)
y
=
−
6 and
Dx
=3
−
(2
D
−
1)
y
.
Then
y
=
c
1
e
2
t
+
c
2
e
t
−
3 and
x
=
−
c
2
e
t
−
3
2
c
1
e
2
t
+
c
3
.
Substituting into (
D
−
2)
x
+(
D
−
2)
y
= 1 gives
c
3
=5
/
2 so that
x
=
−
c
2
e
t
−
3
2
c
1
e
2
t
+
5
2
.
42.
From (
D
−
2)
x
−
y
=
t
−
2 and
−
3
x
+(
D
−
4)
y
=
−
4
t
we obtain (
D
−
1)(
D
−
5)
x
=9
−
8
t
. Then
x
=
c
1
e
t
+
c
2
e
5
t
−
8
5
t
−
3
25
and
y
=(
D
−
2)
x
−
t
+2=
−
c
1
e
t
+3
c
2
e
5
t
+
16
25
+
11
25
t.
43.
From (
D
−
2)
x
−
y
=
−
e
t
and
−
3
x
+(
D
−
4)
y
=
−
7
e
t
we obtain (
D
−
1)(
D
−
5)
x
=
−
4
e
t
so that
x
=
c
1
e
t
+
c
2
e
5
t
+
te
t
.
162

163. Chapter 3 Review Exercises
Then
y
=(
D
−
2)
x
+
e
t
=
−
c
1
e
t
+3
c
2
e
5
t
−
te
t
+2
e
t
.
44.
From (
D
+2)
x
+(
D
+1)
y
= sin 2
t
and 5
x
+(
D
+3)
y
= cos 2
t
we obtain (
D
2
+5)
y
= 2 cos 2
t
−
7 sin 2
t
. Then
y
=
c
1
cos
t
+
c
2
sin
t
−
2
3
cos 2
t
+
7
3
sin 2
t
and
x
=
−
1
5
(
D
+3)
y
+
1
5
cos 2
t
=

1
5
c
1
−
3
5
c
2

sin
t
+

−
1
5
c
2
−
3
5
c
1

cos
t
−
5
3
sin 2
t
−
1
3
cos 2
t.
45.
The period of a spring mass system is given by
T
=2
π/ω
where
ω
2
=
k/m
=
kg/W
, where
k
is the spring
constant,
W
is the weight of the mass attached to the spring, and
g
is the acceleration due to gravity. Thus,
the period of oscillation is
T
=(2
π/
√
kg
)
√
W
. If the weight of the original mass is
W
, then (2
π/
√
kg
)
√
W
=3
and (2
π/
√
kg
)
√
W
−
8 = 2. Dividing, we get
√
W/
√
W
−
8=3
/
2or
W
=
9
4
(
W
−
8). Solving for
W
we find
that the weight of the original mass was 14
.
4 pounds.
46. (a)
Solving
3
8
x
11
+6
x
= 0 subject to
x
(0) = 1 and
x
1
(0) =
−
4 we obtain
x
= cos 4
t
−
sin 4
t
=
√
2 sin (4
t
+3
π/
4)
.
(b)
The amplitude is
√
2, period is
π/
2, and frequency is 2
/π
.
(c)
If
x
= 1 then
t
=
nπ/
2 and
t
=
−
π/
8+
nπ/
2 for
n
=1,2,3,
...
.
(d)
If
x
= 0 then
t
=
π/
16 +
nπ/
4 for
n
=0,1,2,
...
. The motion is upward for
n
even and downward for
n
odd.
(e)
x
1
(3
π/
16) = 0
(f)
If
x
1
= 0 then 4
t
+3
π/
4=
π/
2+
nπ
or
t
=3
π/
16 +
nπ
.
47.
From
mx
11
+4
x
1
+2
x
= 0 we see that non-oscillatory motion results if 16
−
8
m
≥
0or0
<m
≤
2.
48.
From
x
11
+
βx
1
+64
x
= 0 we see that oscillatory motion results if
β
2
−
256
<
0or0
≤|
β
|
<
16.
49.
From
q
11
+10
4
q
= 100 sin 50
t
,
q
(0) = 0, and
q
1
(
t
) = 0 we obtain
q
c
=
c
1
cos 100
t
+
c
2
sin 100
t
,
q
p
=
1
75
sin 50
t
, and
(a)
q
=
−
1
150
sin 100
t
+
1
75
sin 50
t
,
(b)
i
=
−
2
3
cos 100
t
+
2
3
cos 50
t
, and
(c)
q
= 0 when sin 50
t
(1
−
cos 50
t
)=0or
t
=
nπ/
50 for
n
=0,1,2,
...
.
50.
Differentiate
L
d
2
q
dt
2
+
R
dq
dt
+
1
C
q
=
E
(
t
) and use
q
1
(
t
)=
i
(
t
) to obtain the desired result.
51.
For
λ>
0 the general solution is
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx
.Now
y
(0) =
c
1
and
y
(2
π
)=
c
1
cos 2
π
√
λ
+
c
2
sin 2
π
√
λ
, so the condition
y
(0) =
y
(2
π
) implies
c
1
=
c
1
cos 2
π
√
λ
+
c
2
sin 2
π
√
λ
which is true when
√
λ
=
n
or
λ
=
n
2
for
n
=1,2,3,
...
. Since
y
1
=
−
√
λc
1
sin
√
λx
+
√
λc
2
cos
√
λx
=
−
nc
1
sin
nx
+
nc
2
cos
nx
, we see that
y
1
(0) =
nc
2
=
y
1
(2
π
) for
n
=1,2,3,
...
. Thus, the eigenvalues are
n
2
for
n
=1,2,3,
...
, with corresponding eigenfunctions cos
nx
and sin
nx
. When
λ
= 0, the general
solution is
y
=
c
1
x
+
c
2
and the corresponding eigenfunction is
y
=1. For
λ<
0 the general solution is
y
=
c
1
cosh
√
−
λx
+
c
2
sinh
√
−
λx
. In this case
y
(0) =
c
1
and
y
(2
π
)=
c
1
cosh 2
π
√
−
λ
+
c
2
sinh 2
π
√
−
λ
,so
y
(0) =
y
(2
π
) can only be valid for
λ
= 0. Thus, there are no eigenvalues corresponding to
λ<
0.
163

164. 2 4 6 8 10 12 14
t
-20
-10
10
20
r
0
10
15
16
16.1
17
v
0
0 10 15 16.1 17
r
-20 -20 -20 20 20
t
1.55007 2.35494 3.43088 6.11627 4.22339
Chapter 3 Review Exercises
52. (a)
The differential equation is
d
2
r/dt
2
−
ω
2
r
=
−
g
sin
ωt
. The auxiliary equation is
m
2
−
ω
2
=0,so
r
c
=
c
1
e
ωt
+
c
2
e
−
ωt
. A particular solution has the form
r
p
=
A
sin
ωt
+
B
cos
ωt
. Substituting into the differential
equation we find
−
2
Aω
2
sin
ωt
−
2
Bω
2
cos
ωt
=
−
g
sin
ωt
. Thus,
B
=0,
A
=
g/
2
ω
2
, and
r
p
=(
g/
2
ω
2
) sin
ωt
.
The general solution of the differential equation is
r
(
t
)=
c
1
e
ωt
+
c
2
e
−
ωt
+(
g/
2
ω
2
) sin
ωt
. The initial
conditions imply
c
1
+
c
2
=
r
0
and
g/
2
ω
−
ωc
1
+
ωc
2
=
v
0
. Solving for
c
1
and
c
2
we get
c
1
=(2
ω
2
r
0
+2
ωv
0
−
g
)
/
4
ω
2
and
c
2
=(2
ω
2
r
0
−
2
ωv
0
+
g
)
/
4
ω
2
,
so that
r
(
t
)=
2
ω
2
r
0
+2
ωv
0
−
g
4
ω
2
e
ωt
+
2
ω
2
r
0
−
2
ωv
0
+
g
4
ω
2
e
−
ωt
+
g
2
ω
2
sin
ωt.
(b)
The bead will exhibit simple harmonic motion when the exponential terms are missing. Solving
c
1
=0,
c
2
= 0 for
r
0
and
v
0
we find
r
0
= 0 and
v
0
=
g/
2
ω
.
To find the minimum length of rod that will accommodate simple harmonic motion we determine the
amplitude of
r
(
t
) and double it. Thus
L
=
g/ω
2
.
(c)
As
t
increases,
e
ωt
approaches infinity and
e
−
ωt
approaches 0. Since sin
ωt
is bounded, the distance,
r
(
t
), of
the bead from the pivot point increases without bound and the distance of the bead from
P
will eventually
exceed
L/
2.
(d)
(e)
For each
v
0
we want to find the smallest value of
t
for which
r
(
t
)=
±
20. Whether we look for
r
(
t
)=
−
20
or
r
(
t
) = 20 is determined by looking at the graphs in part (d). The total times that the bead stays on the
rod is shown in the table below.
When
v
0
= 16 the bead never leaves the rod.
164

165. 4
The Laplace Transform
Exercises 4.1
1.
{
f
(
t
)
}
=
(
1
0
−
e
−
st
dt
+
(
∞
1
e
−
st
dt
=
1
s
e
−
st
)
)
)
)
1
0
−
1
s
e
−
st
)
)
)
)
∞
1
=
1
s
e
−
s
−
1
s
−
=
0
−
1
s
e
−
s
+
=
2
s
e
−
s
−
1
s
,s>
0
2.
{
f
(
t
)
}
=
(
2
0
4
e
−
st
dt
=
−
4
s
e
−
st
)
)
)
)
2
0
=
−
4
s
(
e
−
2
s
−
1)
,s>
0
3.
{
f
(
t
)
}
=
(
1
0
te
−
st
dt
+
(
∞
1
e
−
st
dt
=
=
−
1
s
te
−
st
−
1
s
2
e
−
st
+
)
)
)
)
1
0
−
1
s
e
−
st
)
)
)
)
∞
1
=
=
−
1
s
e
−
s
−
1
s
2
e
−
s
+
−
=
0
−
1
s
2
+
−
1
s
(0
−
e
−
s
)=
1
s
2
(1
−
e
−
s
)
,s>
0
4.
{
f
(
t
)
}
=
(
1
0
(2
t
+1)
e
−
st
dt
=
=
−
2
s
te
−
st
−
2
s
2
e
−
st
−
1
s
e
−
st
+
)
)
)
)
1
0
=
=
−
2
s
e
−
s
−
2
s
2
e
−
s
−
1
s
e
−
s
+
−
=
0
−
2
s
2
−
1
s
+
=
1
s
(1
−
3
e
−
s
)+
2
s
2
(1
−
e
−
s
)
,s>
0
5.
{
f
(
t
)
}
=
(
π
0
(sin
t
)
e
−
st
dt
=
=
−
s
s
2
+1
e
−
st
sin
t
−
1
s
2
+1
e
−
st
cos
t
+
)
)
)
)
π
0
=
=
0+
1
s
2
+1
e
−
πs
+
−
=
0
−
1
s
2
+1
+
=
1
s
2
+1
(
e
−
πs
+1)
,s>
0
6.
{
f
(
t
)
}
=
(
∞
π/
2
(cos
t
)
e
−
st
dt
=
=
−
s
s
2
+1
e
−
st
cos
t
+
1
s
2
+1
e
−
st
sin
t
+
)
)
)
)
∞
π/
2
=0
−
=
0+
1
s
2
+1
e
−
πs/
2
+
=
−
1
s
2
+1
e
−
πs/
2
,s>
0
7.
f
(
t
)=
1
0
,
0
<t<
1
t, t >
1
{
f
(
t
)
}
=
(
∞
1
te
−
st
dt
=
=
−
1
s
te
−
st
−
1
s
2
e
−
st
+
)
)
)
)
∞
1
=
1
s
e
−
s
+
1
s
2
e
−
s
,
s>
0
8.
f
(
t
)=
1
0
,
0
<t<
1
2
t
−
2
,t>
1
{
f
(
t
)
}
=2
(
∞
1
(
t
−
1)
e
−
st
dt
=2
=
−
1
s
(
t
−
1)
e
−
st
−
1
s
2
e
−
st
+
)
)
)
)
∞
1
=
2
s
2
e
−
s
,
s>
0
9.
f
(
t
)=
1
1
−
t,
0
<t<
1
0
,t>
0
{
f
(
t
)
}
=
(
1
0
(1
−
t
)
e
−
st
dt
=
=
−
1
s
(1
−
t
)
e
−
st
+
1
s
2
e
−
st
+
)
)
)
)
1
0
=
1
s
2
e
−
s
+
1
s
−
1
s
2
,
s>
0
10.
f
(
t
)=



0
,
0
<t<a
c, a<t<b
0
,t>b
;
{
f
(
t
)
}
=
(
b
a
ce
−
st
dt
=
−
c
s
e
−
st
)
)
)
)
b
a
=
c
s
(
e
−
sa
−
e
−
sb
),
s>
0
165

166. Exercises 4.1
11.
{
f
(
t
)
}
=
(
∞
0
e
t
+7
e
−
st
dt
=
e
7
(
∞
0
e
(1
−
s
)
t
dt
=
e
7
1
−
s
e
(1
−
s
)
t
)
)
)
)
∞
0
=0
−
e
7
1
−
s
=
e
7
s
−
1
,s>
1
12.
{
f
(
t
)
}
=
(
∞
0
e
−
2
t
−
5
e
−
st
dt
=
e
−
5
(
∞
0
e
−
(
s
+2)
t
dt
=
−
e
−
5
s
+2
e
−
(
s
+2)
t
)
)
)
)
∞
0
=
e
−
5
s
+2
,s>
−
2
13.
{
f
(
t
)
}
=
(
∞
0
te
4
t
e
−
st
dt
=
(
∞
0
te
(4
−
s
)
t
dt
=
=
1
4
−
s
te
(4
−
s
)
t
−
1
(4
−
s
)
2
e
(4
−
s
)
t
+
)
)
)
)
∞
0
=
1
(4
−
s
)
2
,s>
4
14.
{
f
(
t
)
}
=
(
∞
0
t
2
e
−
2
t
e
−
st
dt
=
(
∞
0
t
2
e
−
(
s
+2)
t
dt
=
=
−
1
s
+2
t
2
e
−
(
s
+2)
t
−
2
(
s
+2)
2
te
−
(
s
+2)
t
−
2
(
s
+2)
3
e
−
(
s
+2)
t
+
)
)
)
)
∞
0
=
2
(
s
+2)
3
,s>
−
2
15.
{
f
(
t
)
}
=
(
∞
0
e
−
t
(sin
t
)
e
−
st
dt
=
(
∞
0
(sin
t
)
e
−
(
s
+1)
t
dt
=
=
−
(
s
+1)
(
s
+1)
2
+1
e
−
(
s
+1)
t
sin
t
−
1
(
s
+1)
2
+1
e
−
(
s
+1)
t
cos
t
+
)
)
)
)
∞
0
=
1
(
s
+1)
2
+1
=
1
s
2
+2
s
+2
,s>
−
1
16.
{
f
(
t
)
}
=
(
∞
0
e
t
(cos
t
)
e
−
st
dt
=
(
∞
0
(cos
t
)
e
(1
−
s
)
t
dt
=
=
1
−
s
(1
−
s
)
2
+1
e
(1
−
s
)
t
cos
t
+
1
(1
−
s
)
2
+1
e
(1
−
s
)
t
sin
t
+
)
)
)
)
∞
0
=
−
1
−
s
(1
−
s
)
2
+1
=
s
−
1
s
2
−
2
s
+2
,s>
1
17.
{
f
(
t
)
}
=
(
∞
0
t
(cos
t
)
e
−
st
dt
=

−
st
s
2
+1
−
s
2
−
1
(
s
2
+1)
2

(cos
t
)
e
−
st
+

t
s
2
+1
+
2
s
(
s
2
+1)
2

(sin
t
)
e
−
st

∞
0
=
s
2
−
1
(
s
2
+1)
2
,s>
0
18.
{
f
(
t
)
}
=
(
∞
0
t
(sin
t
)
e
−
st
dt
=

−
t
s
2
+1
−
2
s
(
s
2
+1)
2

(cos
t
)
e
−
st
−

st
s
2
+1
+
s
2
−
1
(
s
2
+1)
2

(sin
t
)
e
−
st

∞
0
=
2
s
(
s
2
+1)
2
,s>
0
19.
{
2
t
4
}
=2
4!
s
5
20.
{
t
5
}
=
5!
s
6
21.
{
4
t
−
10
}
=
4
s
2
−
10
s
22.
{
7
t
+3
}
=
7
s
2
+
3
s
23.
{
t
2
+6
t
−
3
}
=
2
s
3
+
6
s
2
−
3
s
24.
{−
4
t
2
+16
t
+9
}
=
−
4
2
s
3
+
16
s
2
+
9
s
25.
{
t
3
+3
t
2
+3
t
+1
}
=
3!
s
4
+3
2
s
3
+
3
s
2
+
1
s
26.
{
8
t
3
−
12
t
2
+6
t
−
1
}
=8
3!
s
4
−
12
2
s
3
+
6
s
2
−
1
s
166

167. Exercises 4.1
27.
{
1+
e
4
t
}
=
1
s
+
1
s
−
4
28.
{
t
2
−
e
−
9
t
+5
}
=
2
s
3
−
1
s
+9
+
5
s
29.
{
1+2
e
2
t
+
e
4
t
}
=
1
s
+
2
s
−
2
+
1
s
−
4
30.
{
e
2
t
−
2+
e
−
2
t
}
=
1
s
−
2
−
2
s
+
1
s
+2
31.
{
4
t
2
−
5 sin 3
t
}
=4
2
s
3
−
5
3
s
2
+9
32.
{
cos 5
t
+ sin 2
t
}
=
s
s
2
+25
+
2
s
2
+4
33.
{
sinh
kt
}
=
k
s
2
−
k
2
34.
{
cosh
kt
}
=
s
s
2
−
k
2
35.
{
e
t
sinh
t
}
=
1
e
t
e
t
−
e
−
t
2
c
=
1
1
2
e
2
t
−
1
2
c
=
1
2(
s
−
2)
−
1
2
s
36.
{
e
−
t
cosh
t
}
=
1
e
−
t
e
t
+
e
−
t
2
c
=
1
1
2
+
1
2
e
−
2
t
c
=
1
2
s
+
1
2(
s
+2)
37.
{
sin 2
t
cos 2
t
}
=
1
1
2
sin 4
t
c
=
2
s
2
+16
38.
{
cos
2
t
}
=
1
1
2
+
1
2
cos 2
t
c
=
1
2
s
+
1
2
s
s
2
+4
39. (a)
Using integration byparts for
α>
0,
Γ(
α
+1)=
(
∞
0
t
α
e
−
t
dt
=
−
t
α
e
−
t
)
)
)
∞
0
+
α
(
∞
0
t
α
−
1
e
−
t
dt
=
α
Γ(
α
)
.
(b)
Let
u
=
st
so that
du
=
sdt
. Then
{
t
α
}
=
(
∞
0
e
−
st
t
α
dt
=
(
∞
0
e
−
u
o
u
s
,
α
1
s
du
=
1
s
α
+1
Γ(
α
+1)
,α>
−
1
.
40. (a)
{
t
−
1
/
2
}
=
Γ(1
/
2)
s
1
/
2
=
;
π
s
(b)
{
t
1
/
2
}
=
Γ(3
/
2)
s
3
/
2
=
√
π
2
s
3
/
2
(c)
{
t
3
/
2
}
=
Γ(5
/
2)
s
5
/
2
=
3
√
π
4
s
5
/
2
41.
Identifying
f
(
t
)=
t
n
we have
f
)
(
t
)=
nt
n
−
1
,
n
=1,2,3,
...
. Then, since
f
(0) = 0,
n
{
t
n
−
1
}
=
{
nt
n
−
1
}
=
s
{
t
n
}
and
{
t
n
}
=
n
s
{
t
n
−
1
}
.
For
n
=1,
{
t
}
=
1
s
{
1
}
=
1
s
2
.
For
n
=2,
{
t
2
}
=
2
s
{
t
}
=
2
s
3
.
For
n
=3,
{
t
3
}
=
3
s
{
t
2
}
=
6
s
4
.
42.
Let
F
(
t
)=
t
1
/
3
. Then
F
(
t
) is of exponential order, but
f
(
t
)=
F
)
(
t
)=
1
3
t
−
2
/
3
is unbounded near
t
= 0 and
hence is not of exponential order.
167

168. Exercises 4.1
Let
f
(
t
)=2
te
t
2
cos
e
t
2
=
d
dt
sin
e
t
2
. This function is not of exponential order, but we can show that its Laplace
transform exists. Using integration byparts we have
{
2
te
t
2
cos
e
t
2
}
=
(
∞
0
e
−
st
=
d
dt
sin
e
t
2
+
dt
= lim
a
→∞
!
e
−
st
sin
e
t
2
)
)
)
a
0
+
s
(
a
0
e
−
st
sin
e
t
2
dt
5
=
s
(
∞
0
e
−
st
sin
e
t
2
dt
=
s
{
sin
e
t
2
}
.
Since sin
e
t
2
is continuous and of exponential order,
{
sin
e
t
2
}
exists, and therefore
{
2
te
t
2
cos
e
t
2
}
exists.
43.
The relation will be valid when
s
is greater than the maximum of
c
1
and
c
2
.
44.
Since
e
t
is an increasing function and
t
2
>
ln
M
+
ct
for
M>
0wehave
e
t
2
>e
ln
M
+
ct
=
Me
ct
for
t
sufficiently
large and for any
c
. Thus,
e
t
2
is not of exponential order.
45.
Bypart (c) of Theorem 4
.
1
{
e
(
a
+
ib
)
t
}
=
1
s
−
(
a
+
ib
)
=
1
(
s
−
a
)
−
ib
(
s
−
a
)+
ib
(
s
−
a
)+
ib
=
s
−
a
+
ib
(
s
−
a
)
2
+
b
2
.
ByEuler’s formula,
e
iθ
= cos
θ
+
i
sin
θ
,so
{
e
(
a
+
ib
)
t
}
=
{
e
at
e
ibt
}{
e
at
(cos
bt
+
i
sin
bt
)
}
=
{
e
at
cos
bt
}
+
i
{
e
at
sin
bt
}
=
s
−
a
(
s
−
a
)
2
+
b
2
+
i
b
(
s
−
a
)
2
+
b
2
.
Equating real and imaginaryparts we get
{
e
at
cos
bt
}
=
s
−
a
(
s
−
a
)
2
+
b
2
and
{
e
at
sin
bt
}
=
b
(
s
−
a
)
2
+
b
2
.
46.
We want
f
(
αx
+
βy
)=
αf
(
x
)+
βf
(
y
)or
m
(
αx
+
βy
)+
b
=
α
(
mx
+
b
)+
β
(
my
+
b
)=
m
(
αx
+
βy
)+(
α
+
β
)
b
for all real numbers
α
and
β
. Taking
α
=
β
= 1 we see that
b
=2
b
,so
b
= 0. Thus,
f
(
x
)=
mx
+
b
will be a
linear transformation when
b
=0.
Exercises 4.2
1.
1
1
s
3
c
=
1
2
1
2
s
3
c
=
1
2
t
2
2.
1
1
s
4
c
=
1
6
1
3!
s
4
c
=
1
6
t
3
3.
1
1
s
2
−
48
s
5
c
=
1
1
s
2
−
48
24
·
4!
s
5
c
=
t
−
2
t
4
4.
7
=
2
s
−
1
s
3
+
2
6
=
1
4
·
1
s
2
−
4
6
·
3!
s
4
+
1
120
·
5!
s
6
c
=4
t
−
2
3
t
3
+
1
120
t
5
5.
1
(
s
+1)
3
s
4
c
=
1
1
s
+3
·
1
s
2
+
3
2
·
2
s
3
+
1
6
·
3!
s
4
c
=1+3
t
+
3
2
t
2
+
1
6
t
3
6.
1
(
s
+2)
2
s
3
c
=
1
1
s
+4
·
1
s
2
+2
·
2
s
3
c
=1+4
t
+2
t
2
168

169. Exercises 4.2
7.
1
1
s
2
−
1
s
+
1
s
−
2
c
=
t
−
1+
e
2
t
8.
1
4
s
+
6
s
5
−
1
s
+8
c
=
1
4
·
1
s
+
1
4
·
4!
s
5
−
1
s
+8
c
=4+
1
4
t
4
−
e
−
8
t
9.
1
1
4
s
+1
c
=
1
1
4
·
1
s
+1
/
4
c
=
1
4
e
−
t/
4
10.
1
1
5
s
−
2
c
=
1
1
5
·
1
s
−
2
/
5
c
=
1
5
e
2
t/
5
11.
1
5
s
2
+49
c
=
1
5
7
·
7
s
2
+49
c
=
5
7
sin 7
t
12.
1
10
s
s
2
+16
c
= 10 cos 4
t
13.
1
4
s
4
s
2
+1
c
=
1
s
s
2
+1
/
4
c
= cos
1
2
t
14.
1
1
4
s
2
+1
c
=
1
1
2
·
1
/
2
s
2
+1
/
4
c
=
1
2
sin
1
2
t
15.
1
2
s
−
6
s
2
+9
c
=
1
2
·
s
s
2
+9
−
2
·
3
s
2
+9
c
= 2 cos 3
t
−
2 sin 3
t
16.
1
s
+1
s
2
+2
c
=
7
s
s
2
+2
+
1
√
2
√
2
s
2
+2
6
= cos
√
2
t
+
√
2
2
sin
√
2
t
17.
1
1
s
2
+3
s
c
=
1
1
3
·
1
s
−
1
3
·
1
s
+3
c
=
1
3
−
1
3
e
−
3
t
18.
1
s
+1
s
2
−
4
s
c
=
1
−
1
4
·
1
s
+
5
4
·
1
s
−
4
c
=
−
1
4
+
5
4
e
4
t
19.
1
s
s
2
+2
s
−
3
c
=
1
1
4
·
1
s
−
1
+
3
4
·
1
s
+3
c
=
1
4
e
t
+
3
4
e
−
3
t
20.
1
1
s
2
+
s
−
20
c
=
1
1
9
·
1
s
−
4
−
1
9
·
1
s
+5
c
=
1
9
e
4
t
−
1
9
e
−
5
t
21.
1
0
.
9
s
(
s
−
0
.
1)(
s
+0
.
2)
c
=
1
(0
.
3)
·
1
s
−
0
.
1
+(0
.
6)
·
1
s
+0
.
2
c
=0
.
3
e
0
.
1
t
+0
.
6
e
−
0
.
2
t
22.
1
s
−
3
(
s
−
√
3)(
s
+
√
3)
c
=
7
s
s
2
−
3
−
√
3
·
√
3
s
2
−
3
6
= cosh
√
3
t
−
√
3 sinh
√
3
t
23.
1
s
(
s
−
2)(
s
−
3)(
s
−
6)
c
=
1
1
2
·
1
s
−
2
−
1
s
−
3
+
1
2
·
1
s
−
6
c
=
1
2
e
2
t
−
e
3
t
+
1
2
e
6
t
24.
1
s
2
+1
s
(
s
−
1)(
s
+ 1)(
s
−
2)
c
=
1
1
2
·
1
s
−
1
s
−
1
−
1
3
·
1
s
+1
+
5
6
·
1
s
−
2
c
=
1
2
−
e
t
−
1
3
e
−
t
+
5
6
e
2
t
25.
1
1
s
3
+5
s
c
=
1
1
s
(
s
2
+5)
c
=
1
1
5
·
1
s
−
1
5
s
s
2
+5
c
=
1
5
−
1
5
cos
√
5
t
169

170. Exercises 4.2
26.
1
s
(
s
2
+ 4)(
s
+2)
c
=
1
1
4
·
s
s
2
+4
+
1
4
·
2
s
2
+4
−
1
4
·
1
s
+2
c
=
1
4
cos 2
t
+
1
4
sin 2
t
−
1
4
e
−
2
t
27.
1
2
s
−
4
(
s
2
+
s
)(
s
2
+1)
c
=
1
2
s
−
4
s
(
s
2
+1)
2
c
=
1
−
4
s
+
3
s
+1
+
s
s
2
+1
+
3
s
2
+1
c
=
−
4+3
e
−
t
+ cos
t
+ 3 sin
t
28.
1
1
s
4
−
9
c
=
7
1
6
√
3
·
√
3
s
2
−
3
−
1
6
√
3
·
√
3
s
2
−
3
6
=
1
6
√
3
sinh
√
3
t
−
1
6
√
3
sin
√
3
t
29.
1
s
(
s
2
+ 4)(
s
+2)
c
=
1
1
4
·
s
s
2
+4
+
1
4
·
2
s
2
+4
−
1
4
·
1
s
+2
c
=
1
4
cos 2
t
+
1
4
sin 2
t
−
1
4
e
−
2
t
30.
1
6
s
+3
(
s
2
+ 1)(
s
2
+4)
c
=
1
2
·
s
s
2
+1
+
1
s
2
+1
−
2
·
s
s
2
+4
−
1
2
·
2
s
2
+4
c
= 2 cos
t
+ sin
t
−
2 cos 2
t
−
1
2
sin 2
t
31.
The Laplace transform of the differential equation is
s
{
y
}−
y
(0)
−{
y
}
=
1
s
.
Solving for
{
y
}
we obtain
{
y
}
=
−
1
s
+
1
s
−
1
.
Thus
y
=
−
1+
e
t
.
32.
The Laplace transform of the differential equation is
2
s
{
y
}−
2
y
(0) =
{
y
}
=0
.
Solving for
{
y
}
we obtain
{
y
}
=
6
2
s
+1
=
3
s
+1
/
2
.
Thus
y
=3
e
−
t/
2
.
33.
The Laplace transform of the differential equation is
s
{
y
}−
y
(0)+6
{
y
}
=
1
s
−
4
.
Solving for
{
y
}
we obtain
{
y
}
=
1
(
s
−
4)(
s
+6)
+
2
s
+6
=
1
10
·
1
s
−
4
+
19
10
·
1
s
+6
.
Thus
y
=
1
10
e
4
t
+
19
10
e
−
6
t
.
34.
The Laplace transform of the differential equation is
s
{
y
}− {
y
}
=
2
s
s
2
+25
.
170

171. Exercises 4.2
Solving for
{
y
}
we obtain
{
y
}
=
2
(
s
−
1)(
s
2
+ 25)
=
1
13
·
1
s
−
1
−
1
13
s
s
2
+25
+
5
13
·
5
s
2
+25
.
Thus
y
=
1
13
e
t
−
1
13
cos 5
t
+
5
13
sin 5
t.
35.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)+5[
s
{
y
}−
y
(0)] + 4
{
y
}
=0
.
Solving for
{
y
}
we obtain
{
y
}
=
s
+5
s
2
+5
s
+4
=
4
3
1
s
+1
−
1
3
1
s
+4
.
Thus
y
=
4
3
e
−
t
−
1
3
e
−
4
t
.
36.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)
−
4[
s
{
y
}−
y
(0)] =
6
s
−
3
−
3
s
+1
.
Solving for
{
y
}
we obtain
{
y
}
=
6
(
s
−
3)(
s
2
−
4
s
)
−
3
(
s
+ 1)(
s
2
−
4
s
)
+
s
−
5
s
2
−
4
s
=
5
2
·
1
s
−
2
s
−
3
−
3
5
·
1
s
+1
+
11
10
·
1
s
−
4
.
Thus
y
=
5
2
−
2
e
3
t
−
3
5
e
−
t
+
11
10
e
4
t
.
37.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0) +
{
y
}
=
2
s
2
+2
.
Solving for
{
y
}
we obtain
{
y
}
=
2
(
s
2
+ 1)(
s
2
+2)
+
10
s
s
2
+1
=
10
s
s
2
+1
+
2
s
2
+1
−
2
s
2
+2
.
Thus
y
= 10 cos
t
+ 2 sin
t
−
√
2 sin
√
2
t.
38.
The Laplace transform of the differential equation is
s
2
{
y
}
+9
{
y
}
=
1
s
−
1
.
Solving for
{
y
}
we obtain
{
y
}
=
1
(
s
−
1)(
s
2
+9)
=
1
10
·
1
s
−
1
−
1
10
·
1
s
2
+9
−
1
10
·
s
s
2
+9
.
Thus
y
=
1
10
e
t
−
1
30
sin 3
t
−
1
10
cos 3
t.
39.
The Laplace transform of the differential equation is
2
9
s
3
{
y
}−
s
2
(0)
−
sy
)
(0)
−
y
))
(0)
8
+3
9
s
2
{
y
}−
sy
(0)
−
y
)
(0)
8
−
3[
s
{
y
}−
y
(0)]
−
2
{
y
}
=
1
s
+1
.
171

172. Exercises 4.2
Solving for
{
y
}
we obtain
{
y
}
=
2
s
+3
(
s
+ 1)(
s
−
1)(2
s
+ 1)(
s
+2)
=
1
2
1
s
+1
+
5
18
1
s
−
1
−
8
9
1
s
+1
/
2
+
1
9
1
s
+2
.
Thus
y
=
1
2
e
−
t
+
5
18
e
t
−
8
9
e
−
t/
2
+
1
9
e
−
2
t
.
40.
The Laplace transform of the differential equation is
s
3
{
y
}−
s
2
(0)
−
sy
)
(0)
−
y
))
(0)+2
9
s
2
{
y
}−
sy
(0)
−
y
)
(0)
8
−
[
s
{
y
}−
y
(0)]
−
2
{
y
}
=
3
s
2
+9
.
Solving for
{
y
}
we obtain
{
y
}
=
s
2
+12
(
s
−
1)(
s
+ 1)(
s
+ 2)(
s
2
+9)
=
13
60
1
s
−
1
−
13
20
1
s
+1
+
16
39
1
s
+2
+
3
130
s
s
2
+9
−
1
65
3
s
2
+9
.
Thus
y
=
13
60
e
t
−
13
20
e
−
t
+
16
39
e
−
2
t
+
3
130
cos 3
t
−
1
65
sin 3
t.
41.
For
y
)
+6
y
=
e
4
t
the transfer function is
W
(
s
)=1
/
(
s
+ 6). The zero-input response is
y
0
(
t
)=
1
2
s
+6
c
=2
e
−
6
t
,
and the zero-state response is
y
1
(
t
)=
1
1
(
s
−
4)(
s
+6)
c
=
1
−
1
10
·
1
s
+6
+
1
10
·
1
s
−
4
c
=
−
1
10
e
−
6
t
+
1
10
e
4
t
.
42.
For
y
))
−
4
y
)
=6
e
3
t
−
3
e
−
t
the transfer function is
W
(
s
)=1
/
(
s
2
−
4
s
). The zero-input response is
y
0
(
t
)=
1
s
−
5
s
2
−
4
s
c
=
1
5
4
·
1
s
−
1
4
·
1
s
−
4
c
=
5
4
−
1
4
e
4
t
,
and the zero-state response is
y
1
(
t
)=
1
6
(
s
−
3)(
s
2
−
4
s
)
−
3
(
s
+ 1)(
s
2
−
4
s
)
c
=
1
27
20
·
1
s
−
4
−
2
s
−
3
+
5
4
·
1
s
−
3
5
·
1
s
+1
c
=
27
20
e
4
t
−
2
e
3
t
+
5
4
−
3
5
e
−
t
.
43.
The Laplace transform of the differential equation is
s
{
y
}
+
{
y
}
=
s
+3
(
s
+3)
2
+4
=
s
+3
s
2
+6
s
+13
.
Solving for
{
y
}
we obtain
{
y
}
=
s
+3
(
s
+ 1)(
s
2
+6
s
+ 13)
=
1
4
·
1
s
+1
−
1
4
·
s
+1
s
2
+6
s
+13
=
1
4
·
1
s
+1
−
1
4
=
s
+3
(
s
+3)
2
+4
−
2
(
s
+3)
2
+4
+
.
172

173. Exercises 4.3
Thus
y
=
1
4
e
−
t
−
e
−
3
t
cos 2
t
+
1
4
e
−
3
t
sin 2
t.
44.
Let
f
(
t
) = 1 and
g
(
t
)=
1
1
,t
≥
0,
t

=1
0
,t
=1
. Then
{
f
(
t
)
}
=
{
g
(
t
)
}
= 1, but
f
(
t
)

=
g
(
t
).
Exercises 4.3
1.
h
te
10
t
U
=
1
(
s
−
10)
2
2.
h
te
−
6
t
U
=
1
(
s
+6)
2
3.
h
t
3
e
−
2
t
U
=
3!
(
s
+2)
4
4.
h
t
10
e
−
7
t
U
=
10!
(
s
+7)
11
5.
g
t
t
e
t
+
e
2
t
e
2
r
=
h
te
2
t
+2
te
3
t
+
te
4
t
U
=
1
(
s
−
2)
2
+
2
(
s
−
3)
2
+
1
(
s
−
4)
2
6.
h
e
2
t
(
t
−
1)
2
U
=
h
t
2
e
2
t
−
2
te
2
t
+
e
2
t
U
=
2
(
s
−
2)
3
−
2
(
s
−
2)
2
+
1
s
−
2
7.
h
e
t
sin 3
t
U
=
3
(
s
−
1)
2
+9
8.
h
e
−
2
t
cos 4
t
U
=
s
+2
(
s
+2)
2
+16
9.
{
(1
−
e
t
+3
e
−
4
t
) cos 5
t
}
=
{
cos 5
t
−
e
t
cos 5
t
+3
e
−
4
t
cos 5
t
}
=
s
s
2
+25
−
s
−
1
(
s
−
1)
2
+25
+
3(
s
+4)
(
s
+4)
2
+25
10.
1
e
3
t
=
9
−
4
t
+ 10 sin
t
2
+c
=
1
9
e
3
t
−
4
te
3
t
+10
e
3
t
sin
t
2
c
=
9
s
−
3
−
4
(
s
−
3)
2
+
5
(
s
−
3)
2
+1
/
4
11.
1
1
(
s
+2)
3
c
=
1
1
2
2
(
s
+2)
3
c
=
1
2
t
2
e
−
2
t
12.
1
1
(
s
−
1)
4
c
=
1
1
6
3!
(
s
−
1)
4
c
=
1
6
t
3
e
t
13.
1
1
s
2
−
6
s
+10
c
=
1
1
(
s
−
3)
2
+1
2
c
=
e
3
t
sin
t
14.
1
1
s
2
+2
s
+5
c
=
1
1
2
2
(
s
+1)
2
+2
2
c
=
1
2
e
−
t
sin 2
t
15.
1
s
s
2
+4
s
+5
c
=
1
(
s
+2)
(
s
+2)
2
+1
2
−
2
1
(
s
+2)
2
+1
2
c
=
e
−
2
t
cos
t
−
2
e
−
2
t
sin
t
16.
1
2
s
+5
s
2
+6
s
+34
c
=
1
2
(
s
+3)
(
s
+3)
2
+5
2
−
1
5
5
(
s
+3)
2
+5
2
c
=2
e
−
3
t
cos 5
t
−
1
5
e
−
3
t
sin 5
t
17.
1
s
(
s
+1)
2
c
=
1
s
+1
−
1
(
s
+1)
2
c
=
1
1
s
+1
−
1
(
s
+1)
2
c
=
e
−
t
−
te
−
t
173

174. Exercises 4.3
18.
1
5
s
(
s
−
2)
2
c
=
1
5(
s
−
2) + 10
(
s
−
2)
2
c
=
1
5
s
−
2
+
10
(
s
−
2)
2
c
=5
e
2
t
+10
te
2
t
19.
1
2
s
−
1
s
2
(
s
+1)
3
c
=
1
5
s
−
1
s
2
−
5
s
+1
−
4
(
s
+1)
2
−
3
2
2
(
s
+1)
3
c
=5
−
t
−
5
e
−
t
−
4
te
−
t
−
3
2
t
2
e
−
t
20.
1
(
s
+1)
2
(
s
+2)
4
c
=
1
1
(
s
+2)
2
−
2
(
s
+2)
3
+
1
6
3!
(
s
+2)
4
c
=
te
−
2
t
−
t
2
e
−
2
t
+
1
6
t
3
e
−
2
t
21.
The Laplace transform of the differential equation is
s
{
y
}−
y
(0)+4
{
y
}
=
1
s
+4
.
Solving for
{
y
}
we obtain
{
y
}
=
1
(
s
+4)
2
+
2
s
+4
.Thus
y
=
te
−
4
t
+2
e
−
4
t
.
22.
The Laplace transform of the differential equation is
s
{
y
}− {
y
}
=
1
s
+
1
(
s
−
1)
2
.
Solving for
{
y
}
we obtain
{
y
}
=
1
s
(
s
−
1)
+
1
(
s
−
1)
3
=
−
1
s
+
1
s
−
1
+
1
(
s
−
1)
3
.
Thus
y
=
−
1+
e
t
+
1
2
t
2
e
t
.
23.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)+2
9
s
{
y
}−
y
(0)
8
+
{
y
}
=0
.
Solving for
{
y
}
we obtain
{
y
}
=
s
+3
(
s
+1)
2
=
1
s
+1
+
2
(
s
+1)
2
.
Thus
y
=
e
−
t
+2
te
−
t
.
24.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)
−
4[
s
{
y
}−
y
(0)] + 4
{
y
}
=
6
(
s
−
2)
4
.
Solving for
{
y
}
we obtain
{
y
}
=
1
20
5!
(
s
−
2)
6
. Thus,
y
=
1
20
t
5
e
2
t
.
25.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)
−
6[
s
{
y
}−
y
(0)] + 9
{
y
}
=
1
s
2
.
Solving for
{
y
}
we obtain
{
y
}
=
1+
s
2
s
2
(
s
−
3)
2
=
2
27
1
s
+
1
9
1
s
2
−
2
27
1
s
−
3
+
10
9
1
(
s
−
3)
2
.
Thus
y
=
2
27
+
1
9
t
−
2
27
e
3
t
+
10
9
te
3
t
.
174

175. Exercises 4.3
26.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)
−
4[
s
{
y
}−
y
(0)] + 4
{
y
}
=
6
s
4
.
Solving for
{
y
}
we obtain
{
y
}
=
s
5
−
4
s
4
+6
s
4
(
s
−
2)
2
=
3
4
1
s
+
9
8
1
s
2
+
3
4
2
s
3
+
1
4
3!
s
4
+
1
4
1
s
−
2
−
13
8
1
(
s
−
2)
2
.
Thus
y
=
3
4
+
9
8
t
+
3
4
t
2
+
1
4
t
3
+
1
4
e
2
t
−
13
8
te
2
t
.
27.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)
−
6[
s
{
y
}−
y
(0)] + 13
{
y
}
=0
.
Solving for
{
y
}
we obtain
{
y
}
=
−
3
s
2
−
6
s
+13
=
−
3
2
2
(
s
−
3)
2
+2
2
.
Thus
y
=
−
3
2
e
3
t
sin 2
t.
28.
The Laplace transform of the differential equation is
2
9
s
2
{
y
}−
sy
(0)
8
+20
9
s
{
y
}−
y
(0)
8
+51
{
y
}
=0
.
Solving for
{
y
}
we obtain
{
y
}
=
4
s
+40
2
s
2
+20
s
+51
=
2
s
+20
(
s
+5)
2
+1
/
2
=
2(
s
+5)
(
s
+5)
2
+1
/
2
+
10
(
s
+5)
2
+1
/
2
.
Thus
y
=2
e
−
5
t
cos(
t/
√
2)+10
√
2
e
−
5
t
sin(
t/
√
2)
.
29.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)
−
[
s
{
y
}−
y
(0)] =
s
−
1
(
s
−
1)
2
+1
.
Solving for
{
y
}
we obtain
{
y
}
=
1
s
(
s
2
−
2
s
+2)
=
1
2
1
s
−
1
2
s
−
1
(
s
−
1)
2
+1
+
1
2
1
(
s
−
1)
2
+1
.
Thus
y
=
1
2
−
1
2
e
t
cos
t
+
1
2
e
t
sin
t.
30.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)
−
2[
s
{
y
}−
y
(0)] + 5
{
y
}
=
1
s
+
1
s
2
.
Solving for
{
y
}
we obtain
{
y
}
=
4
s
2
+
s
+1
s
2
(
s
2
−
2
s
+5)
=
7
25
1
s
+
1
5
1
s
2
+
−
7
s/
25 + 109
/
25
s
2
−
2
s
+5
=
7
25
1
s
+
1
5
1
s
2
−
7
25
s
−
1
(
s
−
1)
2
+2
2
+
51
25
2
(
s
−
1)
2
+2
2
.
175

176. Exercises 4.3
Thus
y
=
7
25
+
1
5
t
−
7
25
e
t
cos 2
t
+
51
25
e
t
sin 2
t.
31.
Taking the Laplace transform of both sides of the differential equation and letting
c
=
y
(0) we obtain
{
y
))
}
+
{
2
y
)
}
+
{
y
}
=0
s
2
{
y
}−
sy
(0)
−
y
)
(0)+2
s
{
y
}−
2
y
(0) +
{
y
}
=0
s
2
{
y
}−
cs
−
2+2
s
{
y
}−
2
c
+
{
y
}
=0
t
s
2
+2
s
+1
e
{
y
}
=
cs
+2
c
+2
{
y
}
=
cs
(
s
+1)
2
+
2
c
+2
(
s
+1)
2
=
c
s
+1
−
1
(
s
+1)
2
+
2
c
+2
(
s
+1)
2
=
c
s
+1
+
c
+2
(
s
+1)
2
.
Therefore,
y
(
t
)=
c
1
1
s
+1
c
+(
c
+2)
1
1
(
s
+1)
2
c
=
ce
−
t
+(
c
+2)
te
−
t
.
To find
c
we let
y
(1) = 2. Then 2 =
ce
−
1
+(
c
+2)
e
−
1
=2(
c
+1)
e
−
1
and
c
=
e
−
1. Thus
y
(
t
)=(
e
−
1)
e
−
t
+(
e
+1)
te
−
t
.
32.
Taking the Laplace transform of both sides of the differential equation and letting
c
=
y
)
(0) we obtain
{
y
))
}
+
{
8
y
)
}
+
{
20
y
}
=0
s
2
{
y
}−
y
)
(0)+8
s
{
y
}
+20
{
y
}
=0
s
2
{
y
}−
c
+8
s
{
y
}
+20
{
y
}
=0
(
s
2
+8
s
+ 20)
{
y
}
=
c
{
y
}
=
c
s
2
+8
s
+20
=
c
(
s
+4)
2
+4
.
Therefore,
y
(
t
)=
1
c
(
s
+4)
2
+4
c
=
ce
−
4
t
sin 2
t.
To find
c
we let
y
)
(
π
) = 0. Then 0 =
y
)
(
π
)=
ce
−
4
π
and
c
= 0. Thus,
y
(
t
) = 0. (Since the differential equation
is homogeneous and both boundaryconditions are 0, we can see immediatelythat
y
(
t
) = 0 is a solution. We
have shown that it is the onlysolution.)
33.
Recall from Chapter 3 that
mx
))
=
−
kx
−
βx
)
.Now
m
=
W/g
=4
/
32 =
1
8
slug, and 4 = 2
k
so that
k
= 2 lb/ft.
Thus, the differential equation is
x
))
+7
x
)
+16
x
= 0. The initial conditions are
x
(0) =
−
3
/
2 and
x
)
(0) = 0.
The Laplace transform of the differential equation is
s
2
{
x
}
+
3
2
s
+7
s
{
x
}
+
21
2
+16
{
x
}
=0
.
Solving for
{
x
}
we obtain
{
x
}
=
−
3
s/
2
−
21
/
2
s
2
+7
s
+16
=
−
3
2
s
+7
/
2
(
s
+7
/
2)
2
+(
√
15
/
2)
2
−
7
√
15
10
√
15
/
2
(
s
+7
/
2)
2
+(
√
15
/
2)
2
.
176

177. Exercises 4.3
Thus
x
=
−
3
2
e
−
7
t/
2
cos
√
15
2
t
−
7
√
15
10
e
−
7
t/
2
sin
√
15
2
t.
34.
The differential equation is
d
2
q
dt
2
+20
dq
dt
+ 200
q
= 150
,q
(0) =
q
)
(0) = 0
.
The Laplace transform of this equation is
s
2
{
q
}
+20
s
{
q
}
+ 200
{
q
}
=
150
s
.
Solving for
{
q
}
we obtain
{
q
}
=
150
s
(
s
2
+20
s
+ 200)
=
3
4
1
s
−
3
4
s
+10
(
s
+ 10)
2
+10
2
−
3
4
10
(
s
+ 10)
2
+10
2
.
Thus
q
(
t
)=
3
4
−
3
4
e
−
10
t
cos 10
t
−
3
4
e
−
10
t
sin 10
t
and
i
(
t
)=
q
)
(
t
)=15
e
−
10
t
sin 10
t.
35.
The differential equation is
d
2
q
dt
2
+2
λ
dq
dt
+
ω
2
q
=
E
0
L
,q
(0) =
q
)
(0) = 0
.
The Laplace transform of this equation is
s
2
{
q
}
+2
λs
{
q
}
+
ω
2
{
q
}
=
E
0
L
1
s
or
t
s
2
+2
λs
+
ω
2
e
{
q
}
=
E
0
L
1
s
.
Solving for
{
q
}
and using partial fractions we obtain
{
q
}
=
E
0
L
=
1
/ω
2
s
−
(1
/ω
2
)
s
+2
λ/ω
2
s
2
+2
λs
+
ω
2
+
=
E
0
Lω
2
=
1
s
−
s
+2
λ
s
2
+2
λs
+
ω
2
+
.
For
λ>ω
we write
s
2
+2
λs
+
ω
2
=(
s
+
λ
)
2
−
t
λ
2
−
ω
2
e
, so (recalling that
ω
2
=1
/LC
,)
{
q
}
=
E
0
C
=
1
s
−
s
+
λ
(
s
+
λ
)
2
−
(
λ
2
−
ω
2
)
−
λ
(
s
+
λ
)
2
−
(
λ
2
−
ω
2
)
+
.
Thus for
λ>ω
,
q
(
t
)=
E
0
C
=
1
−
e
−
λt
cosh
a
λ
2
−
ω
2
t
−
λ
√
λ
2
−
ω
2
sinh
a
λ
2
−
ω
2
t
+
.
For
λ<ω
we write
s
2
+2
λs
+
ω
2
=(
s
+
λ
)
2
+
t
ω
2
−
λ
2
e
,so
{
q
}
=
E
0
=
1
s
−
s
+
λ
(
s
+
λ
)
2
+(
ω
2
−
λ
2
)
−
λ
(
s
+
λ
)
2
+(
ω
2
−
λ
2
)
+
.
Thus for
λ<ω
,
q
(
t
)=
E
0
C
=
1
−
e
−
λt
cos
a
ω
2
−
λ
2
t
−
λ
√
λ
2
−
ω
2
sin
a
ω
2
−
λ
2
t
+
.
For
λ
=
ω
,
s
2
+2
λ
+
ω
2
=(
s
+
λ
)
2
and
{
q
}
=
E
0
L
1
s
(
s
+
λ
)
2
=
E
0
L
=
1
/λ
2
s
−
1
/λ
2
s
+
λ
−
1
/λ
(
s
+
λ
)
2
+
=
E
0
Lλ
2
=
1
s
−
1
s
+
λ
−
λ
(
s
+
λ
)
2
+
.
177

178. Exercises 4.3
Thus for
λ
=
ω
,
q
(
t
)=
E
0
C
t
1
−
e
−
λt
−
λte
−
λt
e
.
36.
The differential equation is
R
dq
dt
+
1
C
q
=
E
0
e
−
kt
,q
(0) = 0
.
The Laplace transform of this equation is
R
{
q
}
+
1
C
{
q
}
=
E
0
1
s
+
k
.
Solving for
{
q
}
we obtain
{
q
}
=
E
0
C
(
s
+
k
)(
RC
s
+1)
=
E
0
/R
(
s
+
k
)(
s
+1
/RC
)
.
When 1
/RC

=
k
we have bypartial fractions
{
q
}
=
E
0
R
=
1
/
(1
/RC
−
k
)
s
+
k
−
1
/
(1
/RC
−
k
)
s
+1
/RC
+
=
E
0
R
1
1
/RC
−
k
=
1
s
+
k
−
1
s
+1
/RC
+
.
Thus
q
(
t
)=
E
0
C
1
−
kRC
o
e
−
kt
−
e
−
t/RC
,
.
When 1
/RC
=
k
we have
{
q
}
=
E
0
R
1
(
s
+
k
)
2
.
Thus
q
(
t
)=
E
0
R
te
−
kt
=
E
0
R
te
−
t/RC
.
37.
h
(
t
−
1) (
t
−
1)
U
=
e
−
s
s
2
38.
h
e
2
−
t
(
t
−
2)
U
=
g
e
−
(
t
−
2)
(
t
−
2)
r
=
e
−
2
s
s
+1
39.
h
t
(
t
−
2)
U
=
{
(
t
−
2) (
t
−
2) + 2 (
t
−
2)
}
=
e
−
2
s
s
2
+
2
e
−
2
s
s
40.
h
(3
t
+1) (
t
−
1)
U
=3
h
(
t
−
1) (
t
−
1)
U
+4
h
(
t
−
1)
U
=
e
−
s
s
2
+
4
e
−
s
s
41.
h
cos 2
t
(
t
−
π
)
U
=
{
cos 2(
t
−
π
)(
t
−
π
)
}
=
se
−
πs
s
2
+4
42.
g
sin
t
o
t
−
π
2
,r
=
g
cos
o
t
−
π
2
,o
t
−
π
2
,r
=
se
−
πs
s
2
+1
43.
1
e
−
2
s
s
3
c
=
1
1
2
·
2
s
3
e
−
2
s
c
=
1
2
(
t
−
2)
2
(
t
−
2)
44.
1
(1 +
e
−
2
s
)
2
s
+2
c
=
1
1
s
+2
+
2
e
−
2
s
s
+2
+
e
−
4
s
s
+2
c
=
e
−
2
t
+2
e
−
2(
t
−
2)
(
t
−
2) +
e
−
2(
t
−
4)
(
t
−
4)
45.
1
e
−
πs
s
2
+1
c
= sin(
t
−
π
)(
t
−
π
)
46.
1
se
−
πs/
2
s
2
+4
c
= cos 2
o
t
−
π
2
,o
t
−
π
2
,
178

179. Exercises 4.3
47.
1
e
−
s
s
(
s
+1)
c
=
1
e
−
s
s
−
e
−
s
s
+1
c
=(
t
−
1)
−
e
−
(
t
−
1)
(
t
−
1)
48.
1
e
−
2
s
s
2
(
s
−
1)
c
=
1
−
e
−
2
s
s
−
e
−
2
s
s
2
+
e
−
2
s
s
−
1
c
=
−
(
t
−
2)
−
(
t
−
2) (
t
−
2) +
e
t
−
2
(
t
−
2)
49. (c) 50. (e) 51. (f) 52. (b) 53. (a) 54. (d)
55.
h
2
−
4(
t
−
3)
U
=
2
s
−
4
s
e
−
3
s
56.
h
1
−
(
t
−
4) + (
t
−
5)
U
=
1
s
−
e
−
4
s
s
+
e
−
5
s
s
57.
h
t
2
(
t
−
1)
U
=
h9
(
t
−
1)
2
+2
t
−
1
8
(
t
−
1)
U
=
h9
(
t
−
1)
2
+2(
t
−
1)
−
1
8
(
t
−
1)
U
=
=
2
s
3
+
2
s
2
+
1
s
+
e
−
s
58.
1
sin
t
=
t
−
3
π
2
+c
=
1
−
cos
=
t
−
3
π
2
+=
t
−
3
π
2
+c
=
−
se
−
3
πs/
2
s
2
+1
59.
h
t
−
t
(
t
−
2)
U
=
h
t
−
(
t
−
2) (
t
−
2)
−
2(
t
−
2)
U
=
1
s
2
−
e
−
2
s
s
2
−
2
e
−
2
s
s
60.
h
sin
t
−
sin
t
(
t
−
2
π
)
U
=
h
sin
t
−
sin(
t
−
2
π
)(
t
−
2
π
)
U
=
1
s
2
+1
−
e
−
2
πs
s
2
+1
61.
h
f
(
t
)
U
=
h
(
t
−
a
)
−
(
t
−
b
)
U
=
e
−
as
s
−
e
−
bs
s
62.
h
f
(
t
)
U
=
h
(
t
−
1) + (
t
−
2) + (
t
−
3) +
···
U
=
e
−
s
s
+
e
−
2
s
s
+
e
−
3
s
s
+
···
=
1
s
e
−
s
1
−
e
−
s
63.
The Laplace transform of the differential equation is
s
{
y
}−
y
(0) +
{
y
}
=
5
s
e
−
s
.
Solving for
{
y
}
we obtain
{
y
}
=
5
e
−
s
s
(
s
+1)
=5
e
−
s
!
1
s
−
1
s
+1
5
.
Thus
y
=5 (
t
−
1)
−
5
e
−
(
t
−
1)
(
t
−
1)
.
64.
The Laplace transform of the differential equation is
s
{
y
}−
y
(0) +
{
y
}
=
1
s
−
2
s
e
−
s
.
Solving for
{
y
}
we obtain
{
y
}
=
1
s
(
s
+1)
−
2
e
−
s
s
(
s
+1)
=
1
s
−
1
s
+1
−
2
e
−
s
!
1
s
−
1
s
+1
5
.
Thus
y
=1
−
e
−
t
−
2
b
1
−
e
−
(
t
−
1)

(
t
−
1)
.
65.
The Laplace transform of the differential equation is
s
{
y
}−
y
(0)+2
{
y
}
=
1
s
2
−
e
−
s
s
+1
s
2
.
179

180. Exercises 4.3
Solving for
{
y
}
we obtain
{
y
}
=
1
s
2
(
s
+2)
−
e
−
s
s
+1
s
2
(
s
+1)
=
−
1
4
1
s
+
1
2
1
s
2
+
1
4
1
s
+2
−
e
−
s
!
1
4
1
s
+
1
2
1
s
2
−
1
4
1
s
+2
5
.
Thus
y
=
−
1
4
+
1
2
t
+
1
4
e
−
2
t
−
!
1
4
+
1
2
(
t
−
1)
−
1
4
e
−
2(
t
−
1)
5
(
t
−
1)
.
66.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)+4
{
y
}
=
1
s
−
e
−
s
s
.
Solving for
{
y
}
we obtain
{
y
}
=
1
−
s
s
(
s
2
+4)
−
e
−
s
1
s
(
s
2
+4)
=
1
4
1
s
−
1
4
s
s
2
+4
−
1
2
2
s
2
+4
−
e
−
s
!
1
4
1
s
−
1
4
s
s
2
+4
5
.
Thus
y
=
1
4
−
1
4
cos 2
t
−
1
2
sin 2
t
−
!
1
4
−
1
4
cos 2(
t
−
1)
5
(
t
−
1)
.
67.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)+4
{
y
}
=
e
−
2
πs
1
s
2
+1
.
Solving for
{
y
}
we obtain
{
y
}
=
s
s
2
+4
+
e
−
2
πs
!
1
3
1
s
2
+1
−
1
6
2
s
2
+4
5
.
Thus
y
= cos 2
t
+
!
1
3
sin(
t
−
2
π
)
−
1
6
sin 2(
t
−
2
π
)
5
(
t
−
2
π
)
.
68.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)
−
5[
s
{
y
}−
y
(0)] + 6
{
y
}
=
e
−
s
s
.
Solving for
{
y
}
we obtain
{
y
}
=
e
−
s
1
s
(
s
−
2)(
s
−
3)
+
1
(
s
−
2)(
s
−
3)
=
e
−
s
!
1
6
1
s
−
1
2
1
s
−
2
+
1
3
1
s
−
3
5
−
1
s
−
2
+
1
s
−
3
.
Thus
y
=
!
1
6
−
1
2
e
2(
t
−
1)
+
1
3
e
3(
t
−
1)
5
(
t
−
1) +
e
3
t
−
e
2
t
.
69.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0) +
{
y
}
=
e
−
πs
s
−
e
−
2
πs
s
.
Solving for
{
y
}
we obtain
{
y
}
=
e
−
πs
!
1
s
−
s
s
2
+1
5
−
e
−
2
πs
!
1
s
−
s
s
2
+1
5
+
1
s
2
+1
.
Thus
y
=[1
−
cos(
t
−
π
)] (
t
−
π
)
−
[1
−
cos(
t
−
2
π
)] (
t
−
2
π
) + sin
t.
180

181. Exercises 4.3
70.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)+4
9
s
{
y
}−
y
(0)
8
+3
{
y
}
=
1
s
−
e
−
2
s
s
−
e
−
4
s
s
+
e
−
6
s
s
.
Solving for
{
y
}
we obtain
{
y
}
=
1
3
1
s
−
1
2
1
s
+1
+
1
6
1
s
+3
−
e
−
2
s
!
1
3
1
s
−
1
2
1
s
+1
+
1
6
1
s
+3
5
−
e
−
4
s
!
1
3
1
s
−
1
2
1
s
+1
+
1
6
1
s
+3
5
+
e
−
6
s
!
1
3
1
s
−
1
2
1
s
+1
+
1
6
1
s
+3
5
.
Thus
y
=
1
3
−
1
2
e
−
t
+
1
6
e
−
3
t
−
!
1
3
−
1
2
e
−
(
t
−
2)
+
1
6
e
−
3(
t
−
2)
5
(
t
−
2)
−
!
1
3
−
1
2
e
−
(
t
−
4)
+
1
6
e
−
3(
t
−
4)
5
(
t
−
4) +
!
1
3
−
1
2
e
−
(
t
−
6)
+
1
6
e
−
3(
t
−
6)
5
(
t
−
6)
.
71.
Recall from Chapter 3 that
mx
))
=
−
kx
+
f
(
t
). Now
m
=
W/g
=32
/
32 = 1 slug, and 32 = 2
k
so that
k
= 16 lb/ft. Thus, the differential equation is
x
))
+16
x
=
f
(
t
). The initial conditions are
x
(0)=0,
x
)
(0)=0.
Also, since
f
(
t
)=
7
20
t,
0
≤
t<
5
0
,t
≥
5
and 20
t
= 20(
t
−
5) + 100 we can write
f
(
t
)=20
t
−
20
t
(
t
−
5) = 20
t
−
20(
t
−
5) (
t
−
5)
−
100 (
t
−
5)
.
The Laplace transform of the differential equation is
s
2
{
x
}
+16
{
x
}
=
20
s
2
−
20
s
2
e
−
5
s
−
100
s
e
−
5
s
.
Solving for
{
x
}
we obtain
{
x
}
=
20
s
2
(
s
2
+ 16)
−
20
s
2
(
s
2
+ 16)
e
−
5
s
−
100
s
(
s
2
+ 16)
e
−
5
s
=
=
5
4
·
1
s
2
−
5
16
·
4
s
2
+16
+
t
1
−
e
−
5
s
e
−
=
25
4
·
1
s
−
25
4
·
s
s
2
+16
+
e
−
5
s
.
Thus
x
(
t
)=
5
4
t
−
5
16
sin 4
t
−
!
5
4
(
t
−
5)
−
5
16
sin 4(
t
−
5)
5
(
t
−
5)
−
!
25
4
−
25
4
cos 4(
t
−
5)
5
(
t
−
5)
=
5
4
t
−
5
16
sin 4
t
−
5
4
t
(
t
−
5) +
5
16
sin 4(
t
−
5) (
t
−
5) +
25
4
cos 4(
t
−
5) (
t
−
5)
.
72.
Recall from Chapter 3 that
mx
))
=
−
kx
+
f
(
t
). Now
m
=
W/g
=32
/
32 = 1 slug, and 32 = 2
k
so that
k
= 16 lb/ft. Thus, the differential equation is
x
))
+16
x
=
f
(
t
). The initial conditions are
x
(0)=0,
x
)
(0)=0.
Also, since
f
(
t
)=
1
sin
t,
0
≤
t<
2
π
0
,t
≥
2
π
and sin
t
= sin(
t
−
2
π
) we can write
f
(
t
) = sin
t
−
sin(
t
−
2
π
)(
t
−
2
π
)
.
181

182. Exercises 4.3
The Laplace transform of the differential equation is
s
2
{
x
}
+16
{
x
}
=
1
s
2
+1
−
1
s
2
+1
e
−
2
πs
.
Solving for
{
x
}
we obtain
{
x
}
=
1
(
s
2
+ 16) (
s
2
+1)
−
1
(
s
2
+ 16) (
s
2
+1)
e
−
2
πs
=
−
1
/
15
s
2
+16
+
1
/
15
s
2
+1
−
!
−
1
/
15
s
2
+16
+
1
/
15
s
2
+1
5
e
−
2
πs
.
Thus
x
(
t
)=
−
1
60
sin 4
t
+
1
15
sin
t
+
1
60
sin 4(
t
−
2
π
)(
t
−
2
π
)
−
1
15
sin(
t
−
2
π
)(
t
−
2
π
)
=
7
−
1
60
sin 4
t
+
1
15
sin
t,
0
≤
t<
2
π
0
,t
≥
2
π
.
73.
The differential equation is
2
.
5
dq
dt
+12
.
5
q
=5 (
t
−
3)
.
The Laplace transform of this equation is
s
{
q
}
+5
{
q
}
=
2
s
e
−
3
s
.
Solving for
{
q
}
we obtain
{
q
}
=
2
s
(
s
+5)
e
−
3
s
=
=
2
5
·
1
s
−
2
5
·
1
s
+5
+
e
−
3
s
.
Thus
q
(
t
)=
2
5
(
t
−
3)
−
2
5
e
−
5(
t
−
3)
(
t
−
3)
.
74.
The differential equation is
10
dq
dt
+10
q
=30
e
t
−
30
e
t
(
t
−
1
.
5)
.
The Laplace transform of this equation is
s
{
q
}−
q
0
+
{
q
}
=
3
s
−
1
−
3
e
1
.
5
s
−
1
.
5
e
−
1
.
5
s
.
Solving for
{
q
}
we obtain
{
q
}
=
=
q
0
−
3
2
+
·
1
s
+1
+
3
2
·
1
s
−
1
3
e
1
.
5
=
−
2
/
5
s
+1
+
2
/
5
s
−
1
.
5
+
e
−
1
.
55
.
Thus
q
(
t
)=
=
q
0
−
3
2
+
e
−
t
+
3
2
e
t
+
6
5
e
1
.
5
o
e
−
(
t
−
1
.
5)
−
e
1
.
5(
t
−
1
.
5)
,
(
t
−
1
.
5)
.
75. (a)
The differential equation is
di
dt
+10
i
= sin
t
+ cos
=
t
−
3
π
2
+=
t
−
3
π
2
+
,i
(0) = 0
.
The Laplace transform of this equation is
s
{
i
}
+10
{
i
}
=
1
s
2
+1
+
se
−
3
πs/
2
s
2
+1
.
182

183. 1 2 3 4 5 6
t
-0.2
0.2
i
1 2 3 4 5 6
t
1
q
Exercises 4.3
Solving for
{
i
}
we obtain
{
i
}
=
1
(
s
2
+ 1)(
s
+ 10)
+
s
(
s
2
+ 1)(
s
+ 10)
e
−
3
πs/
2
=
1
101
=
1
s
+10
−
s
s
2
+1
+
10
s
2
+1
+
+
1
101
=
−
10
s
+10
+
10
s
s
2
+1
+
1
s
2
+1
+
e
−
3
πs/
2
.
Thus
i
(
t
)=
1
101
t
e
−
10
t
−
cos
t
+ 10 sin
t
e
+
1
101
!
−
10
e
−
10(
t
−
3
π/
2)
+ 10 cos
=
t
−
3
π
2
+
+ sin
=
t
−
3
π
2
+5 =
t
−
3
π
2
+
.
(b)
The maximum value of
i
(
t
) is approximately0
.
1at
t
=1
.
7, the minimum is approximately
−
0
.
1at4
.
7.
76. (a)
The differential equation is
50
dq
dt
+
1
0
.
01
q
=
E
0
[(
t
−
1)
−
(
t
−
3)]
,q
(0) = 0
or
50
dq
dt
+ 100
q
=
E
0
[(
t
−
1)
−
(
t
−
3)]
,q
(0) = 0
.
The Laplace transform of this equation is
50
s
{
q
}
+ 100
{
q
}
=
E
0
=
1
s
e
−
s
−
1
s
e
−
3
s
+
.
Solving for
{
q
}
we obtain
{
q
}
=
E
0
50
!
e
−
s
s
(
s
+2)
−
e
−
3
s
s
(
s
+2)
5
=
E
0
50
!
1
2
=
1
s
−
1
s
+2
+
e
−
s
−
1
2
=
1
s
−
1
s
+2
+
e
−
3
s
5
.
Thus
q
(
t
)=
E
0
100
bo
1
−
e
−
2(
t
−
1)
,
(
t
−
1)
−
o
1
−
e
−
2(
t
−
3)
,
(
t
−
3)

.
(b)
The maximum value of
q
(
t
) is approximately1 at
t
=3.
77.
The differential equation is
EI
d
4
y
dx
4
=
w
0
[1
−
(
x
−
L/
2)]
.
Taking the Laplace transform of both sides and using
y
(0) =
y
)
(0) = 0 we obtain
s
4
{
y
}−
sy
))
(0)
−
y
)))
(0) =
w
0
EI
1
s
o
1
−
e
−
Ls/
2
,
.
183

184. Exercises 4.3
Letting
y
))
(0) =
c
1
and
y
)))
(0) =
c
2
we have
{
y
}
=
c
1
s
3
+
c
2
s
4
+
w
0
EI
1
s
5
o
1
−
e
−
Ls/
2
,
so that
y
(
x
)=
1
2
c
1
x
2
+
1
6
c
2
x
3
+
1
24
w
0
EI

x
4
−
=
x
−
L
2
+
4
=
x
−
L
2
+

.
To find
c
1
and
c
2
we compute
y
))
(
x
)=
c
1
+
c
2
x
+
1
2
w
0
EI

x
2
−
=
x
−
L
2
+
2
=
x
−
L
2
+

and
y
)))
(
x
)=
c
2
+
w
0
EI
!
x
−
=
x
−
L
2
+=
x
−
L
2
+5
.
Then
y
))
(
L
)=
y
)))
(
L
) = 0 yields the system
c
1
+
c
2
L
+
1
2
w
0
EI

L
2
−
=
L
2
+
2

=
c
1
+
c
2
L
+
3
8
w
0
L
2
EI
=0
c
2
+
w
0
EI
=
L
2
+
=
c
2
+
1
2
w
0
L
EI
=0
.
Solving for
c
1
and
c
2
we obtain
c
1
=
1
8
w
0
L
2
/EI
and
c
2
=
−
1
2
w
0
L/EI
.Thus
y
(
x
)=
w
0
EI

1
16
L
2
x
2
−
1
12
Lx
3
+
1
24
x
4
−
1
24
=
x
−
L
2
+
4
=
x
−
L
2
+

.
78.
The differential equation is
EI
d
4
y
dx
4
=
w
0
[(
x
−
L/
3)
−
(
x
−
2
L/
3)]
.
Taking the Laplace transform of both sides and using
y
(0) =
y
)
(0) = 0 we obtain
s
4
{
y
}−
sy
))
(0)
−
y
)))
(0) =
w
0
EI
1
s
o
e
−
Ls/
3
−
e
−
2
Ls/
3
,
.
Letting
y
))
(0) =
c
1
and
y
)))
(0) =
c
2
we have
{
y
}
=
c
1
s
3
+
c
2
s
4
+
w
0
EI
1
s
5
o
e
−
Ls/
3
−
e
−
2
Ls/
3
,
so that
y
(
x
)=
1
2
c
1
x
2
+
1
6
c
2
x
3
+
1
24
w
0
EI

=
x
−
L
3
+
4
=
x
−
L
3
+
−
=
x
−
2
L
3
+
4
=
x
−
2
L
3
+

.
To find
c
1
and
c
2
we compute
y
))
(
x
)=
c
1
+
c
2
x
+
1
2
w
0
EI

=
x
−
L
3
+
2
=
x
−
L
3
+
−
=
x
−
2
L
3
+
2
=
x
−
2
L
3
+

and
y
)))
(
x
)=
c
2
+
w
0
EI
!=
x
−
L
3
+=
x
−
L
3
+
−
=
x
−
2
L
3
+=
x
−
2
L
3
+5
.
Then
y
))
(
L
)=
y
)))
(
L
) = 0 yields the system
c
1
+
c
2
L
+
1
2
w
0
EI

=
2
L
3
+
2
−
=
L
3
+
2

=
c
1
+
c
2
L
+
1
6
w
0
L
2
EI
=0
c
2
+
w
0
EI
!
2
L
3
−
L
3
5
=
c
2
+
1
3
w
0
L
EI
=0
.
184

185. Exercises 4.3
Solving for
c
1
and
c
2
we obtain
c
1
=
1
6
w
0
L
2
/EI
and
c
2
=
−
1
3
w
0
L/EI
.Thus
y
(
x
)=
w
0
EI

1
12
L
2
x
2
−
1
18
Lx
3
+
1
24

=
x
−
L
3
+
4
=
x
−
L
3
+
−
=
x
−
2
L
3
+
4
=
x
−
2
L
3
+

.
79.
The differential equation is
EI
d
4
y
dx
4
=
2
w
0
L
!
L
2
−
x
+
=
x
−
L
2
+=
x
−
L
2
+5
.
Taking the Laplace transform of both sides and using
y
(0) =
y
)
(0) = 0 we obtain
s
4
{
y
}−
sy
))
(0)
−
y
)))
(0) =
2
w
0
EIL
!
L
2
s
−
1
s
2
+
1
s
2
e
−
Ls/
2
5
.
Letting
y
))
(0) =
c
1
and
y
)))
(0) =
c
2
we have
{
y
}
=
c
1
s
3
+
c
2
s
4
+
2
w
0
EIL
!
L
2
s
5
−
1
s
6
+
1
s
6
e
−
Ls/
2
5
so that
y
(
x
)=
1
2
c
1
x
2
+
1
6
c
2
x
3
+
2
w
0
EIL
!
L
48
x
4
−
1
120
x
5
+
1
120
=
x
−
L
2
+=
x
−
L
2
+5
=
1
2
c
1
x
2
+
1
6
c
2
x
3
+
w
0
60
EIL

5
L
2
x
4
−
x
5
+
=
x
−
L
2
+
5
=
x
−
L
2
+

.
To find
c
1
and
c
2
we compute
y
))
(
x
)=
c
1
+
c
2
x
+
w
0
60
EIL

30
Lx
2
−
20
x
3
+20
=
x
−
L
2
+
3
=
x
−
L
2
+

and
y
)))
(
x
)=
c
2
+
w
0
60
EIL

60
Lx
−
60
x
2
+60
=
x
−
L
2
+
2
=
x
−
L
2
+

.
Then
y
))
(
L
)=
y
)))
(
L
) = 0 yields the system
c
1
+
c
2
L
+
w
0
60
EIL
!
30
L
3
−
20
L
3
+
5
2
L
3
5
=
c
1
+
c
2
L
+
5
w
0
L
2
24
EI
=0
c
2
+
w
0
60
EIL
[60
L
2
−
60
L
2
+15
L
2
]=
c
2
+
w
0
L
4
EI
=0
.
Solving for
c
1
and
c
2
we obtain
c
1
=
w
0
L
2
/
24
EI
and
c
2
=
−
w
0
L/
4
EI
.Thus
y
(
x
)=
w
0
L
2
48
EI
x
2
−
w
0
L
24
EI
+
w
0
60
EIL

5
L
2
x
4
−
x
5
+
=
x
−
L
2
+
5
=
x
−
L
2
+

.
80.
The differential equation is
EI
d
4
y
dx
4
=
w
0
[1
−
(
x
−
L/
2)]
.
Taking the Laplace transform of both sides and using
y
(0) =
y
)
(0) = 0 we obtain
s
4
{
y
}−
sy
))
(0)
−
y
)))
(0) =
w
0
EI
1
s
o
1
−
e
−
Ls/
2
,
.
Letting
y
))
(0) =
c
1
and
y
)))
(0) =
c
2
we have
{
y
}
=
c
1
s
3
+
c
2
s
4
+
w
0
EI
1
s
5
o
1
−
e
−
Ls/
2
,
185

186. Exercises 4.3
so that
y
(
x
)=
1
2
c
1
x
2
+
1
6
c
2
x
3
+
1
24
w
0
EI

x
4
−
=
x
−
L
2
+
4
=
x
−
L
2
+

.
To find
c
1
and
c
2
we compute
y
))
(
x
)=
c
1
+
c
2
x
+
1
2
w
0
EI

x
2
−
=
x
−
L
2
+
2
=
x
−
L
2
+

.
Then
y
(
L
)=
y
))
(
L
) = 0 yields the system
1
2
c
1
L
2
+
1
6
c
2
L
3
+
1
24
w
0
EI

L
4
−
=
L
2
+
4

=
1
2
c
1
L
2
+
1
6
c
2
L
3
+
5
w
0
128
EI
L
4
=0
c
1
+
c
2
L
+
1
2
w
0
EI

L
2
−
=
L
2
+
2

=
c
1
+
c
2
L
+
3
w
0
8
EI
L
2
=0
.
Solving for
c
1
and
c
2
we obtain
c
1
=
9
128
w
0
L
2
/EI
and
c
2
=
−
57
128
w
0
L/EI
.Thus
y
(
x
)=
w
0
EI

9
256
L
2
x
2
−
19
256
Lx
3
+
1
24
x
4
−
1
24
=
x
−
L
2
+
4
=
x
−
L
2
+

.
81.
From Theorem 4.6 we have
{
te
kti
}
=1
/
(
s
−
ki
)
2
. Then, using Euler’s formula,
{
te
kti
}
=
{
t
cos
kt
+
it
sin
kt
}
=
{
t
cos
kt
}
+
i
{
t
sin
kt
}
=
1
(
s
−
ki
)
2
=
(
s
+
ki
)
2
(
s
2
+
k
2
)
2
=
s
2
−
k
2
(
s
2
+
k
2
)
2
+
i
2
ks
(
s
2
+
k
2
)
2
.
Equating real and imaginaryparts we have
{
t
cos
kt
}
=
s
2
−
k
2
(
s
2
+
k
2
)
2
and
{
t
sin
kt
}
=
2
ks
(
s
2
+
k
2
)
2
.
82.
The Laplace transform of the differential equation is
s
2
{
x
}
+
ω
2
{
x
}
=
s
s
2
+
ω
2
.
Solving for
{
x
}
we obtain
{
x
}
=
s/
(
s
2
+
ω
2
)
2
.Thus
x
=(1
/
2
ω
)
t
sin
ωt
.
Exercises 4.4
1.
{
t
cos 2
t
}
=
−
d
ds
=
s
s
2
+4
+
=
s
2
−
4
(
s
2
+4)
2
2.
{
t
sinh 3
t
}
=
−
d
ds
=
3
s
2
−
9
+
=
6
s
(
s
2
−
9)
2
3.
{
t
2
sinh
t
}
=
d
2
ds
2
=
1
s
2
−
1
+
=
6
s
2
+2
(
s
2
−
1)
3
4.
{
t
2
cos
t
}
=
d
2
ds
2
=
s
s
2
+1
+
=
d
ds
=
1
−
s
2
(
s
2
+1)
2
+
=
2
s
t
s
2
−
3
e
(
s
2
+1)
3
5.
h
te
2
t
sin 6
t
U
=
−
d
ds
=
6
(
s
−
2)
2
+36
+
=
12(
s
−
2)
[(
s
−
2)
2
+ 36]
2
186

187. Exercises 4.4
6.
h
te
−
3
t
cos 3
t
U
=
−
d
ds
=
s
+3
(
s
+3)
2
+9
+
=
(
s
+3)
2
−
9
[(
s
+3)
2
+9]
2
7.
h
1
∗
t
3
U
=
1
s
3!
s
4
=
6
s
5
8.
h
t
2
∗
te
t
U
=
2
s
3
(
s
−
1)
2
9.
h
e
−
t
∗
e
t
cos
t
U
=
s
−
1
(
s
+1)[(
s
−
1)
2
+1]
10.
h
e
2
t
∗
sin
t
U
=
1
(
s
−
2)(
s
2
+1)
11.
1
(
t
0
e
τ
dτ
c
=
1
s
{
e
t
}
=
1
s
(
s
−
1)
12.
1
(
t
0
cos
τdτ
c
=
1
s
{
cos
t
}
=
s
s
(
s
2
+1)
=
1
s
2
+1
13.
1
(
t
0
e
−
τ
cos
τdτ
c
=
1
s
h
e
−
t
cos
t
U
=
1
s
s
+1
(
s
+1)
2
+1
=
s
+1
s
(
s
2
+2
s
+2)
14.
1
(
t
0
τ
sin
τdτ
c
=
1
s
{
t
sin
t
}
=
1
s
=
−
d
ds
1
s
2
+1
+
=
−
1
s
−
2
s
(
s
2
+1)
2
=
2
(
s
2
+1)
2
15.
1
(
t
0
τe
t
−
τ
dτ
c
=
{
t
}{
e
t
}
=
1
s
2
(
s
−
1)
16.
1
(
t
0
sin
τ
cos(
t
−
τ
)
dτ
c
=
{
sin
t
}{
cos
t
}
=
s
(
s
2
+1)
2
17.
1
t
(
t
0
sin
τdτ
c
=
−
d
ds
1
(
t
0
sin
τdτ
c
=
−
d
ds
=
1
s
1
s
2
+1
+
=
3
s
2
+1
s
2
(
s
2
+1)
2
18.
1
t
(
t
0
τe
−
τ
dτ
c
=
−
d
ds
1
t
(
t
0
τe
−
τ
dτ
c
=
−
d
ds
=
1
s
1
(
s
+1)
2
+
=
3
s
+1
s
2
(
s
+1)
3
19. (a)
1
1
s
(
s
−
1)
c
=
1
1
/
(
s
−
1)
s
c
=
(
t
0
e
τ
dτ
=
e
t
−
1
(b)
1
1
s
2
(
s
−
1)
c
=
1
1
/s
(
s
−
1)
s
c
=
(
t
0
(
e
τ
−
1)
dτ
=
e
t
−
t
−
1
(c)
1
1
s
3
(
s
−
1)
c
=
1
1
/s
2
(
s
−
1)
s
c
=
(
t
0
(
e
τ
−
τ
−
1)
dτ
=
e
t
−
1
2
t
2
−
t
−
1
20. (a)
The result in (4) is
{
F
(
s
)
G
(
s
)
}
=
f
∗
g
, so identify
F
(
s
)=
2
k
3
(
s
2
+
k
2
)
2
and
G
(
s
)=
4
s
s
2
+
k
2
.
Then
f
(
t
) = sin
kt
−
kt
cos
kt
and
g
(
t
) = 4 cos
kt
so
1
8
k
3
s
(
s
2
+
k
2
)
2
c
=
{
F
(
s
)
G
(
s
)
}
=
f
∗
g
=4
(
t
0
f
(
τ
)
g
(
t
−
τ
)
dt
=4
(
t
0
(sin
kτ
−
kτ
cos
kτ
) cos
k
(
t
−
τ
)
dτ.
187

188. Exercises 4.4
Using a CAS to evaluate the integral we get
1
8
k
3
s
(
s
2
+
k
2
)
3
c
=
t
sin
kt
−
kt
2
cos
kt.
(b)
Observe from part (a) that
h
t
(sin
kt
−
kt
cos
kt
)
U
=
8
k
3
s
(
s
2
+
k
2
)
3
,
and from Theorem 4.8 that
h
tf
(
t
)
U
=
−
F
)
(
s
). We saw in (5) that
{
sin
kt
−
kt
cos
kt
}
=2
k
3
/
(
s
2
+
k
2
)
2
,
so
h
t
(sin
kt
−
kt
cos
kt
)
U
=
−
d
ds
2
k
3
(
s
2
+
k
2
)
2
=
8
k
3
s
(
s
2
+
k
2
)
3
.
21.
f
(
t
)=
−
1
t
1
d
ds
[ln(
s
−
3)
−
ln(
s
+ 1)]
c
=
−
1
t
1
1
s
−
3
−
1
s
+1
c
=
−
1
t
t
e
3
t
−
e
−
t
e
22.
f
(
t
)=
−
1
t
1
d
ds
9
ln
t
s
2
+1
e
−
ln
t
s
2
+4
e8
c
=
−
1
t
1
2
s
s
2
+1
−
2
s
s
2
+2
2
c
=
−
1
t
(2 cos
t
−
2 cos 2
t
)
23.
{
f
(
t
)
}
=
1
1
−
e
−
2
as
!
(
a
0
e
−
st
dt
−
(
2
a
a
e
−
st
dt
5
=
(1
−
e
−
as
)
2
s
(1
−
e
−
2
as
)
=
1
−
e
−
as
s
(1 +
e
−
as
)
24.
{
f
(
t
)
}
=
1
1
−
e
−
2
as
(
a
0
e
−
st
dt
=
1
s
(1 +
e
−
as
)
25.
{
f
(
t
)
}
=
1
1
−
e
−
bs
(
b
0
a
b
te
−
st
dt
=
a
s
=
1
bs
−
1
e
bs
−
1
+
26.
{
f
(
t
)
}
=
1
1
−
e
−
2
s
!
(
1
0
te
−
st
dt
+
(
2
1
(2
−
t
)
e
−
st
dt
5
=
1
−
e
−
s
s
2
(1
−
e
−
2
s
)
27.
{
f
(
t
)
}
=
1
1
−
e
−
πs
(
π
0
e
−
st
sin
tdt
=
1
s
2
+1
·
e
πs/
2
+
e
−
πs/
2
e
πs/
2
−
e
−
πs/
2
=
1
s
2
+1
coth
πs
2
28.
{
f
(
t
)
}
=
1
1
−
e
−
2
πs
(
π
0
e
−
st
sin
tdt
=
1
s
2
+1
·
1
1
−
e
−
πs
29.
The Laplace transform of the differential equation is
s
{
y
}
+
{
y
}
=
2
s
(
s
2
+1)
2
.
Solving for
{
y
}
we obtain
{
y
}
=
2
s
(
s
+ 1)(
s
2
+1)
2
=
−
1
2
1
s
+1
−
1
2
1
s
2
+1
+
1
2
s
s
2
+1
+
1
(
s
2
+1)
2
+
s
(
s
2
+1)
2
.
Thus
y
(
t
)=
−
1
2
e
−
t
−
1
2
sin
t
+
1
2
cos
t
+
1
2
(sin
t
−
t
cos
t
)+
1
2
t
sin
t
=
−
1
2
e
−
t
+
1
2
cos
t
−
1
2
t
cos
t
+
1
2
t
sin
t.
30.
The Laplace transform of the differential equation is
s
{
y
}− {
y
}
=
2(
s
−
1)
((
s
2
−
1)
2
+1)
2
.
188

189. Exercises 4.4
Solving for
{
y
}
we obtain
{
y
}
=
2
((
s
−
1)
2
+1)
2
.
Thus
y
=
e
t
sin
t
−
te
t
cos
t.
31.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)+9
{
y
}
=
s
s
2
+9
.
Letting
y
(0) = 2 and
y
)
(0) = 5 and solving for
{
y
}
we obtain
{
y
}
=
2
s
3
+5
s
2
+19
s
−
45
(
s
2
+9)
2
=
2
s
s
2
+9
+
5
s
2
+9
+
s
(
s
2
+9)
2
.
Thus
y
= 2 cos 3
t
+
5
3
sin 3
t
+
1
6
t
sin 3
t.
32.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0) +
{
y
}
=
1
s
2
+1
.
Solving for
{
y
}
we obtain
{
y
}
=
s
3
−
s
2
+
s
(
s
2
+1)
2
=
s
s
2
+1
−
1
s
2
+1
+
1
(
s
2
+1)
2
.
Thus
y
= cos
t
−
1
2
sin
t
−
1
2
t
cos
t.
33.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0)+16
{
y
}
=
{
cos 4
t
−
cos 4
t
(
t
−
π
)
}
or
(
s
2
+ 16)
{
y
}
=1+
s
s
2
+16
−
e
−
πs
{
cos 4(
t
+
π
)
}
=1+
s
s
2
+16
−
e
−
πs
{
cos 4
t
}
=1+
s
s
2
+16
−
s
s
2
+16
e
−
πs
.
Thus
{
y
}
=
1
s
2
+16
+
s
(
s
2
+ 16)
2
−
s
(
s
2
+ 16)
2
e
−
πs
and
y
=
1
4
sin 4
t
+
1
8
t
sin 4
t
−
1
8
(
t
−
π
) sin 4(
t
−
π
)(
t
−
π
)
.
34.
The Laplace transform of the differential equation is
s
2
{
y
}−
sy
(0)
−
y
)
(0) +
{
y
}
=
g
1
−
o
t
−
π
2
,
+ sin
t
o
t
−
π
2
,r
(
s
2
+1)
{
y
}
=
s
+
1
s
−
1
s
e
−
πs/
2
+
e
−
πs/
2
g
sin
o
t
+
π
2
,r
or
=
s
+
1
s
−
1
s
e
−
πs/
2
+
e
−
πs/
2
{
cos
t
}
=
s
+
1
s
−
1
s
e
−
πs/
2
+
s
s
2
+1
e
−
πs/
2
.
189

190. 1 2 3 4 5 6
t
-1
-0.5
0.5
1
y
1 2 3 4 5 6
t
-4
-2
2
4
y
Exercises 4.4
Thus
{
y
}
=
s
s
2
+1
+
1
s
(
s
2
+1)
−
1
s
(
s
2
+1)
e
−
πs/
2
+
s
(
s
2
+1)
2
e
−
πs/
2
=
s
s
2
+1
+
1
s
−
s
s
2
+1
−
=
1
s
−
s
s
2
+1
+
e
−
πs/
2
+
s
(
s
2
+1)
2
e
−
πs/
2
=
1
s
−
=
1
s
−
s
s
2
+1
+
e
−
πs/
2
+
s
(
s
2
+1)
2
e
−
πs/
2
and
y
=1
−
b
1
−
cos
o
t
−
π
2
,yo
t
−
π
2
,
+
1
2
o
t
−
π
2
,
sin
o
t
−
π
2
,o
t
−
π
2
,
=1
−
(1
−
sin
t
)
o
t
−
π
2
,
+
1
2
o
t
−
π
2
,
cos
t
o
t
−
π
2
,
.
35.
The Laplace transform of the differential equation is
s
2
{
y
}
+
{
y
}
=
1
(
s
2
+1)
+
2
s
(
s
2
+1)
2
.
Thus
{
y
}
=
1
(
s
2
+1)
2
+
2
s
(
s
2
+1)
3
and, using Problem 20,
y
=
1
2
(sin
t
−
t
cos
t
)+
1
4
(
t
sin
t
−
t
2
cos
t
)
.
36. (a)
(b)
37.
The Laplace transform of the given equation is
{
f
}
+
{
t
}{
f
}
=
{
t
}
.
Solving for
{
f
}
we obtain
{
f
}
=
1
s
2
+1
. Thus,
f
(
t
) = sin
t
.
38.
The Laplace transform of the given equation is
{
f
}
=
{
2
t
}−
4
{
sin
t
}{
f
}
.
190

191. Exercises 4.4
Solving for
{
f
}
we obtain
{
f
}
=
2
s
2
+2
s
2
(
s
2
+5)
=
2
5
1
s
2
+
8
5
√
5
√
5
s
2
+5
.
Thus
f
(
t
)=
2
5
t
+
8
5
√
5
sin
√
5
t.
39.
The Laplace transform of the given equation is
{
f
}
=
h
te
t
U
+
{
t
}{
f
}
.
Solving for
{
f
}
we obtain
{
f
}
=
s
2
(
s
−
1)
3
(
s
+1)
=
1
8
1
s
−
1
+
3
4
1
(
s
−
1)
2
+
1
4
2
(
s
−
1)
3
−
1
8
1
s
+1
.
Thus
f
(
t
)=
1
8
e
t
+
3
4
te
t
+
1
4
t
2
e
t
−
1
8
e
−
t
40.
The Laplace transform of the given equation is
{
f
}
+2
{
cos
t
}{
f
}
=4
h
e
−
t
U
+
{
sin
t
}
.
Solving for
{
f
}
we obtain
{
f
}
=
4
s
2
+
s
+5
(
s
+1)
3
=
4
s
+1
−
7
(
s
+1)
2
+4
2
(
s
+1)
3
.
Thus
f
(
t
)=4
e
−
t
−
7
te
−
t
+4
t
2
e
−
t
.
41.
The Laplace transform of the given equation is
{
f
}
+
{
1
}{
f
}
=
{
1
}
.
Solving for
{
f
}
we obtain
{
f
}
=
1
s
+1
. Thus,
f
(
t
)=
e
−
t
.
42.
The Laplace transform of the given equation is
{
f
}
=
{
cos
t
}
+
h
e
−
t
U
{
f
}
.
Solving for
{
f
}
we obtain
{
f
}
=
s
s
2
+1
+
1
s
2
+1
.
Thus
f
(
t
) = cos
t
+ sin
t.
43.
The Laplace transform of the given equation is
{
f
}
=
{
1
}
+
{
t
}−
8
3
{
t
3
}{
f
}
.
Solving for
{
f
}
we obtain
{
f
}
=
s
2
(
s
+1)
s
4
+16
=
s
3
s
4
+16
+
s
2
s
4
+16
.
Thus
f
(
t
) = cos
√
2
t
cosh
√
2
t
+
1
2
√
2
(sin
√
2
t
cosh
√
2
t
+ cos
√
2
t
sinh
√
2
t
)
.
191

192. 0.511.522.53
t
-30
-20
-10
10
20
30
i
Exercises 4.4
44.
The Laplace transform of the given equation is
{
t
}−
2
{
f
}
=
h
e
t
−
e
−
t
U
{
f
}
.
Solving for
{
f
}
we obtain
{
f
}
=
s
2
−
1
2
s
4
=
1
2
1
s
2
−
1
12
3!
s
4
.
Thus
f
(
t
)=
1
2
t
−
1
12
t
3
.
45.
The Laplace transform of the given equation is
s
{
y
}−
y
(0) =
{
1
}− {
sin
t
}− {
1
}{
y
}
.
Solving for
{
f
}
we obtain
{
y
}
=
s
3
−
s
2
+
s
s
(
s
2
+1)
2
=
1
s
2
+1
+
1
2
2
s
(
s
2
+1)
2
.
Thus
y
= sin
t
−
1
2
t
sin
t.
46.
The Laplace transform of the given equation is
s
{
y
}−
y
(0)+6
{
y
}
+9
{
1
}{
y
}
=
{
1
}
.
Solving for
{
f
}
we obtain
{
y
}
=
1
(
s
+3)
2
. Thus,
y
=
te
−
3
t
.
47.
The differential equation is
0
.
1
di
dt
+3
i
+
1
0
.
05
(
t
0
i
(
τ
)
dτ
= 100
9
(
t
−
1)
−
(
t
−
2)
8
or
di
st
+30
i
+ 200
(
t
0
i
(
τ
)
dτ
= 1000
9
(
t
−
1)
−
(
t
−
2)
8
,
where
i
(0) = 0. The Laplace transform of the differential equation is
s
{
i
}−
y
(0)+30
{
i
}
+
200
s
{
i
}
=
1000
s
(
e
−
s
−
e
−
2
s
)
.
Solving for
{
i
}
we obtain
{
i
}
=
1000
e
−
s
−
1000
e
−
2
s
s
2
+30
s
+ 200
=
=
100
s
+10
−
100
s
+20
+
(
e
−
s
−
e
−
2
s
)
.
Thus
i
(
t
) = 100
t
e
−
10(
t
−
1)
−
e
−
20(
t
−
1)
e
(
t
−
1)
−
100
t
e
−
10(
t
−
2)
−
e
−
20(
t
−
2)
e
(
t
−
2)
.
192

193. 0.511.52
t
0.5
1
1.5
2
i
1 2 3 4
t
-1
-0.5
0.5
1
i
Exercises 4.4
48.
The differential equation is
0
.
005
di
dt
+
i
+
1
0
.
02
(
t
0
i
(
τ
)
dτ
= 100
9
t
−
(
t
−
1) (
t
−
1)
8
or
di
st
+ 200
i
+ 10,000
(
t
0
i
(
τ
)
dτ
= 20,000
9
t
−
(
t
−
1) (
t
−
1)
8
,
where
i
(0) = 0. The Laplace transform of the differential equation is
s
{
i
}
+ 200
{
i
}
+
10,000
s
{
i
}
= 20,000
=
1
s
2
−
1
s
2
e
−
s
+
.
Solving for
{
i
}
we obtain
{
i
}
=
20,000
s
(
s
+ 100)
2
(1
−
e
−
s
)=
!
2
s
−
2
s
+ 100
−
200
(
s
+ 100)
2
5
(1
−
e
−
s
)
.
Thus
i
(
t
)=2
−
2
e
−
100
t
−
200
te
−
100
t
−
2(
t
−
1)+2
e
−
100(
t
−
1)
(
t
−
1) + 200(
t
−
1)
e
−
100(
t
−
1)
(
t
−
1)
.
49.
The differential equation is
di
dt
+
i
=
E
(
t
)
,
where
i
(0) = 0. The Laplace transform of this equation is
s
{
i
}
+
{
i
}
=
{
E
(
t
)
}
.
From Problem 23 we have
{
E
(
t
)
}
=
1
−
e
−
s
s
(1 +
e
−
s
)
.
Thus
(
s
+1)
{
i
}
=
1
−
e
−
s
s
(1 +
e
−
s
)
and
{
i
}
=
1
−
e
−
s
s
(
s
+ 1)(1 +
e
−
s
)
=
1
−
e
−
s
s
(
s
+1)
1
1+
e
−
s
=
=
1
s
−
1
s
+1
+
(1
−
e
−
s
)(1
−
e
−
s
+
e
−
2
s
−
e
−
3
s
+
e
−
4
s
−···
)
=
=
1
s
−
1
s
+1
+
(1
−
2
e
−
s
+2
e
−
2
s
−
2
e
−
3
s
+2
e
−
4
s
−···
)
.
Therefore
i
(
t
)=
9
1
−
2(
t
−
1)+2 (
t
−
2)
−
2(
t
−
3)+2 (
t
−
4)
−···
8
−
9
e
−
t
+2
e
−
(
t
−
1)
(
t
−
1)
−
2
e
−
(
t
−
2)
(
t
−
2)
+2
e
−
(
t
−
3)
(
t
−
3)
−
2
e
−
(
t
−
4)
(
t
−
4) +
···
8
=1
−
e
−
t
+2
∞
!
n
=1
(
−
1)
n
t
1
−
e
−
(
t
−
n
)
e
(
t
−
n
)
.
193

194. 1 2 3 4
t
-1
-0.5
0.5
1
i
Exercises 4.4
50.
The differential equation is
di
dt
+
i
=
E
(
t
)
,
where
i
(0) = 0. The Laplace transform of this equation is
s
{
i
}
+
{
i
}
=
{
E
(
t
)
}
.
From Problem 25 we have
{
E
(
t
)
}
=
1
s
=
1
s
−
1
e
s
−
1
+
=
1
s
2
−
1
s
1
e
s
−
1
.
Thus
(
s
+1)
{
i
}
=
1
s
2
−
1
s
1
e
s
−
1
and
{
i
}
=
1
s
2
(
s
+1)
−
1
s
(
s
+1)
1
e
s
−
1
=
=
1
s
2
−
1
s
+
1
s
+1
+
−
=
1
s
−
1
s
+1
+
1
e
s
−
1
=
=
1
s
2
−
1
s
+
1
s
+1
+
−
=
1
s
−
1
s
+1
+
t
e
−
s
+
e
−
2
s
+
e
−
3
s
+
e
−
4
s
−···
e
.
Therefore
i
(
t
)=
t
t
−
1+
e
−
t
)
−
(1
−
e
−
(
t
−
1)
e
(
t
−
1)
−
t
1
−
e
−
(
t
−
2)
e
(
t
−
2)
−
t
1
−
e
−
(
t
−
3)
e
(
t
−
3)
−
t
1
−
e
−
(
t
−
4)
e
(
t
−
4)
−···
=
t
t
−
1+
e
−
t
e
−
∞
!
n
=1
t
1
−
e
−
(
t
−
n
)
e
(
t
−
n
)
.
51.
The differential equation is
x
))
+2
x
)
+10
x
=20
f
(
t
), where
f
(
t
) is the meander function with
a
=
π
. Using the
initial conditions
x
(0) =
x
)
(0) = 0 and taking the Laplace transform we obtain
(
s
2
+2
s
+ 10)
{
x
(
t
)
}
=
20
s
(1
−
e
−
πs
)
1
1+
e
−
πs
=
20
s
(1
−
e
−
πs
)(1
−
e
−
πs
+
e
−
2
πs
−
e
−
3
πs
+
···
)
=
20
s
(1
−
2
e
−
πs
+2
e
−
2
πs
−
2
e
−
3
πs
+
···
)
=
20
s
+
40
s
∞
!
n
=1
(
−
1)
n
e
−
nπs
.
Then
{
x
(
t
)
}
=
20
s
(
s
2
+2
s
+ 10)
+
40
s
(
s
2
+2
s
+ 10)
∞
!
n
=1
(
−
1)
n
e
−
nπs
=
2
s
−
2
s
+4
s
2
+2
s
+10
+
∞
!
n
=1
(
−
1)
n
!
4
s
−
4
s
+8
s
2
+2
s
+10
5
e
−
nπs
=
2
s
−
2(
s
+1)+2
(
s
+1)
2
+9
+4
∞
!
n
=1
(
−
1)
n
!
1
s
−
(
s
+1)+1
(
s
+1)
2
+9
5
e
−
nπs
194

195. π
2π
t
x
−3
3
4π
2π
t
x
−5
5
Exercises 4.4
and
x
(
t
)=2
=
1
−
e
−
t
cos 3
t
−
1
3
e
−
t
sin 3
t
+
+4
∞
!
n
=1
(
−
1)
n
!
1
−
e
−
(
t
−
nπ
)
cos 3(
t
−
nπ
)
−
1
3
e
−
(
t
−
nπ
)
sin 3(
t
−
nπ
)
5
(
t
−
nπ
)
.
The graph of
x
(
t
) on the interval [0
,
2
π
) is shown below.
52.
The differential equation is
x
))
+2
x
)
+
x
=5
f
(
t
), where
f
(
t
) is the square wave function with
a
=
π
. Using the
initial conditions
x
(0) =
x
)
(0) = 0 and taking the Laplace transform, we obtain
(
s
2
+2
s
+1)
{
x
(
t
)
}
=
5
s
1
1+
e
−
πs
=
5
s
(1
−
e
−
πs
+
e
−
2
πs
−
e
−
3
πs
+
e
−
4
πs
−···
)
=
5
s
∞
!
n
=0
(
−
1)
n
e
−
nπs
.
Then
{
x
(
t
)
}
=
5
s
(
s
+1)
2
∞
!
n
=0
(
−
1)
n
e
−
nπs
=5
∞
!
n
=0
(
−
1)
n
=
1
s
−
1
s
+1
−
1
(
s
+1)
2
+
e
−
nπs
and
x
(
t
)=5
∞
!
n
=0
(
−
1)
n
(1
−
e
−
(
t
−
nπ
)
−
(
t
−
nπ
)
e
−
(
t
−
nπ
)
)(
t
−
nπ
)
.
The graph of
x
(
t
) on the interval [0
,
4
π
) is shown below.
53.
Let
u
=
t
−
τ
so that
du
=
dτ
and
f
∗
g
=
(
t
0
f
(
τ
)
g
(
t
−
τ
)
dτ
=
−
(
0
t
f
(
t
−
u
)
g
(
u
)
du
=
g
∗
f.
54.
f
∗
(
g
+
h
)=
(
t
0
f
(
τ
)[
g
(
t
−
τ
)+
h
(
t
−
τ
)]
dτ
=
(
t
0
f
(
τ
)
g
(
t
−
τ
)
dτ
+
(
t
0
f
(
τ
)
h
(
t
−
τ
)
dτ
=
(
t
0
f
(
τ
)[
g
(
t
−
τ
)+
h
(
t
−
τ
)]
dτ
=
f
∗
g
+
f
∗
h
195

196. t
q
−5
5
π3π
Exercises 4.4
55. (a)
The output for the first three lines of the program are
9
y
[
t
]+6
y
)
[
t
]+
y
))
[
t
]==
t
sin[
t
]
1
−
2
s
+9
Y
+
s
2
Y
+6(
−
2+
sY
)==
2
s
(1 +
s
2
)
2
Y
→−
=
−
11
−
4
s
−
22
s
2
−
4
s
3
−
11
s
4
−
2
s
5
(1 +
s
2
)
2
(9+6
s
+
s
2
)
+
The fourth line is the same as the third line with
Y
→
removed. The final line of output shows a solution
involving complex coefficients of
e
it
and
e
−
it
. To get the solution in more standard form write the last line
as two lines:
euler =
{
Eˆ(I t) -
>
Cos[t] + I Sin[t], Eˆ(-I t) -
>
Cos[t] - I Sin[t]
}
InverseLaplaceTransform[Y,s,t]/.euler//Expand
We see that the solution is
y
(
t
)=
=
487
250
+
247
50
t
+
e
−
3
t
+
1
250
(13 cos
t
−
15
t
cos
t
−
9 sin
t
+20
t
sin
t
)
.
(b)
The solution is
y
(
t
)=
1
6
e
t
−
1
6
e
−
t/
2
cos
√
15
t
−
a
3
/
5
6
e
−
t/
2
sin
√
15
t.
(c)
The solution is
q
(
t
)=1
−
cos
t
+(6
−
6 cos
t
)(
t
−
3
π
)
−
(4 + 4 cos
t
)(
t
−
π
)
.
Exercises 4.5
1.
The Laplace transform of the differential equation yields
{
y
}
=
1
s
−
3
e
−
2
s
so that
y
=
e
3(
t
−
2)
(
t
−
2)
.
2.
The Laplace transform of the differential equation yields
{
y
}
=
2
s
+1
+
e
−
s
s
+1
so that
y
=2
e
−
t
+
e
−
(
t
−
1)
(
t
−
1)
.
3.
The Laplace transform of the differential equation yields
{
y
}
=
1
s
2
+1
t
1+
e
−
2
πs
e
196

197. Exercises 4.5
so that
y
= sin
t
+ sin
t
(
t
−
2
π
)
.
4.
The Laplace transform of the differential equation yields
{
y
}
=
1
4
4
s
2
+16
e
−
2
πs
so that
y
=
1
4
sin 4(
t
−
2
π
)(
t
−
2
π
)
.
5.
The Laplace transform of the differential equation yields
{
y
}
=
1
s
2
+1
o
e
−
πs/
2
+
e
−
3
πs/
2
,
so that
y
= sin
o
t
−
π
2
,o
t
−
π
2
,
+ sin
=
t
−
3
π
2
+=
t
−
3
π
2
+
=
−
cos
t
o
t
−
π
2
,
+ cos
t
o
t
−
π
2
,
.
6.
The Laplace transform of the differential equation yields
{
y
}
=
s
s
2
+1
+
1
s
2
+1
(
e
−
2
πs
+
e
−
4
πs
)
so that
y
= cos
t
+ sin
t
[(
t
−
2
π
)+ (
t
−
4
π
)]
.
7.
The Laplace transform of the differential equation yields
{
y
}
=
1
s
2
+2
s
(1 +
e
−
s
)=
!
1
2
1
s
−
1
2
1
s
+2
5
(1 +
e
−
s
)
so that
y
=
1
2
−
1
2
e
−
2
t
+
!
1
2
−
1
2
e
−
2(
t
−
1)
5
(
t
−
1)
.
8.
The Laplace transform of the differential equation yields
{
y
}
=
s
+1
s
2
(
s
−
2)
+
1
s
(
s
−
2)
e
−
2
s
=
3
4
1
s
−
2
−
3
4
1
s
−
1
2
1
s
2
+
!
1
2
1
s
−
2
−
1
2
1
s
5
e
−
2
s
so that
y
=
3
4
e
2
t
−
3
4
−
1
2
t
+
!
1
2
e
2(
t
−
2)
−
1
2
5
(
t
−
2)
.
9.
The Laplace transform of the differential equation yields
{
y
}
=
1
(
s
+2)
2
+1
e
−
2
πs
so that
y
=
e
−
2(
t
−
2
π
)
sin
t
(
t
−
2
π
)
.
10.
The Laplace transform of the differential equation yields
{
y
}
=
1
(
s
+1)
2
e
−
s
so that
y
=(
t
−
1)
e
−
(
t
−
1)
(
t
−
1)
.
197

198. Exercises 4.5
11.
The Laplace transform of the differential equation yields
{
y
}
=
4+
s
s
2
+4
s
+13
+
e
−
πs
+
e
−
3
πs
s
2
+4
s
+13
=
2
3
3
(
s
+2)
2
+3
2
+
s
+2
(
s
+2)
2
+3
2
+
1
3
3
(
s
+2)
2
+3
2
t
e
−
πs
+
e
−
3
πs
e
so that
y
=
2
3
e
−
2
t
sin 3
t
+
e
−
2
t
cos 3
t
+
1
3
e
−
2(
t
−
π
)
sin 3(
t
−
π
)(
t
−
π
)
+
1
3
e
−
2(
t
−
3
π
)
sin 3(
t
−
3
π
)(
t
−
3
π
)
.
12.
The Laplace transform of the differential equation yields
{
y
}
=
1
(
s
−
1)
2
(
s
−
6)
+
e
−
2
s
+
e
−
4
s
(
s
−
1)(
s
−
6)
=
−
1
25
1
s
−
1
−
1
5
1
(
s
−
1)
2
+
1
25
1
s
−
6
+
!
−
1
5
1
s
−
1
+
1
5
1
s
−
6
5
t
e
−
2
s
+
e
−
4
s
e
so that
y
=
−
1
25
e
t
−
1
5
te
t
+
1
25
e
6
t
+
!
−
1
5
e
t
−
2
+
1
5
e
6(
t
−
2)
5
(
t
−
2)
+
!
−
1
5
e
t
−
4
+
1
5
e
6(
t
−
4)
5
(
t
−
4)
.
13.
The Laplace transform of the differential equation yields
{
y
}
=
1
2
2
s
3
y
))
(0) +
1
6
3!
s
4
y
)))
(0) +
1
6
P
0
EI
3!
s
4
e
−
Ls/
2
so that
y
=
1
2
y
))
(0)
x
2
+
1
6
y
)))
(0)
x
3
+
1
6
P
0
EI
=
X
−
L
2
+
3
=
x
−
L
2
+
.
Using
y
))
(
L
)=0and
y
)))
(
L
) = 0 we obtain
y
=
1
4
P
0
L
EI
x
2
−
1
6
P
0
EI
x
3
+
1
6
P
0
EI
=
x
−
L
2
+
3
=
x
−
L
2
+
=









P
0
EI
=
L
4
x
2
−
1
6
x
3
+
,
0
≤
x<
L
2
P
0
L
2
4
EI
=
1
2
x
−
L
12
+
,
L
2
≤
x
≤
L
.
14.
From Problem 13 we know that
y
=
1
2
y
))
(0)
x
2
+
1
6
y
)))
(0)
x
3
+
1
6
P
0
EI
=
X
−
L
2
+
3
=
x
−
L
2
+
.
Using
y
(
L
) = 0 and
y
)
(
L
) = 0 we obtain
y
=
1
16
P
0
L
EI
x
2
−
1
12
P
0
EI
x
3
+
1
6
P
0
EI
=
x
−
L
2
+
3
=
x
−
L
2
+
=









P
0
EI
=
L
16
x
2
−
1
12
x
3
+
,
0
≤
x<
L
2
P
0
EI
=
L
16
x
2
−
1
12
x
3
+
+
1
6
P
0
EI
=
x
−
L
2
+
3
,
L
2
≤
x
≤
L
.
198

199. Exercises 4.6
15.
The Laplace transform of the differential equation yields
{
y
}
=
1
s
2
+
ω
2
so that
y
(
t
) = sin
ωt
. Note that
y
)
(0) = 1, even though the initial condition was
y
)
(0)=0.
Exercises 4.6
1.
Taking the Laplace transform of the system gives
s
{
x
}
=
−{
x
}
+
{
y
}
s
{
y
}−
1=2
{
x
}
{
x
}
=
1
(
s
−
1)(
s
+2)
=
1
3
1
s
−
1
−
1
3
1
s
+2
so that
{
y
}
=
1
s
+
2
s
(
s
−
1)(
s
+2)
=
2
3
1
s
−
1
+
1
3
1
s
+2
.
and
Then
x
=
1
3
e
t
−
1
3
e
−
2
t
and
y
=
2
3
e
t
+
1
3
e
−
2
t
.
2.
Taking the Laplace transform of the system gives
s
{
x
}−
1=2
{
y
}
+
1
s
−
1
s
{
y
}−
1=8
{
x
}−
1
s
2
{
y
}
=
s
3
+7
s
2
−
s
+1
s
(
s
−
1)(
s
2
−
16)
=
1
16
1
s
−
8
15
1
s
−
1
+
173
96
1
s
−
4
−
53
160
1
s
+4
so that
y
=
1
16
−
8
15
e
t
+
173
96
e
4
t
−
53
160
e
−
4
t
.
and
Then
x
=
1
8
y
)
+
1
8
t
=
1
8
t
−
1
15
e
t
+
173
192
e
4
t
+
53
320
e
−
4
t
.
3.
Taking the Laplace transform of the system gives
s
{
x
}
+1=
{
x
}−
2
{
y
}
s
{
y
}−
2=5
{
x
}− {
y
}
{
x
}
=
−
s
−
5
s
2
+9
=
−
s
s
2
+9
−
5
3
3
s
2
+9
so that
x
=
−
cos 3
t
−
5
3
sin 3
t.
and
Then
y
=
1
2
x
−
1
2
x
)
= 2 cos 3
t
−
7
3
sin 3
t.
4.
Taking the Laplace transform of the system gives
(
s
+3)
{
x
}
+
s
{
y
}
=
1
s
(
s
−
1)
{
x
}
+(
s
−
1)
{
y
}
=
1
s
−
1
199

200. Exercises 4.6
{
y
}
=
5
s
−
1
3
s
(
s
−
1)
2
=
−
1
3
1
s
+
1
3
1
s
−
1
+
4
3
1
(
s
−
1)
2
so that
{
x
}
=
1
−
2
s
3
s
(
s
−
1)
2
=
1
3
1
s
−
1
3
1
s
−
1
−
1
3
1
(
s
−
1)
2
.
and
Then
x
=
1
3
−
1
3
e
t
−
1
3
te
t
and
y
=
−
1
3
+
1
3
e
t
+
4
3
te
t
.
5.
Taking the Laplace transform of the system gives
(2
s
−
2)
{
x
}
+
s
{
y
}
=
1
s
(
s
−
3)
{
x
}
+(
s
−
3)
{
y
}
=
2
s
{
x
}
=
−
s
−
3
s
(
s
−
2)(
s
−
3)
=
−
1
2
1
s
+
5
2
1
s
−
2
−
2
s
−
3
so that
{
y
}
=
3
s
−
1
s
(
s
−
2)(
s
−
3)
=
−
1
6
1
s
−
5
2
1
s
−
2
+
8
3
1
s
−
3
.
and
Then
x
=
−
1
2
+
5
2
e
2
t
−
2
e
3
t
and
y
=
−
1
6
−
5
2
e
2
t
+
8
3
e
3
t
.
6.
Taking the Laplace transform of the system gives
(
s
+1)
{
x
}−
(
s
−
1)
{
y
}
=
−
1
s
{
x
}
+(
s
+2)
{
y
}
=1
{
y
}
=
s
+1
/
2
s
2
+
s
+1
=
s
+1
/
2
(
s
+1
/
2)
2
+(
√
3
/
2)
2
so that
{
x
}
=
−
3
/
2
s
2
+
s
+1
=
−
3
/
2
(
s
+1
/
2)
2
+(
√
3
/
2)
2
.
and
Then
y
=
e
−
t/
2
cos
√
3
2
t
and
x
=
e
−
t/
2
sin
√
3
2
t.
7.
Taking the Laplace transform of the system gives
(
s
2
+1)
{
x
}− {
y
}
=
−
2
−{
x
}
+(
s
2
+1)
{
y
}
=1
{
x
}
=
−
2
s
2
−
1
s
4
+2
s
2
=
−
1
2
1
s
2
−
3
2
1
s
2
+2
so that
x
=
−
1
2
t
−
3
2
√
2
sin
√
2
t.
and
Then
y
=
x
))
+
x
=
−
1
2
t
+
3
2
√
2
sin
√
2
t.
8.
Taking the Laplace transform of the system gives
(
s
+1)
{
x
}
+
{
y
}
=1
4
{
x
}−
(
s
+1)
{
y
}
=1
200

201. Exercises 4.6
{
x
}
=
s
+2
s
2
+2
s
+5
=
s
+1
(
s
+1)
2
+2
2
+
1
2
2
(
s
+1)
2
+2
2
so that
{
y
}
=
−
s
+3
s
2
+2
s
+5
=
−
s
+1
(
s
+1)
2
+2
2
+2
2
(
s
+1)
2
+2
2
.
and
Then
x
=
e
−
t
cos 2
t
+
1
2
e
−
t
sin 2
t
and
y
=
−
e
−
t
cos 2
t
+2
e
−
t
sin 2
t.
9.
Adding the equations and then subtracting them gives
d
2
x
dt
2
=
1
2
t
2
+2
t
d
2
y
dt
2
=
1
2
t
2
−
2
t.
Taking the Laplace transform of the system gives
{
x
}
=8
1
s
+
1
24
4!
s
5
+
1
3
3!
s
4
and
{
y
}
=
1
24
4!
s
5
−
1
3
3!
s
4
so that
x
=8+
1
24
t
4
+
1
3
t
3
and
y
=
1
24
t
4
−
1
3
t
3
.
10.
Taking the Laplace transform of the system gives
(
s
−
4)
{
x
}
+
s
3
{
y
}
=
6
s
2
+1
(
s
+2)
{
x
}−
2
s
3
{
y
}
=0
{
x
}
=
4
(
s
−
2)(
s
2
+1)
=
4
5
1
s
−
2
−
4
5
s
s
2
+1
−
8
5
1
s
2
+1
so that
{
y
}
=
2
s
+4
s
3
(
s
−
2)(
s
2
+1)
=
1
s
−
2
s
2
−
2
2
s
3
+
1
5
1
s
−
2
−
6
5
s
s
2
+1
+
8
5
1
s
2
+1
.
and
x
=
4
5
e
2
t
−
4
5
cos
t
−
8
5
sin
t
Then
y
=1
−
2
t
−
2
t
2
+
1
5
e
2
t
−
6
5
cos
t
+
8
5
sin
t.
and
11.
Taking the Laplace transform of the system gives
s
2
{
x
}
+3(
s
+1)
{
y
}
=2
s
2
{
x
}
+3
{
y
}
=
1
(
s
+1)
2
so that
{
x
}
=
−
2
s
+1
s
3
(
s
+1)
=
1
s
+
1
s
2
+
1
2
2
s
3
−
1
s
+1
.
x
=1+
t
+
1
2
t
2
−
e
−
t
Then
y
=
1
3
te
−
t
−
1
3
x
))
=
1
3
te
−
t
+
1
3
e
−
t
−
1
3
.
and
201

202. Exercises 4.6
12.
Taking the Laplace transform of the system gives
(
s
−
4)
{
x
}
+2
{
y
}
=
2
e
−
s
s
−
3
{
x
}
+(
s
+1)
{
y
}
=
1
2
+
e
−
s
s
{
x
}
=
−
1
/
2
(
s
−
1)(
s
−
2)
+
e
−
s
1
(
s
−
1)(
s
−
2)
so that
=
!
1
2
1
s
−
1
−
1
2
1
s
−
2
5
+
e
−
s
!
−
1
s
−
1
+
1
s
−
2
5
{
y
}
=
e
−
s
s
+
s/
4
−
1
(
s
−
1)(
s
−
2)
+
e
−
s
−
s/
2+2
(
s
−
1)(
s
−
2)
and
=
3
4
1
s
−
1
−
1
2
1
s
−
2
+
e
−
s
!
1
s
−
3
2
1
s
−
1
+
1
s
−
2
5
.
x
=
1
2
e
t
−
1
2
e
2
t
+
b
−
e
t
−
1
+
e
2(
t
−
1)

(
t
−
1)
Then
y
=
3
4
e
t
−
1
2
e
2
t
+
!
1
−
3
2
e
t
−
1
+
e
2(
t
−
1)
5
(
t
−
1)
.
and
13.
The system is
x
))
1
=
−
3
x
1
+2(
x
2
−
x
1
)
x
))
2
=
−
2(
x
2
−
x
1
)
x
1
(0) = 0
x
)
1
(0) = 1
x
2
(0) = 1
x
)
2
(0) = 0
.
Taking the Laplace transform of the system gives
(
s
2
+5)
{
x
1
}−
2
{
x
2
}
=1
−
2
{
x
1
}
+(
s
2
+2)
{
x
2
}
=
s
{
x
1
}
=
s
2
+2
s
+2
s
4
+7
s
2
+6
=
2
5
s
s
2
+1
+
1
5
1
s
2
+1
−
2
5
s
s
2
+6
+
4
5
√
6
√
6
s
2
+6
so that
{
x
2
}
=
s
3
+5
s
+2
(
s
2
+ 1)(
s
2
+6)
=
4
5
s
s
2
+1
+
2
5
1
s
2
+1
+
1
5
s
s
2
+6
−
2
5
√
6
√
6
s
2
+6
.
and
x
1
=
2
5
cos
t
+
1
5
sin
t
−
2
5
cos
√
6
t
+
4
5
√
6
sin
√
6
t
Then
x
2
=
4
5
cos
t
+
2
5
sin
t
+
1
5
cos
√
6
t
−
2
5
√
6
sin
√
6
t.
and
14.
In this system
x
1
and
x
2
represent displacements of masses
m
1
and
m
2
from their equilibrium positions. Since
the net forces acting on
m
1
and
m
2
are
−
k
1
x
1
+
k
2
(
x
2
−
x
1
) and
−
k
2
(
x
2
−
x
1
)
−
k
3
x
2
,
202

203. Exercises 4.6
respectively, Newton’s second law of motion gives
m
1
x
))
1
=
−
k
1
x
1
+
k
2
(
x
2
−
x
1
)
m
2
x
))
2
=
−
k
2
(
x
2
−
x
1
)
−
k
3
x
2
.
Using
k
1
=
k
2
=
k
3
=1,
m
1
=
m
2
=1,
x
1
(0) = 0,
x
1
(0) =
−
1,
x
2
(0) = 0, and
x
)
2
(0) = 1, and taking the
Laplace transform of the system, we obtain
(2 +
s
2
)
{
x
1
}− {
x
2
}
=
−
1
{
x
1
}−
(2 +
s
2
)
{
x
2
}
=
−
1
so that
{
x
1
}
=
−
1
s
2
+3
and
{
x
2
}
=
1
s
2
+3
.
Then
x
1
=
−
1
√
3
sin
√
3
t
and
x
2
=
1
√
3
sin
√
3
t.
15. (a)
ByKirchoff’s first law we have
i
1
=
i
2
+
i
3
. ByKirchoff’s second law, on each loop we have
E
(
t
)=
Ri
1
+
L
1
i
)
2
and
E
(
t
)=
Ri
1
+
L
2
i
)
3
or
L
1
i
)
2
+
Ri
2
+
Ri
3
=
E
(
t
) and
L
2
i
)
3
+
Ri
2
+
Ri
3
=
E
(
t
).
(b)
Taking the Laplace transform of the system
0
.
01
i
)
2
+5
i
2
+5
i
3
= 100
0
.
0125
i
)
3
+5
i
2
+5
i
3
= 100
(
s
+ 500)
{
i
2
}
+ 500
{
i
3
}
=
10,000
s
gives
400
{
i
2
}
+(
s
+ 400)
{
i
3
}
=
8,000
s
so that
{
i
3
}
=
8,000
s
2
+ 900
s
=
80
9
1
s
−
80
9
1
s
+ 900
.
Then
i
3
=
80
9
−
80
9
e
−
900
t
and
i
2
=20
−
0
.
0025
i
)
3
−
i
3
=
100
9
−
100
9
e
−
900
t
.
(c)
i
1
=
i
2
+
i
3
=20
−
20
e
−
900
t
16. (a)
Taking the Laplace transform of the system
i
)
2
+
i
)
3
+10
i
2
= 120
−
120 (
t
−
2)
−
10
i
)
2
+5
i
)
3
+5
i
3
=0
(
s
+ 10)
{
i
2
}
+
s
{
i
3
}
=
120
s
t
1
−
e
−
2
s
e
gives
−
10
s
{
i
2
}
+5(
s
+1)
{
i
3
}
=0
so that
{
i
2
}
=
120(
s
+1)
(3
s
2
+11
s
+ 10)
s
t
1
−
e
−
2
s
e
=
!
48
s
+5
/
3
−
60
s
+2
+
12
s
5
t
1
−
e
−
2
s
e
and
{
i
3
}
=
240
3
s
2
+11
s
+10
t
1
−
e
−
2
s
e
=
!
240
s
+5
/
3
−
240
s
+2
5
t
1
−
e
−
2
s
e
.
203

204. Exercises 4.6
i
2
=12+48
e
−
5
t/
3
−
60
e
−
2
t
−
b
12+48
e
−
5(
t
−
2)
/
3
−
60
e
−
2(
t
−
2)

(
t
−
2)
Then
i
3
= 240
e
−
5
t/
3
−
240
e
−
2
t
−
b
240
e
−
5(
t
−
2)
/
3
−
240
e
−
2(
t
−
2)

(
t
−
2)
.
and
(b)
i
1
=
i
2
+
i
3
= 12 + 288
e
−
5
t/
3
−
300
e
−
2
t
−
b
12 + 288
e
−
5(
t
−
2)
/
3
−
300
e
−
2(
t
−
2)

(
t
−
2)
17.
Taking the Laplace transform of the system
i
)
2
+11
i
2
+6
i
3
= 50 sin
t
i
)
3
+6
i
2
+6
i
3
= 50 sin
t
(
s
+ 11)
{
i
2
}
+6
{
i
3
}
=
50
s
2
+1
gives
6
{
i
2
}
+(
s
+6)
{
i
3
}
=
50
s
2
+1
so that
{
i
2
}
=
50
s
(
s
+ 2)(
s
+ 15)(
s
2
+1)
=
−
20
13
1
s
+2
+
375
1469
1
s
+15
+
145
113
s
s
2
+1
+
85
113
1
s
2
+1
.
i
2
=
−
20
13
e
−
2
t
+
375
1469
e
−
15
t
+
145
113
cos
t
+
85
113
sin
t
Then
i
3
=
25
3
sin
t
−
1
6
i
)
2
−
11
6
i
2
=
30
13
e
−
2
t
+
250
1469
e
−
15
t
−
280
113
cos
t
+
810
113
sin
t.
and
18.
Taking the Laplace transform of the system
0
.
5
i
)
1
+50
i
2
=60
0
.
005
i
)
2
+
i
2
−
i
1
=0
s
{
i
1
}
+ 100
{
i
2
}
=
120
s
gives
−
200
{
i
1
}
+(
s
+ 200)
{
i
2
}
=0
so that
{
i
2
}
=
24,000
s
(
s
2
+ 200
s
+ 20,000)
=
6
5
1
s
−
6
5
s
+ 100
(
s
+ 100)
2
+ 100
2
−
6
5
100
(
s
+ 100)
2
+ 100
2
.
i
2
=
6
5
−
6
5
e
−
100
t
cos 100
t
−
6
5
e
−
100
t
sin 100
t
Then
i
1
=0
.
005
i
)
2
+
i
2
=
6
5
−
6
5
e
−
100
t
cos 100
t.
and
19.
Taking the Laplace transform of the system
2
i
)
1
+50
i
2
=60
0
.
005
i
)
2
+
i
2
−
i
1
=0
2
s
{
i
1
}
+50
{
i
2
}
=
60
s
gives
−
200
{
i
1
}
+(
s
+ 200)
{
i
2
}
=0
{
i
2
}
=
6,000
s
(
s
2
+ 200
s
+ 5,000)
so that
=
6
5
1
s
−
6
5
s
+ 100
(
s
+ 100)
2
−
(50
√
2)
2
−
6
√
2
5
50
√
2
(
s
+ 100)
2
−
(50
√
2)
2
.
204

205. Exercises 4.6
i
2
=
6
5
−
6
5
e
−
100
t
cosh 50
√
2
t
−
6
√
2
5
e
−
100
t
sinh 50
√
2
t
Then
i
1
=0
.
005
i
)
2
+
i
2
=
6
5
−
6
5
e
−
100
t
cosh 50
√
2
t
−
9
√
2
10
e
−
100
t
sinh 50
√
2
t.
and
20. (a)
Using Kirchoff’s first law we write
i
1
=
i
2
+
i
3
. Since
i
2
=
dq/dt
we have
i
1
−
i
3
=
dq/dt
. Using Kirchoff’s
second law and summing the voltage drops across the shorter loop gives
E
(
t
)=
iR
1
+
1
C
q,
(
1
)
so that
i
1
=
1
R
1
E
(
t
)
−
1
R
1
C
q.
Then
dq
dt
=
i
1
−
i
3
=
1
R
1
E
(
t
)
−
1
R
1
C
q
−
i
3
and
R
1
dq
dt
+
1
C
q
+
R
1
i
3
=
E
(
t
)
.
Summing the voltage drops across the longer loop gives
E
(
t
)=
i
1
R
1
+
L
di
3
dt
+
R
2
i
3
.
Combining this with (1) we obtain
i
1
R
1
+
L
di
3
dt
+
R
2
i
3
=
i
1
R
1
+
1
C
q
L
di
3
dt
+
R
2
i
3
−
1
C
q
=0
.
or
(b)
Using
L
=
R
1
=
R
2
=
C
=1,
E
(
t
)=50
e
−
t
(
t
−
1) = 50
e
−
1
e
−
(
t
−
1)
(
t
−
1),
q
(0) =
i
3
(0) = 0, and taking
the Laplace transform of the system we obtain
(
s
+1)
{
q
}
+
{
i
3
}
=
50
e
−
1
s
+1
e
−
s
(
s
+1)
{
i
3
}− {
q
}
=0
,
so that
{
q
}
=
50
e
−
1
e
−
s
(
s
+1)
2
+1
and
q
(
t
)=50
e
−
1
e
−
(
t
−
1)
sin(
t
−
1) (
t
−
1) = 50
e
−
t
sin(
t
−
1) (
t
−
1)
.
21. (a)
Taking the Laplace transform of the system
4
θ
))
1
+
θ
))
2
+8
θ
1
=0
θ
))
1
+
θ
))
2
+2
θ
2
=0
4
t
s
2
+2
e
{
θ
1
}
+
s
2
{
θ
2
}
=3
s
gives
s
2
{
θ
1
}
+
t
s
2
+2
e
{
θ
2
}
=0
so that
t
3
s
2
+4
et
s
2
+4
e
{
θ
2
}
=
−
3
s
3
or
{
θ
2
}
=
1
2
s
s
2
+4
/
3
−
3
2
s
s
2
+4
.
205

206. 3π 6π
t
θ1
−2
−1
1
2
3π 6π
t
θ2
−2
−1
1
2
-1-0.50.51
θ1
-2
-1
1
2
θ2
t
θ
1
θ
2
1 -0.2111 0.8263
2 -0.6585 0.6438
3 0.4830 -1.9145
4 -0.1325 0.1715
5 -0.4111 1.6951
6 0.8327 -0.8662
7 0.0458 -0.3186
8 -0.9639 0.9452
9 0.3534 -1.2741
10 0.4370 -0.3502
t=
0
t=
1
t=
2
t=
3
t=
4
t=
5
Exercises 4.6
Then
θ
2
=
1
2
cos
2
√
3
t
−
3
2
cos 2
t
and
θ
..
1
=
−
θ
..
2
−
2
θ
2
so that
θ
1
=
1
4
cos
2
√
3
t
+
3
4
cos 2
t.
(b)
Mass
m
2
has extreme displacements of greater magnitude. Mass
m
1
first passes through its equilibrium
position at about
t
=0
.
87, and mass
m
2
first passes through its equilibrium position at about
t
=0
.
66.
The motion of the pendulums is not periodic since cos 2
t/
√
3 has period
√
3
π
, cos 2
t
has period
π
, and the
ratio of these periods is
√
3 , which is not a rational number.
(c)
The Lissajous curve is plotted for 0
≤
t
≤
30.
(d)
206

207. t=
6
t=
7
t=
8
t=
9
t=
10
t=
0.75
Chapter 4 Review Exercises
(e)
Using a CAS to solve
θ
1
(
t
)=
θ
2
(
t
) we see that
θ
1
=
θ
2
(so that the double pendulum
is straight out) when
t
is about 0
.
75 seconds.
(f)
To make a movie of the pendulum it is necessaryto locate the mass in the plane as a function of time.
Suppose that the upper arm is attached to the origin and that the equilibrium position lies along the
negative
y
-axis. Then mass
m
1
is at (
x,
(
t
)
,y
1
(
t
)) and mass
m
2
is at (
x
2
(
t
)
,y
2
(
t
)), where
x
1
(
t
) = 16 sin
θ
1
(
t
) and
y
1
(
t
)=
−
16 cos
θ
1
(
t
)
and
x
2
(
t
)=
x
1
(
t
)+16sin
θ
2
(
t
) and
y
2
(
t
)=
y
1
(
t
)
−
16 cos
θ
2
(
t
)
.
A reasonable movie can be constructed byletting
t
range from 0 to 10 in increments of 0.1 seconds.
Chapter 4 Review Exercises
1.
{
f
(
t
)
}
=
(
1
0
te
−
st
dt
+
(
∞
1
(2
−
t
)
e
−
st
dt
=
1
s
2
−
2
s
2
e
−
s
2.
{
f
(
t
)
}
=
(
4
2
e
−
st
dt
=
1
s
t
e
−
2
s
−
e
−
4
s
e
3.
False; consider
f
(
t
)=
t
−
1
/
2
.
4.
False, since
f
(
t
)=(
e
t
)
10
=
e
10
t
.
5.
True, since lim
s
→∞
F
(
s
)=1

= 0. (See Theorem 4.5 in the text.)
6.
False; consider
f
(
t
) = 1 and
g
(
t
)=1.
7.
h
e
−
7
t
U
=
1
s
+7
8.
h
te
−
7
t
U
=
1
(
s
+7)
2
9.
{
sin 2
t
}
=
2
s
2
+4
10.
h
e
−
3
t
sin 2
t
U
=
2
(
s
+3)
2
+4
11.
{
t
sin 2
t
}
=
−
d
ds
!
2
s
2
+4
5
=
4
s
(
s
2
+4)
2
207

208. Chapter 4 Review Exercises
12.
{
sin 2
t
(
t
−
π
)
}
=
{
sin 2(
t
−
π
)(
t
−
π
)
}
=
2
s
2
+4
e
−
πs
13.
1
20
s
6
c
=
1
1
6
5!
s
6
c
=
1
6
t
5
14.
1
1
3
s
−
1
c
=
1
1
3
1
s
−
1
/
3
c
=
1
3
e
t/
3
15.
1
1
(
s
−
5)
3
c
=
1
1
2
2
(
s
−
5)
3
c
=
1
2
t
2
e
5
t
16.
1
1
s
2
−
5
c
=
1
−
1
√
5
1
s
+
√
5
+
1
√
5
1
s
−
√
5
c
=
−
1
√
5
e
−
√
5
t
+
1
√
5
e
√
5
t
17.
1
s
s
2
−
10
s
+29
c
=
1
s
−
5
(
s
−
5)
2
+2
2
+
5
2
2
(
s
−
5)
2
+2
2
c
=
e
5
t
cos 2
t
+
5
2
e
5
t
sin 2
t
18.
1
1
s
2
e
−
5
s
c
=(
t
−
5) (
t
−
5)
19.
1
s
+
π
s
2
+
π
2
e
−
s
c
=
1
s
s
2
+
π
2
e
−
s
+
π
s
2
+
π
2
e
−
s
c
= cos
π
(
t
−
1) (
t
−
1) + sin
π
(
t
−
1) (
t
−
1)
20.
1
1
L
2
s
2
+
n
2
π
2
c
=
1
L
2
L
nπ
1
nπ/L
s
2
+(
n
2
π
2
)
/L
2
c
=
1
Lnπ
sin
nπ
L
t
21.
h
e
−
5
t
U
exists for
s>
−
5.
22.
h
te
8
t
f
(
t
)
U
=
−
d
ds
F
(
s
−
8).
23.
{
e
at
f
(
t
−
k
)(
t
−
k
)
}
=
e
−
ks
{
e
a
(
t
+
k
)
f
(
t
)
}
=
e
−
ks
e
ak
{
e
at
f
(
t
)
}
=
e
−
k
(
s
−
a
)
F
(
s
−
a
)
24.
1
(
t
0
e
aτ
f
(
τ
)
dτ
c
=
1
s
{
e
at
f
(
t
)
}
=
F
(
s
−
a
)
s
, whereas
1
e
at
(
t
0
f
(
τ
)
dτ
c
=
1
(
t
0
f
(
τ
)
dτ
c
)
)
)
)
s
→
s
−
a
=
F
(
s
)
s
)
)
)
)
s
→
s
−
a
=
F
(
s
−
a
)
s
−
a
.
25.
f
(
t
)(
t
−
t
0
)
26.
f
(
t
)
−
f
(
t
)(
t
−
t
0
)
27.
f
(
t
−
t
0
)(
t
−
t
0
)
28.
f
(
t
)
−
f
(
t
)(
t
−
t
0
)+
f
(
t
)(
t
−
t
1
)
29.
f
(
t
)=
t
−
[(
t
−
1)+1] (
t
−
1) + (
t
−
1)
−
(
t
−
4) =
t
−
(
t
−
1) (
t
−
1)
−
(
t
−
4)
{
f
(
t
)
}
=
1
s
2
−
1
s
2
e
−
s
−
1
s
e
−
4
s
h
e
t
f
(
t
)
U
=
1
(
s
−
1)
2
−
1
(
s
−
1)
2
e
−
(
s
−
1)
−
1
s
−
1
e
−
4(
s
−
1)
30.
f
(
t
) = sin
t
(
t
−
π
)
−
sin
t
(
t
−
3
π
)=
−
sin(
t
−
π
)(
t
−
π
) + sin(
t
−
3
π
)(
t
−
3
π
)
{
f
(
t
)
}
=
−
1
s
2
+1
e
−
πs
+
1
s
2
+1
e
−
3
πs
h
e
t
f
(
t
)
U
=
−
1
(
s
−
1)
2
+1
e
−
π
(
s
−
1)
+
1
(
s
−
1)
2
+1
e
−
3
π
(
s
−
1)
208

209. Chapter 4 Review Exercises
31.
f
(
t
)=2
−
2(
t
−
2)+[(
t
−
2)+2] (
t
−
2)=2+(
t
−
2) (
t
−
2)
{
f
(
t
)
}
=
2
s
+
1
s
2
e
−
2
s
h
e
t
f
(
t
)
U
=
2
s
−
1
+
1
(
s
−
1)
2
e
−
2(
s
−
1)
32.
f
(
t
)=
t
−
t
(
t
−
1)+(2
−
t
)(
t
−
1)
−
(2
−
t
)(
t
−
2) =
t
−
2(
t
−
1) (
t
−
1)+(
t
−
2) (
t
−
2)
{
f
(
t
)
}
=
1
s
2
−
2
s
2
e
−
s
+
1
s
2
e
−
2
s
h
e
t
f
(
t
)
U
=
1
(
s
−
1)
2
−
2
(
s
−
1)
2
e
−
(
s
−
1)
+
1
(
s
−
1)
2
e
−
2(
s
−
1)
33.
Taking the Laplace transform of the differential equation we obtain
{
y
}
=
5
(
s
−
1)
2
+
1
2
2
(
s
−
1)
3
so that
y
=5
te
t
+
1
2
t
2
e
t
.
34.
Taking the Laplace transform of the differential equation we obtain
{
y
}
=
1
(
s
−
1)
2
(
s
2
−
8
s
+ 20)
=
6
169
1
s
−
1
+
1
13
1
(
s
−
1)
2
−
6
169
s
−
4
(
s
−
4)
2
+2
2
+
5
338
2
(
s
−
4)
2
+2
2
so that
y
=
6
169
e
t
+
1
13
te
t
−
6
169
e
4
t
cos 2
t
+
5
338
e
4
t
sin 2
t.
35.
Taking the Laplace transform of the given differential equation we obtain
{
y
}
=
s
3
+6
s
2
+1
s
2
(
s
+ 1)(
s
+5)
−
1
s
2
(
s
+ 1)(
s
+5)
e
−
2
s
−
2
s
(
s
+ 1)(
s
+5)
e
−
2
s
=
−
6
25
·
1
s
+
1
5
·
1
s
2
+
3
2
·
1
s
+1
−
13
50
·
1
s
+5
−
=
−
6
25
·
1
s
+
1
5
·
1
s
2
+
1
4
·
1
s
+1
−
1
100
·
1
s
+5
+
e
−
2
s
−
=
2
5
·
1
5
−
1
2
·
1
s
+1
+
1
10
·
1
s
+5
+
e
−
2
s
so that
y
=
−
6
25
+
1
5
t
2
+
3
2
e
−
t
−
13
50
e
−
5
t
−
4
25
(
t
−
2)
−
1
5
(
t
−
2)
2
(
t
−
2)
+
1
4
e
−
(
t
−
2)
(
t
−
2)
−
9
100
e
−
5(
t
−
2)
(
t
−
2)
.
36.
Taking the Laplace transform of the differential equation we obtain
{
y
}
=
s
3
+2
s
3
(
s
−
5)
−
2+2
s
+
s
2
s
3
(
s
−
5)
e
−
s
=
−
2
125
1
s
−
2
25
1
s
2
−
1
5
2
s
3
+
127
125
1
s
−
5
−
!
−
37
125
1
s
−
12
25
1
s
2
−
1
5
2
s
3
+
37
125
1
s
−
5
5
e
−
s
so that
y
=
−
2
125
−
2
25
t
−
1
5
t
2
+
127
125
e
5
t
−
!
−
37
125
−
12
25
(
t
−
1)
−
1
5
(
t
−
1)
2
+
37
125
e
5(
t
−
1)
5
(
t
−
1)
.
209

210. Chapter 4 Review Exercises
37.
Taking the Laplace transform of the integral equation we obtain
{
y
}
=
1
s
+
1
s
2
+
1
2
2
s
3
so that
y
(
t
)=1+
t
+
1
2
t
2
.
38.
Taking the Laplace transform of the integral equation we obtain
(
{
f
}
)
2
=6
·
6
s
4
or
{
f
}
=
±
6
·
1
s
2
so that
f
(
t
)=
±
6
t
.
39.
Taking the Laplace transform of the system gives
s
{
x
}
+
{
y
}
=
1
s
2
+1
4
{
x
}
+
s
{
y
}
=2
so that
{
x
}
=
s
2
−
2
s
+1
s
(
s
−
2)(
s
+2)
=
−
1
4
1
s
+
1
8
1
s
−
2
+
9
8
1
s
+2
.
Then
x
=
−
1
4
+
1
8
e
2
t
+
9
8
e
−
2
t
and
y
=
−
x
)
+
t
=
9
4
e
−
2
t
−
1
4
e
2
t
+
t.
40.
Taking the Laplace transform of the system gives
s
2
{
x
}
+
s
2
{
y
}
=
1
s
−
2
2
s
{
x
}
+
s
2
{
y
}
=
−
1
s
−
2
so that
{
x
}
=
2
s
(
s
−
2)
2
=
1
2
1
s
−
1
2
1
s
−
2
+
1
(
s
−
2)
2
and
{
y
}
=
−
s
−
2
s
2
(
s
−
2)
2
=
−
3
4
1
s
−
1
2
1
s
2
+
3
4
1
s
−
2
−
1
(
s
−
2)
2
.
Then
x
=
1
2
−
1
2
e
2
t
+
te
2
t
and
y
=
−
3
4
−
1
2
t
+
3
4
e
2
t
−
te
2
t
.
41.
The integral equation is
10
i
+2
(
t
0
i
(
τ
)
dτ
=2
t
2
+2
t.
Taking the Laplace transform we obtain
{
i
}
=
=
4
s
3
+
2
s
2
+
s
10
s
+2
=
s
+2
s
2
(5
s
+2)
=
−
9
s
+
2
s
2
+
45
5
s
+1
=
−
9
s
+
2
s
2
+
9
s
+1
/
5
.
Thus
i
(
t
)=
−
9+2
t
+9
e
−
t/
5
.
42.
The differential equation is
1
2
d
2
q
dt
2
+10
dq
dt
+ 100
q
=10
−
10 (
t
−
5)
.
210

211. Chapter 4 Review Exercises
Taking the Laplace transform we obtain
{
q
}
=
20
2(
s
2
+20
s
+ 200)
t
1
−
e
−
5
s
e
=
!
1
10
1
s
−
1
10
s
+10
(
s
+ 10)
2
+10
2
−
1
10
10
(
s
+ 10)
2
+10
2
5
t
1
−
e
−
5
s
e
so that
q
(
t
)=
1
10
−
1
10
e
−
10
t
cos 10
t
−
1
10
e
−
10
t
sin 10
t
−
!
1
10
−
1
10
e
−
10(
t
−
5)
cos 10(
t
−
5)
−
1
10
e
−
10(
t
−
5)
sin 10(
t
−
5)
5
(
t
−
5)
.
43.
Taking the Laplace transform of the given differential equation we obtain
{
y
}
=
2
w
0
EIL
=
L
48
·
4!
s
5
−
1
120
·
5!
s
6
+
1
120
·
5!
s
6
e
−
sL/
2
+
+
c
1
2
·
2!
s
3
+
c
2
6
·
3!
s
4
so that
y
=
2
w
0
EIL

L
48
x
4
−
1
120
x
5
+
1
120
=
x
−
L
2
+
5
=
x
−
L
2
+
+
c
1
2
x
2
+
c
2
6
x
3

where
y
))
(0) =
c
1
and
y
)))
(0) =
c
2
. Using
y
))
(
L
) = 0 and
y
)))
(
L
)=0wefind
c
1
=
w
0
L
2
/
24
EI, c
2
=
−
w
0
L/
4
EI.
Hence
y
=
w
0
12
EIL

−
1
5
x
5
+
L
2
x
4
−
L
2
2
x
3
+
L
3
4
x
2
+
1
5
=
x
−
L
2
+
5
=
x
−
L
2
+

.
44.
Taking the Laplace transform of the given differential equation we obtain
{
y
}
=
c
1
2
·
2
s
s
4
+4
+
c
2
4
·
4
s
4
+4
+
w
0
4
EI
·
4
s
4
+4
e
−
sπ/
2
so that
y
=
c
1
2
sin
x
sinh
x
+
c
2
4
(sin
x
cosh
x
−
cos
x
sinh
x
)
+
w
0
4
EI
b
sin
o
x
−
π
2
,
cosh
o
x
−
π
2
,
−
cos
o
x
−
π
2
,
sinh
o
x
−
π
2
,yo
x
−
π
2
,
where
y
))
(0) =
c
1
and
y
)))
(0) =
c
2
. Using
y
(
π
) = 0 and
y
)
(
π
) = 0 we find
c
1
=
w
0
EI
sinh
π
2
sinh
π
,c
2
=
−
w
0
EI
cosh
π
2
sinh
π
.
Hence
y
=
w
0
2
EI
sinh
π
2
sinh
π
sin
x
sinh
x
−
w
0
4
EI
cosh
π
2
sinh
π
(sin
x
cosh
x
−
cos
x
sinh
x
)
+
w
0
4
EI
b
sin
o
x
−
π
2
,
cosh
o
x
−
π
2
,
−
cos
o
x
−
π
2
,
sinh
o
x
−
π
2
,yo
x
−
π
2
,
.
211

212. 5
Series Solutions of Linear Equations
Exercises 5.1
1.
lim
n
→∞
k
k
k
k
a
n
+1
a
n
k
k
k
k
= lim
n
→∞
k
k
k
k
2
n
+1
x
n
+1
/
(
n
+1)
2
n
x
n
/n
k
k
k
k
= lim
n
→∞
2
n
n
+1
|
x
|
=2
|
x
|
The series is absolutely convergent for 2
|
x
|
<
1or
|
x
|
<
1
/
2. At
x
=
−
1
/
2, the series
∞
x
n
=1
(
−
1)
n
n
converges by
the alternating series test. At
x
=1
/
2, the series
∞
x
n
=1
1
n
is the harmonic series which diverges. Thus, the given
series converges on [
−
1
/
2
,
1
/
2).
2.
lim
n
→∞
k
k
k
k
a
n
+1
a
n
k
k
k
k
= lim
n
→∞
k
k
k
k
100
n
+1
(
x
+7)
n
+1
/
(
n
+ 1)!
100
n
(
x
+7)
n
/n
!
k
k
k
k
= lim
n
→∞
100
n
+1
|
x
+7
|
=0
The series is absolutely convergent on (
−∞
,
∞
).
3.
lim
k
→∞
k
k
k
k
a
k
+1
a
k
k
k
k
k
= lim
k
→∞
k
k
k
k
(
x
−
5)
k
+1
/
10
k
+1
(
x
−
5)
k
/
10
k
k
k
k
k
= lim
k
→∞
1
10
|
x
−
5
|
=
1
10
|
x
−
5
|
The series is absolutely convergent for
1
10
|
x
−
5
|
<
1,
|
x
−
5
|
<
10, or on (
−
5
,
15). At
x
=
−
5, the series
∞
x
k
=1
(
−
1)
k
(
−
10)
k
10
k
=
∞
x
k
=1
1 diverges by the
k
-th term test. At
x
= 15, the series
∞
x
k
=1
(
−
1)
k
10
k
10
k
=
∞
x
k
=1
(
−
1)
k
diverges by the
k
-th term test. Thus, the series converges on (
−
5
,
15).
4.
lim
k
→∞
k
k
k
k
a
k
+1
a
k
k
k
k
k
= lim
k
→∞
k
k
k
k
(
k
+ 1)!(
x
−
1)
k
+1
k
!(
x
−
1)
k
k
k
k
k
= lim
k
→∞
(
k
+1)
|
x
−
1
|
=
∞
,x
/
=1
The radius of convergence is 0 and the series converges only for
x
=1.
5.
sin
x
cos
x
=
/
x
−
x
3
6
+
x
5
120
−
x
7
5040
+
···
d/
1
−
x
2
2
+
x
4
24
−
x
6
720
+
···
d
=
x
−
2
x
3
3
+
2
x
5
15
−
4
x
7
315
+
···
6.
e
−
x
cos
x
=
/
1
−
x
+
x
2
2
−
x
3
6
+
x
4
24
−···
d/
1
−
x
2
2
+
x
4
24
−···
d
=1
−
x
+
x
3
3
−
x
4
6
+
···
7.
1
cos
x
=
1
1
−
x
2
2
+
x
4
4!
−
x
6
6!
+
···
=1+
x
2
2
+
5
x
4
4!
+
61
x
6
6!
+
···
Since cos(
π/
2) = cos(
−
π/
2) = 0, the series converges on (
−
π/
2
,π/
2).
8.
1
−
x
2+
x
=
1
2
−
3
4
x
+
3
8
x
2
−
3
16
x
3
+
···
Since the function is undefined at
x
=
−
2, the series converges on (
−
2
,
2).
212

213. Exercises 5.1
9.
∞
x
n
=1
2
nc
n
x
n
−
1
+
∞
x
n
=0
6
c
n
x
n
+1
=2
·
1
·
c
1
x
0
+
∞
x
n
=2
2
nc
n
x
n
−
1
r
νb
k
=
n
−
1
+
∞
x
n
=0
6
c
n
x
n
+1
r
νb
k
=
n
+1
=2
c
1
+
∞
x
k
=1
2(
k
+1)
c
k
+1
x
k
+
∞
x
k
=1
6
c
k
−
1
x
k
=2
c
1
+
∞
x
k
=1
[2(
k
+1)
c
k
+1
+6
c
k
−
1
]
x
k
10.
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
+2
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
+3
∞
x
n
=1
nc
n
x
n
=2
·
2
·
1
c
2
x
0
+2
·
3
·
2
c
3
x
/
+3
·
1
·
c
1
x
/
+
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
r
νb
k
=
n
+2
∞
x
n
=4
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+3
∞
x
n
=2
nc
n
x
n
r
νb
k
=
n
=4
c
2
+ (12
c
3
+ (12
c
3
+3
c
1
)
x
+
∞
x
n
=2
k
(
k
−
1)
c
k
x
k
+2
∞
x
n
=2
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+3
∞
x
n
=2
kc
k
x
k
=4
c
2
+(3
c
1
+12
c
3
)
x
+
∞
x
n
=2

[
k
(
k
−
1) + 3
k
]
c
k
+2(
k
+ 2)(
k
+1)
c
k
+2

x
k
=4
c
2
+(3
c
1
+12
c
3
)
x
+
∞
x
n
=2

k
(
k
+2)
c
k
+2(
k
+ 1)(
k
+2)
c
k
+2

x
k
11.
y
/
=
∞
x
n
=1
(
−
1)
n
+1
x
n
−
1
,y
//
=
∞
x
n
=2
(
−
1)
n
+1
(
n
−
1)
x
n
−
2
(
x
+1)
y
//
+
y
/
=(
x
+1)
∞
x
n
=2
(
−
1)
n
+1
(
n
−
1)
x
n
−
2
+
∞
x
n
=1
(
−
1)
n
+1
x
n
−
1
=
∞
x
n
=2
(
−
1)
n
+1
(
n
−
1)
x
n
−
1
+
∞
x
n
=2
(
−
1)
n
+1
(
n
−
1)
x
n
−
2
+
∞
x
n
=1
(
−
1)
n
+1
x
n
−
1
=
−
x
0
+
x
0
+
∞
x
n
=2
(
−
1)
n
+1
(
n
−
1)
x
n
−
1
r
νb
k
=
n
−
1
+
∞
x
n
=3
(
−
1)
n
+1
(
n
−
1)
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=2
(
−
1)
n
+1
x
n
−
1
r
νb
k
=
n
−
1
=
∞
x
k
=1
(
−
1)
k
+2
kx
k
+
∞
x
k
=1
(
−
1)
k
+3
(
k
+1)
x
k
+
∞
x
k
=1
(
−
1)
k
+2
x
k
=
∞
x
k
=1

(
−
1)
k
+2
k
−
(
−
1)
k
+2
k
−
(
−
1)
k
+2
+(
−
1)
k
+2

x
k
=0
12.
y
/
=
∞
x
n
=1
(
−
1)
n
2
n
2
2
n
(
n
!)
2
x
2
n
−
1
,y
//
=
∞
x
n
=1
(
−
1)
n
2
n
(2
n
−
1)
2
2
n
(
n
!)
2
x
2
n
−
2
xy
//
+
y
/
+
xy
=
∞
x
n
=1
(
−
1)
n
2
n
(2
n
−
1)
2
2
n
(
n
!)
2
x
2
n
−
1
r
νb
k
=
n
+
∞
x
n
=1
(
−
1)
n
2
n
2
2
n
(
n
!)
2
x
2
n
−
1
r
νb
k
=
n
+
∞
x
n
=0
(
−
1)
n
2
2
n
(
n
!)
2
x
2
n
+1
r
νb
k
=
n
+1
213

214. Exercises 5.1
=
∞
x
k
=1
a
(
−
1)
k
2
k
(2
k
−
1)
2
2
k
(
k
!)
2
+
(
−
1)
k
2
k
2
2
k
(
k
!)
2
+
(
−
1)
k
−
1
2
2
k
−
2
[(
k
−
1)!]
2
)
x
2
k
−
1
=
∞
x
k
=1
a
(
−
1)
k
(2
k
)
2
2
2
k
(
k
!)
2
−
(
−
1)
k
2
2
k
−
2
[(
k
−
1)!]
2
)
x
2
k
−
1
=
∞
x
k
=1
(
−
1)
k
a
(2
k
)
2
−
2
2
k
2
2
2
k
(
k
!)
2
)
x
2
k
−
1
=0
13.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
−
xy
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
∞
x
n
=0
c
n
x
n
+1
r
νb
k
=
n
+1
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
∞
x
k
=1
c
k
−
1
x
k
=2
c
2
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
−
c
k
−
1
]
x
k
=0
.
Thus
c
2
=0
(
k
+ 2)(
k
+1)
c
k
+2
−
c
k
−
1
=0
and
c
k
+2
=
1
(
k
+ 2)(
k
+1)
c
k
−
1
,k
=1
,
2
,
3
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
3
=
1
6
c
4
=
c
5
=0
c
6
=
1
180
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
3
=0
c
4
=
1
12
c
5
=
c
6
=0
c
7
=
1
504
and so on. Thus, two solutions are
y
1
=1+
1
6
x
3
+
1
180
x
6
+
···
and
y
2
=
x
+
1
12
x
4
+
1
504
x
7
+
···
.
14.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
+
x
2
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=0
c
n
x
n
+2
r
νb
k
=
n
+2
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=2
c
k
−
2
x
k
=2
c
2
+6
c
3
x
+
∞
x
k
=2
[(
k
+ 2)(
k
+1)
c
k
+2
+
c
k
−
2
]
x
k
=0
.
214

215. Exercises 5.1
Thus
c
2
=
c
3
=0
(
k
+ 2)(
k
+1)
c
k
+2
+
c
k
−
2
=0
and
c
k
+2
=
−
1
(
k
+ 2)(
k
+1)
c
k
−
2
,k
=2
,
3
,
4
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
4
=
−
1
12
c
5
=
c
6
=
c
7
=0
c
8
=
1
672
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
4
=0
c
5
=
−
1
20
c
6
=
c
7
=
c
8
=0
c
9
=
1
1440
and so on. Thus, two solutions are
y
1
=1
−
1
12
x
4
+
1
672
x
8
−···
and
y
2
=
x
−
1
20
x
5
+
1
1440
x
9
−···
.
15.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
−
2
xy
/
+
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
2
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
+
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
2
∞
x
k
=1
kc
k
x
k
+
∞
x
k
=0
c
k
x
k
=2
c
2
+
c
0
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
−
(2
k
−
1)
c
k
]
x
k
=0
.
Thus
2
c
2
+
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
−
(2
k
−
1)
c
k
=0
and
c
2
=
−
1
2
c
0
c
k
+2
=
2
k
−
1
(
k
+ 2)(
k
+1)
c
k
,k
=1
,
2
,
3
,... .
215

216. Exercises 5.1
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
1
2
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
−
1
8
c
6
=
−
7
336
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
1
6
c
5
=
1
24
c
7
=
1
112
and so on. Thus, two solutions are
y
1
=1
−
1
2
x
2
−
1
8
x
4
−
7
336
x
6
−···
and
y
2
=
x
+
1
6
x
3
+
1
24
x
5
+
1
112
x
7
+
···
.
16.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
−
xy
/
+2
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
+2
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
∞
x
k
=1
kc
k
x
k
+2
∞
x
k
=0
c
k
x
k
=2
c
2
+2
c
0
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
−
(
k
−
2)
c
k
]
x
k
=0
.
Thus
2
c
2
+2
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
−
(
k
−
2)
c
k
=0
and
c
2
=
−
c
0
c
k
+2
=
k
−
2
(
k
+ 2)(
k
+1)
c
k
,k
=1
,
2
,
3
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
1
c
3
=
c
5
=
c
7
=
···
=0
c
4
=0
c
6
=
c
8
=
c
10
=
···
=0
.
216

217. Exercises 5.1
For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
−
1
6
c
5
=
−
1
120
and so on. Thus, two solutions are
y
1
=1
−
x
2
and
y
2
=
x
−
1
6
x
3
−
1
120
x
5
−···
.
17.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
+
x
2
y
/
+
xy
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=1
nc
n
x
n
+1
r
νb
k
=
n
+1
+
∞
x
n
=0
c
n
x
n
+1
r
νb
k
=
n
+1
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=2
(
k
−
1)
c
k
−
1
x
k
+
∞
x
k
=1
c
k
−
1
x
k
=2
c
2
+(6
c
3
+
c
0
)
x
+
∞
x
k
=2
[(
k
+ 2)(
k
+1)
c
k
+2
+
kc
k
−
1
]
x
k
=0
.
Thus
c
2
=06
c
3
+
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
+
kc
k
−
1
=0
and
c
3
=
−
1
6
c
0
c
k
+2
=
−
k
(
k
+ 2)(
k
+1)
c
k
−
1
,k
=2
,
3
,
4
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
3
=
−
1
6
c
4
=
c
5
=0
c
6
=
1
45
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
3
=0
c
4
=
−
1
6
c
5
=
c
6
=0
c
7
=
5
252
and so on. Thus, two solutions are
y
1
=1
−
1
6
x
3
+
1
45
x
6
−···
and
y
2
=
x
−
1
6
x
4
+
5
232
x
7
−···
.
217

218. Exercises 5.1
18.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
+2
xy
/
+2
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+2
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
+2
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+2
∞
x
k
=1
kc
k
x
k
+2
∞
x
k
=0
c
k
x
k
=2
c
2
+2
c
0
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
+2(
k
+1)
c
k
]
x
k
=0
.
Thus
2
c
2
+2
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
+2(
k
+1)
c
k
=0
and
c
2
=
−
c
0
c
k
+2
=
−
2
k
+2
c
k
,k
=1
,
2
,
3
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
1
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
1
2
c
6
=
−
1
6
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
−
2
3
c
5
=
4
15
c
7
=
−
8
105
and so on. Thus, two solutions are
y
1
=1
−
x
2
+
1
2
x
4
−
1
6
x
6
+
···
and
y
2
=
x
−
2
3
x
3
+
4
15
x
5
−
8
105
x
7
+
···
.
19.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
(
x
−
1)
y
//
+
y
/
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
1
r
νb
k
=
n
−
1
−
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=1
nc
n
x
n
−
1
r
νb
k
=
n
−
1
=
∞
x
k
=1
(
k
+1)
kc
k
+1
x
k
−
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=0
(
k
+1)
c
k
+1
x
k
=
−
2
c
2
+
c
1
+
∞
x
k
=1
[(
k
+1)
kc
k
+1
−
(
k
+ 2)(
k
+1)
c
k
+2
+(
k
+1)
c
k
+1
]
x
k
=0
.
218

219. Exercises 5.1
Thus
−
2
c
2
+
c
1
=0
(
k
+1)
2
c
k
+1
−
(
k
+ 2)(
k
+1)
c
k
+2
=0
and
c
2
=
1
2
c
1
c
k
+2
=
k
+1
k
+2
c
k
+1
,k
=1
,
2
,
3
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
c
3
=
c
4
=
···
=0. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
1
2
,c
3
=
1
3
,c
4
=
1
4
,
and so on. Thus, two solutions are
y
1
= 1 and
y
2
=
x
+
1
2
x
2
+
1
3
x
3
+
1
4
x
4
+
···
.
20.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
(
x
+2)
y
//
+
xy
/
−
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
1
r
νb
k
=
n
−
1
+
∞
x
n
=2
2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
−
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=1
(
k
+1)
kc
k
+1
x
k
+
∞
x
k
=0
2(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=1
kc
k
x
k
−
∞
x
k
=0
c
k
x
k
=4
c
2
−
c
0
+
∞
x
k
=1

(
k
+1)
kc
k
+1
+2(
k
+ 2)(
k
+1)
c
k
+2
+(
k
−
1)
c
k

x
k
=0
.
Thus
4
c
2
−
c
0
=0
(
k
+1)
kc
k
+1
+2(
k
+ 2)(
k
+1)
c
k
+2
+(
k
−
1)
c
k
=0
,k
=1
,
2
,
3
,...
and
c
2
=
1
4
c
0
c
k
+2
=
−
(
k
+1)
kc
k
+1
+(
k
−
1)
c
k
2(
k
+ 2)(
k
+1)
,k
=1
,
2
,
3
,....
Choosing
c
0
= 1 and
c
1
= 0 we find
c
1
=0
,c
2
=
1
4
,c
3
=
−
1
24
,c
4
=0
,c
5
=
1
480
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=0
c
3
=0
c
4
=
c
5
=
c
6
=
···
=0
.
Thus, two solutions are
y
1
=
c
0
a
1+
1
4
x
2
−
1
24
x
3
+
1
480
x
5
+
···
)
and
y
2
=
c
1
x.
219

220. Exercises 5.1
21.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
−
(
x
+1)
y
/
−
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
−
∞
x
n
=1
nc
n
x
n
−
1
r
νb
k
=
n
−
1
−
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
∞
x
k
=1
kc
k
x
k
−
∞
x
k
=0
(
k
+1)
c
k
+1
x
k
−
∞
x
k
=0
c
k
x
k
=2
c
2
−
c
1
−
c
0
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
−
(
k
+1)
c
k
+1
−
(
k
+1)
c
k
]
x
k
=0
.
Thus
2
c
2
−
c
1
−
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
−
(
k
−
1)(
c
k
+1
+
c
k
)=0
and
c
2
=
c
1
+
c
0
2
c
k
+2
=
c
k
+1
+
c
k
k
+2
c
k
,k
=2
,
3
,
4
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
1
2
,c
3
=
1
6
,c
4
=
1
6
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
1
2
,c
3
=
1
2
,c
4
=
1
4
and so on. Thus, two solutions are
y
1
=1+
1
2
x
2
+
1
6
x
3
+
1
6
x
4
+
···
and
y
2
=
x
+
1
2
x
2
+
1
2
x
3
+
1
4
x
4
+
···
.
22.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have

x
2
+1

y
//
−
6
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
r
νb
k
=
n
+
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
6
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=2
k
(
k
−
1)
c
k
x
k
+
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
6
∞
x
k
=0
c
k
x
k
=2
c
2
−
6
c
0
+(6
c
3
−
6
c
1
)
x
+
∞
x
k
=2

k
2
−
k
−
6

c
k
+(
k
+ 2)(
k
+1)
c
k
+2

x
k
=0
.
Thus
2
c
2
−
6
c
0
=0
6
c
3
−
6
c
1
=0
(
k
−
3)(
k
+2)
c
k
+(
k
+ 2)(
k
+1)
c
k
+2
=0
220

221. Exercises 5.1
and
c
2
=3
c
0
c
3
=
c
1
c
k
+2
=
−
k
−
3
k
+1
c
k
,k
=2
,
3
,
4
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=3
c
3
=
c
5
=
c
7
=
···
=0
c
4
=1
c
6
=
−
1
5
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=1
c
5
=
c
7
=
c
9
=
···
=0
.
Thus, two solutions are
y
1
=1+3
x
2
+
x
4
−
1
5
x
6
+
···
and
y
2
=
x
+
x
3
.
23.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have

x
2
+2

y
//
+3
xy
/
−
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
r
νb
k
=
n
+2
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+3
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
−
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=2
k
(
k
−
1)
c
k
x
k
+2
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+3
∞
x
k
=1
kc
k
x
k
−
∞
x
k
=0
c
k
x
k
=(4
c
2
−
c
0
) + (12
c
3
+2
c
1
)
x
+
∞
x
k
=2

2(
k
+ 2)(
k
+1)
c
k
+2
+

k
2
+2
k
−
1

c
k

x
k
=0
.
Thus
4
c
2
−
c
0
=0
12
c
3
+2
c
1
=0
2(
k
+ 2)(
k
+1)
c
k
+2
+

k
2
+2
k
−
1

c
k
=0
and
c
2
=
1
4
c
0
c
3
=
−
1
6
c
1
c
k
+2
=
−
k
2
+2
k
−
1
2(
k
+ 2)(
k
+1)
c
k
,k
=2
,
3
,
4
,... .
221

222. Exercises 5.1
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
1
4
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
−
7
96
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
−
1
6
c
5
=
7
120
and so on. Thus, two solutions are
y
1
=1+
1
4
x
2
−
7
96
x
4
+
···
and
y
2
=
x
−
1
6
x
3
+
7
120
x
5
−···
.
24.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have

x
2
−
1

y
//
+
xy
/
−
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
r
νb
k
=
n
−
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
−
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=2
k
(
k
−
1)
c
k
x
k
−
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=1
kc
k
x
k
−
∞
x
k
=0
c
k
x
k
=(
−
2
c
2
−
c
0
)
−
6
c
3
x
+
∞
x
k
=2

−
(
k
+ 2)(
k
+1)
c
k
+2
+

k
2
−
1

c
k

x
k
=0
.
Thus
−
2
c
2
−
c
0
=0
−
6
c
3
=0
−
(
k
+ 2)(
k
+1)
c
k
+2
+(
k
−
1)(
k
+1)
c
k
=0
and
c
2
=
−
1
2
c
0
c
3
=0
c
k
+2
=
k
−
1
k
+2
c
k
,k
=2
,
3
,
4
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
1
2
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
−
1
8
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
c
5
=
c
7
=
···
=0
.
222

223. Exercises 5.1
Thus, two solutions are
y
1
=1
−
1
2
x
2
−
1
8
x
4
−···
and
y
2
=
x.
25.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
(
x
−
1)
y
//
−
xy
/
+
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
1
r
νb
k
=
n
−
1
−
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
+
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=1
(
k
+1)
kc
k
+1
x
k
−
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
∞
x
k
=1
kc
k
x
k
+
∞
x
k
=0
c
k
x
k
=
−
2
c
2
+
c
0
+
∞
x
k
=1
[
−
(
k
+ 2)(
k
+1)
c
k
+2
+(
k
+1)
kc
k
+1
−
(
k
−
1)
c
k
]
x
k
=0
.
Thus
−
2
c
2
+
c
0
=0
−
(
k
+ 2)(
k
+1)
c
k
+2
+(
k
−
1)
kc
k
+1
−
(
k
−
1)
c
k
=0
and
c
2
=
1
2
c
0
c
k
+2
=
kc
k
+1
k
+2
−
(
k
−
1)
c
k
(
k
+ 2)(
k
+1)
,k
=1
,
2
,
3
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
1
2
,c
3
=
1
6
,c
4
=0
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
3
=
c
4
=
···
= 0. Thus,
y
=
C
1
/
1+
1
2
x
2
+
1
6
x
3
+
···
d
+
C
2
x
and
y
/
=
C
1
/
x
+
1
2
x
2
+
···
d
+
C
2
.
The initial conditions imply
C
1
=
−
2 and
C
2
=6,so
y
=
−
2
/
1+
1
2
x
2
+
1
6
x
3
+
···
d
+6
x
=8
x
−
2
e
x
.
26.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
(
x
+1)
y
//
−
(2
−
x
)
y
/
+
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
1
r
νb
k
=
n
−
1
+
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
2
∞
x
n
=1
nc
n
x
n
−
1
r
νb
k
=
n
−
1
+
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
+
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=1
(
k
+1)
kc
k
+1
x
k
+
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
2
∞
x
k
=0
(
k
+1)
c
k
+1
x
k
+
∞
x
k
=1
kc
k
x
k
+
∞
x
k
=0
c
k
x
k
=2
c
2
−
2
c
1
+
c
0
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
−
(
k
+1)
c
k
+1
+(
k
+1)
c
k
]
x
k
=0
.
223

224. Exercises 5.1
Thus
2
c
2
−
2
c
1
+
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
−
(
k
+1)
c
k
+1
+(
k
+1)
c
k
=0
and
c
2
=
c
1
−
1
2
c
0
c
k
+2
=
1
k
+2
c
k
+1
−
1
k
+2
c
k
,k
=1
,
2
,
3
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
1
2
,c
3
=
−
1
6
,c
4
=
1
12
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=1
,c
3
=0
,c
4
=
−
1
4
and so on. Thus,
y
=
C
1
/
1
−
1
2
x
2
−
1
6
x
3
+
1
12
x
4
+
···
d
+
C
2
/
x
+
x
2
−
1
4
x
4
+
···
d
and
y
/
=
C
1
/
−
x
−
1
2
x
2
+
1
3
x
3
+
···
d
+
C
2

1+2
x
−
x
3
+
···

.
The initial conditions imply
C
1
= 2 and
C
2
=
−
1, so
y
=2
/
1
−
1
2
x
2
−
1
6
x
3
+
1
12
x
4
+
···
d
−
/
x
+
x
2
−
1
4
x
4
+
···
d
=2
−
x
−
2
x
2
−
1
3
x
3
+
5
12
x
4
+
···
.
27.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
−
2
xy
/
+8
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
2
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
+8
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
2
∞
x
k
=1
kc
k
x
k
+8
∞
x
k
=0
c
k
x
k
=2
c
2
+8
c
0
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
+(8
−
2
k
)
c
k
]
x
k
=0
.
Thus
2
c
2
+8
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
+(8
−
2
k
)
c
k
=0
and
c
2
=
−
4
c
0
c
k
+2
=
2
k
−
8
(
k
+ 2)(
k
+1)
c
k
,k
=1
,
2
,
3
,... .
224

225. Exercises 5.1
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
4
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
4
3
c
6
=
c
8
=
c
10
=
···
=0
.
For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
−
1
c
5
=
1
10
and so on. Thus,
y
=
C
1
/
1
−
4
x
2
+
4
3
x
4
d
+
C
2
/
x
−
x
3
+
1
10
x
5
+
···
d
and
y
/
=
C
1
/
−
8
x
+
16
3
x
3
d
+
C
2
/
1
−
3
x
2
+
1
2
x
4
+
···
d
.
The initial conditions imply
C
1
= 3 and
C
2
=0,so
y
=3
/
1
−
4
x
2
+
4
3
x
4
d
=3
−
12
x
2
+4
x
4
.
28.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
(
x
2
+1)
y
//
+2
xy
/
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
r
νb
k
=
n
+
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=1
2
nc
n
x
n
r
νb
k
=
n
=
∞
x
k
=2
k
(
k
−
1)
c
k
x
k
+
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=1
2
kc
k
x
k
=2
c
2
+(6
c
3
+2
c
1
)
x
+
∞
x
k
=2

k
(
k
+1)
c
k
+(
k
+ 2)(
k
+1)
c
k
+2

x
k
=0
.
Thus
2
c
2
=0
6
c
3
+2
c
1
=0
k
(
k
+1)
c
k
+(
k
+ 2)(
k
+1)
c
k
+2
=0
and
c
2
=0
c
3
=
−
1
3
c
1
c
k
+2
=
−
k
k
+2
c
k
,k
=2
,
3
,
4
,....
225

226. Exercises 5.1
Choosing
c
0
= 1 and
c
1
= 0 we find
c
3
=
c
4
=
c
5
=
···
=0. For
c
0
= 0 and
c
1
= 1 we obtain
c
3
=
−
1
3
c
4
=
c
6
=
c
8
=
···
=0
c
5
=
−
1
5
c
7
=
1
7
and so on. Thus
y
=
c
0
+
c
1
/
x
−
1
3
x
3
+
1
5
x
5
−
1
7
x
7
+
···
d
and
y
/
=
c
1

1
−
x
2
+
x
4
−
x
6
+
···

.
The initial conditions imply
c
0
= 0 and
c
1
=1,so
y
=
x
−
1
3
x
3
+
1
5
x
5
−
1
7
x
7
+
···
.
29.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
+ (sin
x
)
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
+
/
x
−
1
6
x
3
+
1
120
x
5
−···
d

c
0
+
c
1
x
+
c
2
x
2
+
···

=

2
c
2
+6
c
3
x
+12
c
4
x
2
+20
c
5
x
3
+
···

+
a
c
0
x
+
c
1
x
2
+
/
c
2
−
1
6
c
0
d
x
3
+
···
)
=2
c
2
+(6
c
3
+
c
0
)
x
+ (12
c
4
+
c
1
)
x
2
+
/
20
c
5
+
c
2
−
1
6
c
0
d
x
3
+
···
=0
.
Thus
2
c
2
=0
6
c
3
+
c
0
=0
12
c
4
+
c
1
=0
20
c
5
+
c
2
−
1
6
c
0
=0
c
2
=0
and
c
3
=
−
1
6
c
0
c
4
=
−
1
12
c
1
c
5
=
−
1
20
c
2
+
1
120
c
0
.
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=0
,c
3
=
−
1
6
,c
4
=0
,c
5
=
1
120
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=0
,c
3
=0
,c
4
=
−
1
12
,c
5
=0
226

227. Exercises 5.1
and so on. Thus, two solutions are
y
1
=1
−
1
6
x
3
+
1
120
x
5
+
···
and
y
2
=
x
−
1
12
x
4
+
···
.
30.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
+
e
x
y
/
−
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
+
/
1+
x
+
1
2
x
2
+
1
6
x
3
+
···
d

c
1
+2
c
2
x
+3
c
3
x
2
+4
c
4
x
3
+
···

−
∞
x
n
=0
c
n
x
n
=

2
c
2
+6
c
3
x
+12
c
4
x
2
+20
c
5
x
3
+
···

+
a
c
1
+(2
c
2
+
c
1
)
x
+
/
3
c
3
+2
c
2
+
1
2
c
1
d
x
2
+
···
)
−
[
c
0
+
c
1
x
+
c
2
x
2
+
···
]
=(2
c
2
+
c
1
−
c
0
)+(6
c
3
+2
c
2
)
x
+
/
12
c
4
+3
c
3
+
c
2
+
1
2
c
1
d
x
2
+
···
=0
.
Thus
2
c
2
+
c
1
−
c
0
=0
6
c
3
+2
c
2
=0
12
c
4
+3
c
3
+
c
2
+
1
2
c
1
=0
and
c
2
=
1
2
c
0
−
1
2
c
1
c
3
=
−
1
3
c
2
c
4
=
−
1
4
c
3
+
1
12
c
2
−
1
24
c
1
.
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
1
2
,c
3
=
−
1
6
,c
4
=0
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
−
1
2
,c
3
=
1
6
,c
4
=
−
1
24
and so on. Thus, two solutions are
y
1
=1+
1
2
x
2
−
1
6
x
3
+
···
and
y
2
=
x
−
1
2
x
2
+
1
6
x
3
−
1
24
x
4
+
···
.
31.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the first differential equation leads to
y
//
−
xy
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
∞
x
n
=0
c
n
x
n
+1
r
νb
k
=
n
+1
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
∞
x
k
=1
c
k
−
1
x
k
=2
c
2
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
−
c
k
−
1
]
x
k
=1
.
227

228. Exercises 5.1
Thus
2
c
2
=1
(
k
+ 2)(
k
+1)
c
k
+2
−
c
k
−
1
=0
and
c
2
=
1
2
c
k
+2
=
c
k
−
1
(
k
+ 2)(
k
+1)
,k
=1
,
2
,
3
,... .
Let
c
0
and
c
1
be arbitrary and iterate to find
c
2
=
1
2
c
3
=
1
6
c
0
c
4
=
1
12
c
1
c
5
=
1
20
c
2
=
1
40
and so on. The solution is
y
=
c
0
+
c
1
x
+
1
2
x
2
+
1
6
c
0
x
3
+
1
12
c
1
x
4
+
1
40
c
5
+
···
=
c
0
/
1+
1
6
x
3
+
···
d
+
c
1
/
x
+
1
12
x
4
+
···
d
+
1
2
x
2
+
1
40
x
5
+
···
.
Substituting
y
=
S
∞
n
=0
c
n
x
n
into the second differential equation leads to
y
//
−
4
xy
/
−
4
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
∞
x
n
=1
4
nc
n
x
n
r
νb
k
=
n
−
∞
x
n
=0
4
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
∞
x
k
=1
4
kc
k
x
k
−
∞
x
k
=0
4
c
k
x
k
=2
c
2
−
4
c
0
+
∞
x
k
=1

(
k
+ 2)(
k
+1)
c
k
+2
−
4(
k
+1)
c
k

x
k
=
e
x
=1+
∞
x
k
=1
1
k
!
x
k
.
Thus
2
c
2
−
4
c
0
=1
(
k
+ 2)(
k
+1)
c
k
+2
−
4(
k
+1)
c
k
=
1
k
!
and
c
2
=
1
2
+2
c
0
c
k
+2
=
1
(
k
+ 2)!
+
4
k
+2
c
k
,k
=1
,
2
,
3
,....
228

229. Exercises 5.1
Let
c
0
and
c
1
be arbitrary and iterate to find
c
2
=
1
2
+2
c
0
c
3
=
1
3!
+
4
3
c
1
=
1
3!
+
4
3
c
1
c
4
=
1
4!
+
4
4
c
2
=
1
4!
+
1
2
+2
c
0
=
13
4!
+2
c
0
c
5
=
1
5!
+
4
5
c
3
=
1
5!
+
4
5
·
3!
+
16
15
c
1
=
17
5!
+
16
15
c
1
c
6
=
1
6!
+
4
6
c
4
=
1
6!
+
4
·
13
6
·
4!
+
8
6
c
0
=
261
6!
+
4
3
c
0
c
7
=
1
7!
+
4
7
c
5
=
1
7!
+
4
·
17
7
·
5!
+
64
105
c
1
=
409
7!
+
64
105
c
1
and so on. The solution is
y
=
c
0
+
c
1
x
+
/
1
2
+2
c
0
d
x
2
+
/
1
3!
+
4
3
c
1
d
x
3
−
/
13
4!
+2
c
0
d
x
4
+
/
17
5!
+
16
15
c
1
d
x
5
+
/
261
6!
+
4
3
c
0
d
x
6
+
/
409
7!
+
64
105
c
1
d
x
7
+
···
=
c
0
a
1+2
x
2
+2
x
4
+
4
3
x
6
+
···
)
+
c
1
a
x
+
4
3
x
3
+
16
15
x
5
+
64
105
x
7
+
···
)
+
1
2
x
2
+
1
3!
x
3
+
13
4!
x
4
+
17
5!
x
5
+
261
6!
x
6
+
409
7!
x
7
+
···
.
32.
We identify
P
(
x
) = 0 and
Q
(
x
) = sin
x/x
. The Taylor series representation for sin
x/x
is 1
−
x
2
/
3! +
x
4
/
5!
−···
,
for
|
x
|
<
∞
. Thus,
Q
(
x
) is analytic at
x
= 0 and
x
= 0 is an ordinary point of the differential equation.
33. (a)
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
+
xy
/
+
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
+
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=1
kc
k
x
k
+
∞
x
k
=0
c
k
x
k
=(2
c
2
+
c
0
)+
∞
x
k
=1

(
k
+ 2)(
k
+1)
c
k
+2
+(
k
+1)
c
k

x
k
=0
.
Thus
2
c
2
+
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
+(
k
+1)
c
k
=0
and
c
2
=
−
1
2
c
0
c
k
+2
=
−
1
k
+2
c
k
.
Choosing
c
0
= 1 and
c
1
= 0 we find
229

230. -4-2 24
x
-4
-2
2
4
y
N=2
-4-2 24
x
-4
-2
2
4
y
N=4
-4-2 24
x
-4
-2
2
4
y
N=6
-4-2 24
x
-4
-2
2
4
y
N=8
-4-2 24
x
-4
-2
2
4
y
N=10
-4-2 24
x
-4
-2
2
4
y
N=2
-4-2 24
x
-4
-2
2
4
y
N=4
-4-2 24
x
-4
-2
2
4
y
N=6
-4-2 24
x
-4
-2
2
4
y
N=8
-4-2 24
x
-4
-2
2
4
y
N=10
-4-2 24
x
-4
-2
2
4
y1
-4-2 24
x
-4
-2
2
4
y2
Exercises 5.1
c
2
=
−
1
2
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
−
1
4
c
−
1
2
d
=
1
2
2
·
2
c
6
=
−
1
6
c
1
2
2
·
2
d
=
−
1
2
3
·
3!
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
−
1
3
=
−
2
3!
c
5
=
−
1
5
c
−
1
3
d
=
1
5
·
3
=
4
·
2
5!
c
7
=
−
1
7
c
4
·
2
5!
d
=
−
6
·
4
·
2
7!
and so on. Thus, two solutions are
y
1
=
∞
x
k
=0
(
−
1)
k
2
k
·
k
!
x
2
k
and
y
2
=
∞
x
k
=0
(
−
1)
k
2
k
k
!
(2
k
+ 1)!
x
2
k
+1
.
(b)
For
y
1
,
S
3
=
S
2
and
S
5
=
S
4
, so we plot
S
2
,
S
4
,
S
6
,
S
8
, and
S
10
.
For
y
2
,
S
3
=
S
4
and
S
5
=
S
6
, so we plot
S
2
,
S
4
,
S
6
,
S
8
, and
S
10
.
(c)
The graphs of
y
1
and
y
2
obtained from a numerical
solver are shown. We see that the partial sum repre-
sentations indicate the even and odd natures of the so-
lution, but don’t really give a very accurate represen-
tation of the true solution. Increasing
N
to about 20
gives a much more accurate representation on [
−
4
,
4].
230

231. -6-4-2 246
x
-4
-2
2
4
y
-6-4-2 246
x
-4
-2
2
4
y
-6-4-2 246
x
-4
-2
2
4
y
-6-4-2 246
x
-4
-2
2
4
y
-6-4-2 246
x
-4
-2
2
4
y
-6-4-2 246
x
-4
-2
2
4
y
-6-4-2 246
x
-6
-4
-2
2
4
6
y
Exercises 5.1
34. (a)
We have
y
22
+ (cos
x
)
y
=2
c
2
+6
c
3
x
+12
c
4
x
2
+20
c
5
x
3
+30
c
6
x
4
+42
c
7
x
5
+
c
1
−
x
2
2!
+
x
4
4!
−
x
6
6!
+
···
d
(
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
+
···
)
=(2
c
2
+
c
0
)+(6
c
3
+
c
1
)
x
+
c
12
c
4
+
c
2
−
1
2
c
0
d
x
2
+
c
20
c
5
+
c
3
−
1
2
c
1
d
x
3
+
c
30
c
6
+
c
4
+
1
24
c
0
−
1
2
c
2
d
x
4
+
c
42
c
7
+
c
5
+
1
24
c
1
−
1
2
c
3
d
x
5
+
···
.
Then
30
c
6
+
c
4
+
1
24
c
0
−
1
2
c
2
= 0 and 42
c
7
+
c
5
+
1
24
c
1
−
1
2
c
3
=0
,
which gives
c
6
=
−
c
0
/
80 and
c
7
=
−
19
c
1
/
5040. Thus
y
1
(
x
)=1
−
1
2
x
2
+
1
12
x
4
−
1
80
x
6
+
···
and
y
2
(
x
)=
x
−
1
6
x
3
+
1
30
x
5
−
19
5040
x
7
+
···
.
(b)
From part (a) the general solution of the differential equation is
y
=
c
1
y
1
+
c
2
y
2
. Then
y
(0) =
c
1
+
c
2
·
0=
c
1
and
y
2
(0) =
c
1
·
0+
c
2
=
c
2
, so the solution of the initial-value problem is
y
=
y
1
+
y
2
=1+
x
−
1
2
x
2
−
1
6
x
3
+
1
12
x
4
+
1
30
x
5
−
1
80
x
6
−
19
5040
x
7
+
···
.
(c)
(d)
231

232. Exercises 5.1
35.
From
e
x
=
b
∞
k
=0
x
k
/k
! we see that
e
−
x
2
/
2
=
b
∞
k
=0
(
−
x
2
/
2)
k
/k
!=
b
∞
k
=0
(
−
1)
k
x
2
k
/
2
k
k
! . From (5) of Section
3.2 we have
y
2
=
y
1
>
e
−
i
xdx
y
2
1
dx
=
e
−
x
2
/
2
>
e
−
x
2
/
2
(
e
−
x
2
/
2
)
2
dx
=
e
−
x
2
/
2
>
e
−
x
2
/
2
e
−
x
2
dx
=
e
−
x
2
/
2
>
e
x
2
/
2
dx
=
∞
x
k
=0
(
−
1)
k
2
k
k
!
x
2
k
>
∞
x
k
=0
1
2
k
k
!
x
2
k
dx
=
p
∞
x
k
=0
(
−
1)
k
2
k
k
!
x
2
k
qp
∞
x
k
=0
>
1
2
k
k
!
x
2
k
dx
q
=
p
∞
x
k
=0
(
−
1)
k
2
k
k
!
x
2
k
qp
∞
x
k
=0
1
(2
k
+ 1)2
k
k
!
x
2
k
+1
q
=
c
1
−
1
2
x
2
+
1
2
2
·
2
x
4
−
1
2
3
·
3!
x
6
+
···
dc
x
+
1
3
·
2
x
3
+
1
5
·
2
2
·
2
x
5
+
1
7
·
2
3
·
3!
x
7
+
···
d
=
x
−
2
3!
x
3
+
4
·
2
5!
x
5
−
6
·
4
·
2
7!
x
7
+
···
=
∞
x
k
=0
(
−
1)
k
2
k
k
!
(2
k
+ 1)!
x
2
k
+1
.
36.
If
x>
0 and
y>
0, then
y
//
=
−
xy <
0 and the graph of a solution curve is concave down. Thus, whatever
portion of a solution curve lies in the first quadrant is concave down. When
x>
0 and
y<
0,
y
//
=
−
xy >
0,
so whatever portion of a solution curve lies in the fourth quadrant is concave up.
Exercises 5.2
1.
Irregular singular point:
x
=0
2.
Regular singular points:
x
=0,
−
3
3.
Irregular singular point:
x
= 3; regular singular point:
x
=
−
3
4.
Irregular singular point:
x
= 1; regular singular point:
x
=0
5.
Regular singular points:
x
=0,
±
2
i
6.
Irregular singular point:
x
= 5; regular singular point:
x
=0
7.
Regular singular points:
x
=
−
3, 2
8.
Regular singular points:
x
=0,
±
i
9.
Irregular singular point:
x
= 0; regular singular points:
x
=2,
±
5
10.
Irregular singular point:
x
=
−
1; regular singular points:
x
=0,3
11.
Writing the differential equation in the form
y
//
+
5
x
−
1
y
/
+
x
x
+1
y
=0
we see that
x
0
= 1 and
x
0
=
−
1 are regular singular points. For
x
0
= 1 the differential equation can be put in
the form
(
x
−
1)
2
y
//
+5(
x
−
1)
y
/
+
x
(
x
−
1)
2
x
+1
y
=0
.
In this case
p
(
x
) = 5 and
q
(
x
)=
x
(
x
−
1)
2
/
(
x
+ 1). For
x
0
=
−
1 the differential equation can be put in the form
(
x
+1)
2
y
//
+5(
x
+1)
x
+1
x
−
1
y
/
+
x
(
x
+1)
y
=0
.
In this case
p
(
x
)=(
x
+1)
/
(
x
−
1) and
q
(
x
)=
x
(
x
+ 1).
232

233. Exercises 5.2
12.
Writing the differential equation in the form
y
//
+
x
+3
x
y
/
+7
xy
=0
we see that
x
0
= 0 is a regular singular point. Multiplying by
x
2
, the differential equation can be put in the
form
x
2
y
//
+
x
(
x
+3)
y
/
+7
x
3
y
=0
.
We identify
p
(
x
)=
x
+ 3 and
q
(
x
)=7
x
3
.
13.
We identify
P
(
x
)=5
/
3
x
+ 1 and
Q
(
x
)=
−
1
/
3
x
2
, so that
p
(
x
)=
xP
(
x
)=
5
3
+
x
and
q
(
x
)=
x
2
Q
(
x
)=
−
1
3
.
Then
a
0
=
5
3
,
b
0
=
−
1
3
, and the indicial equation is
r
(
r
−
1) +
5
3
r
−
1
3
=
r
2
+
2
3
r
−
1
3
=
1
3
(3
r
2
+2
r
−
1) =
1
3
(3
r
−
1)(
r
+1)=0
.
The indicial roots are
1
3
and
−
1. Since these do not differ by an integer we expect to find two series solutions
using the method of Frobenius.
14.
We identify
P
(
x
)=1
/x
and
Q
(
x
)=10
/x
, so that
p
(
x
)=
xP
(
x
) = 1 and
q
(
x
)=
x
2
Q
(
x
)=10
x
. Then
a
0
=1,
b
0
= 0, and the indicial equation is
r
(
r
−
1) +
r
=
r
2
=0
.
The indicial roots are 0 and 0. Since these are equal, we expect the method of Frobenius to yield a single series
solution.
15.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
2
xy
//
−
y
/
+2
y
=

2
r
2
−
3
r

c
0
x
r
−
1
+
∞
x
k
=1
[2(
k
+
r
−
1)(
k
+
r
)
c
k
−
(
k
+
r
)
c
k
+2
c
k
−
1
]
x
k
+
r
−
1
=0
,
which implies
2
r
2
−
3
r
=
r
(2
r
−
3)=0
and
(
k
+
r
)(2
k
+2
r
−
3)
c
k
+2
c
k
−
1
=0
.
The indicial roots are
r
= 0 and
r
=3
/
2. For
r
= 0 the recurrence relation is
c
k
=
−
2
c
k
−
1
k
(2
k
−
3)
,k
=1
,
2
,
3
,...,
and
c
1
=2
c
0
,c
2
=
−
2
c
0
,c
3
=
4
9
c
0
.
For
r
=3
/
2 the recurrence relation is
c
k
=
−
2
c
k
−
1
(2
k
+3)
k
,k
=1
,
2
,
3
,...,
and
c
1
=
−
2
5
c
0
,c
2
=
2
35
c
0
,c
3
=
−
4
945
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
/
1+2
x
−
2
x
2
+
4
9
x
3
+
···
d
+
C
2
x
3
/
2
/
1
−
2
5
x
+
2
35
x
2
−
4
945
x
3
+
···
d
.
233

234. Exercises 5.2
16.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
2
xy
//
+5
y
/
+
xy
=

2
r
2
+3
r

c
0
x
r
−
1
+

2
r
2
+7
r
+5

c
1
x
r
+
∞
x
k
=2
[2(
k
+
r
)(
k
+
r
−
1)
c
k
+5(
k
+
r
)
c
k
+
c
k
−
2
]
x
k
+
r
−
1
=0
,
which implies
2
r
2
+3
r
=
r
(2
r
+3)=0
,

2
r
2
+7
r
+5

c
1
=0
,
and
(
k
+
r
)(2
k
+2
r
+3)
c
k
+
c
k
−
2
=0
.
The indicial roots are
r
=
−
3
/
2 and
r
=0,so
c
1
=0. For
r
=
−
3
/
2 the recurrence relation is
c
k
=
−
c
k
−
2
(2
k
−
3)
k
,k
=2
,
3
,
4
,...,
and
c
2
=
−
1
2
c
0
,c
3
=0
,c
4
=
1
40
c
0
.
For
r
= 0 the recurrence relation is
c
k
=
−
c
k
−
2
k
(2
k
+3)
,k
=2
,
3
,
4
,...,
and
c
2
=
−
1
14
c
0
,c
3
=0
,c
4
=
1
616
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
x
−
3
/
2
/
1
−
1
2
x
2
+
1
40
x
4
+
···
d
+
C
2
/
1
−
1
14
x
2
+
1
616
x
4
+
···
d
.
17.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
4
xy
//
+
1
2
y
/
+
y
=
/
4
r
2
−
7
2
r
d
c
0
x
r
−
1
+
∞
x
k
=1
a
4(
k
+
r
)(
k
+
r
−
1)
c
k
+
1
2
(
k
+
r
)
c
k
+
c
k
−
1
)
x
k
+
r
−
1
=0
,
which implies
4
r
2
−
7
2
r
=
r
/
4
r
−
7
2
d
=0
and
1
2
(
k
+
r
)(8
k
+8
r
−
7)
c
k
+
c
k
−
1
=0
.
The indicial roots are
r
= 0 and
r
=7
/
8. For
r
= 0 the recurrence relation is
c
k
=
−
2
c
k
−
1
k
(8
k
−
7)
,k
=1
,
2
,
3
,...,
and
c
1
=
−
2
c
0
,c
2
=
2
9
c
0
,c
3
=
−
4
459
c
0
.
234

235. Exercises 5.2
For
r
=7
/
8 the recurrence relation is
c
k
=
−
2
c
k
−
1
(8
k
+7)
k
,k
=1
,
2
,
3
,...,
and
c
1
=
−
2
15
c
0
,c
2
=
2
345
c
0
,c
3
=
−
4
32,085
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
/
1
−
2
x
+
2
9
x
2
−
4
459
x
3
+
···
d
+
C
2
x
7
/
8
/
1
−
2
15
x
+
2
345
x
2
−
4
32,085
x
3
+
···
d
.
18.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
2
x
2
y
//
−
xy
/
+

x
2
+1

y
=

2
r
2
−
3
r
+1

c
0
x
r
+

2
r
2
+
r

c
1
x
r
+1
+
∞
x
k
=2
[2(
k
+
r
)(
k
+
r
−
1)
c
k
−
(
k
+
r
)
c
k
+
c
k
+
c
k
−
2
]
x
k
+
r
=0
,
which implies
2
r
2
−
3
r
+1=(2
r
−
1)(
r
−
1) = 0
,

2
r
2
+
r

c
1
=0
,
and
[(
k
+
r
)(2
k
+2
r
−
3)+1]
c
k
+
c
k
−
2
=0
.
The indicial roots are
r
=1
/
2 and
r
=1,so
c
1
=0. For
r
=1
/
2 the recurrence relation is
c
k
=
−
c
k
−
2
k
(2
k
−
1)
,k
=2
,
3
,
4
,...,
and
c
2
=
−
1
6
c
0
,c
3
=0
,c
4
=
1
168
c
0
.
For
r
= 1 the recurrence relation is
c
k
=
−
c
k
−
2
k
(2
k
+1)
,k
=2
,
3
,
4
,...,
and
c
2
=
−
1
10
c
0
,c
3
=0
,c
4
=
1
360
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
x
1
/
2
/
1
−
1
6
x
2
+
1
168
x
4
+
···
d
+
C
2
x
/
1
−
1
10
x
2
+
1
360
x
4
+
···
d
.
19.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
3
xy
//
+(2
−
x
)
y
/
−
y
=

3
r
2
−
r

c
0
x
r
−
1
+
∞
x
k
=1
[3(
k
+
r
−
1)(
k
+
r
)
c
k
+2(
k
+
r
)
c
k
−
(
k
+
r
)
c
k
−
1
]
x
k
+
r
−
1
=0
,
235

236. Exercises 5.2
which implies
3
r
2
−
r
=
r
(3
r
−
1)=0
and
(
k
+
r
)(3
k
+3
r
−
1)
c
k
−
(
k
+
r
)
c
k
−
1
=0
.
The indicial roots are
r
= 0 and
r
=1
/
3. For
r
= 0 the recurrence relation is
c
k
=
c
k
−
1
(3
k
−
1)
,k
=1
,
2
,
3
,...,
and
c
1
=
1
2
c
0
,c
2
=
1
10
c
0
,c
3
=
1
80
c
0
.
For
r
=1
/
3 the recurrence relation is
c
k
=
c
k
−
1
3
k
,k
=1
,
2
,
3
,...,
and
c
1
=
1
3
c
0
,c
2
=
1
18
c
0
,c
3
=
1
162
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
/
1+
1
2
x
+
1
10
x
2
+
1
80
x
3
+
···
d
+
C
2
x
1
/
3
/
1+
1
3
x
+
1
18
x
2
+
1
162
x
3
+
···
d
.
20.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
x
2
y
//
−
/
x
−
2
9
d
y
=
/
r
2
−
r
+
2
9
d
c
0
x
r
+
∞
x
k
=1
a
(
k
+
r
)(
k
+
r
−
1)
c
k
+
2
9
c
k
−
c
k
−
1
)
x
k
+
r
=0
,
which implies
r
2
−
r
+
2
9
=
/
r
−
2
3
d/
r
−
1
3
d
=0
and
a
(
k
+
r
)(
k
+
r
−
1) +
2
9
)
c
k
−
c
k
−
1
=0
.
The indicial roots are
r
=2
/
3 and
r
=1
/
3. For
r
=2
/
3 the recurrence relation is
c
k
=
3
c
k
−
1
3
k
2
+
k
,k
=1
,
2
,
3
,...,
and
c
1
=
3
4
c
0
,c
2
=
9
56
c
0
,c
3
=
9
560
c
0
.
For
r
=1
/
3 the recurrence relation is
c
k
=
3
c
k
−
1
3
k
2
−
k
,k
=1
,
2
,
3
,...,
and
c
1
=
3
2
c
0
,c
2
=
9
20
c
0
,c
3
=
9
160
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
x
2
/
3
/
1+
3
4
x
+
9
56
x
2
+
9
560
x
3
+
···
d
+
C
2
x
1
/
3
/
1+
3
2
x
+
9
20
x
2
+
9
160
x
3
+
···
d
.
236

237. Exercises 5.2
21.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
2
xy
//
−
(3+2
x
)
y
/
+
y
=

2
r
2
−
5
r

c
0
x
r
−
1
+
∞
x
k
=1
[2(
k
+
r
)(
k
+
r
−
1)
c
k
−
3(
k
+
r
)
c
k
−
2(
k
+
r
−
1)
c
k
−
1
+
c
k
−
1
]
x
k
+
r
−
1
=0
,
which implies
2
r
2
−
5
r
=
r
(2
r
−
5)=0
and
(
k
+
r
)(2
k
+2
r
−
5)
c
k
−
(2
k
+2
r
−
3)
c
k
−
1
=0
.
The indicial roots are
r
= 0 and
r
=5
/
2. For
r
= 0 the recurrence relation is
c
k
=
(2
k
−
3)
c
k
−
1
k
(2
k
−
5)
,k
=1
,
2
,
3
,...,
and
c
1
=
1
3
c
0
,c
2
=
−
1
6
c
0
,c
3
=
−
1
6
c
0
.
For
r
=5
/
2 the recurrence relation is
c
k
=
2(
k
+1)
c
k
−
1
k
(2
k
+5)
,k
=1
,
2
,
3
,...,
and
c
1
=
4
7
c
0
,c
2
=
4
21
c
0
,c
3
=
32
693
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
/
1+
1
3
x
−
1
6
x
2
−
1
6
x
3
+
···
d
+
C
2
x
5
/
2
/
1+
4
7
x
+
4
21
x
2
+
32
693
x
3
+
···
d
.
22.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
x
2
y
//
+
xy
/
+
/
x
2
−
4
9
d
y
=
/
r
2
−
4
9
d
c
0
x
r
+
/
r
2
+2
r
+
5
9
d
c
1
x
r
+1
+
∞
x
k
=2
a
(
k
+
r
)(
k
+
r
−
1)
c
k
+(
k
+
r
)
c
k
−
4
9
c
k
+
c
k
−
2
)
x
k
+
r
=0
,
which implies
r
2
−
4
9
=
/
r
+
2
3
d/
r
−
2
3
d
=0
,
/
r
2
+2
r
+
5
9
d
c
1
=0
,
and
a
(
k
+
r
)
2
−
4
9
)
c
k
+
c
k
−
2
=0
.
The indicial roots are
r
=
−
2
/
3 and
r
=2
/
3, so
c
1
=0. For
r
=
−
2
/
3 the recurrence relation is
c
k
=
−
9
c
k
−
2
3
k
(3
k
−
4)
,k
=2
,
3
,
4
,...,
237

238. Exercises 5.2
and
c
2
=
−
3
4
c
0
,c
3
=0
,c
4
=
9
128
c
0
.
For
r
=2
/
3 the recurrence relation is
c
k
=
−
9
c
k
−
2
3
k
(3
k
+4)
,k
=2
,
3
,
4
,...,
and
c
2
=
−
3
20
c
0
,c
3
=0
,c
4
=
9
1,280
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
x
−
2
/
3
/
1
−
3
4
x
2
+
9
128
x
4
+
···
d
+
C
2
x
2
/
3
/
1
−
3
20
x
2
+
9
1,280
x
4
+
···
d
.
23.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
9
x
2
y
//
+9
x
2
y
/
+2
y
=

9
r
2
−
9
r
+2

c
0
x
r
+
∞
x
k
=1
[9(
k
+
r
)(
k
+
r
−
1)
c
k
+2
c
k
+9(
k
+
r
−
1)
c
k
−
1
]
x
k
+
r
=0
,
which implies
9
r
2
−
9
r
+2=(3
r
−
1)(3
r
−
2)=0
and
[9(
k
+
r
)(
k
+
r
−
1)+2]
c
k
+9(
k
+
r
−
1)
c
k
−
1
=0
.
The indicial roots are
r
=1
/
3 and
r
=2
/
3. For
r
=1
/
3 the recurrence relation is
c
k
=
−
(3
k
−
2)
c
k
−
1
k
(3
k
−
1)
,k
=1
,
2
,
3
,...,
and
c
1
=
−
1
2
c
0
,c
2
=
1
5
c
0
,c
3
=
−
7
120
c
0
.
For
r
=2
/
3 the recurrence relation is
c
k
=
−
(3
k
−
1)
c
k
−
1
k
(3
k
+1)
,k
=1
,
2
,
3
,...,
and
c
1
=
−
1
2
c
0
,c
2
=
5
28
c
0
,c
3
=
−
1
21
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
x
1
/
3
/
1
−
1
2
x
+
1
5
x
2
−
7
120
x
3
+
···
d
+
C
2
x
2
/
3
/
1
−
1
2
x
+
5
28
x
2
−
1
21
x
3
+
···
d
.
24.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
2
x
2
y
//
+3
xy
/
+(2
x
−
1)
y
=

2
r
2
+
r
−
1

c
0
x
r
+
∞
x
k
=1
[2(
k
+
r
)(
k
+
r
−
1)
c
k
+3(
k
+
r
)
c
k
−
c
k
+2
c
k
−
1
]
x
k
+
r
=0
,
238

239. Exercises 5.2
which implies
2
r
2
+
r
−
1=(2
r
−
1)(
r
+1)=0
and
[(
k
+
r
)(2
k
+2
r
+1)
−
1]
c
k
+2
c
k
−
1
=0
.
The indicial roots are
r
=
−
1 and
r
=1
/
2. For
r
=
−
1 the recurrence relation is
c
k
=
−
2
c
k
−
1
k
(2
k
−
3)
,k
=1
,
2
,
3
,...,
and
c
1
=2
c
0
,c
2
=
−
2
c
0
,c
3
=
4
9
c
0
.
For
r
=1
/
2 the recurrence relation is
c
k
=
−
2
c
k
−
1
k
(2
k
+3)
,k
=1
,
2
,
3
,...,
and
c
1
=
−
2
5
c
0
,c
2
=
2
35
c
0
,c
3
=
−
4
945
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
x
−
1
/
1+2
x
−
2
x
2
+
4
9
x
3
+
···
d
+
C
2
x
1
/
2
/
1
−
2
5
x
+
2
35
x
2
−
4
945
x
3
+
···
d
.
25.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
xy
//
+2
y
/
−
xy
=

r
2
+
r

c
0
x
r
−
1
+

r
2
+3
r
+2

c
1
x
r
+
∞
x
k
=2
[(
k
+
r
)(
k
+
r
−
1)
c
k
+2(
k
+
r
)
c
k
−
c
k
−
2
]
x
k
+
r
−
1
=0
,
which implies
r
2
+
r
=
r
(
r
+1)=0
,

r
2
+3
r
+2

c
1
=0
,
and
(
k
+
r
)(
k
+
r
+1)
c
k
−
c
k
−
2
=0
.
The indicial roots are
r
1
= 0 and
r
2
=
−
1, so
c
1
=0. For
r
1
= 0 the recurrence relation is
c
k
=
c
k
−
2
k
(
k
+1)
,k
=2
,
3
,
4
,...,
and
c
2
=
1
3!
c
0
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
1
5!
c
0
c
2
n
=
1
(2
n
+ 1)!
c
0
.
239

240. Exercises 5.2
For
r
2
=
−
1 the recurrence relation is
c
k
=
c
k
−
2
k
(
k
−
1)
,k
=2
,
3
,
4
,...,
and
c
2
=
1
2!
c
0
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
1
4!
c
0
c
2
n
=
1
(2
n
)!
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
∞
x
n
=0
1
(2
n
+ 1)!
x
2
n
+
C
2
x
−
1
∞
x
n
=0
1
(2
n
)!
x
2
n
=
1
x
b
C
1
∞
x
n
=0
1
(2
n
+ 1)!
x
2
n
+1
+
C
2
∞
x
n
=0
1
(2
n
)!
x
2
n
r
=
1
x
[
C
1
sinh
x
+
C
2
cosh
x
]
.
26.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
x
2
y
//
+
xy
/
+
/
x
2
−
1
4
d
y
=
/
r
2
−
1
4
d
c
0
x
r
+
/
r
2
+2
r
+
3
4
d
c
1
x
r
+1
+
∞
x
k
=2
a
(
k
+
r
)(
k
+
r
−
1)
c
k
+(
k
+
r
)
c
k
−
1
4
c
k
+
c
k
−
2
)
x
k
+
r
=0
,
which implies
r
2
−
1
4
=
/
r
−
1
2
d/
r
+
1
2
d
=0
,
/
r
2
+2
r
+
3
4
d
c
1
=0
,
and
a
(
k
+
r
)
2
−
1
4
)
c
k
+
c
k
−
2
=0
.
The indicial roots are
r
1
=1
/
2 and
r
2
=
−
1
/
2, so
c
1
=0. For
r
1
=1
/
2 the recurrence relation is
c
k
=
−
c
k
−
2
k
(
k
+1)
,k
=2
,
3
,
4
,...,
and
c
2
=
−
1
3!
c
0
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
1
5!
c
0
c
2
n
=
(
−
1)
n
(2
n
+ 1)!
c
0
.
240

241. Exercises 5.2
For
r
2
=
−
1
/
2 the recurrence relation is
c
k
=
−
c
k
−
2
k
(
k
−
1)
,k
=2
,
3
,
4
,...,
and
c
2
=
−
1
2!
c
0
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
1
4!
c
0
c
2
n
=
(
−
1)
n
(2
n
)!
c
0
.
The general solution on (0
,
∞
)is
y
=
C
1
x
1
/
2
∞
x
n
=0
(
−
1)
n
(2
n
+ 1)!
x
2
n
+
C
2
x
−
1
/
2
∞
x
n
=0
(
−
1)
n
(2
n
)!
x
2
n
=
C
1
x
−
1
/
2
∞
x
n
=0
(
−
1)
n
(2
n
+ 1)!
x
2
n
+1
+
C
2
x
−
1
/
2
∞
x
n
=0
(
−
1)
n
(2
n
)!
x
2
n
=
x
−
1
/
2
[
C
1
sin
x
+
C
2
cos
x
]
.
27.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
xy
//
−
xy
/
+
y
=

r
2
−
r

c
0
x
r
−
1
+
∞
x
k
=0
[(
k
+
r
+ 1)(
k
+
r
)
c
k
+1
−
(
k
+
r
)
c
k
+
c
k
]
x
k
+
r
=0
which implies
r
2
−
r
=
r
(
r
−
1)=0
and
(
k
+
r
+ 1)(
k
+
r
)
c
k
+1
−
(
k
+
r
−
1)
c
k
=0
.
The indicial roots are
r
1
= 1 and
r
2
=0. For
r
1
= 1 the recurrence relation is
c
k
+1
=
kc
k
(
k
+ 2)(
k
+1)
,k
=0
,
1
,
2
,...,
and one solution is
y
1
=
c
0
x
. A second solution is
y
2
=
x
>
e
−
i
−
dx
x
2
dx
=
x
>
e
x
x
2
dx
=
x
>
1
x
2
/
1+
x
+
1
2
x
2
+
1
3!
x
3
+
···
d
dx
=
x
>
/
1
x
2
+
1
x
+
1
2
+
1
3!
x
+
1
4!
x
2
+
···
d
dx
=
x
a
−
1
x
+ln
x
+
1
2
x
+
1
12
x
2
+
1
72
x
3
+
···
)
=
x
ln
x
−
1+
1
2
x
2
+
1
12
x
3
+
1
72
x
4
+
···
.
The general solution on (0
,
∞
)is
y
=
C
1
x
+
C
2
y
2
(
x
)
.
241

242. Exercises 5.2
28.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
y
//
+
3
x
y
/
−
2
y
=

r
2
+2
r

c
0
x
r
−
2
+

r
2
+4
r
+3

c
1
x
r
−
1
+
∞
x
k
=2
[(
k
+
r
)(
k
+
r
−
1)
c
k
+3(
k
+
r
)
c
k
−
2
c
k
−
2
]
x
k
+
r
−
2
=0
,
which implies
r
2
+2
r
=
r
(
r
+2)=0

r
2
+4
r
+3

c
1
=0
(
k
+
r
)(
k
+
r
+2)
c
k
−
2
c
k
−
2
=0
.
The indicial roots are
r
1
= 0 and
r
2
=
−
2, so
c
1
=0. For
r
1
= 0 the recurrence relation is
c
k
=
2
c
k
−
2
k
(
k
+2)
,k
=2
,
3
,
4
,...,
and
c
2
=
1
4
c
0
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
1
48
c
0
c
6
=
1
1,152
c
0
.
The result is
y
1
=
c
0
/
1+
1
4
x
2
+
1
48
x
4
+
1
1,152
c
6
+
···
d
.
A second solution is
y
2
=
y
1
>
e
−
i
(3
/x
)
dx
y
2
1
dx
=
y
1
>
dx
x
3

1+
1
4
x
2
+
1
48
x
4
+
···

2
=
y
1
>
dx
x
3

1+
1
2
x
2
+
5
48
x
4
+
7
576
x
6
+
···

=
y
1
>
1
x
3
/
1
−
1
2
x
2
+
7
48
x
4
+
19
576
x
6
+
···
d
=
y
1
>
/
1
x
3
−
1
2
x
+
7
48
x
−
19
576
x
3
+
···
d
=
y
1
a
−
1
2
x
2
−
1
2
ln
x
+
7
96
x
2
−
19
2,304
x
4
+
···
)
=
−
1
2
y
1
ln
x
+
y
a
−
1
2
x
2
+
7
96
x
2
−
19
2,304
x
4
+
···
)
.
The general solution on (0
,
∞
)is
y
=
C
1
y
1
(
x
)+
C
2
y
2
(
x
)
.
29.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
xy
//
+(1
−
x
)
y
/
−
y
=
r
2
c
0
x
r
−
1
+
∞
x
k
=0
[(
k
+
r
)(
k
+
r
−
1)
c
k
+(
k
+
r
)
c
k
−
(
k
+
r
)
c
k
−
1
]
x
k
+
r
−
1
=0
,
which implies
r
2
= 0 and
(
k
+
r
)
2
c
k
−
(
k
+
r
)
c
k
−
1
=0
.
242

243. Exercises 5.2
The indicial roots are
r
1
=
r
2
= 0 and the recurrence relation is
c
k
=
c
k
−
1
k
,k
=1
,
2
,
3
,... .
One solution is
y
1
=
c
0
/
1+
x
+
1
2
x
2
+
1
3!
x
3
+
···
d
=
c
0
e
x
.
A second solution is
y
2
=
y
1
>
e
−
i
(1
/x
−
1)
dx
e
2
x
dx
=
e
x
>
e
x
/x
e
2
x
dx
=
e
x
>
1
x
e
−
x
dx
=
e
x
>
1
x
/
1
−
x
+
1
2
x
2
−
1
3!
x
3
+
···
d
dx
=
e
x
>
/
1
x
−
1+
1
2
x
−
1
3!
x
2
+
···
d
dx
=
e
x
a
ln
x
−
x
+
1
2
·
2
x
2
−
1
3
·
3!
x
3
+
···
)
=
e
x
ln
x
−
e
x
∞
x
n
=1
(
−
1)
n
+1
n
·
n
!
x
n
.
The general solution on (0
,
∞
)is
y
=
C
1
e
x
+
C
2
e
x
p
ln
x
−
∞
x
n
=1
(
−
1)
n
+1
n
·
n
!
x
n
q
.
30.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
xy
//
+
y
/
+
y
=
r
2
c
0
x
r
−
1
+
∞
x
k
=1
[(
k
+
r
)(
k
+
r
−
1)
c
k
+(
k
+
r
)
c
k
+
c
k
−
1
]
x
k
+
r
−
1
=0
which implies
r
2
= 0 and
(
k
+
r
)
2
c
k
+
c
k
−
1
=0
.
The indicial roots are
r
1
=
r
2
= 0 and the recurrence relation is
c
k
=
−
c
k
−
1
k
2
,k
=1
,
2
,
3
,... .
One solution is
y
1
=
c
0
/
1
−
x
+
1
2
2
x
2
−
1
(3!)
2
x
3
+
1
(4!)
2
x
4
−···
d
=
c
0
∞
x
n
=0
(
−
1)
n
(
n
!)
2
x
n
.
A second solution is
y
2
=
y
1
>
e
−
i
(1
/x
)
dx
y
2
1
dx
=
y
1
>
dx
x

1
−
x
+
1
4
x
2
−
1
36
x
3
+
···

2
=
y
1
>
dx
x

1
−
2
x
+
3
2
x
2
−
5
9
x
3
+
35
288
x
4
−···

=
y
1
>
1
x
/
1+2
x
+
5
2
x
2
+
23
9
x
3
+
677
288
x
4
+
···
d
dx
=
y
1
>
/
1
x
+2+
5
2
x
+
23
9
x
2
+
677
288
x
3
+
···
d
dx
=
y
1
a
ln
x
+2
x
+
5
4
x
2
+
23
27
x
3
+
677
1,152
x
4
+
···
)
=
y
1
ln
x
+
y
1
/
2
x
+
5
4
x
2
+
23
27
x
3
+
677
1,152
x
4
+
···
d
.
243

244. Exercises 5.2
The general solution on (0
,
∞
)is
y
=
C
1
y
1
(
x
)+
C
2
y
2
(
x
)
.
31.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
xy
//
+(
x
−
6)
y
/
−
3
y
=(
r
2
−
7
r
)
c
0
x
r
−
1
+
∞
x
k
=1

(
k
+
r
)(
k
+
r
−
1)
c
k
+(
k
+
r
−
1)
c
k
−
1
−
6(
k
+
r
)
c
k
−
3
c
k
−
1

x
k
+
r
−
1
=0
,
which implies
r
2
−
7
r
=
r
(
r
−
7)=0
and
(
k
+
r
)(
k
+
r
−
7)
c
k
+(
k
+
r
−
4)
c
k
−
1
=0
.
The indicial roots are
r
1
= 7 and
r
2
=0. For
r
1
= 7 the recurrence relation is
(
k
+7)
kc
k
+(
k
+3)
c
k
−
1
=0
,k
=1
,
2
,
3
,... ,
or
c
k
=
−
k
+3
k
+7
c
k
−
1
,k
=1
,
2
,
3
,... .
Taking
c
0
/
= 0 we obtain
c
1
=
−
1
2
c
0
c
2
=
5
18
c
0
c
3
=
−
1
6
c
0
,
and so on. Thus, the indicial root
r
1
= 7 yields a single solution. Now, for
r
2
= 0 the recurrence relation is
k
(
k
−
7)
c
k
+(
k
−
4)
c
k
−
1
=0
,k
=1
,
2
,
3
,... .
Then
−
6
c
1
−
3
c
0
=0
−
10
c
2
−
2
c
1
=0
−
12
c
3
−
c
2
=0
−
12
c
4
+0
c
3
=0 =
⇒
c
4
=0
−
10
c
5
+
c
4
=0 =
⇒
c
5
=0
−
6
c
6
+2
c
5
=0 =
⇒
c
6
=0
0
c
7
+3
c
6
=0 =
⇒
c
7
is arbitrary
and
c
k
=
−
k
−
4
k
(
k
−
7)
c
k
−
1
,k
=8
,
9
,
10
,... .
Taking
c
0
/
= 0 and
c
7
= 0 we obtain
c
1
=
−
1
2
c
0
c
2
=
1
10
c
0
c
3
=
−
1
120
c
0
c
4
=
c
5
=
c
6
=
···
=0
.
244

245. Exercises 5.2
Taking
c
0
= 0 and
c
7
/
= 0 we obtain
c
1
=
c
2
=
c
3
=
c
4
=
c
5
=
c
6
=0
c
8
=
−
1
2
c
7
c
9
=
5
36
c
7
c
10
=
−
1
36
c
7
.
In this case we obtain the two solutions
y
1
=1
−
1
2
x
+
1
10
x
2
−
1
120
x
3
and
y
2
=
x
7
−
1
2
x
8
+
5
36
x
9
−
1
36
x
10
+
···
.
32.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
x
(
x
−
1)
y
//
+3
y
/
−
2
y
=

4
r
−
r
2

c
0
x
r
−
1
+
∞
x
k
=1
[(
k
+
r
−
1)(
k
+
r
−
12)
c
k
−
1
−
(
k
+
r
)(
k
+
r
−
1)
c
k
+3(
k
+
r
)
c
k
−
2
c
k
−
1
]
x
k
+
r
−
1
=0
,
which implies
4
r
−
r
2
=
r
(4
−
r
)=0
and
−
(
k
+
r
)(
k
+
r
−
4)
c
k
+[(
k
+
r
−
1)(
k
+
r
−
2)
−
2]
c
k
−
1
=0
.
The indicial roots are
r
1
= 4 and
r
2
=0. For
r
1
= 4 the recurrence relation is
−
(
k
+4)
kc
k
+[(
k
+ 3)(
k
+2)
−
2]
c
k
−
1
=0
or
c
k
=(
k
+1)
c
k
−
1
,k
=1
,
2
,
3
,... .
Taking
c
0
/
= 0 we obtain
c
1
=2
c
0
c
2
=3!
c
0
c
3
=4!
c
0
,
and so on. Thus, the indicial root
r
1
= 4 yields a single solution. For
r
2
= 0 the recurrence relation is
−
k
(
k
−
4)
c
k
+
k
(
k
−
3)
c
k
−
1
=0
,k
=1
,
2
,
3
,...,
or
−
(
k
−
4)
c
k
+(
k
−
3)
c
k
−
1
=0
,k
=1
,
2
,
3
,... .
Then
3
c
1
−
2
c
0
=0
2
c
2
−
c
1
=0
c
3
+0
c
2
=0
⇒
c
3
=0
0
c
4
+
c
3
=0
⇒
c
4
is arbitrary
245

246. Exercises 5.2
and
c
k
=
(
k
−
3)
c
k
−
1
c
−
4
,k
=5
,
6
,
7
,... .
Taking
c
0
/
= 0 and
c
4
= 0 we obtain
c
1
=
2
3
c
0
c
2
=
1
3
c
0
c
3
=
c
4
=
c
5
=
···
=0
.
Taking
c
0
= 0 and
c
4
/
= 0 we obtain
c
1
=
c
2
=
c
3
=0
c
5
=2
c
4
c
6
=3
c
4
c
7
=4
c
4
.
In this case we obtain the two solutions
y
1
=1+
2
3
x
+
1
3
x
2
and
y
2
=
x
4
+2
x
5
+3
x
6
+4
x
7
+
···
.
33. (a)
From
t
=1
/x
we have
dt/dx
=
−
1
/x
2
=
−
t
2
. Then
dy
dx
=
dy
dt
dt
dx
=
−
t
2
dy
dt
and
d
2
y
dx
2
=
d
dx
/
dy
dx
d
=
d
dx
/
−
t
2
dy
dt
d
=
−
t
2
d
2
y
dt
2
dt
dx
−
dy
dt
/
2
t
dt
dx
d
=
t
4
d
2
y
dt
2
+2
t
3
dy
dt
.
Now
x
4
d
2
y
dx
2
+
λy
=
1
t
4
/
t
4
d
2
y
dt
2
+2
t
3
dy
dt
d
+
λy
=
d
2
y
dt
2
+
2
t
dy
dt
+
λy
=0
becomes
t
d
2
y
dt
2
+2
dy
dt
+
λty
=0
.
(b)
Substituting
y
=
b
∞
n
=0
c
n
t
n
+
r
into the differential equation and collecting terms, we obtain
t
d
2
y
dt
2
+2
dy
dt
+
λty
=(
r
2
+
r
)
c
0
t
r
−
1
+(
r
2
+3
r
+2)
c
1
t
r
+
∞
x
k
=2
[(
k
+
r
)(
k
+
r
−
1)
c
k
+2(
k
+
r
)
c
k
+
λc
k
−
2
]
t
k
+
r
−
1
=0
,
which implies
r
2
+
r
=
r
(
r
+1)=0
,

r
2
+3
r
+2

c
1
=0
,
and
(
k
+
r
)(
k
+
r
+1)
c
k
+
λc
k
−
2
=0
.
The indicial roots are
r
1
= 0 and
r
2
=
−
1, so
c
1
=0. For
r
1
= 0 the recurrence relation is
c
k
=
−
λc
k
−
2
k
(
k
+1)
,k
=2
,
3
,
4
,...,
246

247. Exercises 5.2
and
c
2
=
−
λ
3!
c
0
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
λ
2
5!
c
0
c
2
n
=(
−
1)
n
λ
n
(2
n
+ 1)!
c
0
.
For
r
2
=
−
1 the recurrence relation is
c
k
=
−
λc
k
−
2
k
(
k
−
1)
,k
=2
,
3
,
4
,...,
and
c
2
=
−
λ
2!
c
0
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
λ
2
4!
c
0
c
2
n
=(
−
1)
n
λ
n
(2
n
)!
c
0
.
The general solution on (0
,
∞
)is
y
(
t
)=
C
1
∞
x
n
=0
(
−
1)
n
(2
n
+ 1)!
(
√
λt
)
2
n
+
C
2
t
−
1
∞
x
n
=0
(
−
1)
n
(2
n
)!
(
√
λt
)
2
n
=
1
t
a
C
1
∞
x
n
=0
(
−
1)
n
(2
n
+ 1)!
(
√
λt
)
2
n
+1
+
C
2
∞
x
n
=0
(
−
1)
n
(2
n
)!
(
√
λt
)
2
n
)
=
1
t
[
C
1
sin
√
λt
+
C
2
cos
√
λt
]
.
(c)
Using
t
=1
/x
, the solution of the original equation is
y
(
x
)
C
1
x
sin
√
λ
x
+
C
2
x
cos
√
λ
x
.
34. (a)
From the boundary conditions
y
(
a
)=0,
y
(
b
)=0wefind
C
1
sin
√
λ
a
+
C
2
cos
√
λ
a
=0
C
1
sin
√
λ
b
+
C
2
cos
√
λ
b
=0
.
Since this is a homogeneous system of linear equations, it will have nontrivial solutions if
k
k
k
k
k
k
k
k
sin
√
λ
a
cos
√
λ
a
sin
√
λ
b
cos
√
λ
b
k
k
k
k
k
k
k
k
= sin
√
λ
a
cos
√
λ
b
−
cos
√
λ
a
sin
√
λ
b
= sin
p
√
λ
a
−
√
λ
b
q
= sin
√
λ
/
b
−
a
ab
d
=0
.
This will be the case if
√
λ
/
b
−
a
ab
d
=
nπ
or
√
λ
=
nλab
b
−
a
=
nλab
L
,n
=1
,
2
,...,
247

248. x
y
5
1
2
Exercises 5.2
or, if
λ
n
=
n
2
π
2
a
2
b
2
L
2
=
P
n
b
4
EI
.
The critical loads are then
P
n
=
n
2
π
2
(
a/b
)
2
EI
0
/L
2
. Using
C
2
=
−
C
1
sin(
√
λ/a
)
/
cos(
√
λ/a
)wehave
y
=
C
1
x
a
sin
√
λ
x
−
sin(
√
λ/a
)
cos(
√
λ/a
)
cos
√
λ
x
)
=
C
3
x
a
sin
√
λ
x
cos
√
λ
a
−
cos
√
λ
x
sin
√
λ
a
)
=
C
3
x
sin
√
λ
/
1
x
−
1
a
d
,
and
y
n
(
x
)=
C
3
x
sin
nπab
L
/
1
x
−
1
a
d
=
C
4
x
sin
nπab
L
c
1
−
a
x
d
.
(b)
When
n
=1,
b
= 11, and
a
=1,wehave,for
C
4
=1,
y
1
(
x
)=
x
sin 1
.
1
π
/
1
−
1
x
d
.
Exercises 5.3
1.
Since
ν
2
=1
/
9 the general solution is
y
=
c
1
J
1
/
3
(
x
)+
c
2
J
−
1
/
3
(
x
).
2.
Since
ν
2
= 1 the general solution is
y
=
c
1
J
1
(
x
)+
c
2
Y
1
(
x
).
3.
Since
ν
2
=25
/
4 the general solution is
y
=
c
1
J
5
/
2
(
x
)+
c
2
J
−
5
/
2
(
x
).
4.
Since
ν
2
=1
/
16 the general solution is
y
=
c
1
J
1
/
4
(
x
)+
c
2
J
−
1
/
4
(
x
).
5.
Since
ν
2
= 0 the general solution is
y
=
c
1
J
0
(
x
)+
c
2
Y
0
(
x
).
6.
Since
ν
2
= 4 the general solution is
y
=
c
1
J
2
(
x
)+
c
2
Y
2
(
x
).
7.
Since
ν
2
= 2 the general solution is
y
=
c
1
J
2
(3
x
)+
c
2
Y
2
(3
x
).
8.
Since
ν
2
=1
/
4 the general solution is
y
=
c
1
J
1
/
2
(6
x
)+
c
2
J
−
1
/
2
(6
x
).
9.
If
y
=
x
−
1
/
2
v
(
x
) then
y
/
=
x
−
1
/
2
v
/
(
x
)
−
1
2
x
−
3
/
2
v
(
x
)
,
y
//
=
x
−
1
/
2
v
//
(
x
)
−
x
−
3
/
2
v
/
(
x
)+
3
4
x
−
5
/
2
v
(
x
)
,
and
x
2
y
//
+2
xy
/
+
λ
2
x
2
y
=
x
3
/
2
v
//
+
x
1
/
2
v
/
+
/
λ
2
x
3
/
2
−
1
4
x
−
1
/
2
d
v.
Multiplying by
x
1
/
2
we obtain
x
2
v
//
+
xv
/
+
/
λ
2
x
2
−
1
4
d
v
=0
,
248

249. Exercises 5.3
whose solution is
v
=
c
1
J
1
/
2
(
λx
)+
c
2
J
−
1
/
2
(
λx
). Then
y
=
c
1
x
−
1
/
2
J
1
/
2
(
λx
)+
c
2
x
−
1
/
2
J
−
1
/
2
(
λx
).
10.
From
y
=
x
n
J
n
(
x
)wefind
y
/
=
x
n
J
/
n
+
nx
n
−
1
J
n
and
y
//
=
x
n
J
//
n
+2
nx
n
−
1
J
/
n
+
n
(
n
−
1)
x
n
−
2
J
n
.
Substituting into the differential equation, we have
x
n
+1
J
//
n
+2
nx
n
J
/
n
+
n
(
n
−
1)
x
n
−
1
J
n
+(1
−
2
n
)(
x
n
J
/
n
+
nx
n
−
1
J
n
)+
x
n
+1
J
n
=
x
n
+1
J
//
n
+(2
n
+1
−
2
n
)
x
n
J
/
n
+(
n
2
−
n
+
n
−
2
n
2
)
x
n
−
1
J
n
+
x
n
+1
J
n
=
x
n
+1
[
x
2
J
//
n
+
xJ
/
n
−
n
2
J
n
+
x
2
J
n
]
=
x
n
+1
[
x
2
J
//
n
+
xJ
/
n
+(
x
2
−
n
2
)
J
n
]
=
x
n
−
1
·
0 (since
J
n
is a solution of Bessel’s equation)
=0
.
Therefore,
x
n
J
n
is a solution of the original equation.
11.
From
y
=
x
−
n
J
n
we find
y
/
=
x
−
n
J
/
n
−
nx
−
n
−
1
J
n
and
y
//
=
x
−
n
J
//
n
−
2
nx
−
n
−
1
J
/
n
+
n
(
n
+1)
x
−
n
−
2
J
n
.
Substituting into the differential equation, we have
xy
//
+(1+2
n
)
y
/
+
xy
=
x
−
n
−
1

x
2
J
//
n
+
xJ
/
n
+

x
2
−
n
2

J
n

=
x
−
n
−
1
·
0 (since
J
n
is a solution of Bessel’s equation)
=0
.
Therefore,
x
−
n
J
n
is a solution of the original equation.
12.
From
y
=
√
xJ
ν
(
λx
)wefind
y
/
=
λ
√
xJ
/
ν
(
λx
)+
1
2
x
−
1
/
2
J
ν
(
λx
)
and
y
//
=
λ
2
√
xJ
//
ν
(
λx
)+
λx
−
1
/
2
J
/
ν
(
λx
)
−
1
4
x
−
3
/
2
J
ν
(
λx
)
.
Substituting into the differential equation, we have
x
2
y
//
+
/
λ
2
x
2
−
ν
2
+
1
4
d
y
=
√
x

λ
2
x
2
J
//
ν
(
λx
)+
λxJ
/
ν
(
λx
)+

λ
2
x
2
−
ν
2

J
ν
(
λx
)

=
√
x
·
0 (since
J
n
is a solution of Bessel’s equation)
=0
.
Therefore,
√
xJ
ν
(
λx
) is a solution of the original equation.
13.
From Problem 10 with
n
=1
/
2 we find
y
=
x
1
/
2
J
1
/
2
(
x
). From Problem 11 with
n
=
−
1
/
2 we find
y
=
x
1
/
2
J
−
1
/
2
(
x
).
14.
From Problem 10 with
n
= 1 we find
y
=
xJ
1
(
x
). From Problem 11 with
n
=
−
1 we find
y
=
xJ
−
1
(
x
)=
−
xJ
1
(
x
).
15.
From Problem 10 with
n
=
−
1wefind
y
=
x
−
1
J
−
1
(
x
). From Problem 11 with
n
= 1 we find
y
=
x
−
1
J
1
(
x
)=
−
x
−
1
J
−
1
(
x
).
16.
From Problem 12 with
λ
= 2 and
ν
= 0 we find
y
=
√
xJ
0
(2
x
).
17.
From Problem 12 with
λ
= 1 and
ν
=
±
3
/
2 we find
y
=
√
xJ
3
/
2
(
x
) and
y
=
√
xJ
−
3
/
2
(
x
).
249

250. Exercises 5.3
18.
From Problem 10 with
n
= 3 we find
y
=
x
3
J
3
(
x
). From Problem 11 with
n
=
−
3wefind
y
=
x
3
J
−
3
(
x
)=
−
x
3
J
3
(
x
).
19. (a)
The recurrence relation follows from
−
νJ
ν
(
x
)+
xJ
ν
−
1
(
x
)=
−
∞
x
n
=0
(
−
1)
n
ν
n
!Γ(1 +
ν
+
n
)
c
x
2
d
2
n
+
ν
+
x
∞
x
n
=0
(
−
1)
n
n
!Γ(
ν
+
n
)
c
x
2
d
2
n
+
ν
−
1
=
−
∞
x
n
=0
(
−
1)
n
ν
n
!Γ(1 +
ν
+
n
)
c
x
2
d
2
n
+
ν
+
∞
x
n
=0
(
−
1)
n
(
ν
+
n
)
n
!Γ(1 +
ν
+
n
)
·
2
c
x
2
dc
x
2
d
2
n
+
ν
−
1
=
∞
x
n
=0
(
−
1)
n
(2
n
+
ν
)
n
!Γ(1 +
ν
+
n
)
c
x
2
d
2
n
+
ν
=
xJ
/
ν
(
x
)
.
(b)
The formula in part (a) is a linear first-order differential equation in
J
ν
(
x
). An integrating factor for this
equation is
x
ν
,so
d
dx
[
x
ν
J
ν
(
x
)] =
x
ν
J
ν
−
1
(
x
)
.
20.
Subtracting the formula in part (a) of Problem 19 from the formula in Example 4 we obtain
0=2
νJ
ν
(
x
)
−
xJ
ν
+1
(
x
)
−
xJ
ν
−
1
(
x
)or2
νJ
ν
(
x
)=
xJ
ν
+1
(
x
)+
xJ
ν
−
1
(
x
)
.
21.
Letting
ν
= 1 in (15) we have
xJ
0
(
x
)=
d
dx
[
xJ
1
(
x
)] so
>
x
0
rJ
0
(
r
)
dr
=
rJ
1
(
r
)
k
k
k
r
=
x
r
=0
=
xJ
1
(
x
)
.
22.
From (14) we obtain
J
/
0
(
x
)=
−
J
1
(
x
), and from (15) we obtain
J
/
0
(
x
)=
J
−
1
(
x
). Thus
J
/
0
(
x
)=
J
−
1
(
x
)=
−
J
1
(
x
).
23.
Since
Γ
/
1
−
1
2
+
n
d
=
(2
n
−
1)!
(
n
−
1)!2
2
n
−
1
we obtain
J
−
1
/
2
(
x
)=
∞
x
n
=0
(
−
1)
n
2
1
/
2
x
−
1
/
2
2
n
(2
n
−
1)!
√
π
x
2
n
=
t
2
πx
cos
x.
24. (a)
By Problem 20, with
ν
=1
/
2, we obtain
J
1
/
2
(
x
)=
xJ
3
/
2
(
x
)+
xJ
−
1
/
2
(
x
) so that
J
3
/
2
(
x
)=
t
2
πx
/
sin
x
x
−
cos
x
d
;
with
ν
=
−
1
/
2 we obtain
−
J
−
1
/
2
(
x
)=
xJ
1
/
2
(
x
)+
xJ
−
3
/
2
(
x
) so that
J
−
3
/
2
(
x
)=
−
t
2
πx
c
cos
x
x
+ sin
x
d
;
and with
ν
=3
/
2 we obtain 3
J
3
/
2
(
x
)=
xJ
5
/
2
(
x
)+
xJ
1
/
2
(
x
) so that
J
5
/
2
(
x
)=
t
2
πx
/
3 sin
x
x
2
−
3 cos
x
x
−
sin
x
d
.
250

251. 5 101520
x
-1
-0.5
0.5
1
y
5 101520
x
-1
-0.5
0.5
1
y
5 101520
x
-1
-0.5
0.5
1
y
5 101520
x
-1
-0.5
0.5
1
y
5 101520
x
-1
-0.5
0.5
1
y
Exercises 5.3
ν
=1
/
2
ν
=
−
1
/
2
ν
=3
/
2
ν
=
−
3
/
2
ν
=5
/
2
25.
Rolle’s theorem states that for a differentiable function
f
(
x
), for which
f
(
a
)=
f
(
b
) = 0, there exists a number
c
between
a
and
b
such that
f
/
(
c
) = 0. From Problem 20 with
ν
= 0 we have
J
/
0
(
x
)=
J
1
(
x
). Thus, if
a
and
b
are successive zeros of
J
0
(
x
), then there exists a
c
between
a
and
b
for which
J
1
(
x
)=
J
/
0
(
x
)=0.
26.
Since
k
k
k
i
b
a
f
(
x
)
dx
k
k
k
≤
i
b
a
|
f
(
x
)
|
dx
, we have
|
J
n
(
x
)
|≤
1
π
>
π
0
|
cos(
x
sin
t
−
nt
)
|
dt
≤
1
π
>
π
0
1
dt
=1
.
27.
Letting
s
=
2
α
t
k
m
e
−
αt/
2
,
we have
dx
dt
=
dx
ds
ds
dt
=
dx
dt
b
2
α
t
k
m
c
−
α
2
d
e
−
αt/
2
r
=
dx
ds
p
−
t
k
m
e
−
αt/
2
q
and
d
2
x
dt
2
=
d
dt
/
dx
dt
d
=
dx
ds
p
α
2
t
k
m
e
−
αt/
2
q
+
d
dt
/
dx
ds
d
p
−
t
k
m
e
−
αt/
2
q
=
dx
ds
p
α
2
t
k
m
e
−
αt/
2
q
+
d
2
x
ds
2
ds
dt
p
−
t
k
m
e
−
αt/
2
q
=
dx
ds
p
α
2
t
k
m
e
−
αt/
2
q
+
d
2
x
ds
2
/
k
m
e
−
αt
d
.
Then
m
d
2
x
dt
2
+
ke
−
αt
x
=
ke
−
αt
d
2
x
ds
2
+
mα
2
t
k
m
e
−
αt/
2
dx
dt
+
ke
−
αt
x
=0
.
Multiplying by 2
2
/α
2
m
we have
2
2
α
2
k
m
e
−
αt
d
2
x
ds
2
+
2
α
t
k
m
e
−
αt/
2
dx
dt
+
2
α
2
k
m
e
−
αt
x
=0
or, since
s
=(2
/α
)
λ
k/m e
−
αt/
2
,
s
2
d
2
x
ds
2
+
s
dx
ds
+
s
2
x
=0
.
28. (a)
We identify
m
=4,
k
= 1, and
α
=0
.
1. Then
x
(
t
)=
c
1
J
0
(10
e
−
0
.
05
t
)+
c
2
Y
0
(10
e
−
0
.
05
t
)
and
x
/
(
t
)=
−
0
.
5
c
1
J
/
0
(10
e
−
0
.
05
t
)
−
0
.
5
c
2
Y
/
0
(10
e
−
0
.
05
t
)
.
251

252. t
x
50 100 150 200
−5
5
10
Exercises 5.3
Now
x
(0) = 1 and
x
/
(0) =
−
1
/
2 imply
c
1
J
0
(10) +
c
2
Y
0
(10) = 1
c
1
J
/
0
(10) +
c
2
Y
/
0
(10) = 1
.
Using Cramer’s rule we obtain
c
1
=
Y
/
0
(10)
−
Y
0
(10)
J
0
(10)
Y
/
0
(10)
−
J
/
0
(10)
Y
0
(10)
and
c
2
=
J
0
(10)
−
J
/
0
(10)
J
0
(10)
Y
/
0
(10)
−
J
/
0
(10)
Y
0
(10)
.
Using
Y
/
0
=
−
Y
1
and
J
/
0
=
−
J
1
and Table 6
.
1wefind
c
1
=
−
4
.
7860 and
c
2
=
−
3
.
1803. Thus
x
(
t
)=
−
4
.
7860
J
0
(10
e
−
0
.
05
t
)
−
3
.
1803
Y
0
(10
e
−
0
.
05
t
)
.
(b)
29.
Differentiating
y
=
x
1
/
2
w

2
3
αx
3
/
2

with respect to
2
3
αx
3
/
2
we obtain
y
/
=
x
1
/
2
w
/
/
2
3
αx
3
/
2
d
αx
1
/
2
+
1
2
x
−
1
/
2
w
/
2
3
αx
3
/
2
d
and
y
//
=
αxw
//
/
2
3
αx
3
/
2
d
αx
1
/
2
+
αw
/
/
2
3
αx
3
/
2
d
+
1
2
αw
/
/
2
3
αx
3
/
2
d
−
1
4
x
−
3
/
2
w
/
2
3
αx
3
/
2
d
.
Then, after combining terms and simplifying, we have
y
//
+
α
2
xy
=
α
a
αx
3
/
2
w
//
+
3
2
w
/
+
/
αx
3
/
2
−
1
4
αx
3
/
2
d
w
)
=0
.
Letting
t
=
2
3
αx
3
/
2
or
αx
3
/
2
=
3
2
t
this differential equation becomes
3
2
α
t
a
t
2
w
//
(
t
)+
tw
/
(
t
)+
/
t
2
−
1
9
d
w
(
t
)
)
=0
,t>
0
.
30.
The general solution of Bessel’s equation is
w
(
t
)=
c
1
J
1
/
3
(
t
)+
c
2
J
−
1
/
3
(
t
)
,t>
0
.
Thus, the general solution of Airy’s equation for
x>
0is
y
=
x
1
/
2
w
/
2
3
αx
3
/
2
d
=
c
1
x
1
/
2
J
1
/
3
/
2
3
αx
3
/
2
d
+
c
2
x
1
/
2
J
−
1
/
3
/
2
3
αx
3
/
2
d
.
31. (a)
Identifying
α
=
1
2
, the general solution of
x
//
+
1
4
tx
=0is
x
(
t
)=
c
1
x
1
/
2
J
1
/
3
/
1
3
x
3
/
2
d
+
c
2
x
1
/
2
J
−
1
/
3
/
1
3
x
3
/
2
d
.
Solving the system
x
(0
.
1) = 1,
x
/
(0
.
1) =
−
1
2
we find
c
1
=
−
0
.
809264 and
c
2
=0
.
782397.
252

253. t
x
50 100 150 200
−1
1
Exercises 5.3
(b)
32. (a)
Letting
t
=
L
−
x
, the boundary-value problem becomes
d
2
θ
dt
2
+
α
2
tθ
=0
,θ
/
(0) = 0
,θ
(
L
)=0
,
where
α
2
=
δg/EI
. This is Airy’s differential equation, so by Problem 30 its solution is
y
=
c
1
t
1
/
2
J
1
/
3
/
2
3
αt
3
/
2
d
+
c
2
t
1
/
2
J
−
1
/
3
/
2
3
αt
3
/
2
d
=
c
1
θ
1
(
t
)+
c
2
θ
2
(
t
)
.
(b)
Looking at the series forms of
θ
1
and
θ
2
we see that
θ
/
1
(0)
/
= 0, while
θ
/
2
(0) = 0. Thus, the boundary
condition
θ
/
(0) = 0 implies
c
1
= 0, and so
θ
(
t
)=
c
2
√
tJ
−
1
/
3
/
2
3
αt
3
/
2
d
.
From
θ
(
L
)=0wehave
c
2
√
LJ
−
1
/
3
/
2
3
αL
3
/
2
d
=0
,
so either
c
2
= 0, in which case
θ
(
t
)=0,or
J
−
1
/
3
(
2
3
αL
3
/
2
) = 0. The column will just start to bend when
L
is the length corresponding to the smallest positive zero of
J
−
1
/
3
. Using
Mathematica
, the first positive
root of
J
−
1
/
3
(
x
)is
x
1
≈
1
.
86635. Thus
2
3
αL
3
/
2
=1
.
86635 implies
L
=
/
3(1
.
86635)
2
α
d
2
/
3
=
a
9
EI
4
δg
(1
.
86635)
2
)
1
/
3
=
a
9(2
.
6
×
10
7
)
π
(0
.
05)
4
/
4
4(0
.
28)
π
(0
.
05)
2
(1
.
86635)
2
)
1
/
3
≈
76
.
9 in.
33.
Setting
y
=
√
xJ
1
(2
√
x
) and differentiating we obtain
y
/
=
√
xJ
/
1
(2
√
x
)
2
2
√
x
+
1
2
√
x
J
1
(2
√
x
)=
J
/
1
(2
√
x
)+
1
2
√
x
J
1
(2
√
x
)
and
y
//
=
J
//
1
(2
√
x
)
2
2
√
x
+
1
2
√
x
J
/
1
(2
√
x
)
2
2
√
x
−
1
4
x
3
/
2
J
1
(2
√
x
)
=
1
√
x
J
//
1
(2
√
x
)+
1
2
x
J
/
1
(2
√
x
)
−
1
4
x
3
/
2
J
1
(2
√
x
)
.
Substituting into the differential equation and letting
t
=2
√
x
we have
xy
//
+
y
=
√
xJ
//
1
(2
√
x
)+
1
2
J
/
1
(2
√
x
)
−
1
4
√
x
J
1
(2
√
x
)+
√
xJ
1
(2
√
x
)
=
1
√
x
a
xJ
//
1
(2
√
x
)+
√
x
2
J
/
1
(2
√
x
)+
/
x
−
1
4
d
J
1
(2
√
x
)
)
=
2
t
a
t
2
4
J
//
1
(
t
)+
t
4
J
/
1
(
t
)+
/
t
2
4
−
1
4
d
J
1
(
t
)
)
=
1
2
t
[
t
2
J
//
1
(
t
)+
tJ
/
1
(
t
)+(
t
2
−
1)
J
1
(
t
)]
.
253

254. x
y
0.2 0.4 0.6 0.8
1
0.1
0.2
0.3
-1-0.5 0.51
x
-1
-0.5
0.5
1
P5
-1-0.5 0.51
x
-1
-0.5
0.5
1
P6
-1-0.5 0.51
x
-1
-0.5
0.5
1
P7
-1-0.5 0.51
x
-1
-0.5
0.5
1
P1
-1-0.5 0.51
x
-1
-0.5
0.5
1
P2
-1-0.5 0.51
x
-1
-0.5
0.5
1
P3
-1-0.5 0.51
x
-1
-0.5
0.5
1
P4
Exercises 5.3
Since
J
1
(
t
) is a solution of
t
2
y
22
+
ty
2
+(
t
2
−
1)
y
= 0, we see that the last expression above is 0 and
y
=
√
xJ
1
(2
√
x
)
is a solution of
xy
22
+
y
=0.
34. (a)
Writing the differential equation in the form
xy
22
+(
PL/M
)
y
= 0, we identify
λ
=
PL/M
. From
Problem 33 the solution of this differential equation is
y
=
c
1
√
xJ
1
c
2
λ
P Lx/M
d
+
c
2
√
xY
1
c
2
λ
P Lx/M
d
.
Now
J
1
(0) = 0, so
y
(0) = 0 implies
c
2
= 0 and
y
=
c
1
√
xJ
1
c
2
λ
P Lx/M
d
.
(b)
From
y
(
L
)=0wehave
y
=
J
1
(2
L
√
PM
) = 0. The first positive zero of
J
1
is 3
.
8317 so, solving 2
L
λ
P
1
/M
=
3
.
8317, we find
P
1
=3
.
6705
M/L
2
. Therefore,
y
1
(
x
)=
c
1
√
xJ
1
p
2
t
3
.
6705
x
L
q
=
c
1
√
xJ
1
2
3
.
8317
L
√
x
3
.
(c)
For
c
1
= 1 and
L
= 1 the graph of
y
1
=
√
xJ
1
(3
.
8317
√
x
) is shown.
35. (a)
Using the expressions for the two linearly independent power series solutions,
y
1
(
x
) and
y
2
(
x
), given in the
text we obtain
P
6
(
x
)=
1
16

231
x
6
−
315
x
4
+ 105
x
2
−
5

and
P
7
(
x
)=
1
16

429
x
7
−
693
x
5
+ 315
x
3
−
35
x

.
(b)
P
6
(
x
) satisfies

1
−
x
2

y
22
−
2
xy
2
+42
y
= 0 and
P
7
(
x
) satisfies

1
−
x
2

y
22
−
2
xy
2
+56
y
=0.
36. (a)
254

255. Exercises 5.3
(b)
Zeros of Legendre polynomials
P
1
(
x
): 0
P
2
(
x
):
±
0
.
57735
P
3
(
x
): 0
,
±
0
.
77460
P
4
(
x
):
±
0
.
33998
,
±
0
.
86115
P
5
(
x
): 0
,
±
0
.
53847
,
±
0
.
90618
P
6
(
x
):
±
0
.
23862
,
±
0
.
66121
,
±
0
.
93247
P
7
(
x
): 0
,
±
0
.
40585
,
±
0
.
74153
,
±
0
.
94911
P
10
(
x
):
±
0
.
14887
,
±
0
.
43340
,
±
0
.
67941
,
±
0
.
86506
,
±
0
.
097391
The zeros of any Legendre polynomial are in the interval (
−
1
,
1) and are symmetric with respect to 0.
37.
The recurrence relation can be written
P
k
+1
(
x
)=
2
k
+1
k
+1
xP
k
(
x
)
−
k
k
+1
P
k
−
1
(
x
)
,k
=2
,
3
,
4
,....
k
=1:
P
2
(
x
)=
3
2
x
2
−
1
2
k
=2:
P
3
(
x
)=
5
3
x
/
3
2
x
2
−
1
2
d
−
2
3
x
=
5
2
x
3
−
3
2
x
k
=3:
P
4
(
x
)=
7
4
x
/
5
2
x
3
−
3
2
x
d
−
3
4
/
3
2
x
2
−
1
2
d
=
35
8
x
4
−
30
8
x
2
+
3
8
k
=4:
P
5
(
x
)=
9
5
x
/
35
8
x
4
−
30
8
x
2
+
3
8
d
−
4
5
/
5
2
x
3
−
3
2
x
d
=
63
8
x
5
−
35
4
x
3
+
15
8
x
k
=5:
P
6
(
x
)=
11
6
x
/
63
8
x
5
−
35
4
x
3
+
15
8
x
d
−
5
6
/
35
8
x
4
−
30
8
x
2
+
3
8
d
=
231
16
x
6
−
315
16
x
4
+
105
16
x
2
−
5
16
k
=6:
P
7
(
x
)=
13
7
x
/
231
16
x
6
−
315
16
x
4
+
105
16
x
2
−
5
16
d
−
5
6
/
63
8
x
5
−
35
4
x
3
+
15
8
x
d
=
429
16
x
7
−
693
16
x
5
+
315
16
x
3
−
35
16
x
38.
If
x
= cos
θ
then
dy
dθ
=
−
sin
θ
dy
dx
,
d
2
y
dθ
2
= sin
2
θ
d
2
y
dx
2
−
cos
dy
dx
,
and
sin
θ
d
2
y
dθ
2
+ cos
θ
dy
dθ
+
n
(
n
+ 1)(sin
θ
)
y
= sin
θ
a

1
−
cos
2
θ

d
2
y
dx
2
−
2 cos
θ
dy
dx
+
n
(
n
+1)
y
)
=0
.
That is,

1
−
x
2

d
2
y
dx
2
−
2
x
dy
dx
+
n
(
n
+1)
y
=0
.
255

256. Chapter 5 Review Exercises
Chapter 5 Review Exercises
1.
The interval of convergence is centered at 4. Since the series converges at
−
2, it converges at least on the interval
[
−
2
,
10). Since it diverges at 13, it converges at most on the interval [
−
5
,
13). Thus at 10 it might converge; at
7 it does converge; at
−
7 it does not converge: and at 11 it might converge.
2.
We have
f
(
x
)=
sin
x
cos
x
=
x
−
x
3
6
+
x
5
120
−···
1
−
x
2
2
+
x
4
24
−···
=
x
+
x
3
3
+
2
x
5
15
+
···
.
3.
Solving
x
2
−
2
x
+ 10 = 0 we obtain
x
=1
±
√
11 , which are singular points. Thus, the minimum radius of
convergence is
|
1
−
√
11
|
=
√
11
−
1.
4.
Setting 1
−
sin
x
= 0 we see the singular points closest to 0 are
−
3
π/
2 and
π/
2. Thus, the minimum radius of
convergence is
π/
2.
5.
The differential equation (
x
3
−
x
2
)
y
//
+
y
/
+
y
= 0 has a regular singular point at
x
= 1 and an irregular singular
point at
x
=1.
6.
The differential equation (
x
−
1)(
x
+3)
y
//
+
y
= 0 has regular singular points at
x
= 1 and
x
=
−
3.
7.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation we obtain
2
xy
//
+
y
/
+
y
=

2
r
2
−
r

c
0
x
r
−
1
+
∞
x
k
=1
[2(
k
+
r
)(
k
+
r
−
1)
c
k
+(
k
+
r
)
c
k
+
c
k
−
1
]
x
k
+
r
−
1
=0
which implies
2
r
2
−
r
=
r
(2
r
−
1)=0
and
(
k
+
r
)(2
k
+2
r
−
1)
c
k
+
c
k
−
1
=0
.
The indicial roots are
r
= 0 and
r
=1
/
2. For
r
= 0 the recurrence relation is
c
k
=
−
c
k
−
1
k
(2
k
−
1)
,k
=1
,
2
,
3
,...,
so
c
1
=
−
c
0
,c
2
=
1
6
c
0
,c
3
=
−
1
90
c
0
.
For
r
=1
/
2 the recurrence relation is
c
k
=
−
c
k
−
1
k
(2
k
+1)
,k
=1
,
2
,
3
,...,
so
c
1
=
−
1
3
c
0
,c
2
=
1
30
c
0
,c
3
=
−
1
630
c
0
.
Two linearly independent solutions are
y
1
=
C
1
/
1
−
x
+
1
6
x
2
−
1
90
x
3
+
···
d
and
y
2
=
C
2
x
1
/
2
/
1
−
1
3
x
+
1
30
x
2
−
1
630
x
3
+
···
d
.
256

257. Chapter 5 Review Exercises
8.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
−
xy
/
−
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
−
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
−
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
−
∞
x
k
=1
kc
k
x
k
−
∞
x
k
=0
c
k
x
k
=2
c
2
−
c
0
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
−
(
k
+1)
c
k
]
x
k
=0
.
Thus
2
c
2
−
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
−
(
k
+1)
c
k
=0
and
c
2
=
1
2
c
0
c
k
+2
=
1
k
+2
c
k
,k
=1
,
2
,
3
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
1
2
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
1
8
c
6
=
1
48
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
1
3
c
5
=
1
15
c
7
=
1
105
and so on. Thus, two solutions are
y
1
=1+
1
2
x
2
+
1
8
x
4
+
1
48
x
6
+
···
and
y
2
=
x
+
1
3
x
3
+
1
15
x
5
+
1
105
x
7
+
···
.
9.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we obtain
(
x
−
1)
y
//
+3
y
=(
−
2
c
2
+3
c
0
)+
∞
x
k
=3
(
k
−
1)(
k
−
2)
c
k
−
1
−
k
(
k
−
1)
c
k
+3
c
k
−
2
]
x
k
−
2
=0
which implies
c
2
=3
c
0
/
2 and
c
k
=
(
k
−
1)(
k
−
2)
c
k
−
1
+3
c
k
−
2
k
(
k
−
1)
,k
=3
,
4
,
5
,... .
257

258. Chapter 5 Review Exercises
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
3
2
,c
3
=
1
2
,c
4
=
5
8
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=0
,c
3
=
1
2
,c
4
=
1
4
and so on. Thus, two solutions are
y
1
=
C
1
/
1+
3
2
x
2
+
1
2
x
3
+
5
8
x
4
+
···
d
and
y
2
=
C
2
/
x
+
1
2
x
3
+
1
4
x
4
+
···
d
.
10.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we obtain
y
//
−
x
2
y
/
+
xy
=2
c
2
+(6
c
3
+
c
0
)
x
+
∞
x
k
=1
[(
k
+ 3)(
k
+2)
c
k
+3
−
(
k
−
1)
c
k
]
x
k
+1
=0
which implies
c
2
=0,
c
3
=
−
c
0
/
6, and
c
k
+3
=
k
−
1
(
k
+ 3)(
k
+2)
c
k
,k
=1
,
2
,
3
,... .
Choosing
c
0
= 1 and
c
1
= 0 we find
c
3
=
−
1
6
c
4
=
c
7
=
c
10
=
···
=0
c
5
=
c
8
=
c
11
=
···
=0
c
6
=
−
1
90
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
3
=
c
6
=
c
9
=
···
=0
c
4
=
c
7
=
c
10
=
···
=0
c
5
=
c
8
=
c
11
=
···
=0
and so on. Thus, two solutions are
y
1
=
c
0
/
1
−
1
6
x
3
−
1
90
x
6
−···
d
and
y
2
=
c
1
x.
11.
Substituting
y
=
b
∞
n
=0
c
n
x
n
+
r
into the differential equation and collecting terms, we obtain
xy
//
+(
x
−
6)
y
/
−
3
y
=(
r
2
−
7
r
)
c
0
x
r
−
1
+
∞
x
k
=1

(
k
+
r
)(
k
+
r
−
1)
c
k
+(
k
+
r
−
1)
c
k
−
1
−
6(
k
+
r
)
c
k
−
3
c
k
−
1

x
k
+
r
−
1
=0
,
which implies
r
2
−
7
r
=
r
(
r
−
7)=0
and
(
k
+
r
)(
k
+
r
−
7)
c
k
+(
k
+
r
−
4)
c
k
−
1
=0
.
258

259. Chapter 5 Review Exercises
The indicial roots are
r
1
= 7 and
r
2
=0. For
r
2
= 0 the recurrence relation is
k
(
k
−
7)
c
k
+(
k
−
4)
c
k
−
1
=0
,k
=1
,
2
,
3
,... .
Then
−
6
c
1
−
3
c
0
=0
−
10
c
2
−
2
c
1
=0
−
12
c
3
−
c
2
=0
−
12
c
4
+0
c
3
=0 =
⇒
c
4
=0
−
10
c
5
+
c
4
=0 =
⇒
c
5
=0
−
6
c
6
+2
c
5
=0 =
⇒
c
6
=0
0
c
7
+3
c
6
=0 =
⇒
c
7
is arbitrary
and
c
k
=
−
k
−
4
k
(
k
−
7)
c
k
−
1
k
=8
,
9
,
10
,... .
Taking
c
0
/
= 0 and
c
7
= 0 we obtain
c
1
=
−
1
2
c
0
c
2
=
1
10
c
0
c
3
=
−
1
120
c
0
c
4
=
c
5
=
c
6
=
···
=0
.
Taking
c
0
= 0 and
c
7
/
= 0 we obtain
c
1
=
c
2
=
c
3
=
c
4
=
c
5
=
c
6
=0
c
8
=
−
1
2
c
7
c
9
=
5
36
c
7
c
10
=
−
1
36
c
7
.
In this case we obtain the two solutions
y
1
=1
−
1
2
x
+
1
10
x
2
−
1
120
x
3
and
y
2
=
x
7
−
1
2
x
8
+
5
36
x
9
−
1
36
x
10
+
···
.
12.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
(cos
x
)
y
//
+
y
=
/
1
−
1
2
x
2
+
1
24
x
4
−
1
720
x
6
+
···
d
(2
c
2
+6
c
3
x
+12
c
4
x
2
+20
c
5
x
3
+30
c
6
x
4
+
···
)
+
∞
x
n
=0
c
n
x
n
=
a
2
c
2
+6
c
3
x
+ (12
c
4
−
c
2
)
x
2
+ (20
c
5
−
3
c
3
)
x
3
+
/
30
c
6
−
6
c
4
+
1
12
c
2
d
x
4
+
···
)
+[
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
···
]
=(
c
0
+2
c
2
)+(
c
1
+6
c
3
)
x
+12
c
4
x
2
+ (20
c
5
−
2
c
3
)
x
3
+
/
30
c
6
−
5
c
4
+
1
12
c
2
d
x
4
+
···
=0
.
259

260. Chapter 5 Review Exercises
Thus
c
0
+2
c
2
=0
c
1
+6
c
3
=0
12
c
4
=0
20
c
5
−
2
c
3
=0
30
c
6
−
5
c
4
+
1
12
c
2
=0
and
c
2
=
−
1
2
c
0
c
3
=
−
1
6
c
1
c
4
=0
c
5
=
1
10
c
3
c
6
=
1
6
c
4
−
1
360
c
2
.
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
1
2
,c
3
=0
,c
4
=0
,c
5
=0
,c
6
=
1
720
and so on. For
c
0
= 0 and
c
1
= 1 we find
c
2
=0
,c
3
=
−
1
6
,c
4
=0
,c
5
=
−
1
60
,c
6
=0
and so on. Thus, two solutions are
y
1
=1
−
1
2
x
2
+
1
720
x
6
+
···
and
y
2
=
x
−
1
6
x
3
−
1
60
x
5
+
···
.
13.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
+
xy
/
+2
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=1
nc
n
x
n
r
νb
k
=
n
+2
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=1
kc
k
x
k
+2
∞
x
k
=0
c
k
x
k
=2
c
2
+2
c
0
+
∞
x
k
=1
[(
k
+ 2)(
k
+1)
c
k
+2
+(
k
+2)
c
k
]
x
k
=0
.
Thus
2
c
2
+2
c
0
=0
(
k
+ 2)(
k
+1)
c
k
+2
+(
k
+2)
c
k
=0
and
c
2
=
−
c
0
c
k
+2
=
−
1
k
+1
c
k
,k
=1
,
2
,
3
,... .
260

261. Chapter 5 Review Exercises
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
1
c
3
=
c
5
=
c
7
=
···
=0
c
4
=
1
3
c
6
=
−
1
15
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=
c
4
=
c
6
=
···
=0
c
3
=
−
1
2
c
5
=
1
8
c
7
=
−
1
48
and so on. Thus, the general solution is
y
=
c
0
/
1
−
x
2
+
1
3
x
4
−
1
15
x
6
+
···
d
+
c
1
/
x
−
1
2
x
3
+
1
8
x
5
−
1
48
x
7
+
···
d
and
y
/
=
c
0
/
−
2
x
+
4
3
x
3
−
2
5
x
5
+
···
d
+
c
1
/
1
−
3
2
x
2
+
5
8
x
4
−
7
48
x
6
+
···
d
.
Setting
y
(0) = 3 and
y
/
(0) =
−
2 we find
c
0
= 3 and
c
1
=
−
2. Therefore, the solution of the initial-value problem
is
y
=3
−
2
x
−
3
x
2
+
x
3
+
x
4
−
1
4
x
5
−
1
5
x
6
+
1
24
x
7
+
···
.
14.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
(
x
+2)
y
//
+3
y
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
1
r
νb
k
=
n
−
1
+2
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+3
∞
x
n
=0
c
n
x
n
r
νb
k
=
n
=
∞
x
k
=1
(
k
+1)
kc
k
+1
x
k
+2
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+3
∞
x
k
=0
c
k
x
k
=4
c
2
+3
c
0
+
∞
x
k
=1
[(
k
+1)
kc
k
+1
+2(
k
+ 2)(
k
+1)
c
k
+2
+3
c
k
]
x
k
=0
.
Thus
4
c
2
+3
c
0
=0
(
k
+1)
kc
k
+1
+2(
k
+ 2)(
k
+1)
c
k
+2
+3
c
k
=0
and
c
2
=
−
3
4
c
0
c
k
+2
=
−
k
2(
k
+2)
c
k
+1
−
3
2(
k
+ 2)(
k
+1)
c
k
,k
=1
,
2
,
3
,... .
261

262. Chapter 5 Review Exercises
Choosing
c
0
= 1 and
c
1
= 0 we find
c
2
=
−
3
4
c
3
=
1
8
c
4
=
1
16
c
5
=
−
9
320
and so on. For
c
0
= 0 and
c
1
= 1 we obtain
c
2
=0
c
3
=
−
1
4
c
4
=
1
16
c
5
=0
and so on. Thus, the general solution is
y
=
c
0
/
1
−
3
4
x
2
+
1
8
x
3
+
1
16
x
4
−
9
320
x
5
+
···
d
+
c
1
/
x
−
1
4
x
3
+
1
16
x
4
+
···
d
and
y
/
=
c
0
/
−
3
2
x
+
3
8
x
2
+
1
4
x
3
−
9
64
x
4
+
···
d
+
c
1
/
1
−
3
4
x
2
+
1
4
x
3
+
···
d
.
Setting
y
(0) = 0 and
y
/
(0) = 1 we find
c
0
= 0 and
c
1
= 1. Therefore, the solution of the initial-value problem is
y
=
x
−
1
4
x
3
+
1
16
x
4
+
···
.
15.
Writing the differential equation in the form
y
//
+
/
1
−
cos
x
x
d
y
/
+
xy
=0
,
and noting that
1
−
cos
x
x
=
x
2
−
x
3
24
+
x
5
720
−···
is analytic at
x
= 0, we conclude that
x
= 0 is an ordinary point of the differential equation.
16.
Writing the differential equation in the form
y
//
+
/
x
e
x
−
1
−
x
d
y
=0
and noting that
x
e
x
−
1
−
x
=
2
x
−
2
3
+
x
18
+
x
2
270
−···
we see that
x
= 0 is a singular point of the differential equation. Since
x
2
/
x
e
x
−
1
−
x
d
=2
x
−
2
x
2
3
+
x
3
18
+
x
4
270
−···
,
we conclude that
x
= 0 is a regular singular point.
262

263. Chapter 5 Review Exercises
17.
Substituting
y
=
b
∞
n
=0
c
n
x
n
into the differential equation we have
y
//
+
x
2
y
/
+2
xy
=
∞
x
n
=2
n
(
n
−
1)
c
n
x
n
−
2
r
νb
k
=
n
−
2
+
∞
x
n
=1
nc
n
x
n
+1
r
νb
k
=
n
+1
+2
∞
x
n
=0
c
n
x
n
+1
r
νb
k
=
n
+1
=
∞
x
k
=0
(
k
+ 2)(
k
+1)
c
k
+2
x
k
+
∞
x
k
=2
(
k
−
1)
c
k
−
1
x
k
+2
∞
x
k
=1
c
k
−
1
x
k
=2
c
2
+(6
c
3
+2
c
0
)
x
+
∞
x
k
=2
[(
k
+ 2)(
k
+1)
c
k
+2
+(
k
+1)
c
k
−
1
]
x
k
=5
−
2
x
+10
x
3
.
Thus
2
c
2
=5
6
c
3
+2
c
0
=
−
2
12
c
4
+3
c
1
=0
20
c
5
+4
c
2
=10
(
k
+ 2)(
k
+1)
c
k
+2
+(
k
+1)
c
k
−
1
=0
,k
=4
,
5
,
6
,...,
and
c
2
=
5
2
c
3
=
−
1
3
c
0
−
1
3
c
4
=
−
1
4
c
1
c
5
=
1
2
−
1
5
c
2
=
1
2
−
1
5
/
5
2
d
=0
c
k
+2
=
−
1
k
+2
c
k
−
1
.
Using the recurrence relation, we find
c
6
=
−
1
6
c
3
=
1
3
·
6
(
c
0
+1)=
1
3
2
·
2!
c
0
+
1
3
2
·
2!
c
7
=
−
1
7
c
4
=
1
4
·
7
c
1
c
8
=
c
11
=
c
14
=
···
=0
c
9
=
−
1
9
c
6
=
−
1
3
3
·
3!
c
0
−
1
3
3
·
3!
c
10
=
−
1
10
c
7
=
−
1
4
·
7
·
10
c
1
c
12
=
−
1
12
c
9
=
1
3
4
·
4!
c
0
+
1
3
4
·
4!
c
13
=
−
1
13
c
0
=
1
4
·
7
·
10
·
13
c
1
263

264. Chapter 5 Review Exercises
and so on. Thus
y
=
c
0
a
1
−
1
3
x
3
+
1
3
2
·
2!
x
6
−
1
3
3
·
3!
x
9
+
1
3
4
·
4!
x
12
−···
)
+
c
1
a
x
−
1
4
x
4
+
1
4
·
7
x
7
−
1
4
·
7
·
10
x
10
+
1
4
·
7
·
10
·
13
x
13
−···
)
+
a
5
2
x
2
−
1
3
x
3
+
1
3
2
·
2!
x
6
−
1
3
3
·
3!
x
9
+
1
3
4
·
4!
x
12
−···
)
.
18. (a)
From
y
=
−
1
u
du
dx
we obtain
dy
dx
=
−
1
u
d
2
u
dx
2
+
1
u
2
/
du
dx
d
2
.
Then
dy/dx
=
x
2
+
y
2
becomes
−
1
u
d
2
u
dx
2
+
1
u
2
/
du
dx
d
2
=
x
2
+
1
u
2
/
du
dx
d
2
,
so
d
2
u
dx
2
+
x
2
u
=0.
(b)
If
u
=
x
1
/
2
w
(
1
2
x
2
) then
u
/
=
x
3
/
2
w
/
/
1
2
x
2
d
+
1
2
x
−
1
/
2
w
/
1
2
x
2
d
and
u
//
=
x
5
/
2
w
//
/
1
2
x
2
d
+2
x
1
/
2
w
/
/
1
2
x
2
d
−
1
4
x
−
3
/
2
w
/
1
2
x
2
d
,
so
u
//
+
x
2
u
=
x
1
/
2
a
x
2
w
//
/
1
2
x
2
d
+2
w
/
/
1
2
x
2
d
+
/
x
2
−
1
4
x
−
2
d
w
/
1
2
x
2
3)
=0
.
Letting
t
=
1
2
x
2
we have
√
2
t
a
2
tw
//
(
t
)+2
w
/
(
t
)+
/
2
t
−
1
4
·
2
t
d
w
(
t
)
)
=0
or
t
2
w
//
(
t
)+
tw
/
(
t
)+
/
t
2
−
1
16
d
w
(
t
)=0
.
This is Bessel’s equation with
ν
=1
/
4, so
w
(
t
)=
c
1
J
1
/
4
(
t
)+
c
2
J
−
1
/
4
(
t
)
.
(c)
We have
y
=
−
1
u
du
dx
=
−
1
x
1
/
2
w
(
t
)
d
dx
x
1
/
2
w
(
t
)
=
−
1
x
1
/
2
w
a
x
1
/
2
dw
dt
dt
dx
+
1
2
x
−
1
/
2
w
)
=
−
1
x
1
/
2
w
a
x
3
/
2
dw
dt
+
1
2
x
1
/
2
w
)
=
−
1
2
xw
a
2
x
2
dw
dt
+
w
)
=
−
1
2
xw
a
4
t
dw
dt
+
w
)
.
264

265. Chapter 5 Review Exercises
Now
4
t
dw
dt
+
w
=4
t
d
dt
[
c
1
J
1
/
4
(
t
)+
c
2
J
−
1
/
4
(
t
)] +
c
1
J
1
/
4
(
t
)+
c
2
J
−
1
/
4
(
t
)
=4
t
a
c
1
/
J
−
3
/
4
(
t
)
−
1
4
t
J
1
/
4
(
t
)
d
+
c
2
/
−
1
4
t
J
−
1
/
4
(
t
)
−
J
3
/
4
(
t
)
3)
+
c
1
J
1
/
4
(
t
)+
c
2
J
−
1
/
4
(
t
)
=4
c
1
tJ
−
3
/
4
(
t
)
−
4
c
2
tJ
3
/
4
(
t
)
=2
c
1
x
2
J
−
3
/
4
/
1
2
x
2
d
−
2
c
2
x
2
J
3
/
4
/
1
2
x
2
d
,
so
y
=
−
2
c
1
x
2
J
−
3
/
4
(
1
2
x
2
)
−
2
c
2
x
2
J
3
/
4
(
1
2
x
2
)
2
x
[
c
1
J
1
/
4
(
1
2
x
2
)+
c
2
J
−
1
/
4
(
1
2
x
2
)]
=
x
−
c
1
J
−
3
/
4
(
1
2
x
2
)+
c
2
J
3
/
4
(
1
2
x
2
)
c
1
J
1
/
4
(
1
2
x
2
)+
c
2
J
−
1
/
4
(
1
2
x
2
)
.
Letting
c
=
c
1
/c
2
we have
y
=
x
J
3
/
4
(
1
2
x
2
)
−
cJ
−
3
/
4
(
1
2
x
2
)
cJ
1
/
4
(
1
2
x
2
)+
J
−
1
/
4
(
1
2
x
2
)
.
19.
Let
y
2
=
1
2
x
[ln(1 +
x
)
−
ln(1
−
x
)]
−
1
so that
y
/
2
=
1
2
x
a
1
1+
x
+
1
1
−
x
)
+
1
2
[ln(1 +
x
)
−
ln(1
−
x
)]
and
y
//
2
=
1
2
x
a
−
1
(1 +
x
)
2
+
1
(1
−
x
)
2
)
+
1
2
a
1
1+
x
+
1
1
−
x
)
+
1
2
a
1
1+
x
+
1
1
−
x
)
=
1
2
x
a
−
1
(1 +
x
)
2
+
1
(1
−
x
)
2
)
+
1
1+
x
+
1
1
−
x
.
Then
(1
−
x
)(1 +
x
)
y
//
2
−
2
xy
/
2
+2
y
2
=0
.
20.
n
=0:
P
0
(
x
)=1
n
=1:
P
1
(
x
)=
1
2
d
dx
(
x
2
−
1) =
x
n
=2:
P
2
(
x
)=
1
8
d
2
dx
2
(
x
2
−
1)
2
=
1
8
d
2
dx
2
(
x
4
−
2
x
2
+1)=
1
8
(12
x
2
−
4) =
3
2
x
2
−
1
2
n
=3:
P
3
(
x
)=
1
48
d
3
dx
3
(
x
2
−
1)
3
=
1
48
d
3
dx
3
(
x
6
−
3
x
4
+3
x
2
−
3) =
1
48
(120
x
3
−
72
x
)=
5
2
x
3
−
3
2
x
265

266. h =
0.1
h =
0.05
xy
x(n) y(n)
1.00 5.0000 1.00 5.0000
1.10 3.9900 1.05 4.4475
1.20 3.2546 1.10 3.9763
1.30 2.7236 1.15 3.5751
1.40 2.3451 1.20 3.2342
1.50 2.0801 1.25 2.9452
1.30 2.7009
1.35 2.4952
1.40 2.3226
1.45 2.1786
1.50 2.0592
h =
0.1
h =
0.05
x(n) y(n) x(n) y(n)
0.00 2.0000 0.00 2.0000
0.10 1.6600 0.05 1.8150
0.20 1.4172 0.10 1.6571
0.30 1.2541 0.15 1.5237
0.40 1.1564 0.20 1.4124
0.50 1.1122 0.25 1.3212
0.30 1.2482
0.35 1.1916
0.40 1.1499
0.45 1.1217
0.50 1.1056
h =
0.1
h =
0.05
x(n) y(n) x(n) y(n)
0.00 0.0000 0.00 0.0000
0.10 0.1005 0.05 0.0501
0.20 0.2030 0.10 0.1004
0.30 0.3098 0.15 0.1512
0.40 0.4234 0.20 0.2028
0.50 0.5470 0.25 0.2554
0.30 0.3095
0.35 0.3652
0.40 0.4230
0.45 0.4832
0.50 0.5465
6
Numerical Solutions of
Ordinary Differential Equations
Exercises 6.1
All tables in this chapter were constructed in a spreadsheet program which does not support subscripts. Consequently,
x
n
and
y
n
will be indicated as
x
(
n
) and
y
(
n
), respectively.
1.
2.
3.
266

267. h =
0.1
h =
0.05
x(n) y(n) x(n) y(n)
0.00 1.0000 0.00 1.0000
0.10 1.1110 0.05 1.0526
0.20 1.2515 0.10 1.1113
0.30 1.4361 0.15 1.1775
0.40 1.6880 0.20 1.2526
0.50 2.0488 0.25 1.3388
0.30 1.4387
0.35 1.5556
0.40 1.6939
0.45 1.8598
0.50 2.0619
h =
0.1
h =
0.05
x(n) y(n) x(n) y(n)
0.00 0.0000 0.00 0.0000
0.10 0.0952 0.05 0.0488
0.20 0.1822 0.10 0.0953
0.30 0.2622 0.15 0.1397
0.40 0.3363 0.20 0.1823
0.50 0.4053 0.25 0.2231
0.30 0.2623
0.35 0.3001
0.40 0.3364
0.45 0.3715
0.50 0.4054
h =
0.1
h =
0.05
x(n) y(n) x(n) y(n)
0.00 0.0000 0.00 0.0000
0.10 0.0050 0.05 0.0013
0.20 0.0200 0.10 0.0050
0.30 0.0451 0.15 0.0113
0.40 0.0805 0.20 0.0200
0.50 0.1266 0.25 0.0313
0.30 0.0451
0.35 0.0615
0.40 0.0805
0.45 0.1022
0.50 0.1266
h =
0.1
h =
0.05
x(n) y(n) x(n) y(n)
0.00 0.5000 0.00 0.5000
0.10 0.5215 0.05 0.5116
0.20 0.5362 0.10 0.5214
0.30 0.5449 0.15 0.5294
0.40 0.5490 0.20 0.5359
0.50 0.5503 0.25 0.5408
0.30 0.5444
0.35 0.5469
0.40 0.5484
0.45 0.5492
0.50 0.5495
Exercises 6.1
4.
5.
6.
7.
267

268. h =
0.1
h =
0.05
x(n) y(n) x(n) y(n)
0.00 1.0000 0.00 1.0000
0.10 1.1079 0.05 1.0519
0.20 1.2337 0.10 1.1079
0.30 1.3806 0.15 1.1684
0.40 1.5529 0.20 1.2337
0.50 1.7557 0.25 1.3043
0.30 1.3807
0.35 1.4634
0.40 1.5530
0.45 1.6503
0.50 1.7560
h =
0.1
h =
0.05
x(n) y(n) x(n) y(n)
1.00 1.0000 1.00 1.0000
1.10 1.0095 1.05 1.0024
1.20 1.0404 1.10 1.0100
1.30 1.0967 1.15 1.0228
1.40 1.1866 1.20 1.0414
1.50 1.3260 1.25 1.0663
1.30 1.0984
1.35 1.1389
1.40 1.1895
1.45 1.2526
1.50 1.3315
h =
0.1
h =
0.05
x(n) y(n} x(n) y(n}
0.00
0.5000
0.00
0.5000
0.10
0.5250
0.05
0.5125
0.20
0.5498
0.10
0.5250
0.30
0.5744
0.15
0.5374
0.40
0.5986
0.20
0.5498
0.50
0.6224
0.25
0.5622
0.30
0.5744
0.35
0.5866
0.40
0.5987
0.45
0.6106
0.50
0.6224
h = 0.1 h = 0.05
x(n) y(n}
exact
x(n) y(n}
exact
0.00
2.0000 2.0000
0.00
2.0000 2.0000
0.10
2.1220 2.1230
0.05
2.0553 2.0554
0.20
2.3049 2.3085
0.10
2.1228 2.1230
0.30
2.5858 2.5958
0.15
2.2056 2.2061
0.40
3.0378 3.0650
0.20
2.3075 2.3085
0.50
3.8254 3.9082
0.25
2.4342 2.4358
0.30
2.5931 2.5958
0.35
2.7953 2.7997
0.40
3.0574 3.0650
0.45
3.4057 3.4189
0.50
3.8840 3.9082
Exercises 6.1
8.
9.
10.
11.
268

269. 1.11.21.31.4
x
5
10
15
20
y
IMPROVED
h=0.1 EULER EULER
x(n) y(n) y(n)
1.00 1.0000 1.0000
1.10 1.2000 1.2469
1.20 1.4938 1.6668
1.30 1.9711 2.6427
1.40 2.9060 8.7988
Exercises 6.1
12. (a) (b)
13. (a)
Using the Euler method we obtain
y
(0
.
1)
≈
y
1
=1
.
2.
(b)
Using
y
≈≈
=4
e
2
x
we see that the local truncation error is
y
≈≈
(
c
)
h
2
2
=4
e
2
c
(0
.
1)
2
2
=0
.
02
e
2
c
.
Since
e
2
x
is an increasing function,
e
2
c
≤
e
2(0
.
1)
=
e
0
.
2
for 0
≤
c
≤
0
.
1. Thus an upper bound for the local
truncation error is 0
.
02
e
0
.
2
=0
.
0244.
(c)
Since
y
(0
.
1)=
e
0
.
2
=1
.
2214, the actual error is
y
(0
.
1)
−
y
1
=0
.
0214, which is less than 0
.
0244.
(d)
Using the Euler method with
h
=0
.
05 we obtain
y
(0
.
1)
≈
y
2
=1
.
21.
(e)
The error in (d) is 1
.
2214
−
1
.
21= 0
.
0114. With global truncation error
O
(
h
), when the step size is halved
we expect the error for
h
=0
.
05 to be one-half the error when
h
=0
.
1. Comparing 0
.
0114 with 0
.
214 we
see that this is the case.
14. (a)
Using the improved Euler method we obtain
y
(0
.
1)
≈
y
1
=1
.
22.
(b)
Using
y
≈≈≈
=8
e
2
x
we see that the local truncation error is
y
≈≈≈
(
c
)
h
3
6
=8
e
2
c
(0
.
1)
3
6
=0
.
001333
e
2
c
.
Since
e
2
x
is an increasing function,
e
2
c
≤
e
2(0
.
1)
=
e
0
.
2
for 0
≤
c
≤
0
.
1. Thus an upper bound for the local
truncation error is 0
.
001333
e
0
.
2
=0
.
001628.
(c)
Since
y
(0
.
1)=
e
0
.
2
=1
.
221403, the actual error is
y
(0
.
1)
−
y
1
=0
.
001403 which is less than 0
.
001628.
(d)
Using the improved Euler method with
h
=0
.
05 we obtain
y
(0
.
1)
≈
y
2
=1
.
221025.
(e)
The error in (d) is 1
.
221403
−
1
.
221025 = 0
.
000378. With global truncation error
O
(
h
2
), when the step size
is halved we expect the error for
h
=0
.
05 to be one-fourth the error for
h
=0
.
1. Comparing 0
.
000378 with
0
.
001403 we see that this is the case.
15. (a)
Using the Euler method we obtain
y
(0
.
1)
≈
y
1
=0
.
8.
(b)
Using
y
≈≈
=5
e
−
2
x
we see that the local truncation error is
5
e
−
2
c
(0
.
1)
2
2
=0
.
025
e
−
2
c
Since
e
−
2
x
is a decreasing function,
e
−
2
c
≤
e
0
= 1for 0
≤
c
≤
0
.
1. Thus an upper bound for the local
truncation error is 0
.
025(1) = 0
.
025.
(c)
Since
y
(0
.
1)=0
.
8234, the actual error is
y
(0
.
1)
−
y
1
=0
.
0234, which is less than 0
.
025.
(d)
Using the Euler method with
h
=0
.
05 we obtain
y
(0
.
1)
≈
y
2
=0
.
8125.
(e)
The error in (d) is 0
.
8234
−
0
.
8125 = 0
.
0109. With global truncation error
O
(
h
), when the step size is
halved we expect the error for
h
=0
.
05 to be one-half the error when
h
=0
.
1. Comparing 0
.
0109 with
0
.
0234 we see that this is the case.
269

270. Exercises 6.1
16. (a)
Using the improved Euler method we obtain
y
(0
.
1)
≈
y
1
=0
.
825.
(b)
Using
y
≈≈≈
=
−
10
e
−
2
x
we see that the local truncation error is
10
e
−
2
c
(0
.
1)
3
6
=0
.
001667
e
−
2
c
.
Since
e
−
2
x
is a decreasing function,
e
−
2
c
≤
e
0
= 1for 0
≤
c
≤
0
.
1. Thus an upper bound for the local
truncation error is 0
.
001667(1) = 0
.
001667.
(c)
Since
y
(0
.
1)=0
.
823413, the actual error is
y
(0
.
1)
−
y
1
=0
.
001587, which is less than 0
.
001667.
(d)
Using the improved Euler method with
h
=0
.
05 we obtain
y
(0
.
1)
≈
y
2
=0
.
823781.
(e)
The error in (d) is
|
0
.
823413
−
0
.
8237181
|
=0
.
000305. With global truncation error
O
(
h
2
), when the step
size is halved we expect the error for
h
=0
.
05 to be one-fourth the error when
h
=0
.
1. Comparing 0
.
000305
with 0
.
001587 we see that this is the case.
17. (a)
Using
y
≈≈
=38
e
−
3(
x
−
1)
we see that the local truncation error is
y
≈≈
(
c
)
h
2
2
=38
e
−
3(
c
−
1)
h
2
2
=19
h
2
e
−
3(
c
−
1)
.
(b)
Since
e
−
3(
x
−
1)
is a decreasing function for 1
≤
x
≤
1
.
5,
e
−
3(
c
−
1)
≤
e
−
3(1
−
1)
= 1for 1
≤
c
≤
1
.
5 and
y
≈≈
(
c
)
h
2
2
≤
19(0
.
1)
2
(1) = 0
.
19
.
(c)
Using the Euler method with
h
=0
.
1we obtain
y
(1
.
5)
≈
1
.
8207. With
h
=0
.
05 we obtain
y
(1
.
5)
≈
1
.
9424.
(d)
Since
y
(1
.
5) = 2
.
0532, the error for
h
=0
.
1is
E
0
.
1
=0
.
2325, while the error for
h
=0
.
05 is
E
0
.
05
=0
.
1109.
With global truncation error
O
(
h
) we expect
E
0
.
1
/E
0
.
05
≈
2. We actually have
E
0
.
1
/E
0
.
05
=2
.
10.
18. (a)
Using
y
≈≈≈
=
−
114
e
−
3(
x
−
1)
we see that the local truncation error is
−
−
−
−
y
≈≈≈
(
c
)
h
3
6
−
−
−
−
=114
e
−
3(
x
−
1)
h
3
6
=19
h
3
e
−
3(
c
−
1)
.
(b)
Since
e
−
3(
x
−
1)
is a decreasing function for 1
≤
x
≤
1
.
5,
e
−
3(
c
−
1)
≤
e
−
3(1
−
1)
= 1for 1
≤
c
≤
1
.
5 and
−
−
−
−
y
≈≈≈
(
c
)
h
3
6
−
−
−
−
≤
19(0
.
1)
3
(1) = 0
.
019
.
(c)
Using the improved Euler method with
h
=0
.
1we obtain
y
(1
.
5)
≈
2
.
080108. With
h
=0
.
05 we obtain
y
(1
.
5)
≈
2
.
059166.
(d)
Since
y
(1
.
5)=2
.
053216, the error for
h
=0
.
1is
E
0
.
1
=0
.
026892, while the error for
h
=0
.
05 is
E
0
.
05
=
0
.
005950. With global truncation error
O
(
h
2
) we expect
E
0
.
1
/E
0
.
05
≈
4. We actually have
E
0
.
1
/E
0
.
05
=
4
.
52.
19. (a)
Using
y
≈≈
=
−
1
(
x
+1)
2
we see that the local truncation error is
−
−
−
−
y
≈≈
(
c
)
h
2
2
−
−
−
−
=
1
(
c
+1)
2
h
2
2
.
(b)
Since
1
(
x
+1)
2
is a decreasing function for 0
≤
x
≤
0
.
5,
1
(
c
+1)
2
≤
1
(0+1)
2
= 1for 0
≤
c
≤
0
.
5 and
−
−
−
−
y
≈≈
(
c
)
h
2
2
−
−
−
−
≤
(1)
(0
.
1)
2
2
=0
.
005
.
270

271. h = 0.1
x(n) y(n}
exact
0.00
2.0000 2.0000
0.10
2.1230 2.1230
0.20
2.3085 2.3085
0.30
2.5958 2.5958
0.40
3.0649 3.0650
0.50
3.9078 3.9082
h = 0.1 R-K IMP EULE R
x(n) y(n) y(n)
0.00
2.0000 2.0000
0.10
2.1205 2.1220
0.20
2.3006 2.3049
0.30
2.5759 2.5858
0.40
3.0152 3.0378
0.50
3.7693 3.8254
x(n) y(n)
1.00 5.0000
1.10 3.9724
1.20 3.2284
1.30 2.6945
1.40 2.3163
1.50 2.0533
x(n) y(n)
0.00 2.0000
0.10 1.6562
0.20 1.4110
0.30 1.2465
0.40 1.1480
0.50 1.1037
Exercises 6.2
(c)
Using the Euler method with
h
=0
.
1we obtain
y
(0
.
5)
≈
0
.
4198. With
h
=0
.
05 we obtain
y
(0
.
5)
≈
0
.
4124.
(d)
Since
y
(0
.
5) = 0
.
4055, the error for
h
=0
.
1is
E
0
.
1
=0
.
0143, while the error for
h
=0
.
05 is
E
0
.
05
=0
.
0069.
With global truncation error
O
(
h
) we expect
E
0
.
1
/E
0
.
05
≈
2. We actually have
E
0
.
1
/E
0
.
05
=2
.
06.
20. (a)
Using
y
≈≈≈
=
2
(
x
+1)
3
we see that the local truncation error is
y
≈≈≈
(
c
)
h
3
6
=
1
(
c
+1)
3
h
3
3
.
(b)
Since
1
(
x
+1)
3
is a decreasing function for 0
≤
x
≤
0
.
5,
1
(
c
+1)
3
≤
1
(0+1)
3
= 1for 0
≤
c
≤
0
.
5 and
y
≈≈≈
(
c
)
h
3
6
≤
(1)
(0
.
1)
3
3
=0
.
000333
.
(c)
Using the improved Euler method with
h
=0
.
1we obtain
y
(0
.
5)
≈
0
.
405281. With
h
=0
.
05 we obtain
y
(0
.
5)
≈
0
.
405419.
(d)
Since
y
(0
.
5)=0
.
405465, the error for
h
=0
.
1is
E
0
.
1
=0
.
000184, while the error for
h
=0
.
05 is
E
0
.
05
=
0
.
000046. With global truncation error
O
(
h
2
) we expect
E
0
.
1
/E
0
.
05
≈
4. We actually have
E
0
.
1
/E
0
.
05
=
3
.
98.
21.
Because
y
∗
n
+1
depends on
y
n
and is used to determine
y
n
+1
, all of the
y
∗
n
cannot be computed at one time
independently of the corresponding
y
n
values. For example, the computation of
y
∗
4
involves the value of
y
3
.
Exercises 6.2
1.
2.
Setting
α
=1
/
4wefind
b
=2,
a
=
−
1and
β
=1
/
4. The resulting second-order Runge-Kutta method is
y
n
+1
=
y
n
−
k
1
+2
k
2
k
1
=
hf
(
x
n
,y
n
)
k
2
=
hf
(
x
n
+
h/
4
,y
n
+
k
1
/
4)
.
3. 4.
271

272. x(n) y(n)
0.00 0.0000
0.10 0.1003
0.20 0.2027
0.30 0.3093
0.40 0.4228
0.50 0.5463
x(n) y(n)
0.00 1.0000
0.10 1.1115
0.20 1.2530
0.30 1.4397
0.40 1.6961
0.50 2.0670
x(n) y(n)
0.00 0.0000
0.10 0.0953
0.20 0.1823
0.30 0.2624
0.40 0.3365
0.50 0.4055
x(n) y(n)
0.00 0.0000
0.10 0.0050
0.20 0.0200
0.30 0.0451
0.40 0.0805
0.50 0.1266
x(n) y(n)
0.00 0.5000
0.10 0.5213
0.20 0.5358
0.30 0.5443
0.40 0.5482
0.50 0.5493
x(n) y(n)
0.00 1.0000
0.10 1.1079
0.20 1.2337
0.30 1.3807
0.40 1.5531
0.50 1.7561
x(n) y(n)
1.00 1.0000
1.10 1.0101
1.20 1.0417
1.30 1.0989
1.40 1.1905
1.50 1.3333
x(n) y(n)
0.00 0.5000
0.10 0.5250
0.20 0.5498
0.30 0.5744
0.40 0.5987
0.50 0.6225
t(n) v(n)
0.0 0.0000
1.0 25.2570
2.0 32.9390
3.0 34.9772
4.0 35.5503
5.0 35.7128
012345
t
10
20
30
40
v(t)
Exercises 6.2
5. 6.
7. 8.
9. 10.
11. 12.
13. (a)
Write the equation in the form
dv
dt
=32
−
0
.
025
v
2
=
f
(
t, v
)
.
(b)
(c)
Separating variables and using partial fractions we have
1
2
√
32
∗
1
√
32
−
√
0
.
025
v
+
1
√
32 +
√
0
.
025
v
√
dv
=
dt
and
1
2
√
32
√
0
.
025
√
ln
|
√
32 +
√
0
.
025
v
|−
ln
|
√
32
−
√
0
.
025
v
|
⇒
=
t
+
c.
Since
v
(0) = 0 we find
c
= 0. Solving for
v
we obtain
v
(
t
)=
16
√
5(
e
√
3
.
2
t
−
1)
e
√
3
.
2
t
+1
and
v
(5)
≈
35
.
7678.
272

273. 012345
t
10
20
30
40
50
A(t)
t
(days) 1 2 3 4 5
A
(observed) 2.78 13.53 36.30 47.50 49.40
A
(approximated) 1.93 12.50 36.46 47.23 49.00
A
(exact) 1.95 12.64 36.63 47.32 49.02
Exercises 6.2
14. (a)
See the table in part (c) of this problem.
(b)
From the graph we estimate
A
(1)
≈
1
.
68,
A
(2)
≈
13
.
2,
A
(3)
≈
36
.
8,
A
(4)
≈
46
.
9, and
A
(5)
≈
48
.
9.
(c)
Let
α
=2
.
128 and
β
=0
.
0432. Separating variables we obtain
dA
A
(
α
−
βA
)
=
dt
1
α
∗
1
A
+
β
α
−
βA
√
dA
=
dt
1
α
[ln
A
−
ln(
α
−
βA
)] =
t
+
c
ln
A
α
−
βA
=
α
(
t
+
c
)
A
α
−
βA
=
e
α
(
t
+
c
)
A
=
αe
α
(
t
+
c
)
−
βAe
α
(
t
+
c
)
→
1+
βe
α
(
t
+
c
)
∞
A
=
αe
α
(
t
+
c
)
.
Thus
A
(
t
)=
αe
α
(
t
+
c
)
1+
βe
α
(
t
+
c
)
=
α
β
+
e
−
α
(
t
+
c
)
=
α
β
+
e
−
αc
e
−
αt
.
From
A
(0) = 0
.
24 we obtain
0
.
24 =
α
β
+
e
−
αc
so that
e
−
αc
=
α/
0
.
24
−
β
≈
8
.
8235 and
A
(
t
)
≈
2
.
128
0
.
0432 + 8
.
8235
e
−
2
.
128
t
.
273

274. h = 0.05 h = 0.1
x(n) y(n) y(n)
1.00
1.0000 1.0000
1.05
1.1112
1.10
1.2511 1.2511
1.15
1.4348
1.20
1.6934 1.6934
1.25
2.1047
1.30
2.9560 2.9425
1.35
7.8981
1.40
1.06E+15 903.0282
1.11.21.31.4
x
5
10
15
20
y
Exercises 6.2
15. (a) (b)
16. (a)
Using the fourth-order Runge-Kutta method we obtain
y
(0
.
1)
≈
y
1
=1
.
2214.
(b)
Using
y
(5)
(
x
)=32
e
2
x
we see that the local truncation error is
y
(5)
(
c
)
h
5
120
=32
e
2
c
(0
.
1)
5
120
=0
.
000002667
e
2
c
.
Since
e
2
x
is an increasing function,
e
2
c
≤
e
2(0
.
1)
=
e
0
.
2
for 0
≤
c
≤
0
.
1. Thus an upper bound for the local
truncation error is 0
.
000002667
e
0
.
2
=0
.
000003257.
(c)
Since
y
(0
.
1)=
e
0
.
2
=1
.
221402758, the actual error is
y
(0
.
1)
−
y
1
=0
.
000002758 which is less than
0
.
000003257.
(d)
Using the fourth-order Runge-Kutta formula with
h
=0
.
05 we obtain
y
(0
.
1)
≈
y
2
=1
.
221402571.
(e)
The error in (d) is 1
.
221402758
−
1
.
221402571 = 0
.
000000187. With global truncation error
O
(
h
4
), when
the step size is halved we expect the error for
h
=0
.
05 to be one-sixteenth the error for
h
=0
.
1. Comparing
0
.
000000187 with 0
.
000002758 we see that this is the case.
17. (a)
Using the fourth-order Runge-Kutta method we obtain
y
(0
.
1)
≈
y
1
=0
.
823416667.
(b)
Using
y
(5)
(
x
)=
−
40
e
−
2
x
we see that the local truncation error is
40
e
−
2
c
(0
.
1)
5
120
=0
.
000003333
.
Since
e
−
2
x
is a decreasing function,
e
−
2
c
≤
e
0
= 1for 0
≤
c
≤
0
.
1. Thus an upper bound for the local
truncation error is 0
.
000003333(1) = 0
.
000003333.
(c)
Since
y
(0
.
1)=0
.
823413441, the actual error is
|
y
(0
.
1)
−
y
1
|
=0
.
000003225, which is less than 0
.
000003333.
(d)
Using the fourth-order Runge-Kutta method with
h
=0
.
05 we obtain
y
(0
.
1)
≈
y
2
=0
.
823413627.
(e)
The error in (d) is
|
0
.
823413441
−
0
.
823413627
|
=0
.
000000185. With global truncation error
O
(
h
4
), when
the step size is halved we expect the error for
h
=0
.
05 to be one-sixteenth the error when
h
=0
.
1.
Comparing 0
.
000000185 with 0
.
000003225 we see that this is the case.
18. (a)
Using
y
(5)
=
−
1026
e
−
3(
x
−
1)
we see that the local truncation error is
−
−
−
−
y
(5)
(
c
)
h
5
120
−
−
−
−
=8
.
55
h
5
e
−
3(
c
−
1)
.
(b)
Since
e
−
3(
x
−
1)
is a decreasing function for 1
≤
x
≤
1
.
5,
e
−
3(
c
−
1)
≤
e
−
3(1
−
1)
= 1for 1
≤
c
≤
1
.
5 and
y
(5)
(
c
)
h
5
120
≤
8
.
55(0
.
1)
5
(1) = 0
.
0000855
.
(c)
Using the fourth-order Runge-Kutta method with
h
=0
.
1we obtain
y
(1
.
5)
≈
2
.
053338827. With
h
=0
.
05
we obtain
y
(1
.
5)
≈
2
.
053222989.
274

275. x
y
2
−5
5
x(n) y(n) x(n) y(n)
0.0 0.0000 1.0 4.2147
0.1 0.1440 1.1 3.8033
0.2 0.5448 1.2 3.1513
0.3 1.1409 1.3 2.3076
0.4 1.8559 1.4 1.3390
0.5 2.6049 1.5 0.3243
0.6 3.3019 1.6 -0.6530
0.7 3.8675 1.7 -1.5117
0.8 4.2356 1.8 -2.1809
0.9 4.3593 1.9 -2.6061
1.0 4.2147 2.0 -2.7539
x
y
2
−5
5
Exercises 6.3
19. (a)
Using
y
(5)
=
24
(
x
+1)
5
we see that the local truncation error is
y
(5)
(
c
)
h
5
120
=
1
(
c
+1)
5
h
5
5
.
(b)
Since
1
(
x
+1)
5
is a decreasing function for 0
≤
x
≤
0
.
5,
1
(
c
+1)
5
≤
1
(0+1)
5
= 1for 0
≤
c
≤
0
.
5 and
y
(5)
(
c
)
h
5
5
≤
(1)
(0
.
1)
5
5
=0
.
000002
.
(c)
Using the fourth-order Runge-Kutta method with
h
=0
.
1we obtain
y
(0
.
5)
≈
0
.
405465168. With
h
=0
.
05
we obtain
y
(0
.
5)
≈
0
.
405465111.
20. (a)
For
y

+
y
= 10 sin 3
x
an integrating factor is
e
x
so that
d
dx
[
e
x
y
]=10
e
x
sin 3
x
=
⇒
e
x
y
=
e
x
sin 3
x
−
3
e
x
cos 3
x
+
c
=
⇒
y
= sin 3
x
−
3 cos 3
x
+
ce
−
x
.
When
x
=0,
y
=0,so0=
−
3+
c
and
c
= 3. The solution is
y
= sin 3
x
−
3 cos 3
x
+3
e
−
x
.
Using Newton’s method we find that
x
=1
.
53235 is the only positive root in [0
,
2].
(b)
Using the fourth-order Runge-Kutta method with
h
=0
.
1we obtain the table of values shown. These values
are used to obtain an interpolating function in
Mathematica
. The graph of the interpolating function is
shown. Using
Mathematica
’s root finding capability we see that the only positive root in [0
,
2] is
x
=1
.
53236.
Exercises 6.3
1.
For
y

−
y
=
x
−
1an integrating factor is
e
−

dx
=
e
−
x
, so that
d
dx
[
e
−
x
y
]=(
x
−
1)
e
−
x
and
y
=
e
x
(
−
xe
−
x
+
c
)=
−
x
+
ce
x
.
From
y
(0) = 1we find
c
= 1and
y
=
−
x
+
e
x
. Comparing exact values with approximations obtained in
Example 1, we find
y
(0
.
2)
≈
1
.
02140276 compared to
y
1
=1
.
02140000,
y
(0
.
4)
≈
1
.
09182470 compared to
275

276. Exercises 6.3
y
2
=1
.
09181796,
y
(0
.
6)
≈
1
.
22211880 compared to
y
3
=1
.
22210646, and
y
(0
.
8)
≈
1
.
42554093 compared to
y
4
=1
.
42552788.
2.
100
REM
ADAMS-BASHFORTH/ADAMS-MOULTON
110
REM
METHOD TO SOLVE Y
≈
=FNF(X,Y)
120
REM
DEFINE FNF(X,Y) HERE
130
REM
GET INPUTS
140
PRINT
150
INPUT
“STEP SIZE=”, H
160
INPUT
“NUMBER OF STEPS (AT LEAST 4)=”,N
170
IF
N
<
4
GOTO
160
180
INPUT
“X0 =”,X
190
INPUT
“Y0 =”,Y
200
PRINT
210
REM
SET UP TABLE
220
PRINT
“X”,“Y”
230
PRINT
240
REM
COMPUTE 3 ITERATES USING RUNGE-KUTTA
250
DIM
Z(4)
260 Z(1)=Y
270
FOR
I=1
TO
3
280 K1=H*FNF(X,Y)
290 K2=H*FNF(X+H/2,Y+K1/2)
300 K3=H*FNF(X+H/2,Y+K2/2)
310 K4=H*FNF(X+H,Y+K3)
320 Y=Y+(K1+2*K2+2*K3+K4)/6
330 Z(I+1)+Y
340 X=X+H
350
PRINT
X,Y
360
NEXT
I
370
REM
COMPUTE REMAINING X AND Y VALUES
380
FOR
I=4
TO
N
390 YP=Y+H*(55*FNF(X,Z(4))-59*FNF(X-H,Z(3))+37*FNF(X-2*H,Z(2))
-9*FNF(X-3*H,Z(1)))/24
400 Y=Y+H*(9*FNF(X+H,YP)+19*FNF(X,Z(4))-5*FNF(X-H,Z(3))+FNF(X-2*H,Z(2)))/24
410 X=X+H
420
PRINT
X,Y
430 Z(1)=Z(2)
440 Z(2)=Z(3)
450 Z(3)=Z(4)
460 Z(4)=Y
470
NEXT
I
480
END
276

277. x(n) y(n)
0.00 1.0000 initial condition
0.20 0.7328 Runge-Kutta
0.40 0.6461 Runge-Kutta
0.60 0.6585 Runge-Kutta
0.7332 predictor
0.80 0.7232 corrector
x(n) y(n)
0.00 2.0000 initial condition
0.20 1.4112 Runge-Kutta
0.40 1.1483 Runge-Kutta
0.60 1.1039 Runge-Kutta
1.2109 predictor
0.80 1.2049 corrector
x(n) y(n) x(n) y(n)
0.00 0.0000 initial condition 0.00 0.0000 initial condition
0.20 0.2027 Runge-Kutta 0.10 0.1003 Runge-Kutta
0.40 0.4228 Runge-Kutta 0.20 0.2027 Runge-Kutta
0.60 0.6841 Runge-Kutta 0.30 0.3093 Runge-Kutta
1.0234 predictor 0.4227 predictor
0.80 1.0297 corrector 0.40 0.4228 corrector
1.5376 predictor 0.5462 predictor
1.00 1.5569 corrector 0.50 0.5463 corrector
0.6840 predictor
0.60 0.6842 corrector
0.8420 predictor
0.70 0.8423 corrector
1.0292 predictor
0.80 1.0297 corrector
1.2592 predictor
0.90 1.2603 corrector
1.5555 predictor
1.00 1.5576 corrector
Exercises 6.3
3.
4.
5.
277

278. x(n) y(n) x(n) y(n)
0.00 1.0000 initial condition 0.00 1.0000 initial condition
0.20 1.4414 Runge-Kutta 0.10 1.2102 Runge-Kutta
0.40 1.9719 Runge-Kutta 0.20 1.4414 Runge-Kutta
0.60 2.6028 Runge-Kutta 0.30 1.6949 Runge-Kutta
3.3483 predictor 1.9719 predictor
0.80 3.3486 corrector 0.40 1.9719 corrector
4.2276 predictor 2.2740 predictor
1.00 4.2280 corrector 0.50 2.2740 corrector
2.6028 predictor
0.60 2.6028 corrector
2.9603 predictor
0.70 2.9603 corrector
3.3486 predictor
0.80 3.3486 corrector
3.7703 predictor
0.90 3.7703 corrector
4.2280 predictor
1.00 4.2280 corrector
x(n) y(n) x(n) y(n)
0.00 0.0000 initial condition 0.00 0.0000 initial condition
0.20 0.0026 Runge-Kutta 0.10 0.0003 Runge-Kutta
0.40 0.0201 Runge-Kutta 0.20 0.0026 Runge-Kutta
0.60 0.0630 Runge-Kutta 0.30 0.0087 Runge-Kutta
0.1362 predictor 0.0201 predictor
0.80 0.1360 corrector 0.40 0.0200 corrector
0.2379 predictor 0.0379 predictor
1.00 0.2385 corrector 0.50 0.0379 corrector
0.0630 predictor
0.60 0.0629 corrector
0.0956 predictor
0.70 0.0956 corrector
0.1359 predictor
0.80 0.1360 corrector
0.1837 predictor
0.90 0.1837 corrector
0.2384 predictor
1.00 0.2384 corrector
Exercises 6.3
6.
7.
278

279. x(n) y(n) x(n) y(n)
0.00 1.0000 initial condition 0.00 1.0000 initial condition
0.20 1.2337 Runge-Kutta 0.10 1.1079 Runge-Kutta
0.40 1.5531 Runge-Kutta 0.20 1.2337 Runge-Kutta
0.60 1.9961 Runge-Kutta 0.30 1.3807 Runge-Kutta
2.6180 predictor 1.5530 predictor
0.80 2.6214 corrector 0.40 1.5531 corrector
3.5151 predictor 1.7560 predictor
1.00 3.5208 corrector 0.50 1.7561 corrector
1.9960 predictor
0.60 1.9961 corrector
2.2811 predictor
0.70 2.2812 corrector
2.6211 predictor
0.80 2.6213 corrector
3.0289 predictor
0.90 3.0291 corrector
3.5203 predictor
1.00 3.5207 corrector
Exercises 6.4
8.
Exercises 6.4
1.
The substitution
y
≈
=
u
leads to the iteration formulas
y
n
+1
=
y
n
+
hu
n
,u
n
+1
=
u
n
+
h
(4
u
n
−
4
y
n
)
.
The initial conditions are
y
0
=
−
2 and
u
0
= 1. Then
y
1
=
y
0
+0
.
1
u
0
=
−
2+0
.
1(1) =
−
1
.
9
u
1
=
u
0
+0
.
1(4
u
0
−
4
y
0
)=1+0
.
1(4+8)=2
.
2
y
2
=
y
1
+0
.
1
u
1
=
−
1
.
9+0
.
1(2
.
2) =
−
1
.
68
.
The general solution of the differential equation is
y
=
c
1
e
2
x
+
c
2
xe
2
x
. From the initial conditions we find
c
1
=
−
2 and
c
2
=5. Thus
y
=
−
2
e
2
x
+5
xe
2
x
and
y
(0
.
2)
≈
1
.
4918.
2.
The substitution
y
≈
=
u
leads to the iteration formulas
y
n
+1
=
y
n
+
hu
n
,u
n
+1
=
u
n
+
h
∗
2
x
u
n
−
2
x
2
y
n
√
.
The initial conditions are
y
0
= 4 and
u
0
= 9. Then
y
1
=
y
0
+0
.
1
u
0
=4+0
.
1(9) = 4
.
9
u
1
=
u
0
+0
.
1
∗
2
1
u
0
−
2
1
y
0
√
=9+0
.
1[2(9)
−
2(4)] = 10
y
2
=
y
1
+0
.
1
u
1
=4
.
9+0
.
1(10) = 5
.
9
.
The general solution of the Cauchy-Euler differential equation is
y
=
c
1
x
+
c
2
x
2
. From the initial conditions we
find
c
1
=
−
1and
c
2
=5. Thus
y
=
−
x
+5
x
2
and
y
(1
.
2) = 6.
3.
The substitution
y
≈
=
u
leads to the system
y
≈
=
u, u
≈
=4
u
−
4
y.
Using formula (4) in the text with
x
corresponding to
t
,
y
corresponding to
x
, and
u
corresponding to
y
,we
obtain
279

280. Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 x y u
0.00 -2.0000 1.0000
0.2000 0.4400 0.5280 0.9072 2.4000 3.2800 3.5360 4.8064 0.20 -1.4928 4.4731
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 x y u
0.00 -2.0000 1.0000
0.1000 0.1600 0.1710 0.2452 1.2000 1.4200 1.4520 1.7124 0.10 -1.8321 2.4427
0.2443 0.3298 0.3444 0.4487 1.7099 2.0031 2.0446 2.3900 0.20 -1.4919 4.4753
Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 x y u
1.00 4.0000 9.0000
1.8000 2.0000 2.0017 2.1973 2.0000 2.0165 1.9865 1.9950 1.20 6.0001 11.0002
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 x y u
1.00 4.0000 9.0000
0.9000 0.9500 0.9501 0.9998 1.0000 1.0023 0.9979 0.9996 1.10 4.9500 10.0000
1.0000 1.0500 1.0501 1.0998 1.0000 1.0019 0.9983 0.9997 1.20 6.0000 11.0000
Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 t y u
0.00 1.0000 2.0000
0.4000 0.4600 0.4660 0.5320 0.6000 0.6599 0.6599 0.7170 0.20 1.4640 2.6594
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 t y u
0.00 1.0000 2.0000
0.2000 0.2150 0.2157 0.2315 0.3000 0.3150 0.3150 0.3298 0.10 1.2155 2.3150
0.2315 0.2480 0.2487 0.2659 0.3299 0.3446 0.3444 0.3587 0.20 1.4640 2.6594
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 t i1 i2
0.00 0.0000 0.0000
10.0000 0.0000 12.5000 -20.0000 0.0000 5.0000 -5.0000 22.5000 0.10 2.5000 3.7500
8.7500 -2.5000 13.4375 -28.7500 -5.0000 4.3750 -10.6250 29.6875 0.20 2.8125 5.7813
10.1563 -4.3750 17.0703 -40.0000 -8.7500 5.0781 -16.0156 40.3516 0.30 2.0703 7.4023
13.2617 -6.3672 22.9443 -55.1758 -12.7344 6.6309 -22.5488 55.3076 0.40 0.6104 9.1919
17.9712 -8.8867 31.3507 -75.9326 -17.7734 8.9856 -31.2024 75.9821 0.50 -1.5619 11.4877
Exercises 6.4
4.
The substitution
y
≈
=
u
leads to the system
y
≈
=
u, u
≈
=
2
x
u
−
2
x
2
y.
Using formula (4) in the text with
x
corresponding to
t
,
y
corresponding to
x
, and
u
corresponding to
y
,we
obtain
5.
The substitution
y
≈
=
u
leads to the system
y
≈
=
u, u
≈
=2
u
−
2
y
+
e
t
cos
t.
Using formula (4) in the text with
y
corresponding to
x
and
u
corresponding to
y
, we obtain
6.
280

281. 12345
t
1
2
3
4
5
6
7
i1,i2
i1
i2
Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 6.0000 2.0000
2.0000 2.2800 2.3160 2.6408 1.2000 1.4000 1.4280 1.6632 0.20 8.3055 3.4199
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 6.0000 2.0000
1.0000 1.0700 1.0745 1.1496 0.6000 0.6500 0.6535 0.7075 0.10 7.0731 2.6524
1.1494 1.2289 1.2340 1.3193 0.7073 0.7648 0.7688 0.8307 0.20 8.3055 3.4199
0.511.52
t
5
10
15
20
x,y
xHtL
yHtL
Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 1.0000 1.0000
0.6000 0.9400 1.1060 1.7788 1.4000 2.0600 2.3940 3.7212 0.20 2.0785 3.3382
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 1.0000 1.0000
0.3000 0.3850 0.4058 0.5219 0.7000 0.8650 0.9068 1.1343 0.10 1.4006 1.8963
0.5193 0.6582 0.6925 0.8828 1.1291 1.4024 1.4711 1.8474 0.20 2.0845 3.3502
Exercises 6.4
As
t
→∞
we see that
i
1
(
t
)
→
6
.
75 and
i
2
(
t
)
→
3
.
4.
7.
8.
281

282. 0.511.52
t
10
20
30
40
50
x,y
yHtL
xHtL
Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 -3.0000 5.0000
-1.0000 -0.9200 -0.9080 -0.8176 -0.6000 -0.7200 -0.7120 -0.8216 0.20 -3.9123 4.2857
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 -3.0000 5.0000
-0.5000 -0.4800 -0.4785 -0.4571 -0.3000 -0.3300 -0.3290 -0.3579 0.10 -3.4790 4.6707
-0.4571 -0.4342 -0.4328 -0.4086 -0.3579 -0.3858 -0.3846 -0.4112 0.20 -3.9123 4.2857
51015202530
t
-5
5
10
15
20
25
30
x,y
xHtL
yHtL
Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 0.5000 0.2000
0.6400 1.2760 1.7028 3.3558 1.3200 1.7720 2.1620 3.5794 0.20 2.1589 2.3279
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 0.5000 0.2000
0.3200 0.4790 0.5324 0.7816 0.6600 0.7730 0.8218 1.0195 0.10 1.0207 1.0115
0.7736 1.0862 1.1929 1.6862 1.0117 1.2682 1.3692 1.7996 0.20 2.1904 2.3592
Exercises 6.4
9.
10.
282

283. 0.10.20.3
t
0.5
1
1.5
2
x,y
xHtL
yHtL
Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 1.0000 -2.0000
-0.8000 -0.5400 -0.6120 -0.3888 0.0000 -0.2000 -0.1680 -0.3584 0.20 0.4179 -2.1824
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 1.0000 -2.0000
-0.4000 -0.3350 -0.3440 -0.2858 0.0000 -0.0500 -0.0460 -0.0934 0.10 0.6594 -2.0476
-0.2866 -0.2376 -0.2447 -0.2010 -0.0929 -0.1362 -0.1335 -0.1752 0.20 0.4173 -2.1821
1234
t
-20
-15
-10
-5
x,y
xHtL
yHtL
Exercises 6.4
11.
Solving for
x
≈
and
y
≈
we obtain the system
x
≈
=
−
2
x
+
y
+5
t
y
≈
=2
x
+
y
−
2
t.
283

284. Runge-Kutta method with h=0.2
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 3.0000 -1.0000
-1.1000 -1.0110 -1.0115 -0.9349 1.1000 1.0910 1.0915 1.0949 0.20 1.9867 0.0933
Runge-Kutta method with h=0.1
m1 m2 m3 m4 k1 k2 k3 k4 t x y
0.00 3.0000 -1.0000
-0.5500 -0.5270 -0.5271 -0.5056 0.5500 0.5470 0.5471 0.5456 0.10 2.4727 -0.4527
-0.5056 -0.4857 -0.4857 -0.4673 0.5456 0.5457 0.5457 0.5473 0.20 1.9867 0.0933
2468
t
-60
-40
-20
20
40
60
x,y
xHtL
yHtL
x
0.0 0.5 1.0 1.5 2.0
y
4.0000 -5.6774 -2.5807 6.3226 1.0000
x
0.00 0.25 0.50 0.75 1.00
y
0.0000 -0.0172 -0.0316 -0.0324 0.0000
Exercises 6.4
12.
Solving for
x
≈
and
y
≈
we obtain the system
x
≈
=
1
2
y
−
3
t
2
+2
t
−
5
y
≈
=
−
1
2
y
+3
t
2
+2
t
+5
.
Exercises 6.5
1.
We identify
P
(
x
)=0,
Q
(
x
)=9,
f
(
x
) = 0, and
h
=(2
−
0)
/
4=0
.
5. Then the finite difference equation is
y
i
+1
+0
.
25
y
i
+
y
i
−
1
=0
.
The solution of the corresponding linear system gives
2.
We identify
P
(
x
)=0,
Q
(
x
)=
−
1,
f
(
x
)=
x
2
, and
h
=(1
−
0)
/
4=0
.
25. Then the finite difference equation is
y
i
+1
−
2
.
0625
y
i
+
y
i
−
1
=0
.
0625
x
2
i
.
The solution of the corresponding linear system gives
284

285. x
0.0 0.2 0.4 0.6 0.8 1.0
y
0.0000 -0.2259 -0.3356 -0.3308 -0.2167 0.0000
x
0.0 0.2 0.4 0.6 0.8 1.0
y
1.0000 1.9600 3.8800 7.7200 15.4000 0.0000
x
0.0000 0.1667 0.3333 0.5000 0.6667 0.8333 1.0000
y
3.0000 3.3751 3.6306 3.6448 3.2355 2.1411 0.0000
x
1.0000 1.1667 1.3333 1.5000 1.6667 1.8333 2.0000
y
1.0000 -0.5918 -1.1626 -1.3070 -1.2704 -1.1541 -1.0000
x
1.000 1.125 1.250 1.375 1.500 1.625 1.750 1.875 2.000
y
5.0000 3.8842 2.9640 2.2064 1.5826 1.0681 0.6430 0.2913 0.0000
x
1.000 1.125 1.250 1.375 1.500 1.625 1.750 1.875 2.000
y
0.0000 -0.1988 -0.4168 -0.6510 -0.8992 -1.1594 -1.4304 -1.7109 -2.0000
Exercises 6.5
3.
We identify
P
(
x
)=2,
Q
(
x
)=1,
f
(
x
)=5
x
, and
h
=(1
−
0)
/
5=0
.
2. Then the finite difference equation is
1
.
2
y
i
+1
−
1
.
96
y
i
+0
.
8
y
i
−
1
=0
.
04(5
x
i
)
.
The solution of the corresponding linear system gives
4.
We identify
P
(
x
)=
−
10,
Q
(
x
) = 25,
f
(
x
) = 1, and
h
=(1
−
0)
/
5=0
.
2. Then the finite difference equation is
−
y
i
+2
y
i
−
1
=0
.
04
.
The solution of the corresponding linear system gives
5.
We identify
P
(
x
)=
−
4,
Q
(
x
)=4,
f
(
x
)=(1+
x
)
e
2
x
, and
h
=(1
−
0)
/
6=0
.
1667. Then the finite difference
equation is
0
.
6667
y
i
+1
−
1
.
8889
y
i
+1
.
3333
y
i
−
1
=0
.
2778(1+
x
i
)
e
2
x
i
.
The solution of the corresponding linear system gives
6.
We identify
P
(
x
)=5,
Q
(
x
)=0,
f
(
x
)=4
√
x
, and
h
=(2
−
1)
/
6=0
.
1667. Then the finite difference equation
is
1
.
4167
y
i
+1
−
2
y
i
+0
.
5833
y
i
−
1
=0
.
2778(4
√
x
i
)
.
The solution of the corresponding linear system gives
7.
We identify
P
(
x
)=3
/x
,
Q
(
x
)=3
/x
2
,
f
(
x
) = 0, and
h
=(2
−
1)
/
8=0
.
125. Then the finite difference equation
is
.
1+
0
.
1875
x
i
2
y
i
+1
+
.
−
2+
0
.
0469
x
2
i
2
y
i
+
.
1
−
0
.
1875
x
i
2
y
i
−
1
=0
.
The solution of the corresponding linear system gives
8.
We identify
P
(
x
)=
−
1
/x
,
Q
(
x
)=
x
−
2
,
f
(
x
)=ln
x/x
2
, and
h
=(2
−
1)
/
8=0
.
125. Then the finite difference
equation is
.
1
−
0
.
0625
x
i
2
y
i
+1
+
.
−
2+
0
.
0156
x
2
i
2
y
i
+
.
1+
0
.
0625
x
i
2
y
i
−
1
=0
.
0156 ln
x
i
.
The solution of the corresponding linear system gives
9.
We identify
P
(
x
)=1
−
x
,
Q
(
x
)=
x
,
f
(
x
)=
x
, and
h
=(1
−
0)
/
10=0
.
1. Then the finite difference equation is
[1+ 0
.
05(1
−
x
i
)]
y
i
+1
+[
−
2+0
.
01
x
i
]
y
i
+[1
−
0
.
05(1
−
x
i
)]
y
i
−
1
=0
.
01
x
i
.
The solution of the corresponding linear system gives
285

286. x
0.0 0.1 0.2 0.3 0.4 0.5 0.6
y
0.0000 0.2660 0.5097 0.7357 0.9471 1.1465 1.3353
0.7 0.8 0.9 1.0
1.5149 1.6855 1.8474 2.0000
x
0.0 0.1 0.2 0.3 0.4 0.5 0.6
y
1.0000 0.8929 0.7789 0.6615 0.5440 0.4296 0.3216
0.7 0.8 0.9 1.0
0.2225 0.1347 0.0601 0.0000
x
0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000
y
0.0000 0.3492 0.7202 1.1363 1.6233 2.2118 2.9386 3.8490 5.0000
r
1.0 1.5 2.0 2.5 3.0 3.5 4.0
u
50.0000 72.2222 83.3333 90.0000 94.4444 97.6190 100.0000
Exercises 6.5
10.
We identify
P
(
x
)=
x
,
Q
(
x
)=1,
f
(
x
)=
x
, and
h
=(1
−
0)
/
10=0
.
1. Then the finite difference equation is
(1+ 0
.
05
x
i
)
y
i
+1
−
1
.
99
y
i
+(1
−
0
.
05
x
i
)
y
i
−
1
=0
.
01
x
i
.
The solution of the corresponding linear system gives
11.
We identify
P
(
x
)=0,
Q
(
x
)=
−
4,
f
(
x
) = 0, and
h
=(1
−
0)
/
8=0
.
125. Then the finite difference equation is
y
i
+1
−
2
.
0625
y
i
+
y
i
−
1
=0
.
The solution of the corresponding linear system gives
12.
We identify
P
(
r
)=2
/r
,
Q
(
r
)=0,
f
(
r
) = 0, and
h
=(4
−
1)
/
6=0
.
5. Then the finite difference equation is
∗
1+
0
.
5
r
i
√
u
i
+1
−
2
u
i
+
∗
1
−
0
.
5
r
i
√
u
i
−
1
=0
.
The solution of the corresponding linear system gives
13. (a)
The difference equation
∗
1+
h
2
P
i
√
y
i
+1
+(
−
2+
h
2
Q
i
)
y
i
+
∗
1
−
h
2
P
i
√
y
i
−
1
=
h
2
f
i
is the same as the one derived on page 383 in the text. The equations are the same because the derivation
was based only on the differential equation, not the boundary conditions. If we allow
i
to range from 0 to
n
−
1we obtain
n
equations in the
n
+ 1unknowns
y
−
1
,
y
0
,
y
1
,
...
,
y
n
−
1
. Since
y
n
is one of the given
boundary conditions, it is not an unknown.
(b)
Identifying
y
0
=
y
(0),
y
−
1
=
y
(0
−
h
), and
y
1
=
y
(0 +
h
) we have from (5) in the text
1
2
h
[
y
1
−
y
−
1
]=
y
≈
(0) = 1or
y
1
−
y
−
1
=2
h.
The difference equation corresponding to
i
=0,
∗
1+
h
2
P
0
√
y
1
+(
−
2+
h
2
Q
0
)
y
0
+
∗
1
−
h
2
P
0
√
y
−
1
=
h
2
f
0
becomes, with
y
−
1
=
y
1
−
2
h
,
∗
1+
h
2
P
0
√
y
1
+(
−
2+
h
2
Q
0
)
y
0
+
∗
1
−
h
2
P
0
√
(
y
1
−
2
h
)=
h
2
f
0
or
2
y
1
+(
−
2+
h
2
Q
0
)
y
0
=
h
2
f
0
+2
h
−
P
0
.
286

287. x
0.0 0.2 0.4 0.6 0.8 1.0
y
-2.2755 -2.0755 -1.8589 -1.6126 -1.3275 -1.0000
x
0.0 0.1 0.2 0.3 0.4 0.5 0.6
y
1.00000 1.04561 1.09492 1.14714 1.20131 1.25633 1.31096
0.7 0.8 0.9 1.0
1.36392 1.41388 1.45962 1.50003
h=0.1
IMPROVED
RUNGE
h=0.05 IMPROVED
RUNGE
x(n)
EULER EULER KUTTA
x(n)
EULER EULER KUTTA
1.00
2.0000 2.0000 2.0000
1.00
2.0000 2.0000 2.0000
1.10
2.1386 2.1549 2.1556
1.05
2.0693 2.0735 2.0736
1.20
2.3097 2.3439 2.3454
1.10
2.1469 2.1554 2.1556
1.30
2.5136 2.5672 2.5695
1.15
2.2328 2.2459 2.2462
1.40
2.7504 2.8246 2.8278
1.20
2.3272 2.3450 2.3454
1.50
3.0201 3.1157 3.1197
1.25
2.4299 2.4527 2.4532
1.30
2.5409 2.5689 2.5695
1.35
2.6604 2.6937 2.6944
1.40
2.7883 2.8269 2.8278
1.45
2.9245 2.9686 2.9696
1.50
3.0690 3.1187 3.1197
h=0.1
IMPROVED
RUNGE
h=0.05 IMPROVED
RUNGE
x(n)
EULER EULER KUTTA
x(n)
EULER EULER KUTTA
0.00
0.0000 0.0000 0.0000
0.00
0.0000 0.0000 0.0000
0.10
0.1000 0.1005 0.1003
0.05
0.0500 0.0501 0.0500
0.20
0.2010 0.2030 0.2026
0.10
0.1001 0.1004 0.1003
0.30
0.3049 0.3092 0.3087
0.15
0.1506 0.1512 0.1511
0.40
0.4135 0.4207 0.4201
0.20
0.2017 0.2027 0.2026
0.50
0.5279 0.5382 0.5376
0.25
0.2537 0.2552 0.2551
0.30
0.3067 0.3088 0.3087
0.35
0.3610 0.3638 0.3637
0.40
0.4167 0.4202 0.4201
0.45
0.4739 0.4782 0.4781
0.50
0.5327 0.5378 0.5376
Chapter 6 Review Exercises
Alternatively, we may simply add the equation
y
1
−
y
−
1
=2
h
to the list of
n
difference equations obtaining
n
+ 1equations in the
n
+ 1unknowns
y
−
1
,
y
0
,
y
1
,
...
,
y
n
−
1
.
(c)
Using
n
= 5 we obtain
14.
Using
h
=0
.
1and, after shooting a few times,
y
≈
(0) = 0
.
43535 we obtain the following table with the fourth-order
Runge-Kutta method.
Chapter 6 Review Exercises
1.
2.
287

288. h=0.1
IMPROVED
RUNGE
h=0.05 IMPROVED
RUNGE
x(n)
EULER EULER KUTTA
x(n)
EULER EULER KUTTA
0.50
0.5000 0.5000 0.5000
0.50
0.5000 0.5000 0.5000
0.60
0.6000 0.6048 0.6049
0.55
0.5500 0.5512 0.5512
0.70
0.7095 0.7191 0.7194
0.60
0.6024 0.6049 0.6049
0.80
0.8283 0.8427 0.8431
0.65
0.6573 0.6609 0.6610
0.90
0.9559 0.9752 0.9757
0.70
0.7144 0.7193 0.7194
1.00
1.0921 1.1163 1.1169
0.75
0.7739 0.7800 0.7801
0.80
0.8356 0.8430 0.8431
0.85
0.8996 0.9082 0.9083
0.90
0.9657 0.9755 0.9757
0.95
1.0340 1.0451 1.0452
1.00
1.1044 1.1168 1.1169
h=0.1
IMPROVED
RUNGE
h=0.05 IMPROVED
RUNGE
x(n)
EULER EULER KUTTA
x(n)
EULER EULER KUTTA
1.00
1.0000 1.0000 1.0000
1.00
1.0000 1.0000 1.0000
1.10
1.2000 1.2380 1.2415
1.05
1.1000 1.1091 1.1095
1.20
1.4760 1.5910 1.6036
1.10
1.2183 1.2405 1.2415
1.30
1.8710 2.1524 2.1909
1.15
1.3595 1.4010 1.4029
1.40
2.4643 3.1458 3.2745
1.20
1.5300 1.6001 1.6036
1.50
3.4165 5.2510 5.8338
1.25
1.7389 1.8523 1.8586
1.30
1.9988 2.1799 2.1911
1.35
2.3284 2.6197 2.6401
1.40
2.7567 3.2360 3.2755
1.45
3.3296 4.1528 4.2363
1.50
4.1253 5.6404 5.8446
x(n) y(n)
0.00 2.0000 initial condition
0.10 2.4734 Runge-Kutta
0.20 3.1781 Runge-Kutta
0.30 4.3925 Runge-Kutta
6.7689
predictor
0.40 7.0783 corrector
Chapter 6 Review Exercises
3.
4.
5.
Using
y
n
+1
=
y
n
+
hu
n
,
u
n
+1
=
u
n
+
h
(2
x
n
+1)
y
n
,
y
0
=3
u
0
=1
we obtain (when
h
=0
.
2)
y
1
=
y
(0
.
2) =
y
0
+
hu
0
=3+(0
.
2)1= 3
.
2. When
h
=0
.
1we have
y
1
=
y
0
+0
.
1
u
0
=3+(0
.
1)1 = 3
.
1
u
1
=
u
0
+0
.
1(2
x
0
+1)
y
0
=1+0
.
1(1)3 = 1
.
3
y
2
=
y
1
+0
.
1
u
1
=3
.
1+0
.
1(1
.
3) = 3
.
23
.
6.
7.
Using
x
0
=1,
y
0
= 2, and
h
=0
.
1we have
x
1
=
x
0
+
h
(
x
0
+
y
0
)=1+0
.
1(1 + 2) = 1
.
3
y
1
=
y
0
+
h
(
x
0
−
y
0
)=2+0
.
1(1
−
2) = 1
.
9
and
x
2
=
x
1
+
h
(
x
1
+
y
1
)=1
.
3+0
.
1(1
.
3+1
.
9) = 1
.
62
y
2
=
y
1
+
h
(
x
1
−
y
1
)=1
.
9+0
.
1(1
.
3
−
1
.
9) = 1
.
84
.
Thus,
x
(0
.
2)
≈
1
.
62 and
y
(0
.
2)
≈
1
.
84.
288

289. x
0.0 0.1 0.2 0.3 0.4 0.5 0.6
y
0.0000 4.1987 8.1049 11.3840 13.7038 14.7770 14.4083
0.7 0.8 0.9 1.0
12.5396 9.2847 4.9450 0.0000
Chapter 6 Review Exercises
8.
We identify
P
(
x
)=0,
Q
(
x
)=6
.
55(1+
x
),
f
(
x
) = 1, and
h
=(1
−
0)
/
10=0
.
1. Then the finite difference
equation is
y
i
+1
+[
−
2+0
.
0655(1+
x
i
)]
y
i
+
y
i
−
1
=0
.
001
or
y
i
+1
+(0
.
0655
x
i
−
1
.
9345)
y
i
+
y
i
−
1
=0
.
001
.
The solution of the corresponding linear system gives
289

290. 7
Vectors
Exercises 7.1
1. (a)
6
i
+12
j (b) i
+8
j (c)
3
i
(d)
√
65
(e)
3
2. (a)
1
3
,
3
.
(b)
1
3
,
4
.
(c)
1(
1
,
−
2
.
(d)
5
(e)
√
5
3. (a)
1
12
,
0
.
(b)
1
4
,
−
5
.
(c)
1
4
,
5
.
(d)
√
41
(e)
√
41
4. (a)
1
2
i
−
1
2
j (b)
2
3
i
+
2
3
j (c)
−
1
3
i
−
j
(d)
2
√
2
/
3
(e)
√
10
/
3
5. (a)
−
9
i
+6
j (b)
−
3
i
+9
j (c)
−
3
i
−
5
j
(d)
3
√
10
(e)
√
34
6. (a)
1
3
,
9
.
(b)
1(
4
,
−
12
.
(c)
1
6
,
18
.
(d)
4
√
10
(e)
6
√
10
7. (a)
−
6
i
+27
j (b) 0 (c)
−
4
i
+18
j (d)
0
(e)
2
√
85
8. (a)
1
21
,
30
.
(b)
1
8
,
12
.
(c)
1
6
,
8
.
(d)
4
√
13
(e)
10
9. (a)
1
4
,
−
12
.(1(
2
,
2
.
=
1
6
,
−
14
.
(b)
1(
3
,
9
.(1(
5
,
5
.
=
1
2
,
4
.
10. (a)
(4
i
+4
j
)
−
(6
i
−
4
j
)=
−
2
i
+8
j (b)
(
−
3
i
−
3
j
)
−
(15
i
−
10
j
)=
−
18
i
+7
j
11. (a)
(4
i
−
4
j
)
−
(
−
6
i
+8
j
)=10
i
−
12
j (b)
(
−
3
i
+3
j
)
−
(
−
15
i
+20
j
)=12
i
−
17
j
12. (a)
1
8
,
0
.(1
0
,
−
6
.
=
1
8
,
6
.
(b)
1(
6
,
0
.(1
0
,
−
15
.
=
1(
6
,
15
.
13. (a)
1
16
,
40
.(1(
4
,
−
12
.
=
1
20
,
52
.
(b)
1(
12
,
−
30
.(1(
10
,
−
30
.
=
1(
2
,
0
.
14. (a)
1
8
,
12
.(1
10
,
6
.
=
1(
2
,
6
.
(b)
1(
6
,
−
9
.(1
25
,
15
.
=
1(
31
,
−
24
.
15.
−
−−
→
P
1
P
2
=
1
2
,
5
.
16.
−
−−
→
P
1
P
2
=
1
6
,
−
4
.
17.
−
−−
→
P
1
P
2
=
1
2
,
2
.
18.
−
−−
→
P
1
P
2
=
1
2
,
−
3
.
290

291. Exercises 7.1
19.
Since
−
−−
→
P
1
P
2
=
−
−
→
OP
2
−
−
−
→
OP
1
,
−
−
→
OP
2
=
−
−−
→
P
1
P
2
+
−
−
→
OP
1
=(4
i
+8
j
)+(
−
3
i
+10
j
)=
i
+18
j
,
and the terminal point is
(1
,
18).
20.
Since
−
−−
→
P
1
P
2
=
−
−
→
OP
2
−
−
−
→
OP
1
,
−
−
→
OP
1
=
−
−
→
OP
2
−
−
−−
→
P
1
P
2
=
f
4
,
7
gnfn
5
,
−
1
g
=
f
9
,
8
g
,
and the initial point is (9
,
8).
21.
a
(=
−
a
),
b
(=
−
1
4
a
),
c
(=
5
2
a
),
e
(= 2
a
), and
f
(=
−
1
2
a
) are parallel to
a
.
22.
We want
−
3
b
=
a
,so
c
=
−
3(9) =
−
27.
23.
f
6
,
15
g
24.
f
5
,
2
g
25.
a
a
a
=
√
4+4=2
√
2;
(a) u
=
1
2
√
2
f
2
,
2
g
=
f
1
√
2
,
1
√
2
g
;
(b)
−
u
=
fn
1
√
2
,
−
1
√
2
g
26.
a
a
a
=
√
9+16=5;
(a) u
=
1
5
fn
3
,
4
g
=
fn
3
5
,
4
5
g
;
(b)
−
u
=
f
3
5
,
−
4
5
g
27.
a
a
a
=5;
(a) u
=
1
5
f
0
,
−
5
g
=
f
0
,
−
1
g
;
(b)
−
u
=
f
0
,
1
g
28.
a
a
a
=
√
1+3=2;
(a) u
=
1
2
f
1
,
−
√
3
g
=
f
1
2
,
−
√
3
2
g
;
(b)
−
u
=
fn
1
2
,
√
3
2
g
29.
a
a
+
b
a
=
af
5
,
12
ga
=
√
25 + 144 = 13;
u
=
1
13
f
5
,
12
g
=
f
5
13
,
12
13
g
30.
a
a
+
b
a
=
afn
5
,
4
ga
=
√
25+16=
√
41 ;
u
=
1
√
41
fn
5
,
4
g
=
fn
5
√
41
,
4
√
41
g
31.
a
a
a
=
√
9+49=
√
58 ;
b
=2(
1
√
58
)(3
i
+7
j
)=
6
√
58
i
+
14
√
58
j
32.
a
a
a
=
f
1
4
+
1
4
=
1
√
2
;
b
=3(
1
1
/
√
2
)(
1
2
i
−
1
2
j
)=
3
√
2
2
i
−
3
√
2
2
j
33.
−
3
4
a
=
fn
3
,
−
15
/
2
g
34.
5(
a
+
b
)=5
f
0
,
1
g
=
f
0
,
5
g
35. 36.
37. x
=
−
(
a
+
b
)=
−
a
−
b 38. x
=2(
a
−
b
)=2
a
−
2
b
39.
b
=(
−
c
)
−
a
;(
b
+
c
)+
a
=
0
;
a
+
b
+
c
=
0
40.
From Problem 39,
e
+
c
+
d
=
0
.But
b
=
e
−
a
and
e
=
a
+
b
,so(
a
+
b
)+
c
+
d
=
0
.
41.
From 2
i
+3
j
=
k
1
b
+
k
2
c
=
k
1
(
i
+
j
)+
k
2
(
i
−
j
)=(
k
1
+
k
2
)
i
+(
k
1
−
k
2
)
j
we obtain the system of equations
k
1
+
k
2
=2,
k
1
−
k
2
= 3.Solving, we find
k
1
=
5
2
and
k
2
=
−
1
2
.Then
a
=
5
2
b
−
1
2
c
.
42.
From 2
i
+3
j
=
k
1
b
+
k
2
c
=
k
1
(
−
2
i
+4
j
)+
k
2
(5
i
+7
j
)=(
−
2
k
1
+5
k
2
)
i
+(4
k
1
+7
k
2
)
j
we obtain the system of
equations
−
2
k
1
+5
k
2
=2,4
k
1
+7
k
2
= 3.Solving, we find
k
1
=
1
34
and
k
2
=
7
17
.
43.
From
y
f
=
1
2
x
we see that the slope of the tangent line at (2
,
2) is 1.A vector with slope 1 is
i
+
j
.A unit vector
is (
i
+
j
)
/
a
i
+
j
a
=(
i
+
j
)
/
√
2=
1
√
2
i
+
1
√
2
j
.Another unit vector tangent to the curve is
−
1
√
2
i
−
1
√
2
j
.
44.
From
y
f
=
−
2
x
+ 3 we see that the slope of the tangent line at (0
,
0) is 3.A vector with slope 3 is
i
+3
j
.A
unit vector is (
i
+3
j
)
/
a
i
+3
j
a
=(
i
+3
j
)
/
√
10 =
1
√
10
i
+
1
√
10
j
.Another unit vector is
−
1
√
10
i
−
1
√
10
j
.
45. (a)
Since
F
f
=
−
F
g
,
a
F
g
a
=
a
F
f
a
=
µ
a
F
n
a
and tan
θ
=
a
F
g
a
/
a
F
n
a
=
µ
a
F
n
a
/
a
F
n
a
=
µ
.
291

292. Exercises 7.1
(b)
θ
= tan
−
1
0
.
6
≈
31
◦
.
46.
Since
w
+
F
1
+
F
2
=
0
,
−
200
j
+
a
F
1
a
cos 20
◦
i
+
a
F
1
a
sin 20
◦
j
na
F
2
a
cos 15
◦
i
+
a
F
2
a
sin 15
◦
j
=
0
or (
a
F
1
a
cos 20
◦
na
F
2
a
cos 15
◦
)
i
+(
a
F
1
a
sin 20
◦
+
a
F
2
a
sin 15
◦
−
200)
j
=
0
.
Thus,
a
F
1
a
cos 20
◦
na
F
2
a
cos 15
◦
=0;
a
F
1
a
sin 20
◦
+
a
F
2
a
sin 15
◦
−
200 = 0.Solving this system for
a
F
1
a
and
a
F
2
a
, we obtain
a
F
1
a
=
200 cos 15
◦
sin 15
◦
cos 20
◦
+ cos 15
◦
sin 20
◦
=
200 cos 15
◦
sin(15
◦
+20
◦
)
=
200 cos 15
◦
sin 35
◦
≈
336
.
8lb
and
a
F
2
a
=
200 cos 20
◦
sin 15
◦
cos 20
◦
+ cos 15
◦
sin 20
◦
=
200 cos 20
◦
sin 35
◦
≈
327
.
7lb
.
47.
Since
y/
2
a
(
L
2
+
y
2
)
3
/
2
is an odd function on [
−
a, a
],
F
y
= 0.Now, using the fact that
L/
(
L
2
+
y
2
)
3
/
2
is an
even function, we have
g
a
−
a
Ldy
2
a
(
L
2
+
y
2
)
3
/
2
=
L
a
g
a
0
dy
(
L
2
+
y
2
)
3
/
2
y
=
L
tan
θ, dy
=
L
sec
2
θdθ
=
L
a
g
tan
−
1
a/L
0
L
sec
2
θdθ
L
3
(1 + tan
2
θ
)
3
/
2
=
1
La
g
tan
−
1
a/L
0
sec
2
θdθ
sec
3
θ
=
1
La
g
tan
−
1
a/L
0
cos
θdθ
=
1
La
sin
θ
n
n
n
n
tan
−
1
a/L
0
=
1
La
a
√
L
2
+
a
2
=
1
L
√
L
2
+
a
2
.
Then
F
x
=
qQ/
4
9u
0
L
√
L
2
+
a
2
and
F
=(
qQ/
4
9u
0
L
√
L
2
+
a
2
)
i
.
48.
Place one corner of the parallelogram at the origin and let two adja-
cent sides be
−
−
→
OP
1
and
−
−
→
OP
2
.Let
M
be the midpoint of the diagonal
connecting
P
1
and
P
2
and
N
be the midpoint of the other diagonal.
By Problem 37,
−
−
→
OM
=
1
2
(
−
−
→
OP
1
+
−
−
→
OP
2
).Since
−
−
→
OP
1
+
−
−
→
OP
2
is the main diagonal of the parallelogram and
N
is
its midpoint,
−
−
→
ON
=
1
2
(
−
−
→
OP
1
+
−
−
→
OP
2
).Thus,
−
−
→
OM
=
−
−
→
ON
and the diagonals bisect each other.
49.
By Problem 39,
−
−
→
AB
+
−
−
→
BC
+
−→
CA
=
0
and
−
−
→
AD
+
−
−
→
DE
+
−
−
→
EC
+
−→
CA
=
0
.From the first equation,
−
−
→
AB
+
−
−
→
BC
=
−
−→
CA
.Since
D
and
E
are midpoints,
−
−
→
AD
=
1
2
−
−
→
AB
and
−
−
→
EC
=
1
2
−
−
→
BC
.Then,
1
2
−
−
→
AB
+
−
−
→
DE
+
1
2
−
−
→
BC
+
−→
CA
=
0
and
−
−
→
DE
=
−
−→
CA
−
1
2
(
−
−
→
AB
+
−
−
→
BC
)=
−
−→
CA
−
1
2
(
−
−→
CA
)=
−
1
2
−→
CA.
Thus, the line segment joining the midpoints
D
and
E
is parallel to the side
AC
and half its length.
50.
We have
−→
OA
= 150 cos 20
◦
i
+150 sin 20
◦
j
,
−
−
→
AB
= 200 cos 113
◦
i
+200 sin 113
◦
j
,
−
−
→
BC
= 240 cos 190
◦
i
+240 sin 190
◦
j
.
Then
r
= (150 cos 20
◦
+ 200 cos 113
◦
+ 240 cos 190
◦
)
i
+ (150 sin 20
◦
+ 200 sin 113
◦
+ 240 sin 190
◦
)
j
≈−
173
.
55
i
+ 193
.
73
j
and
a
r
aL
260
.
09 miles.
292

293. Exercises 7.2
Exercises 7.2
1.
–
6.
7.
A plane perpendicular to the
z
-axis, 5 units above the
xy
-plane
8.
A plane perpendicular to the
x
-axis, 1 unit in front of the
yz
-plane
9.
A line perpendicular to the
xy
-plane at (2
,
3
,
0)
10.
A single point located at (4
,
−
1
,
7)
11.
(2
,
0
,
0), (2
,
5
,
0), (2
,
0
,
8), (2
,
5
,
8), (0
,
5
,
0), (0
,
5
,
8), (0
,
0
,
8), (0
,
0
,
0)
12.
13. (a)
xy
-plane: (
−
2
,
5
,
0),
xz
-plane: (
−
2
,
0
,
4),
yz
-plane: (0
,
5
,
4);
(b)
(
−
2
,
5
,
−
2)
(c)
Since the shortest distance between a point and a plane is a perpendicular line, the point in the plane
x
=3
is (3
,
5
,
4).
14.
We find planes that are parallel to coordinate planes:
(a)
z
=
−
5;
(b)
x
= 1 and
y
=
−
1;
(c)
z
=2
15.
The union of the planes
x
=0,
y
= 0, and
z
=0
16.
The origin (0
,
0
,
0)
17.
The point (
−
1
,
2
,
−
3)
18.
The union of the planes
x
= 2 and
z
=8
19.
The union of the planes
z
= 5 and
z
=
−
5
20.
The line through the points (1
,
1
,
1), (
−
1
,
−
1
,
−
1), and the origin
21.
d
=
a
(3
−
6)
2
+(
−
1
−
4)
2
+(2
−
8)
2
=
√
70
22.
d
=
a
(
−
1
−
0)
2
+(
−
3
−
4)
2
+(5
−
3)
2
=3
√
6
23. (a)
7;
(b)
d
=
a
(
−
3)
2
+(
−
4)
2
=5
24. (a)
2;
(b)
d
=
a
(
−
6)
2
+2
2
+(
−
3)
2
=7
25.
d
(
P
1
,P
2
)=
a
3
2
+6
2
+(
−
6)
2
=9;
d
(
P
1
,P
3
)=
√
2
2
+1
2
+2
2
=3
d
(
P
2
,P
3
)=
a
(2
−
3)
2
+(1
−
6)
2
+(2
−
(
−
6))
2
=
√
90 ; The triangle is a right triangle.
293

294. Exercises 7.2
26.
d
(
P
1
,P
2
)=
√
1
2
+2
2
+4
2
=
√
21 ;
d
(
P
1
,P
3
)=
1
3
2
+2
2
+(2
√
2)
2
=
√
21
d
(
P
2
,P
3
)=
1
(3
−
1)
2
+(2
−
2)
2
+(2
√
2
−
4)
2
=
a
28
−
16
√
2
The triangle is an isosceles triangle.
27.
d
(
P
1
,P
2
)=
a
(4
−
1)
2
+(1
−
2)
2
+(3
−
3)
2
=
√
10
d
(
P
1
,P
3
)=
a
(4
−
1)
2
+(6
−
2)
2
+(4
−
3)
2
=
√
26
d
(
P
2
,P
3
)=
a
(4
−
4)
2
+(6
−
1)
2
+(4
−
3)
2
=
√
26 ; The triangle is an isosceles triangle.
28.
d
(
P
1
,P
2
)=
a
(1
−
1)
2
+(1
−
1)
2
+(1
−
(
−
1))
2
=2
d
(
P
1
,P
3
)=
a
(0
−
1)
2
+(
−
1
−
1)
2
+(1
−
(
−
1))
2
=3
d
(
P
2
,P
3
)=
a
(0
−
1)
2
+(
−
1
−
1)
2
+(1
−
1)
2
=
√
5 ; The triangle is a right triangle.
29.
d
(
P
1
,P
2
)=
a
(
−
2
−
1)
2
+(
−
2
−
2)
2
+(
−
3
−
0)
2
=
√
34
d
(
P
1
,P
3
)=
a
(7
−
1)
2
+ (10
−
2)
2
+(6
−
0)
2
=2
√
34
d
(
P
2
,P
3
)=
a
(7
−
(
−
2))
2
+ (10
−
(
−
2))
2
+(6
−
(
−
3))
2
=3
√
34
Since
d
(
P
1
,P
2
)+
d
(
P
1
,P
3
)=
d
(
P
2
,P
3
), the points
P
1
,
P
2
, and
P
3
are collinear.
30.
d
(
P
1
,P
2
)=
a
(1
−
2)
2
+(4
−
3)
2
+(4
−
2)
2
=
√
6
d
(
P
1
,P
3
)=
a
(5
−
2)
2
+(0
−
3)
2
+(
−
4
−
2)
2
=3
√
6
d
(
P
2
,P
3
)=
a
(5
−
1)
2
+(0
−
4)
2
+(
−
4
−
4)
2
=4
√
6
Since
d
(
P
1
,P
2
)+
d
(
P
1
,P
3
)=
d
(
P
2
,P
3
), the points
P
1
,
P
2
, and
P
3
are collinear.
31.
a
(2
−
x
)
2
+(1
−
2)
2
+(1
−
3)
2
=
√
21 =
⇒
x
2
−
4
x
+9=21 =
⇒
x
2
−
4
x
+4=16
=
⇒
(
x
−
2)
2
=16 =
⇒
x
=2
±
4or
x
=6
,
−
2
32.
a
(0
−
x
)
2
+(3
−
x
)
2
+(5
−
1)
2
=5 =
⇒
2
x
2
−
6
x
+25=25 =
⇒
x
2
−
3
x
=0 =
⇒
x
=0,3
33.
)
1+7
2
,
3+(
−
2)
2
,
1
/
2+5
/
2
2
i
=(4
,
1
/
2
,
3
/
2)
34.
)
0+4
2
,
5+1
2
,
−
8+(
−
6)
2
i
=(2
,
3
,
−
7)
35.
(
x
1
+2)
/
2=
−
1,
x
1
=
−
4; (
y
1
+3)
/
2=
−
4,
y
1
=
−
11; (
z
1
+6)
/
2=8,
z
1
=10
The coordinates of
P
1
are (
−
4
,
−
11
,
10).
36.
(
−
3+(
−
5))
/
2=
x
3
=
−
4; (4 + 8)
/
2=
y
3
=6; (1+3)
/
2=
z
3
=2.
The coordinates of
P
3
are (
−
4
,
6
,
2).
(a)
)
−
3+(
−
4)
2
,
4+6
2
,
1+2
2
i
=(
−
7
/
2
,
5
,
3
/
2)
(b)
)
−
4+(
−
5)
2
,
6+8
2
,
2+3
2
i
=(
−
9
/
2
,
7
,
5
/
2)
37.
−
−−
→
P
1
P
2
=
1(
3
,
−
6
,
1
.
38.
−
−−
→
P
1
P
2
=
1
8
,
−
5
/
2
,
8
.
39.
−
−−
→
P
1
P
2
=
1
2
,
1
,
1
.
40.
−
−−
→
P
1
P
2
1(
3
,
−
3
,
7
.
41. a
+(
b
+
c
)=
1
2
,
4
,
12
.
42.
2
a
−
(
b
−
c
)=
1
2
,
−
6
,
4
.(1(
3
,
−
5
,
−
8
.
=
1
5
,
−
1
,
12
.
294

295. Exercises 7.3
43. b
+2(
a
−
3
c
)=
1(
1
,
1
,
1
.
+2
1(
5
,
−
21
,
−
25
.
=
1(
11
,
−
41
,
−
49
.
44.
4(
a
+2
c
)
−
6
b
=4
1
5
,
9
,
20
.(1(
6
,
6
,
6
.
=
1
26
,
30
,
74
.
45.
)
a
+
c
)
=
)1
3
,
3
,
11
.)
=
√
9+9+121=
√
139
46.
)
c
))
2
b
)
=(
√
4 + 36 + 81 )(2)(
√
1+1+1)=22
√
3
47.
j
j
j
j
a
)
a
)
j
j
j
j
+5
j
j
j
j
b
|
b
)
j
j
j
j
=
1
)
a
)
)
a
)
+5
1
)
b
)
)
b
)
=1+5=6
48.
)
b
)
a
+
)
a
)
b
=
√
1+1+1
1
1
,
−
3
,
2
.
+
√
1+9+4
1(
1
,
1
,
1
.
=
1
√
3
,
−
3
√
3
,
2
√
3
.
+
1(
√
14
,
√
14
,
√
14
.
=
1
√
3
−
√
14
,
−
3
√
3+
√
14
,
2
√
3+
√
14
.
49.
)
a
)
=
√
100 + 25 + 100 = 15;
u
=
−
1
15
1
10
,
−
5
,
10
.
=
1(
2
/
3
,
1
/
3
,
−
2
/
3
.
50.
)
a
)
=
√
1+9+4=
√
14 ;
u
=
1
√
14
(
i
−
3
j
+2
k
)=
1
√
14
i
−
3
√
14
j
+
2
√
14
k
51. b
=4
a
=4
i
−
4
j
+4
k
52.
)
a
)
=
√
36+9+4=7;
b
=
−
1
2
)
1
7
i
1(
6
,
3
,
−
2
.
=
1
3
7
,
−
3
14
,
1
7
.
53.
Exercises 7.3
1. a
·
b
= 10(5) cos(
π/
4) = 25
√
2
2. a
·
b
= 6(12) cos(
π/
6) = 36
√
3
3. a
·
b
=2(
−
1)+(
−
3)2 + 4(5) = 12
4. b
·
c
=(
−
1)3 + 2(6) + 5(
−
1) = 4
5. a
·
c
= 2(3) + (
−
3)6+4(
−
1) =
−
16
6. a
·
(
b
+
c
) = 2(2) + (
−
3)8 + 4(4) =
−
4
7. a
·
(4
b
)=2(
−
4)+(
−
3)8 + 4(20) = 48
8. b
·
(
a
−
c
)=(
−
1)(
−
1)+2(
−
9) + 5(5) = 8
9. a
·
a
=2
2
+(
−
3)
2
+4
2
=29
10.
(2
b
)
·
(3
c
)=(
−
2)9 + 4(18) + 10(
−
3) = 24
11. a
·
(
a
+
b
+
c
) = 2(4) + (
−
3)5 + 4(8) = 25
12.
(2
a
)
·
(
a
−
2
b
) = 4(4) + (
−
6)(
−
7)+8(
−
6) = 10
13.
)
a
·
b
b
·
b
i
b
=

2(
−
1)+(
−
3)2 + 4(5)
(
−
1)
2
+2
2
+5
2

1(
1
,
2
,
5
.
=
12
30
1(
1
,
2
,
5
.
=
1(
2
/
5
,
4
/
5
,
2
.
14.
(
c
·
b
)
a
= [3(
−
1)+6(2)+(
−
1)5]
1
2
,
−
3
,
4
.
=4
1
2
,
−
3
,
4
.
=
1
8
,
−
12
,
16
.
15.
a and f, b and e, c and d
16. (a) a
·
b
=2
·
3+(
−
c
)2 + 3(4) = 0 =
⇒
c
=9
295

296. Exercises 7.3
(b) a
·
b
=
c
(
−
3) +
1
2
(4) +
c
2
=
c
2
−
3
c
+2=(
c
−
2)(
c
−
1) = 0 =
⇒
c
=1,2
17.
Solving the system of equations 3
x
1
+
y
1
−
1=0,
−
3
x
1
+2
y
1
+ 2 = 0 gives
x
1
=4
/
9 and
y
1
=
−
1
/
3.Thus,
v
=
1
4
/
9
,
−
1
/
3
,
1
.
.
18.
If
a
and
b
represent adjacent sides of the rhombus, then
)
a
)
=
)
b
)
, the diagonals of the rhombus are
a
+
b
and
a
−
b
, and
(
a
+
b
)
·
(
a
−
b
)=
a
·
a
−
a
·
b
+
b
·
a
−
b
·
b
=
a
·
a
−
b
·
b
=
)
a
)
2
()
b
)
2
=0
.
Thus, the diagonals are perpendicular.
19.
Since
c
·
a
=
)
b
−
a
·
b
)
a
)
2
a
i
·
a
=
b
·
a
−
a
·
b
)
a
)
2
(
a
·
a
)=
b
·
a
−
a
·
b
)
a
)
2
)
a
)
2
=
b
·
a
−
a
·
b
=0
,
the vectors
c
and
a
are orthogonal.
20. a
·
b
= 1(1) +
c
(1) =
c
+1;
)
a
)
=
√
1+
c
2
,
)
b
)
=
√
2
cos 45
◦
=
1
√
2
=
c
+1
√
1+
c
2
√
2
=
⇒
a
1+
c
2
=
c
+1 =
⇒
1+
c
2
=
c
2
+2
c
+1 =
⇒
c
=0
21. a
·
b
= 3(2) + (
−
1)2 = 4;
)
a
)
=
√
10 ,
)
b
)
=2
√
2
cos
θ
=
4
(
√
10)(2
√
2)
=
1
√
5
=
⇒
θ
= cos
−
1
1
√
5
≈
1
.
11 rad
≈
63
.
43
◦
22. a
·
b
=2(
−
3)+1(
−
4) =
−
10;
)
a
)
=
√
5,
)
b
)
=5
cos
θ
=
−
10
(
√
5)5
=
−
2
√
5
=
⇒
θ
= cos
−
1
(
−
2
/
√
5)
≈
2
.
68 rad
≈
153
.
43
◦
23. a
·
b
=2(
−
1)+4(
−
1) + 0(4) =
−
6;
)
a
)
=2
√
5,
)
b
)
=3
√
2
cos
θ
=
−
6
(2
√
5)(3
√
2)
=
−
1
√
10
=
⇒
θ
= cos
−
1
(
−
1
/
√
10 )
≈
1
.
89 rad
≈
108
.
43
◦
24. a
·
b
=
1
2
(2) +
1
2
(
−
4) +
3
2
(6) = 8;
)
a
)
=
√
11
/
2,
)
b
)
=2
√
14
cos
θ
=
8
(
√
11
/
2)(2
√
14 )
=
8
√
154
=
⇒
θ
= cos
−
1
(8
/
√
154 )
≈
0
.
87 rad
≈
49
.
86
◦
25.
)
a
)
=
√
14 ; cos
α
=1
/
√
14
,α
≈
74
.
50
◦
; cos
β
=2
/
√
14 ,
β
≈
57
.
69
◦
; cos
γ
=3
/
√
14 ,
γ
≈
36
.
70
◦
26.
)
a
)
= 9; cos
α
=2
/
3,
α
≈
48
.
19
◦
; cos
β
=2
/
3,
β
≈
48
.
19
◦
; cos
γ
=
−
1
/
3,
γ
≈
109
.
47
◦
27.
)
a
)
= 2; cos
α
=1
/
2,
α
=60
◦
; cos
β
=0,
β
=90
◦
; cos
γ
=
−
√
3
/
2,
γ
= 150
◦
28.
)
a
)
=
√
78 ; cos
α
=5
/
√
78 ,
α
≈
55
.
52
◦
; cos
β
=7
/
√
78 ,
β
≈
37
.
57
◦
; cos
γ
=2
/
√
78 ,
γ
≈
76
.
91
◦
29.
Let
θ
be the angle between
−
−
→
AD
and
−
−
→
AB
and
a
be the length of an edge of the cube.Then
−
−
→
AD
=
a
i
+
a
j
+
a
k
,
−
−
→
AB
=
a
i
and
cos
θ
=
−
−
→
AD
·
−
−
→
AB
)
−
−
→
AD
))
−
−
→
AB
)
=
a
2
√
3
a
2
√
a
2
=
1
√
3
so
θ
≈
0
.
955317 radian or 54
.
7356
◦
.Letting
φ
be the angle between
−
−
→
AD
and
−→
AC
and noting that
−→
AC
=
a
i
+
a
j
we have
cos
φ
=
−
−
→
AD
·
−→
AC
)
−
−
→
AD
))
−→
AC
)
=
a
2
+
a
2
√
3
a
2
√
2
a
2
=

2
3
so
φ
≈
0
.
61548 radian or 35
.
2644
◦
.
296

297. Exercises 7.3
30.
If
a
and
b
are orthogonal, then
a
·
b
= 0 and
cos
α
1
cos
α
2
+ cos
β
1
cos
β
2
+ cos
γ
1
cos
γ
2
=
a
1
a
a
a
b
1
a
b
a
+
a
2
a
a
a
b
2
a
b
a
+
a
3
a
a
a
b
3
a
b
a
=
1
a
a
aa
b
a
(
a
1
b
1
+
a
2
b
2
+
a
3
b
3
)=
1
a
a
aa
b
a
(
a
·
b
)=0
.
31. a
=
f
5
,
7
,
4
g
;
a
a
a
=3
√
10 ; cos
α
=5
/
3
√
10 ,
α
≈
58
.
19
◦
; cos
β
=7
/
3
√
10 ,
β
≈
42
.
45
◦
; cos
γ
=4
/
3
√
10 ,
γ
≈
65
.
06
◦
32.
We want cos
α
= cos
β
= cos
γ
or
a
1
=
a
2
=
a
3
.Letting
a
1
=
a
2
=
a
3
= 1 we obtain the vector
i
+
j
+
k
.A unit
vector in the same direction is
1
√
3
i
+
1
√
3
j
+
1
√
3
k
.
33.
comp
b
a
=
a
·
b
/
a
b
a
=
f
1
,
−
1
,
3
g,f
2
,
6
,
3
g
/
7=5
/
7
34.
comp
a
b
=
b
·
a
/
a
a
a
=
f
2
,
6
,
3
g,f
1
,
−
1
,
3
g
/
√
11 = 5
/
√
11
35. b
−
a
=
f
1
,
7
,
0
g
; comp
a
(
b
−
a
)=(
b
−
a
)
·
a
/
a
a
a
=
f
1
,
7
,
0
g,f
1
,
−
1
,
3
g
/
√
11 =
−
6
/
√
11
36. a
+
b
=
f
3
,
5
,
6
g
;2
b
=
f
4
,
12
,
6
g
; comp
2
b
(
a
+
b
)
·
2
b
/
|
2
b
|
=
f
3
,
5
,
6
g,f
4
,
12
,
6
g
/
14=54
/
7
37.
−
−
→
OP
=3
i
+10
j
;
a
−
−
→
OP
a
=
√
109 ; comp
−
→
OP
a
=
a
·
−
−
→
OP/
a
−
−
→
OP
a
=(4
i
+6
j
)
·
(3
i
+10
j
)
/
√
109=72
/
√
109
38.
−
−
→
OP
=
f
1
,
−
1
,
1
g
;
a
−
−
→
OP
a
=
√
3 ; comp
−
→
OP
a
=
a
·
−
−
→
OP/
a
−
−
→
OP
a
=
f
2
,
1
,
−
1
g,f
1
,
−
1
,
1
g
/
√
3=0
39. (a)
comp
b
a
=
a
·
b
/
a
b
a
=(
−
5
i
+5
j
)
·
(
−
3
i
+4
j
)
/
5=7
proj
b
a
= (comp
b
a
)
b
/
a
b
a
=7(
−
3
i
+4
j
)
/
5=
−
21
5
i
+
28
5
j
(b)
proj
b
⊥
a
=
a
−
proj
b
a
=(
−
5
i
+5
j
)
−
(
−
21
5
i
+
28
5
j
)=
−
4
5
i
−
3
5
j
40. (a)
comp
b
a
=
a
·
b
/
a
b
a
=(4
i
+2
j
)
·
(
−
3
i
+
j
)
/
√
10 =
−
√
10
proj
b
a
= (comp
b
a
)
b
/
a
b
a
=
−
√
10(
−
3
i
+
j
)
/
√
10=3
i
−
j
(b)
proj
b
⊥
a
=
a
−
proj
b
a
=(4
i
+2
j
)
−
(3
i
−
j
)=
i
+3
j
41. (a)
comp
b
a
=
a
·
b
/
a
b
a
=(
−
i
−
2
j
+7
k
)
·
(6
i
−
3
j
−
2
k
)
/
7=
−
2
proj
b
a
= (comp
b
a
)
b
/
a
b
a
=
−
2(6
i
−
3
j
−
2
k
)
/
7=
−
12
7
i
+
6
7
j
+
4
7
k
(b)
proj
b
⊥
a
=
a
−
proj
b
a
=(
−
i
−
2
j
+7
k
)
−
(
−
12
7
i
+
6
7
j
+
4
7
k
)=
5
7
i
−
20
7
j
+
45
7
k
42. (a)
comp
b
a
=
a
·
b
/
a
b
a
=
f
1
,
1
,
1
g,fn
2
,
2
,
−
1
g
/
3=
−
1
/
3
proj
b
a
= (comp
b
a
)
b
/
a
b
a
=
−
1
3
fn
2
,
2
,
−
1
g
/
3=
f
2
/
9
,
−
2
/
9
,
1
/
9
g
(b)
proj
b
⊥
a
=
a
−
proj
b
a
=
f
1
,
1
,
1
gnf
2
/
9
,
−
2
/
9
,
1
/
9
g
=
f
7
/
9
,
11
/
9
,
8
/
9
g
43. a
+
b
=3
i
+4
j
;
a
a
+
b
a
= 5; comp
(
a
+
b
)
a
=
a
·
(
a
+
b
)
/
a
a
+
b
a
=(4
i
+3
j
)
·
(3
i
+4
j
)
/
5=24
/
5
proj
(
a
+
b
)
a
= (comp
(
a
+
b
)
a
)(
a
+
b
)
/
a
a
+
b
a
=
24
5
(3
i
+4
j
)
/
5=
72
25
i
+
96
25
j
44. a
−
b
=5
i
+2
j
;
a
a
−
b
a
=
√
29 ; comp
(
a
−
b
)
b
=
b
·
(
a
−
b
)
/
a
a
−
b
a
=(
−
i
+
j
)
·
(5
i
+2
j
)
/
√
29 =
−
3
/
√
29
proj
(
a
−
b
)
b
= (comp
(
a
−
b
)
b
)(
a
−
b
)
/
a
a
−
b
a
=
−
3
√
29
(5
i
+2
j
)
/
√
29 =
−
15
29
i
−
6
29
j
proj
(
a
−
b
)
⊥
b
=
b
−
proj
(
a
−
b
)
b
=(
−
i
+
j
)
−
(
−
15
29
i
−
6
29
j
)=
−
14
29
i
+
35
29
j
45.
We identify
a
F
a
= 20,
θ
=60
◦
and
a
d
a
= 100.Then
W
=
a
F
aa
d
a
cos
θ
= 20(100)(
1
2
) = 1000 ft-lb.
46.
We identify
d
=
−
i
+3
j
+8
k
.Then
W
=
F
·
d
=
f
4
,
3
,
5
g,fn
1
,
3
,
8
g
= 45 N-m.
47. (a)
Since
w
and
d
are orthogonal,
W
=
w
·
d
=0.
(b)
We identify
θ
=0
◦
.Then
W
=
a
F
aa
d
a
cos
θ
= 30(
√
4
2
+3
2
) = 150 N-m.
48.
Using
d
=6
i
+2
j
and
F
=3(
3
5
i
+
4
5
j
),
W
=
F
·
d
=
f
9
5
,
12
5
g,f
6
,
2
g
=
78
5
ft-lb.
297

298. Exercises 7.3
49.
Let
a
and
b
be vectors from the center of the carbon atom to the centers of two distinct hydrogen atoms.The
distance between two hydrogen atoms is then
)
b
−
a
)
=
a
(
b
−
a
)
·
(
b
−
a
)=
√
b
·
b
−
2
a
·
b
+
a
·
a
=
a
)
b
)
2
+
)
a
)
2
−
2
)
a
))
b
)
cos
θ
=
a
(1
.
1)
2
+(1
.
1)
2
−
2(1
.
1)(1
.
1) cos 109
.
5
◦
=
a
1
.
21+1
.
21
−
2
.
42(
−
0
.
333807)
≈
1
.
80 angstroms
.
50.
Using the fact that
|
cos
θ
|≤
1, we have
|
a
·
b
|
=
)
a
))
b
)c
cos
θ
|
=
)
a
))
b
)c
cos
θ
ce)
a
))
b
)
.
51.
)
a
+
b
)
2
=(
a
+
b
)
·
(
a
+
b
)=
a
·
a
+2
a
·
b
+
b
·
b
=
)
a
)
2
+2
a
·
b
+
)
b
)
2
e)
a
)
2
+2
|
a
·
b
|
+
)
b
)
2
since
x
≤|
x
|
e)
a
)
2
+2
)
a
))
b
)
+
)
b
)
2
=(
)
a
)
+
)
b
)
)
2
by Problem 50
Thus, since
)
a
+
b
)
and
)
a
)
+
)
b
)
are positive,
)
a
+
b
)e)
a
)
+
)
b
)
.
52.
Let
P
1
(
x
1
,y
1
) and
P
2
(
x
2
,y
2
) be distinct points on the line
ax
+
by
=
−
c
.Then
n
·
−
−−
→
P
1
P
2
=
1
a, b
.d1
x
2
−
x
1
,y
2
−
y
1
.
=
ax
2
−
ax
1
+
by
2
−
by
1
=(
ax
2
+
by
2
)
−
(
ax
1
+
by
1
)=
−
c
−
(
−
c
)=0
,
and the vectors are perpendicular.Thus,
n
is perpendicular to the line.
53.
Let
θ
be the angle between
n
and
−
−−
→
P
2
P
1
.Then
d
=
)
−
−−
→
P
1
P
2
)c
cos
θ
|
=
|
n
·
−
−−
→
P
2
P
1
|
)
n
)
=
c1
a, b
.d1
x
1
−
x
2
,y
1
−
y
2
.c
√
a
2
+
b
2
=
|
ax
1
−
ax
2
+
by
1
−
by
2
|
√
a
2
+
b
2
=
|
ax
1
+
by
1
−
(
ax
2
+
by
2
)
|
√
a
2
+
b
2
=
|
ax
1
+
by
1
−
(
−
c
)
|
√
a
2
+
b
2
=
|
ax
1
+
by
1
+
c
|
√
a
2
+
b
2
.
Exercises 7.4
1. a
×
b
=
(
(
(
(
(
(
(
ijk
1
−
10
035
(
(
(
(
(
(
(
=
(
(
(
(
−
10
35
(
(
(
(
i
−
(
(
(
(
10
05
(
(
(
(
j
+
(
(
(
(
1
−
1
03
(
(
(
(
k
=
−
5
i
−
5
j
+3
k
2. a
×
b
=
(
(
(
(
(
(
(
ij k
21 0
40
−
1
(
(
(
(
(
(
(
=
(
(
(
(
10
0
−
1
(
(
(
(
i
−
(
(
(
(
20
4
−
1
(
(
(
(
j
+
(
(
(
(
21
40
(
(
(
(
k
=
−
i
+2
j
−
4
k
3. a
×
b
=
(
(
(
(
(
(
(
ijk
1
−
31
204
(
(
(
(
(
(
(
=
(
(
(
(
−
31
04
(
(
(
(
i
−
(
(
(
(
11
24
(
(
(
(
j
+
(
(
(
(
1
−
3
20
(
(
(
(
k
=
1(
12
,
−
2
,
6
.
4. a
×
b
=
(
(
(
(
(
(
(
ijk
111
−
523
(
(
(
(
(
(
(
=
(
(
(
(
11
23
(
(
(
(
i
−
(
(
(
(
11
−
53
(
(
(
(
j
+
(
(
(
(
11
−
52
(
(
(
(
k
=
1
1
,
−
8
,
7
.
5. a
×
b
=
(
(
(
(
(
(
(
ijk
2
−
12
−
13
−
1
(
(
(
(
(
(
(
=
(
(
(
(
−
12
3
−
1
(
(
(
(
i
−
(
(
(
(
22
−
1
−
1
(
(
(
(
j
+
(
(
(
(
2
−
1
−
13
(
(
(
(
k
=
−
5
i
+5
k
298

299. Exercises 7.4
6. a
×
b
=
(
(
(
(
(
(
(
ij k
41
−
5
23
−
1
(
(
(
(
(
(
(
=
(
(
(
(
1
−
5
3
−
1
(
(
(
(
i
−
(
(
(
(
4
−
5
2
−
1
(
(
(
(
j
+
(
(
(
(
41
23
(
(
(
(
k
=14
i
−
6
j
+10
k
7. a
×
b
=
(
(
(
(
(
(
(
ijk
1
/
201
/
2
460
(
(
(
(
(
(
(
=
(
(
(
(
01
/
2
60
(
(
(
(
i
−
(
(
(
(
1
/
21
/
2
40
(
(
(
(
j
+
(
(
(
(
1
/
20
46
(
(
(
(
k
=
1(
3
,
2
,
3
.
8. a
×
b
=
(
(
(
(
(
(
(
ijk
050
2
−
34
(
(
(
(
(
(
(
=
(
(
(
(
50
−
34
(
(
(
(
i
−
(
(
(
(
00
24
(
(
(
(
j
+
(
(
(
(
05
2
−
3
(
(
(
(
k
=
1
20
,
0
,
−
10
.
9. a
×
b
=
(
(
(
(
(
(
(
ijk
22
−
4
−
3
−
36
(
(
(
(
(
(
(
=
(
(
(
(
2
−
4
−
36
(
(
(
(
i
−
(
(
(
(
2
−
4
−
36
(
(
(
(
j
+
(
(
(
(
22
−
3
−
3
(
(
(
(
k
=
1
0
,
0
,
0
.
10. a
×
b
=
(
(
(
(
(
(
(
ij k
81
−
6
1
−
210
(
(
(
(
(
(
(
=
(
(
(
(
1
−
6
−
210
(
(
(
(
i
−
(
(
(
(
8
−
6
110
(
(
(
(
j
+
(
(
(
(
81
1
−
2
(
(
(
(
k
=
1(
2
,
−
86
,
−
17
.
11.
−
−−
→
P
1
P
2
=(
−
2
,
2
,
−
4);
−
−−
→
P
1
P
3
=(
−
3
,
1
,
1)
−
−−
→
P
1
P
2
×
−
−−
→
P
1
P
3
=
(
(
(
(
(
(
(
ijk
−
22
−
4
−
31 1
(
(
(
(
(
(
(
=
(
(
(
(
2
−
4
11
(
(
(
(
i
−
(
(
(
(
−
2
−
4
−
31
(
(
(
(
j
+
(
(
(
(
−
22
−
31
(
(
(
(
k
=6
i
+14
j
+4
k
12.
−
−−
→
P
1
P
2
=(0
,
1
,
1);
−
−−
→
P
1
P
3
=(1
,
2
,
2);
−
−−
→
P
1
P
2
×
−
−−
→
P
1
P
3
=
(
(
(
(
(
(
(
ijk
011
122
(
(
(
(
(
(
(
=
(
(
(
(
11
22
(
(
(
(
i
−
(
(
(
(
01
12
(
(
(
(
j
+
(
(
(
(
01
12
(
(
(
(
k
=
j
−
k
13. a
×
b
=
(
(
(
(
(
(
(
ij k
27
−
4
11
−
1
(
(
(
(
(
(
(
=
(
(
(
(
7
−
4
1
−
1
(
(
(
(
i
−
(
(
(
(
2
−
4
1
−
1
(
(
(
(
j
+
(
(
(
(
27
11
(
(
(
(
k
=
−
3
i
−
2
j
−
5
k
is perpendicular to both
a
and
b
.
14. a
×
b
=
(
(
(
(
(
(
(
ijk
−
1
−
24
4
−
10
(
(
(
(
(
(
(
=
(
(
(
(
−
24
−
10
(
(
(
(
i
−
(
(
(
(
−
14
40
(
(
(
(
j
+
(
(
(
(
−
1
−
2
4
−
1
(
(
(
(
k
=
1
4
,
16
,
9
.
is perpendicular to both
a
and
b
.
15. a
×
b
=
(
(
(
(
(
(
(
ij k
5
−
21
20
−
7
(
(
(
(
(
(
(
=
(
(
(
(
−
21
0
−
7
(
(
(
(
i
−
(
(
(
(
51
2
−
7
(
(
(
(
j
+
(
(
(
(
5
−
2
20
(
(
(
(
k
=
1
14
,
37
,
4
.
a
·
(
a
×
b
)=
1
5
,
−
2
,
−
1
.d1
14
,
37
,
4
.
=70
−
74+4=0;
b
·
(
a
×
b
)=
1
2
,
0
,
−
7
.d1
14
,
37
,
4
.
=28+0
−
28 = 0
16. a
×
b
=
(
(
(
(
(
(
(
ijk
1
/
2
−
1
/
40
2
−
26
(
(
(
(
(
(
(
=
(
(
(
(
−
1
/
40
−
26
(
(
(
(
i
−
(
(
(
(
1
/
20
26
(
(
(
(
j
+
(
(
(
(
1
/
2
−
1
/
4
2
−
2
(
(
(
(
k
=
−
3
2
i
−
3
j
−
1
2
k
a
·
(
a
×
b
)=(
1
2
i
−
1
4
j
)
·
(
−
3
2
i
−
3
j
−
1
2
k
)=
−
3
4
+
3
4
+0=0
b
·
(
a
×
b
)=(2
i
−
2
j
+6
k
)
·
(
−
3
2
i
−
3
j
−
1
2
k
)=
−
3+6
−
3=0
299

300. Exercises 7.4
17. (a) b
×
c
=
(
(
(
(
(
(
(
ijk
211
311
(
(
(
(
(
(
(
=
(
(
(
(
11
11
(
(
(
(
i
−
(
(
(
(
21
31
(
(
(
(
j
+
(
(
(
(
21
31
(
(
(
(
k
=
j
−
k
a
×
(
b
×
c
)=
(
(
(
(
(
(
(
ij k
1
−
12
01
−
1
(
(
(
(
(
(
(
=
(
(
(
(
−
12
1
−
1
(
(
(
(
i
−
(
(
(
(
12
0
−
1
(
(
(
(
j
+
(
(
(
(
1
−
1
01
(
(
(
(
k
=
−
i
+
j
+
k
(b) a
·
c
=(
i
−
j
+2
k
)
·
(3
i
+
j
+
k
) = 4; (
a
·
c
)
b
= 4(2
i
+
j
+
k
)=8
i
+4
j
+4
k
a
·
b
=(
i
−
j
+2
k
)
·
(2
i
+
j
+
k
) = 3; (
a
·
b
)
c
= 3(3
i
+
j
+
k
)=9
i
+3
j
+3
k
a
×
(
b
×
c
)=(
a
·
c
)
b
−
(
a
·
b
)
c
=(8
i
+4
j
+4
k
)
−
(9
i
+3
j
+3
k
)=
−
i
+
j
+
k
18. (a) b
×
c
=
(
(
(
(
(
(
(
ijk
12
−
1
−
15 8
(
(
(
(
(
(
(
=
(
(
(
(
2
−
1
58
(
(
(
(
i
−
(
(
(
(
1
−
1
−
18
(
(
(
(
j
+
(
(
(
(
12
−
15
(
(
(
(
k
=21
i
−
7
j
+7
k
a
×
(
b
×
c
)=
(
(
(
(
(
(
(
ijk
30
−
4
21
−
77
(
(
(
(
(
(
(
=
(
(
(
(
0
−
4
−
77
(
(
(
(
i
−
(
(
(
(
3
−
4
21 7
(
(
(
(
j
+
(
(
(
(
30
21
−
7
(
(
(
(
k
=
−
28
i
−
105
j
−
21
k
(b) a
·
c
=(3
i
−
4
k
)
·
(
−
i
+5
j
+8
k
)=
−
35; (
a
·
c
)
b
=
−
35(
i
+2
j
−
k
)=
−
35
i
−
70
j
+35
k
a
·
b
=(3
i
−
4
k
)
·
(
i
+2
j
−
k
) = 7; (
a
·
b
)
c
=7(
−
i
+5
j
+8
k
)=
−
7
i
+35
j
+56
k
a
×
(
b
×
c
)=(
a
·
c
)
b
−
(
a
·
b
)
c
=(
−
35
i
−
70
j
+35
k
)
−
(
−
7
i
+35
j
+56
k
)=
−
28
i
−
105
j
−
21
k
19.
(2
i
)
×
j
=2(
i
×
j
)=2
k
20. i
×
(
−
3
k
)=
−
3(
i
×
k
)=
−
3(
−
j
)=3
j
21. k
×
(2
i
−
j
)=
k
×
(2
i
)+
k
×
(
−
j
)=2(
k
×
i
)
−
(
k
×
j
)=2
j
−
(
−
i
)=
i
+2
j
22. i
×
(
j
×
k
)=
i
×
i
=
0
23.
[(2
k
)
×
(3
j
)]
×
(4
j
)=[2
·
3(
k
×
j
)
×
(4
j
)] = 6(
−
i
)
×
4
j
=(
−
6)(4)(
i
×
j
)=
−
24
k
24.
(2
i
−
j
+5
k
)
×
i
=(2
i
×
i
)+(
−
j
×
i
)+(5
k
×
i
)=2(
i
×
i
)+(
i
×
j
)+5(
k
×
i
)=5
j
+
k
25.
(
i
+
j
)
×
(
i
+5
k
)=[(
i
+
j
)
×
i
]+[(
i
+
j
)
×
5
k
]=(
i
×
i
)+(
j
×
i
)+(
i
×
5
k
)+(
j
×
5
k
)
=
−
k
+5(
−
j
)+5
i
=5
i
−
5
j
−
k
26. i
×
k
−
2(
j
×
i
)=
−
j
−
2(
−
k
)=
−
j
+2
k
27. k
·
(
j
×
k
)=
k
·
i
=0
28. i
·
[
j
×
(
−
k
)] =
i
·
[
−
(
j
×
k
)] =
i
·
(
−
i
)=
−
(
i
·
i
)=
−
1
29.
)
4
j
−
5(
i
×
j
)
)
=
)
4
j
−
5
k
)
=
√
41
30.
(
i
×
j
)
·
(3
j
×
i
)=
k
·
(
−
3
k
)=
−
3(
k
·
k
)=
−
3
31. i
×
(
i
×
j
)=
i
×
k
=
−
j 32.
(
i
×
j
)
×
i
=
k
×
i
=
j
33.
(
i
×
i
)
×
j
=
0
×
j
=
0 34.
(
i
·
i
)(
i
×
j
)=1(
k
)=
k
35.
2
j
·
[
i
×
(
j
−
3
k
)] = 2
j
·
[(
i
×
j
)+(
i
×
(
−
3
k
)] = 2
j
·
[
k
+3(
k
×
i
)] = 2
j
·
(
k
+3
j
)=2
j
·
k
+2
j
·
3
j
=2(
j
·
k
)+6(
j
·
j
) = 2(0) + 6(1) = 6
36.
(
i
×
k
)
×
(
j
×
i
)=(
−
j
)
×
(
−
k
)=(
−
1)(
−
1)(
j
×
k
)=
j
×
k
=
i
37. a
×
(3
b
)=3(
a
×
b
) = 3(4
i
−
3
j
+6
k
)=12
i
−
9
j
+18
k
38. b
×
a
=
−
a
×
b
=
−
(
a
×
b
)=
−
4
i
+3
j
−
6
k
39.
(
−
a
)
×
b
=
−
(
a
×
b
)=
−
4
i
+3
j
−
6
k
300

301. Exercises 7.4
40.
|
a
×
b
|
=
a
4
2
+(
−
3)
2
+6
2
=
√
61
41.
(
a
×
b
)
×
c
=
(
(
(
(
(
(
(
ij k
4
−
36
24
−
1
(
(
(
(
(
(
(
=
(
(
(
(
−
36
4
−
1
(
(
(
(
i
−
(
(
(
(
46
2
−
1
(
(
(
(
j
+
(
(
(
(
4
−
3
2
−
4
(
(
(
(
k
=
−
21
i
+16
j
+22
k
42.
(
a
×
b
)
·
c
= 4(2) + (
−
3)4+6(
−
1) =
−
10
43. a
·
(
b
×
c
)=(
a
×
b
)
·
c
= 4(2) + (
−
3)4+6(
−
1) =
−
10
44.
(4
a
)
·
(
b
×
c
)=(4
a
×
b
)
·
c
=4(
a
×
b
)
·
c
= 16(2) + (
−
12)4 + 24(
−
1) =
−
40
45. (a)
Let
A
=(1
,
3
,
0),
B
=(2
,
0
,
0),
C
=(0
,
0
,
4), and
D
=(1
,
−
3
,
4).Then
−
−
→
AB
=
i
−
3
j
,
−→
AC
=
−
i
−
3
j
+4
k
,
−
−
→
CD
=
i
−
3
j
, and
−
−
→
BD
=
−
i
−
3
j
+4
k
.Since
−
−
→
AB
=
−
−
→
CD
and
−→
AC
=
−
−
→
BD
, the quadrilateral is a parallelogram.
(b)
Computing
−
−
→
AB
×
−→
AC
=
(
(
(
(
(
(
(
ijk
1
−
30
−
1
−
34
(
(
(
(
(
(
(
=
−
12
i
−
4
j
−
6
k
we find that the area is
)(
12
i
−
4
j
−
6
k
)
=
√
144+16+36=14.
46. (a)
Let
A
=(3
,
4
,
1),
B
=(
−
1
,
4
,
2),
C
=(2
,
0
,
2) and
D
=(
−
2
,
0
,
3).Then
−
−
→
AB
=
−
4
i
+
k
,
−→
AC
=
−
i
−
4
j
+
k
,
−
−
→
CD
=
−
4
i
+
k
, and
−
−
→
BD
=
−
i
−
4
j
+
k
.Since
−
−
→
AB
=
−
−
→
CD
and
−→
AC
=
−
−
→
BD
, the quadrilateral is a parallelogram.
(b)
Computing
−
−
→
AB
×
−→
AC
=
(
(
(
(
(
(
(
ijk
−
401
−
1
−
41
(
(
(
(
(
(
(
=4
i
+3
j
+16
k
we find that the area is
)
4
i
+3
j
+16
k
)
=
√
16+9+256=
√
281
≈
16
.
76.
47.
−
−−
→
P
1
P
2
=
j
;
−
−−
→
P
2
P
3
=
−
j
+
k
−
−−
→
P
1
P
2
×
−
−−
→
P
2
P
3
=
(
(
(
(
(
(
(
ijk
010
0
−
11
(
(
(
(
(
(
(
=
(
(
(
(
10
−
11
(
(
(
(
i
−
(
(
(
(
00
01
(
(
(
(
j
+
(
(
(
(
01
0
−
1
(
(
(
(
k
=
i
;
A
=
1
2
)
i
)
=
1
2
sq.unit
48.
−
−−
→
P
1
P
2
=
j
+2
k
;
−
−−
→
P
2
P
3
=2
i
+
j
−
2
k
−
−−
→
P
1
P
2
×
−
−−
→
P
2
P
3
=
(
(
(
(
(
(
(
ij k
01 2
21
−
2
(
(
(
(
(
(
(
=
(
(
(
(
12
1
−
2
(
(
(
(
i
−
(
(
(
(
02
2
−
2
(
(
(
(
j
+
(
(
(
(
01
21
(
(
(
(
k
=
−
4
i
+4
j
−
2
k
A
=
1
2
)(
4
i
+4
j
−
2
k
)
= 3 sq.units
49.
−
−−
→
P
1
P
2
=
−
3
j
−
k
;
−
−−
→
P
2
P
3
=
−
2
i
−
k
−
−−
→
P
1
P
2
×
−
−−
→
P
2
P
3
=
(
(
(
(
(
(
(
ijk
0
−
3
−
1
−
20
−
1
(
(
(
(
(
(
(
=
(
(
(
(
−
3
−
1
0
−
1
(
(
(
(
i
−
(
(
(
(
0
−
1
−
2
−
1
(
(
(
(
j
+
(
(
(
(
0
−
3
−
20
(
(
(
(
k
=3
i
+2
j
−
6
k
A
=
1
2
)
3
i
+2
j
−
6
k
)
=
7
2
sq.units
50.
−
−−
→
P
1
P
2
=
−
i
+3
k
;
−
−−
→
P
2
P
3
=2
i
+4
j
−
k
−
−−
→
P
1
P
2
×
−
−−
→
P
2
P
3
=
(
(
(
(
(
(
(
ijk
−
10 3
24
−
1
(
(
(
(
(
(
(
=
(
(
(
(
03
4
−
1
(
(
(
(
i
−
(
(
(
(
−
13
2
−
1
(
(
(
(
j
+
(
(
(
(
−
10
24
(
(
(
(
k
=
−
12
i
+5
j
−
4
k
A
=
1
2
)(
12
i
+5
j
−
4
k
)
=
√
185
2
sq.units
301

302. Exercises 7.4
51. b
×
c
=
(
(
(
(
(
(
(
ijk
−
140
222
(
(
(
(
(
(
(
=
(
(
(
(
40
22
(
(
(
(
i
−
(
(
(
(
−
10
22
(
(
(
(
j
+
(
(
(
(
−
14
22
(
(
(
(
k
=8
i
+2
j
−
10
k
v
=
|
a
·
(
b
×
c
)
|
=
|
(
i
+
j
)
·
(8
i
+2
j
−
10
k
)
|
=
|
8+2+0
|
= 10 cu.units
52. b
×
c
=
(
(
(
(
(
(
(
ijk
141
115
(
(
(
(
(
(
(
=
(
(
(
(
41
15
(
(
(
(
i
−
(
(
(
(
11
15
(
(
(
(
j
+
(
(
(
(
14
11
(
(
(
(
k
=19
i
−
4
j
−
3
k
v
=
|
a
·
(
b
×
c
)
|
=
|
(3
i
+
j
+
k
)
·
(19
i
−
4
j
−
3
k
)
|
=
|
57
−
4
−
3
|
= 50 cu.units
53. b
×
c
=
(
(
(
(
(
(
(
ijk
−
26
−
6
5
/
231
/
2
(
(
(
(
(
(
(
=
(
(
(
(
6
−
6
32
/
2
(
(
(
(
i
−
(
(
(
(
−
2
−
6
5
/
21
/
2
(
(
(
(
j
+
(
(
(
(
−
26
5
/
23
(
(
(
(
k
=21
i
−
14
j
−
21
k
a
·
(
b
×
c
)=(4
i
+6
j
)
·
(21
i
−
14
j
−
21
k
)=84
−
84 + 0 = 0.The vectors are coplanar.
54.
The four points will be coplanar if the three vectors
−
−−
→
P
1
P
2
=
1
3
,
−
1
,
−
1
.
,
−
−−
→
P
2
P
3
=
1(
3
,
−
5
,
13
.
, and
−
−−
→
P
3
P
4
=
1(
8
,
7
,
−
6
.
are coplanar.
−
−−
→
P
2
P
3
×
−
−−
→
P
3
P
4
=
(
(
(
(
(
(
(
ijk
−
3
−
513
−
87
−
6
(
(
(
(
(
(
(
=
(
(
(
(
−
513
7
−
6
(
(
(
(
i
−
(
(
(
(
−
313
−
8
−
6
(
(
(
(
j
+
(
(
(
(
−
3
−
5
−
87
(
(
(
(
k
=
1(
61
,
−
122
,
−
61
.
−
−−
→
P
1
P
2
·
(
−
−−
→
P
2
P
3
×
−
−−
→
P
3
P
4
)=
1
3
,
−
1
,
−
1
.d1(
61
,
−
122
,
−
61
.
=
−
183 + 122 + 61 = 0
The four points are coplanar.
55. (a)
Since
θ
=90
◦
,
)
a
×
b
)
=
)
a
))
b
)c
sin 90
◦
|
=6
.
4(5) = 32.
(b)
The direction of
a
×
b
is into the fourth quadrant of the
xy
-plane or to the left of the plane determined by
a
and
b
as shown in Figure 6
.
54 in the text.It makes an angle of 30
◦
with the positive
x
-axis.
(c)
We identify
n
=(
√
3
i
−
j
)
/
2.Then
a
×
b
=32
n
=16
√
3
i
−
16
j
.
56.
Using Definition 7
.
4,
a
×
b
=
√
27 (8) sin 120
◦
n
=24
√
3(
√
3
/
2)
n
=36
n
.By the right-hand rule,
n
=
j
or
n
=
−
j
.Thus,
a
×
b
=36
j
or
−
36
j
.
57. (a)
We note first that
a
×
b
=
k
,
b
×
c
=
1
2
(
i
−
k
),
c
×
a
=
1
2
(
j
−
k
),
a
·
(
b
×
c
)=
1
2
,
b
·
(
c
×
a
)=
1
2
, and
c
·
(
a
×
b
)=
1
2
.Then
A
=
1
2
(
i
−
k
)
1
2
=
i
−
k
,
B
=
1
2
(
j
−
k
)
1
2
=
j
−
k
,
and
C
=
k
1
2
=2
k
.
(b)
We need to compute
A
·
(
B
×
C
).Using formula (10) in the text we have
B
×
C
=
(
c
×
a
)
×
(
a
×
b
)
[
b
·
(
c
×
a
)][
c
·
(
a
×
b
)]
=
[(
c
×
a
)
·
b
]
a
−
[(
c
×
a
)
·
a
]
b
[
b
·
(
c
×
a
)][
c
·
(
a
×
b
)]
=
a
c
·
(
a
×
b
)
since (
c
×
a
)
·
a
=0.
Then
A
·
(
B
×
C
)=
b
×
c
a
·
(
b
×
c
)
·
a
c
·
(
a
×
b
)
=
1
c
·
(
a
×
b
)
and the volume of the unit cell of the reciprocal latrice is the reciprocal of the volume of the unit cell of
the original lattice.
302

303. Exercises 7.4
58. a
×
(
b
+
c
)=
(
(
(
(
a
2
a
3
b
2
+
c
2
b
3
+
c
3
(
(
(
(
i
−
(
(
(
(
a
1
a
3
b
1
+
c
1
b
3
+
c
3
(
(
(
(
j
+
(
(
(
(
a
1
a
2
b
1
+
c
1
b
2
+
c
2
(
(
(
(
k
=(
a
2
b
3
−
a
3
b
2
)
i
+(
a
2
c
3
−
a
3
c
2
)
i
−
[(
a
1
b
3
−
a
3
b
1
)
j
+(
a
1
c
3
−
a
3
c
1
)
j
]+(
a
1
b
2
−
a
2
b
1
)
k
+(
a
1
c
2
−
a
2
c
1
)
k
=(
a
2
b
3
−
a
3
b
2
)
i
−
(
a
1
b
3
−
a
3
b
1
)
j
+(
a
1
b
2
−
a
2
b
1
)
k
+(
a
2
c
3
−
a
3
c
2
)
i
−
(
a
1
c
3
−
a
3
c
1
)
j
+(
a
1
c
2
−
a
2
c
1
)
k
=
a
×
b
+
a
×
c
59. b
×
c
=(
b
2
c
3
−
b
3
c
2
)
i
−
(
b
1
c
3
−
b
3
c
1
)
j
+(
b
1
c
2
−
b
2
c
1
)
k
a
×
(
b
×
c
)=[
a
2
(
b
1
c
2
−
b
2
c
1
)+
a
3
(
b
1
c
3
−
b
3
c
1
)]
i
−
[
a
1
(
b
1
c
2
−
b
2
c
1
)
−
a
3
(
b
2
c
3
−
b
3
c
2
)]
j
+[
−
a
1
(
b
1
c
3
−
b
3
c
1
)
−
a
2
(
b
2
c
3
−
b
3
c
2
)]
k
=(
a
2
b
1
c
2
−
a
2
b
2
c
1
+
a
3
b
1
c
3
−
a
3
b
3
c
1
)
i
−
(
a
1
b
1
c
2
−
a
1
b
2
c
1
−
a
3
b
2
c
3
+
a
3
b
3
c
2
)
j
−
(
a
1
b
1
c
3
−
a
1
b
3
c
1
+
a
2
b
2
c
3
−
a
2
b
3
c
2
)
k
(
a
·
c
)
b
−
(
a
·
b
)
c
=(
a
1
c
1
+
a
2
c
2
+
a
3
c
3
)(
b
1
i
+
b
2
j
+
b
3
k
)
−
(
a
1
b
1
+
a
2
b
2
+
a
3
b
3
)(
c
1
i
+
c
2
j
+
c
3
k
)
=(
a
2
b
1
c
2
−
a
2
b
2
c
1
+
a
3
b
1
c
3
−
a
3
b
3
c
1
)
i
−
(
a
1
b
1
c
2
−
a
1
b
2
c
1
−
a
3
b
2
c
3
+
a
3
b
3
c
2
)
j
−
(
a
1
b
1
c
3
−
a
1
b
3
c
1
+
a
2
b
2
c
3
−
a
2
b
3
c
2
)
k
60.
The statement is false since
i
×
(
i
×
j
)=
i
×
k
=
−
j
and (
i
×
i
)
×
j
=
0
×
j
=
0
.
61.
Using equation 9 in the text,
a
·
(
b
×
c
)=
(
(
(
(
(
(
(
a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
(
(
(
(
(
(
(
and (
a
×
b
)
·
c
=
c
·
(
a
×
b
)=
(
(
(
(
(
(
(
c
1
c
2
c
3
a
1
a
2
a
3
b
1
b
2
b
3
(
(
(
(
(
(
(
.
The second determinant can be obtained from the first by an interchange of the second and third rows followed by
an interchange of the new first and second rows.Using Theorem 8
.
11 in the text, we see that
a
·
(
b
×
c
)=(
a
×
b
)
·
c
.
62. a
×
(
b
×
c
)+
b
×
(
c
×
a
)+
c
×
(
a
×
b
)
=(
a
·
c
)
b
−
(
a
·
b
)
c
+(
b
·
a
)
c
−
(
b
·
c
)
a
+(
c
·
b
)
a
−
(
c
·
a
)
b
=[(
a
·
c
)
b
−
(
c
·
a
)
b
]+[(
b
·
a
)
c
−
(
a
·
b
)
c
]+[(
c
·
b
)
a
−
(
b
·
c
)
a
]=
0
63.
Since
)
a
×
b
)
2
=(
a
2
b
3
−
a
3
b
2
)
2
+(
a
1
b
3
−
a
3
b
1
)
2
+(
a
1
b
2
−
a
2
b
1
)
2
=
a
2
2
b
2
3
−
2
a
2
b
3
a
3
b
2
+
a
2
3
b
2
2
+
a
2
1
b
2
3
−
2
a
1
b
3
a
3
b
1
+
a
2
3
b
2
1
+
a
2
1
b
2
2
−
2
a
1
b
2
a
2
b
1
+
a
2
2
b
2
1
and
)
a
)
2
)
b
)
2
−
(
a
·
b
)
2
=(
a
2
1
+
a
2
2
+
a
2
3
)(
b
2
1
+
b
2
2
+
b
2
3
)
−
(
a
1
b
1
+
a
2
b
2
+
a
3
b
3
)
2
=
a
2
1
a
2
2
+
a
2
1
b
2
2
+
a
2
1
b
2
3
+
a
2
2
b
2
1
+
a
2
2
b
2
2
+
a
2
2
b
2
3
+
a
2
3
b
2
1
+
a
2
3
b
2
2
+
a
2
3
b
2
3
−
a
2
1
b
2
1
−
a
2
2
b
2
2
−
a
2
3
b
2
3
−
2
a
1
b
1
a
2
b
2
−
2
a
1
b
1
a
3
b
3
−
2
a
2
b
2
a
3
b
3
=
a
2
1
b
2
2
+
a
2
1
b
2
3
+
a
2
2
b
2
1
+
a
2
2
b
2
3
+
a
2
3
b
2
1
+
a
2
3
b
2
2
−
2
a
1
a
2
b
1
b
2
−
2
a
1
a
3
b
1
b
3
−
2
a
2
a
3
b
2
b
3
we see that
)
a
×
b
)
2
=
)
a
)
2
)
b
)
2
−
(
a
·
b
)
2
.
64.
No.For example
i
×
(
i
+
j
)=
i
×
j
by the distributive law (
iii
) in the text, and the fact that
i
×
i
=
0
.But
i
+
j
does not equal
j
.
65.
By the distributive law (
iii
) in the text:
(
a
+
b
)
×
(
a
−
b
)=(
a
+
b
)
×
a
−
(
a
+
b
)
×
b
=
a
×
a
+
b
×
a
−
a
×
b
−
b
×
b
=2
b
×
a
since
a
×
a
=
0
,
b
×
b
=
0
, and
−
a
×
b
=
b
×
a
.
303

304. Exercises 7.5
Exercises 7.5
The equation of a line through
P
1
and
P
2
in 3-space with
r
1
=
−
−
→
OP
1
and
r
2
=
−
−
→
OP
2
can be expressed as
r
=
r
1
+
t
(
k
a
)
or
r
=
r
2
+
t
(
k
a
) where
a
=
r
2
−
r
1
and k is any non-zero scalar.Thus, the form of the equation of a line is not
unique.(See the alternate solution to Problem 1.)
1. a
=
1
1
−
3
,
2
−
5
,
1
−
(
−
2)
.
=
1(
2
,
−
3
,
3
.
;
1
x, y, z
.
=
1
1
,
2
,
1
.
+
t
1(
2
,
−
3
,
3
.
Alternate Solution:
a
=
1
3
−
1
,
5
−
2
,
−
2
−
1
.
=
1
2
,
3
,
−
3
.
;
1
x, y, z
.
=
1
3
,
5
,
−
2
.
+
t
1
2
,
3
,
−
3
.
2. a
=
1
0
−
(
−
2)
,
4
−
6
,
5
−
3
.
=
1
2
,
−
2
,
2
.
;
1
x, y, z
.
=
1
0
,
4
,
5
.
+
t
1
2
,
−
2
,
2
.
3. a
=
1
1
/
2
−
(
−
3
/
2)
,
−
1
/
2
−
5
/
2
,
1
−
(
−
1
/
2)
.
=
1
2
,
−
3
,
3
/
2
.
;
1
x, y, z
.
=
1
1
/
2
,
−
1
/
2
,
1
.
+
t
1
2
,
−
3
,
3
/
2
.
4. a
=
1
10
−
5
,
2
−
(
−
3)
,
−
10
−
5
.
=
1
5
,
5
,
−
15
.
;
1
x, y, z
.
=
1
10
,
2
,
−
10
.
+
t
1
5
,
5
,
−
15
.
5. a
=
1
1
−
(
−
4)
,
1
−
1
,
−
1
−
(
−
1)
.
=
1
5
,
0
,
0
.
;
1
x, y, z
.
=
1
1
,
1
,
−
1
.
+
t
1
5
,
0
,
0
.
6. a
=
1
3
−
5
/
2
,
2
−
1
,
1
−
(
−
2)
.
=
1
1
/
2
,
1
,
3
.
;
1
x, y, z
.
=
1
3
,
2
,
1
.
+
t
1
1
/
2
,
1
,
3
.
7. a
=
1
2
−
6
,
3
−
(
−
1)
,
5
−
8
.
=
1(
4
,
4
,
−
3
.
;
x
=2
−
4
t
,
y
=3+4
t
,
z
=5
−
3
t
8. a
=
1
2
−
0
,
0
−
4
,
0
−
9
.
=
1
2
,
−
4
,
−
9
.
;
x
=2+2
t
,
y
=
−
4
t
,
z
=
−
9
t
9. a
=
1
1
−
3
,
0
−
(
−
2)
,
0
−
(
−
7)
.
=
1(
2
,
2
,
7
.
;
x
=1
−
2
t
,
y
=2
t
,
z
=7
t
10. a
=
1
0
−
(
−
2)
,
0
−
4
,
5
−
0
.
=
1
2
,
−
4
,
5
.
;
x
=2
t
,
y
=
−
4
t
,
z
=5+5
t
11. a
=
1
4
−
(
−
6)
,
1
/
2
−
(
−
1
/
4)
,
1
/
3
−
1
/
6
.
=
1
10
,
3
/
4
,
1
/
6
.
;
x
=4+10
t
,
y
=
1
2
+
3
4
t
,
z
=
1
3
+
1
6
t
12. a
=
1(
3
−
4
,
7
−
(
−
8)
,
9
−
(
−
1)
.
=
1(
7
,
15
,
10
.
;
x
=
−
3
−
7
t
,
y
=7+15
t
,
z
=9+10
t
13.
a
1
=10
−
1=9,
a
2
=14
−
4 = 10,
a
3
=
−
2
−
(
−
9) = 7;
x
−
10
9
=
y
−
14
10
=
z
+2
7
14.
a
1
=1
−
2
/
3=1
/
3,
a
2
=3
−
0=3,
a
3
=1
/
4
−
(
−
1
/
4)=1
/
2;
x
−
1
1
/
3
=
y
−
3
3
=
z
−
1
/
4
1
/
2
15.
a
1
=
−
7
−
4=
−
11,
a
2
=2
−
2=0,
a
3
=5
−
1=4;
x
+7
−
11
=
z
−
5
4
,
y
=2
16.
a
1
=1
−
(
−
5) = 6,
a
2
=1
−
(
−
2) = 3,
a
3
=2
−
(
−
4) = 6;
x
−
1
6
=
y
−
1
3
=
z
−
2
6
17.
a
1
=5
−
5=0,
a
2
=10
−
1=9,
a
3
=
−
2
−
(
−
14) = 12;
x
=5,
y
−
10
9
=
z
+2
12
18.
a
1
=5
/
6
−
1
/
3=1
/
2;
a
2
=
−
1
/
4
−
3
/
8=
−
5
/
8;
a
3
=1
/
5
−
1
/
10=1
/
10
x
−
5
/
6
1
/
2
=
y
+1
/
4
−
5
/
8
=
z
−
1
/
5
1
/
10
19.
parametric:
x
=4+3
t
,
y
=6+
t/
2,
z
=
−
7
−
3
t/
2; symmetric:
x
−
4
3
=
y
−
6
1
/
2
=
z
+7
−
3
/
2
20.
parametric:
x
=1
−
7
t
,
y
=8
−
8
t
,
z
=
−
2; symmetric:
x
−
1
−
7
=
y
−
8
−
8
,
z
=
−
2
21.
parametric:
x
=5
t
,
y
=9
t
,
z
=4
t
; symmetric:
x
5
=
y
9
=
z
4
22.
parametric:
x
=12
t
,
y
=
−
3
−
5
t
,
z
=10
−
6
t
; symmetric:
x
12
=
y
+3
−
5
=
z
−
10
−
6
23.
Writing the given line in the form
x/
2=(
y
−
1)
/
(
−
3)=(
z
−
5)
/
6, we see that a direction vector is
1
2
,
−
3
,
6
.
.
Parametric equations for the line are
x
=6+2
t
,
y
=4
−
3
t
,
z
=
−
2+6
t
.
304

305. Exercises 7.5
24.
A direction vector is
1
5
,
1
/
3
,
−
2
.
.Symmetric equations for the line are (
x
−
4)
/
5=(
y
+11)
/
(1
/
3) = (
z
+7)
/
(
−
2).
25.
A direction vector parallel to both the
xz
- and
xy
-planes is
i
=
1
1
,
0
,
0
.
.Parametric equations for the line are
x
=2+
t
,
y
=
−
2,
z
= 15.
26. (a)
Since the unit vector
j
=
1
0
,
1
,
0
.
lies along the
y
-axis, we have
x
=1,
y
=2+
t
,
z
=8.
(b)
since the unit vector
k
=
1
0
,
0
,
1
.
is perpendicular to the
xy
-plane, we have
x
=1,
y
=2,
z
=8+
t
.
27.
Both lines go through the points (0
,
0
,
0) and (6
,
6
,
6).Since two points determine a line, the lines are the same.
28. a
and
f
are parallel since
1
9
,
−
12
,
6
.
=
−
3
1(
3
,
4
,
−
2
.
.
c
and
d
are orthogonal since
1
2
,
−
3
,
4
.d1
1
,
4
,
5
/
2
.
=0.
29.
In the
xy
-plane,
z
=9+3
t
= 0 and
t
=
−
3.Then
x
=4
−
2(
−
3) = 10 and
y
=1+2(
−
3) =
−
5.The point is
(10
,
−
5
,
0).In the
xz
-plane,
y
= 1+2
t
= 0 and
t
=
−
1
/
2.Then
x
=4
−
2(
−
1
/
2) = 5 and
z
= 9+3(
−
1
/
2) = 15
/
2.
The point is (5
,
0
,
15
/
2).In the
yz
-plane,
x
=4
−
2
t
= 0 and
t
= 2.Then
y
= 1+2(2) = 5 and
z
= 9+3(2) = 15.
The point is (0
,
5
,
15).
30.
The parametric equations for the line are
x
=1+2
t
,
y
=
−
2+3
t
,
z
=4+2
t
.In the
xy
-plane,
z
=4+2
t
=0
and
t
=
−
2.Then
x
=1+2(
−
2) =
−
3 and
y
=
−
2+3(
−
2) =
−
8.The point is (
−
3
,
−
8
,
0).In the
xz
-plane,
y
=
−
2+3
t
= 0 and
t
=2
/
3.Then
x
= 1 + 2(2
/
3) = 7
/
3 and
z
= 4 + 2(2
/
3) = 16
/
3.The point is (7
/
3
,
0
,
16
/
3).
In the
yz
-plane,
x
=1+2
t
= 0 and
t
=
−
1
/
2.Then
y
=
−
2+3(
−
1
/
2) =
−
7
/
2 and
z
=4+2(
−
1
/
2) = 3.The
point is (0
,
−
7
/
2
,
3).
31.
Solving the system 4 +
t
=6+2
s
,5+
t
=11+4
s
,
−
1+2
t
=
−
3+
s
,or
t
−
2
s
=2,
t
−
4
s
=6,2
t
−
s
=
−
2
yields
s
=
−
2 and
t
=
−
2 in all three equations.Thus, the lines intersect at the point
x
=4+(
−
2)=2,
y
=5+(
−
2) = 3,
z
=
−
1+2(
−
2) =
−
5, or (2
,
3
,
−
5).
32.
Solving the system 1 +
t
=2
−
s
,2
−
t
=1+
s
,3
t
=6
s
,or
t
+
s
=1,
t
+
s
=1,
t
−
2
s
= 0 yields
s
=1
/
3 and
t
=2
/
3 in all three equations.Thus, the lines intersect at the point
x
=1+2
/
3=5
/
3,
y
=2
−
2
/
3=4
/
3,
z
= 3(2
/
3)=2,or(5
/
3
,
4
/
3
,
2).
33.
The system of equations 2
−
t
=4+
s
,3+
t
=1+
s
,1+
t
=1
−
s
,or
t
+
s
=
−
2,
t
−
s
=
−
2,
t
+
s
= 0 has no
solution since
−
2
3
= 0.Thus, the lines do not intersect.
34.
Solving the system 3
−
t
=2+2
s
,2+
t
=
−
2+3
s
, 8+2
t
=
−
2+8
s
,or
t
+2
s
=1,
t
−
3
s
=
−
4, 2
t
−
8
s
=
−
10
yields
s
= 1 and
t
=
−
1 in all three equations.Thus, the lines intersect at the point
x
=3
−
(
−
1)=4,
y
=2+(
−
1) = 1,
z
=8+2(
−
1)=6,or(4
,
1
,
6).
35. a
=
1(
1
,
2
,
−
2
.
,
b
=
1
2
,
3
,
−
6
.
,
a
·
b
= 16,
)
a
)
=3,
)
b
)
= 7; cos
θ
=
a
·
b
)
a
))
b
)
=
16
3
·
7
;
θ
= cos
−
1
16
21
≈
40
.
37
◦
36. a
=
1
2
,
7
,
−
1
.
,
b
=
1(
2
,
1
,
4
.
,
a
·
b
=
−
1,
)
a
)
=3
√
6,
)
b
)
=
√
21 ;
cos
θ
=
a
·
b
)
a
))
b
)
=
−
1
(3
√
6)(
√
21 )
=
−
1
9
√
14
;
θ
= cos
−
1
(
−
1
9
√
14
)
≈
91
.
70
◦
37.
A direction vector perpendicular to the given lines will be
1
1
,
1
,
1
.21(
2
,
1
,
−
5
.
=
1(
6
,
3
,
3
.
.Equations of the
line are
x
=4
−
6
t
,
y
=1+3
t
,
z
=6+3
t
.
38.
The direction vectors of the given lines are
1
3
,
2
,
4
.
and
1
6
,
4
,
8
.
=2
1
3
,
2
,
4
.
.These are parallel, so we need a
third vector parallel to the plane containing the lines which is not parallel to them.The point (1
,
−
1
,
0) is on
the first line and (
−
4
,
6
,
10) is on the second line.A third vector is then
1
1
,
−
1
,
0
.(1(
4
,
6
,
10
.
=
1
5
,
−
7
,
−
10
.
.
Now a direction vector perpendicular to the plane is
1
3
,
2
,
4
.21
5
,
−
7
,
−
10
.
=
1
8
,
50
,
−
31
.
.Equations of the line
through (1
,
−
1
,
0) and perpendicular to the plane are
x
=1+8
t
,
y
=
−
1+50
t
,
z
=
−
31
t
.
305

306. Exercises 7.5
39.
2(
x
−
5)
−
3(
y
−
1)+4(
z
−
3)=0; 2
x
−
3
y
+4
z
=19
40.
4(
x
−
1)
−
2(
y
−
2)+0(
z
−
5)=0; 4
x
−
2
y
=0
41.
−
5(
x
−
6)+0(
y
−
10)+3(
z
+7)=0;
−
5
x
+3
z
=
−
51
42.
6
x
−
y
+3
z
=0
43.
6(
x
−
1
/
2)+8(
y
−
3
/
4)
−
4(
z
+1
/
2) = 0; 6
x
+8
y
−
4
z
=11
44.
−
(
x
+1)+(
y
−
1)
−
(
z
−
0) = 0;
−
x
+
y
−
z
=2
45.
From the points (3
,
5
,
2) and (2
,
3
,
1) we obtain the vector
u
=
i
+2
j
+
k
.From the points (2
,
3
,
1) and
(
−
1
,
−
1
,
4) we obtain the vector
v
=3
i
+4
j
−
3
k
.From the points (
−
1
,
−
1
,
4) and (
x, y, z
) we obtain the vector
w
=(
x
+1)
i
+(
y
+1)
j
+(
z
−
4)
k
.Then, a normal vector is
u
×
v
=
(
(
(
(
(
(
(
ij k
12 1
34
−
3
(
(
(
(
(
(
(
=
−
10
i
+6
j
−
2
k
.
A vector equation of the plane is
−
10(
x
+1)+6(
y
+1)
−
2(
z
−
4) = 0 or 5
x
−
3
y
+
z
=2.
46.
From the points (0
,
1
,
0) and (0
,
1
,
1) we obtain the vector
u
=
k
.From the points (0
,
1
,
1) and (1
,
3
,
−
1) we obtain
the vector
v
=
i
+2
j
−
2
k
.From the points (1
,
3
,
−
1) and (
x, y, z
) we obtain the vector
w
=(
x
−
1)
i
+(
y
−
3)
j
+(
z
+1)
k
.Then, a normal vector is
u
×
v
=
(
(
(
(
(
(
(
ij k
00 1
12
−
2
(
(
(
(
(
(
(
=
−
2
i
+
j
.
A vector equation of the plane is
−
2(
x
−
1)+(
y
−
3)+0(
z
+1)=0or
−
2
x
+
y
=1.
47.
From the points (0
,
0
,
0) and (1
,
1
,
1) we obtain the vector
u
=
i
+
j
+
k
.From the points (1
,
1
,
1) and
(3
,
2
,
−
1) we obtain the vector
v
=2
i
+
j
−
2
k
.From the points (3
,
2
,
−
1) and (
x, y, z
) we obtain the vector
w
=(
x
−
3)
i
+(
y
−
2)
j
+(
z
+1)
k
.Then, a normal vector is
u
×
v
=
(
(
(
(
(
(
(
ij k
11 1
21
−
2
(
(
(
(
(
(
(
=
−
3
i
+4
j
−
k
.
A vector equation of the plane is
−
3(
x
−
3)+4(
y
−
2)
−
(
z
+1)=0or
−
3
x
+4
y
−
z
=0.
48.
The three points are not colinear and all satisfy
x
= 0, which is the equation of the plane.
49.
From the points (1
,
2
,
−
1) and (4
,
3
,
1) we obtain the vector
u
=3
i
+
j
+2
k
.From the points (4
,
3
,
1) and
(7
,
4
,
3) we obtain the vector
v
=3
i
+
j
+2
k
.From the points (7
,
4
,
3) and (
x, y, z
) we obtain the vector
w
=(
x
−
7)
i
+(
y
−
4)
j
+(
z
−
3)
k
.Since
u
×
v
=
0
, the points are colinear.
50.
From the points (2
,
1
,
2) and (4
,
1
,
0) we obtain the vector
u
=2
i
−
2
k
.From the points (4
,
1
,
0) and
(5
,
0
,
−
5) we obtain the vector
v
=
i
−
j
−
5
k
.From the points (5
,
0
,
−
5) and (
x, y, z
) we obtain the vector
w
=(
x
−
5)
i
+
y
j
+(
z
+5)
k
.Then, a normal vector is
u
×
v
=
(
(
(
(
(
(
(
ij k
20
−
2
1
−
1
−
5
(
(
(
(
(
(
(
=
−
2
i
+8
j
−
2
k
.
A vector equation of the plane is
−
2(
x
−
5)+8
y
−
2(
z
+5)=0or
x
−
4
y
+
z
=0.
306

307. Exercises 7.5
51.
A normal vector to
x
+
y
−
4
z
=1is
1
1
,
1
,
−
4
.
.The equation of the parallel plane is
(
x
−
2)+(
y
−
3)
−
4(
z
+5)=0or
x
+
y
−
4
z
= 25.
52.
A normal vector to 5
x
−
y
+
z
=6is
1
5
,
−
1
,
1
,
.
.The equation of the parallel plane is 5(
x
−
0)
−
(
y
−
0)+(
z
−
0) = 0
or 5
x
−
y
+
z
=0.
53.
A normal vector to the
xy
-plane is
1
0
,
0
,
1
.
.The equation of the parallel plane is
z
−
12 = 0 or
z
= 12.
54.
A normal vector is
1
0
,
1
,
0
.
.The equation of the plane is
y
+5=0or
y
=
−
5.
55.
Direction vectors of the lines are
1
3
,
−
1
,
1
.
.and
1
4
,
2
,
1
.
.A normal vector to the plane is
1
3
,
−
1
,
1
.21
4
,
2
,
1
.
=
1(
3
,
1
,
10
.
.A point on the first line, and thus in the plane, is
1
1
,
1
,
2
.
.The equation of the plane is
−
3(
x
−
1)+(
y
−
1) + 10(
z
−
2) = 0 or
−
3
x
+
y
+10
z
= 18.
56.
Direction vectors of the lines are
1
2
,
−
1
,
6
.
and
1
1
,
1
,
−
3
.
.A normal vector to the plane is
1
2
,
−
1
,
6
.21
1
,
1
,
−
3
.
=
1(
3
,
12
,
3
.
.A point on the first line, and thus in the plane, is (1
,
−
1
,
5).The equation of the plane is
−
3(
x
−
1) + 12(
y
+1)+3(
z
−
5) = 0 or
−
x
+4
y
+
z
=0.
57.
A direction vector for the two lines is
1
1
,
2
,
1
.
.Points on the lines are (1
,
1
,
3) and (3
,
0
,
−
2).Thus, another
vector parallel to the plane is
1
1
−
3
,
1
−
0
,
3+2
.
=
1(
2
,
1
,
5
.
.A normal vector to the plane is
1
1
,
2
,
1
.21(
2
,
1
,
5
.
=
1
9
,
−
7
,
5
.
.Using the point (3
,
0
,
−
2) in the plane, the equation of the plane is 9(
x
−
3)
−
7(
y
−
0) + 5(
z
+2)=0
or 9
x
−
7
y
+5
z
= 17.
58.
A direction vector for the line is
1
3
,
2
,
−
2
.
.Letting
t
= 0, we see that the origin is on the line and hence in the
plane.Thus, another vector parallel to the plane is
1
4
−
0
,
0
−
0
,
−
6
−
0
.
=
1
4
,
0
,
−
6
.
.A normal vector to the
plane is
1
3
,
2
,
−
2
.21
4
,
0
,
−
6
.
=
1(
12
,
10
,
−
8
.
.The equation of the plane is
−
12(
x
−
0) + 10(
y
−
0)
−
8(
z
−
0) = 0
or 6
x
−
5
y
+4
z
=0.
59.
A direction vector for the line, and hence a normal vector to the plane, is
1(
3
,
1
,
−
1
/
2
.
.The equation of the
plane is
−
3(
x
−
2) + (
y
−
4)
−
1
2
(
z
−
8) = 0 or
−
3
x
+
y
−
1
2
z
=
−
6.
60.
A normal vector to the plane is
1
2
−
1
,
6
−
0
,
−
3+2
.
=
1
1
,
6
,
−
1
.
.The equation of the plane is
(
x
−
1)+6(
y
−
1)
−
(
z
−
1)=0or
x
+6
y
−
z
=6.
61.
Normal vectors to the planes are
(a)
1
2
,
−
1
,
3
.
,
(b)
1
1
,
2
,
2
.
,
(c)
1
1
,
1
,
−
3
/
2
.
,
(d)
1(
5
,
2
,
4
.
,
(e)
1(
8
,
−
8
,
12
.
,
(f)
1(
2
,
1
,
−
3
.
.Parallel planes are
(c)
and
(e)
, and
(a)
and
(f)
.Perpendicular planes
are
(a)
and
(d)
,
(b)
and
(c)
,
(b)
and
(e)
, and
(d)
and
(f)
.
62.
A normal vector to the plane is
1(
7
,
2
,
3
.
.This is a direction vector for the line and the equations of the line
are
x
=
−
4
−
7
t
,
y
=1+2
t
,
z
=7+3
t
.
63.
A direction vector of the line is
1(
6
,
9
,
3
.
, and the normal vectors of the planes are
(a)
1
4
,
1
,
2
.
,
(b)
1
2
,
−
3
,
1
.
,
(c)
1
10
,
−
15
,
−
5
.
,
(d)
1(
4
,
6
,
2
.
.Vectors
(c)
and
(d)
are multiples of the direction vector and hence the
corresponding planes are perpendicular to the line.
64.
A direction vector of the line is
1(
2
,
4
,
1
.
, and normal vectors to the planes are
(a)
1
1
,
−
1
,
3
.
,
(b)
1
6
,
−
3
,
0
.
,
(c)
1
1
,
−
2
,
5
.
,
(d)
1(
2
,
1
,
−
2
.
.Since the dot product of each normal vector with the direc-
tion vector is non-zero, none of the planes are parallel to the line.
65.
Letting
z
=
t
in both equations and solving 5
x
−
4
y
=8+9
t
,
x
+4
y
=4
−
3
t
, we obtain
x
=2+
t
,
y
=
1
2
−
t
,
z
=
t
.
66.
Letting
y
=
t
in both equations and solving
x
−
z
=2
−
2
t
,3
x
+2
z
=1+
t
, we obtain
x
=1
−
3
5
t
,
y
=
t
,
z
=
−
1+
7
5
t
or, letting
t
=5
s
,
x
=1
−
3
s
,
y
=5
s
,
z
=
−
1+7
s
.
67.
Letting
z
=
t
in both equations and solving 4
x
−
2
y
=1+
t
,
x
+
y
=1
−
2
t
, we obtain
x
=
1
2
−
1
2
t
,
y
=
1
2
−
3
2
t
,
z
=
t
.
307

308. Exercises 7.5
68.
Letting
z
=
t
and using
y
= 0 in the first equation, we obtain
x
=
−
1
2
t
,
y
=0,
z
=
t
.
69.
Substituting the parametric equations into the equation of the plane, we obtain
2(1+2
t
)
−
3(2
−
t
)+2(
−
3
t
)=
−
7or
t
=
−
3.Letting
t
=
−
3 in the equation of the line, we obtain the point of
intersection (
−
5
,
5
,
9).
70.
Substituting the parametric equations into the equation of the plane, we obtain
(3
−
2
t
)+(1+6
t
) + 4(2
−
1
2
t
)=12or2
t
= 0.Letting
t
= 0 in the equation of the line, we obtain the point of
intersection (3
,
1
,
2).
71.
Substituting the parametric equations into the equation of the plane, we obtain 1 + 2
−
(1 +
t
)=8or
t
=
−
6.
Letting
t
=
−
6 in the equation of the line, we obtain the point of intersection (1
,
2
,
−
5).
72.
Substituting the parametric equations into the equation of the plane, we obtain 4 +
t
−
3(2 +
t
)+2(1+5
t
)=0
or
t
= 0.Letting
t
= 0 in the equation of the line, we obtain the point of intersection (4
,
2
,
1).
In Problems 73 and 74, the cross product of the normal vectors to the two planes will be a vector parallel to both
planes, and hence a direction vector for a line parallel to the two planes.
73.
Normal vectors are
1
1
,
1
,
−
4
.
and
1
2
,
−
1
,
1
.
.A direction vector is
1
1
,
1
,
−
4
.21
2
,
−
1
,
1
.
=
1(
3
,
−
9
,
−
3
.
=
−
3
1
1
,
3
,
1
.
.
Equations of the line are
x
=5+
t
,
y
=6+3
t
,
z
=
−
12 +
t
.
74.
Normal vectors are
1
2
,
0
,
1
.
and
1(
1
,
3
,
1
.
.A direction vector is
1
2
,
0
,
1
.21(
1
,
3
,
1
.
=
1(
3
,
−
3
,
6
.
=
−
3
1
1
,
1
,
−
2
.
.
Equations of the line are
x
=
−
3+
t
,
y
=5+
t
,
z
=
−
1
−
2
t
.
In Problems 75 and 76, the cross product of the direction vector of the line with the normal vector of the given plane
will be a normal vector to the desired plane.
75.
A direction vector of the line is
1
3
,
−
1
,
5
.
and a normal vector to the given plane is
1
1
,
1
,
1
.
.A normal vector
to the desired plane is
1
3
,
−
1
,
5
.21
1
,
1
,
1
.
=
1(
6
,
2
,
4
.
.A point on the line, and hence in the plane, is (4
,
0
,
1).
The equation of the plane is
−
6(
x
−
4)+2(
y
−
0)+4(
z
−
1) = 0 or 3
x
−
y
−
2
z
= 10.
76.
A direction vector of the line is
1
3
,
5
,
2
.
and a normal vector to the given plane is
1
2
,
−
4
,
−
1
.
.A normal vector to
the desired plane is
1(
3
,
5
,
2
.21
2
,
−
4
,
−
1
.
=
1
3
,
1
,
2
.
.A point on the line, and hence in the plane, is (2
,
−
2
,
8).
The equation of the plane is 3(
x
−
2)+(
y
+2)+2(
z
−
8) = 0 or 3
x
+
y
+2
z
= 20.
77. 78. 79.
80. 81. 82.
308

309. Exercises 7.6
Exercises 7.6
1.
Not a vector space.Axiom (
vi
) is not satisfied.
2.
Not a vector space.Axiom (
i
) is not satisfied.
3.
Not a vector space.Axiom (
x
) is not satisfied.
4.
A vector space
5.
A vector space
6.
A vector space
7.
Not a vector space.Axiom (
ii
) is not satisfied.
8.
A vector space
9.
A vector space
10.
Not a vector space.Axiom (
i
) is not satisfied.
11.
A subspace
12.
Not a subspace.Axiom (
i
) is not satisfied.
13.
Not a subspace.Axiom (
ii
) is not satisfied.
14.
A subspace
15.
A subspace
16.
A subspace
17.
A subspace
18.
A subspace
19.
Not a subspace.Neither axioms (
i
) nor(
ii
) are satisfied.
20.
A subspace
21.
Let (
x
1
,y
1
,z
1
) and (
x
2
,y
2
,z
2
)bein
S
.Then
(
x
1
,y
1
,z
1
)+(
x
2
,y
2
,z
2
)=(
at
1
,bt
1
,ct
1
)+(
at
2
,bt
2
,ct
2
)=(
a
(
t
1
+
t
2
)
,b
(
t
1
+
t
2
)
,c
(
t
1
+
t
2
))
is in
S
.Also, for (
x, y, z
)in
S
then
k
(
x, y, z
)=(
kx, ky, kz
)=(
a
(
kt
)
,b
(
kt
)
,c
(
kt
)) is also in
S
.
22.
Let (
x
1
,y
1
,z
1
) and (
x
2
,y
2
,z
2
)bein
S
.Then
ax
1
+
by
1
+
cz
1
= 0 and
ax
2
+
by
2
+
cz
2
= 0.Adding gives
a
(
x
1
+
x
2
)+
b
(
y
1
+
y
2
)+
c
(
z
1
+
z
2
) = 0 and so (
x
1
,y
1
,z
1
)+(
x
2
,y
2
,z
2
)=(
x
1
+
x
2
,y
1
+
y
2
,z
1
+
z
2
)isin
S
.Also,
for (
x, y, z
) then
ax
+
by
+
cz
= 0 implies
k
(
ax
+
by
+
cz
)=
k
·
0 = 0 and
a
(
kx
)+
b
(
ky
)+
c
(
kz
) = 0.this means
k
(
x, y, z
)=(
kx
,
ky, kz
)isin
S
.
23. (a)
c
1
u
1
+
c
2
u
2
+
c
3
u
3
=
0
if and only if
c
1
+
c
2
+
c
3
=0,
c
2
+
c
3
=0,
c
3
= 0.The only solution of this system
is
c
1
=0,
c
2
=0,
c
3
=0.
(b)
Solving the system
c
1
+
c
2
+
c
3
=3,
c
2
+
c
3
=
−
4,
c
3
= 8 gives
c
1
=7,
c
2
=
−
12,
c
3
= 8.Thus
a
=7
u
1
−
12
u
2
+8
u
3
.
24. (a)
The assumption
c
1
p
1
+
c
2
p
2
= 0 is equivalent to (
c
1
+
c
2
)
x
+(
c
1
−
c
2
) = 0.Thus
c
1
+
c
2
=0,
c
1
−
c
2
=0.
The only solution of this system is
c
1
=0,
c
2
=0.
(b)
Solving the system
c
1
+
c
2
=5,
c
1
−
c
2
= 2 gives
c
1
=
7
2
,
c
2
=
3
2
.Thus
p
(
x
)=
7
2
p
1
(
x
)+
3
2
p
2
(
x
)
25.
Linearly dependent since
fn
6
,
12
g
=
−
3
2
f
4
,
−
8
g
26.
Linearly dependent since 2
f
1
,
1
g
+3
f
0
,
1
g
+(
−
1)
f
2
,
5
g
=
f
0
,
0
g
27.
Linearly independent
28.
Linearly dependent since for all
x
(1)
·
1+(
−
2)(
x
+ 1) + (1)(
x
+1)
2
+(
−
1)
x
2
=0.
29.
f
is discontinuous at
x
=
−
1 and at
x
=
−
3.
30.
(
x,
sin
x
)=
g
2
π
0
x
sin
xdx
=(
−
x
cos
x
+ sin
x
)
n
n
n
2
π
0
=
−
2
π
309

310. Exercises 7.6
31.
a
x
a
2
=
g
2
π
0
x
2
dx
=
1
3
x
3
n
n
n
n
2
π
0
=
8
3
π
3
and so
a
x
a
=2

2
π
3
3
.Now
a
sin
x
a
2
=
g
2
π
0
sin
2
xdx
=
1
2
g
2
π
0
(1
−
cos 2
x
)
dx
=
1
2
a
x
−
1
2
sin 2
x
L
n
n
n
2
π
0
=
π
and so
a
sin
x
a
=
√
π
.
32.
A basis could be 1,
x
,
e
x
cos 3
x
,
e
x
sin 3
x
.
33.
We need to show that Span
{
x
1
,x
2
,...,x
n
}
is closed under vector addition and scalar multiplication.Suppose
u
and
v
are in Span
{
x
1
,x
2
,...,x
n
}
.Then
u
=
a
1
x
1
+
a
2
x
2
+
···
+
a
n
x
n
and
v
=
b
1
x
1
+
b
2
x
2
+
···
+
b
n
x
n
,so
that
u
+
v
=(
a
1
+
b
1
)
x
1
+(
a
2
+
b
2
)
x
2
+
···
+(
a
n
+
b
n
)
x
n
,
which is in Span
{
x
1
,x
2
,...,x
n
}
.Also, for any real number
k
,
k
u
=
k
(
a
1
x
1
+
a
2
x
2
+
···
+
a
n
x
n
)=
ka
1
x
1
+
ka
2
x
2
+
···
+
ka
n
x
n
,
which is in Span
{
x
1
,x
2
,...,x
n
}
.Thus, Span
{
x
1
,x
2
,...,x
n
}
is a subspace of
V
.
Chapter 7 Review Exercises
1.
True
2.
False; the points must be non-collinear.
3.
False; since a normal to the plane is
f
2
,
3
,
−
4
g
which is not a multiple of the direction vector
f
5
,
−
2
,
1
g
of the
line.
4.
True
5.
True
6.
True
7.
True
8.
True
9.
True
10.
True; since
a
×
b
and
c
×
d
are both normal to the plane and hence parallel (unless
a
×
b
=
0
or
c
×
d
=
0
.)
11.
9
i
+2
j
+2
k 12.
orthogonal
13.
−
5(
k
×
j
)=
−
5(
−
i
)=5
i 14. i
·
(
i
×
j
)=
i
×
k
=
0
15.
y
(
−
12)
2
+4
2
+6
2
=14
16.
(
−
1
−
20)
i
−
(
−
2
−
0)
j
+(8
−
0)
k
=
−
21
i
+2
j
+8
k
17.
−
6
i
+
j
−
7
k
18.
The coordinates of (1
,
−
2
,
−
10) satisfy the given equation.
19.
Writing the line in parametric form, we have
x
=1+
t
,
y
=
−
2+3
t
,
z
=
−
1+2
t
.Substituting into the equation
of the plane yields (1 +
t
)+2(
−
2+3
t
)
−
(
−
1+2
t
) = 13 or
t
= 3.Thus, the point of intersection is
x
= 1+3 = 4,
y
=
−
2 + 3(3) = 7,
z
=
−
1+2(3)=5,or(4
,
7
,
5).
20.
|
a
|
=
y
4
2
+3
2
+(
−
5)
2
=5
√
2;
u
=
−
1
5
√
2
(4
i
+3
j
−
5
k
)=
−
4
5
√
2
i
−
3
5
√
2
j
+
1
√
2
k
21.
x
2
−
2=3,
x
2
=5;
y
2
−
1=5,
y
2
=6;
z
2
−
7=
−
4,
z
2
=3;
P
2
=(5
,
6
,
3)
22.
(5
,
1
/
2
,
5
/
2)
23.
(7
.
2)(10) cos 135
◦
=
−
36
√
2
310

311. Chapter 7 Review Exercises
24.
2
b
=
1(
2
,
4
,
2
a
;4
c
=
b
0
,
−
8
,
8
a
;
a
·
(2
b
+4
c
)=
b
3
,
1
,
0
.d1(
2
,
−
4
,
10
a
=
−
10
25.
12,
−
8, 6
26.
cos
θ
=
a
·
b
|
a
||
b
|
=
1
√
2
√
2
=
1
2
;
θ
=60
◦
27.
A
=
1
2
|
5
i
−
4
j
−
7
k
|
=
3
√
10
2
28.
From 3(
x
−
3)+0(
y
−
6) + (1)(
z
−
(
−
2)) = 0 we obtain 3
x
+
z
=7.
29.
|−
5
−
(
−
3)
|
=2
30.
parallel:
−
2
c
=5,
c
=
−
5
/
2; orthogonal: 1(
−
2)+3(
−
6) +
c
(5) = 0,
c
=4
31. a
×
b
=
(
(
(
(
(
(
(
ijk
110
1
−
21
(
(
(
(
(
(
(
=
(
(
(
(
10
−
21
(
(
(
(
i
−
(
(
(
(
10
11
(
(
(
(
j
+
(
(
(
(
11
1
−
2
(
(
(
(
k
=
i
−
j
−
3
k
A unit vector perpendicular to both
a
and
b
is
a
×
b
)
a
×
b
)
=
1
√
1+1+9
(
i
−
j
−
3
k
)=
1
√
11
i
−
1
√
11
j
−
3
√
11
k
.
32.
)
a
)
=
a
1
/
4+1
/
4+1
/
6=
3
4
; cos
α
=
1
/
2
3
/
4
=
2
3
,
α
≈
48
.
19
◦
; cos
β
=
1
/
2
3
/
4
=
2
3
,
β
≈
48
.
19
◦
;
cos
γ
=
−
1
/
4
3
/
4
=
−
1
3
,
γ
≈
109
.
47
◦
33.
comp
b
a
=
a
·
b
/
)
b
)
=
b
1
,
2
,
−
2
ab
4
,
3
,
0
a
/
5=2
34.
comp
a
b
=
b
·
a
/
)
a
)
=
b
4
,
3
,
0
ab
1
,
2
,
−
2
a
/
3=10
/
3
proj
a
b
= (comp
a
b
)
a
/
)
a
)
= (10
/
3)
b
1
,
2
,
−
2
a
/
3=
b
10
/
9
,
20
/
9
,
−
20
/
9
a
35.
Using the result of Problem 34,
proj
a
⊥
b
=
b
−
proj
a
b
=
b
4
,
3
,
0
.(1
10
/
9
,
20
/
9
,
−
20
/
9
a
=
b
26
/
9
,
7
/
9
,
20
/
9
a
.
36.
comp
b
(
a
−
b
)=(
a
−
b
)
·
b
/
)
b
)
=
1(
3
,
−
1
,
−
2
ab
4
,
3
,
0
a
/
5=
−
3
proj
b
(
a
−
b
) = (comp
b
(
a
−
b
))
b
/
)
b
)
=
−
3
b
4
,
3
,
0
a
/
5=
1(
12
/
5
,
−
9
/
5
,
0
a
proj
b
⊥
(
a
−
b
)=(
a
−
b
)
−
proj
b
(
a
−
b
)=
1(
3
,
−
1
,
−
2
.(1(
12
/
5
,
−
9
/
5
,
0
a
=
1(
3
/
5
,
4
/
5
,
−
10
/
5
a
37.
Let
a
=
b
a, b, c
a
and
r
=
b
x, y, z
a
.Then
(a)
(
r
−
a
)
·
r
=
b
x
−
a, y
−
b, z
−
c
ab
x, y, z
a
=
x
2
−
ax
+
y
2
−
by
+
z
2
−
zc
= 0 implies
(
x
−
a
2
)
2
+(
y
−
b
2
)
2
+(
z
−
c
2
)
2
=
a
2
+
b
2
+
c
2
4
.The surface is a sphere.
(b)
(
r
−
a
)
·
a
=
b
x
−
a, y
−
b, z
−
c
ab
a, b, c
a
=
a
(
x
−
a
)+
b
(
y
−
b
)+
c
(
z
−
c
)=0
The surface is a plane.
38.
b
4
,
2
,
−
2
.(1
2
,
4
,
−
3
a
=
b
2
,
−
2
,
1
a
;
b
2
,
4
,
−
3
.(1
6
,
7
,
−
5
a
=
1(
4
,
−
3
,
2
a
;
b
2
,
−
2
,
1
.d1(
4
,
−
3
,
2
a
=0
The points are the vertices of a right triangle.
39.
A direction vector of the given line is
b
4
,
−
2
,
6
a
.A parallel line containing (7
,
3
,
−
5) is (
x
−
7)
/
4=(
y
−
3)
/
(
−
2) =
(
z
+5)
/
6.
40.
A normal to the plane is
b
8
,
3
,
−
4
a
.The line with this direction vector and through (5
,
−
9
,
3) is
x
=5+8
t
,
y
=
−
9+3
t
,
z
=3
−
4
t
.
311

312. Chapter 7 Review Exercises
41.
The direction vectors are
fn
2
,
3
,
1
g
and
f
2
,
1
,
1
g
.Since
fn
2
,
3
,
1
g,f
2
,
1
,
1
g
= 0, the lines are orthogonal.Solving
1
−
2
t
=
x
=1+2
s
,3
t
=
y
=
−
4+
s
, we obtain
t
=
−
1 and
s
= 1.The point (3
,
−
3
,
0) obtained by letting
t
=
−
1 and
s
= 1 is common to the two lines, so they do intersect.
42.
Vectors in the plane are
f
2
,
3
,
1
g
and
f
1
,
0
,
2
g
.A normal vector is
f
2
,
3
,
1
gf
1
,
0
,
2
g
=
f
6
,
−
3
,
−
3
g
=3
f
2
,
−
1
,
−
1
g
.
An equation of the plane is 2
x
−
y
−
z
=0
43.
The lines are parallel with direction vector
f
1
,
4
,
−
2
g
.Since (0
,
0
,
0) is on the first line and (1
,
1
,
3) is on the second
line, the vector
f
1
,
1
,
3
g
is in the plane.A normal vector to the plane is thus
f
1
,
4
,
−
2
gf
1
,
1
,
3
g
=
f
14
,
−
5
,
−
3
g
.
An equation of the plane is 14
x
−
5
y
−
3
z
=0.
44.
Letting
z
=
t
in the equations of the plane and solving
−
x
+
y
=4+8
t
,3
x
−
y
=
−
2
t
, we obtain
x
=2+3
t
,
y
=6+11
t
,
z
=
t
.Thus, a normal to the plane is
f
3
,
11
,
1
g
and an equation of the plane is
3(
x
−
1) + 11(
y
−
7)+(
z
+1)=0or3
x
+11
y
+
z
=79
.
45. F
=10
a
a
a
a
=
10
√
2
(
i
+
j
)=5
√
2
i
+5
√
2
j
;
d
=
f
7
,
4
,
0
gnf
4
,
1
,
0
g
=3
i
+3
j
W
=
F
·
d
=15
√
2+15
√
2=30
√
2 N-m
46. F
=5
√
2
i
+5
√
2
j
+50
i
=(5
√
2 + 50)
i
+5
√
2
j
;
d
=3
i
+3
j
W
=15
√
2+150+15
√
2=30
√
2 + 150 N-m
≈
192
.
4 N-m
47.
Since
F
2
= 200(
i
+
j
)
/
√
2 = 100
√
2
i
+ 100
√
2
j
,
F
3
=
F
2
−
F
1
= (100
√
2
−
200)
i
+ 100
√
2
j
and
a
F
3
a
=
f
(100
√
2
−
200)
2
+ (100
√
2)
2
= 200
y
2
−
√
2
≈
153 lb.
48.
Let
a
F
1
a
=
F
1
and
a
F
2
a
=
F
2
.Then
F
1
=
F
1
[(cos 45
◦
)
i
+ (sin 45
◦
)
j
] and
F
2
=
F
2
[(cos 120
◦
)
i
+ (sin 120
◦
)
j
], or
F
1
=
F
1
(
1
√
2
i
+
1
√
2
j
) and
F
2
=
F
2
(
−
1
2
i
+
√
3
2
j
).Since
w
+
F
1
+
F
2
=
0
,
F
1
(
1
√
2
i
+
1
√
2
j
)+
F
2
(
−
1
2
i
+
√
3
2
j
)=50
j
,
(
1
√
2
F
1
−
1
2
F
2
)
i
+(
1
√
2
F
1
+
√
3
2
F
2
)
j
=50
j
and
1
√
2
F
1
−
1
2
F
2
=0
,
1
√
2
F
1
+
√
3
2
F
2
=50
.
Solving, we obtain
F
1
= 25(
√
6
−
√
2)
≈
25
.
9 lb and
F
2
= 50(
√
3
−
1)
≈
36
.
6 lb.
49.
Not a vector space.Axiom (
viii
) is not satisfied.
50.
The vectors are linearly independent.The only solution of the system
c
1
=0
,c
1
+2
c
2
+
c
3
=0
,
2
c
1
+3
c
2
−
c
3
=0
is
c
1
=0,
c
2
=0,
c
3
=0.
51.
Let
p
1
and
p
2
be in
P
n
such that
d
2
p
1
dx
2
= 0 and
d
2
p
2
dx
2
= 0.Since
0=
d
2
p
1
dx
2
+
d
2
p
2
dx
2
=
d
2
dx
2
(
p
1
+
p
2
) and 0 =
k
d
2
p
1
dx
2
=
d
2
dx
2
(
kp
1
)
we conclude that the set of polynomials with the given property is a subspace of
P
n
.A basis for the subspace
is 1,
x
.
52.
The intersection
W
1
∩
W
2
is a subspace of
V
.If
x
and
y
are in
W
1
∩
W
2
then
x
and
y
are in each subspace
and so
x
+
y
is in each subspace.That is,
x
+
y
is in
W
1
∩
W
2
.Similarly, if
x
is in
W
1
∩
W
2
then
x
is in each
subspace and so
k
x
is in each subspace.That is,
k
x
is in
W
1
∩
W
2
for any scalar
k
.
The union
W
1
∪
W
2
is generally not a subspace.For example,
W
1
=
41
x, y
g
n
n
y
=
x
}
and
W
2
=
41
x, y
g
n
n
y
=2
x
}
are subspaces of
R
2
.Now
f
1
,
1
g
is in
W
1
and
f
1
,
2
g
is in
W
2
but
f
1
,
1
g
+
f
1
,
2
g
=
f
2
,
3
g
is not in
W
1
∪
W
2
.
312

313. 8
Matrices
Exercises 8.1
1.
2
×
4
2.
3
×
2
3.
3
×
3
4.
1
×
3
5.
3
×
4
6.
8
×
1
7.
Not equal
8.
Not equal
9.
Not equal
10.
Not equal
11.
Solving
x
=
y
−
2,
y
=3
x
−
2 we obtain
x
=2,
y
=4.
12.
Solving
x
2
=9,
y
=4
x
we obtain
x
=3,
y
= 12 and
x
=
−
3,
t
=
−
12.
13.
c
23
= 2(0)
−
3(
−
3) = 9;
c
12
= 2(3)
−
3(
−
2) = 12
14.
c
23
= 2(1)
−
3(0) = 2;
c
12
=2(
−
1)
−
3(0) =
−
2
15. (a) A
+
B
=
2
4
−
25+6
−
6+8 9
−
10
3
=
2
211
2
−
1
3
(b) B
−
A
=
2
−
2
−
46
−
5
8+6
−
10
−
9
3
=
2
−
61
14
−
19
3
(c)
2
A
+3
B
=
2
810
−
12 18
3
+
2
−
618
24
−
30
3
=
2
228
12
−
12
3
16. (a) A
−
B
=



−
2
−
3 0+1
4
−
01
−
2
7+4 3+2



=



−
51
4
−
1
11 5



(b) B
−
A
=



3+2
−
1
−
0
0
−
42
−
1
−
4
−
7
−
2
−
3



=



5
−
1
−
41
−
11
−
5



(c)
2(
A
+
B
)=2



1
−
1
43
31



=



2
−
2
86
62



17. (a) AB
=
2
−
2
−
912
−
6
5+12
−
30 + 8
3
=
2
−
11 6
17
−
22
3
(b) BA
=
2
−
2
−
30 3 + 24
6
−
10
−
9+8
3
=
2
−
32 27
−
4
−
1
3
(c) A
2
=
2
4+15
−
6
−
12
−
10
−
20 15+ 16
3
=
2
19
−
18
−
30 31
3
(d) B
2
=
2
1+18
−
6+12
−
3+6 18+4
3
=
2
19 6
322
3
18. (a) AB
=



−
4+4 6
−
12
−
3+8
−
20 + 10 30
−
30
−
15+ 20
−
32 + 12 48
−
36
−
24+24



=



0
−
65
−
10 0 5
−
20 12 0



313

314. Exercises 8.1
(b) BA
=
2
−
4+30
−
24
−
16+60
−
36
1
−
15+ 16 4
−
30+24
4
=
2
28
2
−
2
4
19. (a) BC
=
2
924
38
4
(b) A
(
BC
)=
2
1
−
2
−
24
42
924
38
4
=
2
38
−
6
−
16
4
(c) C
(
BA
)=
2
02
34
42
00
00
4
=
2
00
00
4
(d) A
(
B
+
C
)=
2
1
−
2
−
24
42
65
55
4
=
2
−
4
−
5
810
4
20. (a) AB
=[5
−
67]



3
4
−
1



=(
−
16)
(b) BA
=



3
4
−
1



[5
−
67]=



15
−
18 21
20
−
24 28
−
56
−
7



(c)
(
BA
)
C
=



15
−
18 21
20
−
24 28
−
56
−
7






12 4
01
−
1
32 1



=



78 54 99
104 72 132
−
26
−
18
−
33



(d)
Since
AB
is 1
×
1 and
C
is 3
×
3 the product (
AB
)
C
is not defined.
21. (a) A
T
A
=[4 8
−
10 ]



4
8
−
10



= (180)
(b) B
T
B
=



2
4
5



[245]=



4810
81620
10 20 25



(c) A
+
B
T
=



4
8
−
10



+



2
4
5



=



6
12
−
5



22. (a) A
+
B
T
=
2
12
24
4
+
2
−
25
37
4
=
2
−
17
511
4
(b)
2
A
T
−
B
T
=
2
24
48
4
−
2
−
25
37
4
=
2
4
−
1
11
4
(c) A
T
(
A
−
B
)=
2
12
24
42
3
−
1
−
3
−
3
4
=
2
−
3
−
7
−
6
−
14
4
23. (a)
(
AB
)
T
=
2
710
38 75
4
T
=
2
738
10 75
4
314

315. Exercises 8.1
(b) B
T
A
T
=
2
5
−
2
10
−
5
32
38
41
3
=
2
738
10 75
3
24. (a) A
T
+
B
=
2
5
−
4
96
3
+
2
−
311
−
72
3
=
2
27
28
3
(b)
2
A
+
B
T
=
2
10 18
−
812
3
+
2
−
3
−
7
11 2
3
=
2
711
314
3
25.
2
−
4
8
3
−
2
4
16
3
+
2
−
6
9
3
=
2
−
14
1
3
26.



6
3
−
3



+



−
5
−
5
15



+



−
6
−
8
10



=



−
5
−
10
22



27.
2
−
19
18
3
−
2
19
20
3
=
2
−
38
−
2
3
28.



−
7
17
−
6



+



−
1
1
4



−



2
8
−
6



=



−
10
10
4



29.
4
×
5
30.
3
×
2
31. A
T
=
2
2
−
3
42
3
;(
A
T
)
T
=
2
24
−
32
3
=
A 32.
(
A
+
B
)
T
=
2
6
−
6
14 10
3
=
A
T
+
B
T
33.
(
AB
)
T
=
2
16 40
−
8
−
20
3
T
=
2
16
−
8
40
−
20
3
;
B
T
A
T
=
2
42
10 5
32
2
−
3
42
3
=
2
16
−
8
40
−
20
3
34.
(6
A
)
T
=
2
12
−
18
24 12
3
=6
A
T
35. B
=
AA
T
=



21
63
25



2
262
135
3
=



515 9
1539 27
92729



=
B
T
36.
Using Problem 33 we have (
AA
T
)
T
=(
A
T
)
T
A
T
=
AA
T
, so that
AA
T
is symmetric.
37.
Let
A
=
2
10
00
3
and
B
=
2
00
01
3
. Then
AB
=
0
.
38.
We see that
A
3
=
B
, but
AC
=



23 4
46 8
6912



=
BC
.
39.
Since (
A
+
B
)
2
=(
A
+
B
)(
A
+
B
)
3
=
A
2
+
AB
+
BA
+
B
2
, and
AB
3
=
BA
in general, (
A
+
B
)
2
3
=
A
2
+2
AB
+
B
2
.
40.
Since (
A
+
B
)(
A
−
B
)=
A
2
−
AB
+
BA
−
B
2
, and
AB
3
=
BA
in general, (
A
+
B
)(
A
−
B
)
3
=
A
2
−
B
2
.
41.
a
11
x
1
+
a
12
x
2
=
b
1
;
a
21
x
1
+
a
22
x
2
=
b
2
42.



26 1
12
−
1
57
−
4






x
1
x
2
x
3



=



7
−
1
9



43.
[
xy
]
2
ab/
2
b/
2
c
32
x
y
3
=[
ax
+
by/
2
bx/
2+
cy
]
2
x
y
3
=[
ax
2
+
bxy/
2+
bxy/
2+
cy
2
]=[
ax
2
+
bxy
+
cy
2
]
44.



0
−
∂/∂z ∂/∂y
∂/∂z
0
−
∂/∂x
−
∂/∂y ∂/∂x
0






P
Q
R



=



−
∂Q/∂z
+
∂R/∂y
∂P/∂z
−
∂R/∂x
−
∂P/∂y
+
∂Q/∂x



= curl
F
315

316. Exercises 8.1
45. (a)
M
Y



x
y
z



=



cos
γ
sin
γ
0
−
sin
γ
cos
γ
0
001






x
y
z



=



x
cos
γ
+
y
sin
γ
−
x
sin
γ
+
y
cos
γ
z



=



x
Y
y
Y
z
Y



(b)
M
R
=



cos
β
0
−
sin
β
01 0
sin
β
0 cos
β



;
M
P



10 0
0 cos
α
sin
α
0
−
sin
α
cos
α



(c)
M
P



1
1
1



=



10 0
0 cos 30
◦
sin 30
◦
0
−
sin 30
◦
cos 30
◦






1
1
1



=



10 0
0
√
3
2
1
2
0
−
1
2
√
3
2






1
1
1



=



1
1
2
(
√
3+1)
1
2
(
√
3
−
1)



M
R
M
P



1
1
1



=



cos 45
◦
0
−
sin 45
◦
01 0
sin 45
◦
0 cos 45
◦






1
1
2
(
√
3+1)
1
2
(
√
3
−
1)



=



√
2
2
0
−
√
2
2
01 0
√
2
2
0
√
2
2






1
1
2
(
√
3+1)
1
2
(
√
3
−
1)



=



1
4
(3
√
2
−
√
6)
1
2
(
√
3+1)
1
4
(
√
2+
√
6)



M
Y
M
R
M
P



1
1
1



=



cos 60
◦
sin 60
◦
0
−
sin 60
◦
cos 60
◦
0
001






1
4
(3
√
2
−
√
6)
1
2
(
√
3+1)
1
4
(
√
2+
√
6)



=



1
2
√
3
2
0
−
√
3
2
1
2
0
001






1
4
(3
√
2
−
√
6)
1
2
(
√
3+1)
1
4
(
√
2+
√
6)



=



1
8
(3
√
2
−
√
6+6+2
√
3)
1
8
(
−
3
√
6+3
√
2+2
√
3+2)
1
4
(
√
2+
√
6)



46. (a) LU
=
2
10
1
2
1
32
2
−
2
03
3
=
2
2
−
2
12
3
=
A
(b) LU
=
2
10
2
3
1
32
62
0
−
1
3
3
=
2
62
41
3
=
A
(c) LU
=



100
010
2101






1
−
21
01 2
00
−
21



=



1
−
21
012
261



=
A
(d) LU
=



100
310
111






111
0
−
2
−
1
001



=



111
312
1
−
11



=
A
47. (a) AB
=
2
A
11
A
12
A
21
A
22
32
B
1
B
2
3
=
2
A
11
B
1
+
A
12
B
2
A
21
B
1
+
A
22
B
2
3
=



17 43
375
−
14 51



since
A
11
B
1
+
A
12
B
2
=
2
13 25
−
949
3
+
2
418
12 26
3
=
2
17 43
375
3
and
A
21
B
1
+
A
22
B
2
=[
−
24 34 ] + [ 10 17 ] = [
−
14 51 ]
.
316

317. Exercises 8.2
(b)
It is easier to enter smaller strings of numbers and the chance of error is decreased. Also, if the large matrix
has submatrices consisting of all zeros or diagonal matrices, these are easily entered without listing all of
the entries.
Exercises 8.2
1.
2
1
−
1
11
43
−
5
3
−
4
R
1
+
R
2
−
−
−
−
−
−→
2
1
−
1
11
07
−
49
3
1
7
R
2
−
−
−
−
−
−→
2
1
−
1
11
01
−
7
3
R
3
+
R
1
−
−
−
−
−
−→
2
10
4
01
−
7
3
The solution is
x
1
=4,
x
2
=
−
7.
2.
2
3
−
2
4
1
−
1
−
2
3
R
12
−
−
−
−
−
−→
2
1
−
1
−
2
3
−
2
4
3
−
3
R
1
+
R
2
−
−
−
−
−
−→
2
1
−
1
−
2
01
10
3
R
2
+
R
1
−
−
−
−
−
−→
2
10
8
01
10
3
The solution is
x
1
=8,
x
2
= 10.
3.
2
93
−
5
2
−
1
−
1
3
1
9
R
1
−
−
−
−
−
−→

1
1
3
−
5
9
21
−
1

−
2
R
1
+
R
2
−
−
−
−
−
−→

1
1
3
−
5
9
0
1
3
1
9

3
R
2
−
−
−
−
−
−→

1
1
3
−
5
9
01
1
3

−
1
3
R
2
+
R
1
−
−
−
−
−
−→

10
−
2
3
01
1
3

The solution is
x
1
=
−
2
3
,
x
2
=
1
3
.
4.
2
10 15
1
32
−
1
3
1
10
R
1
−
−
−
−
−
−→

1
3
2
1
10
32
−
1

−
3
R
1
+
R
2
−
−
−
−
−
−→

1
3
2
1
10
0
−
5
2
−
13
10

−
2
5
R
2
−
−
−
−
−
−→

1
3
2
1
10
01
13
25

−
3
2
R
2
+
R
1
−
−
−
−
−
−→

10
−
17
25
01
13
25

The solution is
x
1
=
−
17
25
,
x
2
=
13
25
.
5.



1
−
1
−
1
−
3
235
7
1
−
23
−
11



−
2
R
1
+
R
2
−
−
−
−
−
−→
−
R
1
+
R
3



1
−
1
−
1
−
3
057
13
0
−
14
−
8



1
5
R
2
−
−
−
−
−
−→



1
−
1
−
1
−
3
01
7
5
13
5
0
−
14
−
8



R
2
+
R
1
−
−
−
−
−
−→
R
2
+
R
3



10
2
5
−
2
5
01
7
5
13
5
00
27
5
−
27
5



5
27
R
3
−
−
−
−
−
−→



10
2
5
−
2
5
01
7
5
13
5
001
−
1



−
2
5
R
3
+
R
1
−
−
−
−
−
−→
−
7
5
R
3
+
R
2



100
0
010
4
001
−
1



The solution is
x
1
=0,
x
2
=4,
x
3
=
−
1.
6.



12
−
1
0
212
9
1
−
11
3



−
2
R
1
+
R
2
−
−
−
−
−
−→
−
R
1
+
R
3



12
−
1
0
0
−
34
9
0
−
32
3



−
1
3
R
2
−
−
−
−
−
−→



12
−
1
0
01
−
4
3
−
3
0
−
32
3



−
2
R
2
+
R
1
−
−
−
−
−
−→
3
R
2
+
R
3



10
5
3
6
01
−
4
3
−
3
00
−
2
−
6



−
1
2
R
3
−
−
−
−
−
−→



10
5
3
6
01
−
4
3
−
3
00 1
3



−
5
3
R
3
+
R
1
−
−
−
−
−
−→
4
3
R
3
+
R
2



100
1
010
1
001
3



The solution is
x
1
=1,
x
2
=1,
x
3
=3.
7.
2
111
0
113
0
3
−
R
1
+
R
2
−
−
−
−
−
−→
2
111
0
002
0
3
317

318. Exercises 8.2
Since
x
3
= 0, setting
x
2
=
t
we obtain
x
1
=
−
t
,
x
2
=
t
,
x
3
=0.
8.
2
12
−
4
9
5
−
12
1
3
−
5
R
1
+
R
2
−
−
−
−
−
−→
2
12
−
4
9
0
−
11 22
−
44
3
−
1
11
R
2
−
−
−
−
−
−→
2
12
−
4
9
01
−
2
4
3
−
2
R
2
+
R
1
−
−
−
−
−
−→
2
10 0
1
01
−
2
4
3
If
x
3
=
t
, the solution is
x
1
=1,
x
2
=4+2
t
,
x
3
=
t
9.



1
−
1
−
1
8
1
−
11
3
−
111
4



row
−
−
−
−
−
−→
operations



1
−
1
−
1
8
002
−
5
000
12



Since the bottom row implies 0 = 12, the system is inconsistent.
10.



31
4
43
−
3
2
−
1
11



row
−
−
−
−
−
−→
operations



1
1
3
4
3
01
−
5
00
0



The solution is
x
1
=3,
x
2
=
−
5.
11.



220
0
−
211
0
301
0



row
−
−
−
−
−
−→
operations



110
0
01
1
3
0
001
0



The solution is
x
1
=
x
2
=
x
3
=0.
12.



1
−
1
−
2
0
245
0
60
−
3
0



row
−
−
−
−
−
−→
operations



1
−
1
−
2
0
01
3
2
0
00 0
0



The solution is
x
1
=
1
2
t
,
x
2
=
−
3
2
t
,
x
3
=
t
.
13.



122
2
111
0
1
−
3
−
1
0



row
−
−
−
−
−
−→
operations



122
2
011
2
001
4



The solution is
x
1
=
−
2,
x
2
−
2,
x
3
=4.
14.



1
−
21
2
3
−
12
5
211
1



row
−
−
−
−
−
−→
operations



1
−
21
2
01
−
1
5
−
1
5
00 0
−
2



Since the bottom row implies 0 =
−
2, the system is inconsistent.
15.



111
3
1
−
1
−
1
−
1
311
5



row
−
−
−
−
−
−→
operations



111
3
011
2
000
0



If
x
3
=
t
the solution is
x
1
=1,
x
2
=2
−
t
,
x
3
=
t
.
16.



1
−
1
−
2
−
1
−
3
−
21
−
7
231
8



row
−
−
−
−
−
−→
operations



1
−
1
−
2
−
1
011
2
000
0



If
x
3
=
t
the solution is
x
1
=1+
t
,
x
2
=2
−
t
,
x
3
=
t
.
318

319. Exercises 8.2
17.





101
−
1
1
0211
3
1
−
10 1
−
1
1111
2





row
−
−
−
−
−
−→
operations





101
−
1
1
01
1
2
1
2
3
2
001
−
5
1
000 1
0





The solution is
x
1
=0,
x
2
=1,
x
3
=1,
x
4
=0.
18.





21 10
3
31 11
4
12 23
3
45
−
21
16





row
−
−
−
−
−
−→
operations





1
1
2
1
2
0
3
2
011
−
2
1
001
−
1
−
1
000 1
0





The solution is
x
1
=1,
x
2
=2,
x
3
=
−
1,
x
4
=0.
19.





135
−
1
1
011
−
1
4
125
−
4
−
2
146
−
2
6





row
−
−
−
−
−
−→
operations





135
−
1
1
011
−
1
4
001
−
4
1
000 0
1





Since the bottom row implies 0 = 1, the system is inconsistent.
20.





120 1
0
49112
0
39621
0
131 9
0





row
−
−
−
−
−
−→
operations





120 1
0
011 8
0
001
−
2
0
000 0
0





If
x
4
=
t
the solution is
x
1
=19
t
,
x
2
=
−
10
t
,
x
3
=2
t
,
x
4
=
t
.
21.



11 1
4
.
280
0
.
2
−
0
.
1
−
0
.
5
−
1
.
978
4
.
10
.
30
.
12
1
.
686



row
−
−
−
−
−
−→
operations



11 1
4
.
28
012
.
333
9
.
447
00 1
4
.
1



The solution is
x
1
=0
.
3,
x
2
=
−
0
.
12,
x
3
=4
.
1.
22.



2
.
51
.
44
.
5
2
.
6170
1
.
350
.
951
.
2
0
.
7545
2
.
73
.
05
−
1
.
44
−
1
.
4292



row
−
−
−
−
−
−→
operations



10
.
561
.
8
1
.
0468
01
−
6
.
3402
−
3
.
3953
00 1
0
.
28



The solution is
x
1
=1
.
45,
x
2
=
−
1
.
62,
x
3
=0
.
28.
23.
From
x
1
Na +
x
2
H
2
O
→
x
3
NaOH +
x
4
H
2
we obtain the system
x
1
=
x
3
,2
x
2
=
x
3
+2
x
4
,
x
2
=
x
3
. We see
that
x
1
=
x
2
=
x
3
, so the second equation becomes 2
x
1
=
x
1
+2
x
4
or
x
1
=2
x
4
. A solution of the system is
x
1
=
x
2
=
x
3
=2
t
,
x
4
=
t
. Letting
t
= 1 we obtain the balanced equation 2Na + 2H
2
O
→
2NaOH + H
2
.
24.
From
x
1
KClO
3
→
x
2
KCl +
x
3
O
2
we obtain the system
x
1
=
x
2
,
x
1
=
x
2
,3
x
1
=2
x
3
. Letting
x
3
=
t
we see that
a solution of the system is
x
1
=
x
2
=
2
3
t
,
x
3
=
t
. Taking
t
= 3 we obtain the balanced equation
2KClO
3
→
2KCl + 3O
2
.
25.
From
x
1
Fe
3
O
4
+
x
2
C
→
x
3
Fe +
x
4
CO we obtain the system 3
x
1
=
x
3
,4
x
1
=
x
4
,
x
2
=
x
4
. Letting
x
1
=
t
we
see that
x
3
=3
t
and
x
2
=
x
4
=4
t
. Taking
t
= 1 we obtain the balanced equation
Fe
3
O
4
+4C
→
3Fe + 4CO
.
319

320. Exercises 8.2
26.
From
x
1
C
5
H
8
+
x
2
O
2
→
x
3
CO
2
+
x
4
H
2
O we obtain the system 5
x
1
=
x
3
,8
x
1
=2
x
4
,2
x
2
=2
x
3
+
x
4
. Letting
x
1
=
t
we see that
x
3
=5
t
,
x
4
=4
t
, and
x
2
=7
t
. Taking
t
= 1 we obtain the balanced equation
C
5
H
8
+7O
2
→
5CO
2
+4H
2
O
.
27.
From
x
1
Cu +
x
2
HNO
3
→
x
3
Cu(NO
3
)
2
+
x
4
H
2
O+
x
5
NO we obtain the system
x
1
=3
,x
2
=2
x
4
,x
2
=2
x
3
+
x
5
,
3
x
2
=6
x
3
+
x
4
+
x
5
.
Letting
x
4
=
t
we see that
x
2
=2
t
and
2
t
=2
x
3
+
x
5
6
t
=6
x
3
+
t
+
x
5
or
2
x
3
+
x
5
=2
t
6
x
3
+
x
5
=5
t.
Then
x
3
=
3
4
t
and
x
5
=
1
2
t
. Finally,
x
1
=
x
3
=
3
4
t
. Taking
t
= 4 we obtain the balanced equation
3Cu + 8HNO
3
→
3Cu(NO
3
)
2
+4H
2
O + 2NO
.
28.
From
x
1
Ca
3
(PO
4
)
2
+
x
2
H
3
PO
4
→
x
3
Ca(H
2
PO
4
)
2
we obtain the system
3
x
1
=
x
3
,
2
x
1
+
x
2
=2
x
3
,
8
x
1
+4
x
2
=8
x
3
,
3
x
2
=4
x
3
.
Letting
x
1
=
t
we see from the first equation that
x
3
=3
t
and from the fourth equation that
x
2
=4
t
. These
choices also satisfy the second and third equations. Taking
t
= 1 we obtain the balanced equation
Ca
3
(PO
4
)
2
+4H
3
PO
4
→
3Ca(H
2
PO
4
)
2
.
29.
The system of equations is
−
i
1
+
i
2
−
i
3
=0
10
−
3
i
1
+5
i
3
=0
27
−
6
i
2
−
5
i
3
=0
or
−
i
1
+
i
2
−
i
3
=0
3
i
1
−
5
i
3
=10
6
i
2
+5
i
3
=27
Gaussian elimination gives



−
11
−
1
0
30
−
5
10
06 5
27



row
−
−
−
−
−
−→
operations



1
−
11
0
01
−
8
/
3
10
/
3
001
1
/
3



.
The solution is
i
1
=
35
9
,
i
2
=
38
9
,
i
3
=
1
3
.
30.
The system of equations is
i
1
−
i
2
−
i
3
=0
52
−
i
1
−
5
i
2
=0
−
10
i
3
+5
i
2
=0
or
i
1
−
i
2
−
i
3
=0
i
1
+5
i
2
=52
5
i
2
−
10
i
3
=0
Gaussian elimination gives



1
−
1
−
1
0
150
52
05
−
10
0



row
−
−
−
−
−
−→
operations



1
−
1
−
1
0
011
/
6
26
/
3
00 1
4



.
The solution is
i
1
= 12,
i
2
=8,
i
3
=4.
31.
Interchange row 1 and row in
I
3
.
32.
Multiply row 3 by
c
in
I
3
.
320

321. Exercises 8.2
33.
Add
c
times row 2 to row 3 in
I
3
.
34.
Add row 4 to row 1 in
I
4
.
35. EA
=



a
21
a
22
a
23
a
11
a
12
a
13
a
31
a
32
a
33



36. EA
=



a
11
a
12
a
13
a
21
a
22
a
23
ca
31
ca
32
ca
33



37. EA
=



a
11
a
12
a
13
a
21
a
22
a
23
ca
21
+
a
31
ca
22
+
a
32
ca
23
+
a
33



38. E
1
E
2
A
=
E
1



a
11
a
12
a
13
a
21
a
22
a
23
ca
21
+
a
31
ca
22
+
a
32
ca
23
+
a
33



=



a
21
a
22
a
23
a
11
a
12
a
13
ca
21
+
a
31
ca
22
+
a
32
ca
23
+
a
33



39.
The system is equivalent to
2
10
1
2
1
32
2
−
2
03
3
X
=
2
2
6
3
.
Letting
Y
=
2
y
1
y
2
3
=
2
2
−
2
03
3
X
we have
2
10
1
2
1
32
y
1
y
2
3
=
2
2
6
3
.
This implies
y
1
= 2 and
1
2
y
1
+
y
2
=1+
y
2
=6or
y
2
= 5. Then
2
2
−
2
03
32
x
1
x
2
3
=
2
2
5
3
,
which implies 3
x
2
=5or
x
2
=
5
3
and 2
x
1
−
2
x
2
=2
x
1
−
10
3
=2or
x
1
=
8
3
. The solution is
X
=

8
3
,
5
3

.
40.
The system is equivalent to
2
10
2
3
1
32
62
0
−
1
3
3
X
=
2
1
−
1
3
.
Letting
Y
=
2
y
1
y
2
3
=
2
62
0
−
1
3
3
X
we have
2
10
2
3
1
32
y
1
y
2
3
=
2
1
−
1
3
.
This implies
y
1
= 1 and
2
3
y
1
+
y
2
=
2
3
+
y
2
=
−
1or
y
2
=
−
5
3
. Then
2
62
0
−
1
3
32
x
1
x
2
3
=
2
1
−
5
3
3
,
which implies
−
1
3
x
2
=
−
5
3
or
x
2
= 5and 6
x
1
+2
x
2
=6
x
1
+10=1or
x
1
=
−
3
2
. The solution is
X
=

−
3
2
,
5

.
41.
The system is equivalent to



100
010
2101






1
−
21
01 2
00
−
21



X
=



2
−
1
1



.
321

322. Exercises 8.2
Letting
Y
=



y
1
y
2
y
3



=



1
−
21
01 2
00
−
21



X
we have



100
010
2101






y
1
y
2
y
3



=



2
−
1
1



.
This implies
y
1
=2,
y
2
=
−
1, and 2
y
1
+10
y
2
+
y
3
=4
−
10 +
y
3
=1or
y
3
= 7. Then



1
−
21
01 2
00
−
21






x
1
x
2
x
3



=



2
−
1
7



,
which implies
−
21
x
3
=7or
x
3
=
−
1
3
,
x
2
+2
x
3
=
x
2
−
2
3
=
−
1or
x
2
=
−
1
3
, and
x
1
−
2
x
2
+
x
3
=
x
1
+
2
3
−
1
3
=2
or
x
1
=
5
3
. The solution is
X
=

5
3
,
−
1
3
,
−
1
3

.
42.
The system is equivalent to



100
310
111






111
0
−
2
−
1
001



X
=



0
1
4



.
Letting
Y
=



y
1
y
2
y
3



=



111
0
−
2
−
1
001



X
we have



100
310
111






y
1
y
2
y
3



=



0
1
4



.
This implies
y
1
=0,3
y
1
+
y
2
=
y
2
= 1, and
y
1
+
y
2
+
y
3
=0+1+
y
3
=4or
y
3
= 3. Then



111
0
−
2
−
1
001






x
1
x
2
x
3



=



0
1
3



,
which implies
x
3
=3,
−
2
x
2
−
x
3
=
−
2
x
2
−
3=1or
x
2
=
−
2, and
x
1
+
x
2
+
x
3
=
x
1
−
2+3=0or
x
1
=
−
1.
The solution is
X
=(
−
1
,
−
2
,
3).
43.
Using the
Solve
function in
Mathematica
we find
(a)
x
1
=
−
0
.
0717393
−
1
.
43084
c
,
x
2
=
−
0
.
332591 + 0
.
855709
c
,
x
3
=
c
, where
c
is any real number
(b)
x
1
=
c/
3,
x
2
=5
c/
6,
x
3
=
c
, where
c
is any real number
(c)
x
1
=
−
3
.
76993,
x
2
=
−
1
.
09071,
x
3
=
−
4
.
50461,
x
4
=
−
3
.
12221
(d)
x
1
=
8
3
−
7
3
b
+
2
3
c
,
x
2
=
2
3
−
1
3
b
−
1
3
c
,
x
3
=
−
3,
x
4
=
b
,
x
5
=
c
, where
b
and
c
are any real numbers.
322

323. Exercises 8.3
Exercises 8.3
1.
2
3
−
1
13
3
row
−
−
−
−
−
−→
operations
2
13
01
3
; The rank is 2.
2.
2
2
−
2
00
3
row
−
−
−
−
−
−→
operations
2
1
−
1
00
3
; The rank is 1.
3.



213
639
−
1
−
1
2
−
3
2



row
−
−
−
−
−
−→
operations



1
1
2
3
2
000
000



; The rank is 1.
4.



112
−
124
−
103



row
−
−
−
−
−
−→
operations



112
015
001



; The rank is 3.
5.



111
104
141



row
−
−
−
−
−
−→
operations



11 1
01
−
3
00 1



; The rank is 3.
6.
2
3
−
120
6 245
3
row
−
−
−
−
−
−→
operations

1
−
1
3
2
3
0
010
5
4

; The rank is 2.
7.





1
−
2
3
−
6
7
−
1
45





row
−
−
−
−
−
−→
operations





1
−
2
01
00
00





; The rank is 2.
8.





1
−
234
1 468
0 100
2568





row
−
−
−
−
−
−→
operations





1
−
234
0100
001
4
3
0000





; The rank is 3.
9.





024 22
410 51
21
2
3
3
1
3
666120





row
−
−
−
−
−
−→
operations





1
1
2
1
3
3
2
1
6
01
4
3
1
−
1
3
0010 2
0000 0





; The rank is 3.
10.







1
−
218
−
11 16
0 013
−
11 15
0 013
−
12108
0 000 01 13
1
−
218
−
11 26







row
−
−
−
−
−
−→
operations







1
−
218
−
1116
0 013
−
1115
0 000 0193
0 000 0010
0 000 0000







; The rank is 4.
11.



123
101
1
−
15



row
−
−
−
−
−
−→
operations



123
011
001



;
Since the rank of the matrix is 3 and there are 3 vectors, the vectors are linearly independent.
323

324. Exercises 8.3
12.





263
1
−
14
321
254





row
−
−
−
−
−
−→
operations





1
−
14
01
−
5
8
001
000





Since the rank of the matrix is 3 and there are 4 vectors, the vectors are linearly dependent.
13.



1
−
13
−
1
1
−
14 2
1
−
157



row
−
−
−
−
−
−→
operations



1
−
13
−
1
0013
0001



Since the rank of the matrix is 3 and there are 3 vectors, the vectors are linearly independent.
14.





2115
2211
3
−
16 1
111
−
1





row
−
−
−
−
−
−→
operations





111
−
1
011
−
7
001
−
3
000 1





Since the rank of the matrix is 4 and there are 4 vectors, the vectors are linearly independent.
15.
Since the number of unknowns is
n
= 8 and the rank of the coefficient matrix is
r
= 3, the solution of the
system has
n
−
r
= 5parameters.
16. (a)
The maximum possible rank of
A
is the number of rows in
A
, which is 4.
(b)
The system is inconsistent if rank(
A
)
<
rank(
A
/
B
) = 2 and consistent if rank(
A
) = rank(
A
/
B
)=2.
(c)
The system has
n
= 6 unknowns and the rank of
A
is
r
= 3, so the solution of the system has
n
−
r
=3
parameters.
17.
Since 2
v
1
+3
v
2
−
v
3
=
0
we conclude that
v
1
,
v
2
, and
v
3
are linearly dependent. Thus, the rank of
A
is at
most 2.
18.
Since the rank of
A
is
r
= 3 and the number of equations is
n
= 6, the solution of the system has
n
−
r
=3
parameters. Thus, the solution of the system is not unique.
19.
The system consists of 4 equations, so the rank of the coefficient matrix is at most 4, and the maximum number
of linearly independent rows is 4. However, the maximum number of linearly independent columns is the same
as the maximum number of linearly independent rows. Thus, the coefficient matrix has at most 4 linearly
independent columns. Since there are 5column vectors, they must be linearly dependent.
20.
Using the
RowReduce
in
Mathematica
we find that the reduced row-echelon form of the augmented matrix is









10000
834
2215
−
261
443
01000
1818
2215
282
443
00100
13
443
−
6
443
00010
4214
2215
−
130
443
00001
−
6079
2215
677
443









.
We conclude that the system is consistent and the solution is
x
1
=
−
226
443
−
834
2215
c
,
x
2
=
282
443
−
1818
2215
c
,
x
3
=
−
6
443
−
13
443
c
,
x
4
=
−
130
443
−
4214
2215
c
,
x
5
=
677
433
+
6079
2215
c
,
x
6
=
c
.
324

325. Exercises 8.4
Exercises 8.4
1.
M
12
=
)
)
)
)
12
−
25
)
)
)
)
=9
2.
M
32
=
)
)
)
)
24
12
)
)
)
)
3.
C
13
=(
−
1)
1+3
)
)
)
)
1
−
1
−
23
)
)
)
)
=1
4.
C
22
=(
−
1)
2+2
)
)
)
)
24
−
25
)
)
)
)
=18
5.
M
33
=
)
)
)
)
)
)
)
020
123
112
)
)
)
)
)
)
)
=2
6.
M
41
=
)
)
)
)
)
)
)
240
2
−
23
10
−
1
)
)
)
)
)
)
)
=24
7.
C
34
=(
−
1)
3+4
)
)
)
)
)
)
)
02 4
12
−
2
11 1
)
)
)
)
)
)
)
=10
8.
C
23
=(
−
1)
2+3
)
)
)
)
)
)
)
02 0
51
−
1
11 2
)
)
)
)
)
)
)
=22
9.
−
7
10.
2
11.
17
12.
−
1
/
2
13.
(1
−
λ
)(2
−
λ
)
−
6=
λ
2
−
3
λ
−
4
14.
(
−
3
−
λ
)(5
−
λ
)
−
8=
λ
2
−
2
λ
−
23
15.
)
)
)
)
)
)
)
020
301
058
)
)
)
)
)
)
)
=
−
3
)
)
)
)
20
58
)
)
)
)
=
−
48
16.
)
)
)
)
)
)
)
500
0
−
30
002
)
)
)
)
)
)
)
=5
)
)
)
)
−
30
02
)
)
)
)
=5(
−
3)(2) =
−
30
17.
)
)
)
)
)
)
)
302
271
264
)
)
)
)
)
)
)
=3
)
)
)
)
71
64
)
)
)
)
+2
)
)
)
)
27
26
)
)
)
)
= 3(22) + 2(
−
2) = 62
18.
)
)
)
)
)
)
)
1
−
1
−
1
22
−
2
119
)
)
)
)
)
)
)
=
)
)
)
)
2
−
2
19
)
)
)
)
−
2
)
)
)
)
−
1
−
1
19
)
)
)
)
+
)
)
)
)
−
1
−
1
2
−
2
)
)
)
)
=20
−
2(
−
8)+4=40
19.
)
)
)
)
)
)
)
453
123
123
)
)
)
)
)
)
)
=4
)
)
)
)
23
23
)
)
)
)
−
5
)
)
)
)
13
13
)
)
)
)
+3
)
)
)
)
12
12
)
)
)
)
=0
20.
)
)
)
)
)
)
)
1
4
60
1
3
80
1
2
90
)
)
)
)
)
)
)
= 0, expanding along the third column.
21.
)
)
)
)
)
)
)
−
2
−
14
−
361
−
348
)
)
)
)
)
)
)
=
−
2
)
)
)
)
61
48
)
)
)
)
+3
)
)
)
)
−
14
48
)
)
)
)
−
3
)
)
)
)
−
14
61
)
)
)
)
=
−
2(44) + 3(
−
24)
−
3(
−
25) =
−
85
22.
)
)
)
)
)
)
)
351
−
125
7
−
410
)
)
)
)
)
)
)
=3
)
)
)
)
25
−
410
)
)
)
)
−
5
)
)
)
)
−
15
710
)
)
)
)
+
)
)
)
)
−
12
7
−
4
)
)
)
)
= 3(40)
−
5(
−
45) + (
−
10) = 335
23.
)
)
)
)
)
)
)
111
xyz
234
)
)
)
)
)
)
)
=
)
)
)
)
yz
34
)
)
)
)
−
)
)
)
)
xz
24
)
)
)
)
+
)
)
)
)
xy
23
)
)
)
)
=(4
y
−
3
z
)
−
(4
x
−
2
z
)+(3
x
−
2
y
)=
−
x
+2
y
−
z
325

326. Exercises 8.4
24.
)
)
)
)
)
)
)
111
xyz
2+
x
3+
y
4+
z
)
)
)
)
)
)
)
=
)
)
)
)
yz
3+
y
4+
z
)
)
)
)
−
)
)
)
)
xz
2+
x
4+
z
)
)
)
)
+
)
)
)
)
xy
2+
x
3+
y
)
)
)
)
=(4
y
+
yz
−
3
z
−
yz
)
−
(4
x
+
xz
−
2
z
−
xz
)+(3
x
+
xy
−
2
y
−
xy
)=
−
x
+2
y
−
z
25.
)
)
)
)
)
)
)
)
)
11
−
30
1532
1
−
210
4800
)
)
)
)
)
)
)
)
)
=2
)
)
)
)
)
)
)
11
−
3
1
−
21
480
)
)
)
)
)
)
)
= 2(4)
)
)
)
)
1
−
3
−
21
)
)
)
)
−
2(8)
)
)
)
)
1
−
3
11
)
)
)
)
=8(
−
5)
−
16(4) =
−
104
26.
)
)
)
)
)
)
)
)
)
21
−
21
0504
1610
5
−
111
)
)
)
)
)
)
)
)
)
=5
)
)
)
)
)
)
)
2
−
21
110
511
)
)
)
)
)
)
)
+4
)
)
)
)
)
)
)
21
−
2
161
5
−
11
)
)
)
)
)
)
)
= 5(0) + 4(80) = 320
27.
Expanding along the first column in the original matrix and each succeeding minor, we obtain 3(1)(2)(4)(2) = 48.
28.
Expanding along the bottom row we obtain
−
1
)
)
)
)
)
)
)
)
)
20 0
−
2
1605
12
−
11
21
−
23
)
)
)
)
)
)
)
)
)
+
)
)
)
)
)
)
)
)
)
220 0
116 0
102
−
1
201
−
2
)
)
)
)
)
)
)
)
)
=
−
1(
−
48)+0=48
.
29.
Solving
λ
2
−
2
λ
−
15
−
20 =
λ
2
−
2
λ
−
35= (
λ
−
7)(
λ
+ 5) = 0 we obtain
λ
= 7 and
−
5.
30.
Solving
−
λ
3
+3
λ
2
−
2
λ
=
−
λ
(
λ
−
2)(
λ
−
1) = 0 we obtain
λ
= 0, 1, and 2.
Exercises 8.5
1.
Theorem 8
.
11
2.
Theorem 8
.
14
3.
Theorem 8
.
14
4.
Theorem 8
.
12 and 8
.
11
5.
Theorem 8
.
12 (twice)
6.
Theorem 8
.
11 (twice)
7.
Theorem 8
.
10
8.
Theorem 8
.
12 and 8
.
9
9.
Theorem 8
.
8
10.
Theorem 8
.
11 (twice)
11.
det
A
=
−
5
12.
det
B
= 2(3)(5) = 30
13.
det
C
=
−
5
14.
det
D
=5
15.
det
A
=6(
2
3
)(
−
4)(
−
5)=80
16.
det
B
=
−
a
13
a
22
a
31
17.
det
C
=(
−
5)(7)(3) =
−
105
18.
det
D
= 4(7)(
−
2) =
−
56
19.
det
A
= 14 = det
A
T
20.
det
A
= 96 = det
T
326

327. Exercises 8.5
21.
det
AB
=
)
)
)
)
)
)
)
0
−
22
10 7 23
8416
)
)
)
)
)
)
)
=
−
80 = 20(
−
4) = det
A
det
B
22.
From Problem 21, (det
A
)
2
= det
A
2
= det
I
=1,sodet
A
=
±
1.
23.
Using Theorems 8
.
14, 8
.
12, and 8
.
9, det
A
=
)
)
)
)
)
)
)
a
12
b
12
c
12
)
)
)
)
)
)
)
=2
)
)
)
)
)
)
)
a
11
b
11
c
11
)
)
)
)
)
)
)
=0.
24.
Using Theorems 8
.
14 and 8
.
9,
det
A
=
)
)
)
)
)
)
)
111
xyz
x
+
y
+
zx
+
y
+
zx
+
y
+
z
)
)
)
)
)
)
)
=(
x
+
y
+
z
)
)
)
)
)
)
)
)
111
xyz
111
)
)
)
)
)
)
)
=0
.
25.
)
)
)
)
)
)
)
115
436
0
−
11
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
11 5
0
−
1
−
14
0
−
11
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
11 5
0
−
1
−
14
0015
)
)
)
)
)
)
)
=1(
−
1)(15) =
−
15
26.
)
)
)
)
)
)
)
24 5
42 0
87
−
2
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
24 5
0
−
6
−
10
0
−
9
−
22
)
)
)
)
)
)
)
=
−
2
)
)
)
)
)
)
)
24 5
03 5
0
−
9
−
22
)
)
)
)
)
)
)
=
−
2
)
)
)
)
)
)
)
24 5
03 5
00
−
7
)
)
)
)
)
)
)
=
−
2(2)(3)(
−
7) = 84
27.
)
)
)
)
)
)
)
−
123
4
−
5
−
2
9
−
96
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
−
12 3
0310
0933
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
−
12 3
0310
00 3
)
)
)
)
)
)
)
=
−
1(3)(3) =
−
9
28.
)
)
)
)
)
)
)
−
22
−
6
501
1
−
22
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
1
−
22
501
−
22
−
6
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
1
−
22
010
−
9
0
−
2
−
2
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
1
−
22
010
−
9
00
−
19
5
)
)
)
)
)
)
)
=
−
1(10)(
−
19
5
)=38
29.
)
)
)
)
)
)
)
)
)
1
−
221
21
−
23
34
−
81
3
−
11 12 2
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
1
−
221
05
−
61
010
−
14
−
2
0
−
56
−
1
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
1
−
221
05
−
61
00
−
2
−
4
0000
)
)
)
)
)
)
)
)
)
= 1(5)(
−
2)(0) = 0
30.
)
)
)
)
)
)
)
)
)
0145
2501
1220
3132
)
)
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
)
)
1220
2501
0145
3132
)
)
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
)
)
1220
01
−
41
0145
0
−
5
−
32
)
)
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
)
)
12 20
01
−
41
00 84
00
−
23 7
)
)
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
)
)
12 2 0
01
−
41
00 8 4
00
−
23
37
2
)
)
)
)
)
)
)
)
)
=
−
(1)(1)(8)(
37
2
)=
−
148
31.
)
)
)
)
)
)
)
)
)
123 4
1357
236 7
15820
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
1234
0123
0
−
10
−
1
03516
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
12 34
01 23
00 22
00
−
17
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
1234
0123
0022
0008
)
)
)
)
)
)
)
)
)
= 1(1)(2)(8) = 16
327

328. Exercises 8.5
32.
)
)
)
)
)
)
)
)
)
2918
1374
0165
3142
)
)
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
)
)
1374
2918
0165
3142
)
)
)
)
)
)
)
)
)
=
−
)
)
)
)
)
)
)
)
)
13 7 4
03
−
13 0
01 6 5
0
−
8
−
17
−
10
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
1374
0165
03
−
13 0
0
−
8
−
17
−
10
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
1374
0165
00
−
31
−
15
0 0 31 30
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
13 7 4
01 6 5
00
−
31
−
15
00 0 15
)
)
)
)
)
)
)
)
)
= 1(1)(
−
31)(15) =
−
465
33.
We first use the second row to reduce the third row. Then we use the first row to reduce the second row.
)
)
)
)
)
)
)
11 1
ab c
0
b
2
−
ab c
2
−
ac
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
11 1
0
b
−
ac
−
a
0
b
(
b
−
a
)
c
(
c
−
a
)
)
)
)
)
)
)
)
=(
b
−
a
)(
c
−
a
)
)
)
)
)
)
)
)
111
011
0
bc
)
)
)
)
)
)
)
.
Expanding along the first row gives (
b
−
a
)(
c
−
a
)(
c
−
b
).
34.
In order, we use the third row to reduce the fourth row, the second row to reduce the third row, and the first
row to reduce the second row. We then pull out a common factor from each column.
)
)
)
)
)
)
)
)
)
1111
abcd
a
2
b
2
c
2
d
2
a
3
b
3
c
3
d
3
)
)
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
)
)
11 1 1
0
b
−
ac
−
ad
−
a
0
b
2
−
ab c
2
−
ac d
2
−
ac
0
b
3
−
ab
2
c
3
−
ac
2
d
3
−
ad
2
)
)
)
)
)
)
)
)
)
=(
b
−
a
)(
c
−
a
)(
d
−
a
)
)
)
)
)
)
)
)
)
)
11 1 1
01 1 1
0
bcd
0
b
2
c
2
d
2
)
)
)
)
)
)
)
)
)
.
Expanding along the first column and using Problem 33 we obtain (
b
−
a
)(
c
−
a
)(
d
−
a
)(
c
−
b
)(
d
−
b
)(
d
−
c
).
35.
Since
C
11
=4,
C
12
= 5, and
C
13
=
−
6, we have
a
21
C
11
+
a
22
C
12
+
a
23
C
13
=(
−
1)(4) + 2(5) + 1(
−
6) = 0. Since
C
12
=5,
C
22
=
−
7, and
C
23
=
−
3, we have
a
13
C
12
+
a
23
C
22
+
a
33
C
32
= 2(5) + 1(
−
7) + 1(
−
3) = 0.
36.
Since
C
11
+
−
7,
C
12
=
−
8, and
C
13
=
−
10 we have
a
21
C
11
+
a
22
C
12
+
a
23
C
13
=
−
2(
−
7)+3(
−
8)
−
1(
−
10) = 0.
Since
C
12
=
−
8,
C
22
=
−
19, and
C
32
=
−
7wehave
a
13
C
12
+
a
23
C
22
+
a
33
C
32
=5(
−
8)
−
1(
−
19)
−
3(
−
7) = 0.
37.
det(
A
+
B
)=
)
)
)
)
10 0
0
−
3
)
)
)
)
=
−
30; det
A
+ det
B
=10
−
31 =
−
21
38.
det(2
A
)=2
5
det
A
= 32(
−
7) =
−
224
39.
Factoring
−
1 out of each row we see that det(
−
A
)=(
−
1)
5
det
A
=
−
det
A
. Then
−
det
A
= det(
−
A
)=
det
A
T
= det
A
and det
A
=0.
40. (a)
Cofactors: 25!
≈
1
.
55(10
25
); Row reduction: 25
3
/
3
≈
5
.
2(10
3
)
(b)
Cofactors: about 90 billion centuries; Row reduction: about
1
10
second
328

329. Exercises 8.6
Exercises 8.6
1. AB
=
2
3
−
2
−
1+1
6
−
6
−
2+3
3
=
2
10
01
3
2. AB
=



2
−
1
−
1+1
−
2+2
6
−
6
−
3+4 6
−
6
2+1
−
3
−
1
−
1+2 2+2
−
3



=



100
010
001



3.
det
A
=9.
A
is nonsingular.
A
−
1
=
1
9
2
11
−
45
3
=

1
9
1
9
−
4
9
5
9

4.
det
A
=5.
A
is nonsingular.
A
−
1
=
1
5
2
31
−
4
1
3
3
=

3
5
1
5
−
4
5
1
15

5.
det
A
= 12.
A
is nonsingular.
A
−
1
=
1
12
2
20
36
3
=

1
6
0
1
4
1
2

6.
det
A
=
−
3
π
2
.
A
is nonsingular.
A
−
1
=
−
1
3
π
2
2
ππ
π
−
2
π
3
=

−
1
3
π
−
1
3
π
−
1
3
π
2
3
π

7.
det
A
=
−
16.
A
is nonsingular.
A
−
1
=
−
1
16



8
−
8
−
8
2
−
46
−
64
−
2



=



−
1
2
1
2
1
2
−
1
8
1
4
−
3
8
3
8
−
1
4
1
8



8.
det
A
=0.
A
is singular.
9.
det
A
=
−
30.
A
is nonsingular.
A
−
1
=
−
1
30



−
14 13 16
−
24
−
2
−
4
−
7
−
4



=



7
15
−
13
30
−
8
15
1
15
−
2
15
1
15
2
15
7
30
2
15



10.
det
A
= 78.
A
is nonsingular.
A
−
1
=
1
78



820 2
−
2
−
519
12
−
93






4
39
10
39
1
39
−
1
39
−
5
78
19
78
2
13
−
3
26
1
26



11.
det
A
=
−
36.
A
is nonsingular.
A
−
1
=
−
1
36



−
12 0 0
0
−
60
0018



=



1
3
00
0
1
6
0
00
−
1
2



12.
det
A
= 16.
A
is nonsingular.
A
−
1
=
1
16



002
800
0160



=



00
1
8
1
2
00
010



13.
det
A
= 27.
A
is nonsingular.
A
−
1
=
1
27





621
−
9
−
36
−
116
−
3
10 17
−
6
−
51
4
−
4312





=





2
9
7
9
−
1
3
−
4
3
−
1
27
1
27
2
9
−
1
9
10
27
17
27
−
2
9
−
17
9
4
27
−
4
27
1
9
4
9





14.
det
A
=
−
6.
A
is nonsingular.
A
−
1
=
−
1
6





01
−
33
013
−
9
0
−
200
−
6
−
1
−
315





=





0
−
1
6
1
2
−
1
2
0
−
1
6
−
1
2
3
2
0
1
3
00
1
1
6
1
2
−
5
2





329

330. Exercises 8.6
15.
2
6
−
2
10
04
01
3
1
6
R
1
−
−
−
−
−
−→
1
4
R
2
2
1
−
1
3
1
6
0
01
0
1
4
3
1
3
R
2
+
R
1
−
−
−
−
−
−→
2
10
1
6
1
12
01
0
1
4
3
;
A
−
1
=

1
6
1
12
0
1
4

16.
2
80
10
0
1
2
01
3
1
8
R
1
−
−
−
−
−
−→
2
R
2
2
10
1
8
0
01
02
3
;
A
−
1
=
2
1
8
0
02
3
17.
2
13
10
53
01
3
−
5
R
1
+
R
2
−
−
−
−
−
−→
2
13
10
0
−
12
−
51
3
−
1
12
R
2
−
−
−
−
−
−→
2
13
10
01
5
12
−
1
12
3
−
3
R
2
+
R
1
−
−
−
−
−
−→

10
−
1
4
1
4
01
5
12
−
1
12

;
A
−
1
=

−
1
4
1
4
5
12
−
1
12

18.
2
2
−
3
10
−
24
01
3
1
2
R
1
−
−
−
−
−
−→
2
1
−
3
2
1
2
0
−
24
01
3
2
R
1
+
R
2
−
−
−
−
−
−→
2
1
−
3
2
1
2
0
01
11
3
3
2
R
2
+
R
1
−
−
−
−
−
−→
2
10
2
3
2
01
11
3
;
A
−
1
=
2
2
3
2
11
3
19.



123
100
456
010
789
001



row
−
−
−
−
−
−→
operations



123
100
012
4
3
−
1
3
0
000
1
−
21



;
A
is singular.
20.



10
−
1
100
0
−
21
010
2
−
13
001



row
−
−
−
−
−
−→
operations



100
5
9
−
1
9
2
9
010
−
2
9
−
5
9
1
9
001
−
4
9
−
1
9
2
9



;
A
−
1
=



5
9
−
1
9
2
9
−
2
9
−
5
9
1
9
−
4
9
−
1
9
2
9



21.



423
100
210
010
−
1
−
20
001



R
13
−
−
−
−
−
−→



−
1
−
20
001
210
010
423
100



row
−
−
−
−
−
−→
operations



100
0
2
3
1
3
010
0
−
1
3
−
2
3
001
1
3
−
2
3
0



;
A
−
1
=



0
2
3
1
3
0
−
1
3
−
2
3
1
3
−
2
3
0



22.



24
−
2
100
42
−
2
010
810
−
6
001



row
−
−
−
−
−
−→
operations



12
−
1
1
2
00
01
−
1
3
1
3
−
1
6
0
00 0
−
2
−
11



;
A
is singular.
23.



−
130
100
3
−
21
010
012
001



row
−
−
−
−
−
−→
operations



1
−
30
−
100
011
110
001
−
1
−
11



row
−
−
−
−
−
−→
operations



100
56
−
3
010
22
−
1
001
−
1
−
11



;
A
−
1
=



56
−
3
22
−
1
−
1
−
11



24.



123
100
014
010
008
001



row
−
−
−
−
−
−→
operations



100
1
−
2
5
8
010
01
−
1
2
001
00
1
8



;
A
−
1
=



1
−
2
5
8
01
−
1
2
00
1
8



330

331. Exercises 8.6
25.





12 31
1000
−
10 21
0100
21
−
30
0010
11 21
0001





row
−
−
−
−
−
−→
operations





123 1
10 00
01
5
2
1
1
2
1
2
00
001
−
2
3
1
3
−
1
−
2
3
0
000 1
−
1
2
1
1
2
1
2





row
−
−
−
−
−
−→
operations






1000
−
1
2
−
2
3
−
1
6
7
6
0100
1
1
3
1
3
−
4
3
0010
0
−
1
3
−
1
3
1
3
0001
−
1
2
1
1
2
1
2






;
A
−
1
=






−
1
2
−
2
3
−
1
6
7
6
1
1
3
1
3
−
4
3
0
−
1
3
−
1
3
1
3
−
1
2
1
1
2
1
2






26.





1000
1000
0010
0100
0001
0010
0100
0001





row
−
−
−
−
−
−→
interchange





1000
1000
0100
0001
0010
0100
0001
0010





;
A
−
1
=





1000
0001
0100
0010





27.
(
AB
)
−
1
=
B
−
1
A
−
1
=

−
1
3
1
3
−
1
10
3

28.
(
AB
)
−
1
=
B
−
1
A
−
1
=



−
1
−
420
26
−
30
36
−
32



29. A
=(
A
−
1
)
−
1
=
2
−
23
3
−
4
3
30. A
T
=
2
12
410
3
;(
A
T
)
−
1
=
2
5
−
1
−
2
1
2
3
;
A
−
1
=
2
5
−
2
−
1
1
2
3
;(
A
−
1
)
T
=
2
5
−
1
−
2
1
2
3
31.
Multiplying
2
4
−
3
x
−
4
32
4
−
3
x
−
4
3
=
2
16
−
3
x
0
016
−
3
x
3
we see that
x
=5.
32. A
−
1
=
2
sin
θ
−
cos
θ
cos
θ
sin
θ
3
33. (a) A
T
=
2
sin
θ
−
cos
θ
cos
θ
sin
θ
3
=
A
−
1
(b) A
T
=



1
√
3
1
√
3
1
√
3
0
1
√
2
−
1
√
2
−
2
√
6
1
√
6
1
√
6



=
A
−
1
34.
Since det
A
·
det
A
−
1
= det
AA
−
1
= det
I
= 1, we see that det
A
−
1
=1
/
det
A
.If
A
is orthogonal, det
A
=
det
A
T
= det
A
−
1
=1
/
det
A
and (det
A
)
2
= 1, so det
A
=
±
1.
35.
Since
A
and
B
are nonsingular, det
AB
= det
A
·
det
B
3
= 0, and
AB
is nonsingular.
36.
Suppose
A
is singular. Then det
A
= 0, det
AB
= det
A
·
det
B
= 0, and
AB
is singular.
37.
Since det
A
·
det
A
−
1
= det
AA
−
1
= det
I
= 1, det
A
−
1
=1
/
det
A
.
38.
Suppose
A
2
=
A
and
A
is nonsingular. Then
A
2
A
−
1
=
AA
−
1
, and
A
=
I
. Thus, if
A
2
=
A
, either
A
is
singular or
A
=
I
.
39.
If
A
is nonsingular, then
A
−
1
exists, and
AB
=
0
implies
A
−
1
AB
=
A
−
1
0
,so
B
=
0
.
40.
If
A
is nonsingular,
A
−
1
exists, and
AB
=
AC
implies
A
−
1
AB
=
A
−
1
AC
,so
B
=
C
.
331

332. Exercises 8.6
41.
No, consider
A
=
2
10
00
3
and
B
=
2
00
01
3
.
42. A
is nonsingular if
a
11
a
22
a
33
=0or
a
11
,
a
22
, and
a
33
are all nonzero.
A
−
1
=



1
/a
11
00
01
/a
22
0
001
/a
33



For any diagonal matrix, the inverse matrix is obtaining by taking the reciprocals of the diagonal entries and
leaving all other entries 0.
43. A
−
1
=

1
3
1
3
2
3
−
1
3

;
A
−
1
2
4
14
3
=
2
6
−
2
3
;
x
1
=6,
x
2
=
−
2
44. A
−
1
=

2
3
1
6
−
1
3
1
6

;
A
−
1
2
2
−
5
3
=
2
1
2
−
3
2
3
;
x
1
=
1
2
,
x
2
=
−
3
2
45. A
−
1
=

1
16
3
8
−
1
8
1
4

;
A
−
1
2
6
1
3
=

3
4
−
1
2

;
x
1
=
3
4
,
x
2
=
−
1
2
46. A
−
1
=
2
−
21
3
2
−
1
2
3
;
A
−
1
2
4
−
3
3
=
2
−
11
15
2
3
;
x
1
=
−
11,
x
2
=
15
2
47. A
−
1
=



−
1
5
1
5
1
5
−
110
6
5
−
1
5
−
1
5



;
A
−
1



−
4
0
6



=



2
4
−
6



;
x
1
=2,
x
2
=4,
x
3
=
−
6
48. A
−
1
=



5
12
−
1
12
1
4
−
2
3
1
3
0
−
1
12
5
12
−
1
4



;
A
−
1



1
2
−
3



=



−
1
2
0
3
2



;
x
1
=
−
1
2
,
x
2
=0,
x
3
=
3
2
49. A
−
1
=



−
2
−
32
1
4
−
1
4
0
5
4
7
4
−
1



;
A
−
1



1
−
3
7



=



21
1
−
11



;
x
1
= 21,
x
2
=1,
x
3
=
−
11
50. A
−
1
=





2
−
111
−
12
−
1
−
1
1
−
111
1
−
110





;
A
−
1





2
1
−
5
3





=





1
2
−
1
−
4





;
x
1
=1,
x
2
=2,
x
3
=
−
1,
x
4
=
−
4
51.
2
7
−
2
32
32
x
1
x
2
3
=
2
b
1
b
2
3
;
A
−
1
=

1
10
1
10
−
3
20
7
20

;
X
=
A
−
1
2
5
4
3
=

9
10
13
20

;
X
=
A
−
1
2
10
50
3
=
2
6
16
3
;
X
=
A
−
1
2
0
−
20
3
=
2
−
2
−
7
3
52.



125
238
−
112






x
1
x
2
x
3



=



b
1
b
2
b
3



;
A
−
1
=



2
−
1
−
1
12
−
7
−
2
−
531



;
X
=
A
−
1



−
1
4
6



=



−
12
−
52
23



;
X
=
A
−
1



3
3
3



=



0
9
−
3



;=
A
−
1



0
−
5
4



=



1
27
−
11



332

333. Exercises 8.7
53.
det
A
=18
3
= 0, so the system has only the trivial solution.
54.
det
A
= 0, so the system has a nontrivial solution.
55.
det
A
= 0, so the system has a nontrivial solution.
56.
det
A
=12
3
= 0, so the system has only the trivial solution.
57. (a)



111
−
R
1
R
2
0
0
−
R
2
R
3






i
1
i
2
i
3



=



0
E
2
−
E
1
E
3
−
E
2



(b)
det
A
=
R
1
R
2
+
R
1
R
3
+
R
2
R
3
>
0, so
A
is nonsingular.
(c) A
−
1
=
1
R
1
R
2
+
R
1
R
3
+
R
2
R
3



R
2
R
3
−
R
2
−
R
3
−
R
2
R
1
R
3
R
3
−
R
1
R
1
R
2
R
2
R
1
+
R
2



;
A
−
1



0
E
2
−
E
1
E
3
−
E
2



=
1
R
1
R
2
+
R
1
R
3
+
R
2
R
3



R
2
E
1
−
R
2
E
3
+
R
3
E
1
−
R
3
E
2
R
1
E
2
−
R
1
E
3
−
R
3
E
1
+
R
3
E
2
−
R
1
E
2
+
R
1
E
3
−
R
2
E
1
+
R
2
E
3



58. (a)
We write the equations in the form
−
4
u
1
+
u
2
+
u
4
=
−
200
u
1
−
4
u
2
+
u
3
=
−
300
u
2
−
4
u
3
+
u
4
=
−
300
u
1
+
u
3
−
4
u
4
=
−
200
.
In matrix form this becomes





−
4101
1
−
410
01
−
41
101
−
4










u
1
u
2
u
3
u
4





=





−
200
−
300
−
300
−
200





.
(b) A
−
1
=






−
7
24
−
1
12
−
1
24
−
1
12
−
1
12
−
7
24
−
1
12
−
1
24
−
1
24
−
1
12
−
7
24
−
1
12
−
1
12
−
1
24
−
1
12
−
7
24






;
A
−
1





−
200
−
300
−
300
−
200





=






225
2
275
2
275
2
225
2






;
u
1
=
u
4
=
225
2
,
u
2
=
u
3
=
275
2
Exercises 8.7
1.
det
A
= 10, det
A
1
=
−
6, det
A
2
= 12;
x
1
=
−
6
10
=
−
3
5
,
x
2
=
12
10
=
6
5
2.
det
A
=
−
3, det
A
1
=
−
6, det
A
2
=
−
6;
x
1
=
−
6
−
3
=2,
x
2
=
−
6
−
3
=2
3.
det
A
=0
.
3, det
A
1
=0
.
03, det
A
2
=
−
0
.
09;
x
1
=
0
.
03
0
.
3
=0
.
1,
x
2
=
−
0
.
09
0
.
3
=
−
0
.
3
4.
det
A
=
−
0
.
015, det
A
1
=
−
0
.
00315, det
A
2
=
−
0
.
00855;
x
1
=
−
0
.
00315
−
0
.
015
=0
.
21,
x
2
=
−
0
.
00855
−
0
.
015
=0
.
57
5.
det
A
= 1, det
A
1
= 4, det
A
2
=
−
7;
x
=4,
y
=
−
7
6.
det
A
=
−
70, det
A
1
=
−
14, det
A
2
=35;
r
=
−
14
−
70
=
1
5
,
s
=
35
−
70
=
−
1
2
7.
det
A
= 11, det
A
1
=
−
44, det
A
2
= 44, det
A
3
=
−
55;
x
1
=
−
44
11
=
−
4,
x
2
=
44
11
=4,
x
3
=
−
55
11
=
−
5
333

334. Exercises 8.7
8.
det
A
=
−
63, det
A
1
= 173, det
A
2
=
−
136, det
A
3
=
−
61
2
;
x
1
=
−
173
63
,
x
2
=
136
63
,
x
3
=
61
126
9.
det
A
=
−
12, det
A
1
=
−
48, det
A
2
=
−
18, det
A
3
=
−
12;
u
=
48
12
=4,
v
=
18
12
=
3
2
,
w
=1
10.
det
A
= 1, det
A
1
=
−
2, det
A
2
= 2, det
A
3
=5;
x
=
−
2,
y
=2,
z
=5
11.
det
A
=6
−
5
k
, det
A
1
=12
−
7
k
, det
A
2
=6
−
7
k
;
x
1
=
12
−
7
k
6
−
5
k
,
x
2
=
6
−
7
k
6
−
5
k
. The system is inconsistent
for
k
=6
/
5.
12. (a)
det
A
=
%
−
1, det
A
1
=
%
−
2, det
A
2
=1;
x
1
=
%
−
2
%
−
1
=
%
−
1
−
1
%
−
1
=1
−
1
%
−
1
,
x
2
=
1
%
−
1
(b)
When
%
=1
.
01,
x
1
=
−
99 and
x
2
= 100. When
%
=0
.
99,
x
1
= 101 and
x
2
=
−
100.
13.
det
A
≈
0
.
6428, det
A
1
≈
289
.
8, det
A
2
≈
271
.
9;
x
1
≈
289
.
8
0
.
6428
≈
450
.
8,
x
2
≈
271
.
9
0
.
6428
≈
423
14.
We have (sin 30
◦
)
F
+ (sin 30
◦
)(0
.
5
N
)+
N
sin 60
◦
= 400 and (cos 30
◦
)
F
+ (cos 30
◦
)(0
.
5
N
)
−
N
cos 60
◦
= 0. The
system is
(sin 30
◦
)
F
+(0
.
5sin 30
◦
+ sin 60
◦
)
N
= 400
(cos 30
◦
)
F
+(0
.
5cos 30
◦
−
cos 60
◦
)
N
=0
.
det
A
≈−
1, det
A
1
≈−
26
.
795, det
A
2
≈−
346
.
41;
F
≈
26
.
795,
N
≈
346
.
41
15.
The system is
i
1
+
i
2
−
i
3
=0
r
1
i
1
−
r
2
i
2
=
E
1
−
E
2
r
2
i
2
+
Ri
3
=
E
2
det
A
=
−
r
1
R
−
r
2
R
−
r
1
r
2
, det
A
3
=
−
r
1
E
2
,
−
r
2
E
1
;
i
3
=
r
1
E
2
+
r
2
E
1
r
1
R
+
r
2
R
+
r
1
r
2
Exercises 8.8
1. K
3
since
2
42
51
32
−
2
5
3
=
2
2
−
5
3
=(
−
1)
2
−
2
5
3
;
λ
=
−
1
2. K
1
and
K
2
since
2
2
−
1
2
−
2
32
1
2
−
√
2
3
=
2
√
2
−
2+2
√
2
3
=
√
2
2
1
2
−
√
2
3
,
λ
=
√
2
2
2
−
1
2
−
2
32
2+
√
2
2
3
=
2
2+2
√
2
2
√
2
3
=
√
2
2
2+
√
2
2
3
;
λ
=
√
2
3. K
3
since
2
63
21
32
−
5
10
3
=
2
0
0
3
=0
2
−
5
10
3
;
λ
=0
4. K
2
since
2
28
−
1
−
2
32
2+2
i
−
1
3
=
2
−
4+4
i
−
2
i
3
=2
i
2
2+2
i
−
1
3
;
λ
=2
i
5. K
2
and
K
3
since



1
−
22
−
21
−
2
221






4
−
4
0



=



12
−
12
0



=3



4
−
4
0



;
λ
=3



1
−
22
−
21
−
2
221






−
1
1
1



=



−
1
1
1



;
λ
=1
334

335. Exercises 8.8
6. K
2
since



−
11 0
12 1
03
−
1






1
4
3



=



3
12
9



=3



1
4
3



;
λ
=3
7.
We solve det(
A
−
λ
I
)=
)
)
)
)
−
1
−
λ
2
−
78
−
λ
)
)
)
)
=(
λ
−
6)(
λ
−
1) = 0.
For
λ
1
= 6 we have
2
−
72
0
−
72
0
3
=
⇒
2
1
−
2
/
7
0
00
0
3
so that
k
1
=
2
7
k
2
.If
k
2
= 7 then
K
1
=
2
2
7
3
.For
λ
2
= 1 we have
2
−
22
0
−
77
0
3
=
⇒
2
1
−
1
0
00
0
3
so that
k
1
=
k
2
.If
k
2
= 1 then
K
2
=
2
1
1
3
.
8.
We solve det(
A
−
λ
I
)=
)
)
)
)
2
−
λ
1
21
−
λ
)
)
)
)
=
λ
(
λ
−
3) = 0.
For
λ
1
= 0 we have
2
21
0
21
0
3
=
⇒
2
11
/
2
0
00
0
3
so that
k
1
=
−
1
2
k
2
.If
k
2
= 2 then
K
1
=
2
−
1
2
3
.For
λ
2
=3wehave
2
−
11
0
2
−
2
0
3
=
⇒
2
1
−
1
0
00
0
3
so that
k
1
=
k
2
.If
k
2
= 1 then
K
2
=
2
1
1
3
.
9.
We solve det(
A
−
λ
I
)=
)
)
)
)
−
8
−
λ
−
1
16
−
λ
)
)
)
)
=(
λ
+4)
2
=0.
For
λ
1
=
λ
2
=
−
4wehave
2
−
4
−
1
0
16 4
0
3
=
⇒
2
11
/
4
0
00
0
3
so that
k
1
=
−
1
4
k
2
.If
k
2
= 4 then
K
1
=
2
−
1
4
3
.
10.
We solve det(
A
−
λ
I
)=
)
)
)
)
1
−
λ
1
1
/
41
−
λ
)
)
)
)
=(
λ
−
3
/
2)(
λ
−
1
/
2) = 0.
For
λ
1
=3
/
2wehave
2
−
1
/
21
0
1
/
4
−
1
/
2
0
3
=
⇒
2
1
−
2
0
00
0
3
so that
k
1
=2
k
2
.If
k
2
= 1 then
K
1
=
2
2
1
3
.If
λ
2
=1
/
2 then
2
1
/
21
0
1
/
41
/
2
0
3
=
⇒
2
12
0
00
0
3
so that
k
1
=
−
2
k
2
.If
k
2
= 1 then
K
2
=
2
−
2
1
3
.
335

336. Exercises 8.8
11.
We solve det(
A
−
λ
I
)=
)
)
)
)
−
1
−
λ
2
−
51
−
λ
)
)
)
)
=
λ
2
+9=(
λ
−
3
i
)(
λ
+3
i
)=0.
For
λ
1
=3
i
we have
2
−
1
−
3
i
2
0
−
51
−
3
i
0
3
=
⇒
2
1
−
(1
/
5) + (3
/
5)
i
0
00
0
3
so that
k
1
=

1
5
−
3
5
i

k
2
.If
k
2
= 5then
K
1
=
2
1
−
3
i
5
3
.For
λ
2
=
−
3
i
we have
2
−
1+3
i
2
0
−
51+3
i
0
3
=
⇒
2
1
−
1
5
−
3
5
i
0
00
0
3
so that
k
1
=

1
5
+
3
5
i

k
2
.If
k
2
= 5then
K
2
=
2
1+3
i
5
3
.
12.
We solve det(
A
−
λ
I
)=
)
)
)
)
1
−
λ
−
1
11
−
λ
)
)
)
)
=
λ
2
−
2
λ
+2=0.
For
λ
1
=1
−
i
we have
2
i
−
1
0
1
i
0
3
=
⇒
2
i
−
1
0
00
0
3
so that
k
1
=
−
ik
2
.If
k
2
= 1 then
K
1
=
2
−
i
1
3
and
K
2
=
K
1
=
2
i
1
3
.
13.
We solve det(
A
−
λ
I
)=
)
)
)
)
4
−
λ
8
0
−
5
−
λ
)
)
)
)
=(
λ
−
4)(
λ
+5)=0.
For
λ
1
= 4 we have
2
08
0
0
−
9
0
3
=
⇒
2
01
0
00
0
3
so that
k
2
=0. If
k
1
= 1 then
K
1
=
2
1
0
3
.For
λ
2
=
−
5we have
2
98
0
00
0
3
=
⇒
2
1
8
9
0
00
0
3
so that
k
1
=
−
8
9
k
2
.If
k
2
= 9 then
K
2
=
2
−
8
9
3
.
14.
We solve det(
A
−
λ
I
)=
)
)
)
)
7
−
λ
0
013
−
λ
)
)
)
)
=(
λ
−
7)(
λ
−
13) = 0.
For
λ
1
= 7 we have
2
00
0
06
0
3
=
⇒
2
01
0
00
0
3
so that
k
2
=0. If
k
1
= 1 then
K
1
=
2
1
0
3
.For
λ
2
=13wehave
2
−
60
0
00
0
3
=
⇒
2
10
0
00
0
3
so that
k
1
=0. If
k
2
= 1 then
K
2
=
2
0
1
3
.
15.
We solve det(
A
−
λ
I
)=
)
)
)
)
)
)
)
5
−
λ
−
10
0
−
5
−
λ
9
5
−
1
−
λ
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
4
−
λ
−
10
4
−
λ
−
5
−
λ
9
4
−
λ
−
1
−
λ
)
)
)
)
)
)
)
=
λ
(4
−
λ
)(
λ
+4)=0.
For
λ
1
= 0 we have



5
−
10
0
0
−
59
0
5
−
10
0



=
⇒



10
−
9
/
25
0
01
−
9
/
5
0
00 0
0



336

337. Exercises 8.8
so that
k
1
=
9
25
k
3
and
k
2
=
9
5
k
3
.If
k
3
= 25then
K
1
=



9
45
25



.If
λ
2
= 4 then



1
−
10
0
0
−
99
0
5
−
1
−
4
0



=
⇒



10
−
1
0
01
−
1
0
00 0
0



so that
k
1
=
k
3
and
k
2
=
k
3
.If
k
3
= 1 then
K
2
=



1
1
1



.If
λ
3
=
−
4 then



9
−
10
0
0
−
19
0
5
−
14
0



=
⇒



10
−
1
0
01
−
9
0
00 0
0



so that
k
1
=
k
3
and
k
2
=9
k
3
.If
k
3
= 1 then
K
3
=



1
9
1



.
16.
We solve det(
A
−
λ
I
)=
)
)
)
)
)
)
)
3
−
λ
00
02
−
λ
0
401
−
λ
)
)
)
)
)
)
)
=(3
−
λ
)(2
−
λ
)(1
−
λ
)=0.
For
λ
1
= 1 we have



200
0
010
0
400
0



=
⇒



100
0
010
0
000
0



so that
k
1
= 0 and
k
2
=0. If
k
3
= 1 then
K
1
=



0
0
1



.If
λ
2
= 2 then



10 0
0
00 0
0
40
−
1
0



=
⇒



100
0
001
0
000
0



so that
k
1
= 0 and
k
3
=0. If
k
2
= 1 then
K
2
=



0
1
0



.If
λ
3
= 3 then



000
0
0
−
10
0
40
−
2
0



=
⇒



10
−
1
/
2
0
01 0
0
00 0
0



so that
k
1
=
1
2
k
3
and
k
2
=0. If
k
3
= 2 then
K
3
=



1
0
2



.
337

338. Exercises 8.8
17.
We solve det(
A
−
λ
I
)=
)
)
)
)
)
)
)
−
λ
40
−
1
−
4
−
λ
0
00
−
2
−
λ
)
)
)
)
)
)
)
=
−
(
λ
+2)
3
=0.
For
λ
1
=
λ
2
=
λ
3
=
−
2wehave



240
0
−
1
−
20
0
000
0



=
⇒



120
0
000
0
000
0



so that
k
1
=
−
2
k
2
.If
k
2
= 1 and
k
3
= 1 then
K
1
=



−
2
1
0



and
K
2
=



0
0
1



.
18.
We solve det(
A
−
λ
I
)=
)
)
)
)
)
)
)
1
−
λ
60
02
−
λ
1
012
−
λ
)
)
)
)
)
)
)
=
)
)
)
)
)
)
)
1
−
λ
60
03
−
λ
3
−
λ
012
−
λ
)
)
)
)
)
)
)
=(3
−
λ
)(1
−
λ
)
2
=0.
For
λ
1
= 3 we have



−
26 0
0
00 0
0
01
−
1
0



=
⇒



10
−
3
0
01
−
1
0
00 0
0



so that
k
1
=3
k
3
and
k
2
=
k
3
.If
k
3
= 1 then
K
1
=



3
1
1



.For
λ
2
=
λ
3
= 1 we have



060
0
011
0
011
0



=
⇒



010
0
001
0
000
0



so that
k
2
= 0 and
k
3
=0. If
k
1
= 1 then
K
2
=



1
0
0



.
19.
We solve det(
A
−
λ
I
)=
)
)
)
)
)
)
)
−
λ
0
−
1
0
−
λ
0
11
−
1
−
λ
)
)
)
)
)
)
)
=
−
(
λ
+ 1)(
λ
2
+1)=0.
For
λ
1
=
−
1wehave



10
−
1
0
11 0
0
11 0
0



=
⇒



10
−
1
0
01 1
0
00 0
0



so that
k
1
=
k
3
and
k
2
=
−
k
3
.If
k
3
= 1 then
K
1
=



1
−
1
1



.For
λ
2
=
i
we have



−
i
0
−
1
0
1
−
i
0
0
11
−
1
−
i
0



=
⇒



10
−
i
0
01
−
1
0
00 0
0



so that
k
1
=
ik
3
and
k
2
=
k
3
.If
k
3
= 1 then
K
2
=



i
1
1



and
K
3
=
K
2
=



−
i
1
1



.
338

339. Exercises 8.8
20.
We solve
det(
A
−
λ
I
)=
)
)
)
)
)
)
)
2
−
λ
−
10
52
−
λ
4
012
−
λ
)
)
)
)
)
)
)
=
−
λ
3
+6
λ
2
−
13
λ
+10=(
λ
−
2)(
−
λ
2
+4
λ
−
5)
=(
λ
−
2)(
λ
−
(2 +
i
))(
λ
−
(2
−
i
)) = 0
.
For
λ
1
= 2 we have



0
−
10
0
504
0
010
0



=
⇒



104
/
5
0
01 0
0
00 0
0



so that
k
1
=
−
4
5
k
3
and
k
2
=0. If
k
3
= 5then
K
1
=



−
4
0
5



.For
λ
2
=2+
i
we have



−
i
−
10
0
5
−
i
4
0
01
−
i
0



=
⇒



1
−
i
0
0
01
−
i
0
000
0



so that
k
1
=
ik
2
and
k
2
=
ik
3
.If
k
3
=
i
then
K
2
=



−
i
−
1
i



.For
λ
3
=2
−
i
we have



i
−
10
0
5
i
4
0
01
i
0



=
⇒



1
i
0
0
01
i
0
000
0



so that
k
1
=
−
ik
2
and
k
2
=
−
ik
3
.If
k
3
=
i
then
K
3
=



−
1
1
i



.
21.
We solve det(
A
−
λ
I
)=
)
)
)
)
)
)
)
1
−
λ
23
05
−
λ
6
00
−
7
−
λ
)
)
)
)
)
)
)
=
−
(
λ
−
1)(
λ
−
5)(
λ
+7)=0.
For
λ
1
= 1 we have



02 3
0
04 6
0
00
−
6
0



=
⇒



010
0
001
0
000
0



so that
k
2
=
k
3
=0. If
k
1
= 1 then
K
1
=



1
0
0



.For
λ
2
= 5we have



−
42 3
0
00 6
0
00
−
12
0



=
⇒



1
−
1
2
0
0
001
0
000
0



339

340. Exercises 8.8
so that
k
3
= 0 and
k
2
=2
k
1
.If
k
1
= 1 then
K
2
=



1
2
0



.For
λ
3
=
−
7wehave



823
0
0126
0
000
0



=
⇒



10
1
4
0
01
1
2
0
000
0



so that
k
1
=
−
1
4
k
3
and
k
2
=
−
1
2
k
3
.If
k
3
= 4 then
K
3
=



−
1
−
2
4



.
22.
We solve det(
A
−
λ
I
)=
)
)
)
)
)
)
)
−
λ
00
0
−
λ
0
001
−
λ
)
)
)
)
)
)
)
=
−
λ
2
(
λ
−
1) = 0.
For
λ
1
=
λ
2
=0wehave



000
0
000
0
001
0



=
⇒



001
0
000
0
000
0



so that
k
3
=0. If
k
1
= 1 and
k
2
= 0 then
K
1
=



1
0
0



and if
k
1
= 0 and
k
2
= 1 then
K
2
=



0
1
0



.For
λ
3
=1we
have



−
100
0
0
−
10
0
000
0



=
⇒



100
0
010
0
000
0



so that
k
1
=
k
2
=0. If
k
3
= 1 then
K
3
=



0
0
1



.
23.
The eigenvalues and eigenvectors of
A
=
2
51
15
3
are
λ
1
=4
,λ
2
=6
,
K
1
=
2
1
−
1
3
,
K
2
=
2
1
1
3
and the eigenvalues and eigenvectors of
A
−
1
=
1
24
2
5
−
1
−
15
3
are
λ
1
=
1
4
,λ
2
=
1
6
,
K
1
=
2
1
−
1
3
,
K
2
=
2
1
1
3
.
24.
The eigenvalues and eigenvectors of
A
=



12
−
1
101
4
−
45



are
λ
1
=1
,λ
2
=2
,λ
3
=3
,
K
1
=



−
1
1
2



,
K
2
=



−
2
1
4



,
K
3
=



−
1
1
4



.
340

341. Exercises 8.8
and the eigenvalues and eigenvectors of
A
−
1
=
1
6



4
−
62
−
19
−
2
−
412
−
2



are
λ
1
=1
,λ
2
=
1
2
,λ
3
=
1
3
,
K
1
=



−
1
1
2



,
K
2
=



−
2
1
4



,
K
3
=



−
1
1
4



.
25.
Since det
A
=
)
)
)
)
60
30
)
)
)
)
= 0 the matrix is singular. Now from
det(
A
−
λ
I
)=
)
)
)
)
6
−
λ
0
3
−
λ
)
)
)
)
=
λ
(
λ
−
6)
we see
λ
= 0 is an eigenvalue.
26.
Since det
A
=
)
)
)
)
)
)
)
101
4
−
45
7
−
48
)
)
)
)
)
)
)
= 0 the matrix is singular. Now from
det(
A
−
λ
I
)=
)
)
)
)
)
)
)
1
−
λ
01
4
−
4
−
λ
5
7
−
48
−
λ
)
)
)
)
)
)
)
=
−
λ
(
λ
2
−
5
λ
−
15)
we see
λ
= 0 is an eigenvalue.
27. (a)
Since
p
+1
−
p
= 1 and
q
+1
−
q
= 1, the first matrix
A
is stochastic. Since
1
2
+
1
4
+
1
4
=1,
1
3
+
1
3
+
1
3
=1,
and
1
6
+
1
3
+
1
2
= 1, the second matrix
A
is stochastic.
(b)
The matrix from part (a) is shown with its eigenvalues and corresponding eigenvectors.



1
2
1
4
1
4
1
3
1
3
1
3
1
6
1
3
1
2



; eigenvalues: 1,
1
6
−
1
12
√
2,
1
6
+
1
12
√
2;
eigenvectors: (1
,
1
,
1),

−
3(
−
1+
√
2)
−
6+
√
2
,
2(2+
√
2)
−
6+
√
2
,
1

,

−
3(1+
√
2)
6+
√
2
,
2(
−
2+
√
2)
6+
√
2
,
1

Further examples indicate that 1 is always an eigenvalue with corresponding eigenvector (1
,
1
,
1). To prove
this, let
A
be a stochastic matrix and
K
=(1
,
1
,
1). Then
AK
=



a
11
···
a
1
n
.
.
.
.
.
.
a
n
1
···
a
nn






1
.
.
.
1



=






a
11
+
···
+
a
1
n
.
.
.
a
n
1
+
···
+
a
nn






=



1
.
.
.
1



=1
K
,
and 1 is an eigenvalue of
A
with corresponding eigenvector (1
,
1
,
1).
(c)
For the 3
×
3 matrix in part (a) we have
A
2
=



3
8
7
24
1
3
1
3
11
36
13
36
5
18
23
72
29
72



,
A
3
=



49
144
29
96
103
288
71
216
11
36
79
216
5
16
67
216
163
432



.
These powers of
A
are also stochastic matrices. To prove that this is true in general for 2
×
2 matrices, we
prove the more general theorem that any product of 2
×
2 stochastic matrices is stochastic. Let
A
=
2
a
11
a
12
a
21
a
22
3
and
B
=
2
b
11
b
12
b
21
b
22
3
341

342. Exercises 8.8
be stochastic matrices. Then
AB
=
2
a
11
b
11
+
a
12
b
21
a
11
b
12
+
a
12
b
22
a
21
b
11
+
a
22
b
21
a
21
b
12
+
a
22
b
22
3
.
The sums of the rows are
a
11
b
11
+
a
12
b
21
+
a
11
b
12
+
a
12
b
22
=
a
11
(
b
11
+
b
12
)+
a
12
(
b
21
+
b
22
)
=
a
11
(1) +
a
12
(1) =
a
11
+
a
12
=1
a
21
b
11
+
a
22
b
21
+
a
21
b
12
+
a
22
b
22
=
a
21
(
b
11
+
b
12
)+
a
22
(
b
21
+
b
22
)
=
a
21
(1) +
a
22
(1) =
a
21
+
a
22
=1
.
Thus, the product matrix
AB
is stochastic. It follows that any power of a 2
×
2 matrix is stochastic. The
proof in the case of an
n
×
n
matrix is very similar.
Exercises 8.9
1.
The characteristic equation is
λ
2
−
6
λ
+ 13 = 0. Then
A
2
−
6
A
+13
I
=
2
−
7
−
12
24 17
3
−
2
6
−
12
24 30
3
+
2
13 0
013
3
=
2
00
00
3
.
2.
The characteristic equation is
−
λ
3
+
λ
2
+4
λ
−
1. Then
−
A
3
+
A
2
+
A
−
I
=
−



2613
4517
159



+



125
045
114



+4



012
103
011



+



100
010
001



=



000
000
000



.
3.
The characteristic equation is
λ
2
−
3
λ
−
10 = 0, with eigenvalues
−
2 and 5. Substituting the eigenvalues into
λ
m
=
c
0
+
c
1
λ
generates
(
−
2)
m
=
c
0
−
2
c
1
5
m
=
c
0
+5
c
1
.
Solving the system gives
c
0
=
1
7
[5(
−
2)
m
+ 2(5)
m
]
,c
1
=
1
7
[
−
(
−
2)
m
+5
m
]
.
Thus
A
m
=
c
0
I
+
c
1
A
=

1
7
[3(
−
1)
m
2
m
+1
+5
m
]
3
7
[
−
(
−
2)
m
+5
m
]
2
7
[
−
(
−
2)
m
+5
m
]
1
7
[(
−
2)
m
+ 6(5)
m
]

and
A
3
=
2
11 57
38 106
3
.
4.
The characteristic equation is
λ
2
−
10
λ
+ 16 = 0, with eigenvalues 2 and 8. Substituting the eigenvalues into
λ
m
=
c
0
+
c
1
λ
generates
2
m
=
c
0
+2
c
1
8
m
=
c
0
+8
c
1
.
Solving the system gives
c
0
=
1
3
(2
m
+2
−
8
m
)
,c
1
=
1
6
(
−
2
m
+8
m
)
.
342

343. Exercises 8.9
Thus
A
m
=
c
0
I
+
c
1
A
=

1
2
(2
m
+8
m
)
1
2
(2
m
−
8
m
)
1
2
(2
m
−
8
m
)
1
2
(2
m
+8
m
)

and
A
4
=
2
2056
−
2040
−
2040 2056
3
.
5.
The characteristic equation is
λ
2
−
8
λ
−
20 = 0, with eigenvalues
−
2 and 10. Substituting the eigenvalues into
λ
m
=
c
0
+
c
1
λ
generates
(
−
2)
m
=
c
0
−
2
c
1
10
m
=
c
0
+10
c
1
.
Solving the system gives
c
0
=
1
6
[5(
−
2)
m
+10
m
]
,c
1
=
1
12
[
−
(
−
2)
m
+10
m
]
.
Thus
A
m
=
c
0
I
+
c
1
A
=

1
6
[(
−
2)
m
+2
m
5
m
+1
]
5
12
[
−
(
−
2)
m
+10
m
]
1
3
[
−
(
−
2)
m
+10
m
]
1
6
[5(
−
2)
m
+10
m
]

and
A
5
=
2
83328 41680
33344 16640
3
.
6.
The characteristic equation is
λ
2
+4
λ
+ 3 = 0, with eigenvalues
−
3 and
−
1. Substituting the eigenvalues into
λ
m
=
c
0
+
c
1
λ
generates
(
−
3)
m
=
c
0
−
3
c
1
(
−
1)
m
=
c
0
−
c
1
.
Solving the system gives
c
0
=
1
2
[
−
(
−
3)
m
+3(
−
1)
m
]
,c
1
=
1
2
[
−
(
−
3)
m
+(
−
1)
m
]
.
Thus
A
m
=
c
0
I
+
c
1
A
=

(
−
1)
m
−
(
−
3)
m
+(
−
1)
m
0(
−
3)
m

and
A
6
=
2
1
−
728
0 729
3
.
7.
The characteristic equation is
−
λ
3
+2
λ
2
+
λ
−
2 = 0, with eigenvalues
−
1, 1, and 2. Substituting the eigenvalues
into
λ
m
=
c
0
+
c
1
λ
+
c
2
λ
2
generates
(
−
1)
m
=
c
0
−
c
1
+
c
2
1=
c
0
+
c
1
+
c
2
2
m
=
c
0
+2
c
1
+4
c
2
.
343

344. Exercises 8.9
Solving the system gives
c
0
=
1
3
[3+(
−
1)
m
−
2
m
]
,
c
1
=
1
2
[1
−
(
−
1)
m
]
,
c
2
=
1
6
[
−
3+(
−
1)
m
+2
m
+1
]
.
Thus
A
m
=
c
0
I
+
c
1
A
+
c
2
A
2
=



1
−
1+2
m
−
1+2
m
0
1
3
[(
−
1)
m
+2
m
+1
]
−
2
3
[(
−
1)
m
−
2
m
]
0
1
3
[
−
(
−
1)
m
+2
m
]
1
3
[2(
−
1)
m
+2
m
]



and
A
10
=



1 1023 1023
0 683 682
0 341 342



.
8.
The characteristic equation is
−
λ
3
−
λ
2
+2
λ
+ 2 = 0, with eigenvalues
−
1,
−
√
2 , and
√
2 . Substituting the
eigenvalues into
λ
m
=
c
0
+
c
1
λ
+
c
2
λ
2
generates
(
−
1)
m
=
c
0
−
c
1
+
c
2
(
−
√
2)
m
=
c
0
−
√
2
c
1
+2
c
2
(
√
2)
m
=
c
0
+
√
2
c
1
+2
c
2
.
Solving the system gives
c
0
=[2
−
(
√
2)
m
−
1
−
(
√
2)
m
−
2
](
−
1)
m
+(
√
2
−
1)(
√
2)
m
−
2
,
c
1
=
1
2
[1
−
(
−
1)
m
](
√
2)
m
−
1
,
c
2
=(
−
1)
m
+1
+
1
2
(1 +
√
2)(
−
1)
m
(
√
2)
m
−
1
+
1
2
(
√
2
−
1)(
√
2)
m
−
1
.
Thus
A
m
=
c
0
I
+
c
1
A
+
c
2
A
2
and
A
6
=



10 7
78
−
7
00 8



.
9.
The characteristic equation is
−
λ
3
+3
λ
2
+6
λ
−
8 = 0, with eigenvalues
−
2, 1, and 4. Substituting the eigenvalues
into
λ
m
=
c
0
+
c
1
λ
+
c
2
λ
2
generates
(
−
2)
m
=
c
0
−
2
c
1
+4
c
2
1=
c
0
+
c
1
+
c
2
4
m
=
c
0
+4
c
1
+16
c
2
.
Solving the system gives
c
0
=
1
9
[8+(
−
1)
m
2
m
+1
−
4
m
]
,
c
1
=
1
18
[4
−
5(
−
2)
m
+4
m
]
,
c
2
=
1
18
[
−
2+(
−
2)
m
+4
m
]
.
344

345. Exercises 8.9
Thus
A
m
=
c
0
I
+
c
1
A
+
c
2
A
2
=



1
9
[(
−
2)
m
+(
−
1)
m
2
m
+1
+3
·
2
2
m
+1
]
1
3
[
−
(
−
2)
m
+4
m
]0
−
2
3
[(
−
2)
m
−
4
m
]
1
3
[(
−
1)
m
2
m
+1
+4
m
]0
1
3
[
−
3+(
−
2)
m
+2
2
m
+1
]
1
3
[
−
(
−
2)
m
+4
m
]1



and
A
10
=



699392 349184 0
698368 350208 0
699391 349184 1



.
10.
The characteristic equation is
−
λ
3
−
3
2
λ
2
+
3
2
λ
+ 1 = 0, with eigenvalues
−
2,
−
1
2
, and 1. Substituting the
eigenvalues into
λ
m
=
c
0
+
c
1
λ
+
c
2
λ
2
generates
(
−
2)
m
=
c
0
−
2
c
1
+4
c
2

−
1
2

m
=
c
0
−
1
2
c
1
+
1
4
c
2
1=
c
0
+
c
1
+
c
2
.
Solving the system gives
c
0
=
1
9
[2
−
m
[(
−
4)
m
+8(
−
1)
m
+2
m
+1
−
(
−
1)
m
2
2
m
+1
]
,
c
1
=
−
1
9
2
−
m
[(
−
4)
m
+4(
−
1)
m
−
5
·
2
m
]
,
c
2
=
2
9
[1+(
−
2)
m
−
(
−
1)
m
2
m
−
1
]
.
Thus
A
m
=
c
0
I
+
c
1
A
+
c
2
A
2
=



1
3
2
−
m
[2(
−
1)
m
+2
m
]
1
3
[
i
−
1+

−
1
2

m
n
0
2
3
i
−
1+

−
1
2

m
n
1
3
i
2+

−
1
2

m
n
0
−
1
9
2
−
m
[7(
−
4)
m
−
6(
−
1)
m
−
3
·
2
m
+(
−
1)
m
2
2
m
+1
]
1
3
i
−
1+

−
1
2

m
n
1
3
[(
−
2)
m
+(
−
1)
m
2
m
+1
]



and
A
8
=



43
128
−
85
256
0
−
85
128
171
256
0
−
32725
128
−
85
256
256



.
11.
The characteristic equation is
λ
2
−
8
λ
+ 16 = 0, with eigenvalues 4 and 4. Substituting the eigenvalues into
λ
m
=
c
0
+
c
1
λ
generates
4
m
=
c
0
+4
c
1
4
m
−
1
m
=
c
1
.
Solving the system gives
c
0
=
−
4
m
(
m
−
1)
,c
1
=4
m
−
1
m.
Thus
345

346. Exercises 8.9
A
m
=
c
0
I
+
c
1
A
=

4
m
−
1
(3
m
+4) 3
·
4
m
−
1
m
−
3
·
4
m
−
1
m
4
m
−
1
(
−
3
m
+4)

and
A
6
=
1
22528 18432
−
18432
−
14336
.
.
12.
The characteristic equation is
−
λ
3
−
λ
2
+21
λ
+ 45= 0, with eigenvalues
−
3,
−
3, and 5. Substituting the
eigenvalues into
λ
m
=
c
0
+
c
1
λ
+
c
2
λ
2
generates
(
−
3)
m
=
c
0
−
3
c
1
+9
c
2
(
−
3)
m
−
1
m
=
c
1
−
6
c
2
5
m
=
c
0
+5
c
1
+25
c
2
.
Solving the system gives
c
0
=
1
64
[73(
−
3)
m
−
2(
−
1)
m
3
m
+2
+9
·
5
m
−
40(
−
3)
m
m
]
,
c
1
=
1
96
[
−
(
−
1)
m
3
m
+2
+9
·
5
m
−
8(
−
3)
m
m
]
,
c
2
=
1
64
[
−
(
−
3)
m
+5
m
−
8(
−
3)
m
−
1
m
]
.
Thus
A
m
=
c
0
I
+
c
1
A
+
c
2
A
2
=






1
32
[31(
−
3)
m
−
(
−
1)
m
3
m
+1
+4
·
5
m
]
1
16
[
−
(
−
3)
m
−
(
−
1)
m
3
m
+1
+4
·
5
m
]
1
32
[(
−
3)
m
+(
−
1)
m
3
m
+1
−
4
·
5
m
]
1
16
[
−
(
−
3)
m
−
(
−
1)
m
3
m
+1
+4
·
5
m
]
1
8
[7(
−
3)
m
−
(
−
1)
m
3
m
+1
+4
·
5
m
]
1
16
[(
−
3)
m
+(
−
1)
m
3
m
+1
−
4
·
5
m
]
3
32
[(
−
3)
m
+(
−
1)
m
3
m
+1
−
4
·
5
m
]
3
16
[(
−
3)
m
+(
−
1)
m
3
m
+1
−
4
·
5
m
]
1
32
[29(
−
3)
m
−
(
−
1)
m
3
m
+2
+12
·
5
m
]






and
A
5
=



178 842
−
421
842 1441
−
842
−
1263
−
2526 1020



.
13. (a)
The characteristic equation is
λ
2
−
4
λ
=
λ
(
λ
−
4) = 0, so 0 is an eigenvalue. Since the matrix satisfies the
characteristic equation,
A
2
=4
A
,
A
3
=4
A
2
=4
2
A
,
A
4
=4
2
A
2
=4
3
A
, and, in general,
A
m
=4
m
A
=
1
4
m
4
m
3(4)
m
3(4)
m
.
.
(b)
The characteristic equation is
λ
2
= 0, so 0 is an eigenvalue. Since the matrix satisfies the characteristic
equation,
A
2
=
0
,
A
3
=
AA
2
=
0
, and, in general,
A
m
=
0
.
(c)
The characteristic equation is
−
λ
3
+5
λ
2
−
6
λ
= 0, with eigenvalues 0, 2, and 3. Substituting
λ
= 0 into
λ
m
=
c
0
+
c
1
λ
+
c
2
λ
2
we find that
c
0
= 0. Using the nonzero eigenvalues, we find
2
m
=2
c
1
+4
c
2
3
m
=3
c
1
+9
c
2
.
Solving the system gives
c
1
=
1
6
[9(2)
m
−
4(3)
m
]
,c
2
=
1
6
[
−
3(2)
m
+ 2(3)
m
]
.
346

347. Exercises 8.9
Thus
A
m
=
c
1
A
+
c
2
A
2
and
A
m
=



2(3)
m
−
1
3
m
−
1
3
m
−
1
1
6
[9(2)
m
−
4(3)
m
]
1
6
[3(2)
m
−
2(3)
m
]
1
6
[
−
3(2)
m
−
2(3)
m
]
1
6
[
−
9(2)
m
+ 8(3)
m
]
1
6
[
−
3(2)
m
+ 4(3)
m
]
1
6
[3(2)
m
+ 4(3)
m
]



.
14. (a)
Let
X
n
−
1
=
2
x
n
−
1
y
n
−
1
3
and
A
=
2
11
01
3
.
Then
X
n
=
AX
n
−
1
=
2
11
10
32
x
n
−
1
y
n
−
1
3
=
2
x
n
−
1
+
y
n
−
1
x
n
−
1
3
.
(b)
The characteristic equation of
A
is
λ
2
−
λ
−
1 = 0, with eigenvalues
λ
1
=
1
2
(1
−
√
5) and
λ
2
=
1
2
(1 +
√
5).
From
λ
m
=
c
0
+
c
1
λ
we get
λ
m
1
=
c
0
+
c
1
λ
1
and
λ
m
2
=
c
0
+
c
1
λ
2
. Solving this system gives
c
0
=(
λ
2
λ
m
1
−
λ
1
λ
m
2
)
/
(
λ
2
−
λ
1
) and
c
1
=(
λ
m
2
−
λ
m
1
)
/
(
λ
2
−
λ
1
)
.
Thus
A
m
=
c
0
I
+
c
1
A
=
1
2
m
+1
√
5
2
(1 +
√
5)
m
+1
−
(1
−
√
5)
m
+1
2(1 +
√
5)
m
−
2(1
−
√
5)
m
2(1 +
√
5)
m
−
2(1
−
√
5)
m
(1 +
√
5)(1
−
√
5)
m
−
(1
−
√
5)(1 +
√
5)
m
3
.
(c)
From part (a),
X
2
=
AX
1
,
X
3
=
AX
2
=
A
2
X
1
,
X
4
=
AX
3
=
A
3
X
1
, and, in general,
X
n
=
A
n
−
1
X
1
.
With
X
1
=
2
1
1
3
we have
X
12
=
A
11
X
1
=
2
144 89
89 55
32
1
1
3
=
2
233
144
3
,
so the number of adult pairs is 233. With
X
1
=
2
1
0
3
we have
A
11
X
1
=
2
144 89
89 55
32
1
0
3
=
2
144
89
3
,
so the number of baby pairs is 144. With
X
1
=
2
2
1
3
we have
A
11
X
1
=
2
144 89
89 55
32
2
1
3
=
2
377
233
3
,
so the total number of pairs is 377.
15.
The characteristic equation of
A
is
λ
2
−
5
λ
+ 10 = 0, so
A
2
−
5
A
+10
I
=
0
and
I
=
−
1
10
A
2
+
1
2
A
. Multiplying
by
A
−
1
we find
A
−
1
=
−
1
10
A
+
1
2
I
=
−
1
10
2
2
−
4
13
3
+
1
2
2
10
01
3
=

3
10
2
5
−
1
10
1
5

.
16.
The characteristic equation of
A
is
−
λ
3
+2
λ
2
+
λ
−
2 = 0, so
−
A
3
+2
A
2
+
A
−
2
I
=
0
and
I
=
−
1
2
A
3
+
A
2
+
1
2
A
.
Multiplying by
A
−
1
we find
A
−
1
=
−
1
2
A
2
+
A
+
1
2
I
=



3
2
1
2
−
5
2
1
2
1
2
−
1
2
1
2
1
2
−
3
2



.
347

348. Exercises 8.9
17. (a)
Since
A
2
=
2
10
−
10
3
we see that
A
m
=
2
10
−
10
3
for all integers
m
≥
2. Thus
A
is not nilpotent.
(b)
Since
A
2
=
0
, the matrix is nilpotent with index 2.
(c)
Since
A
3
=
0
, the matrix is nilpotent with index 3.
(d)
Since
A
2
=
0
, the matrix is nilpotent with index 2.
(e)
Since
A
4
=
0
, the matrix is nilpotent with index 4.
(f)
Since
A
4
=
0
, the matrix is nilpotent with index 4.
18. (a)
If
A
m
=
0
for some
m
, then (det
A
)
m
= det
A
m
= det
0
= 0, and
A
is a singular matrix.
(b)
By (1) of Section 8
.
8wehave
AK
=
λ
K
,
A
2
K
=
λ
AK
=
λ
2
K
,
A
3
K
=
λ
2
AK
=
λ
3
K
, and, in general,
A
m
K
=
λ
m
K
.If
A
is nilpotent with index
m
, then
A
m
=
0
and
λ
m
=0.
Exercises 8.10
1.
(
a
)–(
b
)



00
−
4
0
−
40
−
4015






0
1
0



=



0
−
4
0



=
−
4



0
1
0



;
λ
1
=
−
4



00
−
4
0
−
40
−
4015






4
0
1



=



−
4
0
1



=(
−
1)



4
0
1



;
λ
2
=
−
1



00
−
4
0
−
40
−
4015






1
0
−
4



=



16
0
−
64



=16



1
0
−
4



;
λ
3
=16
(
c
)
K
T
1
K
2
=[010]



4
0
1



=0;
K
T
1
K
3
=[010]



1
0
−
4



=0;
K
T
2
K
3
=[401]



1
0
−
4



=4
−
4=0
2.
(
a
)–(
b
)



1
−
1
−
1
−
11
−
1
−
1
−
11






−
2
1
1



=



−
4
2
2



=2



−
2
1
1



;
λ
1
=2



1
−
1
−
1
−
11
−
1
−
1
−
11






0
1
−
1



=



0
2
−
2



=2



0
1
−
1



;
λ
2
=2



1
−
1
−
1
−
11
−
1
−
1
−
11






1
1
1



=



−
1
−
1
−
1



=(
−
1)



1
1
1



;
λ
3
=
−
1
(
c
)
K
T
1
K
2
=[
−
211]



0
1
−
1



=1
−
1=0;
K
T
1
K
3
=[
−
211]



1
1
1



=
−
2+1+1=0
348

349. Exercises 8.10
K
T
2
K
3
=[0 1
−
1]



1
1
1



=1
−
1=0
3.
(
a
)–(
b
)



513 0
13 50
00
−
8






√
2
2
√
2
2
0



=



9
√
2
9
√
2
0



=18



√
2
2
√
2
2
0



;
λ
1
=18



513 0
13 50
00
−
8






√
3
3
−
√
3
3
√
3
3



=



−
8
√
2
3
8
√
3
3
−
8
√
3
3



=(
−
8)



√
3
3
−
√
3
3
√
3
3



;
λ
2
=
−
8



513 0
13 50
00
−
8






√
6
6
−
√
6
6
−
√
6
3



=



−
8
√
6
6
8
√
6
6
8
√
6
3



=(
−
8)



√
6
6
−
√
6
6
−
√
6
3



;
λ
3
=
−
8
(
c
)
K
T
1
K
2
=[
√
2
2
√
2
2
0]



√
3
3
−
√
3
3
√
3
3



=
√
6
6
−
√
6
6
=0;
K
T
1
K
3
=[
√
2
2
√
2
2
0]



√
6
6
−
√
6
6
−
√
6
3



=
√
12
12
−
√
12
12
=0
K
T
2
K
3
=[
√
3
3
−
√
3
3
√
3
3
]



√
6
6
−
√
6
6
−
√
6
3



=
√
18
18
+
√
18
18
−
√
18
9
=0
4.
(
a
)–(
b
)



322
220
204






−
2
2
1



=



0
0
0



=0



−
2
2
1



;
λ
1
=0



322
220
204






1
2
−
2



=



3
6
−
6



=3



1
2
−
2



;
λ
2
=3



322
220
204






2
1
2



=



12
6
12



=6



2
1
2



;
λ
3
=6
(
c
)
K
T
1
K
2
=[
−
221]



1
2
−
2



=
−
2+4
−
2=0;
K
T
1
K
3
=[
−
221]



2
1
2



=
−
4+2+2=0
K
T
2
K
3
=[1 2
−
2]



2
1
2



=2+2
−
4=0
5.
Orthogonal. Columns form an orthonormal set.
6.
Not orthogonal. Columns one and three are not unit vectors.
7.
Orthogonal. Columns form an orthonormal set.
8.
Not orthogonal. The matrix is singular.
349

350. Exercises 8.10
9.
Not orthogonal. Columns are not unit vectors.
10.
Orthogonal. Columns form an orthogonal set.
11.
λ
1
=
−
8,
λ
2
= 10,
K
1
=
2
1
−
1
3
,
K
2
=
2
1
1
3
,
P
=

1
√
2
1
√
2
−
1
√
2
1
√
2

12.
λ
1
=7,
λ
2
=4,
K
1
=
2
1
0
3
,
K
2
=
2
0
1
3
,
P
=

10
01

13.
λ
1
=0,
λ
2
= 10,
K
1
=
2
3
−
1
3
,
K
2
=
2
1
3
3
,
P
=

3
√
10
1
√
10
−
1
√
10
3
√
10

14.
λ
1
=
1
2
+
√
5
2
,
λ
2
=
1
2
−
√
5
2
,
K
1
=
2
1+
√
5
2
3
,
K
2
=
2
1
−
√
5
2
3
,
P
=


1+
√
5
√
10+2
√
5
1
−
√
5
√
10
−
2
√
5
2
√
10+2
√
5
2
√
10
−
2
√
5


15.
λ
1
=0,
λ
2
=2,
λ
3
=1,
K
1
=



−
1
0
1



,
K
2
=



1
0
1



,
K
3
=



0
1
0



,
P
=



−
1
√
2
1
√
2
0
001
1
√
2
1
√
2
0



16.
λ
1
=
−
1,
λ
2
=1
−
√
2,
λ
3
=1+
√
2,
K
1
=



−
1
0
1



,
K
2
=



1
−
√
2
1



,
K
3
=



1
√
2
1



,
P
=




−
1
√
2
1
2
1
2
0
−
√
2
2
√
2
2
1
√
2
1
2
1
2




17.
λ
1
=
−
11,
λ
2
=0,
λ
3
=6,
K
1
=



−
3
1
1



,
K
2
=



1
−
4
7



,
K
3
=



1
2
1



,
P
=



−
3
√
11
1
√
66
1
√
6
1
√
11
−
4
√
66
2
√
6
1
√
11
7
√
66
1
√
6



18.
λ
1
=
−
18,
λ
2
=0,
λ
3
=9,
K
1
=



1
−
2
2



,
K
2
=



−
2
1
2



,
K
3
=



2
2
1



,
P
=



1
3
−
2
3
2
3
−
2
3
1
3
2
3
2
3
2
3
1
3



19.

3
5
a
4
5
b

2
3
5
4
5
ab
3
=
2
10
01
3
implies
9
25
+
a
2
= 1 and
16
25
+
b
2
= 1. These equations give
a
=
±
4
5
,
b
=
±
3
5
. But
12
25
+
ab
= 0 indicates
a
and
b
must have opposite signs. Therefore choose
a
=
−
4
5
,
b
=
3
5
.
The matrix

3
5
−
4
5
4
5
3
5

is orthogonal.
20.

1
√
5
b
a
1
√
5

2
1
√
5
a
b
1
√
5
3
=
2
10
01
3
implies
1
5
+
b
2
= 1 and
a
2
+
1
5
= 1. These equations give
a
=
±
2
√
5
,
b
=
±
2
√
5
.
But
a
√
5
+
b
√
5
= 0 indicates
a
and
b
must have opposite signs. Therefore choose
a
=
−
2
√
5
,
b
=
2
√
2
.
The matrix

1
√
5
2
√
5
−
2
√
5
1
√
5

is orthogonal.
21. (a) AK
1
=



74
−
4
4
−
8
−
1
−
4
−
1
−
8






4
1
−
1



=



36
9
−
9



=9



4
1
−
1



=
λ
1
K
1
350

351. Exercises 8.10
AK
2
=



74
−
4
4
−
8
−
1
−
4
−
1
−
8






1
0
4



=



−
9
0
−
36



=
−
9



1
0
4



=
λ
2
K
2
AK
3
=



74
−
4
4
−
8
−
1
−
4
−
1
−
8






1
−
4
0



=



−
9
36
0



=
−
9



1
−
4
0



=
λ
3
K
3
(b)
For
λ
2
=
λ
3
=
−
9wehave



16 4
−
4
0
41
−
1
0
−
4
−
11
0



=
⇒



1
1
4
−
1
4
0
00 0
0
00 0
0



so that
k
1
=
−
1
4
k
2
+
1
4
k
3
. The choices
k
2
=0,
k
3
= 4 and
k
2
=
−
4,
k
3
= 0 give the eigenvectors in part (a).
These eigenvectors are not orthogonal. However, by choosing
k
2
=
k
3
= 1 and
k
2
=
−
2,
k
3
= 2 we obtain,
respectively,
K
2
=



0
1
1



,
K
3
=



1
−
2
2



.
Mutually orthogonal eigenvectors are



4
1
−
1



,



0
1
1



,



1
−
2
2



.
22.
The eigenvalues and eigenvectors of
A
are
λ
1
=
λ
2
=
−
1
,λ
3
=
λ
4
=3
,
K
1
=





−
1
1
0
0





,
K
2
=





0
0
−
1
1





,
K
3
=





0
0
1
1





,
K
4
=





1
1
0
0





.
Since

K
1

=

K
2

=

K
3

=

K
4

=
√
2 , an orthogonal matrix is
P
=






−
1
√
2
00
1
√
2
1
√
2
00
1
√
2
0
−
1
√
2
1
√
2
0
0
1
√
2
1
√
2
0






.
23.
Suppose
A
and
B
are orthogonal matrices. Then
A
−
1
=
A
T
and
B
−
1
=
B
T
and
(
AB
)
−
1
=
B
−
1
A
−
1
=
B
T
A
T
=(
AB
)
T
.
Thus
AB
is an orthogonal matrix.
351

352. Exercises 8.11
Exercises 8.11
1.
Taking
X
0
=
2
1
1
3
and computing
X
i
=
AX
i
−
1
for
i
= 1, 2, 3, 4 we obtain
X
1
=
2
2
2
3
,
X
2
=
2
4
4
3
,
X
3
=
2
8
8
3
,
X
4
=
2
16
16
3
.
We conclude that a dominant eigenvector is
K
=
2
1
1
3
with corresponding eigenvalue
λ
=
AK
·
K
K
·
K
=
4
2
=2.
2.
Taking
X
0
=
2
1
1
3
and computing
X
i
=
AX
i
−
1
for
i
= 1, 2, 3, 4, 5we obtain
X
1
=
2
−
5
7
3
,
X
2
=
2
49
−
47
3
,
X
3
=
2
−
437
439
3
,
X
4
=
2
3937
−
3935
3
,
X
5
=
2
−
35429
35431
3
.
We conclude that a dominant eigenvector is
K
=
1
35439
2
−
35429
35431
3
≈
2
−
0
.
99994
1
3
with corresponding eigen-
value
λ
=
AK
·
K
K
·
K
=
−
8
.
9998.
3.
Taking
X
0
=
2
1
1
3
and computing
AX
0
=
2
6
16
3
, we define
X
1
=
1
16
2
6
16
3
=
2
0
.
375
1
3
. Continuing in this
manner we obtain
X
2
=
2
0
.
3363
1
3
,
X
3
=
2
0
.
3335
1
3
,
X
4
=
2
0
.
3333
1
3
.
We conclude that a dominant eigenvector is
K
=
2
0
.
3333
1
3
with corresponding eigenvalue
λ
= 14.
4.
Taking
X
0
=
2
1
1
3
and computing
AX
0
=
2
1
5
3
, we define
X
1
=
1
5
2
1
5
3
=
2
0
.
2
1
3
. Continuing in this manner we
obtain
X
2
=
2
0
.
2727
1
3
,
X
3
=
2
0
.
2676
1
3
,
X
4
=
2
0
.
2680
1
3
,
X
5
=
2
0
.
2679
1
3
.
We conclude that a dominant eigenvector is
K
=
2
0
.
2679
1
3
with corresponding eigenvalue
λ
=6
.
4641.
5.
Taking
X
0
=



1
1
1



and computing
AX
0
=



11
11
6



, we define
X
1
=
1
11



11
11
6



=



1
1
0
.
5455



. Continuing in this
manner we obtain
X
2
=



1
1
0
.
5045



,
X
3
=



1
1
0
.
5005



,
X
4
=



1
1
0
.
5



.
We conclude that a dominant eigenvector is
K
=



1
1
0
.
5



with corresponding eigenvalue
λ
= 10.
352

353. Exercises 8.11
6.
Taking
X
0
=



1
1
1



and computing
AX
0
=



5
2
2



, we define
X
1
=
1
5



5
2
2



=



1
0
.
4
0
.
4



. Continuing in this manner
we obtain
X
2
=



1
0
.
2105
0
.
2105



,
X
3
=



1
0
.
1231
0
.
1231



,
X
4
=



1
0
.
0758
0
.
0758



,
X
5
=



1
0
.
0481
0
.
0481



.
At this point if we restart with
X
0
=



1
0
0



we see that
K
=



1
0
0



is a dominant eigenvector with corresponding
eigenvalue
λ
=3.
7.
Taking
X
0
=
2
1
1
3
and using scaling we obtain
X
1
=
2
0
.
625
1
3
,
X
2
=
2
0
.
5345
1
3
,
X
3
=
2
0
.
5098
1
3
,
X
4
=
2
0
.
5028
1
3
,
X
5
=
2
0
.
5008
1
3
.
Taking
K
=
2
0
.
5
1
3
as the dominant eigenvector we find
λ
1
= 7. Now the normalized eigenvector is
K
1
=
2
0
.
4472
0
.
8944
3
and
B
=
2
1
.
6
−
0
.
8
−
0
.
80
.
4
3
. Taking
X
0
=
2
1
1
3
and using scaling again we obtain
X
1
=
2
1
−
0
.
5
3
,
X
2
=
2
1
−
0
.
5
3
. Taking
K
=
2
1
−
0
.
5
3
we find
λ
2
= 2. The eigenvalues are 7 and 2.
8.
Taking
X
0
=
2
1
1
3
and using scaling we obtain
X
1
=
2
0
.
3333
1
3
,
X
2
=
2
0
.
3333
1
3
. Taking
K
=
2
1
/
3
1
3
as the
dominant eigenvector we find
λ
1
= 10. Now the normalized eigenvector is
K
1
=
2
0
.
3162
0
.
9486
3
and
B
=
2
00
00
3
.An
eigenvector for the zero matrix is
λ
2
= 0. The eigenvalues are 10 and 0.
9.
Taking
X
0
=



1
1
1



and using scaling we obtain
X
1
=



1
0
1



,
X
2
=



1
−
0
.
6667
1



,
X
3
=



1
−
0
.
9091
1



,
X
4
=



1
−
0
.
9767
1



,
X
5
=



1
−
0
.
9942
1



.
Taking
K
=



1
−
1
1



as the dominant eigenvector we find
λ
1
= 4. Now the normalized eigenvector is
K
1
=



0
.
5774
−
0
.
5774
0
.
5774



and
B
=



1
.
6667 0
.
3333
−
1
.
3333
0
.
3333 0
.
6667 0
.
3333
−
1
.
3333 0
.
3333 1
.
6667



.If
X
0
=



1
1
1



is now chosen only one more
353

354. Exercises 8.11
eigenvalue is found. Thus, try
X
0
=



1
1
0



. Using scaling we obtain
X
1
=



1
0
.
5
−
0
.
5



,
X
2
=



1
0
.
2
−
0
.
8



,
X
3
=



1
0
.
0714
−
0
.
9286



,
X
4
=



1
0
.
0244
−
0
.
9756



,
X
5
=



1
0
.
0082
−
0
.
9918



.
Taking
K
=



1
0
−
1



as the eigenvector we find
λ
2
= 3. The normalized eigenvector in this case is
K
2
=



0
.
7071
0
−
0
.
7071



and
C
=



0
.
1667 0
.
3333 0
.
1667
0
.
3333 0
.
6667 0
.
3333
0
.
1667 0
.
3333 0
.
1667



.If
X
0
=



1
1
1



is chosen, and scaling is used we
obtain
X
1
=



0
.
5
1
0
.
5



,
X
2
=



0
.
5
1
0
.
5



. Taking
K
=



0
.
5
1
0
.
5



we find
λ
3
= 1. The eigenvalues are 4, 3, and 1.
The difficulty in choosing
X
0
=



1
1
1



to find the second eigenvector results from the fact that this vector is a
linear combination of the eigenvectors corresponding to the other two eigenvalues, with 0 contribution from the
second eigenvector. When this occurs the development of the power method, shown in the text, breaks down.
10.
Taking
X
0
=



1
1
1



and using scaling we obtain
X
1
=



−
0
.
3636
−
0
.
3636
1



,
X
2
=



−
0
.
2431
0
.
0884
1



,
X
3
=



−
0
.
2504
−
0
.
0221
1



,
X
4
=



−
0
.
2499
−
0
.
0055
1



.
Taking
K
=



−
0
.
25
0
1



as the dominant eigenvector we find
λ
1
= 16. The normalized eigenvector is
K
1
=



−
0
.
2425
0
0
.
9701



and
B
=



−
0
.
9412 0
−
0
.
2353
0
−
40
−
0
.
2353 0
−
0
.
0588



. Taking
X
0
=



1
1
1



and using scaling we obtain
X
1
=



−
0
.
2941
−
1
−
0
.
0735



,
X
2
=



0
.
0735
1
0
.
0184



,
X
3
=



−
0
.
0184
−
1
−
0
.
0046



,
X
4
=



0
.
0046
1
0
.
0011



.
Taking
K
=



0
1
0



as the eigenvector we find
λ
2
=
−
4. The normalized eigenvector in this case is
K
2
=
K
=



0
1
0



and
C
=



−
0
.
9412 0
−
0
.
2353
000
−
0
.
2353 0
−
0
.
0588



. Taking
X
0
=



1
1
1



and using scaling we obtain
X
1
=



−
1
0
−
0
.
25



,
354

355. Exercises 8.11
X
2
=



1
0
0
.
25



. Using
K
=



1
0
0
.
25



we find
λ
3
=
−
1. The eigenvalues are 16,
−
4, and
−
1.
11.
The inverse matrix is
2
4
−
1
−
31
3
. Taking
X
0
=
2
1
1
3
and using scaling we obtain
X
1
=
2
1
−
0
.
6667
3
,
X
2
=
2
1
−
0
.
7857
3
,
X
3
=
2
1
−
0
.
7910
3
,
X
4
=
2
1
−
0
.
7913
3
.
Using
K
=
2
1
−
0
.
7913
3
we find
λ
=4
.
7913. The minimum eigenvalue of
2
11
34
3
is 1
/
4
.
7913
≈
0
.
2087.
12.
The inverse matrix is
2
13
42
3
. Taking
X
0
=
2
1
1
3
and using scaling we obtain
X
1
=
2
0
.
6667
1
3
,
X
2
=
2
0
.
7857
1
3
,...,
X
10
=
2
0
.
75
1
3
.
Using
K
=
2
0
.
75
1
3
we find
λ
= 5. The minimum eigenvalue of
2
−
0
.
20
.
3
0
.
4
−
0
.
1
3
is 1
/
5=0
.
2
13. (a)
Replacing the second derivative with the difference expression we obtain
EI
y
i
+1
−
2
y
i
+
y
i
−
1
h
2
+
Py
i
=0 or
EI
(
y
i
+1
−
2
y
i
+
y
i
−
1
)+
Ph
2
y
i
=0
.
(b)
Expanding the difference equation for
i
= 1, 2, 3 and using
h
=
L/
4,
y
0
= 0, and
y
4
= 0 we obtain
EI
(
y
2
−
2
y
1
+
y
0
)+
PL
2
16
y
1
=0
EI
(
y
3
−
2
y
2
+
y
1
)+
PL
2
16
y
2
=0
EI
(
y
4
−
2
y
3
+
y
2
)+
PL
2
16
y
3
=0
or
2
y
1
−
y
2
=
PL
2
16
EI
y
1
−
y
1
+2
y
2
−
y
3
=
PL
2
16
EI
y
2
−
y
2
+2
y
3
=
PL
2
16
EI
y
3
.
In matrix form this becomes



2
−
10
−
12
−
1
0
−
12






y
1
y
2
y
3



=
PL
2
16
EI



y
1
y
2
y
3



.
(c) A
−
1
=



0
.
750
.
50
.
25
0
.
510
.
5
0
.
250
.
50
.
75



(d)
Taking
X
0
=



1
1
1



and using scaling we obtain
X
1
=



0
.
75
1
0
.
75



,
X
2
=



0
.
7143
1
0
.
7143



,
X
3
=



0
.
7083
1
0
.
7083



,
X
4
=



0
.
7073
1
0
.
7073



,
X
5
=



0
.
7071
1
0
.
7071



.
Using
K
=



0
.
7071
1
0
.
7071



we find
λ
=1
.
7071. Then 1
/λ
=0
.
5859 is the minimum eigenvalue of
A
.
355

356. Exercises 8.11
(e)
Solving
PL
2
16
EI
=0
.
5859 for
P
we obtain
P
=9
.
3726
EI
L
2
. In Example 3 of Section 3
.
9wesaw
P
=
π
2
EI
L
2
≈
9
.
8696
EI
L
2
.
14. (a)
The difference equation is
EI
i
(
y
i
+1
−
2
y
i
+
y
i
−
1
)+
Ph
2
y
i
=0
,i
=1
,
2
,
3
,
where
I
0
=0
.
00200,
I
1
=0
.
00175,
I
2
=0
.
00150,
I
3
=0
.
00125, and
I
4
=0
.
00100. The system of equations
is
0
.
00175
E
(
y
2
−
2
y
1
+
y
0
)+
PL
2
16
y
1
=0
0
.
00150
E
(
y
3
−
2
y
2
+
y
1
)+
PL
2
16
y
2
=0
0
.
00125
E
(
y
4
−
2
y
3
+
y
2
)+
PL
2
16
y
2
=0
or
0
.
0035
y
1
−
0
.
00175
y
2
=
PL
2
16
E
y
1
−
0
.
0015
y
1
+0
.
003
y
2
−
0
.
0015
y
3
=
PL
2
16
E
y
2
−
0
.
00125
y
2
+0
.
0025
y
3
=
PL
2
16
E
y
3
.
In matrix form this becomes



0
.
0035
−
0
.
001750
−
0
.
00150
.
003
−
0
.
0015
0
−
0
.
001250
.
0025






y
1
y
2
y
3



=
PL
2
16
E



y
1
y
2
y
3



.
(b)
The inverse of
A
is
A
−
1
=



428
.
571 333
.
333 200
285
.
714 666
.
667 400
142
.
857 333
.
333 600



.
Taking
X
0
=



1
1
1



and using scaling we obtain
X
1
=



0
.
7113
1
0
.
7958



,
X
2
=



0
.
6710
1
0
.
7679



,
X
3
=



0
.
6645
1
0
.
7635



,
X
4
=



0
.
6634
1
0
.
7628



,
X
5
=



0
.
6632
1
0
.
7627



.
This yields the eigenvalue
λ
= 1161
.
23. The smallest eigenvalue of
A
is then 1
/λ
=0
.
0008612. The lowest
critical load is
P
=
16
E
L
2
(0
.
0008612)
−
0
.
01378
E
L
2
.
15. (a) A
10
=



67,745,349
−
43,691,832 8,258,598
−
43,691,832 28,182,816
−
5,328,720
8,258,598
−
5,328,720 1,008,180



(b) X
10
=
A
10



1
0
0



=



67,745,349
−
43,691,832
8,258,598



≈
67,745,349



1
−
0
.
644942
0
.
121906



356

357. Exercises 8.12
X
12
=
A
12



1
0
0



=



2,680,201,629
−
1,728,645,624
326,775,222



≈
2,680,201,629



1
−
0
.
644968
0
.
121922



.
The vectors appear to be approaching scalar multiples of
K
=(1
,
−
0
.
644968
,
0
.
121922), which approximates
the dominant eigenvector.
(c)
The dominant eigenvalue is
λ
1
=(
AK
·
K
)
/
(
K
·
K
)=6
.
28995.
Exercises 8.12
1.
Distinct eigenvalues
λ
1
=1,
λ
2
= 5imply
A
is diagonalizable.
P
=
2
−
31
11
3
,
D
=
2
10
05
3
2.
Distinct eigenvalues
λ
1
=0,
λ
2
= 6 imply
A
is diagonalizable.
P
=
2
−
5
−
1
42
3
,
D
=
2
00
06
3
3.
For
λ
1
=
λ
2
= 1 we obtain the single eigenvector
K
1
=
2
1
1
3
. Hence
A
is not diagonalizable.
4.
Distinct eigenvalues
λ
1
=
√
5,
λ
2
=
−
√
5imply
A
is diagonalizable.
P
=
2
√
5
−
√
5
11
3
,
D
=
2
√
50
0
−
√
5
3
5.
Distinct eigenvalues
λ
1
=
−
7,
λ
2
= 4 imply
A
is diagonalizable.
P
=
2
13 1
21
3
,
D
=
2
−
70
04
3
6.
Distinct eigenvalues
λ
1
=
−
4,
λ
2
= 10 imply
A
is diagonalizable.
P
=
2
−
31
1
−
5
3
,
D
=
2
−
40
010
3
7.
Distinct eigenvalues
λ
1
=
1
3
,
λ
2
=
2
3
imply
A
is diagonalizable.
P
=
2
11
−
11
3
,
D
=
2
1
3
0
0
2
3
3
8.
For
λ
1
=
λ
2
=
−
3 we obtain the single eigenvector
K
1
=
2
1
1
3
. Hence
A
is not diagonalizable.
9.
Distinct eigenvalues
λ
1
=
−
i
,
λ
2
=
i
imply
A
is diagonalizable.
P
=
2
11
−
ii
3
,
D
=
2
−
i
0
0
i
3
10.
Distinct eigenvalues
λ
1
=1+
i
,
λ
2
=1
−
i
imply
A
is diagonalizable.
P
=
2
22
i
−
i
3
,
D
=
2
1+
i
0
01
−
i
3
357

358. Exercises 8.12
11.
Distinct eigenvalues
λ
1
=1,
λ
2
=
−
1,
λ
3
= 2 imply
A
is diagonalizable.
P
=



101
011
001



,
D
=



100
0
−
10
002



12.
Distinct eigenvalues
λ
1
=3,
λ
2
=4,
λ
3
= 5imply
A
is diagonalizable.
P
=



12 0
02 1
11
−
1



,
D
=



300
040
005



13.
Distinct eigenvalues
λ
1
=0,
λ
2
=1,
λ
3
= 2 imply
A
is diagonalizable.
P
=



111
010
−
111



,
D
=



000
010
002



14.
Distinct eigenvalues
λ
1
=1,
λ
2
=
−
3
i
,
λ
3
=3
i
imply
A
is diagonalizable.
P
=



0
−
3
i
3
i
011
100



,
D
=



100
0
−
3
i
0
003
i



15.
The eigenvalues are
λ
1
=
λ
2
=1,
λ
3
=2. For
λ
1
=
λ
2
= 1 we obtain the single eigenvector
K
1
=



1
0
0



.
Hence
A
is not diagonalizable.
16.
Distinct eigenvalues
λ
1
=1,
λ
2
=2,
λ
3
= 3 imply
A
is diagonalizable.
P
=



110
010
001



,
D
=



100
020
003



17.
Distinct eigenvalues
λ
1
=1,
λ
2
=
√
5,
λ
3
=
−
√
5imply
A
is diagonalizable.
P
=



01+
√
51
−
√
5
02 2
10 0



,
D
=



10 0
0
√
50
00
−
√
5



18.
For
λ
1
=
λ
2
=
λ
3
= 1 we obtain the single eigenvector
K
1
=



1
−
2
1



. Hence
A
is not diagonalizable.
19.
For the eigenvalues
λ
1
=
λ
2
=2,
λ
3
=1,
λ
4
=
−
1 we obtain four linearly independent eigenvectors. Hence
A
is diagonalizable and
P
=





−
3
−
1
−
11
0100
−
3001
1010





,
D
=





200 0
020 0
001 0
000
−
1





.
358

359. Exercises 8.12
20.
The eigenvalues are
λ
1
=
λ
2
=2,
λ
3
=
λ
4
=3. For
λ
3
=
λ
4
= 3 we obtain the single eigenvector
K
1
=





1
0
1
0





.
Hence
A
is not diagonalizable.
21.
λ
1
=0,
λ
2
=2,
K
1
=
3
1
−
1
3
,
K
2
=
3
1
1
3
,
P
=

1
√
2
1
√
2
−
1
√
2
1
√
2

,
D
=
3
00
02
3
22.
λ
1
=
−
1,
λ
2
=4,
K
1
=
3
1
−
2
3
,
K
2
=
3
2
1
3
,
P
=

1
√
5
2
√
5
−
2
√
5
1
√
5

,
D
=
3
−
10
04
3
23.
λ
1
=3,
λ
2
= 10,
K
1
=
3
−
√
10
2
3
,
K
2
=
3
√
10
5
3
,
P
=

−
√
10
√
14
√
10
√
35
2
√
14
15
√
35

,
D
=
3
30
010
3
24.
λ
1
=
−
1,
λ
2
=3,
K
1
=
3
1
1
3
,
K
2
=
3
1
−
1
3
,
P
=

1
√
2
1
√
2
1
√
2
−
1
√
2

,
D
=
3
−
10
03
3
25.
λ
1
=
−
1,
λ
2
=
λ
3
=1,
K
1
=



−
1
1
0



,
K
2
=



1
1
0



,
K
3
=



0
0
1



,
P
=



−
1
√
2
1
√
2
0
1
√
2
1
√
2
0
001



,
D
=



−
100
010
001



26.
λ
1
=
λ
2
=
−
1,
λ
3
=5,
K
1
=



−
1
0
1



,
K
2
=



1
1
0



,
K
3
=



1
−
1
1



,
P
=



−
1
√
2
1
√
2
1
√
3
0
1
√
2
−
1
√
3
1
√
2
0
1
√
3



,
D
=



−
100
0
−
10
005



27.
λ
1
=3,
λ
2
=6,
λ
3
=9,
K
1
=



2
2
1



,
K
2
=



2
−
1
−
2



,
K
3
=



1
−
2
2



,
P
=



2
3
2
3
1
3
2
3
−
1
3
−
2
3
1
3
−
2
3
2
3



,
D
=



300
060
009



28.
λ
1
=1,
λ
2
=2
−
√
2,
λ
3
=2+
√
2,
K
1
=



0
1
0



,
K
2
=



1
−
√
2
0
1



,
K
3
=



1+
√
2
0
1



,
P
=




0
−
√
2
−
√
2
2
√
2+
√
2
2
10 0
0
√
2+
√
2
2
√
2
−
√
2
2




,
D
=



10 0
02
−
2
√
20
002+
√
2



29.
λ
1
=1,
λ
2
=
−
6,
λ
3
=8,
K
1
=



0
1
0



,
K
2
=



1
0
−
1



,
K
3
=



1
0
1



,
P
=



0
1
√
2
1
√
2
100
0
−
1
√
2
1
√
2



,
D
=



100
0
−
60
008



30.
λ
1
=
λ
2
=0,
λ
3
=
−
2,
λ
4
=2,
K
1
=





−
1
0
1
0





,
K
2
=





0
−
1
0
1





,
K
3
=





1
−
1
1
−
1





,
K
4
=





1
1
1
1





359

360. Exercises 8.12
P
=






−
1
√
2
0
1
2
1
2
0
−
1
√
2
−
1
2
1
2
1
√
2
0
1
2
1
2
0
1
√
2
−
1
2
1
2






,
D
=





00 00
00 00
00
−
20
00 02





31.
The given equation can be written as
X
T
AX
= 24: [
xy
]
2
5
−
1
−
15
32
x
y
3
= 24. Using
λ
1
=6,
λ
2
=4,
K
1
=
2
1
−
1
3
,
K
2
=
2
1
1
3
,
P
=

1
√
2
1
√
2
−
1
√
2
1
√
2

and
X
=
PX
1
we find
[
XY
]
2
60
04
32
X
Y
3
=24 or 6
X
2
+4
Y
2
=24
.
The conic section is an ellipse. Now from
X
1
=
P
T
X
we see that the
XY
-coordinates of (1
,
−
1) and (1
,
1)
are (
√
2
,
0) and (0
,
√
2 ), respectively. From this we conclude that the
X
-axis and
Y
-axis are as shown in the
accompanying figure.
32.
The given equation can be written as
X
T
AX
= 288: [
xy
]
2
13
−
5
−
513
32
x
y
3
= 288. Using
λ
1
=8,
λ
2
= 18,
K
1
=
2
1
1
3
,
K
2
=
2
−
1
1
3
,
P
=

1
√
2
−
1
√
2
1
√
2
1
√
2

and
X
=
PX
1
we find
[
XY
]
2
80
018
32
X
Y
3
= 288 or 8
X
2
+18
Y
2
= 288
.
The conic section is an ellipse. Now from
X
1
=
P
T
X
we see that the
XY
-coordinates of (1
,
1) and (1
,
−
1) are
(
√
2
,
0) and (0
,
−
√
2 ), respectively. From this we conclude that the
X
-axis and
Y
-axis are as shown in the
accompanying figure.
33.
The given equation can be written as
X
T
AX
= 20: [
xy
]
2
−
34
43
32
x
y
3
= 20. Using
λ
1
=5,
λ
2
=
−
5,
K
1
=
2
1
2
3
,
K
2
=
2
−
2
1
3
,
P
=

1
√
5
−
2
√
5
2
√
5
1
√
5

and
X
=
PX
1
we find
[
XY
]
2
50
0
−
5
32
X
Y
3
=20 or 5
X
2
−
5
Y
2
=20
.
The conic section is a hyperbola. Now from
X
1
=
P
T
X
we see that the
XY
-coordinates of (1
,
2) and (
−
2
,
1)
are (
√
5
,
0) and (0
,
√
5), respectively. From this we conclude that the
X
-axis and
Y
-axis are as shown in the
accompanying figure.
34.
The given equation can be written as
X
T
AX
= 288:
[
xy
]
2
16 12
12 9
32
x
y
3
+[
−
34]
2
x
y
3
=0
.
Using
λ
1
=25,
λ
2
=0,
K
1
=
2
4
3
3
,
K
2
=
2
−
3
4
3
,
P
=

4
5
3
5
3
5
−
4
5

and
X
=
PX
1
we find
[
XY
]
2
250
00
32
X
Y
3
+[0 5]
2
X
Y
3
= 0 or 25
X
2
+5
Y
=0
.
360

361. Exercises 8.13
The conic section is a parabola. Now from
X
1
=
P
T
X
we see that the
XY
-coordinates of (4
,
3) and (3
,
−
4)
are (5
,
0) and (0
,
−
5), respectively. From this we conclude that the
X
-axis and
Y
-axis are as shown in the
accompanying figure.
35.
Since
D
=
P
−
1
AP
we have
A
=
PDP
−
1
. Hence
A
=
2
11
21
32
20
03
32
−
11
2
−
1
3
=
2
4
−
1
21
3
.
36.
Since eigenvectors are mutually orthogonal we use an orthogonal matrix
P
and
A
=
PDP
T
.
A
=



1
√
3
1
√
2
1
√
6
−
1
√
3
0
2
√
6
1
√
3
−
1
√
2
1
√
6






100
030
005






1
√
3
−
1
√
3
1
√
3
1
√
2
0
−
1
√
2
1
√
6
2
√
6
1
√
6



=



8
3
4
3
−
1
3
4
3
11
3
4
3
−
1
3
4
3
8
3



37.
Since
D
=
P
−
1
AP
we have
A
=
PDP
−
1
A
2
=
PDP
−
1
PDP
−
1
=
PDDP
−
1
=
PD
2
P
−
1
A
3
=
A
2
A
=
PD
2
P
−
1
PDP
−
1
=
PD
2
DP
−
1
=
PD
3
P
−
1
and so on.
38.





2
4
00 0
03
4
00
00(
−
1)
4
0
00 0 (5)
4





=





1600 0
0810 0
0010
0 0 0 625





39.
λ
1
=2,
λ
2
=
−
1,
K
1
=
2
1
1
3
,
K
2
=
2
−
1
2
3
,
P
=
2
1
−
1
12
3
,
P
−
1
=

2
3
1
3
−
1
3
1
3

A
5
=
2
1
−
1
12
32
32 0
0
−
1
3

2
3
1
3
−
1
3
1
3

=
2
21 11
22 10
3
40.
λ
1
=0,
λ
2
=1,
K
1
=
2
5
3
3
,
K
2
=
2
2
1
3
,
P
=
2
52
31
3
,
P
−
1
=
2
−
12
3
−
5
3
A
10
=
2
52
31
32
00
01
32
−
12
3
−
5
3
=
2
6
−
10
3
−
5
3
Exercises 8.13
1. (a)
The message is
M
=
2
19 514 4 0
8 512 16 0
3
. The encoded message is
B
=
AM
=
2
12
11
32
19 514 4 0
8 512 16 0
3
=
2
351538 36 0
27 10 26 20 0
3
.
(b)
The decoded message is
M
=
A
−
1
B
=
2
−
12
1
−
1
32
351538 36 0
27 10 26 20 0
3
=
2
19 514 4 0
8 512 16 0
3
.
361

362. Exercises 8.13
2. (a)
The message is
M
=
2
208 5 0131514525
091908 51850
3
. The encoded message is
B
=
AM
=
2
35
12
32
208 5 0131514525
091908 51850
3
=
2
60 69 110 0 79 70 132 40 75
20 26 43 0 29 2550 1525
3
.
(b)
The decoded message is
M
=
A
−
1
B
=
2
2
−
5
−
13
32
60 69 110 0 79 70 132 40 75
20 26 43 0 29 2550 1525
3
=
2
208 5 0131514525
091908 51850
3
.
3. (a)
The message is
M
=
2
16815145
0 8 1513 5
3
. The encoded message is
B
=
AM
=
2
35
23
32
16815145
0 8 1513 5
3
=
2
48 64 120 107 40
32 40 7567 25
3
.
(b)
The decoded message is
M
=
A
−
1
B
=
2
−
35
2
−
3
3
=
2
48 64 120 107 40
32 40 7567 25
3
=
2
16815145
0 8 1513 5
3
.
4. (a)
The message is
M
=



7150 14151820
8015140131
914019200 0



. The encoded message is
B
=
AM
=



123
112
012






7150 14151820
8015140131
914019200 0



=



5057309975 4422
33 43 1566 5531 21
26 28 1552 40 13 1



.
(b)
The decoded message is
M
=
A
−
1
B
=



01
−
1
2
−
2
−
1
−
111






5057309975 4422
33 43 1566 5531 21
26 28 1552 40 13 1



=



7150 14151820
8015140131
914019200 0



.
5. (a)
The message is
M
=



7150 14151820
8015140131
914019200 0



. The encoded message is
B
=
AM
=



211
111
−
110






7150 14151820
8015140131
914019200 0



=



31 44 1561 50 49 41
24 29 1547 35 31 21
1
−
1515 0
−
15
−
5
−
19



.
(b)
The decoded message is
M
=
A
−
1
B
=



1
−
10
1
−
11
−
23
−
1






31 44 1561 50 49 41
24 29 1547 35 31 21
1
−
1515 0
−
15
−
5
−
19



=



7150 14151820
8015140131
914019200 0



.
362

363. Exercises 8.13
6. (a)
The message is
M
=



4 18010158
140919020
8 50191625



. The encoded message is
B
=
AM
=



530
43
−
1
522






4 18010158
140919020
8 50191625



=



62 90 27 107 75100
50672778 44 67
64 100 18 126 107 130



.
(b)
The decoded message is
M
=
A
−
1
B
=



8
−
6
−
3
−
13 10 5
−
753






62 90 27 107 75100
50672778 44 67
64 100 18 126 107 130



=



4 18010158
140919020
8 50191625



.
7. (a)
The decoded message is
M
=
A
−
1
B
=
2
2
−
3
−
58
32
152 184 171 86 212
95116 107 56 133
3
=
2
19 20 21 4 25
081184
3
.
From correspondence (1) we obtain: STUDY
HARD.
8.
The decoded message is
M
=
A
−
1
B
=
2
1
−
1
1
−
2
32
46
−
7
−
13 22
−
18 1 10
23
−
15
−
14 2
−
18
−
12 5
3
=
2
23 8 1 20 0 13 5
0 23 1518 18 250
3
.
From correspondence (1) we obtain: WHAT
MEWORRY .
9.
The decoded message is
M
=
A
−
1
B
=



00 1
01 0
10
−
1






31 21 21 22 20 9
19 0 9 13 16 15
131208 0 9



=



131208 0 9
19 0 9 13 16 15
1820114200



.
From correspondence (1) we obtain: MATH
ISIMPORTANT.
10.
The decoded message is
M
=
A
−
1
B
=



10
−
1
−
112
0
−
10






36 32 28 61 26 56 10 12
−
9
−
2
−
18
−
1
−
18
−
250 0
23 27 23 41 26 43 512



=



135 52001350
1200208 5012
9 2181182500



.
From correspondence (1) we obtain: MEET
MEATTHELIBRARY .
11.
Let
A
−
1
=
2
uv
xy
3
. Then
A
−
1
B
=
2
uv
xy
32
17 16 18 534 0 34 20 9 525
−
30
−
31
−
32
−
10
−
590
−
54
−
35
−
13
−
6
−
50
3
,
so 17
u
−
30
v
=4,16
u
−
31
v
= 1 and 5
x
−
6
y
=1,25
x
−
50
y
= 25. Then
A
−
1
=
2
21
−
1
−
1
3
and
A
−
1
B
2
4 1 4090145540
131514525020154125
3
.
363

364. Exercises 8.13
From correspondence (1) we obtain: DAD
INEEDMONEY TODAY.
12. (a) M
T
=



22 8 192721 3 3 27211821
13 3 21 22 3 2527 6 7 14 23
22721727521172257



(b) B
T
=
M
=



11 0
10 1
11
−
1



=



37 386156 513351 5030 5751
24 3540 34 48 8 24 44 23 43 28
11
−
24015
−
24 20 6
−
11 5
−
11 16



(c) BA
−
1
=
B



−
111
2
−
1
−
1
10
−
1



=
M
13. (a) B
1
=



15222082362122
10 22 18 23 252 23 25
3 26261423162612



(b)
Using correspondence (1) the encoded message is: OVTHWFUVJVRWYBWYCZZNWPZL.
(c) M
1
A
−
1
B
1
=



14
−
3
23
−
2
−
2
−
43



B
1
=



46 32 14 58 54
−
34 3586
5458425775
−
14 59 95
−
61
−
54
−
34
−
66
−
77 28
−
56
−
108



M
=
M
1
mod 27 =



19 514 4 0 20 8 5
041532113514
20020154 1 250



.
Using correspondence (1) the encoded message is: SEND
THEDOCUMENT TODAY.
Exercises 8.14
1.
[0110]
2.
[1111]
3.
[00011]
4.
[10100]
5.
[10101001]
6.
[01101010]
7.
[1 0 0]
8.
[001]
9.
Parity error
10.
[1010]
11.
[10011]
12.
Parity error
364

365. Exercises 8.14
In Problems 13-18
D
=[
c
1
c
2
c
3
] and
P
=



1101
1011
0111



.
13. D
T
=
P
[1110]
T
=[000]
T
;
C
=[0010110]
14. D
T
=
P
[0011]
T
=[100]
T
;
C
=[1000011]
15. D
T
=
P
[0101]
T
=[010]
T
;
C
=[0100101]
16. D
T
=
P
[0001]
T
=[111]
T
;
C
=[1101001]
17. D
T
=
P
[0110]
T
=[110]
T
;
C
=[1100110]
18. D
T
=
P
[1100]
T
=[011]
T
;
C
=[0111100]
In Problems 19-28
W
represents the correctly decoded message.
19. S
=
HR
T
=
H
[0000000]=[000]
T
; a code word.
W
=[0000]
20. S
=
HR
T
=
H
[1100000]=[011]
T
; not a code word. The error is in the third bit.
W
=[1000]
21. S
=
HR
T
=
H
[1101101]=[101]
T
; not a code word. The error is in the fifth bit.
W
=[0001]
22. S
=
HR
T
=
H
[0101010]=[000]
T
; a code word.
W
=[0010]
23. S
=
HR
T
=
H
[1111111]=[000]
T
; a code word.
W
=[1111]
24. S
=
HR
T
=
H
[1100110]=[000]
T
; a code word.
W
=[0110]
25. S
=
HR
T
=
H
[0111001]=[010]
T
; not a code word. The error is in the second bit.
W
=[1001]
26. S
=
HR
T
=
H
[1001001]=[010]
T
; not a code word. The error is in the second bit.
W
=[0001]
27. S
=
HR
T
=
H
[1011011]=[111]
T
; not a code word. The error is in the seventh bit.
W
=[1010]
28. S
=
HR
T
=
H
[0010011]=[010]
T
; not a code word. The error is in the second bit.
W
=[1011]
29. (a)
2
7
= 128
(b)
2
4
=16
(c)
[0000000],[1101001],[0101010],[1000011],
[1001100],[0100101],[1100110],[0001111],
[1110000],[0011001],[1011010],[0110011],
[0111100],[1010101],[0010110],[1111111]
30. (a)
c
4
=0,
c
3
=1,
c
2
=1,
c
1
=0;[01100110]
(b) H
=





00001111
00110011
01010101
11111111





(c) S
=
HR
T
=
H
[00111100]
T
=[0000]
T
365

366. Exercises 8.15
Exercises 8.15
1.
We have
Y
T
=[1232] and
A
T
=
2
2345
1111
3
.
Now
A
T
A
=
2
5414
14 4
3
and (
A
T
A
)
−
1
=
1
20
2
4
−
14
−
14 54
3
so
X
=(
A
T
A
)
−
1
A
T
Y
=

2
5
3
5

and the least squares line is
y
=0
.
4
x
+0
.
6.
2.
We have
Y
T
=[
−
1357] and
A
T
=
2
0123
1111
3
.
Now
A
T
A
=
2
14 6
64
3
and (
A
T
A
)
−
1
=
1
20
2
4
−
6
−
614
3
so
X
=(
A
T
A
)
−
1
A
T
Y
=

13
5
−
2
5

and the least squares line is
y
=2
.
6
x
−
0
.
4.
3.
We have
Y
T
=[1 1
.
534
.
55] and
A
T
=
2
12345
11111
3
.
Now
A
T
A
=
2
55 15
155
3
and (
A
T
A
)
−
1
=
1
50
2
5
−
15
−
1555
3
so
X
=(
A
T
A
)
−
1
A
T
Y
=
2
1
.
1
−
0
.
3
3
and the least squares line is
y
=1
.
1
x
−
0
.
3.
4.
We have
Y
T
=[0 1
.
534
.
55] and
A
T
=
2
02345
11111
3
.
Now
A
T
A
=
2
5414
14 5
3
and (
A
T
A
)
−
1
=
1
74
2
5
−
14
−
14 54
3
so
X
=(
A
T
A
)
−
1
A
T
Y
=
2
1
.
06757
−
0
.
189189
3
and the least squares line is
y
=1
.
06757
x
−
0
.
189189.
5.
We have
Y
T
=[23559810] and
A
T
=
2
0123456
1111111
3
.
Now
A
T
A
=
2
91 21
21 7
3
and (
A
T
A
)
−
1
=
1
196
2
7
−
21
−
21 91
3
so
X
=(
A
T
A
)
−
1
A
T
Y
=

19
14
27
14

and the least squares line is
y
=1
.
35714
x
+1
.
92857.
6.
We have
Y
T
=[2 2
.
511
.
523
.
2 5] and
A
T
=
2
1234567
1111111
3
.
Now
A
T
A
=
2
140 28
28 7
3
and (
A
T
A
)
−
1
=
1
196
2
7
−
28
−
28 140
3
so
X
=(
A
T
A
)
−
1
A
T
Y
=
2
0
.
407143
0
.
828571
3
and the least squares line is
y
=0
.
407143
x
+0
.
828571.
366

367. Exercises 8.16
7.
We have
Y
T
= [ 220 200 180 170 150 135 ] and
A
T
=
2
20 40 60 80 100 120
1111 1 1
3
.
Now
A
T
A
=
2
36400 420
420 6
3
and (
A
T
A
)
−
1
=
1
42000
2
6
−
420
−
420 36400
3
so
X
=(
A
T
A
)
−
1
A
T
Y
=

−
117
140
703
3

and the least squares line is
v
=
−
0
.
835714
T
+ 234
.
333. At
T
= 140,
v
≈
117
.
333 and at
T
= 160,
v
≈
100
.
619.
8.
We have
Y
T
=[0
.
47 0
.
90 2
.
03
.
77
.
515] and
A
T
=
2
400 450 500 550 600 650
111111
3
.
Now
A
T
A
=
2
1697500 3150
3150 6
3
and (
A
T
A
)
−
1
=
1
262500
2
6
−
3150
−
3150 1697500
3
so
X
=(
A
T
A
)
−
1
A
T
Y
=
2
0
.
0538
−
23
.
3167
3
and the least squares line is
R
=0
.
0538
T
−
23
.
3167. At
T
= 700,
R
≈
14
.
3433.
Exercises 8.16
In Problems 1-5we use the fact that the element
τ
ij
in the transfer matrix
T
is the rate of transfer from compartment
j
to compartment
i
, and the fact that the sum of each column in
T
is 1.
1. (a)
The initial state and the transfer matrix are
X
0
=
2
90
60
3
and
T
=
2
0
.
80
.
4
0
.
20
.
6
3
.
(b)
We have
X
1
=
TX
0
=
2
0
.
80
.
4
0
.
20
.
6
32
90
60
3
=
2
96
54
3
and
X
2
=
TX
1
=
2
0
.
80
.
4
0
.
20
.
6
32
96
54
3
=
2
98
.
4
51
.
6
3
.
(c)
From
T
ˆ
X
−
ˆ
X
=(
T
−
I
)
ˆ
X
=
0
and the fact that the system is closed we obtain
−
0
.
2
x
1
+0
.
4
x
2
=0
x
1
+
x
2
=150
.
The solution is
x
1
= 100,
x
2
= 50, so the equilibrium state is
ˆ
X
=
2
100
50
3
.
2. (a)
The initial state and the transfer matrix are
X
0
=



100
200
150



and
T
=



0
.
700
.
5
0
.
30
.
80
00
.
20
.
5



.
367

368. Year
Bare Space Grasses Small Shrubs
0 10.00 0.00 0.00
1 7.00 3.00 0.00
2 5.05 4.35 0.60
3 3.84 4.78 1.38
4 3.14 4.74 2.13
5 2.75 4.49 2.76
6 2.56 4.19 3.24
Exercises 8.16
(b)
We have
X
1
=
TX
0
=



145
190
115



and
X
2
=
TX
1
=



159
195
.
5
95
.
5



.
(c)
From
T
ˆ
X
−
ˆ
X
=(
T
−
I
)
ˆ
X
=
0
and the fact that the system is closed we obtain
−
0
.
8
x
1
+0
.
5
x
2
=0
0
.
3
x
1
−
0
.
9
x
2
=0
x
1
+
x
2
+
x
3
=450
.
The solution is
x
1
= 145
.
161,
x
2
= 217
.
742,
x
3
=87
.
0968, so the equilibrium state is
ˆ
X
=



145
.
161
217
.
742
87
.
097



.
3. (a)
The initial state and the transfer matrix are
X
0
=



100
0
0



and
T
=



0
.
20
.
50
0
.
30
.
10
0
.
50
.
41



.
(b)
We have
X
1
=
TX
0
=



20
30
50



and
X
2
=
TX
1
=



19
9
72



.
(c)
From
T
ˆ
X
−
ˆ
X
=(
T
−
I
)
ˆ
X
=
0
and the fact that the system is closed we obtain
−
0
.
8
x
1
+0
.
5
x
2
=0
0
.
3
x
1
−
0
.
9
x
2
=0
x
1
+
x
2
+
x
3
= 100
.
The solution is
x
1
=
x
2
=0,
x
3
= 100, so the equilibrium state is
ˆ
X
=



0
0
100



.
4. (a)
The transfer matrix is
T
=



0
.
70
.
050
.
15
0
.
30
.
750
00
.
20
.
85



.
(b)
368

369. Year
Phytoplankton
Water
Zooplankton
0 0.00
100.00
0.00
1 2.00 97.00 1.00
2 3.70 94.26 2.04
3 5.14 91.76 3.10
4 6.36 89.47 4.17
5 7.39 87.37 5.24
6 8.25 85.46 6.30
7 8.97 83.70 7.33
8 9.56 82.10 8.34
9 10.06 80.62 9.32
10 10.46 79.28 10.26
11 10.79 78.04 11.17
12 11.06 76.90 12.04
Chapter 8 Review Exercises
5. (a)
The initial state and the transfer matrix are
X
0
=



0
100
0



and
T
=



0
.
88 0
.
02 0
0
.
06 0
.
97 0
.
05
0
.
06 0
.
01 0
.
95



.
(b)
6.
From
T
ˆ
X
=1
ˆ
X
we see that the equilibrium state vector
ˆ
X
is the eigenvector of the transfer matrix
T
corre-
sponding to the eigenvalue 1. It has the properties that its components add up to the sum of the components
of the initial state vector.
Chapter 8 Review Exercises
1.





234
345
456
567





2.
4
×
3
3. AB
=
2
34
68
3
;
BA
= [ 11 ]
4. A
−
1
=
−
1
2
2
4
−
2
−
31
3
=
2
−
21
3
2
−
1
2
3
5.
False; consider
A
=
2
10
01
3
and
B
=
2
01
10
3
6.
True
7.
det

1
2
A

=

1
2

3
(5) =
5
8
; det(
−
A
T
)=(
−
1)
3
(5) =
−
5
8.
det
AB
−
1
= det
A
/
det
B
=6
/
2=3
9.
0
10.
det
C
=(
−
1)
3
/
det
B
=
−
1
/
10
3
(2) =
−
1
/
2000
11.
False; an eigenvalue can be 0.
369

370. Chapter 8 Review Exercises
12.
True
13.
True
14.
True, since complex roots of real polynomials occur in conjugate pairs.
15.
False; if the characteristic equation of an
n
×
n
matrix has repeated roots, there may not be
n
linearly independent
eigenvectors.
16.
True
17.
True
18.
True
19.
False;
A
is singular and thus not orthogonal.
20.
True
21. A
=
1
2
(
A
+
A
T
)+
1
2
(
A
−
A
T
) where
1
2
(
A
+
A
T
) is symmetric and
1
2
(
A
−
A
T
) is skew-symmetric.
22.
Since det
A
2
= (det
A
)
2
≥
0 and det
2
01
10
3
=
−
1, there is no
A
such that
A
2
=
2
01
10
3
.
23. (a)
2
11
−
1
−
1
3
is nilpotent.
(b)
Since det
A
n
= (det
A
)
n
= 0 we see that det
A
= 0 and
A
is singular.
24. (a)
σ
x
σ
y
=
2
i
0
0
−
i
3
=
−
σ
y
σ
x
;
σ
x
σ
z
=
2
0
−
1
10
3
=
−
σ
z
σ
x
;
σ
y
σ
z
=
2
0
i
i
0
3
=
−
σ
z
σ
y
(b)
We first note that for anticommuting matrices
AB
=
−
BA
,so
C
=2
AB
. Then
C
xy
=
2
2
i
0
0
−
2
i
3
,
C
yz
=
2
02
i
2
i
0
3
, and
C
zx
=
2
02
−
20
3
.
25.



5
−
11
−
9
240
27
115
9



R
13
−
−
−
−
−
−→



115
9
240
27
5
−
11
−
9



row
−
−
−
−
−
−→
operations



11 5
9
01
−
5
9
2
00 1
1
2



row
−
−
−
−
−
−→
operations



100
−
1
2
010
7
001
1
2



.
The solution is
X
=[
−
1
2
7
1
2
]
T
.
26.



111
6
1
−
23
2
20
−
3
3



row
−
−
−
−
−
−→
operations



11 1
6
01
−
2
3
4
3
00 1
1



row
−
−
−
−
−
−→
operations



100
3
010
2
001
1



.
The solution is
x
1
=3,
x
2
=2,
x
3
=1.
27.
Multiplying the second row by
abc
we obtain the third row. Thus the determinant is 0.
28.
Expanding along the first row we see that the result is an expression of the form
ay
+
bx
2
+
cx
+
d
= 0, which
is a parabola since, in this case
a
3
= 0 and
b
3
= 0. Letting
x
= 1 and
y
= 2 we note that the first and second
rows are the same. Similarly, when
x
= 2 and
y
= 3, the first and third rows are the same; and when
x
=3
and
y
= 5, the first and fourth rows are the same. In each case the determinant is 0 and the points lie on the
parabola.
29.
4(
−
2)(3)(
−
1)(2)(5) = 240
30.
(
−
3)(6)(9)(1) =
−
162
370

371. Chapter 8 Review Exercises
31.
Since



1
−
11
51
−
1
121



= 18 = 0, the system has only the trivial solution.
32.
Since



1
−
1
−
1
51
−
1
121



= 0, the system has infinitely many solutions.
33.
From
x
1
I
2
+
x
2
HNO
3
→
x
3
HIO
3
+
x
4
NO
2
+
x
5
H
2
O we obtain the system 2
x
1
=
x
3
,
x
2
=
x
3
+2
x
5
,
x
2
=
x
4
,
3
x
2
=3
x
3
+2
x
4
+
x
5
. Letting
x
4
=
x
2
in the fourth equation we obtain
x
2
=3
x
3
+
x
5
. Taking
x
1
=
t
we see that
x
3
=2
t
,
x
2
=2
t
+2
x
5
, and
x
2
=6
t
+
x
5
. From the latter two equations we get
x
5
=4
t
. Taking
t
= 1 we have
x
1
=1,
x
2
= 10,
x
3
=2,
x
4
= 10, and
x
5
= 4. The balanced equation is I
2
+10HNO
3
→
2HIO
3
+10NO
2
+4H
2
O.
34.
From
x
1
Ca +
x
2
H
3
PO
4
→
x
3
Ca
3
P
2
O
8
+
x
4
H
2
we obtain the system
x
1
=3
x
3
,3
x
2
=2
x
4
,
x
2
=2
x
3
,4
x
2
=8
x
3
.
Letting
x
3
=
t
we see that
x
1
=3
t
,
x
2
=2
t
, and
x
4
=3
t
. Taking
t
= 1 we obtain the balanced equation
3Ca+2H
3
PO
4
→
Ca
3
P
2
O
8
+3H
2
.
35.
det
A
=
−
84, det
A
1
= 42, det
A
2
=
−
21, det
A
3
=
−
56;
x
1
=
42
−
84
=
−
1
2
,
x
2
=
−
21
−
84
=
1
4
,
x
3
=
−
56
−
84
=
2
3
36.
det = 4, det
A
1
= 16, det
A
2
=
−
4, det
A
3
=0;
x
1
=
16
4
=4,
x
2
=
−
4
4
=
−
1,
x
3
=
0
4
=0
37.
det
A
= cos
2
θ
+ sin
2
θ,
det
A
1
=
X
cos
θ
−
Y
sin
θ,
det
A
2
=
Y
cos
θ
+
X
sin
θ
;
x
1
=
X
cos
θ
−
Y
sin
θ
,
y
=
Y
cos
θ
+
X
sin
θ
38. (a)
i
1
−
i
2
−
i
3
−
i
4
=0,
i
2
R
1
=
E
,
i
2
R
1
−
i
3
R
2
=0,
i
3
R
2
−
i
4
R
3
=0
(b)
det
A
=





1
−
1
−
1
−
1
0
R
1
00
0
R
1
−
R
2
0
00
R
2
−
R
3





=
R
1
R
2
R
3
;
det
A
1
=





0
−
1
−
1
−
1
ER
1
00
0
R
1
−
R
2
0
00
R
2
−
R
3





=
−
E
[
−
R
2
R
3
−
R
1
(
R
3
+
R
2
)] =
E
(
R
2
R
3
+
R
1
R
3
+
R
1
R
2
);
i
1
=
det
A
1
det
A
=
E
(
R
2
R
3
+
R
1
R
3
+
R
1
R
2
)
R
1
R
2
R
3
=
E
r
1
R
1
+
1
R
2
+
1
R
3
<
39. AX
=
B
is



23
−
1
1
−
20
−
201






x
1
x
2
x
3



=



6
−
3
9



. Since
A
−
1
=
−
1
3



−
2
−
3
−
2
−
10
−
1
−
4
−
6
−
7



, we have
X
=
A
−
1
B
=



7
5
23



.
40. (a) A
−
1
B
=



3
2
−
1
4
−
9
4
−
1
1
2
3
2
1
2
−
1
4
−
1
4






1
1
1



=



−
1
1
0



(b) A
−
1
B
=



3
2
−
1
4
−
9
4
−
1
1
2
3
2
1
2
−
1
4
−
1
4






−
2
1
3



=



−
10
7
−
2



371

372. Chapter 8 Review Exercises
41.
From the characteristic equation
λ
2
−
4
λ
−
5= 0 we see that the eigenvalues are
λ
1
=
−
1 and
λ
2
=5. For
λ
1
=
−
1 we have 2
k
1
+2
k
2
=0,4
k
1
4
k
2
= 0 and
K
1
=
2
−
1
1
3
.For
λ
2
= 5we have
−
4
k
1
+2
k
2
=0,4
k
1
−
2
k
2
=0
and
K
2
=
2
1
2
3
.
42.
From the characteristic equation
λ
2
= 0 we see that the eigenvalues are
λ
1
=
λ
2
=0. For
λ
1
=
λ
2
=0wehave
4
k
1
= 0 and
K
1
=
2
0
1
3
is a single eigenvector.
43.
From the characteristic equation
−
λ
3
+6
λ
2
+15
λ
+8=
−
(
λ
+1)
2
(
λ
−
8) = 0 we see that the eigenvalues are
λ
1
=
λ
2
=
−
1 and
λ
3
=8. For
λ
1
=
λ
2
=
−
1wehave



424
0
212
0
424
0



row
−
−
−
−
−
−→
operations



1
1
2
1
0
000
0
000
0



.
Thus
K
1
=[1
−
20]
T
and
K
2
=[1 0
−
1]
T
.For
λ
3
= 8 we have



−
524
0
2
−
82
0
42
−
5
0



row
−
−
−
−
−
−→
operations



1
−
2
5
−
4
5
0
01
−
1
2
0
000
0



.
Thus
K
3
=[212]
T
.
44.
From the characteristic equation
−
λ
3
+18
λ
2
−
99
λ
+162 =
−
(
λ
−
9)(
λ
−
6)(
λ
−
3) = 0 we see that the eigenvalues
are
λ
1
=9,
λ
2
= 6, and
λ
3
=3. For
λ
1
= 9 we have



−
2
−
20
0
−
2
−
32
0
02
−
4
0



row
−
−
−
−
−
−→
operations



11 0
0
01
−
2
0
00 0
0



.
Thus
K
1
=[
−
221]
T
.For
λ
2
= 6 we have



1
−
20
0
−
202
0
02
−
1
0



row
−
−
−
−
−
−→
operations



1
−
20
0
01
−
1
2
0
00 0
0



.
Thus
K
2
=[212]
T
.For
λ
3
= 3 we have



4
−
20
0
−
232
0
022
0



row
−
−
−
−
−
−→
operations



1
−
1
2
0
0
011
0
000
0



.
Thus
K
3
=[1 2
−
2]
T
.
45.
From the characteristic equation
−
λ
3
−
λ
2
+21
λ
+45=
−
(
λ
+3)
2
(
λ
−
5) = 0 we see that the eigenvalues are
λ
1
=
λ
2
=
−
3 and
λ
3
=5. For
λ
1
=
λ
2
=
−
3wehave



12
−
3
0
24
−
6
0
−
1
−
23
0



row
−
−
−
−
−
−→
operations



12
−
3
0
00 0
0
00 0
0



.
Thus
K
1
=[
−
210]
T
and
K
2
=[301]
T
.For
λ
3
= 5we have



−
72
−
3
0
2
−
4
−
6
0
−
1
−
2
−
5
0



row
−
−
−
−
−
−→
operations



1
−
2
7
3
7
0
012
0
000
0



.
372

373. Chapter 8 Review Exercises
Thus
K
3
=[
−
1
−
21]
T
.
46.
From the characteristic equation
−
λ
3
+
λ
2
+2
λ
=
−
λ
(
λ
+ 1)(
λ
−
2) = 0 we see that the eigenvalues are
λ
1
=0,
λ
2
=
−
1, and
λ
3
=2. For
λ
1
= 0 we have
k
3
=0,2
k
1
+2
k
2
+
k
3
= 0 and
K
1
=[1
−
10]
T
.For
λ
2
=
−
1we
have



100
0
011
0
222
0



row
−
−
−
−
−
−→
operations



100
0
011
0
000
0



.
Thus
K
2
=[0 1
−
1]
T
.For
λ
3
= 2 we have



−
200
0
0
−
21
0
22
−
1
0



row
−
−
−
−
−
−→
operations



10 0
0
01
−
1
2
0
00 0
0



.
Thus
K
3
=[012]
T
.
47.
Let
X
1
=[
abc
]
T
be the first column of the matrix. Then
X
T
1
[
−
1
√
2
0
1
√
2
]
T
=
1
√
2
(
c
−
a
) = 0 and
X
T
1
[
1
√
3
1
√
3
1
√
3
]
T
=
1
√
3
(
a
+
b
+
c
) = 0. Also
X
T
1
X
1
=
a
2
+
b
2
+
c
2
= 1. We see that
c
=
a
and
b
=
−
2
a
from
the first two equations. Then
a
2
+4
a
2
+
a
2
=6
a
2
= 1 and
a
=
1
√
6
.Thus
X
1
=[
1
√
6
−
2
√
6
1
√
6
]
T
.
48. (a)
Eigenvalues are
λ
1
=
λ
2
= 0 and
λ
3
= 5with corresponding eigenvectors
K
1
=[010]
T
,
K
2
=
[2 0 1]
T
, and
K
3
=[
−
102]
T
. Since

K
1

=1,

K
2

=
√
5, and

K
3

=
√
5, we have
P
=



0
2
√
5
−
1
√
5
10 0
0
1
√
5
2
√
5



and
P
−
1
=
P
T
=



010
2
√
5
0
1
√
5
−
1
√
5
0
2
√
5



.
(b) P
−
1
AP
=



000
000
005



49.
We identify
A
=
2
1
3
2
3
2
1
3
. Eigenvalues are
λ
1
=
−
1
2
and
λ
2
=
5
2
so
D
=
2
−
1
2
0
0
5
2
3
and the equation becomes
[
XY
]
D
2
X
Y
3
=
−
1
2
X
2
+
5
2
Y
2
= 1. The graph is a hyperbola.
50.
We measure years in units of 10, with 0 corresponding to 1890. Then
Y
= [ 63 76 92 106 123 ]
T
and
A
=
2
01234
11111
3
T
,so
A
T
A
=
2
30 10
10 5
3
.Thus
X
=(
A
T
A
)
−
1
A
T
Y
=
1
50
2
5
−
10
−
10 30
3
A
T
Y
=
2
15
62
3
,
and the least squares line is
y
=15
t
+ 62. At
t
= 5(corresponding to 1940) we have
y
= 137. The error in the
predicted population is 5million or 3
.
7%.
51.
The encoded message is
B
=
AM
=
2
10 1
91
32
1912051212 9 205012121
143 8 54 0 151406189 0
3
=
2
204 13 208 55124 120 105214 50 6 138 19 210
18512 188 50 112 108 96 194 456 126 18 189
3
.
373

374. Chapter 8 Review Exercises
52.
The encoded message is
B
=
AM
=
2
10 1
91
32
19 53 0 1 7 14200118
18221902021 5190113
3
=
2
208 72 49 0 30 91 145219 0 11 193
189 67 46 0 29 84 131 199 0 10 175
3
.
53.
The decoded message is
M
=
A
−
1
B
=



−
32
−
1
100
2
−
11






19 0 1514 0 20
35 10 27 53 1 54
515
−
348239



=



8 512 16 0 9
1901514020
85023125



.
From correspondence (1) we obtain: HELP
ISONTHEWAY.
54.
The decoded message is
M
=
A
−
1
B
=



−
32
−
1
100
2
−
11






5221
27 17 40
21 13
−
2



=



18 1519
5221
400



.
From correspondence (1) we obtain: ROSEBUD
.
55. (a)
The parity is even so the decoded message is [
11001]
(b)
The parity is odd; there is a parity error.
56.
From



c
1
c
2
c
3






1101
1011
0111








1
0
0
1





=



0
0
1



we obtain the codeword [
0011001].
374

375. 9
Vector Calculus
Exercises 9.1
1. 2. 3.
4. 5. 6.
7. 8.
9.
Note: the scale is distorted in this graph. For
t
= 0, the graph starts at (1
,
0
,
1). The upper loop shown
intersects the
xz
-plane at about (286751
,
0
,
286751).
10.
375

376. Exercises 9.1
11.
x
=
t
,
y
=
t
,
z
=
t
2
+
t
2
=2
t
2
;
r
(
t
)=
t
i
+
t
j
+2
t
2
k
12.
x
=
t
,
y
=2
t
,
z
=
±
√
t
2
+4
t
2
+1=
±
√
5
t
2
−
1;
r
(
t
)=
t
i
+2
t
j
±
√
5
t
2
−
1
k
13.
x
= 3 cos
t
,
z
=9
−
9 cos
2
t
= 9 sin
2
t
,
y
= 3 sin
t
;
r
(
t
) = 3 cos
t
i
+ 3 sin
t
j
+ 9 sin
2
t
k
14.
x
= sin
t
,
z
=1,
y
= cos
t
;
r
(
t
) = sin
t
i
+ cos
t
j
+
k
15. r
(
t
)=
sin 2
t
t
i
+(
t
−
2)
5
j
+
ln
t
1
/t
k
. Using L’Hˆ
opital’s Rule,
lim
t
→
0
+
r
(
t
)=
N
2 cos 2
t
1
i
+(
t
−
2)
5
j
+
1
/t
−
1
/t
2
k
o
=2
i
−
32
j
.
16. (a)
lim
t
→
α
[
−
4
r
1
(
t
)+3
r
2
(
t
)] =
−
4(
i
−
2
j
+
k
) + 3(2
i
+5
j
+7
k
)=2
i
+23
j
+17
k
(b)
lim
t
→
α
r
1
(
t
)
·
r
2
(
t
)=(
i
−
2
j
+
k
)
·
(2
i
+5
j
+7
k
)=
−
1
17. r
N
(
t
)=
1
t
i
−
1
t
2
j
;
r
NN
(
t
)=
−
1
t
2
i
+
2
t
3
j
18. r
N
(
t
)=
eo
t
sin
t,
1
−
sin
t
:
;
r
NN
(
t
)=
eo
t
cos
t
−
sin
t,
−
cos
t
:
376

377. Exercises 9.1
19. r
N
(
t
)=
e
2
te
2
t
+
e
2
t
,
3
t
2
,
8
t
−
1
:
;
r
NN
(
t
)=
e
4
te
2
t
+4
e
2
t
,
6
t,
8
:
20. r
N
(
t
)=2
t
i
+3
t
2
j
+
1
1+
t
2
k
;
r
NN
(
t
)=2
i
+6
t
j
−
2
t
(1 +
t
2
)
2
k
21. r
N
(
t
)=
−
2 sin
t
i
+ 6 cos
t
j
r
N
(
π/
6) =
−
i
+3
√
3
j
22. r
N
(
t
)=3
t
2
i
+2
t
j
r
N
(
−
1) = 3
i
−
2
j
23. r
N
(
t
)=
j
−
8
t
(1 +
t
2
)
2
k
r
N
(1) =
j
−
2
k
24. r
N
(
t
)=
−
3 sin
t
i
+ 3 cos
t
j
+2
k
r
N
(
π/
4) =
−
3
√
2
2
i
+
3
√
2
2
j
+2
k
25. r
(
t
)=
t
i
+
1
2
t
2
j
+
1
3
t
3
k
;
r
(2) = 2
i
+2
j
+
8
3
k
;
r
N
(
t
)=
i
+
t
j
+
t
2
k
;
r
N
(2) =
i
+2
j
+4
k
Using the point (2
,
2
,
8
/
3) and the direction vector
r
N
(2), we have
x
=2+
t
,
y
=2+2
t
,
z
=8
/
3+4
t
.
26. r
(
t
)=(
t
3
−
t
)
i
+
6
t
t
+1
j
+(2
t
+1)
2
k
;
r
(1) = 3
j
+9
k
;
r
N
(
t
)=(3
t
2
−
1)
i
+
6
(
t
+1)
2
j
+(8
t
+4)
k
;
r
N
(1) = 2
i
+
3
2
j
+12
k
.
Using the point (0
,
3
,
9) and the direction vector
r
N
(1), we have
x
=2
t
,
y
=3+
3
2
t
,
z
=9+12
t
.
27.
d
dt
[
r
(
t
)
×
r
N
(
t
)] =
r
(
t
)
×
r
NN
(
t
)+
r
N
(
t
)
×
r
N
(
t
)=
r
(
t
)
×
r
NN
(
t
)
28.
d
dt
[
r
(
t
)
·
(
t
r
(
t
))] =
r
(
t
)
·
d
dt
(
t
r
(
t
)) +
r
N
(
t
)
·
(
t
r
(
t
)) =
r
(
t
)
·
(
t
r
N
(
t
)+
r
(
t
)) +
r
N
(
t
)
·
(
t
r
(
t
))
=
r
(
t
)
·
(
t
r
N
(
t
)) +
r
(
t
)
·
r
(
t
)+
r
N
(
t
)
·
(
t
r
(
t
)) = 2
t
(
r
(
t
)
·
r
N
(
t
)) +
r
(
t
)
·
r
(
t
)
29.
d
dt
[
r
(
t
)
·
(
r
N
(
t
)
×
r
NN
(
t
))] =
r
(
t
)
·
d
dt
(
r
N
(
t
)
×
r
NN
(
t
)) +
r
N
(
t
)
·
(
r
N
(
t
)
×
r
NN
(
t
))
=
r
(
t
)
·
(
r
N
(
t
)
×
r
NNN
(
t
)+
r
NN
(
t
)
×
r
NN
(
t
)) +
r
N
(
t
)
·
(
r
N
(
t
)
×
r
NN
(
t
))
=
r
(
t
)
·
(
r
N
(
t
)
×
r
NNN
(
t
))
30.
d
dt
[
r
1
(
t
)
×
(
r
2
(
t
)
×
r
3
(
t
))] =
r
1
(
t
)
×
d
dt
(
r
2
(
t
)
×
r
3
(
t
)) +
r
N
1
(
t
)
×
(
r
2
(
t
)
×
r
3
(
t
))
=
r
1
(
t
)
×
(
r
2
(
t
)
×
r
N
3
(
t
)+
r
N
2
(
t
)
×
r
3
(
t
)) +
r
N
1
(
t
)
×
(
r
2
(
t
)
×
r
3
(
t
))
=
r
1
(
t
)
×
(
r
2
(
t
)
×
r
N
3
(
t
)) +
r
1
(
t
)
×
(
r
N
2
(
t
)
×
r
3
(
t
)) +
r
1
(
t
)
×
(
r
2
(
t
)
×
r
3
(
t
))
31.
d
dt
t
r
1
(2
t
)+
r
2
e
1
t
:h
=2
r
N
1
(2
t
)
−
1
t
2
r
N
2
e
1
t
:
377

378. Exercises 9.1
32.
d
dt
[
t
3
r
(
t
2
)] =
t
3
(2
t
)
r
N
(
t
2
)+3
t
2
r
(
t
2
)=2
t
4
r
N
(
t
2
)+3
t
2
r
(
t
2
)
33.
s
2
−
1
r
(
t
)
dt
=
N
s
2
−
1
tdt
o
i
+
N
s
2
−
1
3
t
2
dt
o
j
+
N
s
2
−
1
4
t
3
dt
o
k
=
1
2
t
2



2
−
1
i
+
t
3



2
−
1
j
+
t
4



2
−
1
k
=
3
2
i
+9
j
+15
k
34.
s
4
0
r
(
t
)
dt
=
N
s
4
0
√
2
t
+1
dt
o
i
+
N
s
4
0
−
√
tdt
o
j
+
N
s
4
0
sin
πt dt
o
k
=
1
3
(2
t
+1)
3
/
2




4
0
i
−
2
3
t
3
/
2




4
0
j
−
1
π
cos
πt




4
0
k
=
26
3
i
−
16
3
j
35.
s
r
(
t
)
dt
=
N
s
te
t
dt
o
i
+
N
s
−
e
−
2
t
dt
o
j
+
N
s
te
t
2
dt
o
k
=[
te
t
−
e
t
+
c
1
]
i
+
t
1
2
e
−
2
t
+
c
2
h
j
+
t
1
2
e
t
2
+
c
3
h
k
=
e
t
(
t
−
1)
i
+
1
2
e
−
2
t
j
+
1
2
e
t
2
k
+
c
,
where
c
=
c
1
i
+
c
2
j
+
c
3
k
.
36.
s
r
(
t
)
dt
=
N
s
1
1+
t
2
dt
o
i
+
N
s
t
1+
t
2
dt
o
j
+
N
s
t
2
1+
t
2
dt
o
k
= [tan
−
1
t
+
c
1
]
i
+
t
1
2
ln(1 +
t
2
)+
c
2
h
j
+
N
s

1
−
1
1+
t
2

dt
o
k
= [tan
−
1
t
+
c
1
]
i
+
t
1
2
ln(1 +
t
2
)+
c
2
h
j
+[
t
−
tan
−
1
t
+
c
3
]
k
= tan
−
1
t
i
+
1
2
ln(1 +
t
2
)
j
+(
t
−
tan
−
1
t
)
k
+
c
,
where
c
=
c
1
i
+
c
2
j
+
c
3
k
.
37. r
(
t
)=
s
r
N
(
t
)
dt
=
N
s
6
dt
o
i
+
N
s
6
tdt
o
j
+
N
s
3
t
2
dt
o
k
=[6
t
+
c
1
]
i
+[3
t
2
+
c
2
]
j
+[
t
3
+
c
3
]
k
Since
r
(0) =
i
−
2
j
+
k
=
c
1
i
+
c
2
j
+
c
3
k
,
c
1
=1,
c
2
=
−
2, and
c
3
= 1. Thus,
r
(
t
)=(6
t
+1)
i
+(3
t
2
−
2)
j
+(
t
3
+1)
k
.
38. r
(
t
)=
s
r
N
(
t
)
dt
=
N
s
t
sin
t
2
dt
o
i
+
N
s
−
cos 2
tdt
o
j
=
−
[
1
2
cos
t
2
+
c
1
]
i
+[
−
1
2
sin 2
t
+
c
2
]
j
Since
r
(0) =
3
2
i
=(
−
1
2
+
c
1
)
i
+
c
2
j
,
c
1
= 2 and
c
2
= 0. Thus,
r
(
t
)=
e
−
1
2
cos
t
2
+2
:
i
−
1
2
sin 2
t
j
.
39. r
N
(
t
)=
s
r
NN
(
t
)
dt
=
N
s
12
tdt
o
i
+
N
s
−
3
t
−
1
/
2
dt
o
j
+
N
s
2
dt
o
k
=[6
t
2
+
c
1
]
i
+[
−
6
t
1
/
2
+
c
2
]
j
+[2
t
+
c
3
]
k
Since
r
N
(1) =
j
=(6+
c
1
)
i
+(
−
6+
c
2
)
j
+(2+
c
3
)
k
,
c
1
=
−
6,
c
2
= 7, and
c
3
=
−
2. Thus,
r
N
(
t
)=(6
t
2
−
6)
i
+(
−
6
t
1
/
2
+7)
j
+(2
t
−
2)
k
.
r
(
t
)=
s
r
N
(
t
)
dt
=
N
s
(6
t
2
−
6)
dt
o
i
+
N
s
(
−
6
t
1
/
2
+7)
dt
o
j
+
N
s
(2
t
−
2)
dt
o
k
=[2
t
3
−
6
t
+
c
4
]
i
+[
−
4
t
3
/
2
+7
t
+
c
5
]
j
+[
t
2
−
2
t
+
c
6
]
k
.
Since
r
(1) = 2
i
−
k
=(
−
4+
c
4
)
i
+(3+
c
5
)
j
+(
−
1+
c
6
)
k
,
c
4
=6,
c
5
=
−
3, and
c
6
= 0. Thus,
r
(
t
)=(2
t
3
−
6
t
+6)
i
+(
−
4
t
3
/
2
+7
t
−
3)
j
+(
t
2
−
2
t
)
k
.
378

379. Exercises 9.1
40. r
N
(
t
)=
s
r
NN
(
t
)
dt
=
N
s
sec
2
tdt
o
i
+
N
s
cos
tdt
o
j
+
N
s
−
sin
tdt
o
k
= [tan
t
+
c
1
]
i
+ [sin
t
+
c
2
]
j
+ [cos
t
+
c
3
]
k
Since
r
N
(0) =
i
+
j
+
k
=
c
1
i
+
c
2
j
+(1+
c
3
)
k
,
c
1
=1,
c
2
= 1, and
c
3
= 0. Thus,
r
N
(
t
) = (tan
t
+1)
i
+ (sin
t
+1)
j
+ cos
t
k
.
r
(
t
)=
s
r
N
(
t
)
dt
=
N
s
(tan
t
+1)
dt
o
i
+
N
s
(sin
t
+1)
dt
o
j
+
N
s
cos
tdt
o
k
= [ln
|
sec
t
|
+
t
+
c
4
]
i
+[
−
cos
t
+
t
+
c
5
]
j
+ [sin
t
+
c
6
]
k
.
Since
r
(0) =
−
j
+5
k
=
c
4
i
+(
−
1+
c
5
)
j
+
c
6
k
,
c
4
=0,
c
5
= 0, and
c
6
= 5. Thus,
r
(
t
) = (ln
|
sec
t
|
+
t
)
i
+(
−
cos
t
+
t
)
j
+ (sin
t
+5)
k
.
41. r
N
(
t
)=
−
a
sin
t
i
+
a
cos
t
j
+
c
k
;

r
N
(
t
)

=

(
−
a
sin
t
)
2
+(
a
cos
t
)
2
+
c
2
=
√
a
2
+
c
2
s
=
s
2
π
0

a
2
+
c
2
dt
=

a
2
+
c
2
t



2
π
0
=2
π

a
2
+
c
2
42. r
N
(
t
)=
i
+ (cos
t
−
t
sin
t
)
j
+ (sin
t
+
t
cos
t
)
k

r
N
(
t
)

=

1
2
+ (cos
t
−
t
sin
t
)
2
+ (sin
t
+
t
cos
t
)
2
=
√
2+
t
2
s
=
s
π
0

2+
t
2
dt
=

t
2

2+
t
2
+ln



t
+

2+
t
2








π
0
=
π
2

2+
π
2
+ln(
π
+

2+
π
2
)
−
ln
√
2
43. r
N
(
t
)=(
−
2
e
t
sin 2
t
+
e
t
cos 2
t
)
i
+(2
e
t
cos 2
t
+
e
t
sin 2
t
)
j
+
e
t
k

r
N
(
t
)

=

5
e
2
t
cos
2
2
t
+5
e
2
t
sin
2
2
t
+
e
2
t
=
√
6
e
2
t
=
√
6
e
t
s
=
s
3
π
0
√
6
e
t
dt
=
√
6
e
t



3
π
0
=
√
6(
e
3
π
−
1)
44. r
N
(
t
)=3
i
+2
√
3
t
j
+2
t
2
k
;

r
N
(
t
)

=

3
2
+(2
√
3
t
)
2
+(2
t
2
)
2
=
√
9+12
t
2
+4
t
4
=3+2
t
2
s
=
s
1
0
(3+2
t
2
)
dt
=(3
t
+
2
3
t
3
)




1
0
=3+
2
3
=
11
3
45. r
N
(
t
)=
−
a
sin
t
i
+
a
cos
t
j
;

r
N
(
t
)

=

a
2
sin
2
t
+
a
2
cos
2
t
=
a
,
a>
0;
s
=
s
t
0
adu
=
at
r
(
s
)=
a
cos(
s/a
)
i
+
a
sin(
s/a
)
j
;
r
N
(
s
)=
−
sin(
s/a
)
i
+ cos(
s/a
)
j

r
N
(
s
)

=

sin
2
(
s/a
) + cos
2
(
s/a
)=1
46. r
N
(
s
)=
−
2
√
5
sin(
s/
√
5)
i
+
2
√
5
cos(
s/
√
5)
j
+
1
√
5
k

r
N
(
s
)

=
r
4
5
sin
2
(
s/
√
5)+
4
5
cos
2
(
s/
√
5)+
1
5
=
r
4
5
+
1
5
=1
47.
Since
d
dt
(
r
·
r
)=
d
dt

r

2
=
d
dt
c
2
= 0 and
d
dt
(
r
·
r
)=
r
·
r
N
+
r
N
·
r
=2
r
·
r
N
, we have
r
·
r
N
= 0. Thus,
r
N
is perpendicular to
r
.
48.
Since

r
(
t
)

is the length of
r
(
t
),

r
(
t
)

=
c
represents a curve lying on a sphere of radius
c
centered at the
origin.
379

380. Exercises 9.1
49.
Let
r
1
(
t
)=
x
(
t
)
i
+
y
(
t
)
j
. Then
d
dt
[
u
(
t
)
r
1
(
t
)] =
d
dt
[
u
(
t
)
x
(
t
)
i
+
u
(
t
)
y
(
t
)
j
]=[
u
(
t
)
x
N
(
t
)+
u
N
(
t
)
x
(
t
)]
i
+[
u
(
t
)
y
N
(
t
)+
u
N
(
t
)
y
(
t
)]
j
=
u
(
t
)[
x
N
(
t
)
i
+
y
N
(
t
)
j
]+
u
N
(
t
)[
x
(
t
)
i
+
y
(
t
)
j
]=
u
(
t
)
r
N
1
(
t
)+
u
N
(
t
)
r
1
(
t
)
.
50.
Let
r
1
(
t
)=
x
1
(
t
)
i
+
y
1
(
t
)
j
and
r
2
(
t
)=
x
2
(
t
)
i
+
y
2
(
t
)
j
. Then
d
dt
[
r
1
(
t
)
·
r
2
(
t
)] =
d
dt
[
x
1
(
t
)
x
2
(
t
)+
y
1
(
t
)
y
2
(
t
)] =
x
1
(
t
)
x
N
2
(
t
)+
x
N
1
(
t
)
x
2
(
t
)+
y
1
(
t
)
y
N
2
(
t
)+
y
N
1
(
t
)
y
2
(
t
)
=[
x
1
(
t
)
x
N
2
(
t
)+
y
1
(
t
)
y
N
2
(
t
)]+[
x
N
1
(
t
)
x
2
(
t
)+
y
N
1
(
t
)
y
2
(
t
)] =
r
1
(
t
)
·
r
N
2
(
t
)+
r
N
1
(
t
)
·
r
2
(
t
)
.
51.
d
dt
[
r
1
(
t
)
×
r
2
(
t
)] = lim
h
→
0
r
1
(
t
+
h
)
×
r
2
(
t
+
h
)
−
r
1
(
t
)
×
r
2
(
t
)
h
= lim
h
→
0
r
1
(
t
+
h
)
×
r
2
(
t
+
h
)
−
r
1
(
t
+
h
)
×
r
2
(
t
)+
r
1
(
t
+
h
)
×
r
2
(
t
)
−
r
1
(
t
)
×
r
2
(
t
)
h
= lim
h
→
0
r
1
(
t
+
h
)
×
[
r
2
(
t
+
h
)
−
r
2
(
t
)]
h
+ lim
h
→
0
[
r
1
(
t
+
h
)
−
r
1
(
t
)]
×
r
2
(
t
)
h
=
r
1
(
t
)
×

lim
h
→
0
r
2
(
t
+
h
)
−
r
2
(
t
)
h

+

lim
h
→
0
r
1
(
t
+
h
)
−
r
1
(
t
)
h

×
r
2
(
t
)
=
r
1
(
t
)
×
r
N
2
(
t
)+
r
N
1
(
t
)
×
r
2
(
t
)
52.
Let
v
=
a
i
+
b
j
and
r
(
t
)=
x
(
t
)
i
+
y
(
t
)
j
. Then
s
b
a
v
·
r
(
t
)
dt
=
s
b
a
[
ax
(
t
)+
by
(
t
)]
dt
=
a
s
b
a
x
(
t
)
dt
+
b
s
b
a
y
(
t
)
dt
=
v
·
s
b
a
r
(
t
)
dt.
Exercises 9.2
1. v
(
t
)=2
t
i
+
t
3
j
;
v
(1) = 2
i
+
j
;

v
(1)

=
√
4+1=
√
5;
a
(
t
)=2
i
+3
t
2
j
;
a
(1) = 2
i
+3
j
2. v
(
t
)=2
t
i
−
2
t
3
j
;
v
(1) = 2
i
−
2
j
;

v
(1)

=
√
4+4=2
√
2;
a
(
t
)=2
i
+
6
t
4
j
;
a
(1) = 2
i
+6
j
3. v
(
t
)=
−
2 sinh 2
t
i
+2 cosh 2
t
j
;
v
(0) = 2
j
;

v
(0)

=2;
a
(
t
)=
−
4 cosh 2
t
i
+ 4 sinh 2
t
j
;
a
(0) =
−
4
i
4. v
(
t
)=
−
2 sin
t
i
+ cos
t
j
;
v
(
π/
3) =
−
√
3
i
+
1
2
j
;

v
(
π/
3)

=

3+1
/
4=
√
13
/
2;
a
(
t
)=
−
2 cos
t
i
−
sin
t
j
;
a
(
π/
3) =
−
i
−
√
3
2
j
380

381. Exercises 9.2
5. v
(
t
)=(2
t
−
2)
j
+
k
;
v
(2) = 2
j
+
k

v
(2)

=
√
4+1=
√
5;
a
(
t
)=2
j
;
a
(2) = 2
j
6. v
(
t
)=
i
+
j
+3
t
2
k
;
v
(2) =
i
+
j
+12
k
;

v
(2)

=
√
1+1+144=
√
146 ;
a
(
t
)=6
t
k
;
a
(2) = 12
k
7. v
(
t
)=
i
+2
t
j
+3
t
2
k
;
v
(1) =
i
+2
j
+3
k
;

v
(1)

=
√
1+4+9=
√
14 ;
a
(
t
)=2
j
+6
t
k
;
a
(1) = 2
j
+6
k
8. v
(
t
)=
i
+3
t
2
j
+
k
;
v
(1) =
i
+3
j
+
k
;

v
(1)

=
√
1+9+1=
√
11 ;
a
(
t
)=6
t
j
;
a
(1) = 6
j
9.
The particle passes through the
xy
-plane when
z
(
t
)=
t
2
−
5
t
=0or
t
= 0, 5 which gives us the points (0
,
0
,
0)
and (25
,
115
,
0).
v
(
t
)=2
t
i
+(3
t
2
−
2)
j
+(2
t
−
5)
k
;
v
(0) =
−
2
j
−
5
k
,
v
(5) = 10
i
+73
j
+5
k
;
a
(
t
)=2
i
+6
t
j
+2
k
;
a
(0) = 2
i
+2
k
,
a
(5) = 2
i
+30
j
+2
k
10.
If
a
(
t
)=
0
, then
v
(
t
)=
c
1
and
r
(
t
)=
c
1
t
+
c
2
. The graph of this equation is a straight line.
11.
Initially we are given
s
0
=
0
and
v
0
= (480 cos 30
◦
)
i
+ (480 sin 30
◦
)
j
= 240
√
3
i
+ 240
j
. Using
a
(
t
)=
−
32
j
we
find
v
(
t
)=
s
a
(
t
)
dt
=
−
32
t
j
+
c
240
√
3
i
+ 240
j
=
v
(0) =
c
v
(
t
)=
−
32
t
j
+ 240
√
3
i
+ 240
j
= 240
√
3
i
+ (240
−
32
t
)
j
r
(
t
)=
s
v
(
t
)
dt
= 240
√
3
t
i
+ (240
t
−
16
t
2
)
j
+
b
0
=
r
(0) =
b
.
(a)
The shell’s trajectory is given by
r
(
t
) = 240
√
3
t
i
+ (240
t
−
16
t
2
)
j
or
x
= 240
√
3
t
,
y
= 240
t
−
16
t
2
.
(b)
Solving
dy/dt
= 240
−
32
t
= 0, we see that
y
is maximum when
t
=15
/
2. The maximum
altitude is
y
(15
/
2) = 900 ft.
(c)
Solving
y
(
t
) = 240
t
−
16
t
2
=16
t
(15
−
t
) = 0, we see that the shell is at ground level when
t
= 0 and
t
= 15. The range of the shell is
x
(15) = 3600
√
3
≈
6235 ft.
381

382. Exercises 9.2
(d)
From
(c)
, impact is when
t
= 15. The speed at impact is

v
(15)

=
|
240
√
3
i
+ (240
−
32
·
15)
j
|
=

240
2
·
3+(
−
240)
2
= 480 ft/s.
12.
Initially we are given
s
0
= 1600
j
and
v
0
= (480 cos 30
◦
)
i
+ (480 sin 30
◦
)
j
= 240
√
3
i
+ 240
j
. Using
a
(
t
)=
−
32
j
we find
v
(
t
)=
s
a
(
t
)
dt
=
−
32
t
j
+
c
240
√
3
i
+ 240
j
=
v
(0) =
c
v
(
t
)=
−
32
t
j
+ 240
√
3
i
+ 240
j
= 240
√
3
i
+ (240
−
32
t
)
j
r
(
t
)=
s
v
(
t
)
dt
= 240
√
3
t
i
+ (240
t
−
16
t
2
)
j
+
b
1600
j
=
r
(0) =
b
.
(a)
The shell’s trajectory is given by
r
(
t
) = 240
√
3
t
i
+ (240
t
−
16
t
2
+ 1600)
j
or
x
= 240
√
3
t
,
y
= 240
t
−
16
t
2
+ 1600.
(b)
Solving
dy/dt
= 240
−
32
t
= 0, we see that
y
is maximum when
t
=15
/
2. The maximum
altitude is
y
(15
/
2) = 2500 ft.
(c)
Solving
y
(
t
)=
−
16
t
2
+ 240
t
+ 1600 =
−
16(
t
−
20)(
t
+ 5) = 0, we see that the shell hits the ground
when
t
= 20. The range of the shell is
x
(20) = 4800
√
3
≈
8314 ft.
(d)
From
(c)
, impact is when
t
= 20. The speed at impact is
|
v
(20)

=
|
240
√
3
i
+ (240
−
32
·
20)
j
|
=

240
2
·
3+(
−
400)
2
= 160
√
13
≈
577 ft/s.
13.
We are given
s
0
=81
j
and
v
0
=4
i
. Using
a
(
t
)=
−
32
j
, we have
v
(
t
)=
s
a
(
t
)
dt
=
−
32
t
j
+
c
4
i
=
v
(0) =
c
v
(
t
)=4
i
−
32
t
j
r
(
t
)=
s
v
(
t
)
dt
=4
t
i
−
16
t
2
j
+
b
81
j
=
r
(0) =
b
r
(
t
)=4
t
i
+(81
−
16
t
2
)
j
.
Solving
y
(
t
)=81
−
16
t
2
= 0, we see that the car hits the water when
t
=9
/
4. Then

v
(9
/
4)

=
|
4
i
−
32(9
/
4)
j
|
=

4
2
+72
2
=20
√
13
≈
72
.
11 ft/s.
14.
Let
θ
be the angle of elevation. Then
v
(0) = 98cos
θ
i
+ 98sin
θ
j
. Using
a
(
t
)=
−
9
.
8
j
, we have
v
(
t
)=
s
a
(
t
)
dt
=
−
9
.
8
t
j
+
c
98cos
θ
i
+ 98sin
θ
j
=
v
(0) =
c
v
(
t
)=98cos
θ
i
+ (98sin
θ
−
9
.
8
t
)
j
r
(
t
)=98
t
cos
θ
i
+ (98
t
sin
θ
−
4
.
9
t
2
)
j
+
b
.
Since
r
(0) =
0
,
b
=
0
and
r
(
t
)=98
t
cos
θ
i
+ (98
t
sin
θ
−
4
.
9
t
2
)
j
. Setting
y
(
t
)=98
t
sin
θ
−
4
.
9
t
2
=
t
(98sin
θ
−
4
.
9
t
) = 0, we see that the projectile hits the ground when
t
= 20 sin
θ
. Thus, using
x
(
t
)=98
t
cos
θ
,
490 =
x
(
t
) = 98(20 sin
θ
) cos
θ
or sin 2
θ
=0
.
5. Then 2
θ
=30
◦
or 150
◦
. The angles of elevation are 15
◦
and 75
◦
.
382

383. Exercises 9.2
15.
Let
s
be the initial speed. Then
v
(0) =
s
cos 45
◦
i
+
s
sin 45
◦
j
=
s
√
2
2
i
+
s
√
2
2
j
. Using
a
(
t
)=
−
32
j
,wehave
v
(
t
)=
s
a
(
t
)
dt
=
−
32
t
j
+
c
s
√
2
2
i
+
s
√
2
2
j
=
v
(0) =
c
v
(
t
)=
s
√
2
2
i
+
n
s
√
2
2
−
32
t
g
j
r
(
t
)=
s
√
2
2
t
i
+
n
s
√
2
2
t
−
16
t
2
g
j
+
b
.
Since
r
(0) =
0
,
b
=
0
and
r
(
t
)=
s
√
2
2
t
i
+
n
s
√
2
2
t
−
16
t
2
g
j
.
Setting
y
(
t
)=
s
√
2
t/
2
−
16
t
2
=
t
(
s
√
2
/
2
−
16
t
) = 0 we see that the ball hits the ground when
t
=
√
2
s/
32.
Thus, using
x
(
t
)=
s
√
2
t/
2 and the fact that 100 yd = 300 ft, 300 =
x
(
t
)=
s
√
2
2
(
√
2
s/
32) =
s
2
32
and
s
=
√
9600
≈
97
.
98ft/s.
16.
Let
s
be the initial speed and
θ
the initial angle. Then
v
(0) =
s
cos
θ
i
+
s
sin
θ
j
. Using
a
(
t
)=
−
32
j
, we have
v
(
t
)=
s
a
(
t
)
dt
=
−
32
t
j
+
c
s
cos
θ
i
+
s
sin
θ
j
=
v
(0) =
c
v
(
t
)=
s
cos
θ
i
+(
s
sin
θ
−
32
t
)
j
r
(
t
)=
st
cos
θ
i
+(
st
sin
θ
−
16
t
2
)
j
+
b
.
Since
r
(0) =
0
,
b
=
0
and
r
(
t
)=
st
cos
θ
i
+(
st
sin
θ
−
16
t
2
)
j
. Setting
y
(
t
)=
st
sin
θ
−
16
t
2
=
t
(
s
sin
θ
−
16
t
)=0,
we see that the ball hits the ground when
t
=(
s
sin
θ
)
/
16. Using
x
(
t
)=
st
cos
θ
i
, we see that the range of the
ball is
x

s
sin
θ
16

=
s
2
sin
θ
cos
θ
16
=
s
2
sin 2
θ
32
.
For
θ
=30
◦
, the range is
s
2
sin 60
◦
/
32 =
√
3
s
2
/
64 and for
θ
=60
◦
the range is
s
2
sin 120
◦
/
32 =
√
3
s
2
/
64. In
general, when the angle is 90
◦
−
θ
the range is
[
s
2
sin 2(90
◦
−
θ
)]
/
32 =
s
2
[sin(180
◦
−
2
θ
)]
/
32 =
s
2
(sin 2
θ
)
/
32
.
Thus, for angles
θ
and 90
◦
−
θ
, the range is the same.
383

384. Exercises 9.2
17.
Let the initial speed of the projectile be
s
and let the target be at (
x
0
,y
0
). Then
v
p
(0) =
s
cos
θ
i
+
s
sin
θ
j
and
v
t
(0) =
0
. Using
a
(
t
)=
−
32
j
,wehave
v
p
(
t
)=
s
a
(
t
)
dt
=
−
32
t
j
+
c
s
cos
θ
i
+
s
sin
θ
j
=
v
p
(0) =
c
v
p
(
t
)=
s
cos
θ
i
+(
s
sin
θ
−
32
t
)
j
r
p
(
t
)=
st
cos
θ
i
+(
st
sin
θ
−
16
t
2
)
j
+
b
.
Since
r
p
(0) =
0
,
b
=
0
and
r
p
(
t
)=
st
cos
θ
i
+(
st
sin
θ
−
16
t
2
)
j
. Also,
v
t
(
t
)=
−
32
t
j
+
c
and since
v
t
(0) =
0
,
c
=
0
and
v
t
(
t
)=
−
32
t
j
. Then
r
t
(
t
)=
−
16
t
2
j
+
b
. Since
r
t
(0) =
x
0
i
+
y
0
j
,
b
=
x
0
i
+
y
0
j
and
r
t
(
t
)=
x
0
i
+(
y
0
−
16
t
2
)
j
.
Now, the horizontal component of
r
p
(
t
) will be
x
0
when
t
=
x
0
/s
cos
θ
at which time the vertical component of
r
p
(
t
) will be
(
sx
0
/s
cos
θ
) sin
θ
−
16(
x
0
/s
cos
θ
)
2
=
x
0
tan
θ
−
16(
x
0
/s
cos
θ
)
2
=
y
0
−
16(
x
0
/s
cos
θ
)
2
.
Thus,
r
p
(
x
0
/s
cos
θ
)=
r
t
(
x
0
/s
cos
θ
) and the projectile will strike the target as it falls.
18.
The initial angle is
θ
= 0, the initial height is 1024 ft, and the initial speed is
s
= 180(5280)
/
3600 = 264 ft/s.
Then
x
(
t
) = 264
t
and
y
(
t
)=
−
16
t
2
+1024. Solving
y
(
t
) = 0 we see that the pack hits the ground at
t
= 8seconds
The horizontal distance travelled is
x
(8) = 2112 feet. From the figure in the text, tan
α
= 1024
/
2112 = 16
/
33
and
α
≈
0
.
45 radian or 25
.
87
◦
.
19. r
N
(
t
)=
v
(
t
)=
−
r
0
ω
sin
ωt
i
+
r
0
ω
cos
ωt
j
;
v
=

v
(
t
)

=

r
2
0
ω
2
sin
2
ωt
+
r
2
0
ω
2
cos
2
ωt
=
r
0
ω
ω
=
v/r
0
;
a
(
t
)=
r
NN
(
t
)=
−
r
0
ω
2
cos
ωt
i
−
r
0
ω
2
sin
ωt
j
a
=

a
(
t
)

=

r
2
0
ω
4
cos
2
ωt
+
r
2
0
ω
4
sin
2
ωt
=
r
0
ω
2
=
r
0
(
v/r
0
)
2
=
v
2
/r
0
.
20. (a) v
(
t
)=
−
b
sin
t
i
+
b
cos
t
j
+
c
k
;

v
(
t
)

=

b
2
sin
2
t
+
b
2
cos
2
t
+
c
2
=
√
b
2
+
c
2
(b)
s
=
s
t
0

v
(
u
)

du
s
t
0

b
2
+
c
2
du
=
t

b
2
+
c
2
;
ds
dt
=

b
2
+
c
2
(c)
d
2
s
dt
2
=0;
a
(
t
)=
−
b
cos
t
i
−
b
sin
t
j
;

a
(
t
)

=

b
2
cos
2
t
+
b
2
sin
2
t
=
|
b
|
. Thus,
d
2
s/dt
2

=

a
(
t
)

.
21.
By Problem 19,
a
=
v
2
/r
0
= 1530
2
/
(4000
·
5280)
≈
0
.
1108. We are given
mg
= 192, so
m
= 192
/
32 and
w
e
= 192
−
(192
/
32)(0
.
1108)
≈
191
.
33 lb.
22.
By Problem 19, the centripetal acceleration is
v
2
/r
0
. Then the horizontal force is
mv
2
/r
0
. The vertical force is 32
m
. The resultant force is
U
=(
mv
2
/r
0
)
i
+32
m
j
.
From the figure, we see that tan
φ
=(
mv
2
/r
0
)
/
32
m
=
v
2
/
32
r
0
. Using
r
0
= 60 and
v
= 44 we obtain tan
φ
=44
2
/
32(60)
≈
1
.
0083 and
φ
≈
45
.
24
◦
.
23.
Solving
x
(
t
)=(
v
0
cos
θ
)
t
for
t
and substituting into
y
(
t
)=
−
1
2
gt
2
+(
v
0
sin
θ
)
t
+
s
0
we obtain
y
=
−
1
2
g

x
v
0
cos
θ

2
+(
v
0
sin
θ
)
x
v
0
cos
θ
+
s
0
=
−
g
2
v
2
0
cos
2
θ
x
2
+ (tan
θ
)
x
+
s
0
,
which is the equation of a parabola.
24.
Since the projectile is launched from ground level,
s
0
= 0. To find the maximum height we maximize
y
(
t
)=
−
1
2
gt
2
+(
v
0
sin
θ
)
t
. Solving
y
N
(
t
)=
−
gt
+
v
0
sin
θ
= 0, we see that
t
=(
v
0
/g
) sin
θ
is a critical point.
384

385. Exercises 9.2
Since
y
NN
(
t
)=
−
g<
0,
H
=
y

v
0
sin
θ
g

=
−
1
2
g
v
2
0
sin
2
θ
g
2
+
v
0
sin
θ
v
0
sin
θ
g
=
v
2
0
sin
2
θ
2
g
is the maximum height. To find the range we solve
y
(
t
)=
−
1
2
gt
2
+(
v
0
sin
θ
)
t
=
t
(
v
0
sin
θ
−
1
2
gt
) = 0. The
positive solution of this equation is
t
=(2
v
0
sin
θ
)
/g
. The range is thus
x
(
t
)=(
v
0
cos
θ
)
2
v
0
sin
θ
g
=
v
2
0
sin 2
θ
g
.
25.
Letting
r
(
t
)=
x
(
t
)
i
+
y
(
t
)
j
+
z
(
t
)
k
, the equation
d
r
/dt
=
v
is equivalent to
dx/dt
=6
t
2
x
,
dy/dt
=
−
4
ty
2
,
dz/dt
=2
t
(
z
+ 1). Separating variables and integrating, we obtain
dx/x
=6
t
2
dt
,
dy/y
2
=
−
4
tdt
,
dz/
(
z
+1)=
2
tdt
, and ln
x
=2
t
3
+
c
1
,
−
1
/y
=
−
2
t
2
+
c
2
, ln(
z
+1)=
t
2
+
c
3
. Thus,
r
(
t
)=
k
1
e
2
t
3
i
+
1
2
t
2
+
k
2
j
+(
k
3
e
t
2
−
1)
k
.
26.
We require the fact that
d
r
/dt
=
v
. Then
d
L
dt
=
d
dt
(
r
×
p
)=
r
×
d
p
dt
+
d
r
dt
×
p
=
τ
τ
τ
+
v
×
p
=
τ
τ
τ
+
v
×
m
v
=
τ
τ
τ
+
m
(
v
×
v
)=
τ
τ
τ
+
0
=
τ
τ
τ
.
27. (a)
Since
F
is directed along
r
we have
F
=
c
r
for some constant
c
. Then
τ
τ
τ
=
r
×
F
=
r
×
(
c
r
)=
c
(
r
×
r
)=
0
.
(b)
If
τ
τ
τ
=
0
then
d
L
/dt
=
0
and
L
is constant.
28. (a)
Using Problem 27,
F
=
−
k
(
Mm/r
2
)
u
=
m
a
. Then
a
=
d
2
r
/dt
=
−
k
(
M/r
2
)
u
.
(b)
Using
u
=
r
/r
we have
r
×
r
NN
=
r
×

−
k
M
r
2
u

=
−
kM
r
2
N
r
×
(
1
r
r
)
o
=
−
kM
r
3
(
r
×
r
)=
0
.
(c)
From Theorem 9
.
4(
iv
)wehave
d
dt
(
r
×
v
)=
r
×
d
v
dt
+
d
r
dt
×
v
=
r
×
r
NN
+
v
×
v
=
0
+
0
=
0
.
(d)
Since
r
=
r
u
we have
c
=
r
×
v
=
r
u
×
r
u
N
=
r
2
(
u
×
u
N
).
(e)
Since
u
=(1
/r
)
r
is a unit vector,
u
·
u
= 1 and
d
dt
(
u
·
u
)=
u
·
d
u
dt
+
d
u
dt
·
u
=2
u
·
d
u
dt
=
d
dt
(1) = 0
.
Thus,
u
·
u
N
=0.
(f)
d
dt
(
v
×
c
)=
v
×
d
c
dt
+
d
v
dt
×
c
=
v
×
0
+
a
×
c
=
−
kM
r
2
u
×
c
=
−
kM
r
2
u
×
[
r
2
(
u
×
u
N
)]
=
−
kM
[
u
×
(
u
×
u
N
)] =
−
kM
=
−
kM
[(
u
·
u
N
)
u
−
(
u
·
u
)
u
N
]
by (10) of 7
.
4
=
−
kM
[
0
−
u
N
]=
kM
u
N
=
kM
d
u
dt
(g)
Since
r
·
(
v
×
c
)=(
r
×
v
)
·
c
by Problem 61 in 7
.
4
=
c
·
c
=
c
2
where
c
=

c

385

386. Exercises 9.2
and
(
kM
u
+
d
)
·
r
=(
kM
u
+
d
)
·
r
u
=
kMr
u
·
u
+
r
d
·
u
=
kMr
+
rd
cos
θ
where
d
=

d

we have
c
2
=
kMr
+
rd
cos
θ
or
r
=
c
2
kM
+
d
cos
θ
=
c
2
/kM
1+(
d/kM
) cos
θ
.
(h)
First note that
c>
0 (otherwise there is no orbit) and
d>
0 (since the orbit is not a circle). We recognize
the equation in
(g)
to be that of a conic section with eccentricity
e
=
d/kM
. Since the orbit of the planet
is closed it must be an ellipse.
(i)
At perihelion
c
=

c

=

r
×
v

=
r
0
v
0
sin(
π/r
)=
r
0
v
0
. Since
r
is minimum at this point, we want the
denominator in the equation
r
0
=[
c
2
/kM
]
/
[1 + (
d/kM
) cos
θ
] to be maximum. This occurs when
θ
=0. In
this case
r
0
=
r
2
0
v
2
0
/kM
1+
d/kM
and
d
=
r
0
v
2
0
−
kM.
Exercises 9.3
1. r
N
(
t
)=
−
t
sin
t
i
+
t
cos
t
j
+2
t
k
;
|
r
N
(
t
)
|
=

t
2
sin
2
t
+
t
2
cos
2
t
+4
t
2
=
√
5
t
;
T
(
t
)=
−
sin
t
√
5
i
+
cos
t
√
5
j
+
2
√
5
k
2. r
N
(
t
)=
e
t
(
−
sin
t
+ cos
t
)
i
+
e
t
(cos
t
+ sin
t
)
j
+
√
2
e
t
k
,
|
r
N
(
t
)
|
=[
e
2
t
(sin
2
t
−
2 sin
t
cos
t
+ cos
2
t
)+
e
2
t
(cos
2
t
+ 2 sin
t
cos
t
+ sin
2
t
)+2
e
2
t
]
1
/
2
=
√
4
e
2
t
=2
e
t
;
T
(
t
)=
1
2
(
−
sin
t
+ cos
t
)
i
+
1
2
(cos
t
+ sin
t
)
j
+
√
2
2
k
3.
We assume
a>
0.
r
N
(
t
)=
−
a
sin
t
i
+
a
cos
t
j
+
c
k
;
|
r
N
(
t
)
|
=

a
2
sin
2
t
+
a
2
cos
2
t
+
c
2
=
√
a
2
+
c
2
;
T
(
t
)=
−
a
sin
t
√
a
2
+
c
2
i
+
a
cos
t
√
a
2
+
c
2
j
+
c
√
a
2
+
c
2
k
;
d
T
dt
=
−
a
cos
t
√
a
2
+
c
2
i
−
a
sin
t
√
a
2
+
c
2
j
,




d
T
dt




=
p
a
2
cos
2
t
a
2
+
c
2
+
a
2
sin
2
t
a
2
+
c
2
=
a
√
a
2
+
c
2
;
N
=
−
cos
t
i
−
sin
t
j
;
B
=
T
×
N
=








ijk
−
a
sin
t
√
a
2
+
c
2
a
cos
t
√
a
2
+
c
2
c
√
a
2
+
c
2
−
cos
t
−
sin
t
0








=
c
sin
t
√
a
2
+
c
2
i
−
c
cos
t
√
a
2
+
c
2
j
+
a
√
a
2
+
c
2
k
;
κ
=
|
d
T
/dt
|
|
r
N
(
t
)
|
=
a/
√
a
2
+
c
2
√
a
2
+
c
2
=
a
a
2
+
c
2
4. r
N
(
t
)=
i
+
t
j
+
t
2
k
,
r
N
(1) =
i
+
j
+
k
;
|
r
N
(
t
)
|
=
√
1+
t
2
+
t
4
,
|
r
N
(1)
|
=
√
3;
T
(
t
) = (1 +
t
2
+
t
4
)
−
1
/
2
(
i
+
t
j
+
t
2
k
),
T
(1) =
1
√
3
(
i
+
j
+
k
);
d
T
dt
=
−
1
2
(1 +
t
2
+
t
4
)
−
3
/
2
(2
t
+4
t
3
)
i
+ [(1 +
t
2
+
t
4
)
−
1
/
2
−
t
2
(1 +
t
2
+
t
4
)
−
3
/
2
(2
t
+4
t
3
)]
j
+[2
t
(1 +
t
2
+
t
4
)
−
1
/
2
−
t
2
2
(1 +
t
2
+
t
4
)
−
3
/
2
(2
t
+4
t
3
)]
k
;
d
dt
T
(1) =
−
1
√
3
i
+
1
√
3
k
,




d
dt
T
(1)




=
r
1
3
+
1
3
=
√
2
√
3
;
N
(1) =
−
1
√
2
(
i
−
k
);
386

387. Exercises 9.3
B
(1) =







ijk
1
/
√
31
/
√
31
/
√
3
−
1
/
√
201
/
√
2







=
1
√
6
(
i
−
2
j
+
k
);
κ
=




d
dt
T
(1)




/
|
r
N
(1)
|
=
√
2
/
√
3
√
3
=
√
2
3
5.
From Example 2 in the text, a normal to the osculating plane is
B
(
π/
4) =
1
√
26
(3
i
−
3
j
+2
√
2
k
). The point on the
curve when
t
=
π/
4is(
√
2
,
√
2
,
3
π/
4). An equation of the plane is 3(
x
−
√
2)
−
3(
y
−
√
2)+2
√
2(
z
−
3
π/
4) = 0,
3
x
−
3
y
+2
√
2
z
=3
√
2
π/
2, or 3
√
2
x
−
3
√
2
y
+4
z
=3
π
.
6.
From Problem 4, a normal to the osculating plane is
B
(1) =
1
√
6
(
i
−
2
j
+
k
). The point on the curve when
t
=1
is (1
,
1
/
2
,
1
/
3). An equation of the plane is (
x
−
1)
−
2(
y
−
1
/
2)+(
z
−
1
/
3) = 0 or
x
−
2
y
+
z
=1
/
3.
7. v
(
t
)=
j
+2
t
k
,
|
v
(
t
)
|
=
√
1+4
t
2
;
a
(
t
)=2
k
;
v
·
a
=4
t
,
v
×
a
=2
i
,
|
v
×
a
|
=2;
a
T
=
4
t
√
1+4
t
2
,
a
N
=
2
√
1+4
t
2
8. v
(
t
)=
−
3 sin
t
i
+ 2 cos
t
j
+
k
,
|
v
(
t
)
|
=

9 sin
2
t
+ 4 cos
2
t
+1=

5 sin
2
t
+ 4 sin
2
t
+ 4 cos
2
t
+1=

5

sin
2
t
+1;
a
(
t
)=
−
3 cos
t
i
−
2 sin
t
j
;
v
·
a
= 9 sin
t
cos
t
−
4 sin
t
cos
t
= 5 sin
t
cos
t
,
v
×
a
= 2 sin
t
i
−
3 cos
t
j
+6
k
,
|
v
×
a
|
=

4 sin
2
t
+ 9 cos
2
t
+36=

5

cos
2
t
+8;
a
T
=
√
5 sin
t
cos
t

sin
2
t
+1
,
a
N
=
p
cos
2
t
+8
sin
2
t
+1
9. v
(
t
)=2
t
i
+2
t
j
+4
t
k
,
|
v
(
t
)
|
=2
√
6
t
,
t>
0;
a
(
t
)=2
i
+2
j
+4
k
;
v
·
a
=24
t
,
v
×
a
=
0
;
a
T
=
24
t
2
√
6
t
=2
√
6,
a
N
=0,
t>
0
10. v
(
t
)=2
t
i
−
3
t
2
j
+4
t
3
k
,
|
v
(
t
)
|
=
t
√
4+9
t
2
+16
t
4
,
t>
0;
a
(
t
)=2
i
−
6
t
j
+12
t
2
k
;
v
·
a
=4
t
+18
t
3
+48
t
5
;
v
×
a
=
−
12
t
4
i
−
16
t
3
j
−
6
t
2
k
,
|
v
×
a
|
=2
t
2
√
36
t
4
+64
t
2
+9;
a
T
=
4+18
t
2
+48
t
4
√
4+9
t
2
+16
t
4
,
a
N
=
2
t
√
36
t
4
+64
t
2
+9
√
4+9
t
2
+16
t
4
,
t>
0
11. v
(
t
)=2
i
+2
t
j
,
|
v
(
t
)
|
=2
√
1+
t
2
;
a
(
t
)=2
j
;
v
·
a
=4
t
;
v
×
a
=4
k
,
|
v
×
a
|
=4;
a
T
=
2
t
√
1+
t
2
,
a
N
=
2
√
1+
t
2
12. v
(
t
)=
1
1+
t
2
i
+
t
1+
t
2
j
,
|
v
(
t
)
|
=
√
1+
t
2
1+
t
2
;
a
(
t
)=
−
2
t
(1 +
t
2
)
2
i
+
1
−
t
2
(1 +
t
2
)
2
j
;
v
·
a
=
−
2
t
(1 +
t
2
)
3
+
t
−
t
3
(1 +
t
2
)
3
=
−
t
(1 +
t
2
)
2
;
v
×
a
=
1
(1 +
t
2
)
2
k
,
|
v
×
a
|
=
1
(1 +
t
2
)
2
;
a
T
=
−
t/
(1 +
t
2
)
2
√
1+
t
2
/
(1 +
t
2
)
=
−
t
(1 +
t
2
)
3
/
2
,
a
N
=
1
/
(1 +
t
2
)
2
√
1+
t
2
/
(1 +
t
2
)
=
1
(1 +
t
2
)
3
/
2
13. v
(
t
)=
−
5 sin
t
i
+ 5 cos
t
j
,
|
v
(
t
)
|
=5;
a
(
t
)=
−
5 cos
t
i
−
5 sin
t
j
;
v
·
a
=0,
v
×
a
=25
k
,
|
v
×
a
|
= 25;
a
T
=0,
a
N
=5
14. v
(
t
) = sinh
t
i
+ cosh
t
j
,
|
v
(
t
)
|
=

sinh
2
t
+ cosh
2
t
;
a
(
t
) = cosh
t
i
+ sinh
t
jv
·
a
= 2 sinh
t
cosh
t
;
v
×
a
= (sinh
2
t
−
cosh
2
t
)
k
=
−
k
,
|
v
×
a
|
=1;
a
T
=
2 sinh
t
cosh
t

sinh
2
t
+ cosh
2
t
,
a
N
=
1

sinh
2
t
+ cosh
2
t
15. v
(
t
)=
−
e
−
t
(
i
+
j
+
k
),
|
v
(
t
)
|
=
√
3
e
−
t
;
a
(
t
)=
e
−
t
(
i
+
j
+
k
);
v
·
a
=
−
3
e
−
2
t
;
v
×
a
=
0
,
|
v
×
a
|
=0;
a
T
=
−
√
3
e
−
t
,
a
N
=0
387

388. Exercises 9.3
16. v
(
t
)=
i
+2
j
+4
k
,
|
v
(
t
)
|
=
√
21 ;
a
(
t
)=
0
;
v
·
a
=0,
v
×
a
=
0
,
|
v
×
a
|
=0;
a
T
=0,
a
N
=0
17. v
(
t
)=
−
a
sin
t
i
+
b
cos
t
j
+
c
k
,
|
v
(
t
)
|
=

a
2
sin
2
t
+
b
2
cos
2
t
+
c
2
;
a
(
t
)=
−
a
cos
t
i
−
b
sin
t
j
;
v
×
a
=
bc
sin
t
i
−
ac
cos
t
j
+
ab
k
,
|
v
×
a
|
=

b
2
c
2
sin
2
t
+
a
2
c
2
cos
2
t
+
a
2
b
2
κ
=
|
v
×
a
|
|
v
|
3
=

b
2
c
2
sin
2
t
+
a
2
c
2
cos
2
t
+
a
2
b
2
(
a
2
sin
2
t
+
b
2
cos
2
t
+
c
2
)
3
/
2
;
18. (a) v
(
t
)=
−
a
sin
t
i
+
b
cos
t
j
,
|
v
(
t
)
|
=

a
2
sin
2
t
+
b
2
cos
2
t
;
a
(
t
)=
−
a
cos
t
i
−
b
sin
t
j
;
v
×
a
=
ab
k
;
|
v
×
a
|
=
ab
;
κ
=
ab
(
a
2
sin
2
t
+
b
2
cos
2
t
)
3
/
2
(b)
When
a
=
b
,
|
v
(
t
)
|
=
a
,
|
v
×
a
|
=
a
2
, and
κ
=
a
2
/a
3
=1
/a
.
19.
The equation of a line is
r
(
t
)=
b
+
t
c
, where
b
and
c
are constant vectors.
v
(
t
)=
c
,
|
v
(
t
)
|
=
|
c
|
;
a
(
t
)=
0
;
v
×
a
=
0
,
|
v
×
a
|
=0;
κ
=
|
v
×
a
|
/
|
v
|
3
=0
20. v
(
t
)=
a
(1
−
cos
t
)
i
+
a
sin
t
j
;
v
(
π
)=2
a
i
,
|
v
(
π
)
|
=2
a
;
a
(
t
)=
a
sin
t
i
+
a
cos
t
j
,
a
(
π
)=
−
a
j
;
|
v
×
a
|
=







ijk
2
a
00
0
−
a
0







=
−
2
a
2
k
;
|
v
×
a
|
=2
a
2
;
κ
=
|
v
×
a
|
|
v
|
3
=
2
a
2
8
a
3
=
1
4
a
21. v
(
t
)=
f
N
(
t
)
i
+
g
N
(
t
)
j
,
|
v
(
t
)
|
=

[
f
N
(
t
)]
2
+[
g
N
(
t
)]
2
;
a
(
t
)=
f
NN
(
t
)
i
+
g
NN
(
t
)
j
;
v
×
a
=[
f
N
(
t
)
g
NN
(
t
)
−
g
N
(
t
)
f
NN
(
t
)]
k
,
|
v
×
a
|
=
|
f
N
(
t
)
g
NN
(
t
)
−
g
N
(
t
)
f
NN
(
t
)
|
;
κ
=
|
v
×
a
|
|
v
|
3
=
|
f
N
(
t
)
g
NN
(
t
)
−
g
N
(
t
)
f
NN
(
t
)
|
([
f
N
(
t
)]
2
+[
g
N
(
t
)]
2
)
3
/
2
22.
For
y
=
F
(
x
),
r
(
x
)=
x
i
+
F
(
x
)
j
. We identify
f
(
x
)=
x
and
g
(
x
)=
F
(
x
) in Problem 21. Then
f
N
(
x
)=1,
f
NN
(
x
)=0,
g
N
(
x
)=
F
N
(
x
),
g
NN
(
x
)=
F
NN
(
x
), and
κ
=
|
F
NN
(
x
)
|
/
(1+[
F
N
(
x
)]
2
)
3
/
2
.
23.
F
(
x
)=
x
2
,
F
(0) = 0,
F
(1) = 1;
F
N
(
x
)=2
x
,
F
N
(0) = 0,
F
N
(1) = 2;
F
NN
(
x
)=2,
F
NN
(0)=2,
F
NN
(1) = 2;
κ
(0) =
2
(1+0
2
)
3
/
2
=2;
ρ
(0) =
1
2
;
κ
(1) =
2
(1+2
2
)
3
/
2
=
2
5
√
5
≈
0
.
18;
ρ
(1) =
5
√
5
2
≈
5
.
59; Since 2
>
2
/
5
√
5 , the curve is “sharper” at (0
,
0).
24.
F
(
x
)=
x
3
,
F
(
−
1) =
−
1,
F
(1
/
2) = 1
/
8;
F
N
(
x
)=3
x
2
,
F
N
(
−
1) = 3,
F
N
(1
/
2) = 3
/
4;
F
NN
(
x
)=6
x
,
F
NN
(
−
1) =
−
6,
F
NN
(1
/
2) = 3;
κ
(
−
1) =
|−
6
|
(1+3
2
)
3
/
2
=
6
10
√
10
=
3
5
√
10
≈
0
.
19;
ρ
(
−
1) =
5
√
10
3
≈
5
.
27;
κ
(
1
2
)=
3
[1 + (3
/
4)
2
]
3
/
2
=
3
125
/
64
=
192
125
≈
1
.
54;
ρ
(
1
2
)=
125
192
≈
0
.
65
Since 1
.
54
>
0
.
19, the curve is “sharper” at (1
/
2
,
1
/
8).
25.
At a point of inflection (
x
0
,F
(
x
0
)), if
F
NN
(
x
0
) exists then
F
NN
(
x
0
) = 0. Thus, assuming that lim
x
→
x
0
F
NN
(
x
)
exists,
F
NN
(
x
) and hence
κ
is near 0 for
x
near
x
0
.
26.
We use the fact that
T
·
N
= 0 and
T
·
T
=
N
·
N
= 1. Then
|
a
(
t
)
|
2
=
a
·
a
=(
a
N
N
+
a
T
T
)
·
(
a
N
N
+
a
T
T
)=
a
2
N
N
·
N
+2
a
N
a
T
N
·
T
+
a
2
T
T
·
T
=
a
2
N
+
a
2
T
.
388

389. Exercises 9.4
Exercises 9.4
1.
y
=
−
1
2
x
+
C
2.
x
=
y
2
−
c
3.
x
2
−
y
2
=1+
c
2
4.
4
x
2
+9
y
2
=36
−
c
2
,
−
6
≤
c
≤
6
5.
y
=
x
2
+ln
c
,
c>
0
6.
y
=
x
+ tan
c
,
−
π/x<c<π/
2
7.
x
2
/
9+
z
2
/
4=
c
; elliptical cylinder
8.
x
2
+
y
2
+
z
2
=
c
; sphere
9.
x
2
+3
y
2
+6
z
2
=
c
; ellipsoid
10.
4
y
−
2
z
+1=
c
; plane
11.
12.
Setting
x
=
−
4,
y
= 2, and
z
=
−
3in
x
2
/
16 +
y
2
/
4+
z
2
/
9=
c
we obtain
c
= 3. The equation of the surface is
x
2
/
16 +
y
2
/
4+
z
2
/
9 = 3. Setting
y
=
z
= 0 we find the
x
-intercepts are
±
4
√
3 . Similarly, the
y
-intercepts are
±
2
√
3 and the
z
-intercepts are
±
3
√
3.
13.
z
x
=2
x
−
y
2
;
z
y
=
−
2
xy
+20
y
4
14.
z
x
=
−
3
x
2
+12
xy
3
;
z
y
=18
x
2
y
2
+10
y
15.
z
x
=20
x
3
y
3
−
2
xy
6
+30
x
4
;
z
y
=15
x
4
y
2
−
6
x
2
y
5
−
4
16.
z
x
=3
x
2
y
2
sec
2
(
x
3
y
2
);
z
y
=2
x
3
y
sec
2
(
x
3
y
2
)
389

390. Exercises 9.4
17.
z
x
=
2
√
x
(3
y
2
+1)
;
z
y
=
−
24
y
√
x
(3
y
2
+1)
2
18.
z
x
=12
x
2
−
10
x
+8;
z
y
=0
19.
z
x
=
−
(
x
3
−
y
2
)
−
2
(3
x
2
)=
−
3
x
2
(
x
3
−
y
2
)
−
2
;
z
y
=
−
(
x
3
−
y
2
)
−
2
(
−
2
y
)=2
y
(
x
3
−
y
2
)
−
2
20.
z
x
=6(
−
x
4
+7
y
2
+3
y
)
5
(
−
4
x
3
)=
−
24
x
3
(
−
x
4
+7
y
2
+3
y
)
5
;
z
y
=6(
−
x
4
+7
y
2
+3
y
)
5
(14
y
+3)
21.
z
x
= 2(cos 5
x
)(
−
sin 5
x
)(5) =
−
10 sin 5
x
cos 5
x
;
z
y
= 2(sin 5
y
)(cos 5
y
)(5) = 10 sin 5
y
cos 5
y
22.
z
x
=(2
x
tan
−
1
y
2
)
e
x
2
tan
−
1
y
2
;
z
y
=
2
x
2
y
1+
y
4
e
x
2
tan
−
1
y
2
23.
f
x
=
x
(3
x
2
ye
x
3
y
+
e
x
3
y
=(3
x
3
y
+1)
e
x
3
y
;
f
y
=
x
4
e
x
3
y
24.
f
θ
=
φ
2

cos
θ
φ

1
φ

=
φ
cos
θ
φ
;
f
φ
=
φ
2

cos
θ
φ

−
θ
φ
2

+2
φ
sin
θ
φ
=
−
θ
cos
θ
φ
+2
φ
sin
θ
φ
25.
f
x
=
(
x
+2
y
)3
−
(3
x
−
y
)
(
x
+2
y
)
2
=
7
y
(
x
+2
y
)
2
;
f
y
=
(
x
+2
y
)(
−
1)
−
(3
x
−
y
)(2)
(
x
+2
y
)
2
=
−
7
x
(
x
+2
y
)
2
26.
f
x
=
(
x
2
−
y
2
)
2
y
−
xy
[2(
x
2
−
y
2
)2
x
]
(
x
2
−
y
2
)
4
=
−
3
x
2
y
−
y
3
(
x
2
−
y
2
)
3
;
f
y
=
(
x
2
−
y
2
)
2
x
−
xy
[2(
x
2
−
y
2
)(
−
2
y
)]
(
x
2
−
y
2
)
4
=
3
xy
2
+
x
3
(
x
2
−
y
2
)
3
27.
g
u
=
8
u
4
u
2
+5
v
3
;
g
v
=
15
v
2
4
u
2
+5
v
3
28.
h
r
=
1
2
s
√
r
+
√
s
r
2
;
h
s
=
−
√
r
s
2
−
1
2
r
√
s
29.
w
x
=
y
√
x
;
w
y
=2
√
x
−
y

1
z
e
y/z

−
e
y/z
=2
√
x
−
e
y
z
+1
:
e
y/z
;
w
z
=
−
ye
y/z
e
−
y
z
2
:
=
y
2
z
2
e
y/z
30.
w
x
=
xy

1
x

+ (ln
xz
)
y
=
y
+
y
ln
xz
;
w
y
=
x
ln
xz
;
w
z
=
xy
z
31.
F
u
=2
uw
2
−
v
3
−
vwt
2
sin(
ut
2
);
F
v
=
−
3
uv
2
+
w
cos(
ut
2
);
F
x
= 4(2
x
2
t
)
3
(4
xt
)=16
xt
(2
x
2
t
)
3
= 128
x
7
t
4
;
F
t
=
−
2
uvwt
sin(
ut
2
)+64
x
8
t
3
32.
G
p
=
r
4
s
5
(
p
2
q
3
)
r
4
s
5
−
1
(2
pq
3
)=2
pq
3
r
4
s
5
(
p
2
q
3
)
r
4
s
5
−
1
;
G
q
=
r
4
s
5
(
p
2
q
3
)
r
4
s
5
−
1
(3
p
2
q
2
)=3
p
2
q
2
r
4
s
5
(
p
2
q
3
)
r
4
s
5
−
1
;
G
r
=(
p
2
q
3
)
r
4
s
5
(4
r
3
s
5
) ln(
p
2
q
3
);
G
s
=(
p
2
q
3
)
r
4
s
5
(5
r
4
s
4
) ln(
p
2
q
3
)
33.
∂z
∂x
=
2
x
x
2
+
y
2
,
∂
2
z
∂x
2
=
(
x
2
+
y
2
)2
−
2
x
(2
x
)
(
x
2
+
y
2
)
2
=
2
y
2
−
2
x
2
(
x
2
+
y
2
)
2
;
∂z
∂y
=
2
y
x
2
+
y
2
,
∂
2
z
∂y
2
=
(
x
2
+
y
2
)2
−
2
y
(2
y
)
(
x
2
+
y
2
)
2
=
2
x
2
−
2
y
2
(
x
2
+
y
2
)
2
;
∂
2
z
∂x
2
+
∂
2
z
∂y
2
=
2
y
2
−
2
x
2
+2
x
2
−
2
y
2
(
x
2
+
y
2
)
2
=0
34.
∂z
∂x
=
e
x
2
−
y
2
(
−
2
y
sin 2
xy
)+2
xe
x
2
−
y
2
cos 2
xy
∂
2
z
∂x
2
=
e
x
2
−
y
2
(
−
4
y
2
cos 2
xy
−
8
xy
sin 2
xy
+4
x
2
cos 2
xy
+ 2 cos 2
xy
)
∂z
∂y
=
e
x
2
−
y
2
(
−
2
x
sin 2
xy
)
−
2
ye
x
2
−
y
2
cos 2
xy
∂
2
z
∂y
2
=
e
x
2
−
y
2
(
−
4
x
2
cos 2
xy
+8
xy
sin 2
xy
+4
y
2
cos 2
xy
−
2 cos 2
xy
)
390

391. Exercises 9.4
Adding the second partial derivatives gives
∂
2
z
∂x
2
+
∂
2
z
∂y
2
=[
−
4(
y
2
+
x
2
) cos 2
xy
+4(
x
2
+
y
2
) cos 2
xy
]=0
.
35.
∂u
∂x
= cos
at
cos
x
,
∂
2
u
∂x
2
=
−
cos
at
sin
x
;
∂u
∂t
=
−
a
sin
at
sin
x
,
∂
2
u
∂t
2
=
−
a
2
cos
at
sin
x
;
a
2
∂
2
u
∂x
2
=
a
2
(
−
cos
at
sin
x
)=
∂
2
u
∂t
2
36.
∂u
∂x
=
−
sin(
x
+
at
) + cos(
x
−
at
),
∂
2
u
∂x
2
=
−
cos(
x
+
at
)
−
sin(
x
−
at
);
∂u
∂t
=
−
a
sin(
x
+
at
)
−
a
cos(
x
−
at
),
∂
2
u
∂t
2
=
−
a
2
cos(
x
+
at
)
−
a
2
sin(
x
−
at
);
a
2
∂
2
u
∂x
2
=
−
a
2
cos(
x
+
at
)
−
a
2
sin(
x
−
at
)=
∂
2
u
∂t
2
37.
∂C
∂x
=
−
2
x
kt
t
−
1
/
2
e
−
x
2
/kt
,
∂
2
C
∂x
2
=
4
x
2
k
2
t
2
t
−
1
/
2
e
−
x
2
/kt
−
2
kt
t
−
1
/
2
e
−
x
2
/kt
;
∂C
∂t
=
t
−
1
/
2
x
2
kt
2
e
−
x
2
/kt
−
t
−
3
/
2
2
e
−
x
2
/kt
;
k
4
∂
2
C
∂x
2
=
x
2
kt
2
t
−
1
/
2
e
−
x
2
/kt
−
t
−
1
/
2
2
t
e
−
x
2
/kt
=
∂C
∂t
38. (a)
P
v
=
−
k
(
T/V
2
)
(b)
PV
=
kt
,
PV
T
=
k
,
V
T
=
k/P
(c)
PV
=
kT
,
V
=
kT
p
,
T
p
=
V/k
39.
z
x
=
v
2
e
uv
2
(3
x
2
)+2
uve
uv
2
(1) = 3
x
2
v
2
e
uv
2
+2
uve
uv
2
;
z
y
=
v
2
e
uv
2
(0)+2
uve
uv
2
(
−
2
y
)=
−
4
yuve
uv
2
40.
z
x
=(2
u
cos 4
v
)(2
xy
3
)
−
(4
u
2
sin 4
v
)(3
x
2
)=4
xy
3
u
cos 4
v
−
12
x
2
u
2
sin 4
v
z
y
=(2
u
cos 4
v
)(3
x
2
y
2
)
−
(4
v
2
sin 4
v
)(3
y
2
)=6
x
2
y
2
u
cos 4
v
−
12
y
2
u
2
sin 4
v
41.
z
u
= 4(4
u
3
)
−
10
y
[2(2
u
−
v
)(2)] = 16
u
3
−
40(2
u
−
v
)
y
z
v
=4(
−
24
v
2
)
−
10
y
[2(2
u
−
v
)(
−
1)] =
−
96
v
2
+ 20(2
u
−
v
)
y
42.
z
u
=
2
y
(
x
+
y
)
2

1
v

+
−
2
x
(
x
+
y
)
2

−
v
2
u
2

=
2
y
v
(
x
+
y
)
2
+
2
xv
2
u
2
(
x
+
y
)
2
z
v
=
2
y
(
x
+
y
)
2
e
−
u
v
2
:
+
−
2
x
(
x
+
y
)
2

2
v
u

=
−
2
yu
v
2
(
x
+
y
)
2
−
4
xv
u
(
x
+
y
)
2
43.
w
t
=
3
2
(
u
2
+
v
2
)
1
/
2
(2
u
)(
−
e
−
t
sin
θ
)+
3
2
(
u
2
+
v
2
)
1
/
2
(2
v
)(
−
e
−
t
cos
θ
)
=
−
3
u
(
u
2
+
v
2
)
1
/
2
e
−
t
sin
θ
−
3
v
(
u
2
+
v
2
)
1
/
2
e
−
t
cos
θ
w
θ
=
3
2
(
u
2
+
v
2
)
1
/
2
(2
u
)
e
−
t
cos
θ
+
3
2
(
u
2
+
v
2
)
1
/
2
(2
v
)(
−
e
−
t
sin
θ
)
=3
u
(
u
2
+
v
2
)
1
/
2
e
−
t
cos
θ
−
3
v
(
u
2
+
v
2
)
1
/
2
e
−
t
sin
θ
44.
w
r
=
v/
2
√
uv
1+
uv
(2
r
)+
u/
2
√
uv
1+
uv
(2
rs
2
)=
rv
√
uv
(1 +
uv
)
+
rs
2
u
√
uv
(1 +
uv
)
w
s
=
v/
2
√
uv
1+
uv
(
−
2
s
)+
u/
2
√
uv
1+
uv
(2
r
2
s
)=
−
sv
√
uv
(1 +
uv
)
+
r
2
su
√
uv
(1 +
uv
)
45.
R
u
=
s
2
t
4
(
e
v
2
)+2
rst
4
(
−
2
uve
−
u
2
)+4
rs
2
t
3
(2
uv
2
e
u
2
v
2
)=
s
2
t
4
e
v
2
−
4
uvrst
4
e
−
u
2
+8
uv
2
rs
2
t
3
e
u
2
v
2
R
v
=
s
2
t
4
(2
uve
v
2
)+2
rst
4
(
e
−
u
2
)+4
rs
2
t
3
(2
u
2
ve
u
2
v
2
)=2
s
2
t
4
uve
v
2
+2
rst
4
e
−
u
2
+8
rs
2
t
3
u
2
ve
u
2
v
2
46.
Q
x
=
1
P

t
2
√
1
−
x
2

+
1
q

1
t
2

+
1
r

1
/t
1+(
x/t
)
2

=
t
2
p
√
1
−
x
2
+
1
qt
2
+
t
r
(
t
2
+
x
2
)
391

392. Exercises 9.4
Q
t
=
1
p
(2
t
sin
−
1
x
)+
1
q

−
2
x
t
3

+
1
r

−
x/t
2
1+(
x/t
)
2

=
2
t
sin
−
1
x
p
−
2
x
qt
3
−
x
r
(
t
2
+
x
2
)
47.
w
t
=
2
x
2

x
2
+
y
2
u
rs
+
tu
+
2
y
2

x
2
+
y
2
cosh
rs
u
=
xu

x
2
+
y
2
(
rs
+
tu
)
+
y
cosh
rs
u

x
2
+
y
2
w
r
=
2
x
2

x
2
+
y
2
s
rs
+
tu
+
2
y
2

x
2
+
y
2
st
sinh
rs
u
=
xs

x
2
+
y
2
(
rs
+
tu
)
+
yst
sinh
rs
u

x
2
+
y
2
w
u
=
2
x
2

x
2
+
y
2
t
rs
+
tu
+
2
y
2

x
2
+
y
2
−
t
cosh
rs
u
2
=
xt

x
2
+
y
2
(
rs
+
tu
)
−
yt
cosh
rs
u
2

x
2
+
y
2
48.
s
φ
=2
pe
3
θ
+2
q
[
−
sin(
φ
+
θ
)]
−
2
rθ
2
+ 4(2) = 2
pe
3
θ
−
2
q
sin(
φ
+
θ
)
−
2
rθ
2
+8
s
θ
=2
p
(3
φe
3
θ
)+2
q
[
−
sin(
φ
+
θ
)]
−
2
r
(2
φθ
) + 4(8) = 6
pφe
3
θ
−
2
q
sin(
φ
+
θ
)
−
4
rφθ
+32
49.
dz
dt
=
2
u
u
2
+
v
2
(2
t
)+
2
v
u
2
+
v
2
(
−
2
t
−
3
)=
4
ut
−
4
vt
−
3
u
2
+
v
2
50.
dz
dt
=(3
u
2
v
−
v
4
)(
−
5
e
−
5
t
)+(
u
3
−
4
uv
3
)(5 sec 5
t
tan 5
t
)=
−
5(3
u
2
v
−
v
4
)
e
−
5
t
+5(
u
3
−
4
uv
3
) sec 5
t
tan 5
t
51.
dw
dt
=
−
3 sin(3
u
+4
v
)(2)
−
4 sin(3
u
+4
v
)(
−
1);
u
(
π
)=5
π/
2,
v
(
π
)=
−
5
π/
4
dw
dt




π
=
−
6 sin

15
π
2
−
5
π

+ 4 sin

15
π
2
−
5
π

=
−
2 sin
5
π
2
=
−
2
52.
dw
dt
=
ye
xy
N
−
8
(2
t
+1)
2
o
+
xe
xy
(3);
x
(0) = 4,
y
(0) = 5;
dw
dt




0
=5
e
20
(
−
8)+4
e
20
(3) =
−
28
e
20
53.
With
x
=
r
cos
θ
and
y
=
r
sin
θ
∂u
∂r
=
∂u
∂x
∂x
∂r
+
∂u
∂y
∂y
∂r
=
∂u
∂x
cos
θ
+
∂u
∂y
sin
θ
∂
2
u
∂r
2
=
∂
2
u
∂x
2
∂x
∂r
cos
θ
+
∂
2
u
∂y
2
∂y
∂r
sin
θ
=
∂
2
u
∂x
2
cos
2
θ
+
∂
2
u
∂y
2
sin
2
θ
∂u
∂θ
=
∂u
∂x
∂x
∂θ
+
∂u
∂y
∂y
∂θ
=
∂u
∂x
(
−
r
sin
θ
)+
∂u
∂y
(
r
cos
θ
)
∂
2
u
∂θ
2
=
∂u
∂x
(
−
r
cos
θ
)+
∂
2
u
∂x
2
∂x
∂θ
(
−
r
sin
θ
)+
∂u
∂y
(
−
r
sin
θ
)+
∂
2
u
∂y
2
∂y
∂θ
(
r
cos
θ
)
=
−
r
∂u
∂x
cos
θ
+
r
2
∂
2
u
∂x
2
sin
2
θ
−
r
∂u
∂y
sin
θ
+
r
2
∂
2
u
∂y
2
cos
2
θ.
Using
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0,wehave
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
=
∂
2
u
∂x
2
cos
2
θ
+
∂
2
u
∂y
2
sin
2
θ
+
1
r

∂u
∂x
cos
θ
+
∂u
∂y
sin
θ

+
1
r
2

−
r
∂u
∂x
cos
θ
+
r
2
∂
2
u
∂x
2
sin
2
θ
−
r
∂u
∂y
sin
θ
+
r
2
∂
2
u
∂y
2
cos
2
θ

=
∂
2
u
∂x
2
(cos
2
θ
+ sin
2
θ
)+
∂
2
u
∂y
2
(sin
2
θ
+ cos
2
θ
)+
∂u
∂x

1
r
cos
θ
−
1
r
cos
θ

+
∂u
∂y

1
r
sin
θ
−
1
r
sin
θ

=
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0
.
392

393. Exercises 9.5
54.
dP
dt
=
(
V
−
0
.
0427)(0
.
08)
dT/dt
(
V
−
0
.
0427)
2
−
0
.
08
T
(
dV/dt
)
(
V
−
0
.
0427)
2
+
3
.
6
V
3
dV
dt
=
0
.
08
V
−
0
.
0427
dT
dt
+

3
.
6
V
3
−
0
.
08
T
(
V
−
0
.
0427)
2

dV
dt
55.
Since
dT/dT
= 1 and
∂P/∂T
=0,
0=
F
T
=
∂F
∂P
∂P
∂T
+
∂F
∂V
∂V
∂T
+
∂F
∂T
dT
dT
=
⇒
∂V
∂T
=
−
∂F/∂T
∂F/∂V
=
−
1
∂T/∂V
.
56.
We are given
dE/dt
= 2 and
dR/dt
=
−
1. Then
dI
dt
=
∂I
∂E
dE
dt
+
∂I
∂R
dR
dt
=
1
R
(2)
−
E
R
2
(
−
1), and
when
E
= 60 and
R
= 50,
dI
dt
=
2
50
+
60
50
2
=
1
25
+
3
/
5
25
=
8
125
amp/min.
57.
Since the height of the triangle is
x
sin
θ
, the area is given by
A
=
1
2
xy
sin
θ
. Then
dA
dt
∂A
∂x
dx
dt
+
∂A
∂y
dy
dt
+
∂A
∂θ
dθ
dt
=
1
2
y
sin
θ
dx
dt
+
1
2
x
sin
θ
dy
dt
+
1
2
xy
cos
θ
dθ
dt
.
When
x
= 10,
y
=8,
θ
=
π/
6,
dx/dt
=0
.
3,
dy/dt
=0
.
5, and
dθ/dt
=0
.
1,
dA
dt
=
1
2
(8)

1
2

(0
.
3) +
1
2
(10)

1
2

(0
.
5) +
1
2
(10)(8)
n
√
3
2
g
(0
.
1)
=0
.
6+1
.
25 + 2
√
3=1
.
85+2
√
3
≈
5
.
31 cm
2
/
s.
58.
dw
dt
=
∂w
∂x
dx
dt
+
∂w
∂y
dy
dt
+
∂w
∂z
dz
dt
=
x dx/dt
+
y dy/dt
+
z dz/dt

x
2
+
y
2
+
z
2
=
−
4
x
sin
t
+4
y
cos
t
+5
z

16 cos
2
t
+ 16 sin
2
t
+25
t
2
=
−
16 sin
t
cos
t
+ 16 sin
t
cos
t
+25
t
√
16+25
t
2
=
25
t
√
16+25
t
2
dw
dt




t
=5
π/
2
=
125
π/
2

16 + 625
π
2
/
4
=
125
π
√
64 + 625
π
2
≈
4
.
9743
Exercises 9.5
1.
∇
f
=(2
x
−
3
x
2
y
2
)
i
+(4
y
3
−
2
x
3
y
)
j
2.
∇
f
=4
xye
−
2
x
2
y
i
+(1+2
x
2
e
−
2
x
2
y
)
j
3.
∇
F
=
y
2
z
3
i
+
2
xy
z
3
j
−
3
xy
2
z
4
k
4.
∇
F
=
y
cos
yz
i
+(
x
cos
yz
−
xyz
sin
yz
)
j
−
xy
2
sin
yz
k
5.
∇
f
=2
x
i
−
8
y
j
;
∇
f
(2
,
4) = 4
i
−
32
j
6.
∇
f
=
3
x
2
2

x
3
y
−
y
4
i
+
x
3
−
4
y
3
2

x
3
y
−
y
4
j
;
∇
f
(3
,
2) =
27
√
38
i
−
5
2
√
38
j
7.
∇
F
=2
xz
2
sin 4
y
i
+4
x
2
z
2
cos 4
y
j
+2
x
2
z
sin 4
y
k
∇
F
(
−
2
,π/
3
,
1) =
−
4 sin
4
π
3
i
+ 16 cos
4
π
3
j
+ 8sin
4
π
3
k
=2
√
3
i
−
8
j
−
4
√
3
k
8.
∇
F
=
2
x
x
2
+
y
2
+
z
2
i
+
2
y
x
2
+
y
2
+
z
2
j
+
2
z
x
2
+
y
2
+
z
2
k
;
∇
F
(
−
4
,
3
,
5) =
−
4
25
i
+
3
25
j
+
1
5
k
393

394. Exercises 9.5
9.
D
u
f
(
x, y
) = lim
h
→
0
f
(
x
+
h
√
3
/
2
,y
+
h/
2)
−
f
(
x, y
)
h
= lim
h
→
0
(
x
+
h
√
3
/
2)
2
+(
y
+
h/
2)
2
−
x
2
−
y
2
h
= lim
h
→
0
h
√
3
x
+3
h
2
/
4+
hy
+
h
2
/
4
h
= lim
h
→
0
(
√
3
x
+3
h/
4+
y
+
h/
4) =
√
3
x
+
y
10.
D
u
f
(
x, y
) = lim
h
→
0
f
(
x
+
h
√
2
/
2
,y
+
h
√
2
/
2)
−
f
(
x, y
)
h
= lim
h
→
0
3
x
+3
h
√
2
/
2
−
(
y
+
h
√
2
/
2)
2
−
3
x
+
y
2
h
= lim
h
→
0
3
h
√
2
/
2
−
h
√
2
y
−
h
2
/
2
h
= lim
h
→
0
(3
√
2
/
2
−
√
2
y
−
h/
2) = 3
√
2
/
2
−
√
2
y
11. u
=
√
3
2
i
+
1
2
j
;
∇
f
=15
x
2
y
6
i
+30
x
3
y
5
j
;
∇
f
(
−
1
,
1)=15
i
−
30
j
;
D
u
f
(
−
1
,
1) =
15
√
3
2
−
15 =
15
2
(
√
3
−
2)
12. u
=
√
2
2
i
+
√
2
2
j
;
∇
f
=(4+
y
2
)
i
+(2
xy
−
5)
j
;
∇
f
(3
,
−
1) = 5
i
−
11
j
;
D
u
f
(3
,
−
1) =
5
√
2
2
−
11
√
2
2
=
−
3
√
2
13. u
=
√
10
10
i
−
3
√
10
10
j
;
∇
f
=
−
y
x
2
+
y
2
i
+
x
x
2
+
y
2
j
;
∇
f
(2
,
−
2) =
1
4
i
+
1
4
j
D
u
f
(2
,
−
2) =
√
10
40
−
3
√
10
40
=
−
√
10
20
14. u
=
6
10
i
+
8
10
j
=
3
5
i
+
4
5
j
;
∇
f
=
y
2
(
x
+
y
)
2
i
+
x
2
(
x
+
y
)
2
j
;
∇
f
(2
,
−
1) =
i
+4
j
D
u
f
(2
,
−
1) =
3
5
+
16
5
=
19
5
15. u
=(2
i
+
j
)
/
√
5;
∇
f
=2
y
(
xy
+1)
i
+2
x
(
xy
+1)
j
;
∇
f
(3
,
2) = 28
i
+42
j
D
u
f
(3
,
2) =
2(28)
√
5
+
42
√
5
=
98
√
5
16. u
=
−
i
;
∇
f
=2
x
tan
y
i
+
x
2
sec
2
y
j
;
∇
f
(1
/
2
,π/
3) =
√
3
i
+
j
;
D
u
f
(1
/
2
,π/
3) =
−
√
3
17. u
=
1
√
2
j
+
1
√
2
k
;
∇
F
=2
xy
2
(2
z
+1)
2
i
+2
x
2
y
(2
z
+1)
2
j
+4
x
2
y
2
(2
z
+1)
k
∇
F
(1
,
−
1
,
1)=18
i
−
18
j
+12
k
;
D
u
F
(1
,
−
1
,
1) =
−
18
√
2
+
12
√
2
=
−
6
√
2
=
−
3
√
2
18. u
=
1
√
6
i
−
2
√
6
j
+
1
√
6
k
;
∇
F
=
2
x
z
2
i
−
2
y
z
2
j
+
2
y
2
−
2
x
2
z
3
k
;
∇
F
(2
,
4
,
−
1)=4
i
−
8
j
−
24
k
D
u
F
(2
,
4
,
−
1) =
4
√
6
−
16
√
6
−
24
√
6
=
−
6
√
6
19. u
=
−
k
;
∇
F
=
xy

x
2
y
+2
y
2
z
i
+
x
2
+4
z
2

x
2
y
+2
y
2
z
j
+
y
2

x
2
y
+2
y
2
z
k
∇
F
(
−
2
,
2
,
1) =
−
i
+
j
+
k
;
D
u
F
(
−
2
,
2
,
1) =
−
1
20. u
=
−
(4
i
−
4
j
+2
k
)
/
√
36 =
−
2
3
i
+
2
3
j
−
1
3
k
;
∇
F
=2
i
−
2
y
j
+2
z
k
;
∇
F
(4
,
−
4
,
2) = 2
i
+8
j
+4
k
D
u
F
(4
,
−
4
,
2) =
−
4
3
+
16
3
−
4
3
=
8
3
21. u
=(
−
4
i
−
j
)
/
√
17 ;
∇
f
=2(
x
−
y
)
i
−
2(
x
−
y
)
j
;
∇
f
(4
,
2) = 4
i
−
4
j
;
D
u
F
(4
,
2) =
−
16
√
17
4
√
17
=
−
12
√
17
394

395. Exercises 9.5
22. u
=(
−
2
i
+5
j
)
/
√
29 ;
∇
f
=(3
x
2
−
5
y
)
i
−
(5
x
−
2
y
)
j
;
∇
f
(1
,
1) =
−
2
i
−
3
j
;
D
u
f
(1
,
1) =
4
√
29
−
15
√
29
=
−
11
√
29
23.
∇
f
=2
e
2
x
sin
y
i
+
e
2
x
cos
y
j
;
∇
f
(0
,π/
4) =
√
2
i
+
√
2
2
j
The maximum
D
u
is [(
√
2)
2
+(
√
2
/
2)
2
]
1
/
2
=

5
/
2 in the direction
√
2
i
+(
√
2
/
2)
j
.
24.
∇
f
=(
xye
x
−
y
+
ye
x
−
y
)
i
+(
−
xye
x
−
y
+
xe
x
−
y
)
j
;
∇
f
(5
,
5) = 30
i
−
20
j
The maximum
D
u
is [30
2
+(
−
20)
2
]
1
/
2
=10
√
13 in the direction 30
i
−
20
j
.
25.
∇
F
=(2
x
+4
z
)
i
+2
z
2
j
+(4
x
+4
yz
)
k
;
∇
F
(1
,
2
,
−
1) =
−
2
i
+2
j
−
4
k
The maximum
D
u
is [(
−
2)
2
+2
2
+(
−
4)
2
]
1
/
2
=2
√
6 in the direction
−
2
i
+2
j
−
4
k
.
26.
∇
F
=
yz
i
+
xz
j
+
xy
k
;
∇
F
(3
,
1
,
−
5) =
−
5
i
−
15
j
+3
k
The maximum
D
u
is [(
−
5)
2
+(
−
15)
2
+3
2
]
1
/
2
=
√
259 in the direction
−
5
i
−
15
j
+3
k
.
27.
∇
f
=2
x
sec
2
(
x
2
+
y
2
)
i
+2
y
sec
2
(
x
2
+
y
2
)
j
;
∇
f
(

π/
6
,

π/
6)=2

π/
6 sec
2
(
π/
3)(
i
+
j
)=8

π/
6(
i
+
j
)
The minimum
D
u
is
−
8

π/
6(1
2
+1
2
)
1
/
2
=
−
8

π/
3 in the direction
−
(
i
+
j
).
28.
∇
f
=3
x
2
i
−
3
y
2
j
;
∇
f
(2
,
−
2) = 12
i
−
12
j
= 12(
i
−
j
)
The minimum
D
u
is
−
12[1
2
+(
−
1)
2
]
1
/
2
=
−
12
√
2 in the direction
−
(
i
−
j
)=
−
i
+
j
.
29.
∇
F
=
√
ze
y
2
√
x
i
+
√
xz e
y
j
+
√
x
2
√
z
k
;
∇
F
(16
,
0
,
9) =
3
8
i
+12
j
+
2
3
k
. The minimum
D
u
is
−
[(3
/
8)
2
+12
2
+(2
/
3)
2
]
1
/
2
=
−
√
83,281
/
24 in the direction
−
3
8
i
−
12
j
−
2
3
k
.
30.
∇
F
=
1
x
i
+
1
y
j
−
1
z
k
;
∇
F
(1
/
2
,
1
/
6
,
1
/
3) = 2
i
+6
j
−
3
k
The minimum
D
u
is
−
[2
2
+6
2
+(
−
3)
2
]
1
/
2
=
−
7 in the direction
−
2
i
−
6
j
+3
k
.
31.
Using implicit differentiation on 2
x
2
+
y
2
= 9 we find
y
u
=
−
2
x/y
.At(2
,
1) the slope of the tangent line is
−
2(2)
/
1=
−
4. Thus,
u
=
±
(
i
−
4
j
)
/
√
17 . Now,
∇
f
=
i
+2
y
j
and
∇
f
(3
,
4) =
i
+8
j
. Thus,
D
u
=
±
(1
/
√
17
−
32
√
17 ) =
±
31
/
√
17 .
32.
∇
f
=(2
x
+
y
−
1)
i
+(
x
+2
y
)
j
;
D
u
f
(
x, y
)=
2
x
+
y
−
1
√
2
+
x
+2
y
√
2
=
3
x
+3
y
−
1
√
2
Solving (3
x
+3
y
−
1)
/
√
2 = 0 we see that
D
u
is 0 for all points on the line 3
x
+3
y
=1.
33. (a)
Vectors perpendicular to 4
i
+3
j
are
±
(3
i
−
4
j
). Take
u
=
±

3
5
i
−
4
5
j

.
(b) u
=(4
i
+3
j
)
/
√
16+9=
4
5
i
+
3
5
j
(c) u
=
−
4
5
i
−
3
5
j
34.
D
−
u
f
(
a, b
)=
∇
f
(
a, b
)
·
(
−
u
)=
−∇
f
(
a, b
)
·
u
=
−
D
u
f
(
a, b
)=
−
6
35. (a)
∇
f
=(3
x
2
−
6
xy
2
)
i
+(
−
6
x
2
y
+3
y
2
)
j
D
u
f
(
x, y
)=
3(3
x
2
−
6
xy
2
)
√
10
+
−
6
x
2
y
+3
y
2
√
10
=
9
x
2
−
18
xy
2
−
6
x
2
y
+3
y
2
√
10
(b)
F
(
x, y
)=
3
√
10
(3
x
2
−
6
xy
2
−
2
x
2
y
+
y
2
);
∇
F
=
3
√
10
[(6
x
−
6
y
2
−
4
xy
)
i
+(
−
12
xy
−
2
x
2
+2
y
)
j
]
395

396. Exercises 9.5
D
u
F
(
x, y
)=

3
√
10

3
√
10

(6
x
−
6
y
2
−
4
xy
)+

1
√
10

3
√
10

(
−
12
xy
−
2
x
2
+2
y
)
=
9
5
(3
x
−
3
y
2
−
2
xy
)+
3
5
(
−
6
xy
−
x
2
+
y
)=
1
5
(27
x
−
27
y
2
−
36
xy
−
3
x
2
+3
y
)
36.
∇
U
=
Gmx
(
x
2
+
y
2
)
3
/
2
i
+
Gmy
(
x
2
+
y
2
)
3
/
2
j
=
Gm
(
x
2
+
y
2
)
3
/
2
(
x
i
+
y
j
)
The maximum and minimum values of
D
u
U
(
x, y
) are obtained when
u
is in the directions
∇
U
and
−∇
U
,
respectively. Thus, at a point (
x, y
), not (0
,
0), the directions of maximum and minimum increase in
U
are
x
i
+
y
j
and
−
x
i
−
y
j
, respectively. A vector at (
x, y
) in the direction
±
(
x
i
+
y
j
) lies on a line through the origin.
37.
∇
f
=(3
x
2
−
12)
i
+(2
y
−
10)
j
. Setting
|∇
f
|
= [(3
x
2
−
12)
2
+(2
y
−
10)
2
]
1
/
2
= 0, we obtain 3
x
2
−
12 = 0 and
2
y
−
10 = 0. The points where
|∇
f
|
= 0 are (2
,
5) and (
−
2
,
5).
38.
Let
∇
f
(
a, b
)=
α
i
+
β
j
. Then
D
u
f
(
a, b
)=
∇
f
(
a, b
)
·
u
=
5
13
α
−
12
13
β
= 7 and
D
v
f
(
a, b
)=
∇
f
(
a, b
)
·
v
=
5
13
α
−
12
13
β
=3
.
Solving for
α
and
β
, we obtain
α
= 13 and
β
=
−
13
/
6. Thus,
∇
f
(
a, b
)=13
i
−
(13
/
6)
j
.
39.
∇
T
=4
x
i
+2
y
j
;
∇
T
(4
,
2) = 16
i
+4
j
. The minimum change in temperature (that is, the maximum decrease
in temperature) is in the direction
−∇
T
(4
,
3) =
−
16
i
−
4
j
.
40.
Let
x
(
t
)
i
+
y
(
t
)
j
be the vector equation of the path. At (
x, y
) on this curve, the direction of a tangent vector is
x
u
(
t
)
i
+
y
u
(
t
)
j
. Since we want the direction of motion to be
−∇
T
(
x, y
), we have
x
u
(
t
)
i
+
y
u
(
t
)
j
=
−∇
T
(
x, y
)=
4
x
i
+2
y
j
. Separating variables in
dx/dt
=4
x
, we obtain
dx/x
=4
dt
,ln
x
=4
t
+
c
1
, and
x
=
C
1
e
4
t
. Separating
variables in
dy/dt
=2
y
, we obtain
dy/y
=2
dt
,ln
y
=2
t
+
c
2
, and
y
=
C
2
e
2
t
. Since
x
(0) = 4 and
y
(0) = 2, we
have
x
=4
e
4
t
and
y
=2
e
2
t
. The equation of the path is 4
e
4
t
i
+2
e
2
t
j
for
t
≥
0, or eliminating the parameter,
x
=
y
2
,
y
≥
0.
41.
Let
x
(
t
)
i
+
y
(
t
)
j
be the vector equation of the path. At (
x, y
) on this curve, the direction of a tangent vector
is
x
u
(
t
)
i
+
y
u
(
t
)
j
. Since we want the direction of motion to be
∇
T
(
x, y
), we have
x
u
(
t
)
i
+
y
u
(
t
)
j
=
∇
T
(
x, y
)=
−
4
x
i
−
2
y
j
. Separating variables in
dx/dt
=
−
4
x
we obtain
dx/x
=
−
4
dt
,ln
x
=
−
4
t
+
c
1
and
x
=
C
1
e
−
4
t
.
Separating variables in
dy/dt
=
−
2
y
we obtain
dy/y
=
−
2
dt
,ln
y
=
−
2
t
+
c
2
and
y
=
C
2
e
−
2
t
. Since
x
(0) = 3
and
y
(0) = 4, we have
x
=3
e
−
4
t
and
y
=4
e
−
2
t
. The equation of the path is 3
e
−
4
t
i
+4
e
−
2
t
j
, or eliminating the
parameter, 16
x
=3
y
2
,
y
≥
0.
42.
Substituting
x
=0,
y
=0,
z
= 1, and
T
= 500 into
T
=
k/
(
x
2
+
y
2
+
z
2
) we see that
k
= 500 and
T
(
x, y, z
) = 500
/
(
x
2
+
y
2
+
z
2
).
(a) u
=
1
3
j
1
,
−
2
,
−
2
T
=
1
3
i
−
2
3
j
−
2
3
k
∇
T
=
−
1000
x
(
x
2
+
y
2
+
z
2
)
2
i
−
1000
y
(
x
2
+
y
2
+
z
2
)
2
j
−
1000
z
(
x
2
+
y
2
+
z
2
)
2
k
∇
T
(2
,
3
,
3) =
−
500
121
i
−
750
121
j
−
750
121
k
D
u
T
(2
,
3
,
3) =
1
3

−
500
121

−
2
3

−
750
121

−
2
3

−
750
121

=
2500
363
(b)
The direction of maximum increase is
∇
T
(2
,
3
,
3) =
−
500
121
i
−
750
121
j
−
750
121
k
=
250
121
(
−
2
i
−
3
j
−
3
k
).
(c)
The maximum rate of change of
T
is
|∇
T
(2
,
3
,
3)
|
=
250
121
√
4+9+9=
250
√
22
121
.
396

397. Exercises 9.6
43.
Since
∇
f
=
f
x
(
x, y
)
i
+
f
y
(
x, y
)
j
, we have
∂f/∂x
=3
x
2
+
y
3
+
ye
xy
. Integrating, we obtain
f
(
x, y
)=
x
3
+
xy
3
+
e
xy
+
g
(
y
). Then
f
y
=3
xy
2
+
xe
xy
+
g
u
(
y
)=
−
2
y
2
+3
xy
2
+
xe
xy
. Thus,
g
u
(
y
)=
−
2
y
2
,
g
(
y
)=
−
2
3
y
3
+
c
, and
f
(
x, y
)=
x
3
+
xy
3
+
e
xy
−
2
3
y
3
+
C
.
44.
Let
u
=
u
1
i
+
u
2
j
and
v
=
v
1
i
+
v
2
j
.
D
v
f
=(
f
x
i
+
f
y
j
)
·
v
=
v
1
f
x
+
v
2
f
y
D
u
D
v
f
=
u
∂
∂x
(
v
1
f
x
+
v
2
f
y
)
i
+
∂
∂y
(
v
1
f
x
+
v
2
f
y
)
j
v
·
u
=[(
v
1
f
xx
+
v
2
f
yx
)
i
+(
v
1
f
xy
+
v
2
f
yy
)
j
]
·
u
=
u
1
v
1
f
xx
+
u
1
v
2
f
yx
+
u
2
v
1
f
xy
+
u
2
v
2
f
yy
D
u
f
=(
f
x
i
+
f
y
j
)
·
u
=
u
1
f
x
+
u
2
f
y
D
v
D
u
f
=
u
∂
∂x
(
u
1
f
x
+
u
2
f
y
)
i
+
∂
∂y
(
u
1
f
x
+
u
2
f
y
)
j
v
·
v
=[(
u
1
f
xx
+
u
2
f
yx
)
i
+(
u
1
f
xy
+
u
2
f
yy
)
j
]
·
v
=
u
1
v
1
f
xx
+
u
2
v
1
f
yx
+
u
1
v
2
f
xy
+
u
2
v
2
f
yy
Since the second partial derivatives are continuous,
f
xy
=
f
yx
and
D
u
D
v
f
=
D
v
D
u
f
. [Note that this result is
a generalization of
f
xy
=
f
yx
since
D
i
D
j
f
=
f
yx
and
D
j
D
i
f
=
f
xy
.]
45.
∇
(
cf
)=
∂
∂x
(
cf
)
i
+
∂
∂y
(
cf
)
j
=
cf
x
i
+
cf
y
j
=
c
(
f
x
i
+
f
y
j
)=
c
∇
f
46.
∇
(
f
+
g
)=(
f
x
+
g
x
)
i
+(
f
y
+
g
y
)
j
=(
f
x
i
+
f
y
j
)+(
g
x
i
+
g
y
j
)=
∇
f
+
∇
g
47.
∇
(
fg
)=(
fg
x
+
f
x
g
)
i
+(
fg
y
+
f
y
g
)
j
=
f
(
g
x
i
+
g
y
j
)+
g
(
f
x
i
+
f
y
j
)=
f
∇
g
+
g
∇
f
48.
∇
(
f/g
)=[(
gf
x
−
fg
x
)
/g
2
]
i
+[(
gf
y
−
fg
y
)
/g
2
]
j
=
g
(
f
x
i
+
f
y
j
)
/g
2
−
f
(
g
x
i
+
g
y
j
)
/g
2
=
g
∇
f/g
2
−
f
∇
g/g
2
=(
g
∇
f
−
f
∇
g
)
/g
2
49.
∇×
F
=







ijk
∂/∂x ∂/∂y ∂/∂z
f
1
f
2
f
3







=

∂f
3
∂y
−
∂f
2
∂z

i
+

∂f
1
∂z
−
∂f
3
∂x

j
+

∂f
2
∂x
−
∂f
1
∂y

k
Exercises 9.6
1.
Since
f
(6
,
1) = 4, the level curve is
x
−
2
y
=4.
∇
f
=
i
−
2
j
;
∇
f
(6
,
1) =
i
−
2
j
2.
Since
f
(1
,
3) = 5, the level curve is
y
+2
x
=5
x
or
y
=3
x
,
x

=0.
∇
f
=
−
y
x
2
i
+
1
x
j
;
∇
f
(1
,
3) =
−
3
i
+
j
3.
Since
f
(2
,
5) = 1, the level curve is
y
=
x
2
+1.
∇
f
=
−
2
x
i
+
j
;
∇
f
(2
,
5) =
−
10
i
+
j
4.
Since
f
(
−
1
,
3) = 10, the level curve is
x
2
+
y
2
= 10.
∇
f
=2
x
i
+2
y
j
;
∇
f
(
−
1
,
3) =
−
2
i
+6
j
5.
Since
f
(
−
2
,
−
3) = 2, the level curve is
x
2
/
4+
y
2
/
9=2
or
x
2
/
8+
y
2
/
18= 1.
∇
f
=
x
2
i
+
2
y
9
j
;
∇
f
(
−
2
,
−
3) =
−
i
−
2
3
j
397

398. Exercises 9.6
6.
Since
f
(2
,
2) = 2, the level curve is
y
2
=2
x
,
x

=0.
∇
f
=
−
y
2
x
2
i
+
2
y
x
j
;
∇
f
(2
,
2) =
−
i
+2
j
7.
Since
f
(1
,
1) =
−
1, the level curve is (
x
−
1)
2
−
y
2
=
−
1or
y
2
−
(
x
−
1)
2
=1.
∇
f
=2(
x
−
1)
i
−
2
y
j
;
∇
f
(1
,
1) =
−
2
j
8.
Since
f
(
π/
6
,
3
/
2) = 1, the level curve is
y
−
1 = sin
x
or
y
= 1 + sin
x
, sin
x

=0.
∇
f
=
−
(
y
−
1) cos
x
sin
2
x
i
+
1
sin
x
j
;
∇
f
(
π/
6
,
3
/
2) =
−
√
3
i
+2
j
9.
Since
F
(3
,
1
,
1) = 2, the level surface is
y
+
z
=2.
∇
F
=
j
+
k
;
∇
F
(3
,
1
,
1) =
j
+
k
10.
Since
F
(1
,
1
,
3) =
−
1, the level surface is
x
2
+
y
2
−
z
=
−
1or
z
=1+
x
2
+
y
2
.
∇
F
=2
x
i
+2
y
j
−
k
;
∇
F
(1
,
1
,
3) = 2
i
+2
j
−
k
11.
Since
F
(3
,
4
,
0) = 5, the level surface is
x
2
+
y
2
+
z
2
= 25.
∇
F
=
x

x
2
+
y
2
+
z
2
i
+
y

x
2
+
y
2
+
z
2
j
+
z

x
2
+
y
2
+
z
2
k
;
∇
F
(3
,
4
,
0) =
3
5
i
+
4
5
j
12.
Since
F
(0
,
−
1
,
1) = 0, the level surface is
x
2
−
y
2
+
z
=0or
z
=
y
2
−
x
2
.
∇
F
=2
x
i
−
2
y
j
+
k
;
∇
F
(0
,
−
1
,
1) = 2
j
+
k
13.
F
(
x, y, z
)=
x
2
+
y
2
−
z
;
∇
F
=2
x
i
+2
y
j
−
k
. We want
∇
F
=
c
(4
i
+
j
+
1
2
k
)or2
x
=4
c
,2
y
=
c
,
−
1=
c/
2.
From the third equation
c
=
−
2. Thus,
x
=
−
4 and
y
=
−
1. Since
z
=
x
2
+
y
2
= 16 + 1 = 17, the point on the
surface is (
−
4
,
−
1
,
−
17).
14.
F
(
x, y, z
)=
x
3
+
y
2
+
z
;
∇
F
=3
x
2
i
+2
y
j
+
k
. We want
∇
F
=
c
(27
i
+8
j
+
k
)or3
x
2
=27
c
,2
y
=8
c
,1=
c
.
From
c
= 1 we obtain
x
=
±
3 and
y
= 4. Since
z
=15
−
x
3
−
y
2
=15
−
(
±
3)
3
−
16 =
−
1
∓
27, the points on
the surface are (3
,
4
,
−
28) and (
−
3
,
4
,
26).
398

399. Exercises 9.6
15.
F
(
x, y, z
)=
x
2
+
y
2
+
z
2
;
∇
F
=2
x
i
+2
y
j
+2
z
k
.
∇
F
(
−
2
,
2
,
1) =
−
4
i
+4
j
+2
k
. The equation of the tangent
plane is
−
4(
x
+2)+4(
y
−
2)+2(
z
−
1) = 0 or
−
2
x
+2
y
+
z
=9.
16.
F
(
x, y, z
)=5
x
2
−
y
2
+4
z
2
;
∇
F
=10
x
i
−
2
y
j
+8
z
k
;
∇
F
(2
,
4
,
1) = 20
i
−
8
j
+8
k
. The equation of the tangent
plane is 20(
x
−
2)
−
8(
y
−
4)+8(
z
−
1) = 0 or 5
x
−
2
y
+2
z
=4.
17.
F
(
x, y, z
)=
x
2
−
y
2
−
3
z
2
;
∇
F
=2
x
i
−
2
y
j
−
6
z
k
;
∇
F
(6
,
2
,
3)=12
i
−
4
j
−
18
k
. The equation of the tangent
plane is 12(
x
−
6)
−
4(
y
−
2)
−
18(
z
−
3) = 0 or 6
x
−
2
y
−
9
z
=5.
18.
F
(
x, y, z
)=
xy
+
yz
+
zx
;
∇
F
=(
y
+
z
)
i
+(
x
+
z
)
j
+(
y
+
x
)
k
;
∇
F
(1
,
−
3
,
−
5) =
−
8
i
−
4
j
−
2
k
. The equation
of the tangent plane is
−
8(
x
−
1)
−
4(
y
+3)
−
2(
z
+5)=0or4
x
+2
y
+
z
=
−
7.
19.
F
(
x, y, z
)=
x
2
+
y
2
+
z
;
∇
F
=2
x
i
+2
y
j
+
k
;
∇
F
(3
,
−
4
,
0)=6
i
−
8
j
+
k
. The equation of the tangent plane
is 6(
x
−
3)
−
8(
y
+4)+
z
= 0 or 6
x
−
8
y
+
z
= 50.
20.
F
(
x, y, z
)=
xz
;
∇
F
=
z
i
+
x
k
;
∇
F
(2
,
0
,
3)=3
i
+2
k
. The equation of the tangent plane is 3(
x
−
2)+2(
z
−
3) = 0
or 3
x
+2
z
= 12.
21.
F
(
x, y, z
) = cos(2
x
+
y
)
−
z
;
∇
F
=
−
2 sin(2
x
+
y
)
i
−
sin(2
x
+
y
)
j
−
k
;
∇
F
(
π/
2
,π/
4
,
−
1
/
√
2)=
√
2
i
+
√
2
2
j
−
k
.
The equation of the tangent plane is
√
2
e
x
−
π
2
:
+
√
2
2
e
y
−
π
4
:
−

z
+
1
√
2

=0,
2
e
x
−
π
2
:
+
e
y
−
π
4
:
−
√
2

z
+
1
√
2

=0,or2
x
+
y
−
√
2
z
=
5
π
4
+1.
22.
F
(
x, y, z
)=
x
2
y
3
+6
z
;
∇
F
=2
xy
3
i
+3
x
2
y
2
j
+6
k
;
∇
F
(2
,
1
,
1) = 4
i
+12
j
+6
k
. The equation of the tangent
plane is 4(
x
−
2) + 12(
y
−
1)+6(
z
−
1) = 0 or 2
x
+6
y
+3
z
= 13.
23.
F
(
x, y, z
) = ln(
x
2
+
y
2
)
−
z
;
∇
F
=
2
x
x
2
+
y
2
i
+
2
y
x
2
+
y
2
j
−
k
;
∇
F
(1
/
√
2
,
1
/
√
2
,
0) =
√
2
i
+
√
2
j
−
k
.
The equation of the tangent plane is
√
2

x
−
1
√
2

+
√
2

y
−
1
√
2

−
(
z
−
0) = 0,
2

x
−
1
√
2

+2

y
−
1
√
2

−
√
2
z
=0,or2
x
+2
y
−
√
2
z
=2
√
2.
24.
F
(
x, y, z
)=8
e
−
2
y
sin 4
x
−
z
;
∇
F
=32
e
−
2
y
cos 4
x
i
−
16
e
−
2
y
sin 4
x
j
−
k
;
∇
F
(
π/
24
,
0
,
4) = 16
√
3
i
−
8
j
−
k
.
The equation of the tangent plane is
16
√
3(
x
−
π/
24)
−
8(
y
−
0)
−
(
z
−
4) = 0 or 16
√
3
x
−
8
y
−
z
=
2
√
3
π
3
−
4
.
25.
The gradient of
F
(
x, y, z
)=
x
2
+
y
2
+
z
2
is
∇
F
=2
x
i
+2
y
j
+2
z
k
, so the normal vector to the surface at
(
x
0
,y
0
,z
0
)is2
x
0
i
+2
y
0
j
+2
z
0
k
. A normal vector to the plane 2
x
+4
y
+6
z
= 1 is 2
i
+4
j
+6
k
. Since we want
the tangent plane to be parallel to the given plane, we find
c
so that 2
x
0
=2
c
,2
y
0
=4
c
,2
z
0
=6
c
or
x
0
=
c
,
y
0
=2
c
,
z
0
=3
c
.Now,(
x
0
,y
0
,z
0
) is on the surface, so
c
2
+(2
c
)
2
+(3
c
)
2
=14
c
2
= 7 and
c
=
±
1
/
√
2 . Thus,
the points on the surface are (
√
2
/
2
,
√
2
,
3
√
2
/
2) and
−
√
2
/
2
,
−
√
2
,
−
3
√
2
/
2).
26.
The gradient of
F
(
x, y, z
)=
x
2
−
2
y
2
−
3
z
2
is
∇
F
(
x, y, z
)=2
x
i
−
4
y
j
−
6
z
k
, so a normal vector to the surface at
(
x
0
,y
0
,z
0
)is
∇
F
(
x
0
,y
0
,z
0
)=2
x
0
i
−
4
y
0
j
−
6
z
0
k
. A normal vector to the plane 8
x
+4
y
+6
z
= 5 is 8
i
+4
j
+6
k
. Since
we want the tangent plane to be parallel to the given plane, we find
c
so that 2
x
0
=8
c
,
−
4
y
0
=4
c
,
−
6
z
0
=6
c
or
x
0
=4
c
,
y
0
=
−
c
,
z
0
=
−
c
.Now,(
x
0
,y
0
,z
0
) is on the surface, so (4
c
)
2
−
2(
−
c
)
2
−
3(
−
c
)
2
=11
c
2
= 33 and
c
=
±
√
3 . Thus, the points on the surface are (4
√
3
,
−
√
3
,
−
√
3) and (
−
4
√
3
,
√
3
,
√
3).
399

400. Exercises 9.6
27.
The gradient of
F
(
x, y, z
)=
x
2
+4
x
+
y
2
+
z
2
−
2
z
is
∇
F
=(2
x
+4)
i
+2
y
j
+(2
z
−
2)
k
, so a normal to the
surface at (
x
0
,y
0
,z
0
)is(2
x
0
+4)
i
+2
y
0
j
+(2
z
0
−
2)
k
. A horizontal plane has normal
c
k
for
c

= 0. Thus, we
want 2
x
0
+4=0, 2
y
0
=0,2
z
0
−
2=
c
or
x
0
=
−
2,
y
0
=0,
z
0
=
c
+ 1. since (
x
0
,y
0
,z
0
) is on the surface,
(
−
2)
2
+4(
−
2)+(
c
+1)
2
−
2(
c
+1)=
c
2
−
5 = 11 and
c
=
±
4. The points on the surface are (
−
2
,
0
,
5) and
(
−
2
,
0
,
−
3).
28.
The gradient of
F
(
x, y, z
)=
x
2
+3
y
2
+4
z
2
−
2
xy
is
∇
F
=(2
x
−
2
y
)
i
+(6
y
−
2
x
)
j
+8
z
k
, so a normal to the
surface at (
x
0
,y
0
,z
0
)is2(
x
0
−
y
0
)
i
+ 2(3
y
0
−
x
0
)
j
+8
z
0
k
.
(a)
A normal to the
xz
plane is
c
j
for
c

= 0. Thus, we want 2(
x
0
−
y
0
)=0,2(3
y
0
−
x
0
)=
c
,8
z
0
=0or
x
0
=
y
0
,
3
y
0
−
x
0
=
c/
2,
z
0
= 0. Solving the first two equations, we obtain
x
0
=
y
0
=
c/
4. Since (
x
0
,y
0
,z
0
)ison
the surface, (
c/
4)
2
+3(
c/
4)
2
+ 4(0)
2
−
2(
c/
4)(
c/
4)=2
c
2
/
16 = 16 and
c
=
±
16
/
√
2 . Thus, the points on
the surface are (4
/
√
2
,
4
/
√
2
,
0) and (
−
4
/
√
2
,
−
4
/
√
2
,
0).
(b)
A normal to the
yz
-plane is
c
i
for
c

= 0. Thus, we want 2(
x
0
−
y
0
)=
c
, 2(3
y
0
−
x
0
) = 0, 8
z
0
=0or
x
0
−
y
0
=
c/
2,
x
0
=3
y
0
,
z
0
= 0. Solving the first two equations, we obtain
x
0
=3
c/
4 and
y
0
=
c/
4. Since (
x
0
,y
0
,z
0
)
is on the surface, (3
c/
4)
2
+3(
c/
4)
2
+ 4(0)
2
−
2(3
c/
4)(
c/
4) = 6
c
2
/
16 = 16 and
c
=
±
16
√
6 . Thus, the points
on the surface are (12
/
√
6
,
4
/
√
6
,
0) on the surface are (12
/
√
6
,
4
/
√
6
,
0) and (
−
12
/
√
6
,
−
4
/
√
6
,
0).
(c)
A normal to the
xy
-plane is
c
k
for
c

= 0. Thus, we want 2(
x
0
−
y
0
) = 0, 2(3
y
0
−
x
0
) = 0, 8
z
0
=
c
or
x
0
=
y
0
,3
y
0
−
x
0
=0,
z
0
=
c/
8. Solving the first two equations, we obtain
x
0
=
y
0
= 0. Since (
x
0
,y
0
,z
0
)
is on the surface, 0
2
+ 3(0)
2
+4(
c/
8)
2
−
2(0)(0) =
c
2
/
16 = 16 and
c
=
±
16. Thus, the points on the surface
are (0
,
0
,
2) and (0
,
0
,
−
2).
29.
If (
x
0
,y
0
,z
0
)ison
x
2
/a
2
+
y
2
/b
2
+
z
2
/c
2
= 1, then
x
2
0
/a
2
+
y
2
0
/b
2
+
z
2
0
/c
2
= 1 and (
x
0
,y
0
,z
0
) is on the plane
xx
0
/a
2
+
yy
0
/b
2
+
zz
0
/c
2
= 1. A normal to the surface at (
x
0
,y
0
,z
0
)is
∇
F
(
x
0
,y
0
,z
0
)=(2
x
0
/a
2
)
i
+(2
y
0
/b
2
)
j
+(2
z
0
/c
2
)
k
.
A normal to the plane is (
x
0
/a
2
)
i
+(
y
0
/b
2
)
j
+(
z
0
/c
2
)
k
. Since the normal to the surface is a multiple of the
normal to the plane, the normal vectors are parallel and the plane is tangent to the surface.
30.
If (
x
0
,y
0
,z
0
)ison
x
2
/a
2
−
y
2
/b
2
+
z
2
/c
2
= 1, then
x
2
0
/a
2
−
y
2
0
/b
2
+
z
2
0
/c
2
= 1 and (
x
0
,y
0
,z
0
) is on the plane
xx
0
/a
2
−
yy
0
/b
2
+
zz
0
/c
2
= 1. A normal to the surface at (
x
0
,y
0
,z
0
)is
∇
F
(
x
0
,y
0
,z
0
)=(2
x
0
/a
2
)
i
−
(2
y
0
/b
2
)
j
+(2
z
0
/c
2
)
k
.
A normal to the plane is (
x
0
/a
2
)
i
−
(
y
0
/b
2
)
j
+(
z
0
/c
2
)
k
. Since the normal to the surface is a multiple of the
normal to the plane, the normal vectors are parallel, and the plane is tangent to the surface.
31.
Let
F
(
x, y, z
)=
x
2
+
y
2
−
z
2
. Then
∇
F
=2
x
i
+2
y
j
−
2
z
k
and a normal to the surface at (
x
0
,y
0
,z
0
)is
x
0
i
+
y
0
j
−
z
0
k
. An equation of the tangent plane at (
x
0
,y
0
,z
0
)is
x
0
(
x
−
x
0
)+
y
0
(
y
−
y
0
)
−
z
0
(
z
−
z
0
)=0or
x
0
x
+
y
0
y
−
z
0
z
=
x
2
0
+
y
2
0
−
z
2
0
. Since (
x
0
,y
0
,z
0
) is on the surface,
z
2
0
=
x
2
0
+
y
2
0
and
x
2
0
+
y
2
0
−
z
2
0
= 0. Thus,
the equation of the tangent plane is
x
0
x
+
y
0
y
−
z
0
z
= 0, which passes through the origin.
32.
Let
F
(
x, y, z
)=
√
x
+
√
y
+
√
z
. Then
∇
F
=
1
2
√
x
i
+
1
2
√
y
j
+
1
2
√
z
k
and a normal to the surface at (
x
0
,y
0
,z
0
)
is
1
2
√
x
0
i
+
1
2
√
y
0
j
+
1
2
√
z
0
k
. An equation of the tangent plane at (
x
0
,y
0
,z
0
)is
1
2
√
x
0
(
x
−
x
0
)+
1
2
√
y
0
(
y
−
y
0
)+
1
2
√
z
0
(
z
−
z
0
)=0
or
1
√
x
0
x
+
1
√
y
0
y
+
1
√
z
0
z
=
√
x
0
+
√
y
0
+
√
z
0
=
√
a.
400

401. -3 3
x
-3
3
y
-3 3
x
-3
3
y
Exercises 9.7
The sum of the intercepts is
√
x
0
√
a
+
√
y
0
√
a
+
√
z
0
√
a
=(
√
x
0
+
√
y
0
+
√
z
0
)
√
a
=
√
a
·
√
a
=
a
.
33.
F
(
x, y, z
)=
x
2
+2
y
2
+
z
2
;
∇
F
=2
x
i
+4
y
j
+2
z
k
;
∇
F
(1
,
−
1
,
1)=2
i
−
4
j
+2
k
. Parametric equations of the
line are
x
=1+2
t
,
y
=
−
1
−
4
t
,
z
=1+2
t
.
34.
F
(
x, y, z
)=2
x
2
−
4
y
2
−
z
;
∇
F
=4
x
i
−
8
y
j
−
k
;
∇
F
(3
,
−
2
,
2) = 12
i
+16
j
−
k
. Parametric equations of the
line are
x
=3+12
t
,
y
=
−
2+16
t
,
z
=2
−
t
.
35.
F
(
x, y, z
)=4
x
2
+9
y
2
−
z
;
∇
F
=8
x
i
+18
y
j
−
k
;
∇
F
(1
/
2
,
1
/
3
,
3) = 4
i
+6
j
−
k
. Symmetric equations of the
line are
x
−
1
/
2
4
=
y
−
1
/
3
6
=
z
−
3
−
1
.
36.
F
(
x, y, z
)=
x
2
+
y
2
−
z
2
;
∇
F
=2
x
i
+2
y
j
−
2
z
k
;
∇
F
(3
,
4
,
5)=6
i
+8
j
−
10
k
. Symmetric equations of the line
are
x
−
3
6
=
y
−
4
8
=
z
−
5
−
10
.
37.
A normal to the surface at (
x
0
,y
0
,z
0
)is
∇
F
(
x
0
,y
0
,z
0
)=2
x
0
i
+2
y
0
j
+2
z
0
k
. Parametric equations of the normal
line are
x
=
x
0
+2
x
0
t
,
y
=
y
0
+2
y
0
t
,
z
=
z
0
+2
z
0
t
. Letting
t
=
−
1
/
2, we see that the normal line passes
through the origin.
38.
The normal lines to
F
(
x, y, z
)=0and
G
(
x, y, z
)=0are
F
x
i
+
F
y
j
+
F
z
k
and
G
x
i
+
G
y
j
+
G
z
k
, respectively.
These vectors are orthogonal if and only if their dot product is 0. Thus, the surfaces are orthogonal at
P
if and
only if
F
x
G
x
+
F
y
G
y
+
F
z
G
z
=0.
39.
Let
F
(
x, y, z
)=
x
2
+
y
2
+
z
2
−
25 and
G
(
x, y, z
)=
−
x
2
+
y
2
+
z
2
. Then
F
x
G
x
+
F
y
G
y
+
F
z
g
z
=(2
x
)(
−
2
x
)+(2
y
)(2
y
)+(2
z
)(2
z
)=4(
−
x
2
+
y
2
+
z
2
)
.
For (
x, y, z
) on both surfaces,
F
(
x, y, z
)=
G
(
x, y, z
) = 0. Thus,
F
x
G
x
+
F
y
G
y
+
F
z
G
z
= 4(0) = 0 and the
surfaces are orthogonal at points of intersection.
40.
Let
F
(
x, y, z
)=
x
2
−
y
2
+
z
2
−
4 and
G
(
x, y, z
)=1
/xy
2
−
z
. Then
F
x
G
x
+
F
y
G
y
+
F
z
G
z
=(2
x
)(
−
1
/x
2
y
2
)+(
−
2
y
)(
−
2
/xy
3
)+(2
z
)(
−
1)
=
−
2
/xy
2
+4
/xy
2
−
2
z
= 2(1
/xy
2
−
z
)
.
For (
x, y, z
) on both surfaces,
F
(
x, y, z
)=
G
(
x, y, z
)=0. Thus,
F
x
G
x
+
F
y
G
y
+
F
z
G
z
= 2(0) and the surfaces
are orthogonal at points of intersection.
Exercises 9.7
1. 2.
401

402. -3 3
x
-3
3
y
-3 3
x
-3
3
y
-3 3
x
-3
3
y
-3 3
x
-3
3
y
Exercises 9.7
3. 4.
5. 6.
7.
curl
F
=(
x
−
y
)
i
+(
x
−
y
)
j
; div
F
=2
z
8.
curl
F
=
−
2
x
2
i
+ (10
y
−
18
x
2
)
j
+(4
xz
−
10
z
)
k
; div
F
=0
9.
curl
F
=
0
; div
F
=4
y
+8
z
10.
curl
F
=(
xe
2
y
+
ye
−
yz
+2
xye
2
y
)
i
−
ye
2
y
j
+3(
x
−
y
)
2
k
; div
F
=3(
x
−
y
)
2
−
ze
−
yz
11.
curl
F
=(4
y
3
−
6
xz
2
)
i
+(2
z
3
−
3
x
2
)
k
; div
F
=6
xy
12.
curl
F
=
−
x
3
z
i
+(3
x
2
yz
−
z
)
j
+
.
3
2
x
2
y
2
−
y
−
15
y
2
F
k
; div
F
=(
x
3
y
−
x
)
−
(
x
3
y
−
x
)=0
13.
curl
F
=(3
e
−
z
−
8
yz
)
i
−
xe
−
z
j
; div
F
=
e
−
z
+4
z
2
−
3
ye
−
z
14.
curl
F
=(2
xyz
3
+3
y
)
i
+(
y
ln
x
−
y
2
z
3
)
j
+(2
−
z
ln
x
)
k
; div
F
=
yz
x
−
3
z
+3
xy
2
z
2
15.
curl
F
=(
xy
2
e
y
+2
xye
y
+
x
3
ye
z
+
x
3
yze
z
)
i
−
y
2
e
y
j
+(
−
3
x
2
yze
z
−
xe
x
)
k
; div
F
=
xye
x
+
ye
x
−
x
3
ze
z
16.
curl
F
=(5
xye
5
xy
+
e
5
xy
+3
xz
3
sin
xz
3
−
cos
xz
3
)
i
+(
x
2
y
cos
yz
−
5
y
2
e
5
xy
)
j
+(
−
z
4
sin
xz
3
−
x
2
z
cos
yz
)
k
; div
F
=2
x
sin
yz
17.
div
r
=1+1+1=3
18.
curl
r
=







ijk
∂/∂x ∂/∂y ∂/∂z
xyz







=0
i
−
0
j
+0
k
=
0
19. a
×∇
=







ijk
a
1
a
2
a
3
∂/∂x ∂/∂y ∂/∂z







=

a
2
∂
∂z
−
a
3
∂
∂y

i
+

a
3
∂
∂x
−
a
1
∂
∂z

j
+

a
1
∂
∂y
−
a
2
∂
∂x

k
402

403. Exercises 9.7
(
a
×∇
)
×
r
=








ij k
a
2
∂
∂z
−
a
3
a
3
∂
∂x
−
a
1
∂
∂z
a
1
∂
∂y
−
a
2
∂
∂x
xy z








=(
−
a
1
−
a
1
)
i
−
(
a
2
+
a
2
)
j
+(
−
a
3
−
a
3
)
k
=
−
2
a
20.
∇×
(
a
×
r
)=(
∇·
r
)
a
−
(
∇·
a
)
r
=(1+1+1)
a
−

a
1
∂
∂x
+
a
2
∂
∂y
+
a
3
∂
∂z

r
=3
a
−
(
a
1
i
+
a
2
j
+
a
3
k
)=2
a
21.
∇·
(
a
×
r
)=







∂/∂x ∂/∂y ∂/∂z
a
1
a
2
a
3
xyz







=
∂
∂x
(
a
2
z
−
a
3
y
)
−
∂
∂y
(
a
1
z
−
a
3
x
)+
∂
∂z
(
a
1
y
−
a
2
x
)=0
22.
∇×
r
=







ijk
∂/∂x ∂/∂y ∂/∂z
xyz







=
0
;
a
×
(
∇×
r
)=
a
×
0
=
0
23. r
×
a
=







ijk
xyz
a
1
a
2
a
3







=(
a
3
y
−
a
2
z
)
i
−
(
a
3
x
−
a
1
z
)
j
+(
a
2
x
−
a
1
y
)
k
;
r
·
r
=
x
2
+
y
2
+
z
2
∇×
[(
r
·
r
)
a
]=







ijk
∂/∂x ∂/∂y ∂/∂z
(
r
·
r
)
a
1
(
r
·
r
)
a
2
(
r
·
r
)
a
3







=(2
ya
3
−
2
za
2
)
i
−
(2
xa
3
−
2
za
1
)
j
+(2
xa
2
−
2
ya
1
)
k
=2(
r
×
a
)
24. r
·
a
=
a
1
x
+
a
2
y
+
a
3
z
;
r
·
r
=
x
2
+
y
2
+
z
2
;
∇·
[(
r
·
r
)
a
]=2
xa
1
+2
ya
2
+2
za
3
=2(
r
·
a
)
25.
Let
F
=
P
(
x, y, z
)
i
+
Q
(
x, y, z
)
j
+
R
(
x, y, z
)
k
and
G
=
S
(
x, y, z
)
i
+
T
(
x, y, z
)
j
+
U
(
x, y, z
)
k
.
∇·
(
F
+
G
)=
∇·
[(
P
+
S
)
i
+(
Q
+
T
)
j
+(
R
+
U
)
k
]=
P
x
+
S
x
+
Q
y
+
T
y
+
R
z
+
U
z
=(
P
x
+
Q
y
+
R
z
)+(
S
x
+
T
y
+
U
z
)=
∇·
F
+
∇·
G
26.
Let
F
=
P
(
x, y, z
)
i
+
Q
(
x, y, z
)
j
+
R
(
x, y, z
)
k
and
G
=
S
(
x, y, z
)
i
+
T
(
x, y, z
)
j
+
U
(
x, y, z
)
k
.
∇×
(
F
+
G
)=







ijk
∂/∂x ∂/∂y ∂/∂z
P
+
SQ
+
TR
+
U







=(
R
y
+
U
y
−
Q
z
−
T
z
)
i
−
(
R
x
+
U
x
−
P
z
−
S
z
)
j
+(
Q
x
+
T
x
−
P
y
−
S
y
)
k
=(
R
y
−
Q
z
)
i
−
(
R
x
−
P
z
)
j
+(
Q
x
−
P
y
)
k
+(
U
y
−
T
z
)
i
−
(
U
x
−
S
z
)
j
+(
T
x
−
S
y
)
k
=
∇×
F
+
∇×
G
27.
∇·
(
f
F
)=
∇·
(
fP
i
+
fQ
j
+
fR
k
)=
fP
x
+
Pf
x
+
fQ
y
+
Qf
y
+
fR
z
+
Rf
z
=
f
(
P
x
+
Q
y
+
R
z
)+(
Pf
x
+
Qf
y
+
Rf
z
)=
f
(
∇·
F
)+
F
·
(
∇
f
)
403

404. Exercises 9.7
28.
∇×
(
f
F
)=







ijk
∂/∂x ∂/∂y ∂/∂z
fP fQ fR







=(
fR
y
+
Rf
y
−
fQ
z
−
Qf
z
)
i
−
(
fR
x
+
Rf
x
−
fP
z
−
Pf
z
)
j
+(
fQ
x
+
Qf
x
−
fP
y
−
Pf
y
)
k
=(
fR
y
−
fQ
z
)
i
−
(
fR
x
−
fP
z
)
j
+(
fQ
x
−
fP
y
)
k
+(
Rf
y
−
Qf
z
)
i
−
(
Rf
x
−
Pf
z
)
j
+(
Qf
x
−
Pf
y
)
k
=
f
[(
R
y
−
Q
z
)
i
−
(
R
x
−
P
z
)
j
+(
Q
x
−
P
y
)
k
+







ijk
f
x
f
y
f
z
PQR







=
f
(
∇×
F
)+(
∇
f
)
×
F
29.
Assuming continuous second partial derivatives,
curl (grad
f
)=
∇×
(
f
x
i
+
f
y
j
+
f
z
k
)=







ijk
∂/∂x ∂/∂y ∂/∂z
f
x
f
y
f
z







=(
f
zy
−
f
yz
)
i
−
(
f
zx
−
f
xz
)
j
+(
f
yx
−
f
xy
)
k
=
0
.
30.
Assuming continuous second partial derivatives,
div (curl
F
)=
∇·
[(
R
y
−
Q
z
)
i
−
(
R
x
−
P
z
)
j
+(
Q
x
−
P
y
)
k
]
=(
R
yx
−
Q
zx
−
(
R
xy
−
P
zy
)+(
Q
xz
−
P
yz
)=0
.
31.
Let
F
=
P
(
x, y, z
)
i
+
Q
(
x, y, z
)
j
+
R
(
x, y, z
)
k
and
G
=
S
(
x, y, z
)
i
+
T
(
x, y, z
)
j
+
U
(
x, y, z
)
k
.
F
×
G
=







ijk
PQR
STU







=(
QU
−
RT
)
i
−
(
PU
−
RS
)
j
+(
PT
−
QS
)
k
div (
F
×
G
)=(
QU
x
+
Q
x
U
−
RT
x
−
R
x
T
)
−
(
PU
y
+
P
y
U
−
RS
y
−
R
y
S
)+(
PT
z
+
P
z
T
−
QS
z
−
Q
z
S
)
=
S
(
R
y
−
Q
z
)+
T
(
P
z
−
R
x
)+
U
(
Q
x
−
P
y
)
−
P
(
U
y
−
T
z
)
−
Q
(
S
z
−
U
x
)
−
R
(
T
x
−
S
y
)
=
G
·
(curl
F
)
−
F
·
(curl
G
)
32.
Using Problems 26 and 29,
curl (curl
F
+ grad
f
)=
∇×
(curl
F
+ grad
f
)=
∇×
(curl
F
)+
∇×
(grad
f
)
= curl (curl
F
) + curl (grad
f
) = curl (curl
F
)+
0
= curl (curl
F
)
.
33.
∇·∇
f
=
∇·
(
f
x
i
+
f
y
j
+
f
z
k
)=
f
xx
+
f
yy
+
f
zz
34.
Using Problem 27,
∇·
(
f
∇
f
)=
f
(
∇·∇
f
)+
∇
f
·∇
f
=
f
(
∇
2
f
)+
|∇
f
|
2
.
35.
curl
F
=
−
8
yz
i
−
2
z
j
−
x
k
; curl (curl
F
)=2
i
−
(8
y
−
1)
j
+8
z
k
36. (a)
For
F
=
P
i
+
Q
j
+
R
k
,
curl (curl
F
)=(
Q
xy
−
P
yy
−
P
zz
+
R
xz
)
i
+(
R
yz
−
Q
zz
−
Q
xx
+
P
yx
)
j
+(
P
zx
−
R
xx
−
R
yy
+
Q
zy
)
k
and
404

405. Exercises 9.7
−∇
2
F
+ grad (div
F
)=
−
(
P
xx
+
P
yy
+
P
zz
)
i
−
(
Q
xx
+
Q
yy
+
Q
zz
)
j
−
(
R
xx
+
R
yy
+
R
zz
)
k
+ grad (
P
x
+
Q
y
+
R
z
)
=
−
P
xx
i
−
Q
yy
j
−
R
zz
k
+(
−
P
yy
−
P
zz
)
i
+(
−
Q
xx
−
Q
zz
)
j
+(
−
R
xx
−
R
yy
)
k
+(
P
xx
+
Q
yx
+
R
zx
)
i
+(
P
xy
+
Q
yy
+
R
zy
)
j
+(
P
xz
+
Q
yz
+
R
zz
)
k
=(
−
P
yy
−
P
zz
+
Q
yx
+
R
zx
)
i
+(
−
Q
xx
−
Q
zz
+
P
xy
+
R
zy
)
j
+(
−
R
xx
−
R
yy
+
P
xz
+
Q
yz
)
k
.
Thus, curl (curl
F
)=
−∇
2
F
+ grad (div
F
).
(b)
For
F
=
xy
i
+4
yz
2
j
+2
xz
k
,
∇
2
F
=0
i
+8
y
j
+0
k
, div
F
=
y
+4
z
2
+2
x
, and grad (div
F
)=2
i
+
j
+8
z
k
.
Then curl (curl
F
)=
−
8
y
j
+2
i
+
j
+8
z
k
=2
i
+(1
−
8
y
)
j
+8
z
k
.
37.
∂f
∂x
=
−
x
(
x
2
+
y
2
+
z
2
)
−
3
/
2
∂f
∂y
=
−
y
(
x
2
+
y
2
+
z
2
)
−
3
/
2
∂f
∂z
=
−
z
(
x
2
+
y
2
+
z
2
)
−
3
/
2
∂
2
f
∂x
2
=3
x
2
(
x
2
+
y
2
+
z
2
)
−
5
/
2
−
(
x
2
+
y
2
+
z
2
)
−
3
/
2
∂
2
f
∂y
2
=3
y
2
(
x
2
+
y
2
+
z
2
)
−
5
/
2
−
(
x
2
+
y
2
+
z
2
)
−
3
/
2
∂
2
f
∂z
2
=3
z
2
(
x
2
+
y
2
+
z
2
)
−
5
/
2
−
(
x
2
+
y
2
+
z
2
)
−
3
/
2
Adding the second partial derivatives gives
∂
2
f
∂x
2
+
∂
2
f
∂y
2
+
∂
2
f
∂z
2
=
3(
x
2
+
y
2
+
z
2
)
(
x
2
+
y
2
+
z
2
)
5
/
2
−
3(
x
2
+
y
2
+
z
2
)
−
3
/
2
=3(
x
2
+
y
2
+
z
2
)
−
3
/
2
−
3(
x
2
+
y
2
+
z
2
)
−
3
/
2
=0
except when
x
=
y
=
z
=0.
38.
f
x
=
1
1+
4
y
2
(
x
2
+
y
2
−
1)
2

−
4
xy
(
x
2
+
y
2
−
1)
2

=
−
4
xy
(
x
2
+
y
2
−
1)
2
+4
y
2
f
xx
=
−
[(
x
2
+
y
2
−
1)
2
+4
y
2
]4
y
−
4
xy
[4
x
(
x
2
+
y
2
−
1)]
[(
x
2
+
y
2
−
1)
2
+4
y
2
]
2
=
12
x
4
y
−
4
y
5
+8
x
2
y
3
−
8
x
2
y
−
8
y
3
−
4
y
[(
x
2
+
y
2
−
1)
2
+4
y
2
]
2
f
y
=
1
1+
4
y
2
(
x
2
+
y
2
−
1)
2
N
2(
x
2
+
y
2
−
1)
−
4
y
2
(
x
2
+
y
2
−
1)
2
o
=
2(
x
2
−
y
2
−
1)
(
x
2
+
y
2
−
1)
2
+4
y
2
f
yy
=
[(
x
2
+
y
2
−
1)
2
+4
y
2
](
−
4
y
)
−
2(
x
2
−
y
2
−
1)[4
y
(
x
2
+
y
2
−
1)+8
y
]
[(
x
2
+
y
2
−
1)
2
+4
y
2
]
2
=
−
12
x
4
y
+4
y
5
−
8
x
2
y
3
+8
x
2
y
+8
y
3
+4
y
[(
x
2
+
y
2
−
1)
2
+4
y
2
]
2
∇
2
f
=
f
xx
+
f
yy
=0
405

406. Exercises 9.7
39.
curl
F
=
−
Gm
1
m
2







ijk
∂/∂x ∂/∂y ∂/∂z
x/
|
r
|
3
y/
|
r
|
3
z/
|
r
|
3







=
−
Gm
1
m
2
[(
−
3
yz/
|
r
|
5
+3
yz/
|
r
|
5
)
i
−
(
−
3
xz/
|
r
|
5
+3
xz/
|
r
|
5
)
j
+(
−
3
xy/
|
r
|
5
+3
xy/
|
r
|
5
)
k
]
=
0
div
F
=
−
Gm
1
m
2
N
−
2
x
2
+
y
2
+
z
2
|
r
|
5
/
2
+
x
2
−
2
y
2
+
z
2
|
r
|
5
/
2
+
x
2
+
y
2
−
2
z
2
|
r
|
5
/
2
o
=0
40.
1
2
curl
v
=
1
2
curl (
ω
ω
ω
×
r
)=
1
2







ijk
∂/∂x ∂/∂y ∂/∂z
ω
2
z
−
ω
3
yω
3
x
−
ω
1
zω
1
y
−
ω
2
x







=
1
2
[(
ω
1
+
ω
1
)
i
−
(
−
ω
2
−
ω
2
)
j
+(
ω
3
+
ω
3
)
k
]=
ω
1
i
+
ω
2
j
+
ω
3
k
=
ω
ω
ω
41.
Using Problems 31 and 29,
∇·
F
= div (
∇
f
×∇
g
)=
∇
g
·
(curl
∇
f
)
−∇
f
·
(curl
∇
g
)=
∇
g
·
0
−∇
f
·
0
=0
.
42.
Recall that
a
·
(
a
×
b
) = 0. Then, using Problems 31, 29, and 28,
∇·
F
= div (
∇
f
×
f
∇
g
)=
f
∇
g
·
(curl
∇
f
)
−∇
f
·
(curl
f
∇
g
)=
f
∇
g
·
0
−∇
f
·
(
∇×
f
∇
g
)
=
−∇
f
·
[
f
(
∇×∇
g
)+(
∇
f
×∇
g
)] =
−∇
f
·
[
f
curl
∇
g
+(
∇
f
×∇
g
]
=
−∇
f
·
[
f
0
+(
∇
f
×∇
g
)] =
−∇
f
·
(
∇
f
×∇
g
)=0
.
43. (a)
Expressing the vertical component of
V
in polar coordinates, we have
2
xy
(
x
2
+
y
2
)
2
=
2
r
2
sin
θ
cos
θ
r
4
=
sin 2
θ
r
2
.
Similarly,
x
2
−
y
2
(
x
2
+
y
2
)
2
=
r
2
(cos
2
θ
−
sin
2
θ
)
r
4
=
cos 2
θ
r
2
.
Since lim
r
→∞
(sin 2
θ
)
/r
2
= lim
r
→∞
(cos 2
θ
)
/r
2
=0,
V
≈
A
i
for
r
large or (
x, y
) far from the origin.
(b)
Identifying
P
(
x, y
)=
A
N
1
−
x
2
−
y
2
(
x
2
−
y
2
)
2
o
,
Q
(
x, y
)=
−
2
Axy
(
x
2
+
y
2
)
2
, and
R
(
x, y
) = 0, we have
P
y
=
2
Ay
(3
x
2
−
y
2
)
(
x
2
+
y
2
)
3
,Q
x
=
2
Ay
(3
x
2
−
y
2
)
(
x
2
+
y
2
)
3
,
and
P
z
=
Q
z
=
R
x
=
R
y
=0
.
Thus, curl
V
=(
R
y
−
Q
z
)
i
+(
P
z
−
R
x
)
j
+(
Q
x
−
P
y
)
k
= 0 and
V
is irrotational.
(c)
Since
P
x
=
2
Ax
(
x
2
−
3
y
2
)
(
x
2
+
y
2
)
3
,
Q
y
=
2
Ax
(3
y
2
−
x
2
)
(
x
2
+
y
2
)
3
, and
R
z
=0,
∇·
F
=
P
x
+
Q
y
+
R
z
= 0 and
V
is
incompressible.
44.
We first note that curl (
∂
H
/∂t
)=
∂
(curl
H
)
/∂t
and curl (
∂
E
/∂t
)=
∂
(curl
E
)
/∂t
. Then, from Problem 30,
−∇
2
E
=
−∇
2
E
+
0
=
−∇
2
E
+ grad 0 =
−∇
2
E
+ grad (div
E
) = curl (curl
E
)
= curl

−
1
c
∂
H
∂t

=
−
1
c
∂
∂t
curl
H
=
−
1
c
∂
∂t

1
c
∂
E
∂t

=
−
1
c
2
∂
2
E
∂t
406

407. Exercises 9.8
and
∇
2
E
=
1
c
2
∂
2
E
/∂t
2
. Similarly,
−∇
2
H
=
−∇
2
H
+ grad (div
H
) = curl (curl
H
) = curl

1
c
∂
E
∂t

=
1
c
∂
∂t
curl
E
=
1
c
∂
∂t

−
1
c
∂
H
∂t

=
−
1
c
2
∂
2
H
∂t
2
and
∇
2
H
=
1
c
2
∂
2
H
/∂t
2
.
45.
We note that div
F
=2
xyz
−
2
xyz
+1=1

=0. If
F
= curl
G
, then div (curl
G
) = div
F
= 1. But, by
Problem 30, for any vector field
G
, div (curl
G
)=0. Thus,
F
cannot be the curl of
G
.
Exercises 9.8
1.
s
C
2
xy dx
=
s
π/
4
0
2(5 cos
t
)(5 sin
t
)(
−
5 sin
t
)
dt
=
−
250
s
π/
4
0
sin
2
t
cos
tdt
=
−
250

1
3
sin
3
t





π/
4
0
=
−
125
√
2
6
s
C
2
xy dy
=
s
π/
4
0
2(5 cos
t
)(5 sin
t
)(5 cos
t
)
dt
= 250
s
π/
4
0
cos
2
t
sin
tdt
= 250

−
1
3
cos
3
t





π/
4
0
=
250
3
n
1
−
√
2
4
g
=
125
6
(4
−
√
2)
s
C
2
xy ds
=
s
π/
4
0
2(5 cos
t
)(5 sin
t
)

25 sin
2
t
+ 25 cos
2
tdt
= 250
s
π/
4
0
sin
t
cos
tdt
= 250

1
2
sin
2
t





π/
4
0
=
125
2
2.
s
C
(
x
3
+2
xy
2
+2
x
)
dx
=
s
1
0
[8
t
3
+ 2(2
t
)(
t
4
) + 2(2
t
)]2
dt
=2
s
1
0
(8
t
3
+4
t
5
+4
t
)
dt
=2

2
t
4
+
2
3
t
6
+2
t
2





1
0
=
28
3
s
C
(
x
3
+2
xy
2
+2
x
)
dy
=
s
1
0
[8
t
3
+ 2(2
t
)(
t
4
) + 2(2
t
)]2
tdt
=2
s
1
0
(8
t
4
+4
t
6
+4
t
2
)
dt
=2

8
5
t
5
+
4
7
t
7
+
4
3
t
3





1
0
=
736
105
s
C
(
x
3
+2
xy
2
+2
x
)
ds
=
s
1
0
[8
t
3
+ 2(2
t
)(
t
4
) + 2(2
t
)]

4+4
t
2
dt
=8
s
1
0
t
(1 +
t
2
)
5
/
2
dt
=8

1
7
(1 +
t
2
)
7
/
2





1
0
=
8
7
(2
7
/
2
−
1)
3.
s
C
(3
x
2
+6
y
2
)
dx
=
s
0
−
1
[3
x
2
+ 6(2
x
+1)
2
]
dx
=
s
0
−
1
(27
x
2
+24
x
+6)
dx
=(9
x
3
+12
x
2
+6
x
)



0
−
1
=
−
(
−
9+12
−
6) = 3
s
C
(3
x
2
+6
y
2
)
dy
=
s
0
−
1
[3
x
2
+ 6(2
x
+1)
2
]2
dx
=6
s
C
(3
x
2
+6
y
2
)
ds
=
s
0
−
1
[3
x
2
+ 6(2
x
+1)
2
]
√
1+4
dx
=3
√
5
407

408. Exercises 9.8
4.
s
C
x
2
y
3
dx
=
s
8
1
x
2
27
x
2
/
8
dx
=
8
27
s
8
1
dx
=
56
27
s
C
x
2
y
3
dy
=
s
8
1
x
2
27
x
2
/
8
x
−
1
/
3
dx
=
8
27
s
8
1
x
−
1
/
3
dx
=
4
9
x
2
/
3




8
1
=
4
3
s
C
x
2
y
3
ds
=
s
8
1
x
2
27
x
2
/
8

1+
x
−
2
/
3
dx
=
8
27
s
8
1
x
−
1
/
3

1+
x
2
/
3
dx
=
8
27
(1 +
x
2
/
3
)
3
/
2




8
1
=
8
27
(5
3
/
2
−
2
3
/
2
)
5.
s
C
zdx
=
s
π/
2
0
t
(
−
sin
t
)
dt
Integration by parts
=(
t
cos
t
−
sin
t
)



π/
2
0
=
−
1
s
C
zdy
=
s
π/
2
0
t
cos
tdt
Integration by parts
=(
t
sin
t
+ cos
t
)



π/
2
0
=
π
2
−
1
s
C
zdz
=
s
π/
2
0
tdt
=
1
2
t
2




π/
2
0
=
π
2
8
s
C
zds
=
s
π/
2
0
t

sin
2
t
+ cos
2
t
+1
dt
=
√
2
s
π/
2
0
tdt
=
π
2
√
2
8
6.
s
C
4
xyz dx
=
s
1
0
4

1
3
t
3

(
t
2
)(2
t
)
t
2
dt
=
8
3
s
1
0
t
8
dt
=
8
27
t
9




1
0
=
8
27
s
C
4
xyz dy
=
s
1
0
4

1
3
t
3

(
t
2
)(2
t
)2
tdt
=
16
3
s
1
0
t
7
dt
=
2
3
t
8




1
0
=
2
3
s
C
4
xyz dz
=
s
1
0
4

1
3
t
3

(
t
2
)(2
t
)2
dt
=
16
3
s
1
0
t
6
dt
=
16
21
t
7




1
0
=
16
21
s
C
4
xyz ds
=
s
1
0
4

1
3
t
3

(
t
2
)(2
t
)

t
4
+4
t
2
+4
dt
=
8
3
s
1
0
t
6
(
t
2
+2)
dt
=
8
3

1
9
t
9
+
2
7
t
7





1
0
=
200
189
7.
Using
x
as the parameter,
dy
=
dx
and
s
C
(2
x
+
y
)
dx
+
xy dy
=
s
2
−
1
(2
x
+
x
+3+
x
2
+3
x
)
dx
=
s
2
−
1
(
x
2
+6
x
+3)
dx
=

1
3
x
3
+3
x
2
+3
x





2
−
1
=21
.
8.
Using
x
as the parameter,
dy
=2
xdx
and
s
C
(2
x
+
y
)
dx
+
xy dy
=
s
2
−
1
(2
x
+
x
2
+1)
dx
+
s
2
−
1
x
(
x
2
+1)2
xdx
=
s
2
−
1
(2
x
4
+3
x
2
+2
x
+1)
dx
=

2
5
x
5
+
x
3
+
x
2
+
x





2
−
1
=
141
5
.
9.
From (
−
1
,
2) to (2
,
2) we use
x
as a parameter with
y
= 2 and
dy
= 0. From (2
,
2) to (2
,
5) we use
y
as a
parameter with
x
= 2 and
dx
=0.
s
C
(2
x
+
y
)
dx
+
xy dy
=
s
2
−
1
(2
x
+2)
dx
+
s
5
2
2
ydy
=(
x
2
+2
x
)



2
−
1
+
y
2



5
2
=9+21=30
408

409. Exercises 9.8
10.
From (
−
1
,
2) to (
−
1
,
0) we use
y
as a parameter with
x
=
−
1 and
dx
= 0. From (
−
1
,
0) to (2
,
0) we use
x
as a
parameter with
y
=
dy
= 0. From (2
,
0) to (2
,
5) we use
y
as a parameter with
x
= 2 and
dx
=0.
s
C
(2
x
+
y
)
dx
+
xy dy
=
s
0
2
(
−
1)
ydy
+
s
2
−
1
2
xdx
+
s
5
0
2
ydy
=
−
1
2
y
2



0
2
+
x
2



2
−
1
+
y
2



5
0
=2+3+25=30
11.
Using
x
as a the parameter,
dy
=2
xdx
.
s
C
ydx
+
xdy
=
s
1
0
x
2
dx
+
s
1
0
x
(2
x
)
dx
=
s
1
0
3
x
2
dx
=
x
3



1
0
=1
12.
Using
x
as the parameter,
dy
=
dx
.
s
C
ydx
+
xdy
=
s
1
0
xdx
+
s
1
0
xdx
=
s
1
0
2
xdx
=
x
2



1
0
=1
13.
From (0
,
0) to (0
,
1) we use
y
as a parameter with
x
=
dx
= 0. From (0
,
1) to (1
,
1) we use
x
as a parameter
with
y
= 1 and
dy
=0.
s
C
ydx
+
xdy
=0+
s
1
0
1
dx
=1
14.
From (0
,
0) to (1
,
0) we use
x
as a parameter with
y
=
dy
= 0. From (1
,
0) to (1
,
1) we use
y
as a parameter
with
x
= 1 and
dx
=0.
s
C
ydx
+
xdy
=0+
s
1
0
1
dy
=1
15.
s
C
(6
x
2
+2
y
2
)
dx
+4
xy dy
=
s
9
4
(6
t
+2
t
2
)
1
2
t
−
1
/
2
dt
+
s
9
4
4
√
ttdt
=
s
9
4
(3
t
1
/
2
+5
t
3
/
2
)
dt
=(2
t
3
/
2
+2
t
5
/
2
)



9
4
= 460
16.
s
C
(
−
y
2
)
dx
+
xy dy
=
s
2
0
(
−
t
6
)2
dt
+
s
2
0
(2
t
)(
t
3
)3
t
2
dt
=
s
2
0
4
t
6
dt
=
4
7
t
7




2
0
=
512
7
17.
s
C
2
x
3
ydx
+(3
x
+
y
)
dy
=
s
1
−
1
2(
y
6
)
y
2
ydy
+
s
1
−
1
(3
y
2
+
y
)
dy
=
s
1
−
1
(4
y
8
+3
y
2
+
y
)
dy
=

4
9
y
9
+
y
3
+
1
2
y
2





1
−
1
=
26
9
18.
s
C
4
xdx
+2
ydy
=
s
2
−
1
4(
y
3
+ 1)3
y
2
dy
+
s
2
−
1
2
ydy
=
s
2
−
1
(12
y
5
+12
y
2
+2
y
)
dy
=(2
y
6
+4
y
3
+
y
2
)



2
−
1
= 165
19.
From (
−
2
,
0) to (2
,
0) we use
x
as a parameter with
y
=
dy
= 0. From (2
,
0) to (
−
2
,
0) we parameterize the
semicircle as
x
= 2 cos
θ
and
y
= 2 sin
θ
for 0
≤
θ
≤
π
.
=
ˇ
C
(
x
2
+
y
2
)
dx
−
2
xy dy
=
s
2
−
2
x
2
dx
+
s
π
0
4(
−
2 sin
θdθ
)
−
s
π
0
8cos
θ
sin
θ
(2 cos
θdθ
)
=
1
3
x
3




2
−
2
−
8
s
π
0
(sin
θ
+ 2 cos
2
θ
sin
θ
)
dθ
=
16
3
−
8

−
cos
θ
−
2
3
cos
3
θ





π
0
=
16
3
−
80
3
=
−
64
3
409

410. Exercises 9.8
20.
We start at (0
,
0) and use
x
as a parameter.
=
ˇ
C
(
x
2
+
y
2
)
dx
−
2
xy dy
=
s
1
0
(
x
2
+
x
4
)
dx
−
2
s
1
0
xx
2
(2
xdx
)+
s
0
1
(
x
2
+
x
)
dx
−
2
s
0
1
x
√
x

1
2
x
−
1
/
2

dx
=
s
1
0
(
x
2
−
3
x
4
)
dx
+
s
0
1
x
2
dx
=
s
1
0
(
−
3
x
4
)
dx
=
−
3
5
x
5




1
0
=
−
3
5
21.
From (1
,
1) to (
−
1
,
1) and (
−
1
,
−
1) to (1
,
−
1) we use
x
as a parameter with
y
= 1 and
y
=
−
1, respectively,
and
dy
= 0. From (
−
1
,
1) to (
−
1
,
−
1) and (1
,
−
1) to (1
,
1) we use
y
as a parameter with
x
=
−
1 and
x
=1,
respectively, and
dx
=0.
=
ˇ
C
x
2
y
3
dx
−
xy
2
dy
=
s
−
1
1
x
2
(1)
dx
+
s
−
1
1
−
(
−
1)
y
2
dy
+
s
1
−
1
x
2
(
−
1)
3
dx
+
s
1
−
1
−
(1)
y
2
dy
=
1
3
x
3




−
1
1
+
1
3
y
3




−
1
1
−
1
3
x
3




1
−
1
−
1
3
y
3




1
−
1
=
−
8
3
22.
From (2
,
4) to (0
,
4) we use
x
as a parameter with
y
= 4 and
dy
= 0. From (0
,
4) to (0
,
0) we use
y
as a parameter
with
x
=
dx
= 0. From (0
,
0) to (2
,
4) we use
y
=2
x
and
dy
=2
dx
.
=
ˇ
C
x
2
y
3
dx
−
xy
2
dy
=
s
0
2
x
2
(64)
dx
−
s
0
4
0
dy
+
s
2
0
x
2
(8
x
3
)
dx
−
s
2
0
x
(4
x
2
)2
dx
=
64
3
x
3




0
2
+
4
3
x
6




2
0
−
2
x
4




2
0
=
−
512
3
+
256
3
−
32 =
−
352
3
23.
=
ˇ
C
(
x
2
−
y
2
)
ds
=
s
2
π
0
(25 cos
2
θ
−
25 sin
2
θ
)

25 sin
2
θ
+ 25 cos
2
θdθ
= 125
s
2
π
0
(cos
2
θ
−
sin
2
θ
)
dθ
= 125
s
2
π
0
cos 2
θdθ
=
125
2
sin 2
θ




2
π
0
=0
24.
=
ˇ
C
ydx
−
xdy
=
s
π
0
3 sin
t
(
−
2 sin
t
)
dt
−
s
π
0
2 cos
t
(3 cos
t
)
dt
=
−
6
s
π
0
(sin
2
t
+ cos
2
t
)
dt
=
−
6
s
π
0
dt
=
−
6
π
Thus,
s
−
C
ydx
−
xdy
=6
π
.
25.
We parameterize the line segment from (0
,
0
,
0) to (2
,
3
,
4) by
x
=2
t
,
y
=3
t
,
z
=4
t
for 0
≤
t
≤
1. We
parameterize the line segment from (2
,
3
,
4) to (6
,
8
,
5) by
x
=2+4
t
,
y
=3+5
t
,
z
=4+
t
,0
≤
t
≤
1.
=
ˇ
C
ydx
+
zdy
+
xdz
=
s
1
0
3
t
(2
dt
)+
s
1
0
4
t
(3
dt
)+
s
1
0
2
t
(4
dt
)+
s
1
0
(3+5
t
)(4
dt
)
+
s
1
0
(4 +
t
)(5
dt
)+
s
1
0
(2+4
t
)
dt
=
s
1
0
(55
t
+ 34)
dt
=

55
2
t
2
+34
t





1
0
=
123
2
26.
s
C
ydx
+
zdy
+
xdz
=
s
2
0
t
3
(3
dt
)+
s
2
0

5
4
t
2

(3
t
2
dt
)+
s
2
0
(3
t
)

5
2
tdt

=
s
2
0

3
t
3
+
15
4
t
4
+
15
2
t
2

dt
=

3
4
t
4
+
3
4
t
5
+
5
2
t
3





2
0
=56
410

411. Exercises 9.8
27.
From (0
,
0
,
0) to (6
,
0
,
0) we use
x
as a parameter with
y
=
dy
= 0 and
z
=
dz
= 0. From (6
,
0
,
0) to (6
,
0
,
5) we
use
z
as a parameter with
x
= 6 and
dx
= 0 and
y
=
dy
= 0. From (6
,
0
,
5) to (6
,
8
,
5) we use
y
as a parameter
with
x
= 6 and
dx
= 0 and
z
= 5 and
dz
=0.
s
C
ydx
+
zdy
+
xdz
=
s
6
0
0+
s
5
0
6
dz
+
s
8
0
5
dy
=70
28.
We parametrize the line segment from (0
,
0
,
0) to (6
,
8
,
0) by
x
=6
t
,
y
=8
t
,
z
= 0 for 0
≤
t
≤
1. From (6
,
8
,
0)
to (6
,
8
,
5) we use
z
as a parameter with
x
=6,
dx
= 0, and
y
=8,
dy
=0.
s
C
ydx
+
zdy
+
xdz
=
s
1
0
8
t
(6
dt
)+
s
5
0
6
dz
=24
t
2



1
0
+30=54
29. F
=
e
3
t
i
−
(
e
−
4
t
)
e
t
j
=
e
3
t
i
−
e
−
3
t
j
;
d
r
=(
−
2
e
−
2
t
i
+
e
t
j
)
dt
;
F
·
d
r
=(
−
2
e
t
−
e
−
2
t
)
dt
;
s
C
F
·
d
r
=
s
ln 2
0
(
−
2
e
t
−
e
−
2
t
)
dt
=

−
2
e
t
+
1
2
e
−
2
t





ln 2
0
=
−
31
8
−

−
3
2

=
−
19
8
30. F
=
e
t
i
+
te
t
3
j
+
t
3
e
t
6
k
;
d
r
=(
i
+2
t
j
+3
t
2
k
)
dt
;
s
C
F
·
d
r
=
s
1
0
(
e
t
+2
t
2
e
t
3
+3
t
5
e
t
6
)
dt
=

e
t
+
2
3
e
t
3
+
1
2
e
t
6





1
0
=
13
6
(
e
−
1)
31.
Using
x
as a parameter,
r
(
x
)=
x
i
+ln
x
j
. Then
F
=ln
x
i
+
x
j
,
d
r
=(
i
+
1
x
j
)
dx
, and
W
=
s
C
F
·
d
r
=
s
e
1
(ln
x
+1)
dx
=(
x
ln
x
)



e
1
=
e.
32.
Let
r
1
=(
−
2+2
t
)
i
+(2
−
2
t
)
j
and
r
2
=2
t
i
+3
t
j
for 0
≤
t
≤
1. Then
d
r
1
=2
i
−
2
j
,d
r
2
=2
i
+3
j
,
F
1
=2(
−
2+2
t
)(2
−
2
t
)
i
+ 4(2
−
2
t
)
2
j
=(
−
8
t
2
+16
t
−
8)
i
+ (16
t
2
−
32
t
+ 16)
j
,
F
2
= 2(2
t
)(3
t
)
i
+ 4(3
t
)
2
j
=12
t
2
i
+36
t
2
j
,
and
W
=
s
C
1
F
1
·
d
r
1
+
s
cC
2
F
2
·
d
r
2
=
s
1
0
(
−
16
t
2
+32
t
−
16
−
32
t
2
+64
t
−
32)
dt
+
s
1
0
(24
t
2
+ 108
t
2
)
dt
=
s
1
0
(84
t
2
+96
t
−
48)
dt
= (28
t
3
+48
t
2
−
48
t
)



1
0
=28
.
33.
Let
r
1
=(1+2
t
)
i
+
j
,
r
2
=3
i
+(1+
t
)
j
, and
r
3
=(3
−
2
t
)
i
+(2
−
t
)
j
for 0
≤
t
≤
1.
Then
d
r
1
=2
i
,d
r
2
=
j
,d
r
3
=
−
2
i
−
j
,
F
1
=(1+2
t
+2)
i
+(6
−
2
−
4
t
)
j
=(3+2
t
)
i
+(4
−
4
t
)
j
,
F
2
=(3+2+2
t
)
i
+(6+6
t
−
6)
j
=(5+2
t
)
i
+6
t
j
,
F
3
=(3
−
2
t
+4
−
2
t
)
i
+ (12
−
6
t
−
6+4
t
)
j
=(7
−
4
t
)
i
+(6
−
2
t
)
j
,
411

412. Exercises 9.8
and
W
=
s
C
1
F
1
·
d
r
1
+
s
C
2
F
2
·
d
r
2
+
s
C
3
F
3
·
d
r
3
=
s
1
0
(6+4
t
)
dt
+
s
1
0
6
tdt
+
s
1
0
(
−
14+8
t
−
6+2
t
)
dt
=
s
1
0
(
−
14+20
t
)
dt
=(
−
14
t
+10
t
2
)



1
0
=
−
4
.
34. F
=
t
3
i
+
t
4
j
+
t
5
k
;
d
r
=3
t
2
i
+2
t
j
+
k
;
W
=
s
C
F
·
d
r
=
s
3
1
(3
t
5
+2
t
5
+
t
5
)
dt
=
s
3
1
6
t
5
dt
=
t
6



3
1
= 728
35. r
= 3 cos
t
i
+ 3 sin
t
j
,0
≤
t
≤
2
π
;
d
r
=
−
3 sin
t
i
+ 3 cos
t
j
;
F
=
a
i
+
b
j
;
W
=
s
C
F
·
d
r
=
s
2
π
0
(
−
3
a
sin
t
+3
b
cos
t
)
dt
=(3
a
cos
t
+3
b
sin
t
)



2
π
0
=0
36.
Let
r
=
t
i
+
t
j
+
t
k
for 1
≤
t
≤
3. Then
d
r
=
i
+
j
+
k
, and
F
=
c
|
r
|
3
(
t
i
+
t
j
+
t
k
)=
ct
(
√
3
t
2
)
3
(
i
+
j
+
k
)=
c
3
√
3
t
2
(
i
+
j
+
k
)
,
W
=
s
C
F
·
d
r
=
s
3
1
c
3
√
3
t
2
(1+1+1)
dt
=
c
√
3
s
3
1
1
t
2
dt
=
c
√
3

−
1
t





3
1
=
c
√
3

−
1
3
+1

=
2
c
3
√
3
.
37.
s
C
1
y
2
dx
+
xy dy
=
s
1
0
(4
t
+2)
2
2
dt
+
s
1
0
(2
t
+ 1)(4
t
+ 2)4
dt
=
s
1
0
(64
t
2
+64
t
+ 16)
dt
=

64
3
t
3
+32
t
2
+16
t





1
0
=
64
3
+ 32 + 16 =
208
3
s
C
2
y
2
dx
+
xy dy
=
s
√
3
1
4
t
4
(2
t
)
dt
+
s
√
3
1
2
t
4
(4
t
)
dt
=
s
√
3
1
16
t
5
dt
=
8
3
t
6




√
3
1
=72
−
8
3
=
208
3
s
C
3
y
2
dx
+
xy dy
=
s
e
3
e
4(ln
t
)
2
1
t
dt
+
s
e
3
e
2(ln
t
)
2
2
t
dt
=
s
e
3
e
8
t
(ln
t
)
2
dt
=
8
3
(ln
t
)
3




e
3
e
=
8
3
(27
−
1) =
208
3
38.
s
C
1
xy ds
=
s
2
0
t
(2
t
)
√
1+4
dt
=2
√
5
s
2
0
t
2
dt
=2
√
5

1
3
t
3





2
0
=
16
√
5
3
s
C
2
xy ds
=
s
2
0
t
(
t
2
)

1+4
t
2
dt
=
s
2
0
t
3

1+4
t
2
dt
u
=1+4
t
2
,du
=8
t dt, t
2
=
1
4
(
u
−
1)
=
s
17
1
1
4
(
u
−
1)
u
1
/
2
1
8
du
=
1
32
s
17
1
(
u
3
/
2
−
u
1
/
2
)
du
=
1
32

2
5
u
5
/
2
−
2
3
u
3
/
2





17
1
=
391
√
17+1
120
s
C
3
xy ds
=
s
3
2
(2
t
−
4)(4
t
−
8)
√
4+16
dt
=16
√
5
s
3
2
(
t
−
2)
2
dt
=16
√
5
N
1
3
(
t
−
2)
3
o




3
2
=
16
√
5
3
C
1
and
C
3
are different parameterizations of the same curve, while
C
1
and
C
2
are different curves.
39.
Since
v
·
v
=
v
2
,
d
dt
v
2
=
d
dt
(
v
·
v
)=
v
·
d
v
dt
+
d
v
dt
·
v
=2
d
v
dt
·
v
. Then
W
=
s
C
F
·
d
r
=
s
b
a
m
a
·

d
r
dt
dt

=
m
s
b
a
d
v
dt
·
v
dt
=
m
s
b
a
1
2

d
dt
v
2

dt
=
1
2
m
(
v
2
)



b
a
=
1
2
m
[
v
(
b
)]
2
−
1
2
m
[
v
(
a
)]
2
.
412

413. Exercises 9.9
40.
We are given
ρ
=
kx
. Then
m
=
s
C
ρds
=
s
π
0
kx ds
=
k
s
π
0
(1 + cos
t
)

sin
2
t
+ cos
2
tdt
=
k
s
π
0
(1 + cos
t
)
dt
=
k
(
t
+ sin
t
)



π
0
=
kπ.
41.
From Problem 40,
m
=
kπ
and
ds
=
dt
.
M
x
=
s
C
yρ ds
=
s
C
kxy ds
=
k
s
π
0
(1 + cos
t
) sin
tdt
=
k

−
cos
t
+
1
2
sin
2
t





π
0
=2
k
M
y
=
s
C
xρ ds
=
s
C
kx
2
ds
=
k
s
π
0
(1 + cos
t
)
2
dt
=
k
s
π
0
(1 + 2 cos
t
+ cos
2
t
)
dt
=
k

t
+ 2 sin
t
+
1
2
t
+
1
4
sin 2
t





π
0
=
3
2
kπ
¯
x
=
M
y
/m
=
3
kπ/
2
kπ
=
3
2
;¯
y
=
M
x
/m
=
2
k
kπ
=
2
π
. The center of mass is (3
/
2
,
2
/π
).
42.
On
C
1
,
T
=
i
and
F
·
T
= comp
T
F
≈
1. On
C
2
,
T
=
−
j
and
F
·
T
= comp
T
F
≈
2. On
C
3
,
T
=
−
i
and
F
·
T
= comp
T
F
≈
1
.
5. Using the fact that the lengths of
C
1
,
C
2
, and
C
3
are 4, 5, and 5, respectively, we have
W
=
s
C
F
·
T
ds
=
s
C
1
F
·
T
ds
+
s
C
2
F
·
T
ds
+
s
C
3
F
·
T
ds
≈
1(4) + 2(5) + 1
.
5(5) = 21
.
5 ft-lb.
Exercises 9.9
1. (a)
P
y
=0=
Q
x
and the integral is independent of path.
φ
x
=
x
2
,
φ
=
1
3
x
3
+
g
(
y
),
φ
y
=
g
u
(
y
)=
y
2
,
g
(
y
)=
1
3
y
3
,
φ
=
1
3
x
3
+
1
3
y
3
,
s
(2
,
2)
(0
,
0)
x
2
dx
+
y
2
dy
=
1
3
(
x
3
+
y
3
)



(2
,
2)
(0
,
0)
=
16
3
(b)
Use
y
=
x
for 0
≤
x
≤
2.
s
(2
,
2)
(0
,
0)
x
2
dx
+
y
2
dy
=
s
2
0
(
x
2
+
x
2
)
dx
=
2
3
x
3



2
0
=
16
3
2. (a)
P
y
=2
x
=
Q
x
and the integral is independent of path.
φ
x
=2
xy
,
φ
=
x
2
y
+
g
(
y
),
φ
y
=
x
2
+
g
u
(
y
)=
x
2
,
g
(
y
)=0,
φ
=
x
2
y
,
s
(2
,
4)
(1
,
1)
2
xy dx
+
x
2
dy
=
x
2
y



(2
,
4)
(1
,
1)
=16
−
1=15
(b)
Use
y
=3
x
−
2 for 1
≤
x
≤
2.
s
(2
,
4)
(1
,
1)
2
xy dx
+
x
2
dy
=
s
2
1
[2
x
(3
x
−
2) +
x
2
(3)]
dx
=
s
2
1
(9
x
2
−
4
x
)
dx
=(3
x
3
−
2
x
2
)



2
1
=15
3. (a)
P
y
=2=
Q
x
and the integral is independent of path.
φ
x
=
x
+2
y
,
φ
=
1
2
x
2
+2
xy
+
g
(
y
),
φ
y
=2
x
+
g
u
(
y
)=2
x
−
y
,
g
(
y
)=
−
1
2
y
2
,
φ
=
1
2
x
2
+2
xy
−
1
2
y
2
,
s
(3
,
2)
(1
,
0)
(
x
+2
y
)
dx
+(2
x
−
y
)
dy
=

1
2
x
2
+2
xy
−
1
2
y
2





(3
,
2)
(1
,
0)
=14
413

414. Exercises 9.9
(b)
Use
y
=
x
−
1 for 1
≤
x
≤
3.
s
(3
,
2)
(1
,
0)
(
x
+2
y
)
dx
+(2
x
−
y
)
dy
=
s
3
1
[
x
+2(
x
−
1)+2
x
−
(
x
−
1)
dx
=
s
3
1
(4
x
−
1)
dx
=(2
x
2
−
x
)



3
1
=14
4. (a)
P
y
=
−
cos
x
sin
y
=
Q
x
and the integral is independent of path.
φ
x
= cos
x
cos
y
,
φ
= sin
x
cos
y
+
g
(
y
),
φ
y
=
−
sin
x
sin
y
+
g
N
(
y
)=1
−
sin
x
sin
y
,
g
(
y
)=
y
,
φ
= sin
x
cos
y
+
y
,
s
(
π/
2
,
0)
(0
,
0)
cos
x
cos
ydx
+(1
−
sin
x
sin
y
)
dy
= (sin
x
cos
y
+
y
)



(
π/
2
,
0)
(0
,
0)
=1
(b)
Use
y
= 0 for 0
≤
x
≤
π/
2.
s
(
π/
2
,
0)
(0
,
0)
cos
x
cos
ydx
+(1
−
sin
x
sin
y
)
dy
=
s
π/
2
0
cos
xdx
= sin
x



π/
2
0
=1
5. (a)
P
y
=1
/y
2
=
Q
x
and the integral is independent of path.
φ
x
=
−
1
y
,
φ
=
−
x
y
+
g
(
y
),
φ
y
=
x
y
2
+
g
N
(
x
)=
x
y
2
,
g
(
y
)=0,
φ
=
−
x
y
,
s
(4
,
4)
(4
,
1)
−
1
y
dx
+
x
y
2
dy
=(
−
x
y
)




(4
,
4)
(4
,
1)
=3
(b)
Use
x
= 4 for 1
≤
y
≤
4.
s
(4
,
4)
(4
,
1)
−
1
y
dx
+
x
y
2
dy
=
s
4
1
4
y
2
dy
=
−
4
y




4
1
=3
6. (a)
P
y
=
−
xy
(
x
2
+
y
2
)
−
3
/
2
=
Q
x
and the integral is independent of path.
φ
x
=
x

x
2
+
y
2
,
φ
=

x
2
+
y
2
+
g
(
y
),
φ
y
=
y

x
2
+
y
2
+
g
N
(
y
)=
y

x
2
+
y
2
,
g
(
y
)=0,
φ
=

x
2
+
y
2
,
s
(3
,
4)
(1
,
0)
xdx
+
ydy

x
2
+
y
2
=

x
2
+
y
2



(3
,
4)
(1
,
0)
=4
(b)
Use
y
=2
x
−
2 for 1
≤
x
≤
3.
s
(3
,
4)
(1
,
0)
xdx
+
ydy

x
2
+
y
2
=
s
3
1
x
+(2
x
−
2)2

x
2
+(2
x
−
2)
2
dx
=
s
3
1
5
x
−
4
√
5
x
2
−
8
x
+4
=

5
x
2
−
8
x
+4



3
1
=4
7. (a)
P
y
=4
xy
=
Q
x
and the integral is independent of path.
φ
x
=2
y
2
x
−
3,
φ
=
x
2
y
2
−
3
x
+
g
(
y
),
φ
y
=2
x
2
y
+
g
N
(
y
)=2
x
2
y
+4,
g
(
y
)=4
y
,
φ
=
x
2
y
2
−
3
x
+4
y
,
s
(3
,
6)
(1
,
2)
(2
y
2
x
−
3)
dx
+(2
yx
2
+4)
dy
=(
x
2
y
2
−
3
x
+4
y
)



(3
,
6)
(1
,
2)
= 330
(b)
Use
y
=2
x
for 1
≤
x
≤
3.
s
(3
,
6)
(1
,
2)
(2
y
2
x
−
3)
dx
+(2
yx
2
+4)
dy
=
s
3
1
.
[2(2
x
)
2
x
−
3] + [2(2
x
)
x
2
+ 4]2
F
dx
=
s
3
1
(16
x
3
+5)
dx
=(4
x
4
+5
x
)



3
1
= 330
8. (a)
P
y
=4=
Q
x
and the integral is independent of path.
φ
x
=5
x
+4
y
,
φ
=
5
2
x
2
+4
xy
+
g
(
y
),
φ
y
=4
x
+
g
N
(
y
)=4
x
−
8
y
3
,
g
(
y
)=
−
2
y
4
,
φ
=
5
2
x
2
+4
xy
−
2
y
4
,
414

415. Exercises 9.9
s
(0
,
0)
(
−
1
,
1)
(5
x
+4
y
)
dx
+(4
x
−
8
y
3
)
dy
=

5
2
x
2
+4
xy
−
2
y
4





(0
,
0)
(
−
1
,
1)
=
7
2
(b)
Use
y
=
−
x
for
−
1
≤
x
≤
0.
s
(0
,
0)
(
−
1
,
1)
(5
x
+4
y
)
dx
+(4
x
−
8
y
3
)
dy
=
s
0
−
1
[(5
x
−
4
x
)+(4
x
+8
x
3
)(
−
1)]
dx
=
s
0
−
1
(
−
3
x
−
8
x
3
)
dx
=

−
3
2
x
2
−
2
x
4





0
−
1
=
7
2
9. (a)
P
y
=3
y
2
+3
x
2
=
Q
x
and the integral is independent of path.
φ
x
=
y
3
+3
x
2
y
,
φ
=
xy
3
+
x
3
y
+
g
(
y
),
φ
y
=3
xy
2
+
x
3
+
g
N
(
y
)=
x
3
+3
y
2
x
+1,
g
(
y
)=
y
,
φ
=
xy
3
+
x
3
y
+
y
,
s
(2
,
8)
(0
,
0)
(
y
3
+3
x
2
y
)
dx
+(
x
3
+3
y
2
x
+1)
dy
=(
xy
3
+
x
3
y
+
y
)



(2
,
8)
(0
,
0)
= 1096
(b)
Use
y
=4
x
for 0
≤
x
≤
2.
s
(2
,
8)
(0
,
0)
(
y
3
+3
x
2
y
)
dx
+(
x
3
+3
y
2
x
+1)
dy
=
s
2
0
[(64
x
3
+12
x
3
)+(
x
3
+48
x
3
+ 1)(4)]
dx
=
s
2
0
(272
x
3
+4)
dx
= (68
x
4
+4
x
)



2
0
= 1096
10. (a)
P
y
=
−
xy
cos
xy
−
sin
xy
−
20
y
3
=
Q
x
and the integral is independent of path.
φ
x
=2
x
−
y
sin
xy
−
5
y
4
,
φ
=
x
2
+ cos
xy
−
5
xy
4
+
g
(
y
),
φ
y
=
−
x
sin
xy
−
20
xy
3
+
g
N
(
y
)=
−
20
xy
3
−
x
sin
xy
,
g
(
y
)=0,
φ
=
x
2
+ cos
xy
−
5
xy
4
,
s
(1
,
0)
(
−
2
,
0)
(2
x
−
y
sin
xy
−
5
y
4
)
dx
−
(20
xy
3
+
x
sin
xy
)
dy
=(
x
2
+ cos
xy
−
5
xy
4
)



(1
,
0)
(
−
2
,
0)
=
−
3
(b)
Use
y
= 0 for
−
2
≤
x
≤
1.
s
(1
,
0)
(
−
2
,
0)
(2
x
−
y
sin
xy
−
5
y
4
)
dx
−
(20
xy
3
+
x
sin
xy
)
dy
=
s
1
−
2
2
xdx
=
x
2



1
−
2
=
−
3
11.
P
y
=12
x
3
y
2
=
Q
x
and the vector field is a gradient field.
φ
x
=4
x
3
y
3
+3,
φ
=
x
4
y
3
+3
x
+
g
(
y
),
φ
y
=3
x
4
y
2
+
g
N
(
y
)=3
x
4
y
2
+1,
g
(
y
)=
y
,
φ
=
x
4
y
3
+3
x
+
y
12.
P
y
=6
xy
2
=
Q
x
and the vector field is a gradient field.
φ
x
=2
xy
3
,
φ
=
x
2
y
3
+
g
(
y
),
φ
y
=3
x
2
y
2
+
g
N
(
y
)=3
x
2
y
2
+3
y
2
,
g
(
y
)=
y
3
,
φ
=
x
2
y
3
+
y
3
13.
P
y
=
−
2
xy
3
sin
xy
2
+2
y
cos
xy
2
,
Q
x
=
−
2
xy
3
cos
xy
2
−
2
y
sin
xy
2
and the vector field is not a gradient field.
14.
P
y
=
−
4
xy
(
x
2
+
y
2
+1)
−
3
=
Q
x
and the vector field is a gradient field.
φ
x
=
x
(
x
2
+
y
2
+1)
−
2
,
φ
=
−
1
2
(
x
2
+
y
2
+1)
−
1
+
g
(
y
),
φ
y
=
y
(
x
2
+
y
2
+1)
−
2
+
g
N
(
y
)=
y
(
x
2
+
y
2
+1)
−
2
,
g
(
y
)=0,
φ
=
−
1
2
(
x
2
+
y
2
+1)
−
1
15.
P
y
=1=
Q
x
and the vector field is a gradient field.
φ
x
=
x
3
+
y
,
φ
=
1
4
x
4
+
xy
+
g
(
y
),
φ
y
=
x
+
g
N
(
y
)=
x
+
y
3
,
g
(
y
)
−
1
4
y
4
,
φ
=
1
4
x
4
+
xy
+
1
4
y
4
16.
P
y
=4
e
2
y
,
Q
x
=
e
2
y
and the vector field is not a gradient field.
415

416. Exercises 9.9
17.
Since
P
y
=
−
e
−
y
=
Q
x
,
F
is conservative and
s
C
F
·
d
r
is independent of the path. Thus, instead of the given
curve we may use the simpler curve
C
1
:
y
=
x
,0
≤
x
≤
1. Then
W
=
s
C
1
(2
x
+
e
−
y
)
dx
+(4
y
−
xe
−
y
)
dy
=
s
1
0
(2
x
+
e
−
x
)
dx
+
s
1
0
(4
x
−
xe
−
x
)
dx
Integration by parts
=(
x
2
−
e
−
x
)



1
0
+(2
x
2
+
xe
−
x
+
e
−
x
)



1
0
= [(1
−
e
−
1
)
−
(
−
1)] + [(2 +
e
−
1
+
e
−
1
)
−
(1)] = 3 +
e
−
1
.
18.
Since
P
y
=
−
e
−
y
=
Q
x
,
F
is conservative and
s
C
F
·
d
r
is independent of the path. Thus, instead of the given
curve we may use the simpler curve
C
1
:
y
=0,
−
2
≤−
x
≤
2. Then
dy
= 0 and
W
=
s
C
1
(2
x
+
e
−
y
)
dx
+(4
y
−
xe
−
y
)
dy
=
s
−
2
2
(2
x
+1)
dx
=(
x
2
+
x
)



−
2
2
=(4
−
2)
−
(4 + 2) =
−
4
.
19.
P
y
=
z
=
Q
x
,
Q
z
=
x
=
R
y
,
R
x
=
y
=
P
z
, and the integral is independent of path. Parameterize the line
segment between the points by
x
=1+
t
,
y
=1+3
t
,
z
=1+7
t
, for 0
≤
t
≤
1. Then
dx
=
dt
,
dy
=3
dt
,
dz
=7
dt
and
s
(2
,
4
,
8)
(1
,
1
,
1)
yz dx
+
xz dy
+
xy dz
=
s
1
0
[(1 + 3
t
)(1+7
t
)+(1+
t
)(1+7
t
)(3)+(1+
t
)(1+3
t
)(7)]
dt
=
s
1
0
(11+62
t
+63
t
2
)
dt
= (11
t
+31
t
2
+21
t
3
)



1
0
=63
.
20.
P
y
=0=
Q
x
,
Q
z
=0=
R
y
,
R
x
=0=
P
z
and the integral is independent of path. Parameterize the line
segment between the points by
x
=
t
,
y
=
t
,
z
=
t
, for 0
≤
t
≤
1. Then
dx
=
dy
=
dz
=
dt
and
s
(1
,
1
,
1)
(0
,
0
,
0)
2
xdx
+3
y
2
dy
+4
z
3
dz
=
s
1
0
(2
t
+3
t
2
+4
t
3
)
dt
=(
t
2
+
t
3
+
t
4
)



1
0
=3
.
21.
P
y
=2
x
cos
y
=
Q
x
,
Q
z
=0=
R
y
,
R
x
=3
e
3
z
=
P
z
, and the integral is independent of path. Integrating
φ
x
=2
x
sin
y
+
e
3
z
we find
φ
=
x
2
sin
y
+
xe
3
z
+
g
(
y, z
). Then
φ
y
=
x
2
cos
y
+
g
y
=
Q
=
x
2
cos
y
,so
g
y
=0,
g
(
y, z
)=
h
(
z
), and
φ
=
x
2
sin
y
+
xe
3
z
+
h
(
z
). Now
φ
z
=3
xe
3
z
+
h
N
(
z
)=
R
=3
xe
3
z
+5, so
h
N
(
z
) = 5 and
h
(
z
)=5
z
.Thus
φ
=
x
2
sin
y
+
xe
3
z
+5
z
and
s
(2
,π/
2
,
1)
(1
,
0
,
0)
(2
x
sin
y
+
e
3
z
)
dx
+
x
2
cos
ydy
+(3
xe
3
z
+5)
dz
=(
x
2
sin
y
+
xe
3
z
+5
z
)



(2
,π/
2
,
1)
(1
,
0
,
0)
= [4(1) + 2
e
3
+5]
−
[0+1+0]=8+2
e
3
.
22.
P
y
=0=
Q
x
,
Q
z
=0=
R
y
,
R
x
=0=
P
z
, and the integral is independent of path. Parameterize the line
segment between the points by
x
=1+2
t
,
y
=2+2
t
,
z
= 1, for 0
≤
t
≤
1. Then
dx
=2
dt
,
dz
= 0 and
s
(3
,
4
,
1)
(1
,
2
,
1)
(2
x
+1)
dx
+3
y
2
dy
+
1
z
dz
=
s
1
0
[(2 + 4
t
+1)2+3(2+2
t
)
2
2]
dt
=
s
1
0
(24
t
2
+56
t
+ 30)
dt
=(8
t
3
+28
t
2
+30
t
)



1
0
=66
.
416

417. Exercises 9.9
23.
P
y
=0=
Q
x
,
Q
z
=0=
R
y
,
R
x
=2
e
2
z
=
P
z
and the integral is independent of path. Parameterize the line
segment between the points by
x
=1+
t
,
y
=1+
t
,
z
= ln 3, for 0
≤
t
≤
1. Then
dx
=
dy
=
dt
,
dz
= 0 and
s
(2
,
2ln3)
(1
,
1
,
ln 3)
e
2
z
dx
+3
y
2
dy
+2
xe
2
z
dz
=
s
1
0
[
e
2ln3
+ 3(1 +
t
)
2
]
dt
=[9
t
+(1+
t
)
3
]



1
0
=16
.
24.
P
y
=0=
Q
x
,
Q
z
=2
y
=
R
y
,
R
x
=2
x
=
P
z
and the integral is independent of path. Parameterize the line
segment between the points by
x
=
−
2(1
−
t
),
y
= 3(1
−
t
),
z
=1
−
t
, for 0
≤
t
≤
1. Then
dx
=2
dt
,
dy
=
−
3
dt
,
dz
=
−
dt
, and
s
(0
,
0
,
0)
(
−
2
,
3
,
1)
2
xz dx
+2
yz dy
+(
x
2
+
y
2
)
dz
=
s
1
0
[
−
4(1
−
t
)
2
(2) + 6(1
−
t
)
2
(
−
3) + 4(1
−
t
)
2
(
−
1) + 9(1
−
t
)
2
(
−
1)]
dt
=
s
1
0
−
39(1
−
t
)
2
dt
= 13(1
−
t
)
3



1
0
=
−
13
.
25.
P
y
=1
−
z
sin
x
=
Q
x
,
Q
z
= cos
x
=
R
y
,
R
x
=
−
y
sin
x
=
P
z
and the integral is independent of path. Integrating
φ
x
=
y
−
yz
sin
x
we find
φ
=
xy
+
yz
cos
x
+
g
(
y, z
). Then
φ
y
=
x
+
z
cos
x
+
g
y
(
y, z
)=
Q
=
x
+
z
cos
x
,so
g
y
=0,
g
(
y, z
)=
h
(
z
), and
φ
=
xy
+
yz
cos
x
+
h
(
z
). Now
φ
z
=
y
cos
x
+
h
(
z
)=
R
=
y
cos
x
,so
h
(
z
) = 0 and
φ
=
xy
+
yz
cos
x
. Since
r
(0) = 4
j
and
r
(
π/
2) =
π
i
+
j
+4
k
,
s
C
F
·
d
r
=(
xy
+
yz
cos
x
)




(
π,
1
,
4)
(0
,
4
,
0)
=(
π
−
4)
−
(0+0)=
π
−
4
.
26.
P
y
=0=
Q
x
,
Q
z
=0=
R
y
R
x
=
−
e
z
=
P
z
and the integral is independent of path. Integrating
φ
x
=2
−
e
z
we
find
φ
=2
x
−
xe
z
+
g
(
y, z
). Then
φ
y
=
g
y
=2
y
−
1, so
g
(
y, z
)=
y
2
−
y
+
h
(
z
) and
φ
=2
x
−
xe
z
+
y
2
−
y
+
h
(
z
).
Now
φ
z
=
−
xe
z
+
h
N
(
z
)=
R
=2
−
xe
z
,so
h
N
(
z
)=2,
h
(
z
)=2
z
, and
φ
=2
x
−
xe
z
+
y
2
−
y
+2
z
.Thus
s
C
F
·
d
r
=(2
x
−
xe
z
+
y
2
−
y
+2
z
)




(2
,
2
,
2)
(
−
1
,
1
,
−
1)
=(4
−
2
e
2
+4
−
2+4)
−
(
−
2+
e
−
1
+1
−
1
−
2) = 14
−
2
e
2
−
e
−
1
.
27.
Since
P
y
=
Gm
1
m
2
(2
xy/
|
r
|
5
)=
Q
x
,
Q
z
=
Gm
1
m
2
(2
yz/
|
r
|
5
)=
R
y
, and
R
x
=
Gm
1
m
2
(2
xz/
|
r
|
5
)=
P
z
, the
force field is conservative.
φ
x
=
−
Gm
1
m
2
x
(
x
2
+
y
2
+
z
2
)
3
/
2
,
φ
=
Gm
1
m
2
(
x
2
+
y
2
+
z
2
)
−
1
/
2
+
g
(
y, z
),
φ
y
=
−
Gm
1
m
2
y
(
x
2
+
y
2
+
z
2
)
3
/
2
+
g
y
(
y, z
)=
−
Gm
1
m
2
y
(
x
2
+
y
2
+
z
2
)
3
/
2
,
g
(
y, z
)=
h
(
z
),
φ
=
Gm
1
m
2
(
x
2
+
y
2
+
z
2
)
−
1
/
2
+
h
(
z
),
φ
z
=
−
Gm
1
m
2
z
(
x
2
+
y
2
+
z
2
)
3
/
2
+
h
N
(
z
)=
−
Gm
1
m
2
z
(
x
2
+
y
2
+
z
2
)
3
/
2
,
h
(
z
)=0,
φ
=
Gm
1
m
2

x
2
+
y
2
+
z
2
=
Gm
1
m
2
|
r
|
28.
Since
P
y
=24
xy
2
z
=
Q
x
,
Q
z
=12
x
2
y
2
=
R
y
, and
R
x
=8
xy
3
=
P
z
,
F
is conservative. Thus, the work done
between two points is independent of the path. From
φ
x
=8
xy
3
z
we obtain
φ
=4
x
2
y
3
z
which is a potential
function for
F
. Then
W
=
s
(1
,
√
3
,π/
3)
(2
,
0
,
0)
F
·
d
r
=4
x
2
y
3
z



(1
,
√
3
,π/
3)
(2
,
0
,
0)
=4
√
3
π
and
W
=
s
(0
,
2
,π/
2)
(2
,
0
,
0)
F
·
d
r
=0
.
417

418. Exercises 9.9
29.
Since
F
is conservative,
s
C
1
F
·
d
r
=
s
−
C
2
F
·
d
r
. Then, since the simply closed curve
C
is composed of
C
1
and
C
2
,
=
ˇ
C
F
·
d
r
=
s
C
1
F
·
d
r
+
s
C
2
F
·
d
r
=
s
C
1
F
·
d
r
−
s
−
C
2
F
·
d
r
=0
.
30.
From
F
=(
x
2
+
y
2
)
n/
2
(
x
i
+
y
j
) we obtain
P
y
=
nxy
(
x
2
+
y
2
)
n/
2
−
1
=
Q
x
, so that
F
is conservative. From
φ
x
=
x
(
x
2
+
y
2
)
n/
2
we obtain the potential function
φ
=(
x
2
+
y
2
)
(
n
+2)
/
2
/
(
n
+ 2). Then
W
=
s
(
x
2
,y
2
)
(
x
1
,y
1
)
F
·
d
r
=

(
x
2
+
y
2
)
(
n
+2)
/
2
n
+2





(
x
2
,y
2
)
(
x
1
,y
1
)
=
1
n
+2
t
(
x
2
2
+
y
2
2
)
(
n
+2)
/
2
−
(
x
2
1
+
y
2
1
)
(
n
+2)
/
2
h
.
31.
From Problem 39 in Exercises 9
.
8,
d
v
dt
·
d
r
dt
=
d
v
dt
·
v
=
1
2
d
dt
v
2
. Then, using
dp
dt
=
∂p
∂x
dx
dt
+
∂p
∂y
dy
dt
=
∇
p
·
d
r
dt
,
we have
s
m
d
v
dt
·
d
r
dt
dt
+
s
∇
p
·
d
r
dt
=
s
0
dt
1
2
m
s
d
dt
v
2
dt
+
s
dp
dt
dt
= constant
1
2
mv
2
+
p
= constant.
32.
By Problem 31, the sum of kinetic and potential energies in a conservative force field is constant. That is, it is
independent of points
A
and
B
,so
p
(
B
)+
K
(
B
)=
p
(
A
)+
K
(
A
).
Exercises 9.10
1.
s
3
−
1
(6
xy
−
5
e
y
)
dx
=(3
x
2
y
−
5
xe
y
)



3
−
1
= (27
y
−
15
e
y
)
−
(3
y
+5
e
y
)=24
y
−
20
e
y
2.
s
2
1
tan
xy dy
=
1
x
ln
|
sec
xy
|



2
1
=
1
x
ln
|
sec 2
x
−
sec
x
|
3.
s
3
x
1
x
3
e
xy
dy
=
x
2
e
xy



3
x
1
=
x
2
(
e
3
x
2
−
e
x
)
4.
s
y
3
√
y
(8
x
3
y
−
4
xy
2
)
dx
=(2
x
4
y
−
2
x
2
y
2
)



y
3
√
y
=(2
y
13
−
2
y
8
)
−
(2
y
3
−
2
y
3
)=2
y
13
−
2
y
8
5.
s
2
x
0
xy
x
2
+
y
2
dy
=
x
2
ln(
x
2
+
y
2
)



2
x
0
=
x
2
[ln(
x
2
+4
x
2
)
−
ln
x
2
]=
x
2
ln 5
6.
s
x
x
3
e
2
y/x
dy
=
x
2
e
2
y/x



x
x
3
=
x
2
(
e
2
x/x
−
e
2
x
3
/x
)=
x
2
(
e
2
−
e
2
x
2
)
418

419. Exercises 9.10
7.
s
sec
y
tan
y
(2
x
+ cos
y
)
dx
=(
x
2
+
x
cos
y
)



sec
y
tan
y
= sec
2
y
+ sec
y
cos
y
−
tan
2
y
−
tan
y
cos
y
= sec
2
y
+1
−
tan
2
y
−
sin
y
=2
−
sin
y
8.
s
1
√
y
y
ln
xdx
Integration by parts
=
y
(
x
ln
x
−
x
)



1
√
y
=
y
(0
−
1)
−
y
(
√
y
ln
√
y
−
√
y
)=
−
y
−
y
√
y

1
2
ln
y
−
1

9. 10. 11. 12.
13.
ss
R
x
3
y
2
dA
=
s
1
0
s
x
0
x
3
y
2
dy dx
=
s
1
0
1
3
x
3
y
3




x
0
dx
=
1
3
s
1
0
x
6
dx
=
1
21
x
7




1
0
=
1
21
14.
ss
R
(
x
+1)
dA
=
s
2
0
s
4
−
x
x
(
x
+1)
dy dx
=
s
2
0
(
xy
+
y
)



4
−
x
x
dx
=
s
2
0
[(4
x
−
x
2
+4
−
x
)
−
(
x
2
+
x
)]
dx
=
s
2
0
(2
x
−
2
x
2
+4)
dx
=

x
2
−
2
3
x
3
+4
x





2
0
=
20
3
15.
ss
R
(2
x
+4
y
+1)
dA
=
s
1
0
s
x
2
x
3
(2
x
+4
y
+1)
dy dx
=
s
1
0
(2
xy
+2
y
2
+
y
)



x
2
x
3
dx
=
s
1
0
[(2
x
3
+2
x
4
+
x
2
)
−
(2
x
4
+2
x
6
+
x
3
)]
dx
=
s
1
0
(
x
3
+
x
2
−
2
x
6
)
dx
=

1
4
x
4
+
1
3
x
3
−
2
7
x
7





1
0
=
1
4
+
1
3
−
2
7
=
25
84
16.
ss
R
xe
y
dA
=
s
1
0
s
x
0
xe
y
dy dx
=
s
1
0
xe
y



x
0
dx
=
s
1
0
(
xe
x
−
x
)
dx
Integration by parts
=

xe
x
−
e
x
−
1
2
x
2





1
0
=

e
−
e
−
1
2

−
(
−
1) =
1
2
17.
ss
R
2
xy dA
=
s
2
0
s
8
x
3
2
xy dy dx
=
s
2
0
xy
2



8
x
3
dx
=
s
2
0
(64
x
−
x
7
)
dx
=

32
x
2
−
1
8
x
8





2
0
=96
419

420. Exercises 9.10
18.
ss
R
x
√
y
dA
=
s
1
−
1
s
3
−
x
2
x
2
+1
xy
−
1
/
2
dy dx
=
s
1
−
1
2
x
√
y



3
−
x
2
x
2
+1
dx
=2
s
1
−
1
(
x

3
−
x
2
−
x

x
2
+1)
dx
=2
N
−
1
3
(3
−
x
2
)
3
/
2
−
1
3
(
x
2
+1)
3
/
2
o




1
−
1
=
−
2
3
[(2
3
/
2
+2
3
/
2
)
−
(2
3
/
2
+2
3
/
2
)] = 0
19.
ss
R
y
1+
xy
dA
=
s
1
0
s
1
0
y
1+
xy
dx dy
=
s
1
0
ln(1 +
xy
)



1
0
dy
=
s
1
0
ln(1 +
y
)
dy
= [(1 +
y
)ln(1+
y
)
−
(1 +
y
)]



1
0
=(2ln2
−
2)
−
(
−
1) = 2 ln 2
−
1
20.
ss
R
sin
πx
y
dA
=
s
2
1
s
y
2
0
sin
πx
y
dx dy
=
s
2
1

−
y
π
cos
πx
y





y
2
0
dy
=
s
2
1
e
−
y
π
cos
πy
+
y
π
:
dy
Integration by parts
=

−
y
π
2
sin
πy
−
1
π
3
cos
πy
+
y
2
2
π





2
1
=

−
1
π
3
+
2
π

−

1
π
3
+
1
2
π

=
3
π
2
−
4
2
π
3
21.
ss
R

x
2
+1
dA
=
s
√
3
0
s
x
−
x

x
2
+1
dy dx
=
s
√
3
0
y

x
2
+1



x
−
x
dx
=
s
√
3
0
(
x

x
2
+1+
x

x
2
+1)
dx
=
s
√
3
0
2
x

x
2
+1
dx
=
2
3
(
x
2
+1)
3
/
2




√
3
0
=
2
3
(4
3
/
2
−
1
3
/
2
)=
14
3
22.
ss
R
xdA
=
s
π/
4
0
s
1
tan
y
xdxdy
=
s
π/
4
0
1
2
x
2




1
tan
y
dy
=
1
2
s
π/
4
0
(1
−
tan
2
y
)
dy
=
1
2
s
π/
4
0
(2
−
sec
2
y
)
dy
=
1
2
(2
y
−
tan
y
)



π/
4
0
=
1
2
e
π
2
−
1
:
=
π
4
−
1
2
23.
The correct integral is
(c)
.
V
=2
s
2
−
2
s
√
4
−
y
2
0
(4
−
y
)
dx dy
=2
s
2
−
2
(4
−
y
)
x



√
4
−
y
2
0
dy
=2
s
2
−
2
(4
−
y
)

4
−
y
2
dy
=2
N
2
y

4
−
y
2
+ 8sin
−
1
y
2
+
1
3
(4
−
y
2
)
3
/
2
o




2
−
2
= 2(4
π
−
(
−
4
π
)] = 16
π
24.
The correct integral is
(b)
.
A
=8
s
2
0
s
√
4
−
y
2
0
(4
−
y
2
)
1
/
2
dx dy
=8
s
2
0
(4
−
y
2
)
1
/
2
x



√
4
−
y
2
0
dy
=8
s
2
0
(4
−
y
2
)
dy
=8

4
y
−
1
3
y
3





2
0
=
128
3
420

421. Exercises 9.10
25.
Setting
z
=0wehave
y
=6
−
2
x
.
V
=
s
3
0
s
6
−
2
x
0
(6
−
2
x
−
y
)
dy dx
=
s
3
0

6
y
−
2
xy
−
1
2
y
2





6
−
2
x
0
dx
=
s
3
0
[6(6
−
2
x
)
−
2
x
(6
−
2
x
)
−
1
2
(6
−
2
x
)
2
]
dx
=
s
3
0
(18
−
12
x
+2
x
2
)
dx
=

18
x
−
6
x
2
+
2
3
x
3





3
0
=18
26.
Setting
z
=0wehave
y
=
±
2.
V
=
s
3
0
s
2
0
(4
−
y
2
)
dy dx
=
s
3
0

4
y
−
1
3
y
3





2
0
dx
=
s
3
0
16
3
dx
=16
27.
Solving for
z
, we have
x
=2
−
1
2
x
+
1
2
y
. Setting
z
= 0, we see that this surface (plane) intersects
the
xy
-plane in the line
y
=
x
−
4. since
z
(0
,
0) = 2
>
0, the surface lies above the
xy
-plane over
the quarter-circular region.
V
=
s
2
0
s
√
4
−
x
2
0

2
−
1
2
x
+
1
2
y

dy dx
=
s
2
0

2
y
−
1
2
xy
+
1
4
y
2





√
4
−
x
2
0
dx
=
s
2
0

2

4
−
x
2
−
1
2
x

4
−
x
2
+1
−
1
4
x
2

dx
=
N
x

4
−
x
2
+ 4 sin
−
1
x
2
+
1
6
(4
−
x
2
)
3
/
2
+
x
−
1
12
x
3
o




2
0
=

2
π
+2
−
2
3

−
4
3
=2
π
28.
Setting
z
=0wehave
y
= 3. Using symmetry,
V
=2
s
√
3
0
s
3
x
2
(3
−
y
)
dy dx
=2
s
√
3
0

3
y
−
1
2
y
2





3
x
2
dx
=2
s
√
3
0

9
2
−
3
x
2
+
1
2
x
4

dx
=2

9
2
x
−
x
3
+
1
10
x
5





√
3
0
=2

9
2
√
3
−
3
√
3+
9
10
√
3

=
24
√
3
5
.
29.
Note that
z
=1+
x
2
+
y
2
is always positive. Then
V
=
s
1
0
s
3
−
3
x
0
(1 +
x
2
+
y
2
)
dy dx
=
s
1
0

y
+
x
2
y
+
1
3
y
3





3
−
3
x
0
dx
=
s
1
0
[(3
−
3
x
)+
x
2
(3
−
3
x
) + 9(1
−
x
)
3
]
dx
=
s
1
0
(12
−
30
x
+30
x
2
−
12
x
3
)
dx
= (12
x
−
15
x
2
+10
x
3
−
3
x
4
)



1
0
=4
.
30.
In the first octant,
z
=
x
+
y
is nonnegative. Then
V
=
s
3
0
s
√
9
−
x
2
0
(
x
+
y
)
dy dx
=
s
3
0

xy
+
1
2
y
2





√
9
−
x
2
0
dx
=
s
3
0

x

9
−
x
2
+
9
2
−
1
2
x
2

dx
=
N
−
1
3
(9
−
x
2
)
3
/
2
+
9
2
x
−
1
6
x
3
o




3
0
=

27
2
−
9
2

−
(
−
9) = 18
.
421

422. Exercises 9.10
31.
In the first octant
z
=6
/y
is positive. Then
V
=
s
6
1
s
5
0
6
y
dx dy
=
s
6
1
6
x
y




5
0
dy
=30
s
6
1
dy
y
=30ln
y



6
1
=30ln6
.
32.
Setting
z
=0,wehave
x
2
/
4+
y
2
/
16 = 1. Using symmetry,
V
=4
s
2
0
s
2
√
4
−
x
2
0

4
−
x
2
−
1
4
y
2

dy dx
=4
s
2
0

4
y
−
x
2
y
−
1
12
y
3





2
√
4
−
x
2
0
dx
=4
s
2
0
N
8

4
−
x
2
−
2
x
2

4
−
x
2
−
2
3
(4
−
x
2
)
3
/
2
o
dx
Trig substitution
=4
N
4
x

4
−
x
2
+ 16 sin
−
1
x
2
−
1
4
x
(2
x
2
−
4)

4
−
x
2
−
4 sin
−
1
x
2
+
1
12
x
(2
x
2
−
20)

4
−
x
2
−
4 sin
x
2
o




2
0
=4

16
π
2
−
4
π
2
−
4
π
2

−
(0) = 16
π.
33.
Note that
z
=4
−
y
2
is positive for
|
y
|≤
1. Using symmetry,
V
=2
s
2
0
s
√
2
x
−
x
2
0
(4
−
y
2
)
dy dx
=2
s
2
0

4
y
−
1
3
y
3





√
2
x
−
x
2
0
dx
=2
s
2
0
N
4

2
x
−
x
2
−
1
3
(2
x
−
x
2
)

2
x
−
x
2
o
dx
=2
s
2
0

4

1
−
(
x
−
1)
2
−
1
3
[1
−
(
x
−
1)
2
]

1
−
(
x
−
1)
2

dx
u
=
x
−
1
,du
=
dx
=2
s
1
−
1
N
4

1
−
u
2
−
1
3
(1
−
u
2
)

1
−
u
2
o
du
=2
s
1
−
1

11
3

1
−
u
2
+
1
3
u
2

1
−
u
2

du
Trig substitution
=2
N
11
6
u

1
−
u
2
+
11
6
sin
u
+
1
24
x
(2
x
2
−
1)

1
−
u
2
+
1
24
sin
−
1
u
o




1
−
1
=2
Na
11
6
π
2
+
1
24
π
2

−

−
11
6
π
2
−
1
24
π
2
lo
=
15
π
4
.
34.
From
z
=1
−
x
2
and
z
=1
−
y
2
we have 1
−
x
2
=1
−
y
2
or
y
=
x
(in the first octant).
Thus, the surfaces intersect in the plane
y
=
x
. Using symmetry,
V
=2
s
1
0
s
1
x
(1
−
y
2
)
dy dx
=2
s
1
0

y
−
1
3
y
3





1
x
dx
=2
s
1
0

2
3
−
x
+
1
3
x
3

dx
=2

2
3
x
−
1
2
x
2
+
1
12
x
4





1
0
=
1
2
.
35.
s
1
0
s
1
x
x
2

1+
y
4
dy dx
=
s
1
0
s
y
0
x
2

1+
y
4
dx dy
=
s
1
0
1
3
x
3

1+
y
4




y
0
dy
=
1
3
s
1
0
y
3

1+
y
4
dy
=
1
3
N
1
6
(1 +
y
4
)
3
/
2
o




1
0
=
1
18
(2
√
2
−
1)
422

423. Exercises 9.10
36.
s
1
0
s
2
2
y
e
−
y/x
dx dy
=
s
2
0
s
x/
2
0
e
−
y/x
dy dx
=
s
2
0
−
xe
−
y/x



x/
2
0
=
s
2
0
(
−
xe
−
1
/
2
+
x
)
dx
=
s
2
0
(1
−
e
−
1
/
2
)
xdx
=
1
2
(1
−
e
−
1
/
2
)
x
2




2
0
= 2(1
−
e
−
1
/
2
)
37.
s
2
0
s
4
y
2
cos
x
3
/
2
dx dy
=
s
4
0
s
√
x
0
cos
x
3
/
2
dy dx
=
s
4
0
y
cos
x
3
/
2



√
x
0
dx
=
s
4
0
√
x
cos
x
3
/
2
dx
=
2
3
sin
x
3
/
2




4
0
=
2
3
sin 8
38.
s
1
−
1
s
√
1
−
x
2
−
√
1
−
x
2
x

1
−
x
2
−
y
2
dy dx
=
s
1
−
1
s
√
1
−
y
2
−
√
1
−
y
2
x

1
−
x
2
−
y
2
dx dy
=
s
1
−
1
N
−
1
3
(1
−
x
2
−
y
2
)
3
/
2
o




√
1
−
y
2
−
√
1
−
y
2
dy
=
−
1
3
s
1
−
1
(0
−
0)
dy
=0
39.
s
1
0
s
1
x
1
1+
y
4
dy dx
=
s
1
0
s
y
0
1
1+
y
4
dx dy
=
s
1
0
x
1+
y
4




y
0
dy
=
s
1
0
y
1+
y
4
dy
=
1
2
tan
−
1
y
2




1
0
=
π
8
40.
s
4
0
s
2
√
y

x
3
+1
dx dy
=
s
2
0
s
x
2
0

x
3
+1
dy dx
=
s
2
0
y

x
3
+1



x
2
0
dx
=
s
2
0
x
2

x
3
+1
dx
=
2
9
(
x
3
+1)
3
/
2




2
0
=
2
9
(9
3
/
2
−
1
3
/
2
)=
52
9
41.
m
=
s
3
0
s
4
0
xy dx dy
=
s
3
0
1
2
x
2
y




4
0
dy
=
s
3
0
8
ydy
=4
y
2



3
0
=36
M
y
=
s
3
0
s
4
0
x
2
ydxdy
=
s
3
0
1
3
x
3
y




4
0
dy
=
s
3
0
64
3
ydy
=
32
3
y
2




3
0
=96
M
x
=
s
3
0
s
4
0
xy
2
dx dy
=
s
3
0
1
2
x
2
y
2




4
0
=
s
3
0
8
y
2
dy
=
8
3
y
3




3
0
=72
¯
x
=
M
y
/m
=96
/
36 = 8
/
3; ¯
y
=
M
x
/m
=72
/
36 = 2. The center of mass is (8
/
3
,
2).
42.
m
=
s
2
0
s
4
−
2
x
0
x
2
dy dx
=
s
2
0
x
2
y



4
−
2
x
0
dx
=
s
2
0
x
2
(4
−
2
x
)
dx
=
s
2
0
(4
x
2
−
2
x
3
)
dx
=

4
3
x
3
−
1
2
x
4





2
0
=
32
3
−
8=
8
3
M
y
=
s
2
0
s
4
−
2
x
0
x
3
dy dx
=
s
2
0
x
3
y



4
−
2
x
0
dx
=
s
2
0
x
3
(4
−
2
x
)
dx
=
s
2
0
(4
x
3
−
2
x
4
)
dx
=

x
4
−
2
5
x
5





2
0
=16
−
64
5
=
16
5
M
x
=
s
2
0
s
4
−
2
x
0
x
2
ydydx
=
s
2
0
1
2
x
2
y
2




s
4
−
2
x
0
dx
=
1
2
s
2
0
x
2
(4
−
2
x
)
2
dx
=
1
2
s
2
0
(16
x
2
−
16
x
3
+4
x
4
)
dx
=2
s
2
0
(4
x
2
−
4
x
3
+
x
4
)
dx
=2

4
3
x
3
−
x
4
+
1
5
x
5





2
0
=2

32
3
−
16 +
32
5

=
32
15
423

424. Exercises 9.10
¯
x
=
M
y
/m
=
16
/
5
8
/
3
=6
/
5; ¯
y
=
M
x
/m
=
32
/
15
8
/
3
=4
/
5. The center of mass is (6
/
5
,
4
/
5).
43.
Since both the region and
ρ
are symmetric with respect to the line
x
=3,¯
x
=3.
m
=
s
3
0
s
6
−
y
y
2
ydxdy
=
s
3
0
2
xy



6
−
y
y
dy
=
s
3
0
2
y
(6
−
y
−
y
)
dy
=
s
3
0
(12
y
−
4
y
2
)
dy
=

6
y
2
−
4
3
y
3





3
0
=18
M
x
=
s
3
0
s
6
−
y
y
2
y
2
dx dy
=
s
3
0
2
xy
2



6
−
y
y
dx dy
=
s
3
0
2
y
2
(6
−
y
−
y
)
dy
=
s
3
0
(12
y
2
−
4
y
3
)
dy
=(4
y
3
−
y
4
)



3
0
=27
¯
y
=
M
x
/m
=27
/
18= 3
/
2. The center of mass is (3
,
3
/
2).
44.
Since both the region and
ρ
are symmetric with respect to the
y
-axis, ¯
x
= 0. Using
symmetry,
m
=
s
3
0
s
y
0
(
x
2
+
y
2
)
dx dy
=
s
3
0

1
3
x
3
+
xy
2





y
0
dy
=
s
3
0

1
3
y
3
+
y
3

dy
=
4
3
s
3
0
y
3
dy
=
1
3
y
4




3
0
=27
.
M
x
=
s
3
0
s
y
0
(
x
2
y
+
y
3
)
dx dy
=
s
3
0

1
3
x
3
y
+
xy
3





y
0
dy
=
s
3
0

1
3
y
4
+
y
4

dy
=
4
3
s
3
0
y
4
dy
=
4
15
y
5




3
0
=
324
5
¯
y
=
M
x
/m
=
324
/
5
27
=12
/
5. The center of mass is (0
,
12
/
5).
45.
m
=
s
1
0
s
x
2
0
(
x
+
y
)
dy dx
=
s
1
0

xy
+
1
2
y
2





x
2
0
dx
=
s
1
0

x
3
+
1
2
x
4

dx
=

1
4
x
4
+
1
10
x
5





1
0
=
7
20
M
y
=
s
1
0
s
x
2
0
(
x
2
+
xy
)
dy dx
=
s
1
0

x
2
y
+
1
2
xy
2





x
2
0
dx
=
s
1
0

x
4
+
1
2
x
5

dx
=

1
5
x
5
+
1
12
x
6





1
0
=
17
60
M
x
=
s
1
0
s
x
2
0
(
xy
+
y
2
)
dy dx
=
s
1
0

1
2
xy
2
+
1
3
y
3





x
2
0
=
s
1
0

1
2
x
5
+
1
3
x
6

dx
=

1
12
x
6
+
1
21
x
7





1
0
=
11
84
¯
x
=
M
y
/m
=
17
/
60
7
/
20
=17
/
21; ¯
y
=
M
x
/m
=
11
/
84
7
/
20
=55
/
147. The center of mass is (17
/
21
,
55
/
147).
46.
m
=
s
4
0
s
√
x
0
(
y
+5)
dy dx
=
s
4
0

1
2
y
2
+5
y





√
x
0
dx
=
s
4
0

1
2
x
+5
√
x

dx
=

1
4
x
2
+
10
3
x
3
/
2





4
0
=
92
3
424

425. Exercises 9.10
M
y
=
s
4
0
s
√
x
0
(
xy
+5
x
)
dy dx
=
s
4
0

1
2
xy
2
+5
xy





√
x
0
dx
=
s
4
0

1
2
x
2
+5
x
3
/
2

dx
=

1
6
x
3
+2
x
5
/
2





4
0
=
224
3
M
x
=
s
4
0
s
√
x
0
(
y
2
+5
y
)
dy dx
=
s
4
0

1
3
y
3
+
5
2
y
2





√
x
0
dx
=
s
4
0

1
3
x
3
/
2
+
5
2
x

dx
=

2
15
x
5
/
2
+
5
4
x
2





4
0
=
364
15
¯
x
=
M
y
/m
=
224
/
3
92
/
3
=56
/
23; ¯
y
=
M
x
/m
=
364
/
15
92
/
3
=91
/
115. The center of mass is (56
/
23
,
91
/
115).
47.
The density is
ρ
=
ky
. Since both the region and
ρ
are symmetric with respect to the
y
-axis, ¯
x
= 0. Using symmetry,
m
=2
s
1
0
s
1
−
x
2
0
ky dy dx
=2
k
s
1
0
1
2
y
2




1
−
x
2
0
dx
=
k
s
1
0
(1
−
x
2
)
2
dx
=
k
s
1
0
(1
−
2
x
2
+
x
4
)
dx
=
k

x
−
2
3
x
3
+
1
5
x
5





1
0
=
k

1
−
2
3
+
1
5

=
8
15
k
M
x
=2
s
1
0
s
1
−
x
2
0
ky
2
dy dx
=2
k
s
1
0
1
3
y
3




1
−
x
2
0
dx
=
2
3
k
s
1
0
(1
−
x
2
)
3
dx
=
2
3
k
s
1
0
(1
−
3
x
2
+3
x
4
−
x
6
)
dx
=
2
3
k

x
−
x
3
+
3
5
x
5
−
1
7
x
7





1
0
=
2
3
k

1
−
1+
3
5
−
1
7

=
32
105
k
¯
y
=
M
x
/m
=
32
k/
105
8
k/
15
=4
/
7. The center of mass is (0
,
4
/
7).
48.
The density is
ρ
=
kx
.
m
=
s
π
0
s
sin
x
0
kx dy dx
=
s
π
0
kxy



sin
x
0
dx
=
s
π
0
kx
sin
xdx
Integration by parts
=
k
(sin
x
−
x
cos
x
)



π
0
=
kπ
M
y
=
s
π
0
s
sin
x
0
kx
2
dy dx
=
s
π
0
kx
2
y



sin
x
0
dx
=
s
π
0
kx
2
sin
xdx
Integration by parts
=
k
(
−
x
2
cos
x
+ 2 cos
x
+2
x
sin
x
)



π
0
=
k
[(
π
2
−
2)
−
2] =
k
(
π
2
−
4)
M
x
=
s
π
0
s
sin
x
0
kxydydx
=
s
π
0
1
2
kxy
2




sin
x
0
dx
=
s
π
0
1
2
kx
sin
2
xdx
=
s
π
0
1
4
kx
(1
−
cos 2
x
)
dx
=
1
4
k
N
s
π
0
xdx
−
s
π
0
x
cos 2
xdx
o
Integration by parts
=
1
4
k
N
1
2
x
2



π
0
−
1
4
(cos 2
x
+2
x
sin 2
x
)



π
0
o
=
1
4
k

1
2
π
2

=
1
8
kπ
2
¯
x
=
M
y
/m
=
k
(
π
2
−
4)
kπ
=
π
−
4
/π
;¯
y
=
M
x
/m
=
kπ
2
/
8
kπ
=
π/
8. The center of mass is (
π
−
4
/π, π/
8).
425

426. Exercises 9.10
49.
m
=
s
1
0
s
e
x
0
y
3
dy dx
=
s
1
0
1
4
y
4




e
x
0
dx
=
s
1
0
1
4
e
4
x
dx
=
1
16
e
4
x




1
0
=
1
16
(
e
4
−
1)
M
y
=
s
1
0
s
e
x
0
xy
3
dy dx
=
s
1
0
1
4
xy
4




e
x
0
dx
=
s
1
0
1
4
xe
4
x
dx
Integration by parts
=
1
4

1
4
xe
4
x
−
1
16
e
4
x





1
0
=
1
4

3
16
e
4
+
1
16

=
1
64
(3
e
4
+1)
M
x
=
s
1
0
s
e
x
0
y
4
dy dx
=
s
1
0
1
5
y
5




e
x
0
dx
=
s
1
0
1
5
e
5
x
dx
=
1
25
e
5
x




1
0
=
1
25
(
e
5
−
1)
¯
x
=
M
y
/m
=
(3
e
4
+1)
/
64
(
e
4
−
1)
/
16
=
3
e
4
+1
4(
e
4
−
1)
;¯
y
=
M
x
/m
=
(
e
5
−
1)
/
25
(
e
4
−
1)
/
16
=
16(
e
5
−
1)
25(
e
4
−
1)
The center of mass is

3
e
4
+1
4(
e
4
−
1)
,
16(
e
5
−
1)
25(
e
4
−
1)

≈
(0
.
77
,
1
.
76).
50.
Since both the region and
ρ
are symmetric with respect to the
y
-axis, ¯
x
= 0. Using
symmetry,
m
=2
s
3
0
s
√
9
−
x
2
0
x
2
dy dx
=2
s
3
0
x
2
y



√
9
−
x
2
0
dx
=2
s
3
0
x
2

9
−
x
2
dx
Trig substitution
=2
n
x
8
(2
x
2
−
9)

9
−
x
2
+
81
8
sin
−
1
x
3
/




3
0
=
81
4
π
2
=
81
π
2
.
M
x
=2
s
3
0
s
√
9
−
x
2
0
x
2
ydydx
=2
s
3
0
1
2
x
2
y
2




√
9
−
x
2
0
dy dx
=
s
3
0
x
2
(9
−
x
2
)
dx
=

3
x
2
−
1
5
x
5





3
0
=
162
5
¯
y
=
M
x
/m
=
162
/
5
81
π/
8
=16
/
5
π
. The center of mass is (0
,
16
/
5
π
).
51.
I
x
=
s
1
0
s
y
−
y
2
0
2
xy
2
dx dy
=
s
1
0
x
2
y
2



y
−
y
2
0
dy
=
s
1
0
(
y
−
y
2
)
2
y
2
dy
=
s
1
0
(
y
4
−
2
y
5
+
y
6
)
dy
=

1
5
y
5
−
1
3
y
6
+
1
7
y
7





1
0
=
1
105
52.
I
x
=
s
1
0
s
√
x
x
2
x
2
y
2
dy dx
=
s
1
0
1
3
x
2
y
3




√
x
x
2
dx
=
1
3
s
1
0
(
x
7
/
2
−
x
8
)
dx
=
1
3

2
9
x
9
/
2
−
1
9
x
9





1
0
=
1
27
53.
Using symmetry,
I
x
=2
s
π/
2
0
s
cos
x
0
ky
2
dy dx
=2
k
s
π/
2
0
1
3
y
3




cos
x
0
dx
=
2
3
k
s
π/
2
0
cos
3
xdx
=
2
3
k
s
π/
2
0
cos
x
(1
−
sin
2
x
)
dx
=
2
3
k

sin
x
−
1
3
sin
3
x





π/
2
0
=
4
9
k.
54.
I
x
=
s
2
0
s
√
4
−
x
2
0
y
3
dy dx
=
s
2
0
1
4
y
4




√
4
−
x
2
0
dx
=
1
4
s
2
0
(4
−
x
2
)
2
dx
=
1
4
s
2
0
(16
−
8
x
2
+
x
4
)
dx
=
1
4

16
x
−
8
3
x
3
+
1
5
x
5





2
0
=
1
4

32
−
64
3
+
32
5

426

427. Exercises 9.10
=8

1
−
2
3
+
1
5

=
64
15
55.
I
y
=
s
4
0
s
√
y
0
x
2
ydxdy
=
s
4
0
1
3
x
3
y




√
y
0
dy
=
1
3
s
4
0
y
3
/
2
ydy
=
1
3
s
4
0
y
5
/
2
dy
=
1
3

2
7
y
7
/
2





4
0
=
2
21
(4
7
/
2
)=
256
21
56.
I
y
=
s
1
0
s
√
x
x
2
x
4
dy dx
=
s
1
0
x
4
y



√
x
x
2
dx
=
s
1
0
(
x
9
/
2
−
x
6
)
dx
=

2
11
x
11
/
2
−
1
7
x
7





1
0
=
3
77
57.
I
y
=
s
1
0
s
3
y
(4
x
3
+3
x
2
y
)
dx dy
=
s
1
0
(
x
4
+
x
3
y
)



3
y
dy
=
s
1
0
(81+27
y
−
2
y
4
)
dy
=

81
y
+
27
2
y
2
−
2
5
y
5





1
0
=
941
10
58.
The density is
ρ
=
ky
. Using symmetry,
I
y
=2
s
1
0
s
1
−
x
2
0
kx
2
ydydx
=2
s
1
0
1
2
kx
2
y
2




1
−
x
2
0
dx
=
k
s
1
0
x
2
(1
−
x
2
)
2
dx
=
k
s
1
0
(
x
2
−
2
x
4
+
x
6
)
dx
=
k

1
3
x
3
−
2
5
x
5
+
1
7
x
7





1
0
=
8
k
105
.
59.
Using symmetry,
m
=2
s
a
0
s
√
a
2
−
y
2
0
xdxdy
=2
s
a
0
1
2
x
2




√
a
2
−
y
2
0
dy
=
s
a
0
(
a
2
−
y
2
)
dy
=

a
2
y
−
1
3
y
3





a
0
=
2
3
a
3
.
I
y
=2
s
a
0
s
√
a
2
−
y
2
0
x
3
dx dy
=2
s
a
0
1
4
x
4




√
a
2
−
y
2
0
dy
=
1
2
s
a
0
(
a
2
−
y
2
)
2
dy
=
1
2
s
a
0
(
a
4
−
2
a
2
y
2
+
y
4
)
dy
=
1
2

a
4
y
−
2
3
a
2
y
3
+
1
5
y
5





a
0
=
4
15
a
5
R
g
=
r
I
y
m
=
p
4
a
5
/
15
2
a
3
/
3
=
r
2
5
a
60.
m
=
s
a
0
s
a
−
x
0
kdydx
=
s
a
0
ky



a
−
x
0
dx
=
k
s
a
0
(
a
−
x
)
dx
=
k

ax
−
1
2
x
2





a
0
=
1
2
ka
2
I
x
=
s
a
0
s
a
−
x
0
ky
2
dy dx
=
s
a
0
1
3
ky
3




a
−
x
0
dx
=
1
3
k
s
a
0
(
a
−
x
)
3
dx
=
1
3
k
s
a
0
(
a
3
−
3
a
2
x
+3
ax
2
−
x
3
)
dx
=
1
3
k

a
3
x
−
3
2
a
2
x
2
+
ax
3
−
1
4
x
4





a
0
=
1
12
ka
4
R
g
=
r
I
x
m
=
p
ka
4
/
12
ka
2
/
2
=
r
1
6
a
427

428. Exercises 9.10
61. (a)
Using symmetry,
I
x
=4
s
a
0
s
b
√
a
2
−
x
2
/a
0
y
2
dy dx
=
4
b
3
3
a
3
s
a
0
(
a
2
−
x
2
)
3
/
2
dx
x
=
a
sin
θ, dx
=
a
cos
θdθ
=
4
3
ab
3
s
π/
2
0
cos
4
θdθ
=
4
3
ab
3
s
π/
2
0
1
4
(1 + cos 2
θ
)
2
dθ
=
1
3
ab
3
s
π/
2
0

1 + cos 2
θ
+
1
2
+
1
2
cos 4
θ

dθ
=
1
3
ab
3

3
2
θ
+
1
2
sin 2
θ
+
1
8
sin 4
θ





π/
2
0
=
ab
3
π
4
.
(b)
Using symmetry,
I
y
=4
s
a
0
s
b
√
a
2
−
x
2
/a
0
x
2
dy dx
=
4
b
a
s
a
0
x
2

a
2
−
x
2
dx
x
=
a
sin
θ, dx
=
a
cos
θdθ
=4
a
3
b
s
π/
2
0
sin
2
θ
cos
2
θdθ
=4
a
3
b
s
π/
2
0
1
4
(1
−
cos
2
2
θ
)
dθ
=
a
3
b
s
π/
2
0

1
−
1
2
−
1
2
cos 4
θ

dθ
=
a
3
b

1
2
θ
−
1
8
sin 4
θ





π/
2
0
=
a
3
bπ
4
.
(c)
Using
m
=
πab
,
R
g
=

I
x
/m
=
1
2

ab
3
π/πab
=
1
2
b
.
(d)
R
g
=

I
y
/m
=
1
2

a
3
bπ/πab
=
1
2
a
62.
The equation of the ellipse is 9
x
2
/a
2
+4
y
2
/b
2
= 1 and the equation of the parabola is
y
=
±
(9
bx
2
/
8
a
2
−
b/
2). Letting
I
e
and
I
p
represent the moments of inertia of the
ellipse and parabola, respectively, about the
x
-axis, we have
I
e
=2
s
0
−
a/
3
s
b
√
a
2
−
9
x
2
/
2
a
0
y
2
dy dx
=
b
3
12
a
3
s
0
−
a/
3
(
a
2
−
9
x
2
)
3
/
2
dx
x
=
a
3
sin
θ, dx
=
a
3
cos
θdθ
=
b
3
12
a
3
a
4
3
s
0
−
π/
3
cos
4
θdθ
=
b
3
a
36
3
π
16
=
ab
3
π
192
and
I
p
=2
s
2
a/
3
0
s
b/
2
−
9
bx
2
/
8
a
2
0
y
2
dy dx
=
2
3
s
2
a/
3
0

b
2
−
9
b
8
a
2
x
2

3
dx
=
2
3
b
3
8
s
2
a/
3
0

1
−
9
4
a
2
x
2

3
dx
=
b
3
12
s
2
a/
3
0

1
−
27
4
a
2
x
2
+
243
16
a
4
x
4
−
729
64
a
6
x
6

dx
=
b
3
12

x
−
9
4
a
2
x
3
+
243
80
a
4
x
5
−
729
64
a
6
x
7





2
a/
3
0
=
b
3
12
32
a
105
=
8
ab
3
315
.
Then
I
x
=
I
e
+
I
p
=
ab
3
π
192
+
8
ab
3
315
.
63.
From Problem 60,
m
=
1
2
ka
2
and
I
x
=
1
12
ka
4
.
I
y
=
s
a
0
s
a
−
x
0
kx
2
dy dx
=
s
a
0
kx
2
y



a
−
x
0
dx
=
k
s
a
0
x
2
(
a
−
x
)
dx
=
k

1
3
ax
3
−
1
4
x
4





a
0
=
1
12
ka
4
I
0
=
I
x
+
I
y
=
1
12
ka
4
+
1
12
ka
4
=
1
6
ka
4
64.
From Problem 52,
I
x
=
1
27
, and from Problem 56,
I
y
=
3
77
. Thus,
I
0
=
I
x
+
I
y
=
1
27
+
3
77
=
158
2079
.
428

429. Exercises 9.11
65.
The density is
ρ
=
k/
(
x
2
+
y
2
). Using symmetry,
I
0
=2
s
√
2
0
s
6
−
y
2
y
2
+2
(
x
2
+
y
2
)
k
x
2
+
y
2
dx dy
=2
s
√
2
0
kx



6
−
y
2
y
2
+2
dy
=2
k
s
√
2
0
(6
−
y
2
−
y
2
−
2)
dy
=2
k

4
y
−
2
3
y
3





√
2
0
=2
k

8
3
√
2

=
16
√
2
3
k.
66.
I
0
=
s
3
0
s
4
y
k
(
x
2
+
y
2
)
dx dy
=
k
s
3
0

1
3
x
3
+
xy
2





4
y
dy
=
k
s
3
0

64
3
+4
y
2
−
1
3
y
3
−
y
3

dy
=
k

64
3
y
+
4
3
y
3
−
1
3
y
4





3
0
=73
k
67.
From Problem 60,
m
=
1
2
ka
2
, and from Problem 61,
I
0
=
1
6
ka
4
. Then
R
g
=

I
0
/m
=
p
ka
4
/
6
ka
2
/
2
=
r
1
3
a
.
68.
Since the plate is homogeneous, the density is
ρ
=
m/:ω
. Using symmetry,
I
0
=4
s
-o
2
0
s
ω/
2
0
m
:ω
(
x
2
+
y
2
)
dy dx
=
4
m
:ω
s
-o
2
0

x
2
y
+
1
3
y
3





ω/
2
0
dx
=
4
m
:ω
s
-o
2
0

ω
2
x
2
+
ω
3
24

dx
=
4
m
:ω

ω
6
x
3
+
ω
3
24
x





-o
2
0
=
4
m
:ω

ω:
3
48
+
:ω
3
48

=
m
:
2
+
ω
2
12
.
Exercises 9.11
1.
Using symmetry,
A
=2
s
π/
2
−
π/
2
s
3+3 sin
θ
0
rdrdθ
=2
s
π/
2
−
π/
2
1
2
r
2




3+3 sin
θ
0
dθ
=
s
π/
2
−
π/
2
9(1 + sin
θ
)
2
dθ
=9
s
π/
2
−
π/
2
(1 + 2 sin
θ
+ sin
2
θ
)
dθ
=9

θ
−
2 cos
θ
+
1
2
θ
−
1
4
sin 2
θ





π/
2
−
π/
2
=9
N
3
2
π
2
−
3
2
e
−
π
2
:
o
=
27
π
2
2.
Using symmetry,
A
=2
s
π
0
s
2+cos
θ
0
rdrdθ
=2
s
π
0
1
2
r
2




2+cos
θ
0
dθ
=
s
π
0
(2 + cos
θ
)
2
dθ
=
s
π
0
(4 + 4 cos
θ
+ cos
2
θ
)
dθ
=

4
θ
+ 4 sin
θ
+
1
2
θ
+
1
4
cos 2
θ





π
0
=

4
π
+
π
2
+
1
4

−

1
4

=
9
π
2
.
429

430. Exercises 9.11
3.
Solving
r
= 2 sin
θ
and
r
= 1, we obtain sin
θ
=1
/
2or
θ
=
π/
6. Using symmetry,
A
=2
s
π/
6
0
s
2 sin
θ
0
rdrdθ
+2
s
π/
2
π/
6
s
1
0
rdrdθ
=2
s
π/
6
0
1
2
r
2




2 sin
θ
0
dθ
+2
s
π/
2
π/
6
1
2
r
2




1
0
dθ
=
s
π/
6
0
4 sin
2
θdθ
+
s
π/
2
π/
6
dθ
=(2
θ
−
sin 2
θ
)



π/
6
0
+
e
π
2
−
π
6
:
=
π
3
−
√
3
2
+
π
3
=
4
π
−
3
√
3
6
4.
A
=
s
π/
4
0
s
8 sin 4
θ
0
rdrdθ
=
s
π/
4
0
1
2
r
2




8 sin 4
θ
0
dθ
=
1
2
s
π/
4
0
64 sin
2
4
θdθ
=32

1
2
θ
−
1
16
sin 8
θ





π/
4
0
=4
π
5.
Using symmetry,
V
=2
s
π/
6
0
s
5 cos 3
θ
0
4
rdrdθ
=4
s
π/
6
0
r
2



5 cos 3
θ
0
dθ
=4
s
π/
6
0
25 cos
2
3
θdθ
= 100

1
2
θ
+
1
12
sin 6
θ





π/
6
0
=
25
π
3
6.
V
=
s
2
π
0
s
2
0

9
−
r
2
rdrdθ
=
s
2
π
0
−
1
3
(9
−
r
2
)
3
/
2




2
0
dθ
=
−
1
3
s
2
π
0
(5
3
/
2
−
27)
dθ
=
1
3
(27
−
5
3
/
2
)2
π
=
2
π
(27
−
5
√
5)
3
7.
V
=
s
2
π
0
s
3
1

16
−
r
2
rdrdθ
=
s
2
π
0
−
1
3
(16
−
r
2
)
3
/
2




3
1
dθ
=
−
1
3
s
2
π
0
(7
3
/
2
−
15
3
/
2
)
dθ
=
1
3
(15
3
/
2
−
7
3
/
2
)2
π
=
2
π
(15
√
15
−
7
√
7)
3
8.
V
=
s
2
π
0
s
5
0
√
r
2
rdrdθ
=
s
2
π
0
1
3
r
3




5
0
dθ
=
s
2
π
0
125
3
dθ
=
250
π
3
9.
V
=
s
π/
2
0
s
1+cos
θ
0
(
r
sin
θ
)
rdrdθ
=
s
π/
2
0
1
3
r
3
sin
θ




1+cos
θ
0
dθ
=
1
3
s
π/
2
0
(1 + cos
θ
)
3
sin
θdθ
=
1
3
N
−
1
4
(1 + cos
θ
)
4
o




π/
2
0
=
−
1
12
(1
−
2
4
)=
5
4
430

431. Exercises 9.11
10.
Using symmetry,
V
=2
s
π/
2
0
s
cos
θ
0
(2 +
r
2
)
rdrdθ
=
s
π/
2
0

r
2
+
1
4
r
4





cos
θ
0
dθ
=2
s
π/
2
0

cos
2
θ
+
1
4
cos
4
θ

dθ
=2
s
π/
2
0
0
cos
2
θ
+
1
4

1 + cos 2
θ
2

2
,
dθ
=
s
π/
2
0

2 cos
2
θ
+
1
8
+
1
4
cos 2
θ
+
1
8
cos
2
2
θ

dθ
=

θ
+
1
2
sin 2
θ
+
1
8
θ
+
1
8
sin 2
θ
+
1
16
θ
+
1
64
sin 4
θ





π/
2
0
=
19
π
32
.
11.
m
=
s
π/
2
0
s
3
1
kr dr dθ
=
k
s
π/
2
0
1
2
r
2




3
1
dθ
=
1
2
k
s
π/
2
0
8
dθ
=2
kπ
M
y
=
s
π/
2
0
s
3
1
kxrdrdθ
=
k
s
π/
2
0
s
3
1
r
2
cos
θdrdθ
=
k
s
π/
2
0
1
3
r
3
cos
θ




3
1
dθ
=
1
3
k
s
π/
2
0
26 cos
θdθ
=
26
3
k
sin
θ




π/
2
0
=
26
3
k
¯
x
=
M
y
/m
=
26
k/
3
2
kπ
=
13
3
π
. Since the region and density function are symmetric about the ray
θ
=
π/
4,
¯
y
=¯
x
=13
/
3
π
and the center of mass is (13
/
3
π,
13
/
3
π
).
12.
The interior of the upper-half circle is traced from
θ
=0to
π/
2. The density is
kr
.
Since both the region and the density are symmetric about the polar axis, ¯
y
=0.
m
=
s
π/
2
0
s
cos
θ
0
kr
2
dr dθ
=
k
s
π/
2
0
1
3
r
3




cos
θ
0
dθ
=
k
3
s
π/
2
0
cos
3
θdθ
=
k
3

2
3
+
1
3
cos
2
θ

sin
θ




π/
2
0
=
2
k
9
M
y
=
k
s
π/
2
0
s
cos
θ
0
(
r
cos
θ
)(
r
)(
rdrdθ
)=
k
s
π/
2
0
s
cos
θ
0
r
3
cos
θdrdθ
=
k
s
π/
2
0
1
4
r
4
cos
θ




cos
θ
0
dθ
=
k
4
s
π/
2
0
cos
5
θdθ
=
k
4

sin
θ
−
2
3
sin
3
θ
+
1
5
sin
5
θ





π/
2
0
=
2
k
15
Thus, ¯
x
=
2
k/
15
2
k/
9
=3
/
5 and the center of mass is (3
/
5
,
0).
13.
In polar coordinates the line
x
= 3 becomes
r
cos
θ
=3or
r
= 3 sec
θ
. The angle of
inclination of the line
y
=
√
3
x
is
π/
3.
m
=
s
π/
3
0
s
3 sec
θ
0
r
2
rdrdθ
=
s
π/
3
0
1
4
r
4




3 sec
θ
0
dθ
=
81
4
s
π/
3
0
sec
4
θdθ
=
81
4
s
π/
3
0
(1 + tan
2
θ
) sec
2
θdθ
=
81
4

tan
θ
+
1
3
tan
3
θ





π/
3
0
=
81
4
(
√
3+
√
3)=
81
2
√
3
431

432. Exercises 9.11
M
y
=
s
π/
3
0
s
3 sec
θ
0
xr
2
rdrdθ
=
s
π/
3
0
s
3 sec
θ
0
r
4
cos
θdrdθ
=
s
π/
3
0
1
5
r
5
cos
θ




3 sec
θ
0
dθ
=
243
5
s
π/
3
0
sec
5
θ
cos
θdθ
=
243
5
s
π/
3
0
sec
4
θdθ
=
243
5
(2
√
3)=
486
5
√
3
M
x
=
s
π/
3
0
s
3 sec
θ
0
yr
2
rdrdθ
=
s
π/
3
0
s
3 sec
θ
0
r
4
sin
θdθ
=
s
π/
3
0
1
5
r
5
sin
θ




3 sec
θ
0
=
243
5
s
π/
3
0
sec
5
θ
sin
θdθ
=
243
5
s
π/
3
0
tan
θ
sec
4
θdθ
=
243
5
s
π/
3
0
tan
θ
(1 + tan
2
θ
) sec
2
θdθ
=
243
5
s
π/
3
0
(tan
θ
+ tan
3
θ
) sec
2
θdθ
=
243
5

1
2
tan
2
θ
+
1
4
tan
4
θ





π/
3
0
=
243
5

3
2
+
9
4

=
729
4
¯
x
=
M
y
/m
=
486
√
3
/
5
81
√
3
/
2
=12
/
5; ¯
y
=
M
x
/m
=
729
/
4
81
√
3
/
2
=3
√
3
/
2. The center of mass is (12
/
5
,
3
√
3
/
2).
14.
Since both the region and the density are symmetric about the
x
-axis, ¯
y
= 0. Using
symmetry,
m
=2
s
π/
4
0
s
4 cos 2
θ
0
kr dr dθ
=2
k
s
π/
4
0
1
2
r
2




4 cos 2
θ
0
dθ
=16
k
s
π/
4
0
cos
2
2
θdθ
=16
k

1
2
θ
+
1
8
sin 4
θ





π/
4
0
=2
kπ
M
y
=2
s
π/
4
0
s
4 cos 2
θ
0
kxr dr dθ
=2
k
s
π/
4
0
s
4 cos 2
θ
0
r
2
cos
θdrdθ
=2
k
s
π/
4
0
1
3
r
3
cos
θ




4 cos 2
θ
0
dθ
=
128
3
k
s
π/
4
0
cos
3
2
θ
cos
θdθ
=
128
3
k
s
π/
4
0
(1
−
2 sin
2
θ
)
3
cos
θdθ
=
128
3
k
s
π/
4
0
(1
−
6 sin
2
θ
+ 12 sin
4
θ
−
8sin
6
θ
) cos
θdθ
=
128
3
k

sin
θ
−
2 sin
3
θ
+
12
5
sin
5
θ
−
8
7
sin
7
θ





π/
4
0
=
128
3
k
n
√
2
2
−
√
2
2
+
3
√
2
10
−
√
2
14
g
=
1024
105
√
2
k
¯
x
=
M
y
/m
=
1024
√
2
k/
105
2
kπ
=
512
√
2
105
π
. The center of mass is (512
√
2
/
105
π,
0) or approximately (2
.
20
,
0).
15.
The density is
ρ
=
k/r
.
m
=
s
π/
2
0
s
2+2 cos
θ
2
k
r
rdrdθ
=
k
s
π/
2
0
s
2+2 cos
θ
2
dr dθ
=
k
s
π/
2
0
2 cos
θdθ
=2
k
(sin
θ
)



π/
2
0
=2
k
M
y
=
s
π/
2
0
s
2+2 cos
θ
2
x
k
r
rdrdθ
=
k
s
π/
2
0
s
2+2 cos
θ
2
r
cos
θdrdθ
=
k
s
π/
2
0
1
2
r
2




2+2 cos
θ
2
cos
θdθ
=
1
2
k
s
π/
2
0
(8cos
θ
+ 4 cos
2
θ
) cos
θdθ
=2
k
s
π/
2
0
(2 cos
2
θ
+ cos
θ
−
sin
2
θ
cos
θ
)
dθ
=2
k

θ
+
1
2
sin 2
θ
+ sin
θ
−
1
3
sin
3
θ





π/
2
0
=2
k

π
2
+
2
3

=
3
π
+4
3
k
432

433. Exercises 9.11
M
x
=
s
π/
2
0
s
2+2 cos
θ
2
y
k
r
rdrdθ
=
k
s
π/
2
0
s
2+2 cos
θ
2
r
sin
θdrdθ
=
k
s
π/
2
0
1
2
r
2




2+2 cos
θ
2
sin
θdθ
=
1
2
k
s
π/
2
0
(8cos
θ
+ 4 cos
2
θ
) sin
θdθ
=
1
2
k

−
4 cos
2
θ
−
4
3
cos
3
θ





π/
2
0
=
1
2
k
N
−

−
4
−
4
3
lo
=
8
3
k
¯
x
=
M
y
/m
=
(3
π
+4)
k/
3
2
k
=
3
π
+4
6
;¯
y
=
M
x
/m
=
8
k/
3
2
k
=
4
3
. The center of mass is ((3
π
+4)
/
6
,
4
/
3).
16.
m
=
s
π
0
s
2+2 cos
θ
0
kr dr dθ
=
k
s
π
0
1
2
r
2




2+2 cos
θ
0
dθ
=2
k
s
π
0
(1 + cos
θ
)
2
dθ
=2
k
s
π
0
(1 + 2 cos
θ
+ cos
2
θ
)
dθ
=2
k

θ
+ 2 sin
θ
+
1
2
θ
+
1
4
sin 2
θ





π
0
=3
πk
M
y
=
s
π
0
s
2+2 cos
θ
0
kxr dr dθ
=
k
s
π
0
s
2+2 cos
θ
0
r
2
cos
θdrdθ
=
k
s
π
0
1
3
r
3




2+2 cos
θ
0
cos
θdθ
=
8
3
k
s
π
0
(1 + cos
θ
)
3
cos
θdθ
=
8
3
k
s
π
0
(cos
θ
+ 3 cos
2
θ
+ 3 cos
3
θ
+ cos
4
θ
)
dθ
=
8
3
k
N
sin
θ
+

3
2
θ
+
3
4
sin 2
θ

+ (3 sin
θ
−
sin
3
θ
)+

3
8
θ
+
1
4
sin 2
θ
+
1
32
sin 4
θ
lo




π
0
=
8
3
k

15
8
π

=5
πk
M
x
=
s
π
0
s
2+2 cos
θ
0
kyrdrdθ
=
k
s
π
0
s
2+2 cos
θ
0
r
2
sin
θdrdθ
=
k
s
π
0
1
3
r
3




2+2 cos
θ
0
sin
θdθ
=
8
3
k
s
π
0
(1 + cos
θ
)
3
sin
θdθ
=
8
3
k
s
π
0
(1 + 3 cos
θ
+ 3 cos
2
θ
+ cos
3
θ
) sin
θdθ
=
8
3
k

−
cos
θ
−
3
2
cos
2
θ
−
cos
3
θ
−
1
4
cos
4
θ





π
0
=
8
3
k
N
1
4
−

−
15
4
lo
=
32
3
k
¯
x
=
M
y
/m
=
5
πk
3
πk
=5
/
3; ¯
y
=
M
x
/m
=
32
k/
3
3
πk
=32
/
9
π
. The center of mass is (5
/
3
,
32
/
9
π
).
17.
I
x
=
s
2
π
0
s
a
0
y
2
kr dr dθ
=
k
s
2
π
0
s
a
0
r
3
sin
2
θdrdθ
=
k
s
2
π
0
1
4
r
4
sin
2
θ




a
0
dθ
=
ka
4
4
s
2
π
0
sin
2
θdθ
=
ka
4
4

1
2
θ
−
1
4
sin 2
θ





2
π
0
=
kπa
4
4
18.
I
x
=
s
2
π
0
s
a
0
y
2
1
1+
r
4
rdrdθ
=
s
2
π
0
s
a
0
r
3
1+
r
4
sin
2
θdrdθ
=
s
2
π
0
1
4
ln(1 +
r
4
)




a
0
sin
2
θdθ
=
1
4
ln(1 +
a
4
)

1
2
θ
−
1
4
sin 2
θ





2
π
0
=
π
4
ln(1 +
a
4
)
19.
Solving
a
=2
a
cos
θ
, cos
θ
=1
/
2or
θ
=
π/
3. The density is
k/r
3
. Using symmetry,
I
y
=2
s
π/
3
0
s
2
a
cos
θ
a
x
2
k
r
3
rdrdθ
=2
k
s
π/
3
0
s
2
a
cos
θ
a
cos
2
θdrdθ
=2
k
s
π/
3
0
(2
a
cos
3
θ
−
a
cos
2
θ
)
dθ
=2
ak

2 sin
θ
−
2
3
sin
3
θ
−
1
2
θ
−
1
4
sin 2
θ





π/
3
0
=2
ak
n
√
3
−
√
3
4
−
π
6
−
√
3
8
g
=
5
ak
√
3
4
−
akπ
3
433

434. Exercises 9.11
20.
Solving 1 = 2 sin 2
θ
, we obtain sin 2
θ
=1
/
2or
θ
=
π/
12 and
θ
=5
π/
12.
I
y
=
s
5
π/
12
π/
12
s
2 sin 2
θ
1
x
2
sec
2
θ r dr dθ
=
s
5
π/
12
π/
12
s
2 sin 2
θ
1
r
3
dr dθ
=
s
5
π/
12
π/
12
1
4
r
4




2 sin 2
θ
1
dθ
=4
s
5
π/
12
π/
12
sin
4
2
θdθ
=2

3
4
θ
−
1
4
sin 4
θ
+
1
32
sin 8
θ





5
π/
12
π/
12
=2
0n
5
π
16
+
√
3
8
−
√
3
64
g
−
n
π
16
−
√
3
8
+
√
3
64
g,
=
8
π
+7
√
3
16
21.
From Problem 17,
I
x
=
kπa
4
/
4. By symmetry,
I
y
=
I
x
.Thus
I
0
=
kπa
4
/
2.
22.
The density is
ρ
=
kr
.
I
0
=
s
π
0
s
θ
0
r
2
(
kr
)
rdrdθ
=
k
s
π
0
s
θ
0
r
4
dr dθ
=
k
s
π
0
1
5
r
5




θ
0
dθ
=
1
5
k
s
π
0
θ
5
dθ
=
1
5
k

1
6
θ
6





π
0
=
kπ
6
30
23.
The density is
ρ
=
k/r
.
I
0
=
s
3
1
s
1
/r
0
r
2
k
r
rdθdr
=
k
s
3
1
s
1
/r
0
r
2
dθ dr
=
k
s
3
1
r
2

1
r

dr
=
k

1
2
r
2





3
1
=4
k
24.
I
0
=
s
π
0
s
2
a
cos
θ
0
r
2
kr dr dθ
=
k
s
π
0
1
4
r
4




2
a
cos
θ
0
dθ
=4
ka
4
s
π
0
cos
4
θdθ
=4
ka
4

3
8
θ
+
1
4
sin 2
θ
+
1
32
sin 4
θ





π
0
=4
ka
4

3
π
8

=
3
kπa
4
2
25.
s
3
−
3
s
√
9
−
x
2
0

x
2
+
y
2
dy dx
=
s
π
0
s
3
0
|
r
|
rdrdθ
=
s
π
0
1
3
r
3




3
0
dθ
=9
s
π
0
dθ
=9
π
26.
s
√
2
/
2
0
s
√
1
−
y
2
y
y
2

x
2
+
y
2
dx dy
=
s
π/
4
0
s
1
0
r
2
sin
2
θ
|
r
|
rdrdθ
=
s
π/
4
0
s
1
0
r
2
sin
2
θdrdθ
=
s
π/
4
0
1
3
r
3
sin
2
θ




1
0
dθ
=
1
3
s
π/
4
0
sin
2
θdθ
=
1
3

1
2
θ
−
1
4
sin 2
θ





π/
4
0
=
π
−
2
24
27.
s
1
0
s
√
1
−
y
2
0
e
x
2
+
y
2
dx dy
=
s
π/
2
0
s
1
0
e
r
2
rdrdθ
=
s
π/
2
0
1
2
e
r
2




1
0
dθ
=
1
2
s
π/
2
0
(
e
−
1)
dθ
=
π
(
e
−
1)
4
434

435. Exercises 9.11
28.
s
√
π
−
√
π
s
√
π
−
x
2
0
sin(
x
2
+
y
2
)
dy dx
=
s
π
0
s
√
π
0
(sin
r
2
)
rdrdθ
=
s
π
0
−
1
2
cos
r
2




√
π
0
dθ
=
−
1
2
s
π
0
(
−
1
−
1)
dθ
=
π
29.
s
1
0
s
√
4
−
x
2
√
1
−
x
2
x
2
x
2
+
y
2
dy dx
+
s
2
1
s
√
4
−
x
2
0
x
2
x
2
+
y
2
dy dx
=
s
π/
2
0
s
2
1
r
2
cos
2
θ
r
2
rdrdθ
=
s
π/
2
0
s
2
1
r
cos
2
θdrdθ
=
s
π/
2
0
1
2
r
2




2
1
cos
2
θdθ
=
3
2
s
π/
2
0
cos
2
θdθ
=
3
2

1
2
θ
+
1
4
sin 2
θ





π/
2
0
=
3
π
8
30.
s
1
0
s
√
2
y
−
y
2
0
(1
−
x
2
−
y
2
)
dx dy
=
s
π/
4
0
s
2 sin
θ
0
(1
−
r
2
)
rdrdθ
+
s
π/
2
π/
4
s
csc
θ
0
(1
−
r
2
)
rdrdθ
=
s
π/
4
0

1
2
r
2
−
1
4
r
4





2 sin
θ
0
dθ
+
s
π/
2
π/
4

1
2
r
2
−
1
4
r
4





csc
θ
0
dθ
=
s
π/
4
0
(2 sin
2
θ
−
4 sin
4
θ
)
dθ
+
s
π/
2
π/
4

1
2
csc
2
θ
−
1
4
csc
4
θ

dθ
=
N
θ
−
1
2
sin 2
θ
−

3
2
θ
−
sin 2
θ
+
1
8
sin 4
θ
lo
+
N
−
1
2
cot
θ
−
1
4

−
cot
θ
−
1
3
cot
3
θ
lo




π/
2
π/
4
=

−
π
8
+
1
2

+
N
0
−

−
1
4
+
1
12
lo
=
16
−
3
π
24
31.
s
5
−
5
s
√
25
−
x
2
0
(4
x
+3
y
)
dy dx
=
s
π
0
s
5
0
(4
r
cos
θ
+3
r
sin
θ
)
rdrdθ
=
s
π
0
s
5
0
(4
r
2
cos
θ
+3
r
2
sin
θ
)
dr dθ
=
s
π
0

4
3
r
3
cos
θ
+
r
3
sin
θ





5
0
dθ
=
s
π
0

500
3
cos
θ
+ 125 sin
θ

dθ
=

500
3
sin
θ
−
125 cos
θ





π
0
= 250
32.
s
1
0
s
√
1
−
y
2
0
1
1+

x
2
+
y
2
dx dy
=
s
π/
2
0
s
1
0
1
1+
r
rdrdθ
=
s
π/
2
0
s
1
0

1
−
1
1+
r

dr dθ
=
s
π/
2
0
[
r
−
ln(1 +
r
)]



1
0
dθ
=
s
π/
2
0
(1
−
ln 2)
dθ
=
π
2
(1
−
ln 2)
33.
The volume of the cylindrical portion of the tank is
V
c
=
π
(4
.
2)
2
19
.
3
≈
1069
.
56 m
3
. We take the equation of
the ellipsoid to be
x
2
(4
.
2)
2
+
z
2
(5
.
15)
2
=1 or
z
=
±
5
.
15
4
.
2

(4
.
2)
2
−
x
2
−
y
2
.
435

436. Exercises 9.11
The volume of the ellipsoid is
V
e
=2

5
.
15
4
.
2

ss
R

(4
.
2)
2
−
x
2
−
y
2
dx dy
=
10
.
3
4
.
2
s
2
π
0
s
4
.
2
0
[(4
.
2)
2
−
r
2
]
1
/
2
rdrdθ
=
10
.
3
4
.
2
s
2
π
0
0

−
1
2

2
3
[(4
.
2)
2
−
r
2
]
3
/
2




4
.
2
0
,
dθ
=
10
.
3
4
.
2
1
3
s
2
π
0
(4
.
2)
3
dθ
=
2
π
3
10
.
3
4
.
2
(4
.
2)
3
≈
380
.
53
.
The volume of the tank is approximately 1069
.
56+380
.
53 = 1450
.
09 m
3
.
34.
ss
R
(
x
+
y
)
dA
=
s
π/
2
0
s
2
2 sin
θ
(
r
cos
θ
+
r
sin
θ
)
rdrdθ
=
s
π/
2
0
s
2
2 sin
θ
r
2
(cos
θ
+ sin
θ
)
dr dθ
=
s
π/
2
0
1
3
r
3
(cos
θ
+ sin
θ
)




2
2 sin
θ
dθ
=
8
3
s
π/
2
0
(cos
θ
+ sin
θ
−
sin
3
θ
cos
θ
−
sin
4
θ
)
dθ
=
8
3

sin
θ
−
cos
θ
−
1
4
sin
4
θ
+
1
4
sin
3
θ
cos
θ
−
3
8
θ
+
3
16
sin 2
θ





π/
2
0
=
8
3
Na
1
−
1
4
−
3
π
16

−
(
−
1)
o
=
28
−
3
π
6
35.
I
2
=
s
∞
0
s
∞
0
e
−
(
x
2
+
y
2
)
dx dy
=
s
π/
2
0
s
∞
0
e
−
r
2
rdrdθ
=
s
π/
2
0
lim
t
→∞
−
1
2
e
−
r
2




t
0
dθ
=
s
π/
2
0
lim
t
→∞

−
1
2
e
−
t
2
+
1
2

dθ
=
s
π/
2
0
1
2
dθ
=
π
4
;
I
=
√
π
2
Exercises 9.12
1.
The sides of the triangle are
C
1
:
y
=0,0
≤
x
≤
1;
C
2
:
x
=1,0
≤
y
≤
3;
C
3
:
y
=3
x
,
0
≤−
x
≤
1.
=
ˇ
C
(
x
−
y
)
dx
+
xy dy
=
s
1
0
xdx
+
s
3
0
ydy
+
s
0
1
(
x
−
3
x
)
dx
+
s
0
1
x
(3
x
)3
dx
=

1
2
x
2





1
0
+

1
2
y
2





3
0
+(
−
x
2
)



1
0
+(3
x
2
)



0
1
=
1
2
+
9
2
+1
−
3=3
ss
R
(
y
+1)
dA
=
s
1
0
s
3
x
0
(
y
+1)
dy dx
=
s
1
0

1
2
y
2
+
y





3
x
0
dx
=
s
1
0

9
2
x
2
+3
x

dx
=

3
2
x
3
+
3
2
x
2





1
0
=3
2.
The sides of the rectangle are
C
1
:
y
=0,
−
1
≤
x
≤
1;
C
2
:
x
=1,0
≤
y
≤
1;
C
3
:
y
=1,1
≥
x
≥−
1;
C
4
:
x
=
−
1, 1
≥
y
≥
0.
=
ˇ
C
3
x
2
ydx
+(
x
2
−
5
y
)
dy
=
s
1
−
1
0
dx
+
s
1
0
(1
−
5
y
)
dy
=
s
1
−
1
0
dx
+
s
1
0
(1
−
5
y
)
dy
=
s
−
1
1
3
x
2
dx
+
s
0
1
(1
−
5
y
)
dy
=

y
−
5
2
y
2





1
0
+
x
3



−
1
1
+

y
−
5
2
y
2





0
1
=
−
2
ss
R
(2
x
−
3
x
2
)
dA
=
s
1
0
s
1
−
1
(2
x
−
3
x
2
)
dx dy
=
s
1
0
(
x
2
−
x
3
)



1
−
1
dy
=
s
1
0
(
−
2)
dy
=
−
2
436

437. Exercises 9.12
3.
=
ˇ
C
−
y
2
dx
+
x
2
dy
=
s
2
π
0
(
−
9 sin
2
t
)(
−
3 sin
t
)
dt
+
s
2
π
0
9 cos
2
t
(3 cos
t
)
dt
=27
s
2
π
0
[(1
−
cos
2
t
) sin
t
+(1
−
sin
2
t
) cos
t
]
dt
=27

−
cos
t
+
1
3
cos
3
t
+ sin
t
−
1
3
sin
3
t





2
π
0
= 27(0) = 0
ss
R
(2
x
+2
y
)
dA
=2
s
2
π
0
s
3
0
(
r
cos
θ
+
r
sin
θ
)
rdrdθ
=2
s
2
π
0
s
3
0
r
2
(cos
θ
+ sin
θ
)
dr dθ
=2
s
2
π
0
N
1
3
r
3
(cos
θ
+ sin
θ
)
o




3
0
dθ
=18
s
2
π
0
(cos
θ
+ sin
θ
)
dθ
= 18(sin
θ
−
cos
θ
)



2
π
0
= 18(0) = 0
4.
The sides of the region are
C
1
:
y
=0,0
≤
x
≤
2;
C
2
:
y
=
−
x
+2,2
≥
x
≥
1;
C
3
:
y
=
√
x
,1
≥
x
≥
0.
=
ˇ
C
−
2
y
2
dx
+4
xy dy
=
s
2
0
0
dx
+
s
1
2
−
2(
−
x
+2)
2
dx
+
s
1
2
4
x
(
−
x
+ 2)(
−
dx
)
+
s
0
1
−
2
xdx
+
s
0
1
4
x
√
x

1
2
√
x

dx
=0+
2
3
+
8
3
+1
−
1=
10
3
ss
R
8
ydA
=
s
1
0
s
2
−
y
y
2
8
ydxdy
=
s
1
0
8
y
(2
−
y
−
y
2
)
dy
=

8
y
2
−
8
3
y
3
−
2
y
4





1
0
=
10
3
5.
P
=2
y
,
P
y
=2,
Q
=5
x
,
Q
x
=5
=
ˇ
C
2
ydx
+5
xdy
=
ss
R
(5
−
2)
dA
=3
ss
R
dA
= 3(25
π
)=75
π
6.
P
=
x
+
y
2
,
P
y
=2
y
,
Q
=2
x
2
−
y
,
Q
x
=4
x
=
ˇ
C
(
x
+
y
2
)
dx
+(2
x
2
−
y
)
dy
=
ss
R
(4
x
−
2
y
)
dA
=
s
2
−
2
s
4
x
2
(4
x
−
2
y
)
dy dx
=
s
2
−
2
(4
xy
−
y
2
)



4
x
2
dx
=
s
2
−
2
(16
x
−
16
−
4
x
3
+
x
4
)
dx
=

8
x
2
−
16
x
−
x
4
+
1
5
x
5





2
−
2
=
−
96
5
7.
P
=
x
4
−
2
y
3
,
P
y
=
−
6
y
2
,
Q
=2
x
3
−
y
4
,
Q
x
=6
x
2
. Using polar coordinates,
=
ˇ
C
(
x
4
−
2
y
3
)
dx
+(2
x
3
−
y
4
)
dy
=
ss
R
(6
x
2
+6
y
2
)
dA
=
s
2
π
0
s
2
0
6
r
2
rdrdθ
=
s
2
π
0

3
2
r
4





2
0
dθ
=
s
2
π
0
24
dθ
=48
π.
437

438. Exercises 9.12
8.
P
=
x
−
3
y
,
P
y
=
−
3,
Q
=4
x
+
y
,
Q
x
=4
=
ˇ
C
(
x
−
3
y
)
dx
+4(
x
+
y
)
dy
=
ss
R
(4+3)
dA
= 7(10) = 70
9.
P
=2
xy
,
P
y
=2
x
,
Q
=3
xy
2
,
Q
x
=3
y
2
=
ˇ
C
2
xy dx
+3
xy
2
dy
=
ss
R
(3
y
2
−
2
x
)
dA
=
s
2
1
s
2
x
2
(3
y
2
−
2
x
)
dy dx
=
s
2
1
(
y
3
−
2
xy
)



2
x
2
dx
=
s
2
1
(8
x
3
−
4
x
2
−
8+4
x
)
dx
=

2
x
4
−
4
3
x
3
−
8
x
+2
x
2





2
1
=
40
3
−

−
16
3

=
56
3
10.
P
=
e
2
x
sin 2
y
,
P
y
=2
e
2
x
cos 2
y
,
Q
=
e
2
x
cos 2
y
,
Q
x
=2
e
2
x
cos 2
y
=
ˇ
C
=
e
2
x
sin 2
ydx
+
e
2
x
cos 2
ydy
=
ss
R
0
dA
=0
11.
P
=
xy
,
P
y
=
x
,
Q
=
x
2
,
Q
x
=2
x
. Using polar coordinates,
=
ˇ
C
xy dx
+
x
2
dy
=
ss
R
(2
x
−
x
)
dA
=
s
π/
2
−
π/
2
s
1
0
r
cos
θ r dr dθ
=
s
π/
2
−
π/
2

1
3
r
3
cos
θ





1
0
dθ
=
s
π/
2
−
π/
2
1
3
cos
θdθ
=
1
3
sin
θ




π/
2
−
π/
2
=
2
3
12.
P
=
e
x
2
,
P
y
=0,
Q
= 2 tan
−
1
x
,
Q
x
=
2
1+
x
2
=
ˇ
C
e
x
2
dx
+ 2 tan
−
1
xdy
=
ss
R
2
1+
x
2
dA
=
s
0
−
1
s
1
−
x
2
1+
x
2
dy dx
=
s
0
−
1

2
y
1+
x
2





1
−
x
dx
=
s
0
−
1

2
1+
x
2
+
2
x
1+
x
2

dx
= [2 tan
−
1
x
+ ln(1 +
x
2
)]



0
−
1
=0
−
e
−
π
2
+ln2
:
=
π
2
−
ln 2
13.
P
=
1
3
y
3
,
P
y
=
y
2
,
Q
=
xy
+
xy
2
,
Q
x
=
y
+
y
2
=
ˇ
C
1
3
y
3
dx
+(
xy
+
xy
2
)
dy
=
ss
R
ydA
=
s
1
/
√
2
0
s
1
−
y
2
y
2
ydxdy
=
s
1
/
√
2
0
(
xy
)



1
−
y
2
y
2
dy
=
s
1
/
√
2
0
(
y
−
y
3
−
y
3
)
dy
=

1
2
y
2
−
1
2
y
4





1
/
√
2
0
=
1
4
−
1
8
=
1
8
438

439. Exercises 9.12
14.
P
=
xy
2
,
P
y
=2
xy
,
Q
= 3 cos
y
,
Q
x
=0
=
ˇ
C
xy
2
dx
+ 3 cos
ydy
=
ss
R
(
−
2
xy
)
dA
=
−
s
1
0
s
x
2
x
3
2
xy dy dx
=
−
s
1
0
(
xy
)



x
2
x
3
dx
=
−
s
1
0
(
x
3
−
x
4
)
dx
=

1
4
x
4
−
1
5
x
5





1
0
=
−
1
20
15.
P
=
ay
,
P
y
=
a
,
Q
=
bx
,
Q
x
=
b
.
=
ˇ
C
ay dx
+
bx dy
=
ss
R
(
b
−
a
)
dA
=(
b
−
a
)
×
(area bounded by
C
)
16.
P
=
P
(
x
),
P
y
=0,
Q
=
Q
(
y
),
Q
x
=0.
=
ˇ
C
P
(
x
)
dx
+
Q
(
y
)
dy
=
ss
R
0
dA
=0
17.
For the first integral:
P
=0,
P
y
=0,
Q
=
x
,
Q
x
=1;
=
ˇ
C
xdy
=
ss
R
1
dA
= area of
R
.
For the second integral:
P
=
y
,
P
y
=1,
Q
=0,
Q
x
=0;
−
=
ˇ
C
ydx
=
−
ss
R
−
1
dA
= area of
R
.
Thus,
=
ˇ
C
xdy
=
−
=
ˇ
C
ydx
.
18.
P
=
−
y
,
P
y
=
−
1,
Q
=
x
,
Q
x
=1.
1
2
=
ˇ
C
−
ydx
+
xdy
=
1
2
ss
R
2
dA
=
ss
R
dA
= area of
R
19.
A
=
ss
R
dA
=
=
ˇ
C
xdy
=
s
2
π
0
a
cos
3
t
(3
a
sin
2
t
cos
tdt
)=3
a
2
s
2
π
0
sin
2
t
cos
4
tdt
=3
a
2

1
16
t
−
1
64
sin 4
t
+
1
48
sin
3
2
t





2
π
0
=
3
8
πa
2
20.
A
=
ss
R
dA
=
=
ˇ
C
xdy
=
s
2
π
0
a
cos
t
(
b
cos
tdt
)=
ab
s
2
π
0
cos
2
tdt
=
ab

1
2
t
+
1
4
sin 2
t





2
π
0
=
πab
21. (a)
Parameterize
C
by
x
=
x
1
+(
x
2
−
x
1
)
t
and
y
=
y
1
+(
y
2
−
y
1
)
t
for 0
≤
t
≤
1. Then
s
C
−
ydx
+
xdy
=
s
1
0
−
[
y
1
+(
y
2
−
y
1
)
t
](
x
2
−
x
1
)
dt
+
s
1
0
[
x
1
+(
x
2
−
x
1
)
t
](
y
2
−
y
1
)
dt
=
−
(
x
2
−
x
1
)
n
y
1
t
+
1
2
(
y
2
−
y
1
)
t
2
/




1
0
+(
y
2
−
y
1
)
n
x
1
t
+
1
2
(
x
2
−
x
1
)
t
2
/




1
0
=
−
(
x
2
−
x
1
)
n
y
1
+
1
2
(
y
2
−
y
1
)
/
+(
y
2
−
y
1
)
n
x
1
+
1
2
(
x
2
−
x
1
)
/
=
x
1
y
2
−
x
2
y
1
.
(b)
Let
C
i
be the line segment from (
x
i
,y
i
)to(
x
i
+1
,y
i
+1
) for
i
=1,2,
...
,
n
−
1, and
C
2
the line segment
from (
x
n
,y
n
)to(
x
1
,y
1
). Then
A
=
1
2
=
ˇ
C
−
ydx
+
xdy
Using Problem 18
=
1
2
0
s
C
1
−
ydx
+
xdy
+
s
C
2
−
ydx
+
xdy
+
···
+
s
C
n
−
1
−
ydx
+
xdy
+
s
C
n
−
ydx
+
xdy
,
=
1
2
(
x
1
y
2
−
x
2
y
1
)+
1
2
(
x
2
y
3
−
x
3
y
2
)+
1
2
(
x
n
−
1
y
n
−
x
n
y
n
−
1
)+
1
2
(
x
n
y
1
−
x
1
y
n
)
.
22.
From part
(b)
of Problem 21
A
=
1
2
[(
−
1)(1)
−
(1)(3)] +
1
2
[(1)(2)
−
(4)(1)] +
1
2
[(4)(5)
−
(3)(2)] +
1
2
[(3)(3)
−
(
−
1)(5)]
=
1
2
(
−
4
−
2+14+14)=11
.
439

440. Exercises 9.12
23.
P
=4
x
2
−
y
3
,
P
y
=
−
3
y
2
;
Q
=
x
3
+
y
2
,
Q
x
=3
x
2
.
=
ˇ
C
(4
x
2
−
y
3
)
dx
+(
x
3
+
y
2
)
dy
=
ss
R
(3
x
2
+3
y
2
)
dA
=
s
2
π
0
s
2
1
3
r
2
(
rdrdθ
)=
s
2
π
0

3
4
r
4





2
1
dθ
=
s
2
π
0
45
4
dθ
=
45
π
2
24.
P
= cos
x
2
−
y
,
P
y
=
−
1;
Q
=

y
3
+1,
Q
x
=0
=
ˇ
C
(cos
x
2
−
y
)
dx
+

y
3
+1
dy
=
ss
R
(0+1)
dA
ss
R
dA
=(6
√
2)
2
−
π
(2)(4) = 72
−
8
π
25.
We first observe that
P
y
=(
y
4
−
3
x
2
y
2
)
/
(
x
2
=
y
2
)
3
=
Q
x
. Letting
C
N
be the circle
x
2
+
y
2
=
1
4
we have
=
ˇ
C
−
y
3
dx
+
xy
2
dy
(
x
2
+
y
2
)
2
=
=
ˇ
C
N
−
y
3
dx
+
xy
2
dy
(
x
2
+
y
2
)
2
x
=
1
4
cos
t, dx
=
−
1
4
sin
t dt, y
=
1
4
sin
t, dy
=
1
4
cos
tdt
=
s
2
π
0
−
1
64
sin
3
t
(
−
1
4
sin
tdt
)+
1
4
cos
t
(
1
16
sin
2
t
)(
1
4
cos
tdt
)
1
/
256
=
s
2
π
0
(sin
4
t
+ sin
2
t
cos
2
t
)
dt
=
s
2
π
0
(sin
4
t
+ (sin
2
t
−
sin
4
t
)
dt
=
s
2
π
0
sin
2
tdt
=

1
2
t
−
1
4
sin 2
t





2
π
0
=
π
26.
We first observe that
P
y
=[4
y
2
−
(
x
+1)
2
]
/
[(
x
+1)
2
+4
y
2
]
2
=
Q
x
. Letting
C
N
be the ellipse (
x
+1)
2
+4
y
2
=4
we have
=
ˇ
C
−
y
(
x
+1)
2
+4
y
2
dx
+
x
+1
(
x
+1)
2
+4
y
2
dy
=
=
ˇ
C
N
−
y
(
x
+1)
2
+4
y
2
dx
+
x
+1
(
x
+1)
2
+4
y
2
dy
x
+ 1 = 2 cos
t
,
dx
=
−
2 sin
t dt, y
= sin
t, dy
= cos
tdt
=
s
2
π
0
N
−
sin
t
4
(
−
2 sin
t
)+
2 cos
t
4
cos
t
o
dt
=
1
2
s
2
π
0
(sin
2
t
+ cos
2
t
)
dt
=
π.
27.
Writing
ss
R
x
2
dA
=
ss
R
(
Q
x
−
P
y
)
dA
we identify
Q
= 0 and
P
=
−
x
2
y
. Then, with
C
:
x
= 3 cos
t
,
y
= 2 sin
t
,0
≤
t
≤
2
π
, we have
ss
R
x
2
dA
=
=
ˇ
C
Pdx
+
Qdy
=
=
ˇ
C
−
x
2
ydx
=
−
s
2
π
0
9 cos
2
t
(2 sin
t
)(
−
3 sin
t
)
dt
=
54
4
s
2
π
0
4 sin
2
t
cos
2
tdt
=
27
2
s
2
π
0
sin
2
2
tdt
=
27
4
s
2
π
0
(1
−
cos 4
t
)
dt
=
27
4

t
−
1
4
sin 4
t





2
π
0
=
27
π
2
.
28.
Writing
ss
R
[1
−
2(
y
−
1)]
dA
=
ss
R
(
Q
x
−
P
y
)
dA
we identify
Q
=
x
and
P
=(
y
−
1)
2
. Then, with
440

441. Exercises 9.12
C
1
:
x
= cos
t
,
y
−
1 = sin
t
,
−
π/
2
≤
t
≤
π/
2, and
C
2
:
x
=0,2
≥
y
≥
0,
ss
R
[1
−
2(
y
−
1)]
dA
=
s
C
1
Pdx
+
Qdy
+
s
C
2
Pdx
+
Qdy
=
s
C
1
(
y
−
1)
2
dx
+
xdy
+
s
C
2
0
dy
=
s
π/
2
−
π/
2
[sin
2
t
(
−
sin
t
) + cos
t
cos
t
]
dt
=
s
π/
2
−
π/
2
[cos
2
t
−
(1
−
cos
2
t
) sin
t
]
dt
=
s
π/
2
−
π/
2
N
1
2
(1 + cos 2
t
)
−
sin
t
+ cos
2
t
sin
t
o
dt
=

1
2
t
+
1
4
sin 2
t
+ cos
t
−
1
3
cos
3
t





π/
2
−
π/
2
=
π
4
−
e
−
π
4
:
=
π
2
.
29.
P
=
x
−
y
,
P
y
=
−
1,
Q
=
x
+
y
,
Q
x
=1;
W
=
=
ˇ
C
F
·
dr
=
ss
R
2
dA
=2
×
area = 2

3
π
4

=
3
2
π
30.
P
=
−
xy
2
,
P
y
=
−
2
xy
,
Q
=
x
2
y
,
Q
x
=2
xy
. Using polar coordinates,
W
=
=
ˇ
C
F
·
d
r
=
ss
R
4
xy dA
=
s
π/
2
0
s
2
1
4(
r
cos
θ
)(
r
sin
θ
)
rdrdθ
=
s
π/
2
0
(
r
4
cos
θ
sin
θ
)



2
1
dθ
=15
s
π/
2
0
sin
θ
cos
θdθ
=
15
2
sin
2
θ




π/
2
0
=
15
2
.
31.
Since
s
B
A
Pdx
+
Qdy
is independent of path,
P
y
=
Q
x
by Theorem 9
.
9. Then, by Green’s Theorem
=
ˇ
C
Pdx
+
Qdy
=
ss
R
(
Q
x
−
P
y
)
dA
=
ss
R
0
dA
=0
.
32.
Let
P
= 0 and
Q
=
x
2
. Then
Q
x
−
P
y
=2
x
and
1
2
A
=
ˇ
C
x
2
dy
=
1
2
A
ss
R
2
xdA
=
((
R
xdA
A
=¯
x.
Let
P
=
y
2
and
Q
= 0. Then
Q
x
−
P
y
=
−
2
y
and
−
1
2
A
=
ˇ
C
y
2
dx
=
−
1
2
A
ss
R
−
2
ydA
=
((
R
ydA
A
=¯
y.
33.
Using Green’s Theorem,
W
=
=
ˇ
C
F
·
d
r
=
=
ˇ
C
−
ydx
+
xdy
=
ss
R
2
dA
=2
s
2
π
0
s
1+cos
θ
0
rdrdθ
=2
s
2
π
0

1
2
r
2





1+cos
θ
0
dθ
=
s
2
π
0
(1 + 2 cos
θ
+ cos
2
θ
)
dθ
=

θ
+ 2 sin
θ
+
1
2
θ
+
1
4
sin 2
θ





2
π
0
=3
π.
441

442. Exercises 9.13
Exercises 9.13
1.
Letting
z
= 0, we have 2
x
+3
y
= 12. Using
f
(
x, y
)=
z
=3
−
1
2
x
−
3
4
y
we have
f
x
=
−
1
2
,
f
y
=
−
3
4
,1+
f
2
x
+
f
2
y
=
29
16
. Then
A
=
s
6
0
s
4
−
2
x/
3
0

29
/
16
dy dx
=
√
29
4
s
6
0

4
−
2
3
x

dx
=
√
29
4

4
x
−
1
3
x
2





6
0
=
√
29
4
(24
−
12) = 3
√
29
.
2.
We see from the graph in Problem 1 that the plane is entirely above the region bounded by
r
= sin 2
θ
in the first octant. Using
f
(
x, y
)=
z
=3
−
1
2
x
−
3
4
y
we have
f
x
=
−
1
2
,
f
y
=
−
3
4
,
1+
f
2
x
+
f
2
y
=
29
16
. Then
A
=
s
π/
2
0
s
sin 2
θ
0

29
/
16
rdrdθ
=
√
29
4
s
π/
2
0
1
2
r
2




sin 2
θ
0
dθ
=
√
29
8
s
π/
2
0
sin
2
2
θdθ
=
√
29
8

1
2
θ
−
1
8
sin 4
θ





π/
2
0
=
√
29
π
32
.
3.
Using
f
(
x, y
)=
z
=
√
16
−
x
2
we see that for 0
≤
x
≤
2 and 0
≤
y
≤
5,
z>
0.
Thus, the surface is entirely above the region. Now
f
x
=
−
x
√
16
−
x
2
,
f
y
=0,
1+
f
2
x
+
f
2
y
=1+
x
2
16
−
x
2
=
16
16
−
x
2
and
A
=
s
5
0
s
2
0
4
√
16
−
x
2
dx dy
=4
s
5
0
sin
−
1
x
4




2
0
dy
=4
s
5
0
π
6
dy
=
10
π
3
.
4.
The region in the
xy
-plane beneath the surface is bounded by the graph of
x
2
+
y
2
=2.
Using
f
(
x, y
)=
z
=
x
2
+
y
2
we have
f
x
=2
x
,
f
y
=2
y
,1+
f
2
x
+
f
2
y
=1+4(
x
2
+
y
2
).
Then,
A
=
s
2
π
0
s
√
2
0

1+4
r
2
rdrdθ
=
s
2
π
0
1
12
(1+4
r
2
)
3
/
2




√
2
0
dθ
=
1
12
s
2
π
0
(27
−
1)
dθ
=
13
π
3
.
5.
Letting
z
= 0 we have
x
2
+
y
2
= 4. Using
f
(
x, y
)=
z
=4
−
(
x
2
+
y
2
)wehave
f
x
=
−
2
x
,
f
y
=
−
2
y
,1+
f
2
x
+
f
2
y
=1+4(
x
2
+
y
2
). Then
A
=
s
2
π
0
s
2
0

1+4
r
2
rdrdθ
=
s
2
π
0
1
3
(1+4
r
2
)
3
/
2




2
0
dθ
=
1
12
s
2
π
0
(17
3
/
2
−
1)
dθ
=
π
6
(17
3
/
2
−
1)
.
442

443. Exercises 9.13
6.
The surfaces
x
2
+
y
2
+
z
2
= 2 and
z
2
=
x
2
+
y
2
intersect on the cylinder 2
x
2
+2
y
2
=2
or
x
2
+
y
2
= 1. There are portions of the sphere within the cone both above and
below the
xy
-plane. Using
f
(
x, y
)=

2
−
x
2
−
y
2
we have
f
x
=
−
x

2
−
x
2
−
y
2
,
f
y
=
−
y

2
−
x
2
−
y
2
,1+
f
2
x
+
f
2
y
=
2
2
−
x
2
−
y
2
. Then
A
=2
0
s
2
π
0
s
1
0
√
2
√
2
−
r
2
rdrdθ
,
=2
√
2
s
2
π
0
−

2
−
r
2



1
0
dθ
=2
√
2
s
2
π
0
(
√
2
−
1)
dθ
=4
π
√
2(
√
2
−
1)
.
7.
Using
f
(
x, y
)=
z
=

25
−
x
2
−
y
2
we have
f
x
=
−
x

25
−
x
2
−
y
2
,
f
y
=
−
y

25
−
x
2
−
y
2
,1+
f
2
x
+
f
2
y
=
25
25
−
x
2
−
y
2
. Then
A
=
s
5
0
s
√
25
−
y
2
/
2
0
5

25
−
x
2
−
y
2
dx dy
=5
s
5
0
sin
−
1
x

25
−
y
2




√
25
−
y
2
/
2
0
dy
=5
s
5
0
π
6
dy
=
25
π
6
.
8.
In the first octant, the graph of
z
=
x
2
−
y
2
intersects the
xy
-plane in the line
y
=
x
. The
surface is in the firt octant for
x>y
. Using
f
(
x, y
)=
z
=
x
2
−
y
2
we have
f
x
=2
x
,
f
y
=
−
2
y
,1+
f
2
x
+
f
2
y
=1+4
x
2
+4
y
2
. Then
A
=
s
π/
4
0
s
2
0

1+4
r
2
rdrdθ
=
s
π/
4
0
1
12
(1+4
r
2
)
3
/
2




2
0
dθ
=
1
12
s
π/
4
0
(17
3
/
2
−
1)
dθ
=
π
48
(17
3
/
2
−
1)
.
9.
There are portions of the sphere within the cylinder both above and below the
xy
-plane.
Using
f
(
x, y
)=
z
=

a
2
−
x
2
−
y
2
we have
f
x
=
−
x

1
2
−
x
2
−
y
2
,
f
y
=
−
y

a
2
−
x
2
−
y
2
,
1+
f
2
x
+
f
2
y
=
a
2
a
2
−
x
2
−
y
2
. Then, using symmetry,
A
=2
0
2
s
π/
2
0
s
a
sin
θ
0
a
√
a
2
−
r
2
rdrdθ
,
=4
a
s
π/
2
0
−

a
2
−
r
2



a
sin
θ
0
dθ
=4
a
s
π/
2
0
(
a
−
a

1
−
sin
2
θ
)
dθ
=4
a
2
s
π/
2
0
(1
−
cos
θ
)
dθ
=4
a
2
(
θ
−
sin
θ
)



π/
2
0
=4
a
2
e
π
2
−
1
:
=2
a
2
(
π
−
2)
.
443

444. Exercises 9.13
10.
There are portions of the cone within the cylinder both above and below the
xy
-plane. Using
f
(
x, y
)=
1
2

x
2
+
y
2
, we have
f
x
=
x
2

x
2
+
y
2
,
f
y
=
y
2

x
2
+
y
2
,1+
f
2
x
+
f
2
y
=
5
4
.
Then, using symmetry,
A
=2
0
2
s
π/
2
0
s
2 cos
θ
0
r
5
4
rdrdθ
,
=2
√
5
s
π/
2
0
1
2
r
2




2 cos
θ
0
dθ
=4
√
5
s
π/
2
0
cos
2
θdθ
=4
√
5

1
2
θ
+
1
4
sin 2
θ





π/
2
0
=
√
5
π.
11.
There are portions of the surface in each octant with areas equal to the area of the portion
in the first octant. Using
f
(
x, y
)=
z
=

a
2
−
y
2
we have
f
x
=0,
f
y
=
y

a
2
−
y
2
,
1+
f
2
x
+
f
2
y
=
a
2
a
2
−
y
2
. Then
A
=8
s
a
0
s
√
a
2
−
y
2
0
a

a
2
−
y
2
dx dy
=8
a
s
a
0
x

a
2
−
y
2




√
a
2
−
y
2
0
dy
=8
a
s
a
0
dy
=8
a
2
.
12.
From Example 1, the area of the portion of the hemisphere within
x
2
+
y
2
=
b
2
is 2
πa
(
a
−
√
a
2
−
b
2
). Thus,
the area of the sphere is
A
= 2 lim
b
→
a
2
πa
(
a
−

a
2
−
b
2
) = 2(2
πa
2
)=4
πa
2
.
13.
The projection of the surface onto the
xz
-plane is shown in the graph. Using
f
(
x, z
)=
y
=
√
a
2
−
x
2
−
z
2
we have
f
x
=
−
x
√
a
2
−
x
2
−
z
2
,
f
z
=
−
z
√
a
2
−
x
2
−
z
2
,1+
f
2
x
+
f
2
z
=
a
2
a
2
−
x
2
−
z
2
. Then
A
=
s
2
π
0
s
√
a
2
−
c
2
1
√
a
2
−
c
2
2
a
√
a
2
−
r
2
rdrdθ
=
a
s
2
π
0
−

a
2
−
r
2




√
a
2
−
c
2
1
√
a
2
−
c
2
2
dθ
=
a
s
2
π
0
(
c
2
−
c
1
)
dθ
=2
πa
(
c
2
−
c
1
)
.
14.
The surface area of the cylinder
x
2
+
z
2
=
a
2
from
y
=
c
1
to
y
=
c
2
is the area of a cylinder of radius
a
and
height
c
2
−
c
1
. This is 2
πa
(
c
2
−
c
1
).
15.
z
x
=
−
2
x
,
z
y
=0;
dS
=
√
1+4
x
2
dA
ss
S
xdS
=
s
4
0
s
√
2
0
x

1+4
x
2
dx dy
=
s
4
0
1
12
(1 + 4
x
2
)
3
/
2




√
2
0
dy
=
s
4
0
13
6
dy
=
26
3
16.
See Problem 15.
ss
S
xy
(9
−
4
z
)
dS
=
ss
S
xy
(1+4
x
2
)
dS
=
s
4
0
s
√
2
0
xy
(1+4
x
2
)
3
/
2
dx dy
=
s
4
0
y
20
(1+4
x
2
)
5
/
2




√
2
0
dy
=
s
4
0
242
20
ydy
=
121
10
s
4
0
ydy
=
121
10

1
2
y
2





4
0
=
484
5
444

445. Exercises 9.13
17.
z
x
=
x

x
2
+
y
2
,
z
y
=
y

x
2
+
y
2
;
dS
=
√
2
dA
.
Using polar coordinates,
ss
S
xz
3
dS
=
ss
R
x
(
x
2
+
y
2
)
3
/
2
√
2
dA
=
√
2
s
2
π
0
s
1
0
(
r
cos
θ
)
r
3
/
2
rdrdθ
=
√
2
s
2
π
0
s
1
0
r
7
/
2
cos
θdrdθ
=
√
2
s
2
π
0
2
9
r
9
/
2
cos
θ




1
0
dθ
=
√
2
s
2
π
0
2
9
cos
θdθ
=
2
√
2
9
sin
θ




2
π
0
=0
.
18.
z
x
=
x

x
2
+
y
2
,
z
y
=
y

x
2
+
y
2
;
dS
=
√
2
dA
.
Using polar coordinates,
ss
S
(
x
+
y
+
z
)
dS
=
ss
R
(
x
+
y
+

x
2
+
y
2
)
√
2
dA
=
√
2
s
2
π
0
s
4
1
(
r
cos
θ
+
r
sin
θ
+
r
)
rdrdθ
=
√
2
s
2
π
0
s
4
1
r
2
(1 + cos
θ
+ sin
θ
)
dr dθ
=
√
2
s
2
π
0
1
3
r
3
(1 + cos
θ
+ sin
θ
)




4
1
dθ
=
63
√
2
3
s
2
π
0
(1 + cos
θ
+ sin
θ
)
dθ
=21
√
2(
θ
+ sin
θ
−
cos
θ
)



2
π
0
=42
√
2
π.
19.
z
=

36
−
x
2
−
y
2
,
z
x
=
−
x

36
−
x
2
−
y
2
,
z
y
=
−
y

36
−
x
2
−
y
2
;
dS
=
p
1+
x
2
36
−
x
2
−
y
2
+
y
2
36
−
x
2
−
y
2
dA
=
6

36
−
x
2
−
y
2
dA
.
Using polar coordinates,
ss
S
(
x
2
+
y
2
)
zdS
=
ss
R
(
x
2
+
y
2
)

36
−
x
2
−
y
2
6

36
−
x
2
−
y
2
dA
=6
s
2
π
0
s
6
0
r
2
rdrdθ
=6
s
2
π
0
1
4
r
4




6
0
dθ
=6
s
2
π
0
324
dθ
= 972
π.
20.
z
x
=1,
z
y
=0;
dS
=
√
2
dA
ss
S
z
2
dS
=
s
1
−
1
s
1
−
x
2
0
(
x
+1)
2
√
2
dy dx
=
√
2
s
1
−
1
y
(
x
+1)
2



1
−
x
2
0
dx
=
√
2
s
1
−
1
(1
−
x
2
)(
x
+1)
2
dx
=
√
2
s
1
−
1
(1+2
x
−
2
x
3
−
x
4
)
dx
=
√
2

x
+
x
2
−
1
2
x
4
−
1
5
x
5





1
−
1
=
8
√
2
5
445

446. Exercises 9.13
21.
z
x
=
−
x
,
z
y
=
−
y
;
dS
=

1+
x
2
+
y
2
dA
ss
S
xy dS
=
s
1
0
s
1
0
xy

1+
x
2
+
y
2
dx dy
=
s
1
0
1
3
y
(1 +
x
2
+
y
2
)
3
/
2




1
0
dy
=
s
1
0
N
1
3
y
(2 +
y
2
)
3
/
2
−
1
3
y
(1 +
y
2
)
3
/
2
o
dy
=
N
1
15
(2 +
y
2
)
5
/
2
−
1
15
(1 +
y
2
)
5
/
2
o




1
0
=
1
15
(3
5
/
2
−
2
7
/
2
+1)
22.
z
=
1
2
+
1
2
x
2
+
1
2
y
2
,
z
x
=
x
,
z
y
=
y
;
dS
=

1+
x
2
+
y
2
dA
.
Using polar coordinates,
ss
S
2
zdS
=
ss
R
(1 +
x
2
+
y
2
)

1+
x
2
+
y
2
dA
=
s
π/
2
π/
3
s
1
0
(1 +
r
2
)

1+
r
2
rdrdθ
=
s
π/
2
π/
3
s
1
0
(1 +
r
2
)
3
/
2
rdrdθ
=
s
π/
2
π/
3
1
5
(1 +
r
2
)
5
/
2




1
0
dθ
=
1
5
s
π/
2
π/
3
(2
5
/
2
−
1)
dθ
=
4
√
2
−
1
5
e
π
2
−
π
3
:
=
(4
√
2
−
1)
π
30
.
23.
y
x
=2
x
,
y
z
=0;
dS
=
√
1+4
x
2
dA
ss
S
24
√
yzdS
=
s
3
0
s
2
0
24
xz

1+4
x
2
dx dz
=
s
3
0
2
z
(1+4
x
2
)
3
/
2



2
0
dz
= 2(17
3
/
2
−
1)
s
3
0
zdz
= 2(17
3
/
2
−
1)

1
2
z
2





3
0
= 9(17
3
/
2
−
1)
24.
x
y
=
−
2
y
,
x
z
=
−
2
z
;
dS
=

1+4
y
2
+4
z
2
dA
.
Using polar coordinates,
ss
S
(1+4
y
2
+4
z
2
)
1
/
2
dS
=
s
π/
2
0
s
2
1
(1+4
r
2
)
rdrdθ
=
s
π/
2
0
1
16
(1+4
r
2
)
2



2
1
dθ
=
1
16
s
π/
2
0
12
dθ
=
3
π
8
.
25.
Write the equation of the surface as
y
=
1
2
(6
−
x
−
3
z
).
y
x
=
−
1
2
,
y
z
=
−
3
2
;
dS
=

1+1
/
4+9
/
4=
√
14
2
.
ss
S
(3
z
2
+4
yz
)
dS
=
s
2
0
s
6
−
3
z
0
N
3
z
2
+4
z
1
2
(6
−
x
−
3
z
)
o
√
14
2
dx dz
=
√
14
2
s
2
0
[3
z
2
x
−
z
(6
−
x
−
3
z
)
2
]



6
−
3
z
0
dz
=
√
14
2
s
2
0
.
[3
z
2
(6
−
3
z
)
−
0]
−
[0
−
z
(6
−
3
z
)
2
]
F
dz
=
√
14
2
s
2
0
(36
z
−
18
z
2
)
dz
=
√
14
2
(18
z
2
−
6
z
3
)



2
0
=
√
14
2
(72
−
48) = 12
√
14
446

447. Exercises 9.13
26.
Write the equation of the surface as
x
=6
−
2
y
−
3
z
. Then
x
y
=
−
2,
x
z
=
−
3;
dS
=
√
1+4+9=
√
14 .
ss
S
(3
z
2
+4
yz
)
dS
=
s
2
0
s
3
−
3
z/
2
0
(3
z
2
+4
yz
)
√
14
dy dz
=
√
14
s
2
0
(3
yz
+2
y
2
z
)



3
−
3
z/
2
0
dz
=
√
14
s
2
0
N
9
z
e
1
−
z
2
:
+18
z
e
1
−
z
2
:
2
o
dz
=
√
14
s
2
0

27
z
−
45
2
z
2
+
9
2
z
3

dz
=
√
14

27
2
z
2
−
15
2
z
3
+
9
8
z
4





2
0
=
√
14(54
−
60 + 18) = 2
√
14
27.
The density is
ρ
=
kx
2
. The surface is
z
=1
−
x
−
y
. Then
z
x
=
−
1,
z
y
=
−
1;
dS
=
√
3
dA
.
m
=
ss
S
kx
2
dS
=
k
s
1
0
s
1
−
x
0
x
2
√
3
dy dx
=
√
3
k
s
1
0
1
3
x
3




1
−
x
0
dx
=
√
3
3
k
s
1
0
(1
−
x
)
3
dx
=
√
3
3
k
N
−
1
4
(1
−
x
)
4
o




1
0
=
√
3
12
k
28.
z
x
=
−
x

4
−
x
2
−
y
2
,
z
y
=
−
y

4
−
x
2
−
y
2
;
dS
=
p
1+
x
2
4
−
x
2
−
y
2
+
y
2
4
−
x
2
−
y
2
dA
=
2

4
−
x
2
−
y
2
dA
.
Using symmetry and polar coordinates,
m
=4
ss
S
|
xy
|
dS
=4
s
π/
2
0
s
2
0
(
r
2
cos
θ
sin
θ
)
2
√
4
−
r
2
rdrdθ
=4
s
π/
2
0
s
2
0
r
2
(4
−
r
2
)
−
1
/
2
sin 2
θ
(
rdr
)
dθ
u
=4
−
r
2
,du
=
−
2
rdr, r
2
=4
−
u
=4
s
π/
2
0
s
0
4
(4
−
u
)
u
−
1
/
2
sin 2
θ

−
1
2
du

dθ
=
−
2
s
π/
2
0
s
0
4
(4
u
−
1
/
2
−
u
1
/
2
) sin 2
θdudθ
=
−
2
s
π/
2
0

8
u
1
/
2
−
2
3
u
3
/
2





0
4
sin 2
θdθ
=
−
2
s
π/
2
0

−
32
3
sin 2
θ

dθ
=
64
3

−
1
2
cos 2
θ





π/
2
0
=
64
3
.
29.
The surface is
g
(
x, y, z
)=
y
2
+
z
2
−
4=0.
∇
g
=2
y
j
+2
z
k
,
|∇
g
|
=2

y
2
+
z
2
;
n
=
y
j
+
z
k

y
2
+
z
2
;
F
·
n
=
2
yz

y
2
+
z
2
+
yz

y
2
+
z
2
=
3
yz

y
2
+
z
2
;
z
=

4
−
y
2
,
z
x
=0,
z
y
=
−
y

4
−
y
2
;
dS
=
p
1+
y
2
4
−
y
2
dA
=
2

4
−
y
2
dA
Flux =
ss
S
F
·
n
dS
=
ss
R
3
yz

y
2
+
z
2
2

4
−
y
2
dA
=
ss
R
3
y

4
−
y
2

y
2
+4
−
y
2
2

4
−
y
2
dA
=
s
3
0
s
2
0
3
ydydx
=
s
3
0
3
2
y
2




2
0
dx
=
s
3
0
6
dx
=18
447

448. Exercises 9.13
30.
The surface is
g
(
x, y, z
)=
x
2
+
y
2
+
z
−
5=0.
∇
g
=2
x
i
+2
y
j
+
k
,
|∇
g
|
=

1+4
x
2
+4
y
2
;
n
=
2
x
i
+2
y
j
+
k

1+4
x
2
+4
y
2
;
F
·
n
=
z

1+4
x
2
+4
y
2
;
z
x
=
−
2
x
,
z
y
=
−
2
y
,
dS
=

1+4
x
2
+4
y
2
dA
. Using polar coordinates,
Flux =
ss
S
F
·
n
dS
=
ss
R
z

1+4
x
2
+4
y
2

1+4
x
2
+4
y
2
dA
=
ss
R
(5
−
x
2
−
y
2
)
dA
=
s
2
π
0
s
2
0
(5
−
r
2
)
rdrdθ
=
s
2
π
0

5
2
r
2
−
1
4
r
4





2
0
dθ
=
s
2
π
0
6
dθ
=12
π.
31.
From Problem 30,
n
=
2
x
i
+2
y
j
+
k

1+4
x
2
+4
y
2
. Then
F
·
n
=
2
x
2
+2
y
2
+
z

1+4
x
2
+4
y
2
. Also, from Problem 30,
dS
=

1+4
x
2
+4
y
2
dA
. Using polar coordinates,
Flux =
ss
S
F
·
n
dS
=
ss
R
2
x
2
+2
y
2
+
z

1+4
x
2
+4
y
2

1+4
x
2
+4
y
2
dA
=
ss
R
(2
x
2
+2
y
2
+5
−
x
2
−
y
2
)
dA
=
s
2
π
0
s
2
0
(
r
2
+5)
rdrdθ
=
s
2
π
0

1
4
r
4
+
5
2
r
2





2
0
dθ
=
s
2
π
0
14
dθ
=28
π.
32.
The surface is
g
(
x, y, z
)=
z
−
x
−
3=0.
∇
g
=
−
i
+
k
,
|∇
g
|
=
√
2;
n
=
−
i
+
k
√
2
;
F
·
n
=
1
√
2
x
3
y
+
1
√
2
xy
3
;
z
x
=1,
z
y
=0,
dS
=
√
2
dA
. Using polar coordinates,
Flux =
ss
S
F
·
n
dS
=
ss
R
1
√
2
(
x
3
y
+
xy
3
)
√
2
dA
=
ss
R
xy
(
x
2
+
y
2
)
dA
=
s
π/
2
0
s
2 cos
θ
0
(
r
2
cos
θ
sin
θ
)
r
2
rdrdθ
=
s
π/
2
0
s
2 cos
θ
0
r
5
cos
θ
sin
θdrdθ
=
s
π/
2
0
1
6
r
6
cos
θ
sin
θ




2 cos
θ
0
dθ
=
1
6
s
π/
2
0
64 cos
7
θ
sin
θdθ
=
32
3

−
1
8
cos
8
θ





π/
2
0
=
4
3
.
33.
The surface is
g
(
x, y, z
)=
x
2
+
y
2
+
z
−
4.
∇
g
=2
x
i
+2
y
j
+
k
,
|∇
g
|
=

4
x
2
+4
y
2
+1;
n
=
2
x
i
+2
y
j
+
k

4
x
2
+4
y
2
+1
;
F
·
n
=
x
3
+
y
3
+
z

4
x
2
+4
y
2
+1
;
z
x
=
−
2
x
,
z
y
=
−
2
y
,
dS
=

1+4
x
2
+4
y
2
dA
. Using polar coordinates,
Flux =
ss
S
F
·
n
dS
=
ss
R
(
x
3
+
y
3
+
z
)
dA
=
ss
R
(4
−
x
2
−
y
2
+
x
3
+
y
3
)
dA
=
s
2
π
0
s
2
0
(4
−
r
2
+
r
3
cos
3
θ
+
r
3
sin
3
θ
)
rdrdθ
=
s
2
π
0

2
r
2
−
1
4
r
4
+
1
5
r
5
cos
3
θ
+
1
5
r
5
sin
3
θ





2
0
dθ
=
s
2
π
0

4+
32
5
cos
3
θ
+
32
5
sin
3
θ

dθ
=4
θ



2
π
0
+0+0=8
π.
448

449. Exercises 9.13
34.
The surface is
g
(
x, y, z
)=
x
+
y
+
z
−
6.
∇
g
=
i
+
j
+
k
,
|∇
g
|
=
√
3;
n
=(
i
+
j
+
k
)
/
√
3;
F
·
n
=(
e
y
+
e
x
+18
y
)
/
√
3;
z
x
=
−
1,
z
y
=
−
1,
dS
=
√
1+1+1
dA
=
√
3
dA
.
Flux =
ss
S
F
·
n
dS
=
ss
r
(
e
y
+
e
x
+18
y
)
dA
=
s
6
0
s
6
−
x
0
(
e
y
+
e
x
+18
y
)
dy dx
=
s
6
0
(
e
y
+
ye
x
+9
y
2
)



6
−
x
0
dx
=
s
6
0
[
e
6
−
x
+(6
−
x
)
e
x
+ 9(6
−
x
)
2
−
1]
dx
=[
−
e
6
−
x
+6
e
x
−
xe
x
+
e
x
−
3(6
−
x
)
3
−
x
]



6
0
=(
−
1+6
e
6
−
6
e
6
+
e
6
−
6)
−
(
−
e
6
+6+1
−
648) = 2
e
6
+ 634
≈
1440
.
86
35.
For
S
1
:
g
(
x, y, z
)=
x
2
+
y
2
−
z
,
∇
g
=2
x
i
+2
y
j
−
k
,
|∇
g
|
=

4
x
2
+4
y
2
+1;
n
1
=
2
x
i
+2
y
j
−
k

4
x
2
+4
y
2
+1
;
F
·
n
1
=
2
xy
2
+2
x
2
y
−
5
z

4
x
2
+4
y
2
+1
;
z
x
=2
x
,
z
y
=2
y
,
dS
1
=

1+4
x
2
+4
y
2
dA
.For
S
2
:
g
(
x, y, z
)=
z
−
1,
∇
g
=
k
,
|∇
g
|
=1;
n
2
=
k
;
F
·
n
2
=5
z
;
z
x
=0,
z
y
=0,
dS
2
=
dA
. Using polar coordinates and
R
:
x
2
+
y
2
≤
1
we have
Flux =
ss
S
1
F
·
n
1
dS
1
+
ss
S
2
F
·
n
2
dS
2
=
ss
R
(2
xy
2
+2
x
2
y
−
5
z
)
dA
+
ss
R
5
zdA
=
ss
R
[2
xy
2
+2
x
2
y
−
5(
x
2
+
y
2
) + 5(1)]
dA
=
s
2
π
0
s
1
0
(2
r
3
cos
θ
sin
2
θ
+2
r
3
cos
2
θ
sin
θ
−
5
r
2
+5)
rdrdθ
=
s
2
π
0

2
5
r
5
cos
θ
sin
2
θ
+
2
5
r
5
cos
2
θ
sin
θ
−
5
4
r
4
+
5
2
r
2





1
0
dθ
=
s
2
π
0
N
2
5
(cos
θ
sin
2
θ
+ cos
2
θ
sin
θ
)+
5
4
o
dθ
=
2
5

1
3
sin
3
θ
−
1
3
cos
3
θ





2
π
0
+
5
4
θ




2
π
0
=
2
5
N
−
1
3
−

−
1
3
lo
+
5
2
π
=
5
2
π.
36.
For
S
1
:
g
(
x, y, z
)=
x
2
+
y
2
+
z
−
4,
∇
g
=2
x
i
+2
y
j
+
k
,
|∇
g
|
=

4
x
2
+4
y
2
+1;
n
1
=
2
x
i
+2
y
j
+
k

4
x
2
+4
y
2
+1
;
F
·
n
1
=6
z
2
/

4
x
2
+4
y
2
+1;
z
x
=
−
2
x
,
z
y
=
−
2
y
,
dS
1
=

1+4
x
2
+4
y
2
dA
.For
S
2
:
g
(
x, y, z
)=
x
2
+
y
2
−
z
,
∇
g
=2
x
i
+2
y
j
−
k
,
|∇
g
|
=

4
x
2
+4
y
2
+1;
n
2
=
2
x
i
+2
y
j
−
k

4
x
2
+
y
2
+1
;
F
·
n
2
=
−
6
z
2
/

4
x
2
+4
y
2
+1;
z
x
=2
x
,
z
y
=2
y
,
dS
2
=

1+4
x
2
+4
y
2
dA
. Using polar coordinates and
R
:
x
2
+
y
2
≤
2wehave
Flux =
ss
S
1
F
·
n
1
dS
1
+
ss
S
1
F
·
n
2
dS
2
=
ss
R
6
z
2
dA
+
ss
−
6
z
2
dA
=
ss
R
[6(4
−
x
2
−
y
2
)
2
−
6(
x
2
+
y
2
)
2
]
dA
=6
s
2
π
0
s
√
2
0
[(4
−
r
2
)
2
−
r
4
]
rdrdθ
=6
s
2
π
0
N
−
1
6
(4
−
r
2
)
3
−
1
6
r
6
o




√
2
0
dθ
=
−
s
2
π
0
[(2
3
−
4
3
)+(
√
2)
6
]
dθ
=
s
2
π
0
48
dθ
=96
π.
449

450. Exercises 9.13
37.
The surface is
g
(
x, y, z
)=
x
2
+
y
2
+
z
2
−
a
2
=0.
∇
g
=2
x
i
+2
y
j
+2
z
k
,
|∇
g
|
=2

x
2
+
y
2
+
z
2
;
n
=
x
i
+
y
j
+
z
k

x
2
+
y
2
+
z
2
;
F
·
n
=
−
(2
x
i
+2
y
j
+2
z
k
)
·
x
i
+
y
j
+
z
k

x
2
+
y
2
+
z
2
=
−
2
x
2
+2
y
2
+2
z
2

x
2
+
y
2
+
z
2
=
−
2

x
2
+
y
2
+
z
2
=
−
2
a.
Flux =
ss
S
−
2
adS
=
−
2
a
×
area =
−
2
a
(4
πa
2
)=
−
8
πa
3
38. n
1
=
k
,
n
2
=
−
i
,
n
3
=
j
,
n
4
=
−
k
,
n
5
=
i
,
n
6
=
−
j
;
F
·
n
1
=
z
=1,
F
·
n
2
=
−
x
=0,
F
·
n
3
=
y
=1,
F
·
n
4
=
−
z
=0,
F
·
n
5
=
x
=1,
F
·
n
6
=
−
y
= 0; Flux =
ss
S
1
1
dS
+
ss
S
3
1
dS
+
ss
S
5
1
dS
=3
39.
Refering to the solution to Problem 37, we find
n
=
x
i
+
y
j
+
z
k

x
2
+
y
2
+
z
2
and
dS
=
a

a
2
−
x
2
−
y
2
dA
.
Now
F
·
n
=
kq
r
|
r
|
3
·
r
|
r
|
=
kq
|
r
|
4
|
r
|
2
=
kq
|
r
|
2
=
kq
x
2
+
y
2
+
z
2
=
kq
a
2
and
Flux =
ss
S
F
·
n
dS
=
ss
S
kq
a
2
dS
=
kq
a
2
×
area =
kq
a
2
(4
πa
2
)=4
πkq.
40.
We are given
σ
=
kz
.Now
z
x
−
x

16
−
x
2
−
y
2
,
z
y
=
−
y

16
−
x
2
−
y
2
;
dS
=
p
1+
x
2
16
−
x
2
−
y
2
+
y
2
16
−
x
2
−
y
2
dA
=
4

16
−
x
2
−
y
2
dA
Using polar coordinates,
Q
=
ss
S
kz dS
=
k
ss
R

16
−
x
2
−
y
2
4

16
−
x
2
−
y
2
dA
=4
k
s
2
π
0
s
3
0
rdrdθ
=4
k
s
2
π
0
1
2
r
2




3
0
dθ
=4
k
s
2
π
0
9
2
dθ
=36
πk.
41.
The surface is
z
=6
−
2
x
−
3
y
. Then
z
x
=
−
2,
z
y
=
−
3,
dS
=
√
1+4+9=
√
14
dA
.
The area of the surface is
A
(
s
)=
ss
S
dS
=
s
3
0
s
2
−
2
x/
3
0
√
14
dy dx
=
√
14
s
3
0

2
−
2
3
x

dx
=
√
14

2
x
−
1
3
x
2





3
0
=3
√
14
.
¯
x
=
1
3
√
14
ss
S
xdS
=
1
3
√
14
s
3
0
s
2
−
2
x/
3
0
√
14
xdydx
=
1
3
s
3
0
xy




2
−
2
x/
3
0
dx
=
1
3
s
3
0

2
x
−
2
3
x
2

dx
=
1
3

x
2
−
2
9
x
3





3
0
=1
¯
y
=
1
3
√
14
ss
S
ydS
=
1
3
√
14
s
3
0
s
2
−
2
x/
3
0
√
14
ydydx
=
1
3
s
3
0
1
2
y
2




2
−
2
x/
3
0
dx
=
1
6
s
3
0

2
−
2
3
x

2
dx
=
1
6
0
−
1
2

2
−
2
3
x

3
,




3
0
=
2
3
450

451. Exercises 9.14
¯
z
=
1
3
√
14
ss
S
zdS
=
1
3
√
14
s
3
0
s
2
−
2
x/
3
0
(6
−
2
x
−
3
y
)
√
14
dy dx
=
1
3
s
3
0

6
y
−
2
xy
−
3
2
y
2





2
−
2
x/
3
0
dx
=
1
3
s
3
0

6
−
4
x
+
2
3
x
2

dx
=
1
3

6
x
−
2
x
2
+
2
9
x
3





3
0
=2
The centroid is (1
,
2
/
3
,
2).
42.
The area of the hemisphere is
A
(
s
)=2
πa
2
. By symmetry, ¯
x
=¯
y
=0.
z
x
=
−
x

a
2
−
x
2
−
y
2
,
z
y
=
−
y

a
2
−
x
2
−
y
2
;
dS
=
p
1+
x
2
a
2
−
x
2
−
y
2
+
y
2
a
2
−
x
2
−
y
2
dA
=
a

a
2
−
x
2
−
y
2
dA
Using polar coordinates,
z
=
ss
S
zdS
2
πa
2
=
1
2
πa
2
ss
R

a
2
−
x
2
−
y
2
a

a
2
−
x
2
−
y
2
dA
=
1
2
πa
s
2
π
0
s
a
0
rdrdθ
=
1
2
πa
s
2
π
0
1
2
r
2




a
0
dθ
=
1
2
πa
s
2
π
0
1
2
s
2
dθ
=
a
2
.
The centroid is (0
,
0
,a/
2).
43.
The surface is
g
(
x, y, z
)=
z
−
f
(
x, y
)=0.
∇
g
=
−
f
x
i
−
f
y
j
+
k
,
|∇
g
|
=

f
2
x
+
f
2
y
+1;
n
=
−
f
x
i
−
f
y
j
+
k

1+
f
2
x
+
f
2
y
;
F
·
n
=
−
Pf
x
−
Qf
y
+
R

1+
f
2
x
+
f
2
y
;
dS
=

1+
f
2
x
+
f
2
y
dA
ss
S
F
·
n
dS
=
ss
R
−
Pf
x
−
Qf
y
+
R

1+
f
2
x
+
f
2
y

1+
f
2
x
+
f
2
y
dA
=
ss
R
(
−
Pf
x
−
Qf
y
+
R
)
dA
Exercises 9.14
1. Surface Integral
: curl
F
=
−
10
k
. Letting
g
(
x, y, z
)=
z
−
1, we have
∇
g
=
k
and
n
=
k
. Then
ss
S
(curl
F
)
·
n
dS
=
ss
S
(
−
10)
dS
=
−
10
×
(area of
S
)=
−
10(4
π
)=
−
40
π.
Line Integral
: Parameterize the curve
C
by
x
= 2 cos
t
,
y
= 2 sin
t
,
z
= 1, for 0
≤
t
≤
2
π
. Then
=
ˇ
C
F
·
d
r
=
=
ˇ
C
5
ydx
−
5
xdy
+3
dz
=
s
2
π
0
[10 sin
t
(
−
2 sin
t
)
−
10 cos
t
(2 cos
t
)]
dt
=
s
2
π
0
(
−
20 sin
2
t
−
20 cos
2
t
)
dt
=
s
2
π
0
−
20
dt
=
−
40
π.
2. Surface Integral
: curl
F
=4
i
−
2
j
−
3
k
. Letting
g
(
x, y, z
)=
x
2
+
y
2
+
z
−
16,
∇
g
=2
x
i
+2
y
j
+
k
, and
n
=(2
x
i
+2
y
j
+
k
)
/

4
x
2
+4
y
2
+ 1 . Thus,
ss
S
(curl
F
)
·
n
dS
=
ss
S
8
x
−
4
y
−
3

4
x
2
+4
y
2
+1
dS.
Letting the surface be
z
=16
−
x
2
−
y
2
, we have
z
x
=
−
2
x
,
z
y
=
−
2
y
, and
dS
=

1+4
x
2
+4
y
2
dA
. Then, using polar coordinates,
451

452. Exercises 9.14
ss
S
(curl
F
)
·
n
dS
=
ss
R
(8
x
−
4
y
−
3)
dA
=
s
2
π
0
s
4
0
(8
r
cos
θ
−
4
r
sin
θ
−
3)
rdrdθ
=
s
2
π
0

8
3
r
3
cos
θ
−
4
3
r
3
sin
θ
−
3
2
r
2





4
0
dθ
=
s
2
π
0

512
3
cos
θ
−
256
3
sin
θ
−
24

dθ
=

512
3
sin
θ
+
256
3
cos
θ
−
24
θ





2
π
0
=
−
48
π.
Line Integral
: Parameterize the curve
C
by
x
= 4 cos
t
,
y
= 4 sin
t
,
z
= 0, for 0
≤
t
≤
2
π
. Then,
=
ˇ
C
F
·
d
r
=
=
ˇ
C
2
zdx
−
3
xdy
+4
ydz
=
s
2
π
0
[
−
12 cos
t
(4 cos
t
)]
dt
=
s
2
π
0
−
48cos
2
tdt
=(
−
24
t
−
12 sin 2
t
)



2
π
0
=
−
48
π.
3. Surface Integral
: curl
F
=
i
+
j
+
k
. Letting
g
(
x, y, z
)=2
x
+
y
+2
z
−
6, we have
∇
g
=2
i
+
j
+2
k
and
n
=(2
i
+
j
+2
k
)
/
3. Then
ss
S
(curl
F
)
·
n
dS
=
ss
S
5
3
dS
. Letting
the surface be
z
=3
−
1
2
y
−
x
we have
z
x
=
−
1,
z
y
=
−
1
2
, and
dS
=

1+(
−
1)
2
+(
−
1
2
)
2
dA
=
3
2
dA
. Then
ss
S
(curl
F
)
·
n
dS
=
ss
R
5
3

3
2

dA
=
5
2
×
(area of
R
)=
5
2
(9) =
45
2
.
Line Integral
:
C
1
:
z
=3
−
x
,0
≤
x
≤
3,
y
=0;
C
2
:
y
=6
−
2
x
,3
≥
x
≥
0,
z
=0;
C
3
:
z
=3
−
y/
2, 6
≥
y
≥
0,
x
=0.
=
ˇ
C
zdx
+
xdy
+
ydz
=
ss
C
1
zdx
+
s
C
2
xdy
+
s
C
3
ydz
=
s
3
0
(3
−
x
)
dx
+
s
0
3
x
(
−
2
dx
)+
s
0
6
y
(
−
dy/
2)
=

3
x
−
1
2
x
2





3
0
−
x
2




0
3
−
1
4
y
2




0
6
=
9
2
−
(0
−
9)
−
1
4
(0
−
36) =
45
2
4. Surface Integral
: curl
F
=
0
and
ss
S
(curl
F
)
·
n
dS
=0.
Line Integral
: the curve is
x
= cos
t
,
y
= sin
t
,
z
=0,0
≤
t
≤
2
π
.
=
ˇ
C
xdx
+
ydy
+
zdz
=
s
2
π
0
[cos
t
(
−
sin
t
) + sin
t
(cos
t
)]
dt
=0
.
5.
curl
F
=2
i
+
j
. A unit vector normal to the plane is
n
=(
i
+
j
+
k
)
/
√
3 . Taking the
equation of the plane to be
z
=1
−
x
−
y
,wehave
z
x
=
z
y
=
−
1. Thus,
dS
=
√
1+1+1
dA
=
√
3
dA
and
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
S
√
3
dS
=
√
3
ss
R
√
3
dA
=3
×
(area of
R
) = 3(1
/
2) = 3
/
2
.
6.
curl
F
=
−
2
xz
i
+
z
2
k
. A unit vector normal to the plane is
n
=(
j
+
k
)
/
√
2 . From
z
=1
−
y
, we have
z
x
=0
and
z
y
=
−
1. Thus,
dS
=
√
1+1
dA
=
√
2
dA
and
452

453. Exercises 9.14
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
R
1
√
2
z
2
√
2
dA
=
ss
R
(1
−
y
)
2
dA
=
s
2
0
s
1
0
(1
−
y
)
2
dy dx
=
s
2
0
−
1
3
(1
−
y
)
3




1
0
dx
=
s
2
0
1
3
dx
=
2
3
.
7.
curl
F
=
−
2
y
i
−
z
j
−
x
k
. A unit vector normal to the plane is
n
=(
j
+
k
)
/
√
2 . From
z
=1
−
y
we have
z
x
=0
and
z
y
=
−
1. Then
dS
=
√
1+1
dA
=
√
2
dA
and
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
R
N
−
1
√
2
(
z
+
x
)
o
√
2
dA
=
ss
R
(
y
−
x
−
1)
dA
=
s
2
0
s
1
0
(
y
−
x
−
1)
dy dx
=
s
2
0

1
2
y
2
−
xy
−
y





1
0
dx
=
s
2
0

−
x
−
1
2

dx
=

−
1
2
x
2
−
1
2
x





2
0
=
−
3
.
8.
curl
F
=2
i
+2
j
+3
k
. Letting
g
(
x, y, z
)=
x
+2
y
+
z
−
4, we have
∇
g
=
i
+2
j
+
k
and
n
=(
i
+2
j
+
k
)
/
√
6 . From
z
=4
−
x
−
2
y
we have
z
x
=
−
1 and
z
y
=
−
2. Then
dS
=
√
6
dA
and
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
R
1
√
6
(9)
√
6
dA
=
ss
R
9
dA
= 9(4) = 36
.
9.
curl
F
=(
−
3
x
2
−
3
y
2
)
k
. A unit vector normal to the plane is
n
=(
i
+
j
+
k
)
/
√
3 . From
z
=1
−
x
−
y
, we have
z
x
=
z
y
=
−
1 and
dS
=
√
3
dA
. Then, using polar coordinates,
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
R
(
−
√
3
x
2
−
√
3
y
2
)
√
3
dA
=3
ss
R
(
−
x
2
−
y
2
)
dA
=3
s
2
π
0
s
1
0
(
−
r
2
)
rdrdθ
=3
s
2
π
0
−
1
4
r
4




1
0
dθ
=3
s
2
π
0
−
1
4
dθ
=
−
3
π
2
.
10.
curl
F
=2
xyz
i
−
y
2
z
j
+(1
−
x
2
)
k
. A unit vector normal to the surface is
n
=
2
y
j
+
k

4
y
2
+1
. From
z
=9
−
y
2
we have
z
x
=0,
z
y
=
−
2
y
and
dS
=

1+4
y
2
dA
. Then
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
R
(
−
2
y
3
z
+1
−
x
2
)
dA
=
s
3
0
s
y/
2
0
[
−
2
y
3
(9
−
y
2
)+1
−
x
2
]
dx dy
=
s
3
0

−
18
y
3
x
+2
y
5
x
+
x
−
1
3
x
3





y/
2
0
dy
=
s
3
0

−
9
y
4
+
y
6
+
1
2
y
−
1
24
y
3

dy
=

−
9
5
y
5
+
1
7
y
7
+
1
4
y
2
−
1
96
y
4





3
0
≈
123
.
57
.
11.
curl
F
=3
x
2
y
2
k
. A unit vector normal to the surface is
n
=
8
x
i
+2
y
j
+2
z
k

64
x
2
+4
y
2
+4
z
2
=
4
x
i
+
y
j
+
z
k

16
x
2
+
y
2
+
z
2
.
453

454. Exercises 9.14
From
z
x
=
−
4
x

4
−
4
x
2
−
y
2
,
z
y
=
−
y

4
−
4
x
2
−
y
2
we obtain
dS
=2
p
1+3
x
2
4
−
4
x
2
−
y
2
dA
. Then
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
R
3
x
2
y
2
z

16
x
2
+
y
2
+
z
2
n
2
p
1+3
x
2
4
−
4
x
2
−
y
2
g
dA
=
ss
R
3
x
2
y
2
dA
Using symmetry
=12
s
1
0
s
2
√
1
−
x
2
0
x
2
y
2
dy dx
=12
s
1
0

1
3
x
2
y
3





2
√
1
−
x
2
0
dx
=32
s
1
0
x
2
(1
−
x
2
)
3
/
2
dx
x
= sin
t, dx
= cos
tdt
=32
s
π/
2
0
sin
2
t
cos
4
tdt
=
π.
12.
curl
F
=
i
+
j
+
k
. Taking the surface
S
bounded by
C
to be the portion of the plane
x
+
y
+
z
= 0 inside
C
,wehave
n
=(
i
+
j
+
k
)
/
√
3 and
dS
=
√
3
dA
.
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
S
√
3
dS
=
√
3
ss
R
√
3
dA
=3
×
(area of
R
)
The region
R
is obtained by eliminating
z
from the equations of the plane and the sphere.
This gives
x
2
+
xy
+
y
2
=
1
2
. Rotating axes, we see that
R
is enclosed by the ellipse
X
2
/
(1
/
3) +
Y
2
/
1 = 1 in a rotated coordinate system. Thus,
=
ˇ
C
F
·
d
r
=3
×
(area of
R
)=3

π
1
√
3
1

=
√
3
π.
13.
Parameterize
C
by
x
= 4 cos
t
,
y
= 2 sin
t
,
z
= 4, for 0
≤
t
≤
2
π
. Then
ss
S
(curl
F
)
·
n
dS
=
=
ˇ
C
F
·
d
r
=
=
ˇ
C
6
yz dx
+5
xdy
+
yze
x
2
dz
=
s
2
π
0
[6(2 sin
t
)(4)(
−
4 sin
t
) + 5(4 cos
t
)(2 cos
t
)+0]
dt
=8
s
2
π
0
(
−
24 sin
2
t
+ 5 cos
2
t
)
dt
=8
s
2
π
0
(5
−
29 sin
2
t
)
dt
=
−
152
π.
14.
Parameterize
C
by
x
= 5 cos
t
,
y
= 5 sin
t
,
z
= 4, for 0
≤
t
≤
2
π
. Then,
ss
S
(curl
F
)
·
n
dS
=
=
ˇ
C
F
·
r
=
=
ˇ
C
ydx
+(
y
−
x
)
dy
+
z
2
dz
=
s
2
π
0
[(5 sin
t
)(
−
5 sin
t
)+(5sin
t
−
5 cos
t
)(5 cos
t
)]
dt
=
s
2
π
0
(25 sin
t
cos
t
−
25)
dt
=

25
2
sin
2
t
−
25
t





2
π
0
=
−
50
π.
454

455. Exercises 9.14
15.
Parameterize
C
by
C
1
:
x
=0,
z
=0,2
≥
y
≥
0;
C
2
:
z
=
x
,
y
=0,0
≤
x
≤
2;
C
3
:
x
=2,
z
=2,0
≤
y
≤
2;
C
4
:
z
=
x
,
y
=2,2
≥
x
≥
0. Then
ss
S
(curl
F
)
·
n
dS
=
=
ˇ
C
F
·
r
=
=
ˇ
C
3
x
2
dx
+8
x
3
ydy
+3
x
2
ydz
=
s
C
1
0
dx
+0
dy
+0
dz
+
s
C
2
3
x
2
dx
+
s
C
3
64
dy
+
s
C
4
3
x
2
dx
+6
x
2
dx
=
s
2
0
3
x
2
dx
+
s
2
0
64
dy
+
s
0
2
9
x
2
dx
=
x
3



2
0
+64
y



2
0
+3
x
3



0
2
= 112
.
16.
Parameterize
C
by
x
= cos
t
,
y
= sin
t
,
z
= sin
t
,0
≤
t
≤
2
π
. Then
ss
S
(curl
F
)
·
n
dS
=
=
ˇ
C
F
·
r
=
=
ˇ
C
2
xy
2
zdx
+2
x
2
yz dy
+(
x
2
y
2
−
6
x
)
dz
=
s
2
π
0
[2 cos
t
sin
2
t
sin
t
(
−
sin
t
)+2cos
2
t
sin
t
sin
t
cos
t
+ (cos
2
t
sin
2
t
−
6 cos
t
) cos
t
]
dt
=
s
2
π
0
(
−
2 cos
t
sin
4
t
+ 3 cos
3
t
sin
2
t
−
6 cos
2
t
)
dt
=
−
6
π.
17.
We take the surface to be
z
= 0. Then
n
=
k
and
dS
=
dA
. Since curl
F
=
1
1+
y
2
i
+2
ze
x
2
j
+
y
2
k
,
=
ˇ
C
z
2
e
x
2
dx
+
xy dy
+ tan
−
1
ydz
=
ss
S
(curl
F
)
·
n
dS
=
ss
S
y
2
dS
=
ss
R
y
2
dA
=
s
2
π
0
s
3
0
r
2
sin
2
θ r dr dθ
=
s
2
π
0
1
4
r
4
sin
2
θ




3
0
dθ
=
81
4
s
2
π
0
sin
2
θdθ
=
81
π
4
.
18. (a)
curl
F
=
xz
i
−
yz
j
. A unit vector normal to the surface is
n
=
2
x
i
+2
y
j
+
k

4
x
2
+4
y
2
+1
and
dS
=

1+4
x
2
+4
y
2
dA
. Then, using
x
= cos
t
,
y
= sin
t
,0
≤
t
≤
2
π
, we have
ss
S
(curl
F
)
·
n
dS
=
ss
R
(2
x
2
z
−
2
y
2
z
)
dA
=
ss
R
(2
x
2
−
2
y
2
)(1
−
x
2
−
y
2
)
dA
=
ss
R
(2
x
2
−
2
y
2
−
2
x
4
+2
y
4
)
dA
=
s
2
π
0
s
1
0
(2
r
2
cos
2
θ
−
2
r
2
sin
2
θ
−
2
r
4
cos
4
θ
+2
r
4
cos
4
θ
)
rdrdθ
=2
s
2
π
0
s
1
0
[
r
3
cos 2
θ
−
r
5
(cos
2
θ
−
sin
2
θ
)(cos
2
θ
+ sin
2
θ
)]
dr dθ
=2
s
2
π
0
s
1
0
(
r
3
cos 2
θ
−
r
5
cos 2
θ
)
dr dθ
=2
s
2
π
0
cos 2
θ

1
4
r
4
−
1
6
r
6





1
0
dθ
=
1
6
s
2
π
0
cos 2
θdθ
=0
.
(b)
We take the surface to be
z
= 0. Then
n
=
k
, curl
F
·
n
= curl
F
·
k
= 0 and
ss
S
(curl
F
)
·
n
dS
=0.
455

456. Exercises 9.14
(c)
By Stoke’s Theorem, using
z
=0,wehave
ss
S
(curl
F
)
·
n
dS
=
=
ˇ
C
F
·
d
r
=
=
ˇ
C
xyz dz
=
=
ˇ
C
xy
(0)
dz
=0
.
Exercises 9.15
1.
s
4
2
s
2
−
2
s
1
−
1
(
x
+
y
+
z
)
dx dy dz
=
s
4
2
s
2
−
2

1
2
x
2
+
xy
+
xz





1
−
1
dy dz
=
s
4
2
s
2
−
2
(2
y
+2
z
)
dy dz
=
s
4
2
(
y
2
+2
yz
)



2
−
2
dz
=
s
4
2
8
zdz
=4
z
2



4
2
=48
2.
s
3
1
s
x
1
s
xy
2
24
xy dz dy dx
=
s
3
1
s
x
1
24
xyz



xy
2
dy dx
=
s
3
1
s
x
1
(24
x
2
y
2
−
48
xy
)
dy dx
=
s
3
1
(8
x
2
y
3
−
24
xy
2
)



x
1
dx
=
s
3
1
(8
x
5
−
24
x
3
−
8
x
2
+24
x
)
dx
=

4
3
x
6
−
6
x
4
−
8
3
x
3
+12
x
2





3
1
= 522
−
14
3
=
1552
3
3.
s
6
0
s
6
−
x
0
s
6
−
x
−
z
0
dy dz dx
=
s
6
0
s
6
−
x
0
(6
−
x
−
z
)
dz dx
=
s
6
0

6
z
−
xz
−
1
2
z
2





6
−
x
0
dx
=
s
6
0
n
6(6
−
x
)
−
x
(6
−
x
)
−
1
2
(6
−
x
)
2
/
dx
=
s
6
0

18
−
6
x
+
1
2
x
2

dx
=

18
x
−
3
x
2
+
1
6
x
3





6
0
=36
4.
s
1
0
s
1
−
x
0
s
√
y
0
4
x
2
z
3
dz dy dx
=
s
1
0
s
1
−
x
0
x
2
z
4



√
y
0
dy dx
=
s
1
0
s
1
−
x
0
x
2
y
2
dy dx
=
s
1
0
1
3
x
2
y
3




1
−
x
0
dx
=
1
3
s
1
0
x
2
(1
−
x
)
3
dx
=
1
3
s
1
0
(
x
2
−
3
x
3
+3
x
4
−
x
5
)
dx
=
1
3

1
3
x
3
−
3
4
x
4
+
3
5
x
5
−
1
6
x
6





1
0
=
1
180
5.
s
π/
2
0
s
y
2
0
s
y
0
cos
x
y
dz dx dy
=
s
π/
2
0
s
y
2
0
y
cos
x
y
dx dy
=
s
π/
2
0
y
2
sin
x
y




y
2
0
dy
=
s
π/
2
0
y
2
sin
ydy
Integration by parts
=(
−
y
2
cos
y
+ 2 cos
y
+2
y
sin
y
)



π/
2
0
=
π
−
2
6.
s
√
2
0
s
2
√
y
s
e
x
2
0
xdzdxdy
=
s
√
2
0
s
2
√
y
xe
x
2
dx dy
=
s
√
2
0
1
2
e
x
2




2
√
y
dy
=
1
2
s
√
2
0
(
e
4
−
e
y
)
dy
=
1
2
(
ye
4
−
e
y
)



√
2
0
=
1
2
[(
e
4
√
2
−
e
√
2
)
−
(
−
1)] =
1
2
(1 +
e
4
√
2
−
e
√
2
)
456

457. Exercises 9.15
7.
s
1
0
s
1
0
s
2
−
x
2
−
y
2
0
xye
z
dz dx dy
=
s
1
0
s
1
0
xye
z



2
−
x
2
−
y
2
0
dx dy
=
s
1
0
s
1
0
(
xye
2
−
x
2
−
y
2
−
xy
)
dx dy
=
s
1
0

−
1
2
ye
2
−
x
2
−
y
2
−
1
2
x
2
y





1
0
dy
=
s
1
0

−
1
2
ye
1
−
y
2
−
1
2
y
+
1
2
ye
2
−
y
2

dy
=

1
4
e
1
−
y
2
−
1
4
y
2
−
1
4
e
2
−
y
2





1
0
=

1
4
−
1
4
−
1
4
e

−

1
4
e
−
1
4
e
2

=
1
4
e
2
−
1
2
e
8.
s
4
0
s
1
/
2
0
s
x
2
0
1

x
2
−
y
2
dy dx dz
=
s
4
0
s
1
/
2
0
sin
−
1
y
x




x
2
0
dx dz
=
s
4
0
s
1
/
2
0
sin
−
1
xdxdz
Integration by parts
=
s
4
0
(
x
sin
−
1
x
+

1
−
x
2
)



1
/
2
0
dz
=
s
4
0
n
1
2
π
6
+
√
3
2
−
1
g
dz
=
π
3
+2
√
3
−
4
9.
sss
D
zdV
=
s
5
0
s
3
1
s
y
+2
y
zdxdydz
=
s
5
0
s
3
1
xz



y
+2
y
dy dz
=
s
5
0
s
3
1
2
zdydz
=
s
5
0
2
yz



3
1
dz
=
s
5
0
4
zdz
=2
z
2



5
0
=50
10.
Using symmetry,
sss
D
(
x
2
+
y
2
)
dV
=2
s
2
0
s
4
x
2
s
4
−
y
0
(
x
2
+
y
2
)
dz dy dx
=2
s
2
0
s
4
x
2
(
x
2
+
y
2
)
z



4
−
y
0
dy dx
=2
s
2
0
s
4
x
2
(4
x
2
−
x
2
y
+4
y
2
−
y
3
)
dy dx
=2
s
2
0

4
x
2
y
−
1
2
x
2
y
2
+
4
3
y
3
−
1
4
y
4





4
x
2
dx
=2
s
2
0
Na
8
x
2
+
64
3

−

4
x
4
+
5
6
x
6
−
1
4
x
8
lo
dx
=2

8
3
x
3
+
64
3
x
−
4
5
x
5
−
5
42
x
7
+
1
36
x
9





2
0
=
23,552
315
.
11.
The other five integrals are
s
4
0
s
2
−
x/
2
0
s
4
x
+2
y
F
(
x, y, z
)
dz dy dx
,
s
4
0
s
z
0
s
(
z
−
x
)
/
2
0
F
(
x, y, z
)
dy dx dz
,
s
4
0
s
4
x
s
(
z
−
x
)
/
2
0
F
(
x, y, z
)
dy dz dx
,
s
4
0
s
z/
2
0
s
z
−
2
y
0
F
(
x, y, z
)
dx dy dz
,
s
2
0
s
4
2
y
s
z
−
2
y
0
F
(
x, y, z
)
dx dz dy
.
12.
The other five integrals are
s
3
0
s
√
36
−
4
y
2
/
3
0
s
3
1
F
(
x, y, z
)
dz dx dy
,
s
3
1
s
2
0
s
√
36
−
9
x
2
/
2
0
F
(
x, y, z
)
dy dx dz
,
s
3
1
s
3
0
s
√
36
−
4
y
2
/
3
0
F
(
x, y, z
)
dx dy dz
,
s
3
0
s
3
1
s
√
36
−
4
y
2
/
3
0
F
(
x, y, z
)
dx dz dy
,
s
2
0
s
3
1
s
√
36
−
9
x
2
/
2
0
F
(
x, y, z
)
dy dz dx
.
457

458. Exercises 9.15
13. (a)
V
=
s
2
0
s
8
x
3
s
4
0
dz dy dx
(b)
V
=
s
8
0
s
4
0
s
y
1
/
3
0
dx dz dy
(c)
V
=
s
4
0
s
2
0
s
8
x
2
dy dx dz
14.
Solving
z
=
√
x
and
x
+
z
= 2, we obtain
x
=1,
z
=1.
(a)
V
=
s
3
0
s
1
0
s
2
−
z
z
2
dx dz dy
(b)
V
=
s
1
0
s
2
−
z
z
2
s
3
0
dy dx dz
(c)
V
=
s
3
0
s
1
0
s
√
x
0
dz dx dy
+
s
3
0
s
2
1
s
2
−
x
0
dz dx dy
15.
16.
The region in the first
octant is shown.
17. 18.
19. 20.
21.
Solving
x
=
y
2
and 4
−
x
=
y
2
, we obtain
x
=2,
y
=
±
√
2 . Using symmetry,
V
=2
s
3
0
s
√
2
0
s
4
−
y
2
y
2
dx dy dz
=2
s
3
0
s
√
2
0
(4
−
2
y
2
)
dy dz
=2
s
3
0

4
y
−
2
3
y
3





√
2
0
dz
=2
s
3
0
8
√
2
3
dz
=16
√
2
.
22.
V
=
s
2
0
s
√
4
−
x
2
0
s
x
+
y
0
dz dy dx
=
s
2
0
s
√
4
−
x
2
0
z



x
+
y
0
dy dx
=
s
2
0
s
√
4
−
x
2
0
(
x
+
y
)
dy dx
=
s
2
0

xy
+
1
2
y
2





√
4
−
x
2
0
dx
=
s
2
0
n
x

4
−
x
2
+
1
2
(4
−
x
2
)
/
dx
=
n
−
1
3
(4
−
x
2
)
3
/
2
+2
x
−
1
6
x
3
/




2
0
=

4
−
4
3

−

−
8
3

=
16
3
458

459. Exercises 9.15
23.
Adding the two equations, we obtain 2
y
= 8. Thus, the paraboloids intersect
in the plane
y
= 4. Their intersection is a circle of radius 2. Using symmetry,
V
=4
s
2
0
s
√
4
−
x
2
0
s
8
−
x
2
−
z
2
x
2
+
z
2
dy dz dx
=4
s
2
0
s
√
4
−
x
2
0
(8
−
2
x
2
−
2
z
2
)
dz dx
=4
s
2
0
n
2(4
−
x
2
)
z
−
2
3
z
3
/




√
4
−
x
2
0
dx
=4
s
2
0
4
3
(4
−
x
2
)
3
/
2
dx
Trig substitution
=
16
3
a
−
x
8
(2
x
2
−
20)

4
−
x
2
+ 6 sin
−
1
x
2
h




2
0
=16
π.
24.
Solving
x
=2,
y
=
x
, and
z
=
x
2
+
y
2
, we obtain the point (2
,
2
,
8).
V
=
s
2
0
s
x
0
s
x
2
+
y
2
0
dz dy dx
=
s
2
0
s
x
0
(
x
2
+
y
2
)
dy dx
=
s
2
0

x
2
y
+
1
3
y
3





x
0
dx
=
s
2
0
4
3
x
3
dx
=
1
3
x
4




2
0
=
16
3
.
25.
We are given
ρ
(
x, y, z
)=
kz
.
m
=
s
8
0
s
4
0
s
y
1
/
3
0
kz dx dz dy
=
k
s
8
0
s
4
0
xz



y
1
/
3
0
dz dy
=
k
s
8
0
s
4
0
y
1
/
3
zdzdy
=
k
s
8
0
1
2
y
1
/
3
z
2




4
0
dy
=8
k
s
8
0
y
1
/
3
dy
=8
k

3
4
y
4
/
3





8
0
=96
k
M
xy
=
s
8
0
s
4
0
s
y
1
/
3
0
kz
2
dx dz dy
=
k
s
8
0
s
4
0
xz
2



y
1
/
3
0
dz dy
=
k
s
8
0
s
4
0
y
1
/
3
z
2
dz dy
=
k
s
8
0
1
3
y
1
/
3
z
3




4
0
dy
=
64
3
k
s
8
0
y
1
/
3
dy
=
64
3
k

3
4
y
4
/
3





8
0
= 256
k
M
xz
=
s
8
0
s
4
0
s
y
1
/
3
0
kyzdxdzdy
=
k
s
8
0
s
4
0
xyz



y
1
/
3
0
dz dy
=
k
s
8
0
s
4
0
y
4
/
3
zdzdy
=
k
s
8
0
1
2
y
4
/
3
z
2




4
0
dy
=8
k
s
8
0
y
4
/
3
dy
=8
k

3
7
y
7
/
3





8
0
=
3072
7
k
M
yz
=
s
8
0
s
4
0
s
y
1
/
3
0
kxz dx dz dy
=
k
s
8
0
s
4
0
1
2
x
2
z




y
1
/
3
0
dz dy
=
1
2
k
s
8
0
s
4
0
y
2
/
3
zdzdy
=
1
2
k
s
8
0
1
2
y
2
/
3
z
2




4
0
dy
=4
k
s
8
0
y
2
/
3
dy
=4
k

3
5
y
5
/
3





8
0
=
384
5
k
¯
x
=
M
yz
/m
=
384
k/
5
96
k
=4
/
5; ¯
y
=
M
xz
/m
=
3072
k/
7
96
k
=32
/
7; ¯
z
=
M
xy
/m
=
256
k
96
k
=8
/
3
The center of mass is (4
/
5
,
32
/
7
,
8
/
3).
26.
We use the form of the integral in Problem 14
(b)
of this section. Without loss of generality, we take
ρ
=1.
m
=
s
1
0
s
2
−
z
z
2
s
3
0
dy dx dz
=
s
1
0
s
2
−
z
z
2
3
dx dz
=3
s
1
0
(2
−
z
−
z
2
)
dz
=3

2
z
−
1
2
z
2
−
1
3
z
3





1
0
=
7
2
459

460. Exercises 9.15
M
xy
=
s
1
0
s
2
−
z
z
2
s
3
0
zdydxdz
=
s
1
0
s
2
−
z
z
2
yz



3
0
dx dz
=
s
1
0
s
2
−
z
z
2
3
zdxdz
=3
s
1
0
xz



2
−
z
z
2
dz
=3
s
1
0
(2
z
−
z
2
−
z
3
)
dz
=3

z
2
−
1
3
z
3
−
1
4
z
4





1
0
=
5
4
M
xz
=
s
1
0
s
2
−
z
z
2
s
3
0
ydydxdz
=
s
1
0
s
2
−
z
z
2
1
2
y
2




3
0
dx dz
=
9
2
s
1
0
s
2
−
z
z
2
dx dz
=
9
2
s
1
0
(2
−
z
−
z
2
)
dz
=
9
2

2
z
−
1
2
z
2
−
1
3
z
3





1
0
=
21
4
M
yz
=
s
1
0
s
2
−
z
z
2
s
3
0
xdydxdz
=
s
1
0
s
2
−
z
z
2
xy



3
0
dx dz
=
s
1
0
s
2
−
x
z
2
3
xdxdz
=3
s
1
0
1
2
x
2




2
−
z
z
2
dz
=
3
2
s
1
0
(4
−
4
z
+
z
2
−
z
4
)
dz
=
3
2

4
z
−
2
z
2
+
1
3
z
3
−
1
5
z
5





1
0
=
16
5
¯
x
=
M
yz
/m
=
16
/
5
7
/
2
=32
/
35, ¯
y
=
M
xz
/m
=
21
/
4
7
/
2
=3
/
2, ¯
z
=
M
xy
/m
=
5
/
4
7
/
2
=5
/
14.
The centroid is (32
/
35
,
3
/
2
,
5
/
14).
27.
The density is
ρ
(
x, y, z
)=
ky
. Since both the region and the density function are symmetric
with respect to the
xy
-and
yz
-planes, ¯
x
=¯
z
= 0. Using symmetry,
m
=4
s
3
0
s
2
0
s
√
4
−
x
2
0
ky dz dx dy
=4
k
s
3
0
s
2
0
yz



√
4
−
x
2
0
dx dy
=4
k
s
3
0
s
2
0
y

4
−
x
2
dx dy
=4
k
s
3
0
y
e
x
2

4
−
x
2
+ 2 sin
−
1
x
2
:




2
0
dy
=4
k
s
3
0
πy dy
=4
πk

1
2
y
2





3
0
=18
πk
M
xz
=4
s
3
0
s
2
0
s
√
4
−
x
2
0
ky
2
dz dx dy
=4
k
s
3
0
s
2
0
y
2
z



√
4
−
x
2
0
dx dy
=4
k
s
3
0
s
2
0
y
2

4
−
x
2
dx dy
=4
k
s
3
0
y
2
e
x
2

4
−
x
2
+ 2 sin
−
1
x
2
:




2
0
dy
=4
k
s
3
0
πy
2
dy
=4
πk

1
3
y
3





3
0
=36
πk.
¯
y
=
M
xz
/m
=
36
πk
18
πk
= 2. The center of mass is (0
,
2
,
0).
28.
The density is
ρ
(
x, y, z
)=
kz
.
m
=
s
1
0
s
x
x
2
s
y
+2
0
kz dz dy dx
=
k
s
1
0
s
x
x
2
1
2
z
2




y
+2
0
dy dx
=
1
2
k
s
1
0
s
x
x
2
(
y
+2)
2
dy dx
=
1
2
k
s
1
0
1
3
(
y
+2)
3




x
x
2
dx
=
1
6
k
s
1
0
[(
x
+2)
3
−
(
x
2
+2)
3
]
dx
=
1
6
k
s
1
0
[(
x
+2)
3
−
(
x
6
+6
x
4
+12
x
2
+8)]
dx
=
1
6
k
N
1
4
(
x
+2)
4
−
1
7
x
7
−
6
5
x
5
−
4
x
3
−
8
x
o




1
0
=
407
840
k
460

461. Exercises 9.15
M
xy
=
s
1
0
s
x
x
2
s
y
+2
0
kz
2
dz dy dx
=
k
s
1
0
s
x
x
2
1
3
z
3




y
+2
0
dy dx
=
1
3
k
s
1
0
s
x
x
2
(
y
+2)
3
dy dx
=
1
3
k
s
1
0
1
4
(
y
+2)
4




x
x
2
dx
=
1
12
k
s
1
0
[(
x
+2)
4
−
(
x
2
+2)
4
]
dx
=
1
12
k
s
1
0
[(
x
+2)
4
−
(
x
8
+8
x
6
+24
x
4
+32
x
2
+ 16)]
dx
=
1
12
k
N
1
5
(
x
+2)
5
−
1
9
x
9
−
8
7
x
7
−
24
5
−
32
3
x
3
−
16
x
o




1
0
=
1493
1890
k
M
xz
=
s
1
0
s
x
x
2
s
y
+2
0
kyz dz dy dx
=
k
s
1
0
s
x
x
2
1
2
yz
2




y
+2
0
dy dx
=
1
2
k
s
1
0
s
x
x
2
y
(
y
+2)
2
dy dx
=
1
2
k
s
1
0
s
x
x
2
(
y
3
+4
y
2
+4
y
)
dy dx
=
1
2
k
s
1
0

1
4
y
4
+
4
3
y
3
+2
y
2





x
x
2
dx
=
1
2
k
s
1
0

−
1
4
x
8
−
4
3
x
6
−
74
x
4
+
4
3
x
3
+2
x
2

dx
=
1
2
k

−
1
36
x
9
−
4
21
x
7
−
7
20
x
5
+
1
3
x
4
+
2
3
x
3





1
0
=
68
315
k
M
yz
=
s
1
0
s
x
x
2
s
y
+2
0
kxz dz dy dx
=
k
s
1
0
s
x
x
2
1
2
xz
2




y
+2
0
dy dx
=
1
2
k
s
1
0
s
x
x
2
x
(
y
+2)
2
dy dx
=
1
2
k
s
1
0
1
3
x
(
y
+2)
3




x
x
2
dx
=
1
6
k
s
1
0
[
x
(
x
+2)
3
−
x
(
x
2
+2)
3
]
dx
=
1
6
k
s
1
0
[
x
4
+6
x
3
+12
x
2
+8
x
−
x
(
x
2
+2)
3
]
dx
=
1
6
k
N
1
5
x
5
+
3
2
x
4
+4
x
3
+4
x
2
−
1
8
(
x
2
+2)
4
o




1
0
=
21
80
k
¯
x
=
M
yz
/m
=
21
k/
80
407
k/
840
= 441
/
814, ¯
y
=
M
xz
/m
=
68
k/
315
407
k/
840
= 544
/
1221,
¯
z
=
M
xy
/m
=
1493
k/
1890
407
k/
840
= 5972
/
3663. The center of mass is (441
/
814
,
544
/
1221
,
5972
/
3663).
29.
m
=
s
1
−
1
s
√
1
−
x
2
−
√
1
−
x
2
s
8
−
y
2+2
y
(
x
+
y
+4)
dz dy dx
30.
Both the region and the density function are symmetric with respect to the
xz
- and
yz
-planes. Thus,
m
=4
s
2
−
1
s
√
1+
z
2
0
s
√
1+
z
2
−
y
2
0
z
2
dx dy dz
.
461

462. Exercises 9.15
31.
We are given
ρ
(
x, y, z
)=
kz
.
I
y
=
s
8
0
s
4
0
s
y
1
/
3
0
kz
(
x
2
+
z
2
)
dx dz dy
=
k
s
8
0
s
4
0

1
3
x
3
z
+
xz
3





y
1
/
3
0
dz dy
=
k
s
8
0
s
4
0

1
3
yz
+
y
1
/
3
z
3

dz dy
=
k
s
8
0

1
6
yz
2
+
1
4
y
1
/
3
z
4





4
0
dy
=
k
s
8
0

8
3
y
+64
y
1
/
3

dy
=
k

4
3
y
2
+48
y
4
/
3





8
0
=
2560
3
k
From Problem 25,
m
=96
k
. Thus,
R
g
=

I
y
/m
=
r
2560
k/
3
96
k
=
4
√
5
3
.
32.
We are given
ρ
(
x, y, z
)=
k
.
I
x
=
s
1
0
s
2
−
z
z
2
s
3
0
k
(
y
2
+
z
2
)
dy dx dz
=
k
s
1
0
s
2
−
z
z
2

1
3
y
3
+
yz
2





3
0
dx dz
=
k
s
1
0
s
2
−
z
z
2
(9+3
z
2
)
dx dz
=
k
s
1
0
(9
x
+3
xz
2
)



2
−
z
z
2
dz
=
k
s
1
0
(18
−
9
z
−
3
z
2
−
3
z
3
−
3
z
4
)
dz
=
k

18
z
−
9
2
z
2
−
z
3
−
3
4
z
4
−
3
5
z
5





1
0
=
223
20
k
m
=
s
1
0
s
2
−
z
z
2
s
3
0
kdydxdz
=
k
s
1
0
s
2
−
z
z
2
3
dx dz
=3
k
s
1
0
(2
−
z
−
z
2
)
dz
=3
k

2
z
−
1
2
z
2
−
1
3
z
3





1
0
=
7
2
k
R
g
=
r
I
x
m
=
p
223
k/
20
7
k/
2
=
r
223
70
33.
I
z
=
k
s
1
0
s
1
−
x
0
s
1
−
x
−
y
0
(
x
2
+
y
2
)
dz dy dx
=
k
s
1
0
s
1
−
x
0
(
x
2
+
y
2
)(1
−
x
−
y
)
dy dx
=
k
s
1
0
s
1
−
x
0
(
x
2
−
x
3
−
x
2
y
+
y
2
−
xy
2
−
y
3
)
dy dx
=
k
s
1
0
n
(
x
2
−
x
3
)
y
−
1
2
x
2
y
2
+
1
3
(1
−
x
)
y
3
−
1
4
y
4
/




1
−
x
0
dx
=
k
s
1
0
n
1
2
x
2
−
x
3
+
1
2
x
4
+
1
12
(1
−
x
)
4
/
dx
=
k
n
1
6
x
6
−
1
4
x
4
+
1
10
x
5
−
1
60
(1
−
x
)
5
/




1
0
=
k
30
34.
We are given
ρ
(
x, y, z
)=
kx
.
I
y
=
s
1
0
s
2
0
s
4
−
z
z
kx
(
x
2
+
z
2
)
dy dx dz
=
k
s
1
0
s
2
0
(
x
3
+
xz
2
)
y



4
−
z
z
dx dz
=
k
s
1
0
s
2
0
(
x
3
+
xz
2
)(4
−
2
z
)
dx dz
=
k
s
1
0

1
4
x
4
+
1
2
x
2
z
2

(4
−
2
z
)




2
0
dz
=
k
s
1
0
(4 + 2
z
2
)(4
−
2
z
)
dz
=4
k
s
1
0
(4
−
2
z
+2
z
2
−
z
3
)
dz
=4
k

4
z
−
z
2
+
2
3
z
3
−
1
4
z
4





1
0
=
41
3
k
35.
x
= 10 cos 3
π/
4=
−
5
√
2;
y
= 10 sin 3
π/
4=5
√
2; (
−
5
√
2
,
5
√
2
,
5)
36.
x
= 2 cos 5
π/
6=
−
√
3;
y
= 2 sin 5
π/
6=1; (
−
√
3
,
1
,
−
3)
37.
x
=
√
3 cos
π/
3=
√
3
/
2;
y
=
√
3 sin
π/
3=3
/
2; (
√
3
/
2
,
3
/
2
,
−
4)
38.
x
= 4 cos 7
π/
4=2
√
2;
y
= 4 sin 7
π/
4=
−
2
√
2; (2
√
2
,
−
2
√
2
,
0)
39.
With
x
= 1 and
y
=
−
1wehave
r
2
= 2 and tan
θ
=
−
1. The point is (
√
2
,
−
π/
4
,
−
9).
462

463. Exercises 9.15
40.
With
x
=2
√
3 and
y
=2wehave
r
2
= 16 and tan
θ
=1
/
√
3 . The point is (4
,π/
6
,
17).
41.
With
x
=
−
√
2 and
y
=
√
6wehave
r
2
= 8and tan
θ
=
−
√
3 . The point is (2
√
2
,
2
π/
3
,
2).
42.
With
x
= 1 and
y
= 2 we have
r
2
= 5 and tan
θ
= 2. The point is (
√
5
,
tan
−
1
2
,
7).
43.
r
2
+
z
2
=25
44.
r
cos
θ
+
r
sin
θ
−
z
=1
45.
r
2
−
z
2
=1
46.
r
2
cos
2
θ
+
z
2
=16
47.
z
=
x
2
+
y
2
48.
z
=2
y
49.
r
cos
θ
=5,
x
=5
50.
tan
θ
=1
/
√
3,
y/x
=1
/
√
3,
x
=
√
3
y
,
x>
0
51.
The equations are
r
2
=4,
r
2
+
z
2
= 16, and
z
=0.
V
=
s
2
π
0
s
2
0
s
√
16
−
r
2
0
rdzdrdθ
=
s
2
π
0
s
2
0
r

16
−
r
2
dr dθ
=
s
2
π
0
−
1
3
(16
−
r
2
)
3
/
2




2
0
dθ
=
s
2
π
0
1
3
(64
−
24
√
3)
dθ
=
2
π
3
(64
−
24
√
3)
52.
The equation is
z
=10
−
r
2
.
V
=
s
2
π
0
s
3
0
s
10
−
r
2
1
rdzdrdθ
=
s
2
π
0
s
3
0
r
(9
−
r
2
)
dr dθ
=
s
2
π
0

9
2
r
2
−
1
4
r
4





3
0
dθ
=
s
2
π
0
81
4
dθ
=
81
π
2
.
53.
The equations are
z
=
r
2
,
r
= 5, and
z
=0.
V
=
s
2
π
0
s
5
0
s
r
2
0
rdzdrdθ
=
s
2
π
0
s
5
0
r
3
dr dθ
=
s
2
π
0
1
4
r
4




5
0
dθ
=
s
2
π
0
625
4
dθ
=
625
π
2
54.
Substituting the first equation into the second, we see that the surfaces intersect
in the plane
y
= 4. Using polar coordinates in the
xz
-plane, the equations of the
surfaces become
y
=
r
2
and
y
=
1
2
r
2
+2.
V
=
s
2
π
0
s
2
0
s
r
2
/
2+2
r
2
rdydrdθ
=
s
2
π
0
s
2
0
r

r
2
2
+2
−
r
2

dr dθ
=
s
2
π
0
s
2
0

2
r
−
1
2
r
3

dr dθ
=
s
2
π
0

r
2
−
1
8
r
4





2
0
dθ
=
s
2
π
0
2
dθ
=4
π
55.
The equation is
z
=
√
a
2
−
r
2
. By symmetry, ¯
x
=¯
y
=0.
m
=
s
2
π
0
s
a
0
s
√
a
2
−
r
2
0
rdzdrθ
=
s
2
π
0
s
a
0
r

a
2
−
r
2
dr dθ
=
s
2
π
0
−
1
3
(
a
2
−
r
2
)
3
/
2




a
0
dθ
=
s
2
π
0
1
3
a
3
dθ
=
2
3
πa
3
463

464. Exercises 9.15
M
xy
=
s
2
π
0
s
a
0
s
√
a
2
−
r
2
0
zr dz dr dθ
=
s
2
π
0
s
a
0
1
2
rz
2




√
a
2
−
r
2
0
dr dθ
=
1
2
s
2
π
0
s
a
0
r
(
a
2
−
r
2
)
dr dθ
=
1
2
s
2
π
0

1
2
a
2
r
2
−
1
4
r
4





a
0
dθ
=
1
2
s
2
π
0
1
4
a
4
dθ
=
1
4
πa
4
¯
z
=
M
xy
/m
=
πa
4
/
4
2
πa
3
/
3
=3
a/
8. The centroid is (0
,
0
,
3
a/
8).
56.
We use polar coordinates in the
yz
-plane. The density is
ρ
(
x, y, z
)=
kz
. By symmetry,
¯
y
=¯
z
=0.
m
=
s
2
π
0
s
4
0
s
5
0
kxr dx dr dθ
=
k
s
2
π
0
s
4
0
1
2
rz
2




5
0
dr dθ
=
k
2
s
2
π
0
s
4
0
25
rdrdθ
=
25
k
2
s
2
π
0
1
2
r
2




4
0
dθ
=
25
k
2
s
2
π
0
8
dθ
= 200
kπ
M
yz
=
s
2
π
0
s
4
0
s
5
0
kx
2
rdxdrdθ
=
k
s
2
π
0
s
4
0
1
3
rx
3




5
0
dr dθ
=
1
3
k
s
2
π
0
s
4
0
125
rdrdθ
=
1
3
k
s
2
π
0
125
2
r
2




4
0
dθ
=
1
3
k
s
2
π
0
1000
dθ
=
2000
3
kπ
¯
x
=
M
yz
/m
=
2000
kπ/
3
200
kπ
=10
/
3. The center of mass of the given solid is (10
/
3
,
0
,
0).
57.
The equation is
z
=
√
9
−
r
2
and the density is
ρ
=
k/r
2
. When
z
=2,
r
=
√
5.
I
z
=
s
2
π
0
s
√
5
0
s
√
9
−
r
2
2
r
2
(
k/r
2
)
rdzdrdθ
=
k
s
2
π
0
s
√
5
0
rz



√
9
−
r
2
2
dr dθ
=
k
s
2
π
0
s
√
5
0
(
r

9
−
r
2
−
2
r
)
dr dθ
=
k
s
2
π
0
N
−
1
3
(9
−
r
2
)
3
/
2
−
r
2
o




√
5
0
dθ
=
k
s
2
π
0
4
3
dθ
=
8
3
πk
58.
The equation is
z
=
r
and the density is
ρ
=
kr
.
I
x
=
s
2
π
0
s
1
0
s
1
r
(
y
2
+
z
2
)(
kr
)
rdzdrdθ
=
k
s
2
π
0
s
1
0
s
1
r
(
r
4
sin
2
θ
+
r
2
z
2
)
dz dr dθ
=
k
s
2
π
0
s
1
0
N
(
r
4
sin
2
θ
)
z
+
1
3
r
2
z
3
o




1
r
dr dθ
=
k
s
2
π
0
s
1
0

r
4
sin
2
θ
+
1
3
r
2
−
r
5
sin
2
θ
−
1
3
r
5

dr dθ
=
k
s
2
π
0

1
5
r
5
sin
2
θ
+
1
9
r
3
−
1
6
r
6
sin
2
θ
−
1
18
r
6





1
0
dθ
=
k
s
2
π
0

1
30
sin
2
θ
+
1
18

dθ
=
k

1
60
θ
−
1
120
sin 2
θ
+
1
18
θ





2
π
0
=
13
90
πk
59. (a)
x
=(2
/
3) sin(
π/
2) cos(
π/
6) =
√
3
/
3;
y
=(2
/
3) sin(
π/
2) sin(
π/
6) = 1
/
3;
z
=(2
/
3) cos(
π/
2) = 0; (
√
3
/
3
,
1
/
3
,
0)
(b)
With
x
=
√
3
/
3 and
y
=1
/
3wehave
r
2
=4
/
9 and tan
θ
=
√
3
/
3. The point is (2
/
3
,π/
6
,
0).
60. (a)
x
= 5 sin(5
π/
4) cos(2
π/
3)=5
√
2
/
4;
y
= 5 sin(5
π/
4) sin(2
π/
3) =
−
5
√
6
/
4;
z
= 5 cos(5
π/
4) =
−
5
√
2
/
2; (5
√
2
/
4
,
−
5
√
6
/
4
,
−
5
√
2
/
2)
464

465. Exercises 9.15
(b)
With
x
=5
√
2
/
4 and
y
=
−
5
√
6
/
4wehave
r
2
=25
/
2 and tan
θ
=
−
√
3.
The point is (5
/
√
2
,
2
π/
3
,
−
5
√
2
/
2).
61. (a)
x
= 8sin(
π/
4) cos(3
π/
4) =
−
4;
y
= 8sin(
π/
4) sin(3
π/
4) = 4;
z
= 8cos(
π/
4) = 4
√
2;
(
−
4
,
4
,
4
√
2)
(b)
With
x
=
−
4 and
y
= 4 we have
r
2
= 32 and tan
θ
=
−
1. The point is (4
√
2
,
3
π/
4
,
4
√
2).
62. (a)
x
=(1
/
3) sin(5
π/
3) cos(
π/
6) =
−
1
/
4;
y
=(1
/
3) sin(5
π/
3) sin(
π/
6) =
−
√
3
/
12;
z
=(1
/
3) cos(5
π/
3) = 1
/
6; (
−
1
/
4
,
−
√
3
/
12
,
1
/
6)
(b)
With
x
=
−
1
/
4 and
y
=
−
√
3
/
12 we have
r
2
=1
/
12 and tan
θ
=
√
3
/
3.
The point is (1
/
2
√
3
,π/
6
,
1
/
6).
63.
With
x
=
−
5,
y
=
−
5, and
z
= 0, we have
ρ
2
= 50, tan
θ
= 1, and cos
φ
= 0. The point is (5
√
2
,π/
2
,
5
π/
4).
64.
With
x
=1,
y
=
−
√
3 , and
z
= 1, we have
ρ
2
= 5, tan
θ
=
−
√
3 , and cos
φ
=1
/
√
5.
The point is (
√
5
,
cos
−
1
1
/
√
5
,
−
π/
3).
65.
With
x
=
√
3
/
2,
y
=1
/
2, and
z
=1,wehave
ρ
2
= 2, tan
θ
=1
/
√
3 , and cos
φ
=1
/
√
2.
The point is (
√
2
,π/
4
,π/
6).
66.
With
x
=
−
√
3
/
2,
y
= 0, and
z
=
−
1
/
2, we have
ρ
2
= 1, tan
θ
= 0, and cos
φ
=
−
1
/
2.
The point is (1
,
2
π/
3
,
0).
67.
ρ
=8
68.
ρ
2
=4
ρ
cos
φ
;
ρ
= 4 cos
φ
69.
4
z
2
=3
x
2
+3
y
2
+3
z
2
;4
ρ
2
cos
2
φ
=3
ρ
2
; cos
φ
=
±
√
3
/
2;
φ
=
π/
6, 5
π/
6
70.
−
x
2
−
y
2
−
z
2
=1
−
2
z
2
;
−
ρ
2
=1
−
2
ρ
2
cos
2
φ
;
ρ
2
(2 cos
2
φ
−
1) = 1
71.
x
2
+
y
2
+
z
2
= 100
72.
cos
φ
=1
/
2;
ρ
2
cos
2
φ
=
ρ
2
/
4; 4
z
2
=
x
2
+
y
2
+
z
2
;
x
2
+
y
2
=3
z
2
73.
ρ
cos
φ
=2;
z
=2
74.
ρ
(1
−
cos
2
φ
) = cos
φ
;
ρ
2
−
ρ
2
cos
2
φ
=
ρ
cos
φ
;
x
2
+
y
2
+
z
2
−
z
2
=
z
;
z
=
x
2
+
y
2
75.
The equations are
φ
=
π/
4 and
ρ
=3.
V
=
s
2
π
0
s
π/
4
0
s
3
0
ρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π/
4
0
1
3
ρ
3
sin
φ




3
0
dφ dθ
=
s
2
π
0
s
π/
4
0
9 sin
φdφdθ
=
s
2
π
0
−
9 cos
φ



π/
4
0
dθ
=
−
9
s
2
π
0
n
√
2
2
−
1
g
dθ
=9
π
(2
−
√
2)
76.
The equations are
ρ
=2,
θ
=
π/
4, and
θ
=
π/
3.
s
π/
3
π/
4
s
π/
2
0
s
2
0
ρ
2
sin
φdρdφdθ
=
s
π/
3
π/
4
s
π/
2
0
1
3
ρ
3
sin
φ




2
0
dφ dθ
=
s
π/
3
π/
4
s
π/
2
0
8
3
sin
φdφdθ
=
8
3
s
π/
3
π/
4
−
cos
φ



π/
2
0
dθ
=
8
3
s
π/
3
π/
4
(0+1)
dθ
=
2
π
9
465

466. Exercises 9.15
77.
Using Problem 69, the equations are
φ
=
π/
6,
θ
=
π/
2,
θ
= 0, and
ρ
cos
φ
=2.
V
=
s
π/
2
0
s
π/
6
0
s
2 sec
φ
0
ρ
2
sin
φdρdφdθ
=
s
π/
2
0
s
π/
6
0
1
3
ρ
3
sin
φ




2 sec
φ
0
dφ dθ
=
8
3
s
π/
2
0
s
π/
6
0
sec
3
φ
sin
φdφdθ
=
8
3
s
π/
2
0
s
π/
6
0
sec
2
φ
tan
φdφdθ
=
8
3
s
π/
2
0
1
2
tan
2
φ




π/
6
0
dθ
=
4
3
s
π/
2
0
1
3
dθ
=
2
9
π
78.
The equations are
ρ
= 1 and
φ
=
π/
4. We find the volume above the
xy
-plane and double.
V
=2
s
2
π
0
s
π/
2
π/
4
s
1
0
ρ
2
sin
φdρdφdθ
=2
s
2
π
0
s
π/
2
π/
4
1
3
ρ
3
sin
φ




1
0
dφ dθ
=
2
3
s
2
π
0
s
π/
2
π/
4
sin
φdφdθ
=
2
3
s
2
π
0
−
cos
φ



π/
2
π/
4
dθ
=
2
3
s
2
π
0
√
2
2
dθ
=
2
π
√
2
3
79.
By symmetry, ¯
x
=¯
y
= 0. The equations are
φ
=
π/
4 and
ρ
= 2 cos
φ
.
m
=
s
2
π
0
s
π/
4
0
s
2 cos
φ
0
ρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π/
4
0
1
3
ρ
3
sin
φ




2 cos
φ
0
dφ dθ
=
8
3
s
2
π
0
s
π/
4
0
sin
φ
cos
3
φdφdθ
=
8
3
s
2
π
0
−
1
4
cos
4
φ




π/
4
0
dθ
=
−
2
3
s
2
π
0

1
4
−
1

dθ
=
π
M
xy
=
s
2
π
0
s
π/
4
0
s
2 cos
φ
0
zρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π/
4
0
s
2 cos
φ
0
ρ
3
sin
φ
cos
φdρdφdθ
=
s
2
π
0
s
π/
4
0
1
4
ρ
4
sin
φ
cos
φ




2 cos
φ
0
dφ dθ
=4
s
2
π
0
s
π/
4
0
cos
5
φ
sin
φdφdθ
=4
s
2
π
0
−
1
6
cos
6
φ




π/
4
0
dθ
=
−
2
3
s
2
π
0

1
8
−
1

dθ
=
7
6
π
¯
z
=
M
xy
/m
=
7
π/
6
π
=7
/
6. The centroid is (0
,
0
,
7
/
6).
80.
We are given density =
kz
. By symmetry, ¯
x
=¯
y
= 0. The equation is
ρ
=1.
m
=
s
2
π
0
s
π/
2
0
s
1
0
kzρ
2
sin
φdρdφdθ
=
k
s
2
π
0
s
π/
2
0
s
1
0
ρ
3
sin
φ
cos
φdρdφdθ
=
k
s
2
π
0
s
π/
2
0
1
4
ρ
4
sin
φ
cos
φ




1
0
dφ dθ
=
1
4
k
s
2
π
0
s
π/
2
0
sin
φ
cos
φdφdθ
=
1
4
k
s
2
π
0
1
2
sin
2
φ




π/
2
0
dθ
=
1
8
k
s
2
π
0
dθ
=
kπ
4
M
xy
=
s
2
π
0
s
π/
2
0
s
1
0
kz
2
ρ
2
sin
φdρdφdθ
=
k
s
2
π
0
s
π/
2
0
s
1
0
ρ
4
cos
2
φ
sin
φdρdφdθ
=
k
s
2
π
0
s
π/
2
0
1
5
ρ
5
cos
2
φ
sin
φ




1
0
dφ dθ
=
1
5
k
s
2
π
0
s
π/
2
0
cos
2
φ
sin
φdφdθ
=
1
5
k
s
2
π
0
−
1
3
cos
3
φ




π/
2
0
dθ
=
−
1
15
k
s
2
π
0
(0
−
1)
dθ
=
2
15
kπ
466

467. Exercises 9.16
¯
z
=
M
xy
/m
=
2
kπ/
15
kπ/
4
=8
/
15. The center of mass is (0
,
0
,
8
/
15).
81.
We are given density =
k/ρ
.
m
=
s
2
π
0
s
cos
−
1
4
/
5
0
s
5
4 sec
φ
k
ρ
ρ
2
sin
φdρdφdθ
=
k
s
2
π
0
s
cos
−
1
4
/
5
0
1
2
ρ
2
sin
φ




5
4 sec
φ
dφ dθ
=
1
2
k
s
2
π
0
s
cos
−
1
4
/
5
0
(25 sin
φ
−
16 tan
φ
sec
φ
)
dφ dθ
=
1
2
k
s
2
π
0
(
−
25 cos
φ
−
16 sec
φ
)



cos
−
1
4
/
5
0
dθ
=
1
2
k
s
2
π
0
[
−
25(4
/
5)
−
16(5
/
4)
−
(
−
25
−
16)]
dθ
=
1
2
k
s
2
π
0
dθ
=
kπ
82.
We are given density =
kρ
.
I
z
=
s
2
π
0
s
π
0
s
a
0
(
x
2
+
y
2
)(
kρ
)
ρ
2
sin
φdρdφdθ
=
k
s
2
π
0
s
π
0
s
a
0
(
ρ
2
sin
2
φ
cos
2
θ
+
ρ
2
sin
2
φ
sin
2
θ
)
ρ
3
sin
φdρdφdθ
=
k
s
2
π
0
s
π
0
s
a
0
ρ
5
sin
3
φdρdφdθ
=
k
s
2
π
0
s
π
0
1
6
ρ
6
sin
3
φ




a
0
dφ dθ
=
1
6
ka
6
s
2
π
0
s
π
0
sin
3
φdφdθ
=
1
6
ka
3
s
2
π
0
s
π
0
(1
−
cos
2
φ
) sin
φdφdθ
=
1
6
ka
3
s
2
π
0

−
cos
φ
+
1
3
cos
3
φ





π
0
dθ
=
1
6
ka
3
s
2
π
0
4
3
dθ
=
4
π
9
ka
6
Exercises 9.16
1.
div
F
=
y
+
z
+
x
The Triple Integral
:
sss
D
div
F
dV
=
s
1
0
s
1
0
s
1
0
(
x
+
y
+
z
)
dx dy dz
=
s
1
0
s
1
0

1
2
x
2
+
xy
+
xz





1
0
dy dz
=
s
1
0
s
1
0

1
2
+
y
+
z

dy dz
=
s
1
0

1
2
y
+
1
2
y
2
+
yz





1
0
dz
=
s
1
0
(1 +
z
)
dz
=
1
2
(1 +
z
2
)



1
0
=2
−
1
2
=
3
2
The Surface Integral
: Let the surfaces be
S
1
in
z
=0,
S
2
in
z
=1,
S
3
in
y
=0,
S
4
in
y
=1,
S
5
in
x
=0,
and
S
6
in
x
= 1. The unit outward normal vectors are
−
k
,
k
,
−
j
,
j
,
−
i
and
i
, respectively. Then
ss
S
F
·
n
dS
=
ss
S
1
F
·
(
−
k
)
dS
1
+
ss
S
2
F
·
k
dS
2
+
ss
S
3
F
·
(
−
j
)
dS
3
+
ss
S
4
F
·
j
dS
4
+
ss
S
5
F
·
(
−
i
)
dS
5
+
ss
S
6
F
·
i
dS
6
467

468. Exercises 9.16
=
ss
S
1
(
−
xz
)
dS
1
+
ss
S
2
xz dS
2
+
ss
S
3
(
−
yz
)
dS
3
+
ss
S
4
yz dS
4
+
ss
S
5
(
−
xy
)
dS
5
+
ss
S
6
xy dS
6
=
ss
S
2
xdS
2
+
ss
S
4
zdS
4
+
ss
S
6
ydS
6
=
s
1
0
s
1
0
xdxdy
+
s
1
0
s
1
0
zdzdx
+
s
1
0
s
1
0
ydydz
=
s
1
0
1
2
dy
+
s
1
0
1
2
dx
+
s
1
0
1
2
dz
=
3
2
.
2.
div
F
=6
y
+4
z
The Triple Integral
:
sss
D
div
F
dV
=
s
1
0
s
1
−
x
0
s
1
−
x
−
y
0
(6
y
+4
z
)
dz dy dx
=
s
1
0
s
1
−
x
0
(6
yz
+2
z
2
)



1
−
x
−
y
0
dy dx
=
s
1
0
s
1
−
x
0
(
−
4
y
2
+2
y
−
2
xy
+2
x
2
−
4
x
+2)
dy dx
=
s
1
0

−
4
3
y
3
+
y
2
−
xy
2
+2
x
2
y
−
4
xy
+2
y





1
−
x
0
dx
=
s
1
0

−
5
3
x
3
+5
x
2
−
5
x
+
5
3

dx
=

−
5
12
x
4
+
5
3
x
3
−
5
2
x
2
+
5
3
x





1
0
=
5
12
The Surface Integral
: Let the surfaces be
S
1
in the plane
x
+
y
+
z
=1,
S
2
in
z
=0,
S
3
in
x
= 0, and
S
4
in
y
= 0. The unit outward normal vectors are
n
1
=(
i
+
j
+
k
)
/
√
3,
n
2
=
−
k
,
n
3
=
−
i
, and
n
4
=
−
j
, respectively.
Now on
S
1
,
dS
1
=
√
3
dA
1
,on
S
3
,
x
= 0, and on
S
4
,
y
=0,so
ss
S
F
·
n
dS
=
ss
S
1
F
·
n
1
dS
1
+
ss
S
2
F
·
(
−
k
)
dS
2
+
ss
S
3
F
·
(
−
j
)
dS
3
+
ss
S
4
F
·
(
−
i
)
dS
4
=
s
1
0
s
1
−
x
0
(6
xy
+4
y
(1
−
x
−
y
)+
xe
−
y
)
dy dx
+
s
1
0
s
1
−
x
0
(
−
xe
−
y
)
dy dx
+
ss
S
3
(
−
6
xy
)
dS
3
+
ss
S
4
(
−
4
yz
)
dS
4
=
s
1
0

xy
2
+2
y
2
−
4
3
y
3
−
xe
−
y





1
−
x
0
dx
+
s
1
0
xe
−
y



1
−
x
0
dx
+0+0
=
s
1
0
[
x
(1
−
x
)
2
+ 2(1
−
x
)
2
−
4
3
(1
−
x
)
3
−
xe
x
−
1
+
x
]
dx
+
s
1
0
(
xe
x
−
1
−
x
)
dx
=
N
1
2
x
2
−
2
3
x
3
+
1
4
x
4
−
2
3
(1
−
x
)
3
+
1
3
(1
−
x
)
4
o




1
0
=
5
12
.
468

469. Exercises 9.16
3.
div
F
=3
x
2
+3
y
2
+3
z
2
. Using spherical coordinates,
ss
S
F
·
n
dS
=
sss
D
3(
x
2
+
y
2
+
z
2
)
dV
=
s
2
π
0
s
π
0
s
a
0
3
ρ
2
ρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π
0
3
5
ρ
5
sin
φ




a
0
dφ dθ
=
3
a
5
5
s
2
π
0
s
π
0
sin
φdφdθ
=
3
a
5
5
s
2
π
0
−
cos
φ



π
0
dθ
=
6
a
5
5
s
2
π
0
dθ
=
12
πa
5
5
.
4.
div
F
=4+1+4=9.
Using the formula for the volume of a sphere,
ss
S
F
·
n
dS
=
sss
D
9
dV
=9

4
3
π
2
3

=96
π.
5.
div
F
=2(
z
−
1). Using cylindrical coordinates,
ss
S
F
·
n
dS
=
sss
D
2(
z
−
1)
V
=
s
2
π
0
s
4
0
s
5
1
2(
z
−
1)
dz r dr dθ
=
s
2
π
0
s
4
0
(
z
−
1)
2



5
1
rdrdθ
=
s
2
π
0
s
4
0
16
rdrdθ
=
s
2
π
0
8
r
2



4
0
dθ
= 128
s
2
π
0
dθ
= 256
π.
6.
div
F
=2
x
+2
z
+12
z
2
.
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
s
3
0
s
2
0
s
1
0
(2
x
+2
z
+12
z
2
)
dx dy dz
=
s
3
0
s
2
0
(
x
2
+2
xz
+12
xz
2
)



1
0
dy dz
=
s
3
0
s
2
0
(1+2
z
+12
z
2
)
dy dz
=
s
3
0
2(1+2
z
+12
z
2
)
dz
=(2
z
+2
z
2
+8
z
3
)



3
0
= 240
7.
div
F
=3
z
2
. Using cylindrical coordinates,
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
s
2
π
0
s
√
3
0
s
√
4
−
r
2
0
3
z
2
rdzdrdθ
=
s
2
π
0
s
√
3
0
rz
3



√
4
−
r
2
0
dr dθ
=
s
2
π
0
s
√
3
0
r
(4
−
r
2
)
3
/
2
dr dθ
=
s
2
π
0
−
1
5
(4
−
r
2
)
5
/
2




√
3
0
dθ
=
s
2
π
0
−
1
5
(1
−
32)
dθ
=
s
2
π
0
31
5
dθ
=
62
π
5
.
8.
div
F
=2
x
.
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
s
3
0
s
9
x
2
s
9
−
y
0
2
xdzdydx
=
s
3
0
s
9
x
2
2
x
(9
−
y
)
dy dx
=
s
3
0
−
x
(9
−
y
)
2



9
x
2
dx
=
s
3
0
x
(9
−
x
)
2
dx
=
s
3
0
(
x
3
−
18
x
2
+81
x
)
dx
=

1
4
x
4
−
6
x
3
+
81
2
x
2





3
0
=
891
4
469

470. Exercises 9.16
9.
div
F
=
1
x
2
+
y
2
+
z
2
. Using spherical coordinates,
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
s
2
π
0
s
π
0
s
b
a
1
ρ
2
ρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π
0
(
b
−
a
) sin
φdφdθ
=(
b
−
a
)
s
2
π
0
−
cos
φ



π
0
dθ
=(
b
−
a
)
s
2
π
0
2
dθ
=4
π
(
b
−
a
)
.
10.
Since div
F
=0,
ss
S
F
·
n
dS
=
sss
D
0
dV
=0.
11.
div
F
=2
z
+10
y
−
2
z
=10
y
.
ss
S
F
·
n
dS
=
sss
D
10
ydV
=
s
2
0
s
2
−
x
2
/
2
0
s
4
−
z
z
10
ydydzdx
=
s
2
0
s
2
−
x
2
/
2
0
5
y
2



4
−
z
z
dz dx
=
s
2
0
s
2
−
x
2
/
2
0
(80
−
40
z
)
dz dx
=
s
2
0
(80
z
−
20
z
2
)



2
−
x
2
/
2
0
dx
=
s
2
0
(80
−
5
x
4
)
dx
=(80
x
−
x
5
)



2
0
= 128
12.
div
F
=30
xy
.
ss
S
F
·
n
dS
=
sss
D
30
xy dV
=
s
2
0
s
2
−
x
0
s
3
x
+
y
30
xy dz dy dx
=
s
2
0
s
2
−
x
0
30
xyz



3
x
+
y
dy dx
=
s
2
0
s
2
−
x
0
(90
xy
−
30
x
2
y
−
30
xy
2
)
dy dx
=
s
2
0
(45
xy
2
−
15
x
2
y
2
−
10
xy
3
)



2
−
x
0
dx
=
s
2
0
(
−
5
x
4
+45
x
3
−
120
x
2
+ 100
x
)
dx
=

−
x
5
+
45
4
x
4
−
40
x
3
+50
x
2





2
0
=28
13.
div
F
=6
xy
2
+1
−
6
xy
2
= 1. Using cylindrical coordinates,
ss
S
F
·
n
dS
=
sss
D
dV
=
s
π
0
s
2 sin
θ
0
s
2
r
sin
θ
r
2
dz r dr dθ
=
s
π
0
s
2 sin
θ
0
(2
r
sin
θ
−
r
2
)
rdrdθ
=
s
π
0

2
3
r
3
sin
θ
−
1
4
r
4





2 sin
θ
0
dθ
=
s
π
0

16
3
sin
4
θ
−
4 sin
4
θ

dθ
=
4
3
s
π
0
sin
4
θdθ
=
4
3

3
8
θ
−
1
4
sin 2
θ
+
1
32
sin 4
θ





π
0
=
π
2
14.
div
F
=
y
2
+
x
2
. Using spherical coordinates, we have
x
2
+
y
2
=
ρ
2
sin
2
φ
and
z
=
ρ
cos
φ
or
ρ
=
z
sec
φ
. Then
ss
S
F
·
n
dS
=
sss
D
(
x
2
+
y
2
)
dS
=
s
2
π
0
s
π/
4
0
s
4 sec
φ
2 sec
φ
ρ
2
sin
2
φρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π/
4
0
1
5
ρ
5
sin
3
φ




4 sec
φ
2 sec
φ
dφ dθ
=
s
2
π
0
s
π/
4
0
992
5
sec
5
φ
sin
3
φdφdθ
=
s
2
π
0
s
π/
4
0
992
5
tan
3
φ
sec
2
φdφdθ
=
992
5
s
2
π
0
1
4
tan
4
φ




π/
4
0
dθ
=
992
5
s
2
π
0
1
4
dθ
=
496
π
5
.
470

471. Exercises 9.16
15. (a)
div
E
=
q
n
−
2
x
2
+
y
2
+
z
2
(
x
2
+
y
2
+
z
2
)
5
/
2
+
x
2
−
2
y
2
+
z
2
(
x
2
+
y
2
+
z
2
)
5
/
2
+
x
2
+
y
2
−
2
z
2
(
x
2
+
y
2
+
z
2
)
5
/
2
/
=0
ss
S
∪
S
a
(
E
·
n
)
dS
=
sss
D
div
E
dV
=
sss
D
0
dV
=0
(b)
From
(a)
,
ss
S
(
E
·
n
)
dS
+
ss
S
a
(
E
·
n
)
dS
= 0 and
ss
S
(
E
·
n
)
dS
=
−
ss
S
a
(
E
·
n
)
dS
.On
S
a
,
|
r
|
=
a
,
n
=
−
(
x
i
+
y
j
+
z
k
)
/a
=
−
r
/a
and
E
·
n
=(
q
r
/a
3
)
·
(
−
r
/a
)=
−
qa
2
/a
4
=
−
q/a
2
.Thus
ss
S
(
E
·
n
)
dS
=
−
ss
S
a
e
−
q
a
2
:
dS
=
q
a
2
ss
S
a
dS
=
q
a
2
×
(area of
S
a
)=
q
a
2
(4
πa
2
)=4
πq.
16. (a)
By Gauss’ Law
ss
(
E
·
n
)
dS
=
sss
D
4
πρ dV
, and by the Divergence Theorem
ss
S
(
E
·
n
)
dS
=
sss
D
div
E
dV
.Thus
sss
D
4
πρ dV
=
sss
D
div
E
dV
and
sss
D
(4
πρ
−
div
E
)
dV
=0.
Since this holds for all regions
D
,4
πρ
−
div
E
= 0 and div
E
=4
πρ
.
(b)
Since
E
is irrotational,
E
=
∇
φ
and
∇
2
φ
=
∇·∇
φ
=
∇
E
= div
E
=4
πρ
.
17.
Since div
a
= 0, by the Divergence Theorem
ss
S
(
a
·
n
)
dS
=
sss
D
div
a
dV
=
sss
D
0
dV
=0
.
18.
By the Divergence Theorem and Problem 30 in Section 9
.
7,
ss
S
(curl
F
·
n
)
dS
=
sss
D
div (curl
F
)
dV
=
sss
D
0
dV
=0
.
19.
By the Divergence Theorem and Problem 27 in Section 9
.
7,
ss
S
(
f
∇
g
)
·
n
dS
=
sss
D
div (
f
∇
g
)
dV
=
sss
D
∇·
(
f
∇
g
)
dV
=
sss
D
[
f
(
∇·∇
g
)+
∇
g
·∇
f
]
dV
=
sss
D
(
f
∇
2
g
+
∇
g
·∇
f
)
dV.
20.
By the Divergence Theorem and Problems 25 and 27 in Section 8
.
7,
ss
S
(
f
∇
g
−
g
∇
f
)
·
n
dS
=
sss
D
div (
f
∇
g
−
g
∇
f
)
dV
=
sss
D
∇·
(
f
∇
g
−
g
∇
f
)
dV
=
sss
D
[
f
(
∇·∇
g
)+
∇
g
·∇
f
−
g
(
∇·∇
f
)
−∇
f
·∇
g
]
dV
=
sss
D
(
f
∇
2
g
−
g
∇
2
f
)
dV.
21.
If
G
(
x, y, z
) is a vector valued function then we define surface integrals and triple integrals of
G
component-wise.
In this case, if
a
is a constant vector it is easily shown that
ss
S
a
·
G
dS
=
a
·
ss
S
G
dS
and
sss
D
a
·
G
dV
=
a
·
sss
D
G
dV.
Now let
F
=
f
a
. Then
ss
S
F
·
n
dS
=
ss
S
(
f
a
)
·
n
dS
=
ss
S
a
·
(
f
n
)
dS
and, using Problem 27 in Section 8
.
7 and the fact that
∇·
a
= 0, we have
sss
D
div
F
dV
=
sss
D
∇·
(
f
a
)
dV
=
sss
D
[
f
(
∇·
a
)+
a
·∇
f
]
dV
=
sss
D
a
·∇
fdV.
471

472. 1 2
1
2
v=0
u=2
v=u
S
u
v
3 6
-4
-2
2
y=14-3x
y=x/2
y=-2x/3
x
y
-2 2 4
3
6
v=1
u=4
v=5
u=-1
S
u
v
-2 3
3
6
x+2y=1
x+2y=5
x-y=-1
x-y=4
x
y
Exercises 9.16
By the Divergence Theorem,
ss
S
a
·
(
f
n
)
dS
=
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
sss
D
a
·∇
fdV
and
a
·

ss
S
f
n
dS

=
a
·

sss
D
∇
fdV

or
a
·

ss
S
f
n
dS
−
sss
D
∇
fdV

=0
.
Since
a
is arbitrary,
ss
S
f
n
dS
−
sss
D
∇
fdV
= 0 and
ss
S
f
n
dS
=
sss
D
∇
fdV.
22. B
+
W
=
−
ss
S
p
n
dS
+
m
g
=
m
g
−
sss
D
∇
pdV
=
m
g
−
sss
D
ρ
g
dV
=
m
g
−

sss
D
ρdV

g
=
m
g
−
m
g
=
0
Exercises 9.17
1.
T
:(0
,
0)
→
(0
,
0); (0
,
2)
→
(
−
2
,
8); (4
,
0)
→
(16
,
20); (4
,
2)
→
(14
,
28)
2.
Writing
x
2
=
v
−
u
and
y
=
v
+
u
and solving for
u
and
v
, we obtain
u
=(
y
−
x
2
)
/
2 and
v
=(
x
2
+
y
)
/
2.
Then the images under
T
−
1
are (1
,
1)
→
(0
,
1); (1
,
3)
→
(1
,
2); (
√
2
,
2)
→
(0
,
2).
3.
The
uv
-corner points (0
,
0), (2
,
0), (2
,
2) corre-
spond to
xy
-points (0
,
0), (4
,
2), (6
,
−
4).
v
=0:
x
=2
u
,
y
=
u
=
⇒
y
=
x/
2
u
=2:
x
=4+
v
,
y
=2
−
3
v
=
⇒
y
=2
−
3(
x
−
4) =
−
3
x
+14
v
=
u
:
x
=3
u
,
y
=
−
2
u
=
⇒
y
=
−
2
x/
3
4.
Solving for
x
and
y
we see that the transformation
is
x
=2
u/
3+
v/
3,
y
=
−
u/
3+
v/
3. The
uv
-corner
points (
−
1
,
1), (4
,
1), (4
,
5), (
−
1
,
5) correspond to
the
xy
-points (
−
1
/
3
,
2
/
3),
(3
,
−
1), (13
/
3
,
1
/
3), (1
,
2).
v
=1:
x
+2
y
=1;
v
=5:
x
+2
y
=5;
u
=
−
1:
x
−
y
=
−
1;
u
=4:
x
−
y
=4
472

473. -4 -2 2
2
v=0
u=1
v=2
u=0
S
u
v
-4 -2 2
2
y=0
x=1-y
x=y /4-4
2
2
x
y
2 4
2
4
v=1
u=2
v=2
u=1
S
u
v
2 4
2
4
y=1
y=4
y=x /4
y=x
2
2
x
y
v=0
u=1
v=1
u=0
S
u
v
y=0
y=1-x
x=0
x
y
Exercises 9.17
5.
The
uv
-corner points (0
,
0),
(1
,
0), (1
,
2), (0
,
2) correspond
to the
xy
-points (0
,
0), (1
,
0),
(
−
3
,
2), (
−
4
,
0).
v
=0:
x
=
u
2
,
y
=0 =
⇒
y
= 0 and 0
≤
x
≤
1
u
=1:
x
=1
−
v
2
,
y
=
v
=
⇒
x
=1
−
y
2
v
=2:
x
=
u
2
−
4,
y
=2
u
=
⇒
x
=
y
2
/
4
−
4
u
=0:
x
=
−
v
2
,
y
=0 =
⇒
y
= 0 and
−
4
≤
x
≤
0
6.
The
uv
-corner points (1
,
1), (2
,
1), (2
,
2), (1
,
2)
correspond to the
xy
-points (1
,
1), (2
,
1), (4
,
4), (2
,
4).
v
=1:
x
=
u
,
y
=1 =
⇒
y
=1,1
≤
x
≤
2
u
=2:
x
=2
v
,
y
=
v
2
=
⇒
y
=
x
2
/
4
v
=2:
x
=2
u
,
y
=4 =
⇒
y
=4,2
≤
x
≤
4
u
=1:
x
=
v
,
y
=
v
2
=
⇒
y
=
x
2
7.
∂
(
x, y
)
∂
(
u, v
)
=




−
ve
−
u
e
−
u
ve
u
e
u




=
−
2
v
8.
∂
(
x, y
)
∂
(
u, v
)
=




3
e
3
u
sin
ve
3
u
cos
v
3
e
3
u
cos
v
−
e
3
u
sin
v




=
−
3
e
6
u
9.
∂
(
u, v
)
∂
(
x, y
)
=




−
2
y/x
3
1
/x
2
−
y
2
/x
2
2
y/x




=
−
3
y
2
x
4
=
−
3
e
y
x
2
:
2
=
−
3
u
2
;
∂
(
x, y
)
∂
(
u, v
)
=
1
−
3
u
2
=
−
1
3
u
2
10.
∂
(
u, v
)
∂
(
x, y
)
=








2(
y
2
−
x
2
)
(
x
2
+
y
2
)
2
−
4
xy
(
x
2
+
y
2
)
2
4
xy
(
x
2
+
y
2
)
2
2(
y
2
−
x
2
)
(
x
2
+
y
2
)
2








=
4
(
x
2
+
y
2
)
2
From
u
=2
x/
(
x
2
+
y
2
) and
v
=
−
2
y
(
x
2
+
y
2
) we obtain
u
2
+
v
2
=4
/
(
x
2
+
y
2
). Then
x
2
+
y
2
=4
/
(
u
2
+
v
2
) and
∂
(
x, y
)
/∂
(
u, v
)=(
x
2
+
y
2
)
2
/
4=4
/
(
u
2
+
v
2
)
2
.
11. (a)
The
uv
-corner points (0
,
0), (1
,
0), (1
,
1), (0
,
1) corre-
spond to the
xy
-points (0
,
0), (1
,
0), (0
,
1), (0
,
0).
v
=0:
x
=
u
,
y
=0 =
⇒
y
=0,0
≤
x
≤
1
u
=1:
x
=1
−
v
,
y
=
v
=
⇒
y
=1
−
x
v
=1:
x
=0,
y
=
u
=
⇒
x
=0,0
≤
y
≤
1
u
=0:
x
=0,
y
=0
(b)
Since the segment
u
=0,0
≤
v
≤
1 in the
uv
-plane maps to the origin in the
xy
-plane, the transformation
is not one-to-one.
12.
∂
(
x, y
)
∂
(
u, v
)
=




1
−
vv
−
uu




=
u
. The transformation is 0 when
u
is 0, for 0
≤
v
≤
1.
473

474. -6 -3 3 6
-3
3
R1
R2
R3
R4
R
x
y
-3 3
-2
2
S
u
v
3
-2
2
R1
R2
R3
R4
R
x
y
2 4
3
6
S
u
v
2
R1
R2
R3
R4
R
x
y
2
S
u
v
3
3
R1
R2
R3
R4
R
x
y
2
2
S
u
v
Exercises 9.17
13.
R
1:
x
+
y
=
−
1=
⇒
v
=
−
1
R
2:
x
−
2
y
=6 =
⇒
u
=6
R
3:
x
+
y
=3 =
⇒
v
=3
R
4:
x
−
2
y
=
−
6=
⇒
u
=
−
6
∂
(
u, v
)
∂
(
x, y
)
=




1
−
2
11




=3 =
⇒
∂
(
x, y
)
∂
(
u, v
)
=
1
3
ss
R
(
x
+
y
)
dA
=
ss
S
v

1
3

dA
N
=
1
3
s
3
−
1
s
6
−
6
vdudv
=
1
3
(12)
s
3
−
1
vdv
=4

1
2

v
2




3
−
1
=16
14.
R
1:
y
=
−
3
x
+3 =
⇒
v
=3
R
2:
y
=
x
−
π
=
⇒
u
=
π
R
3:
y
=
−
3
x
+6 =
⇒
v
=6
R
4:
y
=
x
=
⇒
u
=0
∂
(
u, v
)
∂
(
x, y
)
=




1
−
1
31




=4 =
⇒
∂
(
x, y
)
∂
(
u, v
)
=
1
4
ss
R
cos
1
2
(
x
−
y
)
3
x
+
y
dA
=
ss
S
cos
u/
2
v

1
4

dA
N
=
1
4
s
6
3
s
π
0
cos
u/
2
v
du dv
=
1
4
s
6
3
2 sin
u/
2
v




π
0
dv
=
1
2
s
6
3
dv
v
=
1
2
ln
v




6
3
=
1
2
ln 2
15.
R
1:
y
=
x
2
=
⇒
u
=1
R
2:
x
=
y
2
=
⇒
v
=1
R
3:
y
=
1
2
x
2
=
⇒
u
=2
R
4:
x
=
1
2
y
2
=
⇒
v
=2
∂
(
u, v
)
∂
(
x, y
)
=





2
x/y
−
x
2
/y
2
−
y
2
/x
2
2
y/x





=3 =
⇒
∂
(
x, y
)
∂
(
u, v
)
=
1
3
ss
R
y
2
x
dA
=
ss
S
v

1
3

dA
N
=
1
3
s
2
1
s
2
1
vdudv
=
1
3
s
2
1
vdv
=
1
6
v
2




2
1
=
1
2
16.
R
1:
x
2
+
y
2
=2
y
=
⇒
v
=1
R
2:
x
2
+
y
2
=2
x
=
⇒
u
=1
R
3:
x
2
+
y
2
=6
y
=
⇒
v
=1
/
3
R
4:
x
2
+
y
2
=4
x
=
⇒
u
=1
/
2
∂
(
u, v
)
∂
(
x, y
)
=








2(
y
2
−
x
2
)
(
x
2
+
y
2
)
2
−
4
xy
(
x
2
+
y
2
)
2
−
4
xy
(
x
2
+
y
2
)
2
2(
x
2
−
y
2
)
(
x
2
+
y
2
)
2








=
−
4
(
x
2
+
y
2
)
2
474

475. a
b
c
d
R1
R2
R3
R4
R
u
v
a
b
c
d
S
u
v
2 4
-2
2
R1
R2
R3
R4
R
x
y
5 10
-2
2
S
u
v
1
2
R1
R2
R3
R
x
y
1
2
S
u
v
Exercises 9.17
Using
u
2
+
v
2
=4
/
(
x
2
+
y
2
) we see that
∂
(
x, y
)
/∂
(
u, v
)=
−
4
/
(
u
2
+
v
2
)
2
.
ss
R
(
x
2
+
y
2
)
−
3
dA
=
ss
S

4
u
2
+
v
2

−
3




−
4
(
u
2
+
v
2
)
2




dA
N
=
1
16
s
1
1
/
3
s
1
1
/
2
(
u
2
+
v
2
)
du dv
=
115
5184
17.
R
1: 2
xy
=
c
=
⇒
v
=
c
R
2:
x
2
−
y
2
=
b
=
⇒
u
=
b
R
3: 2
xy
=
d
=
⇒
v
=
d
R
4:
x
2
−
y
2
=
a
=
⇒
u
=
a
∂
(
u, v
)
∂
(
x, y
)
=




2
x
−
2
y
2
y
2
x




=4(
x
2
+
y
2
)
=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
1
4(
x
2
+
y
2
)
ss
R
(
x
2
+
y
2
)
dA
=
ss
S
(
x
2
+
y
2
)
1
4(
x
2
+
y
2
)
dA
N
=
1
4
s
d
c
s
b
a
du dv
=
1
4
(
b
−
a
)(
d
−
c
)
18.
R
1:
xy
=
−
2=
⇒
v
=
−
2
R
2:
x
2
−
y
2
=9 =
⇒
u
=9
R
3:
xy
=2 =
⇒
v
=2
R
4:
x
2
−
y
2
=1 =
⇒
u
=1
∂
(
u, v
)
∂
(
x, y
)
=




2
x
−
2
y
yx




=2(
x
2
+
y
2
)
=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
1
2(
x
2
+
y
2
)
ss
R
(
x
2
+
y
2
) sin
xy dA
=
ss
S
(
x
2
+
y
2
) sin
v

1
2(
x
2
+
y
2
)

dA
N
=
1
2
s
2
−
2
s
9
1
sin
vdudv
=
1
2
s
2
−
2
8sin
vdv
=0
19.
R
1:
y
=
x
2
=
⇒
v
+
u
=
v
−
u
=
⇒
u
=0
R
2:
y
=4
−
x
2
=
⇒
v
+
u
=4
−
(
v
−
u
)
=
⇒
v
+
u
=4
−
v
+
u
=
⇒
v
=2
R
3:
x
=1 =
⇒
v
−
u
=1 =
⇒
v
=1+
u
∂
(
x, y
)
∂
(
u, v
)
=






−
1
2
√
v
−
u
1
2
√
v
−
u
11






=
−
1
√
v
−
u
ss
R
x
y
+
x
2
dA
=
ss
S
√
v
−
u
2
v




−
1
√
v
−
u




dA
N
=
1
2
s
1
0
s
2
1+
u
1
v
dv du
=
1
2
s
1
0
[ln 2
−
ln(1 +
u
)]
du
=
1
2
ln 2
−
1
2
[(1 +
u
) ln(1 +
u
)
−
(1 +
u
)]



1
0
=
1
2
ln 2
−
1
2
[2 ln 2
−
2
−
(0
−
1)] =
1
2
−
1
2
ln 2
475

476. -4 -2 2
3
R1
R2
R3
R
x
y
2
2
S
u
v
2 4
2
4
R1
R2
R3
R4
R
x
y
2 4
2
4
S
u
v
-1 1
1
R1
R2
R3
R
x
y
-1 1
1
S
u
v
Exercises 9.17
20.
Solving
x
=2
u
−
4
v
,
y
=3
u
+
v
for
u
and
v
we obtain
u
=
1
14
x
+
2
7
y
,
v
=
−
3
14
x
+
1
7
y
. The
xy
-corner points (
−
4
,
1),
(0
,
0), (2
,
3) correspond to the
uv
-points (0
,
1), (0
,
0), (1
,
0).
∂
(
x, y
)
∂
(
u, v
)
=




2
−
4
31




=14
ss
R
ydA
=
ss
S
(3
u
+
v
)(14)
dA
N
=14
s
1
0
s
1
−
u
0
(3
u
+
v
)
dv du
=14
s
1
0

3
uv
+
1
2
v
2





1
−
u
0
du
=14
s
1
0

1
2
+2
u
−
5
2
u
2

du
=

7
u
+14
u
2
−
35
3
u
3





1
0
=
28
3
21.
R
1:
y
=1
/x
=
⇒
u
=1
R
2:
y
=
x
=
⇒
v
=1
R
3:
y
=4
/x
=
⇒
u
=4
R
4:
y
=4
x
=
⇒
v
=4
∂
(
u, v
)
∂
(
x, y
)
=




yx
−
y/x
2
1
/x




=
2
y
x
=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
x
2
y
8
ss
R
y
4
dA
=
ss
S
u
2
v
2

1
2
v

du dv
=
1
2
s
4
1
u
2
vdudv
=
1
2
s
4
1
1
3
u
3
v




4
1
dv
=
1
6
s
4
1
63
vdv
=
21
4
v
2




4
1
=
315
4
22.
Under the transformation
u
=
y
+
z
,
v
=
−
y
+
z
,
w
=
x
−
y
the parallelepiped
D
is mapped to the parallelepiped
E
:1
≤
u
≤
3,
−
1
≤
v
≤
1, 0
≤
w
≤
3.
∂
(
u, v, w
)
∂
(
x, y, z
)
=







011
0
−
11
1
−
10







=2 =
⇒
∂
(
x, y, z
)
∂
(
u, v, w
)
=
1
2
ss
D
(4
z
+2
x
−
2
y
)
dV
=
ss
E
(2
u
+2
v
+2
w
)
1
2
dV
N
=
1
2
s
3
0
s
1
−
1
s
3
1
(2
u
+2
v
+2
w
)
du dv dw
=
1
2
s
3
0
s
1
−
1
(
u
2
+2
uv
+2
uw
)



3
1
dv dw
=
1
2
s
3
0
s
1
−
1
(8+ 4
v
+4
w
)
dv dw
=
s
3
0
(4
v
+
v
2
+2
vw
)



1
−
1
dw
=
s
3
0
(8+ 4
w
)
dw
=(8
w
+2
w
2
)



3
0
=42
23.
We let
u
=
y
−
x
and
v
=
y
+
x
.
R
1:
y
=0 =
⇒
u
=
−
x
,
v
=
x
=
⇒
v
=
−
u
R
2:
x
+
y
=1 =
⇒
v
=1
R
3:
x
=0 =
⇒
u
=
y
,
v
=
y
=
⇒
v
=
u
∂
(
u, v
)
∂
(
x, y
)
=




−
11
11




=
−
2=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
−
1
2
476

477. -2
2
R1
R
2
R3
R
x
y
2
2
S
u
v
4
4
R1
R2
R3
R4
R
x
y
2 4
4
8
S
u
v
4
-2
2
R1 R2
R3
R4
R
x
y
4
-2
2
S
u
v
Exercises 9.17
ss
R
e
(
y
−
x
)
/
(
y
+
x
)
dA
=
ss
S
e
u/v




−
1
2




dA
N
=
1
2
s
1
0
s
v
−
v
e
u/v
du dv
=
1
2
s
1
0
ve
u/v



v
−
v
dv
=
1
2
s
1
0
v
(
e
−
e
−
1
)
dv
=
1
2
(
e
−
e
−
1
)
1
2
v
2




1
0
=
1
4
(
e
−
e
−
1
)
24.
We let
u
=
y
−
x
and
v
=
y
.
R
1:
y
=0 =
⇒
v
=0,
u
=
−
x
=
⇒
v
=0,0
≤
u
≤
2
R
2:
x
=0 =
⇒
v
=
u
R
3:
y
=
x
+2 =
⇒
u
=2
∂
(
u, v
)
∂
(
x, y
)
=




−
11
01




=
−
1=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
−
1
ss
R
e
y
2
−
2
xy
+
x
2
dA
=
ss
S
e
u
2
|−
1
|
dA
N
=
s
2
0
s
u
0
e
u
2
dv du
=
s
2
0
ue
u
2
du
=
1
2
e
u
2




2
0
=
1
2
(
e
4
−
1)
25.
Noting that
R
2,
R
3, and
R
4 have equations
y
+2
x
=8,
y
−
2
x
=0,
and
y
+2
x
= 2, we let
u
=
y/x
and
v
=
y
+2
x
.
R
1:
y
=0 =
⇒
u
=0,
v
=2
x
=
⇒
u
=0,2
≤
v
≤
8
R
2:
y
+2
x
=8=
⇒
v
=8
R
3:
y
−
2
x
=0 =
⇒
u
=2
R
4:
y
+2
x
=2 =
⇒
v
=2
∂
(
u, v
)
∂
(
x, y
)
=




−
y/x
2
1
/x
21




=
−
y
+2
x
x
2
=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
−
x
2
y
+2
x
ss
R
(6
x
+3
y
)
dA
=3
ss
S
(
y
+2
x
)




−
x
2
y
+2
x




dA
N
=3
ss
S
x
2
dA
N
From
y
=
ux
we see that
v
=
ux
+2
x
and
x
=
v/
(
u
+ 2). Then
3
ss
S
x
2
dA
N
=3
s
2
0
s
8
2
v
2
(
u
+2)
2
dv du
=
s
2
0
v
3
(
u
+2)
2




8
2
du
= 504
s
2
0
du
(
u
+2)
2
=
−
504
u
+2




2
0
= 126
.
26.
We let
u
=
x
+
y
and
v
=
x
−
y
.
R
1:
x
+
y
=1 =
⇒
u
=1
R
2:
x
−
y
=1 =
⇒
v
=1
R
3:
x
+
y
=3 =
⇒
u
=3
R
4:
x
−
y
=
−
1=
⇒
v
=
−
1
∂
(
u, v
)
∂
(
x, y
)
=




11
1
−
1




=
−
2=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
−
1
2
ss
R
(
x
+
y
)
4
e
x
−
y
dA
=
ss
S
u
4
e
v




−
1
2




dA
N
=
1
2
s
3
1
s
1
−
1
u
4
e
v
dv du
=
1
2
s
3
1
u
4
e
v



1
−
1
du
=
e
−
e
−
1
2
s
3
1
u
4
du
=
e
−
e
−
1
10
u
5




3
1
=
242(
e
−
e
−
1
)
10
=
121
5
(
e
−
e
−
1
)
27.
Let
u
=
xy
and
v
=
xy
1
.
4
. Then
xy
1
.
4
=
c
=
⇒
v
=
c
;
xy
=
b
=
⇒
u
=
b
;
xy
1
.
4
=
d
=
⇒
v
=
d
;
xy
=
a
=
⇒
u
=
a
.
∂
(
u, v
)
∂
(
x, y
)
=




yx
y
1
.
4
1
.
4
xy
0
.
4




=0
.
4
xy
1
.
4
=0
.
4
v
=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
5
2
v
477

478. Exercises 9.17
ss
R
dA
=
ss
S
5
2
v
dA
N
=
s
d
c
s
b
a
5
2
v
du dv
=
5
2
(
b
−
a
)
s
d
c
dv
v
=
5
2
(
b
−
a
)(ln
d
−
ln
c
)
28.
The image of the ellipsoid
x
2
/a
2
+
y
2
/b
2
+
z
2
/c
2
= 1 under the transformation
u
=
x/a
,
v
=
y/b
,
w
=
z/c
,is
the unit sphere
u
2
+
v
2
+
w
2
= 1. The volume of this sphere is
4
3
π
.Now
∂
(
x, y, z
)
∂
(
u, v, w
)
=







a
00
0
b
0
00
c







=
abc
and
sss
D
dV
=
sss
E
abc dV
N
=
abc
sss
E
dV
N
=
abc

4
3
π

=
4
3
πabc.
29.
The image of the ellipse is the unit circle
x
2
+
y
2
= 1. From
∂
(
x, y
)
∂
(
u, v
)
=




50
03




= 15 we obtain
ss
R

x
2
25
+
y
2
9

dA
=
ss
S
(
u
2
+
v
2
)15
dA
N
=15
s
2
π
0
s
1
0
r
2
rdrdθ
=
15
4
s
2
π
0
r
4



1
0
dθ
=
15
4
s
2
π
0
dθ
=
15
π
2
.
30.
∂
(
x, y, z
)
∂
(
ρ, φ, θ
)
=







sin
φ
cos
θρ
cos
φ
cos
θ
−
ρ
sin
φ
sin
θ
sin
φ
sin
θρ
cos
φ
sin
θρ
sin
φ
cos
θ
cos
φ
−
ρ
sin
φ
0







= cos
φ
(
ρ
2
sin
φ
cos
φ
cos
2
θ
+
ρ
2
sin
φ
cos
φ
sin
2
θ
)+
ρ
sin
φ
(
ρ
sin
2
φ
cos
2
θ
+
ρ
sin
2
φ
sin
2
θ
)
=
ρ
2
sin
φ
cos
2
φ
(cos
2
θ
+ sin
2
θ
)+
ρ
2
sin
3
φ
(cos
2
θ
+ sin
2
θ
)=
ρ
2
sin
φ
(cos
2
φ
+ sin
2
φ
)=
ρ
2
sin
φ
Chapter 9 Review Exercises
1.
True;
|
v
(
t
)
|
=
√
2
2.
True; for all
t
,
y
=4.
3.
True
4.
False; consider
r
(
t
)=
t
2
i
. In this case,
v
(
t
)=2
t
i
and
a
(
t
)=2
i
. Since
v
·
a
=4
t
, the velocity and acceleration
vectors are not orthogonal for
t

=0.
5.
False;
∇
f
is perpendicular to the level curve
f
(
x, y
)=
c
.
6.
False; consider
f
(
x, y
)=
xy
at (0
,
0).
7.
True; the value is 4
/
3.
8.
True; since 2
xy dx
−
x
2
dy
is not exact.
9.
False;
s
C
xdx
+
x
2
dy
= 0 from (
−
1
,
0) to (1
,
0) along the
x
-axis and along the semicircle
y
=
√
1
−
x
2
, but
since
xdx
+
x
2
dy
is not exact, the integral is not independent of path.
10.
True
11.
False; unless the first partial derivatives are continuous.
12.
True
478

479. Chapter 9 Review Exercises
13.
True
14.
True; since curl
F
=
0
when
F
is a conservative vector field.
15.
True
16.
True
17.
True
18.
True
19. F
=
∇
φ
=
−
x
(
x
2
+
y
2
)
−
3
/
2
i
−
y
(
x
2
+
y
2
)
−
3
/
2
j
20.
curl
F
=







ijk
∂/∂x ∂/∂y ∂/∂z
f
(
x
)
g
(
y
)
h
(
z
)







=
0
21. v
(
t
)=6
i
+
j
+2
t
k
;
a
(
t
)=2
k
. To find when the particle passes through the plane, we solve
−
6
t
+
t
+
t
2
=
−
4
or
t
2
−
5
t
+ 4 = 0. This gives
t
= 1 and
t
=4.
v
(1) = 6
i
+
j
+2
k
,
a
(1) = 2
k
;
v
(4) = 6
i
+
j
+8
k
,
a
(4) = 2
k
22.
We are given
r
(0) =
i
+2
j
+3
k
.
r
(
t
)=
s
v
(
t
)
dt
=
s
(
−
10
t
i
+(3
t
2
−
4
t
)
j
+
k
)
dt
=
−
5
t
2
i
+(
t
3
−
2
t
2
)
j
+
t
k
+
c
i
+2
j
+3
k
=
r
(0) =
c
r
(
t
)=(1
−
5
t
2
)
i
+(
t
3
−
2
t
2
+2)
j
+(
t
+3)
k
r
(2) =
−
19
i
+2
j
+5
k
23. v
(
t
)=
s
a
(
t
)
dt
=
s
(
√
2 sin
t
i
+
√
2 cos
t
j
)
dt
=
−
√
2 cos
t
i
+
√
2 sin
t
j
+
c
;
−
i
+
j
+
k
=
v
(
π/
4) =
−
i
+
j
+
c
,
c
=
k
;
v
(
t
)=
−
√
2 cos
t
i
+
√
2 sin
t
j
+
k
;
r
(
t
)=
−
√
2 sin
t
i
−
√
2 cos
t
j
+
t
k
+
b
;
i
+2
j
+(
π/
4)
k
=
r
(
π/
4) =
−
i
−
j
+(
π/
4)
k
+
b
,
b
=2
i
+3
j
;
r
(
t
)=(2
−
√
2 sin
t
)
i
+(3
−
√
2 cos
t
)
j
+
t
k
;
r
(3
π/
4) =
i
+4
j
+(3
π/
4)
k
24. v
(
t
)=
t
i
+
t
2
j
−
t
k
;
|
v
|
=
t
√
t
2
+2,
t>
0;
a
(
t
)=
i
+2
t
j
−
k
;
v
·
a
=
t
+2
t
3
+
t
=2
t
+2
t
3
;
v
×
a
=
t
2
i
+
t
2
k
,
|
v
×
a
|
=
t
2
√
2;
a
T
=
2
t
+2
t
3
t
√
t
2
+2
=
2+2
t
2
√
t
2
+2
,
a
N
=
t
2
√
2
t
√
t
2
+2
=
√
2
t
√
t
2
+2
;
κ
=
t
2
√
2
t
3
(
t
2
+2)
3
/
2
=
√
2
t
(
t
2
+2)
3
/
2
25.
26. r
N
(
t
) = sinh
t
i
+ cosh
t
j
+
k
,
r
N
(1) = sinh 1
i
+ cosh 1
j
+
k
;
|
r
N
(
t
)
|
=

sinh
2
t
+ cosh
2
t
+1=

2 cosh
2
t
=
√
2 cosh
t
;
|
r
N
(1)
|
=
√
2 cosh 1;
T
(
t
)=
1
√
2
tanh
t
i
+
1
√
2
j
+
1
√
2
sech
t
k
,
T
(1) =
1
√
2
(tanh 1
i
+
j
+ sech 1
k
);
d
T
dt
=
1
√
2
sech
2
t
i
−
1
√
2
sech
t
tanh
t
k
;
d
dt
T
(1) =
1
√
2
sech
2
1
i
−
1
√
2
sech 1 tanh 1
k
,




d
dt
T
(1)




=
sech 1
√
2

sech
2
1 + tanh
2
1=
1
√
2
sech 1;
N
(1) = sech 1
i
−
tanh 1
k
;
479

480. Chapter 9 Review Exercises
B
(1) =
T
(1)
×
N
(1) =
−
1
√
2
tanh 1
i
+
1
√
2
(tanh
2
1 + sech
2
1)
j
−
1
√
2
sech 1
k
=
1
√
2
(
−
tanh 1
i
+
j
−
sech 1
k
)
κ
=




d
dt
T
(1)




/
|
r
u
(1)
|
=
(sech 1)
/
√
2
√
2 cosh 1
=
1
2
sech
2
1
27.
∇
f
=(2
xy
−
y
2
)
i
+(
x
2
−
2
xy
)
j
;
u
=
2
√
40
i
+
6
√
40
j
=
1
√
10
(
i
+3
j
);
D
u
f
=
1
√
10
(2
xy
−
y
2
+3
x
2
−
6
xy
)=
1
√
10
(3
x
2
−
4
xy
−
y
2
)
28.
∇
F
=
2
x
x
2
+
y
2
+
z
2
i
+
2
y
x
2
+
y
2
+
z
2
j
+
2
z
x
2
+
y
2
+
z
2
k
;
u
=
−
2
3
i
+
1
3
j
+
2
3
k
;
D
u
F
=
−
4
x
+2
y
+4
z
3(
x
2
+
y
2
+
z
2
)
29.
f
x
=2
xy
4
,
f
y
=4
x
2
y
3
.
(a) u
=
i
,
D
u
(1
,
1) =
f
x
(1
,
1) = 2
(b) u
=(
i
−
j
)
/
√
2,
D
u
(1
,
1) = (2
−
4)
/
√
2=
−
2
/
√
2
(c) u
=
j
,
D
u
(1
,
1) =
f
y
(1
,
1) = 4
30. (a)
dw
dt
=
∂w
∂x
dx
dt
+
∂w
∂y
dy
dt
+
∂w
∂z
dz
dt
=
x

x
2
+
y
2
+
z
2
6 cos 2
t
+
y

x
2
+
y
2
+
z
2
(
−
8sin 2
t
)+
z

x
2
+
y
2
+
z
2
15
t
2
=
(6
x
cos 2
t
−
8
y
sin 2
t
+15
zt
2
)

x
2
+
y
2
+
z
2
(b)
∂w
∂t
=
∂w
∂x
∂x
∂t
+
∂w
∂y
∂y
∂t
+
∂w
∂z
∂z
∂t
=
x

x
2
+
y
2
+
z
2
6
r
cos
2
t
r
+
y

x
2
+
y
2
+
z
2

8
r
t
2
sin
2
r
t

+
z

x
2
+
y
2
+
z
2
15
t
2
r
3
=

6
x
r
cos
2
t
r
+
8
yr
t
2
sin
2
r
t
+15
zt
2
r
3


x
2
+
y
2
+
z
2
31.
F
(
x, y, z
) = sin
xy
−
z
;
∇
F
=
y
cos
xy
i
+
x
cos
xy
j
−
k
;
∇
F
(1
/
2
,
2
π/
3
,
√
3
/
2) =
π
3
i
+
1
4
j
−
k
.
The equation of the tangent plane is
π
3

x
−
1
2

+
1
4

y
−
2
π
3

−
n
z
−
√
3
2
g
= 0 or 4
πx
+3
y
−
12
z
=4
π
−
6
√
3.
32.
We want to find a normal to the surface that is parallel to
k
.
∇
F
=(
y
−
2)
i
+(
x
−
2
y
)
j
+2
z
k
. We need
y
−
2=0
and
x
−
2
y
= 0. The tangent plane is parallel to
z
= 2 when
y
= 2 and
x
= 4. In this case
z
2
= 5. The points
are (4
,
2
,
√
5 ) and (4
,
2
,
−
√
5).
33. (a)
V
=
s
1
0
s
2
x
x

1
−
x
2
dy dx
=
s
1
0
y

1
−
x
2



2
x
x
dx
=
s
1
0
x

1
−
x
2
dx
=
−
1
3
(1
−
x
2
)
3
/
2




1
0
=
1
3
(b)
V
=
s
1
0
s
y
y/
2

1
−
x
2
dx dy
+
s
2
1
s
1
y/
2

1
−
x
2
dx dy
480

481. Chapter 9 Review Exercises
34.
We are given
ρ
=
k
(
x
2
+
y
2
).
m
=
s
1
0
s
x
2
x
3
k
(
x
2
+
y
2
)
dy dx
=
k
s
1
0

x
2
y
+
1
3
y
3





x
2
x
3
dx
=
k
s
1
0

x
4
+
1
3
x
6
−
x
5
−
1
3
x
9

dx
=
k

1
5
x
5
+
1
21
x
7
−
1
6
x
6
−
1
30
x
10





1
0
=
k
21
M
y
=
s
1
0
s
x
2
x
3
k
(
x
3
+
xy
2
)
dy dx
=
k
s
1
0

x
3
y
+
1
3
xy
3





x
2
x
3
dx
=
k
s
1
0

x
5
+
1
3
x
7
−
x
6
−
1
3
x
10

dx
=
k

1
6
x
6
+
1
24
x
8
−
1
7
x
7
−
1
33
x
11





1
0
=
65
k
1848
M
x
=
s
1
0
s
x
2
x
3
k
(
x
2
y
+
y
3
)
dy dx
=
k
s
1
0

1
2
x
2
y
2
+
1
4
y
4





x
2
x
3
dx
=
k
s
1
0

1
2
x
6
+
1
4
x
8
−
1
2
x
8
−
1
4
x
12

dx
=
k

1
14
x
7
−
1
36
x
9
−
1
52
x
13





1
0
=
20
k
819
¯
x
=
M
y
/m
=
65
k/
1848
k/
21
=65
/
88; ¯
y
=
M
x
/m
=
20
k/
819
k/
21
=20
/
39 The center of mass is (65
/
88
,
20
/
39).
35.
I
y
=
s
1
0
s
x
2
x
3
k
(
x
4
+
x
2
y
2
)
dy dx
=
k
s
1
0

x
4
y
+
1
3
x
2
y
3





x
2
x
3
dx
=
k
s
1
0

x
6
+
1
3
x
8
−
x
7
−
1
3
x
11

dx
=
k

1
7
x
7
+
1
27
x
9
−
1
8
x
8
−
1
36
x
12





1
0
=
41
1512
k
36. (a)
Using symmetry,
V
=8
s
a
0
s
√
a
2
−
x
2
0
s
√
a
2
−
x
2
−
y
2
0
dz dy dx
=8
s
a
0
s
√
a
2
−
x
2
0

a
2
−
x
2
−
y
2
dy dx
Trig substitution
=8
s
a
0

y
2

a
2
−
x
2
−
y
2
+
a
2
−
x
2
2
sin
−
1
y
√
a
2
−
x
2
o




√
a
2
−
x
2
0
dx
=8
s
a
0
π
2
a
2
−
x
2
2
dx
=2
π

a
2
x
−
1
3
x
3





a
0
=
4
3
πa
3
(b)
Using symmetry,
V
=2
s
2
π
0
s
a
0
s
√
a
2
−
r
2
0
rdzdrdθ
=2
s
2
π
0
s
a
0
r

a
2
−
r
2
dr dθ
=2
s
2
π
0
−
1
3
(
a
2
−
r
2
)
3
/
2




a
0
dθ
=
2
3
s
2
π
0
a
3
dθ
=
4
3
πa
3
(c)
V
=
s
2
π
0
s
π
0
s
a
0
ρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π
0
1
3
ρ
3
sin
φ




a
0
dφ dθ
=
1
3
s
2
π
0
s
π
0
a
3
sin
φdφdθ
=
1
3
s
2
π
0
−
a
3
cos
φ



π
0
dθ
=
1
3
s
2
π
0
2
a
3
dθ
=
4
3
πa
3
481

482. Chapter 9 Review Exercises
37.
We use spherical coordinates.
V
=
s
2
π
0
s
π/
4
tan
−
1
1
/
3
s
3 sec
φ
0
ρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π/
4
tan
−
1
1
/
3
1
3
ρ
3
sin
φ




3 sec
φ
0
dφ dθ
=
1
3
s
2
π
0
s
π/
4
tan
−
1
1
/
3
27 sec
3
φ
sin
φdφdθ
=9
s
2
π
0
s
π/
4
tan
−
1
1
/
3
tan
φ
sec
2
φdφdθ
=9
s
2
π
0
1
2
tan
2
φ




π/
4
tan
−
1
1
/
3
dθ
=
9
2
s
2
π
0

1
−
1
9

dθ
=8
π
38.
V
=
s
2
π
0
s
π/
6
0
s
2
1
ρ
2
sin
φdρdφdθ
=
s
2
π
0
s
π/
6
0
1
3
ρ
3
sin
φ




2
1
dφ dθ
=
s
2
π
0
s
π/
6
0

8
3
sin
φ
−
1
3
sin
φ

dφ dθ
=
7
3
s
2
π
0
s
π/
6
0
sin
φdφdθ
=
7
3
s
2
π
0
−
cos
φ



π/
6
0
dθ
=
7
3
s
2
π
0
0
−
√
3
2
−
(
−
1)
,
dθ
=
7
3
n
1
−
√
3
2
g
2
π
=
7
π
3
(2
−
√
3)
39.
2
xy
+2
xy
+2
xy
=6
xy
40.







ijk
∂/∂x ∂/∂y ∂/∂z
x
2
yxy
2
2
xyz







=2
xz
i
−
2
yz
j
+(
y
2
−
x
2
)
k
41.
∂
∂x
(2
xz
)
−
∂
∂y
(2
yz
)+
∂
∂z
(
y
2
−
x
2
)=0
42.
∇
(6
xy
)=6
y
i
+6
x
j
43.
s
C
z
2
x
2
+
y
2
ds
=
s
2
π
π
4
t
2
cos
2
2
t
+ sin
2
2
t

4 sin
2
2
t
+ 4 cos
2
2
t
+4
dt
=
s
2
π
π
8
√
2
t
2
dt
=
8
√
2
3
t
3




2
π
π
=
56
√
2
π
3
3
44.
s
C
(
xy
+4
x
)
ds
=
s
0
1
[
x
(2
−
2
x
)+4
x
]
√
1+4
dx
=
√
5
s
0
1
(6
x
−
2
x
2
)
dx
=
√
5

3
x
2
−
2
3
x
3





0
1
=
−
7
√
5
3
45.
Since
P
y
=6
x
2
y
=
Q
x
, the integral is independent of path.
φ
x
=3
x
2
y
2
,
φ
=
x
3
y
2
+
g
(
y
),
φ
y
=2
x
3
y
+
g
N
(
y
)=2
x
3
y
−
3
y
2
;
g
(
y
)=
−
y
3
;
φ
=
x
3
y
2
−
y
3
;
s
(1
,
−
2)
(0
,
0)
3
x
2
y
2
dx
+(2
x
3
y
−
3
y
2
)
dy
=(
x
3
y
2
−
y
3
)



(1
,
−
2)
(0
,
0)
=12
46.
Let
x
=
a
cos
t
,
y
=
a
sin
t
,0
≤
t
≤
2
π
. Then using
dx
=
−
a
sin
tdt
,
dy
=
a
cos
tdt
,
x
2
+
y
2
=
a
2
we have
=
ˇ
C
−
ydx
+
xdy
x
2
+
y
2
=
s
2
π
0
1
a
2
[
−
a
sin
t
(
−
a
sin
t
)+
a
cos
t
(
a
cos
t
)]
dt
=
s
2
π
0
(sin
2
t
+ cos
2
t
)
dt
=
s
2
π
0
dt
=2
π.
47.
s
C
y
sin
πz dx
+
x
2
e
y
dy
+3
xyz dz
=
s
1
0
[
t
2
sin
πt
3
+
t
2
e
t
2
(2
t
)+3
tt
2
t
3
(3
t
2
)]
dt
=
s
1
0
(
t
2
sin
πt
3
+2
t
3
e
t
2
+9
t
8
)
dt
=

−
1
3
π
cos
πt
3
+
t
9





1
0
+2
s
1
0
t
3
e
t
2
dt
Integration by parts
=
2
3
π
+1+(
t
2
e
t
2
−
e
t
2
)



1
0
=
2
3
π
+2
482

483. Chapter 9 Review Exercises
48.
Parameterize
C
by
x
= cos
t
,
y
= sin
t
;0
≤
t
≤
2
π
. Then
=
ˇ
C
F
·
d
r
=
s
2
π
0
[4 sin
t
(
−
sin
tdt
)+6cos
t
(cos
t
)
dt
]=
s
2
π
0
(6 cos
2
t
−
4 sin
2
t
)
dt
=
s
2
π
0
(10 cos
2
t
−
4)
dt
=

5
t
+
5
2
sin 2
t
−
4
t





2
π
0
=2
π.
Using Green’s Theorem,
Q
x
−
P
y
=6
−
4 = 2 and
=
ˇ
C
F
·
d
r
=
ss
R
2
dA
=2(
π
·
1
2
)=2
π
.
49.
Let
r
1
=
π
2
t
i
and
r
2
=
π
2
i
+
πt
j
for 0
≤
t
≤
1. Then
d
r
1
=
π
2
i
,
d
r
2
=
π
j
,
F
1
=
0
,
F
2
=
π
2
sin
πt
i
+
πt
sin
π
2
j
=
π
2
sin
πt
i
+
πt
j
,
and
W
=
s
C
1
F
1
·
d
r
1
+
s
C
2
F
2
·
d
r
2
=
s
1
0
π
2
tdt
=
1
2
π
2
t
2




1
0
=
π
2
2
.
50.
Parameterize the line segment from (
−
1
/
2
,
1
/
2) to (
−
1
,
1) using
y
=
−
x
as
x
goes from
−
1
/
2to
−
1. Parameterize
the line segment from (
−
1
,
1) to (1
,
1) using
y
=1as
x
goes from
−
1 to 1. Parameterize the line segment from
(1
,
1) to (1
,
√
3 ) using
x
=1as
y
goes from 1 to
√
3 . Then
W
=
s
C
F
·
d
r
=
s
−
1
−
1
/
2
F
·
(
dx
i
−
dx
j
)+
s
1
−
1
F
·
(
dx
i
)+
s
√
3
1
F
·
(
dy
j
)
=
s
−
1
−
1
/
2

2
x
2
+(
−
x
)
2
−
1
x
2
+(
−
x
)
2

dx
+
s
1
−
1
2
x
2
+1
dx
+
s
√
3
1
1
1+
y
2
dy
=
s
−
1
−
1
/
2
1
2
x
2
dx
+
s
1
−
1
2
1+
x
2
dx
+
s
√
3
1
1
1+
y
2
dy
=
−
1
2
x




−
1
−
1
/
2
+ 2 tan
−
1
x



1
−
1
+ tan
−
1
y



√
3
1
=
−
1
2
+2
e
π
2
:
+
π
12
=
13
π
−
6
12
.
51.
z
x
=2
x
,
z
y
=0;
dS
=
√
1+4
x
2
dA
ss
S
z
xy
dS
=
s
3
1
s
2
1
x
2
xy

1+4
x
2
dx dy
=
s
3
1
1
y
N
1
12
(1+4
x
2
)
3
/
2
o




2
1
dy
=
1
12
s
3
1
17
3
/
2
−
5
3
/
2
y
dy
=
17
√
17
−
5
√
5
12
ln
y




3
1
=
17
√
17
−
5
√
5
12
ln 3
52. n
=
k
,
F
·
n
=3;
flux =
((
S
F
·
n
dS
=3
((
S
dS
=3
×
(area of
S
) = 3(1) = 3
53.
The surface is
g
(
x, y, z
)=
x
2
+
y
2
+
z
2
−
a
2
=0.
∇
g
=2(
x
i
+
y
j
+
z
k
)=2
r
,
n
=
r
/
|
r
|
,
F
=
c
∇
(1
/
|
r
|
)+
c
∇
(
x
2
+
y
2
+
z
2
)
−
1
/
2
=
c
−
x
i
−
y
j
−
z
k
(
x
2
+
y
2
+
z
2
)
3
/
2
=
−
c
r
/
|
r
|
3
F
·
n
=
−
r
|
r
|
3
·
r
|
r
|
=
−
c
r
·
r
|
r
|
4
=
−
c
|
r
|
2
|
r
|
4
=
−
c
|
r
|
2
=
−
c
a
2
483

484. Chapter 9 Review Exercises
flux =
ss
S
F
·
n
dS
=
−
c
a
2
ss
S
dS
=
−
c
a
2
×
(area of
S
)=
−
c
a
2
(4
πa
2
)=
−
4
πc
54.
In Problem 53,
F
is not continuous at (0
,
0
,
0) which is in any acceptable region containing the sphere.
55.
Since
F
=
c
∇
(1
/r
), div
F
=
∇·
(
c
∇
(1
/r
)) =
c
∇
2
(1
/r
)=
c
∇
2
[(
x
2
+
y
2
+
z
2
)
−
1
/
2
] = 0 by Problem 37 in Section
9
.
7. Then, by the Divergence Theorem,
flux
F
=
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
sss
D
0
dV
=0
.
56.
Parameterize
C
by
x
= 2 cos
t
,
y
= 2 sin
t
,
z
= 5, for 0
≤
t
≤
2
π
. Then
ss
S
(curl
F
·
n
)
dS
=
=
ˇ
C
F
·
d
r
=
=
ˇ
C
6
xdx
+7
zdy
+8
ydz
=
s
2
π
0
[12 cos
t
(
−
2 sin
t
) + 35(2 cos
t
)]
dt
=
s
2
π
0
(70 cos
t
−
24 sin
t
cos
t
)
dt
= (70 sin
t
−
12 sin
2
t
)



2
π
0
=0
.
57.
Identify
F
=
−
2
y
i
+3
x
j
+10
z
k
. Then curl
F
=5
k
. The curve
C
lies in the plane
z
=3,
so
n
=
k
and
dS
=
dA
. Thus,
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
)
·
n
dS
=
ss
R
5
dA
=5
×
(area of
R
) = 5(25
π
) = 125
π.
58.
Since curl
F
=
0
,
=
ˇ
C
F
·
d
r
=
ss
S
(curl
F
·
n
)
dS
=
ss
S
0
dS
=0.
59.
div
f
=1+1=1=3;
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
sss
D
3
dV
=3
×
(volume of
D
)=3
π
60.
div
F
=
x
2
+
y
2
+
z
2
. Using cylindrical coordinates,
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
sss
D
(
x
2
+
y
2
+
z
2
)
dV
=
s
2
π
0
s
1
0
s
1
0
(
r
2
+
z
2
)
rdzdrdθ
=
s
2
π
0
s
1
0

r
3
z
+
1
3
rz
3





1
0
dr dθ
=
s
2
π
0
s
1
0

r
3
+
1
3
r

dr dθ
=
s
2
π
0

1
4
r
4
+
1
6
r
2





1
0
dθ
=
s
2
π
0
5
12
dθ
=
5
π
6
.
61.
div
F
=2
x
+2(
x
+
y
)
−
2
y
=4
x
ss
S
F
·
n
dS
=
sss
D
div
F
dV
=
sss
D
4
xdV
=
s
1
0
s
1
−
x
2
0
s
2
−
z
0
4
xdydzdx
=
s
1
0
s
1
−
x
2
0
4
x
(2
−
z
)
dz dx
=
s
1
0
s
1
−
x
2
0
(8
x
−
4
xz
)
dz dx
=
s
1
0
(8
xz
−
2
xz
2
)



1
−
x
2
0
dx
=
s
1
0
[8
x
(1
−
x
2
)
−
2
x
(1
−
x
2
)
2
]
dx
=
N
−
2(1
−
x
2
)
2
+
1
3
(1
−
x
2
)
3
o




1
0
=
5
3
484

485. Chapter 9 Review Exercises
62.
For
S
1
,
n
=(
x
i
+
y
j
)
/

x
2
+
y
2
; for
S
2
,
n
2
=
−
k
and
z
= 0; and for
S
3
,
n
3
=
k
and
z
=
c
.
Then
ss
S
F
·
n
dS
=
ss
S
1
F
·
n
1
dS
1
+
ss
S
2
F
·
n
2
dS
2
+
ss
S
3
F
·
n
3
dS
3
=
ss
S
1
x
2
+
y
2

x
2
+
y
2
dS
1
+
ss
S
2
(
−
z
2
−
1)
dS
2
+
ss
S
3
(
z
2
+1)
dS
3
=
ss
S
1

x
2
+
y
2
dS
1
+
ss
S
2
(
−
1)
dS
2
+
ss
S
3
(
c
2
+1)
dS
3
=
a
ss
S
1
dS
1
−
ss
S
2
dS
2
+(
c
2
+1)
ss
S
3
dS
3
=
a
(2
πac
)
−
πa
2
+(
c
2
+1)
πa
2
=2
πa
2
c
+
πa
2
c
2
.
63.
x
=0 =
⇒
u
=0,
v
=
−
y
2
=
⇒
u
=0,
−
1
≤
v
≤
0
x
=1 =
⇒
u
=2
y
,
v
=1
−
y
2
=1
−
u
2
/
4
y
=0 =
⇒
u
=0,
v
=
x
2
=
⇒
u
=0,0
≤
v
≤
1
y
=1 =
⇒
u
=2
x
,
v
=
x
2
−
1=
u
2
/
4
−
1
∂
(
u, v
)
∂
(
x, y
)
=




2
y
2
x
2
x
−
2
y




=
−
4(
x
2
+
y
2
)=
⇒
∂
(
x, y
)
∂
(
u, v
)
=
−
1
4(
x
2
+
y
2
)
ss
R
(
x
2
+
y
2
)
3

x
2
−
y
2
dA
=
ss
S
(
x
2
+
y
2
)
3
√
v




−
1
4(
x
2
+
y
2
)




dA
N
=
1
4
s
2
0
s
1
−
u
2
/
4
u
2
/
4
−
1
v
1
/
3
dv du
=
1
4
s
2
0
3
4
v
4
/
3




1
−
u
2
/
4
u
2
/
4
−
1
du
=
3
16
s
2
0
t
(1
−
u
2
/
4)
4
/
3
−
(
u
2
/
4
−
1)
4
/
3
h
du
=
3
16
s
2
0
t
(1
−
u
2
/
4)
4
/
3
−
(1
−
u
2
/
4)
4
/
3
h
du
=0
64.
y
=
x
=
⇒
u
+
uv
=
v
+
uv
=
⇒
v
=
u
x
=2 =
⇒
u
+
uv
=2 =
⇒
v
=(2
−
u
)
/u
y
=0 =
⇒
v
+
uv
=0 =
⇒
v
=0or
u
=
−
1
(we take
v
=0)
∂
(
x, y
)
∂
(
u, v
)
=




1+
wu
v
1+
u




=1+
u
+
v
Using
x
=
u
+
uv
and
y
=
v
+
uv
we find
(
x
−
y
)
2
=(
u
+
uv
−
v
−
uv
)
2
=(
u
−
v
)
2
=
u
2
−
2
uv
+
v
2
x
+
y
=
u
+
uv
+
v
+
uv
=
u
+
v
+2
uv
(
x
−
y
)
2
+2(
x
+
y
)+1=
u
2
+2
uv
+
v
2
+2(
u
+
v
)+1=(
u
+
v
)
2
+2(
u
+
v
)+1=(
u
+
v
+1)
2
.
Then
ss
R
1

(
x
−
y
)
2
+2(
x
+
y
)+1
dA
=
ss
S
1
u
+
v
+1
(
u
+
v
+1)
dA
N
=
s
1
0
s
2
/
(1+
v
)
v
du dv
=
s
1
0

2
1+
v
−
v

dv
=
N
2ln(1+
v
)
−
1
2
v
2
o




1
0
= 2 ln 2
−
1
2
.
65.
The equations of the spheres are
x
2
+
y
2
+
z
2
=
a
2
and
x
2
+
y
2
+(
z
−
a
)
2
= 1. Subtracting these equations, we
485

486. Chapter 9 Review Exercises
obtain (
z
−
a
)
2
−
z
2
=1
−
a
2
or
−
2
az
+
a
2
=1
−
a
2
. Thus, the spheres intersect on the plane
z
=
a
−
1
/
2
a
.
The region of integration is
x
2
+
y
2
+(
a
−
1
/
2
a
)
2
=
a
2
or
r
2
=1
−
1
/
4
a
2
. The area is
A
=
a
s
2
π
0
s
√
1
−
1
/
4
a
2
0
(
a
2
−
r
2
)
−
1
/r
rdrdθ
=2
πa
[
−
(
a
2
−
r
2
)
1
/
2
]



√
1
−
1
/
4
a
2
0
=2
πa
n
a
−
N
a
2
−

1
−
1
4
a
2
lo
1
/
2
g
=2
πa

a
−
0

a
−
1
2
a

2
,
1
/
2

=
π.
66. (a)
Both states span 7 degrees of longitude and 4 degrees of latitude, but Colorado is larger because it lies
to the south of Wyoming. Lines of longitude converge as they go north, so the east-west dimensions of
Wyoming are shorter than those of Colorado.
(b)
We use the function
f
(
x, y
)=

R
2
−
x
2
−
y
2
to describe the northern
hemisphere, where
R
≈
3960 miles is the radius of the Earth. We need to
compute the surface area over a polar rectangle
P
of the form
θ
1
≤
θ
≤
θ
2
,
R
cos
φ
2
≤
r
≤
R
cos
φ
1
. We have
f
x
=
−
x

R
2
−
x
2
−
y
2
and
f
y
=
−
y

R
2
−
x
2
−
y
2
so that

1+
f
2
x
+
f
2
y
=
p
1+
x
2
+
y
2
R
2
−
x
2
−
y
2
=
R
√
R
2
−
r
2
.
Thus
A
=
ss
P

1+
f
2
x
+
f
2
y
dA
=
s
θ
2
θ
1
s
R
cos
φ
1
R
cos
φ
2
R
√
R
2
−
r
2
rdrdθ
=(
θ
2
−
θ
1
)
R

R
2
−
r
2




R
cos
φ
2
R
cos
φ
1
=(
θ
2
−
θ
1
)
R
2
(sin
φ
2
−
sin
φ
1
)
.
The ratio of Wyoming to Colorado is then
sin 45
◦
−
sin 41
◦
sin 41
◦
−
sin 37
◦
≈
0
.
941. Thus Wyoming is about 6%
smaller
than Colorado.
(c)
97,914
/
104,247
≈
0
.
939, which is close to the theoretical value of 0
.
941. (Our formula for the area says that
the area of Colorado is approximately 103,924 square miles, while the area of Wyoming is approximately
97,801 square miles.)
486

487. 10
Systems of Linear Differential Equations
Exercises 10.1
1.
Let
X
=
L
x
y
e
. Then
X

=
L
3
−
5
48
e
X
.
2.
Let
X
=
L
x
y
e
. Then
X

=
L
4
−
7
50
e
X
.
3.
Let
X
=



x
y
z



. Then
X

=



−
34
−
9
6
−
10
10 4 3



X
.
4.
Let
X
=



x
y
z



. Then
X

=



1
−
10
102
−
101



X
.
5.
Let
X
=



x
y
z



. Then
X

=



1
−
11
21
−
1
111



X
+



0
−
3
t
2
t
2



+



t
0
−
t



+



−
1
0
2



.
6.
Let
X
=



x
y
z



. Then
X

=



−
34 0
50 9
01 6



X
+



e
−
t
sin 2
t
4
e
−
t
cos 2
t
−
e
−
t



7.
dx
dt
=4
x
+2
y
+
e
t
;
dy
dt
=
−
x
+3
y
−
e
t
8.
dx
dt
=7
x
+5
y
−
9
z
−
8
e
−
2
t
;
dy
dt
=4
x
+
y
+
z
+2
e
5
t
;
dz
dt
=
−
2
y
+3
z
+
e
5
t
−
3
e
−
2
t
487

488. Exercises 10.1
9.
dx
dt
=
x
−
y
+2
z
+
e
−
t
−
3
t
;
dy
dt
=3
x
−
4
y
+
z
+2
e
−
t
+
t
;
dz
dt
=
−
2
x
+5
y
+6
z
+2
e
−
t
−
t
10.
dx
dt
=3
x
−
7
y
+ 4 sin
t
+(
t
−
4)
e
4
t
;
dy
dt
=
x
+
y
+ 8 sin
t
+(2
t
+1)
e
4
t
11.
Since
X

=
L
−
5
−
10
e
e
−
5
t
and
L
3
−
4
4
−
7
e
X
=
L
−
5
−
10
e
e
−
5
t
we see that
X

=
L
3
−
4
4
−
7
e
X
.
12.
Since
X

=
L
5 cos
t
−
5 sin
t
2 cos
t
−
4 sin
t
e
e
t
and
L
−
25
−
24
e
X
=
L
5 cos
t
−
5 sin
t
2 cos
t
−
4 sin
t
e
e
t
we see that
X

=
L
−
25
−
24
e
X
.
13.
Since
X

=
L
3
/
2
−
3
e
e
−
3
t/
2
and
L
−
11
/
4
1
−
1
e
X
=
L
3
/
2
−
3
e
e
−
3
t/
2
we see that
X

=
L
−
11
/
4
1
−
1
e
X
.
14.
Since
X

=
L
5
−
1
e
e
t
+
L
4
−
4
e
te
t
and
L
21
−
10
e
X
=
L
5
−
1
e
e
t
+
L
4
−
4
e
te
t
we see that
X

=
L
21
−
10
e
X
.
15.
Since
X

=



0
0
0



and



121
6
−
10
−
1
−
2
−
1



X
=



0
0
0



we see that
X

=



121
6
−
10
−
1
−
2
−
1



X
.
16.
Since
X

=



cos
t
1
2
sin
t
−
1
2
cos
t
−
cos
t
−
sin
t



and



10 1
11 0
−
20
−
1



X
=



cos
t
1
2
sin
t
−
1
2
cos
t
−
cos
t
−
sin
t



we see that
X

=



10 1
11 0
−
20
−
1



X
.
17.
Yes, since
W
(
X
1
,
X
2
)=
−
2
e
−
8
t
L
= 0 and
X
1
and
X
2
are linearly independent on
−∞
<t<
∞
.
18.
Yes, since
W
(
X
1
,
X
2
)=8
e
2
t
L
= 0 and
X
1
and
X
2
are linearly independent on
−∞
<t<
∞
.
488

489. Exercises 10.1
19.
No, since
W
(
X
1
,
X
2
,
X
3
) = 0 and
X
1
,
X
2
, and
X
3
are linearly dependent on
−∞
<t<
∞
.
20.
Yes, since
W
(
X
1
,
X
2
,
X
3
)=
−
84
e
−
t
L
= 0 and
X
1
,
X
2
, and
X
3
are linearly independent on
−∞
<t<
∞
.
21.
Since
X

p
=
L
2
−
1
e
and
L
14
32
e
X
p
+
L
2
−
4
e
t
+
L
−
7
−
18
e
=
L
2
−
1
e
we see that
X

p
=
L
14
32
e
X
p
+
L
2
−
4
e
t
+
L
−
7
−
18
e
.
22.
Since
X

p
=
L
0
0
e
and
L
21
1
−
1
e
X
p
+
L
−
5
2
e
=
L
0
0
e
we see that
X

p
=
L
21
1
−
1
e
X
p
+
L
−
5
2
e
.
23.
Since
X

p
=
L
2
0
e
e
t
+
L
1
−
1
e
te
t
and
L
21
34
e
X
p
−
L
1
7
e
e
t
=
L
2
0
e
e
t
+
L
1
−
1
e
te
t
we see that
X

p
=
L
21
34
e
X
p
−
L
1
7
e
e
t
.
24.
Since
X

p
=



3 cos 3
t
0
−
3 sin 3
t



and



123
−
420
−
610



X
p
+



−
1
4
3



sin 3
t
=



3 cos 3
t
0
−
3 sin 3
t



we see that
X

p
=



123
−
420
−
610



X
p
+



−
1
4
3



sin 3
t.
25.
Let
X
1
=



6
−
1
−
5



e
−
t
,
X
2
=



−
3
1
1



e
−
2
t
,
X
3
=



2
1
1



e
3
t
,
and
A
=



060
101
110



.
Then
X

1
=



−
6
1
5



e
−
t
=
AX
1
,
X

2
=



6
−
2
−
2



e
−
2
t
=
AX
2
,
X

3
=



6
3
3



e
3
t
=
AX
3
,
and
W
(
X
1
,
X
2
,
X
3
)=20
L
= 0 so that
X
1
,
X
2
, and
X
3
form a fundamental set for
X

=
AX
on
−∞
<t<
∞
.
489

490. Exercises 10.1
26.
Let
X
1
=
L
1
−
1
−
√
2
e
e
√
2
t
,
X
2
=
L
1
−
1+
√
2
e
e
−
√
2
t
,
X
p
=
L
1
0
e
t
2
+
L
−
2
4
e
t
+
L
1
0
e
,
and
A
=
L
−
1
−
1
−
11
e
.
Then
X

1
=
L
√
2
−
2
−
√
2
e
e
√
2
t
=
AX
1
,
X

2
=
L
−
√
2
−
2
−
√
2
e
e
−
√
2
t
=
AX
2
,
X

p
=
L
2
0
e
t
+
L
−
2
4
e
=
AX
p
+
L
1
1
e
t
2
+
L
4
−
6
e
t
+
L
−
1
5
e
,
and
W
(
X
1
,
X
2
)=2
√
2
L
= 0 so that
X
p
is a particular solution and
X
1
and
X
2
form a fundamental set on
−∞
<t<
∞
.
Exercises 10.2
1.
The system is
X

=
L
12
43
e
X
and det(
A
−
λ
I
)=(
λ
−
5)(
λ
+ 1) = 0. For
λ
1
= 5 we obtain
L
−
42
0
4
−
2
0
e
=
⇒
L
1
−
1
/
2
0
00
0
e
so that
K
1
=
L
1
2
e
.
For
λ
2
=
−
1 we obtain
L
22
0
44
0
e
=
⇒
L
11
0
00
0
e
so that
K
2
=
L
−
1
1
e
.
Then
X
=
c
1
L
1
2
e
e
5
t
+
c
2
L
−
1
1
e
e
−
t
.
2.
The system is
X

=
L
22
13
e
X
and det(
A
−
λ
I
)=(
λ
−
1)(
λ
−
4) = 0. For
λ
1
= 1 we obtain
L
12
0
12
0
e
=
⇒
L
12
0
00
0
e
so that
K
1
=
L
−
2
1
e
.
For
λ
2
= 4 we obtain
L
−
22
0
1
−
1
0
e
=
⇒
L
−
11
0
00
0
e
so that
K
2
=
L
1
1
e
.
490

491. Exercises 10.2
Then
X
=
c
1
L
−
2
1
e
e
t
+
c
2
L
1
1
e
e
4
t
.
3.
The system is
X

=
L
−
42
−
5
/
22
e
X
and det(
A
−
λ
I
)=(
λ
−
1)(
λ
+ 3) = 0. For
λ
1
= 1 we obtain
L
−
52
0
−
5
/
21
0
e
=
⇒
L
−
52
0
00
0
e
so that
K
1
=
L
2
5
e
.
For
λ
2
=
−
3 we obtain
L
−
12
0
−
5
/
25
0
e
=
⇒
L
−
12
0
00
0
e
so that
K
2
=
L
2
1
e
.
Then
X
=
c
1
L
2
5
e
e
t
+
c
2
L
2
1
e
e
−
3
t
.
4.
The system is
X

=
L
−
5
/
22
3
/
4
−
2
e
X
and det(
A
−
λ
I
)=(
λ
+ 1)(2
λ
+ 7) = 0. For
λ
1
=
−
7
/
2 we obtain
L
12
0
3
/
43
/
2
0
e
=
⇒
L
−
12
0
00
0
e
so that
K
1
=
L
−
2
1
e
.
For
λ
2
=
−
1 we obtain
L
−
3
/
22
0
3
/
4
−
1
0
e
=
⇒
L
−
34
0
00
0
e
so that
K
2
=
L
4
3
e
.
Then
X
=
c
1
L
−
2
1
e
e
−
7
t/
2
+
c
2
L
4
3
e
e
−
t
.
5.
The system is
X

=
L
10
−
5
8
−
12
e
X
and det(
A
−
λ
I
)=(
λ
−
8)(
λ
+ 10) = 0. For
λ
1
= 8 we obtain
L
2
−
5
0
8
−
20
0
e
=
⇒
L
1
−
5
/
2
0
00
0
e
so that
K
1
=
L
5
2
e
.
For
λ
2
=
−
10 we obtain
L
20
−
5
0
8
−
2
0
e
=
⇒
L
1
−
1
/
4
0
00
0
e
so that
K
2
=
L
1
4
e
.
Then
X
=
c
1
L
5
2
e
e
8
t
+
c
2
L
1
4
e
e
−
10
t
.
6.
The system is
X

=
L
−
62
−
31
e
X
491

492. Exercises 10.2
and det(
A
−
λ
I
)=
λ
(
λ
+ 5) = 0. For
λ
1
= 0 we obtain
L
−
62
0
−
31
0
e
=
⇒
L
1
−
1
/
3
0
00
0
e
so that
K
1
=
L
1
3
e
.
For
λ
2
=
−
5 we obtain
L
−
12
0
−
36
0
e
=
⇒
L
1
−
2
0
00
0
e
so that
K
2
=
L
2
1
e
.
Then
X
=
c
1
L
1
3
e
+
c
2
L
2
1
e
e
−
5
t
.
7.
The system is
X

=



11
−
1
02 0
01
−
1



X
and det(
A
−
λ
I
)=(
λ
−
1)(2
−
λ
)(
λ
+1)=0. For
λ
1
=1,
λ
2
= 2, and
λ
3
=
−
1 we obtain
K
1
=



1
0
0



,
K
2
=



2
3
1



,
and
K
3
=



1
0
2



,
so that
X
=
c
1



1
0
0



e
t
+
c
2



2
3
1



e
2
t
+
c
3



1
0
2



e
−
t
.
8.
The system is
X

=



2
−
70
5104
052



X
and det(
A
−
λ
I
)=(2
−
λ
)(
λ
−
5)(
λ
−
7)=0. For
λ
1
=2,
λ
2
= 5, and
λ
3
= 7 we obtain
K
1
=



4
0
−
5



,
K
2
=



−
7
3
5



,
and
K
3
=



−
7
5
5



,
so that
X
=
c
1



4
0
−
5



e
2
t
+
c
2



−
7
3
5



e
5
t
+
c
3



−
7
5
5



e
7
t
.
9.
We have det(
A
−
λ
I
)=
−
(
λ
+ 1)(
λ
−
3)(
λ
+ 2) = 0. For
λ
1
=
−
1,
λ
2
= 3, and
λ
3
=
−
2 we obtain
K
1
=



−
1
0
1



,
K
2
=



1
4
3



,
and
K
3
=



1
−
1
3



,
so that
X
=
c
1



−
1
0
1



e
−
t
+
c
2



1
4
3



e
3
t
+
c
3



1
−
1
3



e
−
2
t
.
492

493. Exercises 10.2
10.
We have det(
A
−
λ
I
)=
−
λ
(
λ
−
1)(
λ
−
2) = 0. For
λ
1
=0,
λ
2
= 1, and
λ
3
= 2 we obtain
K
1
=



1
0
−
1



,
K
2
=



0
1
0



,
and
K
3
=



1
0
1



,
so that
X
=
c
1



1
0
−
1



+
c
2



0
1
0



e
t
+
c
3



1
0
1



e
2
t
.
11.
We have det(
A
−
λ
I
)=
−
(
λ
+ 1)(
λ
+1
/
2)(
λ
+3
/
2) = 0. For
λ
1
=
−
1,
λ
2
=
−
1
/
2, and
λ
3
=
−
3
/
2 we obtain
K
1
=



4
0
−
1



,
K
2
=



−
12
6
5



,
and
K
3
=



4
2
−
1



,
so that
X
=
c
1



4
0
−
1



e
−
t
+
c
2



−
12
6
5



e
−
t/
2
+
c
3



4
2
−
1



e
−
3
t/
2
.
12.
We have det(
A
−
λ
I
)=(
λ
−
3)(
λ
+ 5)(6
−
λ
) = 0. For
λ
1
=3,
λ
2
=
−
5, and
λ
3
= 6 we obtain
K
1
=



1
1
0



,
K
2
=



1
−
1
0



,
and
K
3
=



2
−
2
11



,
so that
X
=
c
1



1
1
0



e
3
t
+
c
2



1
−
1
0



e
−
5
t
+
c
3



2
−
2
11



e
6
t
.
13.
We have det(
A
−
λ
I
)=(
λ
+1
/
2)(
λ
−
1
/
2) = 0. For
λ
1
=
−
1
/
2 and
λ
2
=1
/
2 we obtain
K
1
=
L
0
1
e
and
K
2
=
L
1
1
e
,
so that
X
=
c
1
L
0
1
e
e
−
t/
2
+
c
2
L
1
1
e
e
t/
2
.
If
X
(0) =
L
3
5
e
then
c
1
= 2 and
c
2
=3.
14.
We have det(
A
−
λ
I
)=(2
−
λ
)(
λ
−
3)(
λ
+ 1) = 0. For
λ
1
=2,
λ
2
= 3, and
λ
3
=
−
1 we obtain
K
1
=



5
−
3
2



,
K
2
=



2
0
1



,
and
K
3
=



−
2
0
1



,
so that
X
=
c
1



5
−
3
2



e
2
t
+
c
2



2
0
1



e
3
t
+
c
3



−
2
0
1



e
−
t
.
493

494. Exercises 10.2
If
X
(0) =



1
3
0



then
c
1
=
−
1,
c
2
=5
/
2, and
c
3
=
−
1
/
2.
15. X
=
c
1



0
.
382175
0
.
851161
0
.
359815



e
8
.
58979
t
+
c
2



0
.
405188
−
0
.
676043
0
.
615458



e
2
.
25684
t
+
c
3



−
0
.
923562
−
0
.
132174
0
.
35995



e
−
0
.
0466321
t
16. X
=
c
1







0
.
0312209
0
.
949058
0
.
239535
0
.
195825
0
.
0508861







e
5
.
05452
t
+
c
2







−
0
.
280232
−
0
.
836611
−
0
.
275304
0
.
176045
0
.
338775







e
4
.
09561
t
+
c
3







0
.
262219
−
0
.
162664
−
0
.
826218
−
0
.
346439
0
.
31957







e
−
2
.
92362
t
+
c
4







0
.
313235
0
.
64181
0
.
31754
0
.
173787
−
0
.
599108







e
2
.
02882
t
+
c
5







−
0
.
301294
0
.
466599
0
.
222136
0
.
0534311
−
0
.
799567







e
−
0
.
155338
t
17. (a)
(b)
Letting
c
1
= 1 and
c
2
= 0 we get
x
=5
e
8
t
,
y
=2
e
8
t
. Eliminating the parameter we find
y
=
2
5
x
,
x>
0.
When
c
1
=
−
1 and
c
2
= 0 we find
y
=
2
5
x
,
x<
0. Letting
c
1
= 0 and
c
2
= 1 we get
x
=
e
−
10
t
,
y
=4
e
−
10
t
.
Eliminating the parameter we find
y
=4
x
,
x>
0. Letting
c
1
= 0 and
c
2
=
−
1wefind
y
=4
x
,
x<
0.
(c)
The eigenvectors
K
1
=(5
,
2) and
K
2
=(1
,
4) are shown in the figure in part (
a
).
18. (a)
Letting
c
1
= 1 and
c
2
= 0 we get
x
=
−
2
e
t
,
y
=
e
t
. Eliminating
the parameter we find
y
=
−
1
2
x
,
x<
0. When
c
1
=
−
1 and
c
2
=0
we find
y
=
−
1
2
x
,
x>
0. Letting
c
1
= 0 and
c
2
= 1 we get
x
=
e
4
t
,
y
=
e
4
t
. Eliminating the parameter we find
y
=
x
,
x>
0. When
c
1
= 0 and
c
2
=
−
1 we find
y
=
x
,
x<
0.
494

495. Exercises 10.2
Letting
c
1
= 1 and
c
2
= 0 we get
x
=
−
2
e
−
7
t/
2
,
y
=
e
−
7
t/
2
. Eliminating the parameter we find
y
=
−
1
2
x
,
x<
0.
When
c
1
=
−
1 and
c
2
= 0 we find
y
=
−
1
2
x
,
x>
0. Letting
c
1
= 0 and
c
2
= 1 we get
x
=4
e
−
t
,
y
=3
e
−
t
.
Eliminating the parameter we find
y
=
3
4
x
,
x>
0. When
c
1
= 0 and
c
2
=
−
1 we find
y
=
3
4
x
,
x<
0.
19.
We have det(
A
−
λ
I
)=
λ
2
=0. For
λ
1
= 0 we obtain
K
=
L
1
3
e
.
A solution of (
A
−
λ
1
I
)
P
=
K
is
P
=
L
1
2
e
so that
X
=
c
1
L
1
3
e
+
c
2
3L
1
3
e
t
+
L
1
2
e5
.
20.
We have det(
A
−
λ
I
)=(
λ
+1)
2
=0. For
λ
1
=
−
1 we obtain
K
=
L
1
1
e
.
A solution of (
A
−
λ
1
I
)
P
=
K
is
P
=
L
0
1
/
5
e
so that
X
=
c
1
L
1
1
e
e
−
t
+
c
2
3L
1
1
e
te
−
t
+
L
0
1
/
5
e
e
−
t

.
21.
We have det(
A
−
λ
I
)=(
λ
−
2)
2
=0. For
λ
1
= 2 we obtain
K
=
L
1
1
e
.
A solution of (
A
−
λ
1
I
)
P
=
K
is
P
=
L
−
1
/
3
0
e
so that
X
=
c
1
L
1
1
e
e
2
t
+
c
2
3L
1
1
e
te
2
t
+
L
−
1
/
3
0
e
e
2
t

.
22.
We have det(
A
−
λ
I
)=(
λ
−
6)
2
=0. For
λ
1
= 6 we obtain
K
=
L
3
2
e
.
495

496. Exercises 10.2
A solution of (
A
−
λ
1
I
)
P
=
K
is
P
=
L
1
/
2
0
e
so that
X
=
c
1
L
3
2
e
e
6
t
+
c
2
3L
3
2
e
te
6
t
+
L
1
/
2
0
e
e
6
t

.
23.
We have det(
A
−
λ
I
)=(1
−
λ
)(
λ
−
2)
2
=0. For
λ
1
= 1 we obtain
K
1
=



1
1
1



.
For
λ
2
= 2 we obtain
K
2
=



1
0
1



and
K
3
=



1
1
0



.
Then
X
=
c
1



1
1
1



e
t
+
c
2



1
0
1



e
2
t
+
c
3



1
1
0



e
2
t
.
24.
We have det(
A
−
λ
I
)=(
λ
−
8)(
λ
+1)
2
=0. For
λ
1
= 8 we obtain
K
1
=



2
1
2



.
For
λ
2
=
−
1 we obtain
K
2
=



0
−
2
1



and
K
3
=



1
−
2
0



.
Then
X
=
c
1



2
1
2



e
8
t
+
c
2



0
−
2
1



e
−
t
+
c
3



1
−
2
0



e
−
t
.
25.
We have det(
A
−
λ
I
)=
−
λ
(5
−
λ
)
2
=0. For
λ
1
= 0 we obtain
K
1
=



−
4
−
5
2



.
For
λ
2
= 5 we obtain
K
=



−
2
0
1



.
A solution of (
A
−
λ
1
I
)
P
=
K
is
P
=



5
/
2
1
/
2
0



496

497. Exercises 10.2
so that
X
=
c
1



−
4
−
5
2



+
c
2



−
2
0
1



e
5
t
+
c
3






−
2
0
1



te
5
t
+



5
/
2
1
/
2
0



e
5
t



.
26.
We have det(
A
−
λ
I
)=(1
−
λ
)(
λ
−
2)
2
=0. For
λ
1
= 1 we obtain
K
1
=



1
0
0



.
For
λ
2
= 2 we obtain
K
=



0
−
1
1



.
A solution of (
A
−
λ
2
I
)
P
=
K
is
P
=



0
−
1
0



so that
X
=
c
1



1
0
0



e
t
+
c
2



0
−
1
1



e
2
t
+
c
3






0
−
1
1



te
2
t
+



0
−
1
0



e
2
t



.
27.
We have det(
A
−
λ
I
)=
−
(
λ
−
1)
3
=0. For
λ
1
= 1 we obtain
K
=



0
1
1



.
Solutions of (
A
−
λ
1
I
)
P
=
K
and (
A
−
λ
1
I
)
Q
=
P
are
P
=



0
1
0



and
Q
=



1
/
2
0
0



so that
X
=
c
1



0
1
1



+
c
2






0
1
1



te
t
+



0
1
0



e
t



+
c
3






0
1
1



t
2
2
e
t
+



0
1
0



te
t
+



1
/
2
0
0



e
t



.
28.
We have det(
A
−
λ
I
)=(
λ
−
4)
3
=0. For
λ
1
= 4 we obtain
K
=



1
0
0



.
Solutions of (
A
−
λ
1
I
)
P
=
K
and (
A
−
λ
1
I
)
Q
=
P
are
P
=



0
1
0



and
Q
=



0
0
1



497

498. Exercises 10.2
so that
X
=
c
1



1
0
0



e
4
t
+
c
2






1
0
0



te
4
t
+



0
1
0



e
4
t



+
c
3






1
0
0



t
2
2
e
4
t
+



0
1
0



te
4
t
+



0
0
1



e
4
t



.
29.
We have det(
A
−
λ
I
)=(
λ
−
4)
2
=0. For
λ
1
= 4 we obtain
K
=
L
2
1
e
.
A solution of (
A
−
λ
1
I
)
P
=
K
is
P
=
L
1
1
e
so that
X
=
c
1
L
2
1
e
e
4
t
+
c
2
3L
2
1
e
te
4
t
+
L
1
1
e
e
4
t

.
If
X
(0) =
L
−
1
6
e
then
c
1
=
−
7 and
c
2
= 13.
30.
We have det(
A
−
λ
I
)=
−
(
λ
+ 1)(
λ
−
1)
2
=0. For
λ
1
=
−
1 we obtain
K
1
=



−
1
0
1



.
For
λ
2
= 1 we obtain
K
2
=



1
0
1



and
K
3
=



0
1
0



so that
X
=
c
1



−
1
0
1



e
−
t
+
c
2



1
0
1



e
t
+
c
3



0
1
0



e
t
.
If
X
(0) =



1
2
5



then
c
1
=2,
c
2
= 3, and
c
3
=2.
31.
In this case det(
A
−
λ
I
)=(2
−
λ
)
5
, and
λ
1
= 2 is an eigenvalue of multiplicity 5. Linearly independent
eigenvectors are
K
1
=







1
0
0
0
0







,
K
2
=







0
0
1
0
0







,
and
K
3
=







0
0
0
1
0







.
498

499. Exercises 10.2
32.
Letting
c
1
= 1 and
c
2
= 0 we get
x
=
e
t
,
y
=
e
t
. Eliminating the
parameter we find
y
=
x
,
x>
0. When
c
1
=
−
1 and
c
2
= 0 we find
y
=
x
,
x<
0.
Letting
c
1
= 1 and
c
2
= 0 we get
x
=
e
2
t
,
y
=
e
2
t
. Eliminating the parameter we find
y
=
x
,
x>
0. When
c
1
=
−
1 and
c
2
= 0 we find
y
=
x
,
x<
0.
In Problems 33-46 the form of the answer will vary according to the choice of eigenvector. For example, in
Problem 33, if
K
1
is chosen to be
L
1
2
−
i
e
the solution has the form
X
=
c
1
L
cos
t
2 cos
t
+ sin
t
e
e
4
t
+
c
2
L
sin
t
2 sin
t
−
cos
t
e
e
4
t
.
33.
We have det(
A
−
λ
I
)=
λ
2
−
8
λ
+ 17 = 0. For
λ
1
=4+
i
we obtain
K
1
=
L
2+
i
5
e
so that
X
1
=
L
2+
i
5
e
e
(4+
i
)
t
=
L
2 cos
t
−
sin
t
5 cos
t
e
e
4
t
+
i
L
cos
t
+ 2 sin
t
5 sin
t
e
e
4
t
.
Then
X
=
c
1
L
2 cos
t
−
sin
t
5 cos
t
e
e
4
t
+
c
2
L
2 sin
t
+ cos
t
5 sin
t
e
e
4
t
.
499

500. Exercises 10.2
34.
We have det(
A
−
λ
I
)=
λ
2
+1=0. For
λ
1
=
i
we obtain
K
1
=
L
−
1
−
i
2
e
so that
X
1
=
L
−
1
−
i
2
e
e
it
=
L
sin
t
−
cos
t
2 cos
t
e
+
i
L
−
cos
t
−
sin
t
2 sin
t
e
.
Then
X
=
c
1
L
sin
t
−
cos
t
2 cos
t
e
+
c
2
L
−
cos
t
−
sin
t
2 sin
t
e
.
35.
We have det(
A
−
λ
I
)=
λ
2
−
8
λ
+ 17 = 0. For
λ
1
=4+
i
we obtain
K
1
=
L
−
1
−
i
2
e
so that
X
1
=
L
−
1
−
i
2
e
e
(4+
i
)
t
=
L
sin
t
−
cos
t
2 cos
t
e
e
4
t
+
i
L
−
sin
t
−
cos
t
2 sin
t
e
e
4
t
.
Then
X
=
c
1
L
sin
t
−
cos
t
2 cos
t
e
e
4
t
+
c
2
L
−
sin
t
−
cos
t
2 sin
t
e
e
4
t
.
36.
We have det(
A
−
λ
I
)=
λ
2
−
10
λ
+ 34 = 0. For
λ
1
=5+3
i
we obtain
K
1
=
L
1
−
3
i
2
e
so that
X
1
=
L
1
−
3
i
2
e
e
(5+3
i
)
t
=
L
cos 3
t
+ 3 sin 3
t
2 cos 3
t
e
e
5
t
+
i
L
sin 3
t
−
3 cos 3
t
2 cos 3
t
e
e
5
t
.
Then
X
=
c
1
L
cos 3
t
+ 3 sin 3
t
2 cos 3
t
e
e
5
t
+
c
2
L
sin 3
t
−
3 cos 3
t
2 cos 3
t
e
e
5
t
.
37.
We have det(
A
−
λ
I
)=
λ
2
+9=0. For
λ
1
=3
i
we obtain
K
1
=
L
4+3
i
5
e
so that
X
1
=
L
4+3
i
5
e
e
3
it
=
L
4 cos 3
t
−
3 sin 3
t
5 cos 3
t
e
+
i
L
4 sin 3
t
+ 3 cos 3
t
5 sin 3
t
e
.
Then
X
=
c
1
L
4 cos 3
t
−
3 sin 3
t
5 cos 3
t
e
+
c
2
L
4 sin 3
t
+ 3 cos 3
t
5 sin 3
t
e
.
38.
We have det(
A
−
λ
I
)=
λ
2
+2
λ
+5=0. For
λ
1
=
−
1+2
i
we obtain
K
1
=
L
2+2
i
1
e
so that
X
1
=
L
2+2
i
1
e
e
(
−
1+2
i
)
t
= ( 2 cos 2
t
−
2 sin 2
t
cos 2
t
)
e
−
t
+
i
L
2 cos 2
t
+ 2 sin 2
t
sin 2
t
e
e
−
t
.
500

501. Exercises 10.2
Then
X
=
c
1
L
2 cos 2
t
−
2 sin 2
t
cos 2
t
e
e
−
t
+
c
2
L
2 cos 2
t
+ 2 sin 2
t
sin 2
t
e
e
−
t
.
39.
We have det(
A
−
λ
I
)=
−
λ
1
λ
2
+1
2
=0. For
λ
1
= 0 we obtain
K
1
=



1
0
0



.
For
λ
2
=
i
we obtain
K
2
=



−
i
i
1



so that
X
2
=



−
i
i
1



e
it
=



sin
t
−
sin
t
cos
t



+
i



−
cos
t
cos
t
sin
t



.
Then
X
=
c
1



1
0
0



+
c
2



sin
t
−
sin
t
cos
t



+
c
3



−
cos
t
cos
t
sin
t



.
40.
We have det(
A
−
λ
I
)=
−
(
λ
+ 3)(
λ
2
−
2
λ
+ 5) = 0. For
λ
1
=
−
3 we obtain
K
1
=



0
−
2
1



.
For
λ
2
=1+2
i
we obtain
K
2
=



−
2
−
i
−
3
i
2



so that
X
2
=



−
2 cos 2
t
+ sin 2
t
3 sin 2
t
2 cos 2
t



e
t
+
i



−
cos 2
t
−
2 sin 2
t
−
3 cos 2
t
2 sin 2
t



e
t
.
Then
X
=
c
1



0
−
2
1



e
−
3
t
+
c
2



−
2 cos 2
t
+ sin 2
t
3 sin 2
t
2 cos 2
t



e
t
+
c
3



−
cos 2
t
−
2 sin 2
t
−
3 cos 2
t
2 sin 2
t



e
t
.
41.
We have det(
A
−
λ
I
)=(1
−
λ
)(
λ
2
−
2
λ
+ 2) = 0. For
λ
1
= 1 we obtain
K
1
=



0
2
1



.
For
λ
2
=1+
i
we obtain
K
2
=



1
i
i



501

502. Exercises 10.2
so that
X
2
=



1
i
i



e
(1+
i
)
t
=



cos
t
−
sin
t
−
sin
t



e
t
+
i



sin
t
cos
t
cos
t



e
t
.
Then
X
=
c
1



0
2
1



e
t
+
c
2



cos
t
−
sin
t
−
sin
t



e
t
+
c
3



sin
t
cos
t
cos
t



e
t
.
42.
We have det(
A
−
λ
I
)=
−
(
λ
−
6)(
λ
2
−
8
λ
+ 20) = 0. For
λ
1
= 6 we obtain
K
1
=



0
1
0



.
For
λ
2
=4+2
i
we obtain
K
2
=



−
i
0
2



so that
X
2
=



−
i
0
2



e
(4+2
i
)
t
=



sin 2
t
0
2 cos 2
t



e
4
t
+
i



−
cos 2
t
0
2 sin 2
t



e
4
t
.
Then
X
=
c
1



0
1
0



e
6
t
+
c
2



sin 2
t
0
2 cos 2
t



e
4
t
+
c
3



−
cos 2
t
0
2 sin 2
t



e
4
t
.
43.
We have det(
A
−
λ
I
)=(2
−
λ
)(
λ
2
+4
λ
+ 13) = 0. For
λ
1
= 2 we obtain
K
1
=



28
−
5
25



.
For
λ
2
=
−
2+3
i
we obtain
K
2
=



4+3
i
−
5
0



so that
X
2
=



4+3
i
−
5
0



e
(
−
2+3
i
)
t
=



4 cos 3
t
−
3 sin 3
t
−
5 cos 3
t
0



e
−
2
t
+
i



4 sin 3
t
+ 3 cos 3
t
−
5 sin 3
t
0



e
−
2
t
.
Then
X
=
c
1



28
−
5
25



e
2
t
+
c
2



4 cos 3
t
−
3 sin 3
t
−
5 cos 3
t
0



e
−
2
t
+
c
3



4 sin 3
t
+ 3 cos 3
t
−
5 sin 3
t
0



e
−
2
t
.
502

503. Exercises 10.2
44.
We have det(
A
−
λ
I
)=
−
(
λ
+ 2)(
λ
2
+4)=0. For
λ
1
=
−
2 we obtain
K
1
=



0
−
1
1



.
For
λ
2
=2
i
we obtain
K
2
=



−
2
−
2
i
1
1



so that
X
2
=



−
2
−
2
i
1
1



e
2
it
=



−
2 cos 2
t
+ 2 sin 2
t
cos 2
t
cos 2
t



+
i



−
2 cos 2
t
−
2 sin 2
t
sin 2
t
sin 2
t



.
Then
X
=
c
1



0
−
1
1



e
−
2
t
+
c
2



−
2 cos 2
t
+ 2 sin 2
t
cos 2
t
cos 2
t



+
c
3



−
2 cos 2
t
−
2 sin 2
t
sin 2
t
sin 2
t



.
45.
We have det(
A
−
λ
I
)=(1
−
λ
)(
λ
2
+ 25) = 0. For
λ
1
= 1 we obtain
K
1
=



25
−
7
6



.
For
λ
2
=5
i
we obtain
K
2
=



1+5
i
1
1



so that
X
2
=



1+5
i
1
1



e
5
it
=



cos 5
t
−
5 sin 5
t
cos 5
t
cos 5
t



+
i



sin 5
t
+ 5 cos 5
t
sin 5
t
sin 5
t



.
Then
X
=
c
1



25
−
7
6



e
t
+
c
2



cos 5
t
−
5 sin 5
t
cos 5
t
cos 5
t



+
c
3



sin 5
t
+ 5 cos 5
t
sin 5
t
sin 5
t



.
If
X
(0) =



4
6
−
7



then
c
1
=
c
2
=
−
1 and
c
3
=6.
46.
We have det(
A
−
λ
I
)=
λ
2
−
10
λ
+ 29 = 0. For
λ
1
=5+2
i
we obtain
K
1
=
L
1
1
−
2
i
e
so that
X
1
=
L
1
1
−
2
i
e
e
(5+2
i
)
t
=
L
cos 2
t
cos 2
t
+ 2 sin 2
t
e
e
5
t
+
i
L
sin 2
t
sin 2
t
−
2 cos 2
t
e
e
5
t
.
503

504. Exercises 10.2
and
X
=
c
1
L
cos 2
t
cos 2
t
+ 2 sin 2
t
e
e
5
t
+
c
3
L
sin 2
t
sin 2
t
−
2 cos 2
t
e
e
5
t
.
If
X
(0) =
L
−
2
8
e
then
c
1
=
−
2 and
c
2
=5.
47.
48. (a)
From det(
A
−
λ
I
)=
λ
(
λ
−
2) = 0 we get
λ
1
= 0 and
λ
2
=2. For
λ
1
= 0 we obtain
L
11
0
11
0
e
=
⇒
L
11
0
00
0
e
so that
K
1
=
L
−
1
1
e
.
For
λ
2
= 2 we obtain
L
−
11
0
1
−
1
0
e
=
⇒
L
−
11
0
00
0
e
so that
K
2
=
L
1
1
e
.
Then
X
=
c
1
L
−
1
1
e
+
c
2
L
1
1
e
e
2
t
.
The line
y
=
−
x
is not a trajectory of the system.
Trajectories are
x
=
−
c
1
+
c
2
e
2
t
,
y
=
c
1
+
c
2
e
2
t
or
y
=
x
+2
c
1
. This is a family of lines perpendicular
to the line
y
=
−
x
. All of the constant solutions
of the system do, however, lie on the line
y
=
−
x
.
(b)
From det(
A
−
λ
I
)=
λ
2
= 0 we get
λ
1
= 0 and
K
=
L
−
1
1
e
.
A solution of (
A
−
λ
1
I
)
P
=
K
is
P
=
L
−
1
0
e
so that
X
=
c
1
L
−
1
1
e
+
c
2
3L
−
1
1
e
t
+
L
−
1
0
e5
.
504

505. Exercises 10.2
All trajectories are parallel to
y
=
−
x
, but
y
=
−
x
is not a trajectory. There are constant
solutions of the system, however, that do lie
on the line
y
=
−
x
.
49.
The system of differential equations is
x

1
=2
x
1
+
x
2
x

2
=2
x
2
x

3
=2
x
3
x

4
=2
x
4
+
x
5
x

5
=2
x
5
.
We see immediately that
x
2
=
c
2
e
2
t
,
x
3
=
c
3
e
2
t
, and
x
5
=
c
5
e
2
t
. Then
x

1
=2
x
1
+
c
2
e
2
t
=
⇒
x
1
=
c
2
te
2
t
+
c
1
e
2
t
and
x

4
=2
x
4
+
c
5
e
2
t
=
⇒
x
4
=
c
5
te
2
t
+
c
4
e
2
t
.
The general solution of the system is
X
=







c
2
te
2
t
+
c
1
e
2
t
c
2
e
2
t
c
3
e
2
t
c
5
te
2
t
+
c
4
e
2
t
c
5
e
2
t







=
c
1







1
0
0
0
0







e
2
t
+
c
2














1
0
0
0
0







te
2
t
+







0
1
0
0
0







e
2
t







+
c
3







0
0
1
0
0







e
2
t
+
c
4







0
0
0
1
0







e
2
t
+
c
5














0
0
0
1
0







te
2
t
+







0
0
0
0
1







e
2
t







505

506. Exercises 10.2
=
c
1
K
1
e
2
t
+
c
2







K
1
te
2
t
+







0
1
0
0
0







e
2
t







+
c
3
K
2
e
2
t
+
c
4
K
3
e
2
t
+
c
5







K
3
te
2
t
+







0
0
0
0
1







e
2
t







.
There are three solutions of the form
X
=
K
e
2
t
, where
K
is an eigenvector, and two solutions of the form
X
=
K
te
2
t
+
P
e
2
t
. See (12) in the text. From (13) and (14) in the text
(
A
−
2
I
)
K
1
=
0
and
(
A
−
2
I
)
K
2
=
K
1
.
This implies







01000
00000
00000
00001
00000














p
1
p
2
p
3
p
4
p
5







=







1
0
0
0
0







,
so
p
2
= 1 and
p
5
= 0, while
p
1
,
p
3
, and
p
4
are arbitrary. Choosing
p
1
=
p
3
=
p
4
= 0 we have
P
=







1
0
0
0
0







.
Therefore a solution is
X
=







1
0
0
0
0







te
2
t
+







1
0
0
0
0







e
2
t
.
Repeating for
K
3
we find
P
=







0
0
0
0
1







,
506

507. Exercises 10.2
so another solution is
X
=







0
0
0
1
0







te
2
t
+







0
0
0
0
1







e
2
t
.
50.
From
x
= 2 cos 2
t
−
2 sin 2
t
,
y
=
−
cos 2
t
we find
x
+2
y
=
−
2 sin 2
t
. Then
(
x
+2
y
)
2
= 4 sin
2
2
t
= 4(1
−
cos
2
2
t
)=4
−
4 cos
2
2
t
=4
−
4
y
2
and
x
2
+4
xy
+4
y
2
=4
−
4
y
2
or
x
2
+4
xy
+8
y
2
=4
.
This is a rotated conic section and, from the discriminant
b
2
−
4
ac
=16
−
32
<
0, we see that the curve is an
ellipse.
51.
Suppose the eigenvalues are
α
±
iβ
,
β>
0. In Problem 36 the eigenvalues are 5
±
3
i
, in Problem 37 they are
±
3
i
, and in Problem 38 they are
−
1
±
2
i
. From Problem 47 we deduce that the phase portrait will consist of a
family of closed curves when
α
= 0 and spirals when
α
L
= 0. The origin will be a repellor when
α>
0, and an
attractor when
α<
0.
52. (a)
The given system can be written as
x

1
=
−
k
1
+
k
2
m
1
x
1
+
k
2
m
1
x
2
,x

2
=
k
2
m
1
x
1
−
k
2
m
1
x
2
.
In terms of matrices this is
X

=
AX
where
X
=

x
1
x
2

and
A
=



−
k
1
+
k
2
m
1
k
2
m
1
k
2
m
1
−
k
2
m
1



.
(b)
If
X
=
K
e
ωt
then
X

=
ω
2
K
e
ωt
and
AX
=
AK
e
ωt
so that
X

=
AX
becomes
ω
2
K
e
ωt
=
AK
e
ωt
or
(
A
−
ω
2
I
)
K
= 0. Now let
ω
2
=
λ
.
(c)
When
m
1
=1,
m
2
=1,
k
1
= 3, and
k
2
= 2 we obtain

−
52
2
−
2

. The eigenvalues and corresponding
eigenvectors of
A
are
λ
1
=
−
1,
λ
2
=
−
6,
K
1
=

1
2

,
K
2
=

−
2
1

. Since
ω
1
=
i
,
ω
2
=
−
i
,
ω
3
=
√
6
i
, and
ω
4
=
−
√
6
i
a solution is
X
=
c
1

1
2

e
it
+
c
2

1
2

e
−
it
+
c
3

−
2
1

e
√
6
it
+
c
4

−
2
1

e
−
√
6
it
.
Using
e
it
= cos
t
+
i
sin
t
and
e
√
6
it
= cos
√
6
t
+
i
sin
√
6
t
the preceding solution can be rewritten as
X
=
b
1

1
2

cos
t
+
b
2

1
2

sin
t
+
b
3

−
2
1

cos
√
6
t
+
b
4

−
2
1

sin
√
6
t
where
b
1
,
b
2
,
b
3
, and
b
4
are constants.
507

508. Exercises 10.3
Exercises 10.3
1.
λ
1
=7,
λ
2
=
−
4,
K
1
=

3
1

,
K
2
=

−
2
3

,
P
=

3
−
2
13

;
X
=
PY
=

3
−
2
13

c
1
e
7
t
c
2
e
−
4
t

=

3
c
1
e
7
t
−
2
c
2
e
−
4
t
c
1
e
7
t
+3
c
2
e
−
4
t

=
c
1

3
1

e
7
t
+
c
2

−
2
3

e
−
4
t
2.
λ
1
=0,
λ
2
=1,
K
1
=

1
−
1

,
K
2
=

1
1

,
P
=

11
−
11

;
X
=
PY
=

11
−
11

c
1
c
2
e
t

=

c
1
+
c
2
e
t
−
c
1
+
c
2
e
t

=
c
1

1
−
1

+
c
2

1
1

e
t
3.
λ
1
=
1
2
,
λ
2
=
3
2
,
K
1
=

1
−
2

,
K
2
=

1
2

,
P
=

11
−
22

;
X
=
PY
=

11
−
22

c
1
e
t/
2
c
2
e
3
t/
2

=

c
1
e
t/
2
+
c
2
e
3
t/
2
−
2
c
1
e
t/
2
+2
c
2
e
3
t/
2

=
c
1

1
−
2

e
t/
2
+
c
2

1
2

e
3
t/
2
4.
λ
1
=
−
√
2,
λ
2
=
√
2,
K
1
=

−
1
1+
√
2

,
K
2
=

−
1
1
−
√
2

,
P
=

−
1
−
1
1+
√
21
−
√
2

;
X
=
PY
=

−
1
−
1
1+
√
21
−
√
2

c
1
e
−
√
2
t
c
2
e
√
2
t

=

−
c
1
e
−
√
2
t
−
c
2
e
√
2
t
(1 +
√
2)
c
1
e
−
√
2
t
+(1
−
√
2)
c
2
e
√
2
t

=
c
1

−
1
1+
√
2

e
−
√
2
t
+
c
2

−
1
1
−
√
2

e
√
2
t
5.
λ
1
=
−
4,
λ
2
=2,
λ
3
=6,
K
1
=



−
1
1
0



,
K
2
=



1
1
1



,
K
3
=



0
0
1



,
P
=



−
110
110
011



;
X
=
PY
=



−
110
110
011






c
1
e
−
4
t
c
2
e
2
t
c
3
e
6
t



=



−
c
1
e
−
4
t
+
c
2
e
2
t
c
1
e
−
4
t
+
c
2
e
2
t
c
2
e
2
t
+
c
3
e
6
t



=
c
1



−
1
1
0



e
−
4
t
+
c
2



1
1
0



e
2
t
+
c
3



0
0
1



e
6
t
6.
λ
1
=
−
1,
λ
2
=1,
λ
3
=4,
K
1
=



−
1
0
1



,
K
2
=



1
−
2
1



,
K
3
=



1
1
1



,
P
=



−
111
0
−
21
111



;
X
=
PY
=



−
111
0
−
21
111






c
1
e
−
t
c
2
e
t
c
3
e
4
t



=



−
c
1
+
c
2
e
t
+
c
3
e
4
t
−
2
c
2
e
t
+
c
3
e
4
t
c
1
e
−
t
+
c
2
e
t
+
c
3
e
4
t



=
c
1



−
1
0
1



e
−
t
+
c
2



1
−
2
1



e
t
+
c
3



1
1
1



e
4
t
7.
λ
1
=
−
1,
λ
2
=2,
λ
3
=2,
K
1
=



1
1
1



,
K
2
=



−
1
1
0



,
K
3
=



−
1
0
1



,
P
=



1
−
1
−
1
110
101



;
X
=
PY
=



1
−
1
−
1
110
101






c
1
e
−
t
c
2
e
2
t
c
3
e
2
t



=



c
1
e
−
t
−
c
2
e
2
t
−
c
3
e
2
t
c
1
e
−
t
+
c
2
e
2
t
c
1
e
−
t
+
c
3
e
2
t



=
c
1



1
1
1



e
−
t
+
c
2



−
1
1
0



e
2
t
+
c
3



−
1
0
1



e
2
t
508

509. Exercises 10.4
8.
λ
1
=0,
λ
2
=0,
λ
3
=0,
λ
4
=4,
K
1
=





−
1
1
0
0





,
K
2
=





−
1
0
1
0





,
K
3
=





−
1
0
0
1





,
K
4
=





1
1
1
1





,
P
=





−
1
−
1
−
11
1001
0101
0011





;
X
=
PY
=





−
1
−
1
−
11
1001
0101
0011










c
1
c
2
c
3
c
4
e
4
t





=





−
c
1
−
c
2
−
c
3
+
c
4
e
4
t
c
1
+
c
4
e
4
t
c
2
+
c
4
e
4
t
c
3
+
c
4
e
4
t





=
c
1





−
1
1
0
0





+
c
2





−
1
0
1
0





+
c
3





−
1
0
0
1





+
c
4





1
1
1
1





e
4
t
9.
λ
1
=1,
λ
2
=2,
λ
3
=3,
K
1
=



1
1
1



,
K
2
=



2
2
3



,
K
3
=



3
4
5



,
P
=



123
124
135



;
X
=
PY
=



123
124
135






c
1
e
t
c
2
e
2
t
c
3
e
3
t



=



c
1
e
t
+2
c
2
e
2
t
+3
c
3
e
3
t
c
1
e
t
+2
c
2
e
2
t
+4
c
3
e
3
t
c
1
e
t
+3
c
2
e
2
t
+5
c
3
e
3
t



=
c
1



1
1
1



e
t
+
c
2



2
2
3



e
2
t
+
c
3



3
4
5



e
3
t
10.
λ
1
=0,
λ
2
=
−
2
√
2,
λ
3
=2
√
2,
K
1
=



1
0
−
1



,
K
2
=



1
−
√
2
1



,
K
3
=



1
√
2
1



,
P
=



111
0
−
√
2
√
2
−
111



;
X
=
PY
=



111
0
−
√
2
√
2
−
111






c
1
c
2
e
−
2
√
2
t
c
3
e
2
√
2
t



=



c
1
+
c
2
e
−
2
√
2
t
+
c
3
e
2
√
2
t
−
√
2
c
2
e
−
2
√
2
t
+
√
2
c
3
e
2
√
2
t
−
c
1
+
c
2
e
−
2
√
2
t
+
c
3
e
2
√
2
t



=
c
1



1
0
−
1



+
c
2



1
−
√
2
1



e
−
2
√
2
t
+
c
3



1
√
2
1



e
2
√
2
t
Exercises 10.4
1.
From
X

=
L
3
−
3
2
−
2
e
X
+
L
4
−
1
e
we obtain
X
c
=
c
1
L
1
1
e
+
c
2
L
3
2
e
e
t
.
Then
Φ
=
L
13
e
t
12
e
t
e
and
Φ
−
1
=
L
−
23
e
−
t
−
e
−
t
e
509

510. Exercises 10.4
so that
U
=
+
Φ
−
1
F
dt
=
+
L
−
11
5
e
−
t
e
dt
=
L
−
11
t
−
5
e
−
t
e
and
X
p
=
ΦU
=
L
−
11
−
11
e
t
+
L
−
15
−
10
e
.
2.
From
X

=
L
2
−
1
3
−
2
e
X
+
L
0
4
e
t
we obtain
X
c
=
c
1
L
1
1
e
e
t
+
c
2
L
1
3
e
e
−
t
.
Then
Φ
=
L
e
t
e
−
t
e
t
3
e
−
t
e
and
Φ
−
1
=
L
3
2
e
−
t
−
1
2
e
−
t
−
1
2
e
t
1
2
e
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
−
2
te
−
t
2
te
t
e
dt
=
L
2
te
−
t
+2
e
−
t
2
te
t
−
2
e
t
e
and
X
p
=
ΦU
=
L
4
8
e
t
+
L
0
−
4
e
.
3.
From
X

=
L
3
−
5
3
/
4
−
1
e
X
+
L
1
−
1
e
e
t/
2
we obtain
X
c
=
c
1
L
10
3
e
e
3
t/
2
+
c
2
L
2
1
e
e
t/
2
.
Then
Φ
=
L
10
e
3
t/
2
2
e
t/
2
3
e
3
t/
2
e
t/
2
e
and
Φ
−
1
=
L
1
4
e
−
3
t/
2
−
1
2
e
−
3
t/
2
−
3
4
e
−
t/
2
5
2
e
−
t/
2
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
3
4
e
−
t
−
13
4
e
dt
=
L
−
3
4
e
−
t
−
13
4
t
e
and
X
p
=
ΦU
=
L
−
13
/
2
−
13
/
4
e
te
t/
2
+
L
−
15
/
2
−
9
/
4
e
e
t/
2
.
4.
From
X

=
L
2
−
1
42
e
X
+
L
sin 2
t
2 cos 2
t
e
we obtain
X
c
=
c
1
L
−
sin 2
t
2 cos 2
t
e
e
2
t
+
c
2
L
cos 2
t
2 sin 2
t
e
e
2
t
.
Then
Φ
=
L
−
e
2
t
sin 2
te
2
t
cos 2
t
2
e
2
t
cos 2
t
2
e
2
t
sin 2
t
e
and
Φ
−
1
=
s
−
1
2
e
−
2
t
sin 2
t
1
4
e
−
2
t
cos 2
t
1
2
e
−
2
t
cos 2
t
1
4
e
−
2
t
sin 2
t
i
so that
U
=
+
Φ
−
1
F
dt
=
+
L
1
2
cos 4
t
1
2
sin 4
t
e
dt
=
L
1
8
sin 4
t
−
1
8
cos 4
t
e
510

511. Exercises 10.4
and
X
p
=
ΦU
=
s
−
1
8
sin 2
t
cos 4
t
−
1
8
cos 2
t
cos 4
t
1
4
cos 2
t
sin 4
t
−
1
4
sin 2
t
cos 4
t
i
e
2
t
.
5.
From
X

=
L
02
−
13
e
X
+
L
1
−
1
e
e
t
we obtain
X
c
=
c
1
L
2
1
e
e
t
+
c
2
L
1
1
e
e
2
t
.
Then
Φ
=
L
2
e
t
e
2
t
e
t
e
2
t
e
and
Φ
−
1
=
L
e
−
t
−
e
−
t
−
e
−
2
t
2
e
−
2
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
2
−
3
e
−
t
e
dt
=
L
2
t
3
e
−
t
e
and
X
p
=
ΦU
=
L
4
2
e
te
t
+
L
3
3
e
e
t
.
6.
From
X

=
L
02
−
13
e
X
+
L
2
e
−
3
t
e
we obtain
X
c
=
c
1
L
2
1
e
e
t
+
c
2
L
1
1
e
e
2
t
.
Then
Φ
=
L
2
e
t
e
2
t
e
t
e
2
t
e
and
Φ
−
1
=
L
e
−
t
−
e
−
t
−
e
−
2
t
2
e
−
2
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
2
e
−
t
−
e
−
4
t
−
2
e
−
2
t
+2
e
−
5
t
e
dt
=
s
−
2
e
−
t
+
1
4
e
−
4
t
e
−
2
t
−
2
5
e
−
5
t
i
and
X
p
=
ΦU
=
s
1
10
e
−
3
t
−
3
−
3
20
e
−
3
t
−
1
i
.
7.
From
X

=
L
18
1
−
1
e
X
+
L
12
12
e
t
we obtain
X
c
=
c
1
L
4
1
e
e
3
t
+
c
2
L
−
2
1
e
e
−
3
t
.
Then
Φ
=
L
4
e
3
t
−
2
e
−
3
t
e
3
t
e
−
3
t
e
and
Φ
−
1
=
s
1
6
e
−
3
t
1
3
e
−
3
t
−
1
6
e
3
t
2
3
e
3
t
i
so that
U
=
+
Φ
−
1
F
dt
=
+
L
6
te
−
3
t
6
te
3
t
e
dt
=
L
−
2
te
−
3
t
−
2
3
e
−
3
t
2
te
3
t
−
2
3
e
3
t
e
and
X
p
=
ΦU
=
L
−
12
0
e
t
+
L
−
4
/
3
−
4
/
3
e
.
511

512. Exercises 10.4
8.
From
X

=
L
18
1
−
1
e
X
+
L
e
−
t
te
t
e
we obtain
X
c
=
c
1
L
4
1
e
e
3
t
+
c
2
L
−
2
1
e
e
−
3
t
.
Then
Φ
=
L
4
e
3
t
−
2
e
3
t
e
3
t
e
−
3
t
e
and
Φ
−
1
=
s
1
6
e
−
3
t
1
3
e
−
3
t
−
1
6
e
3
t
2
3
e
3
t
i
so that
U
=
+
Φ
−
1
F
dt
=
+
s
1
6
e
−
4
t
+
1
3
te
−
2
t
−
1
6
e
2
t
+
2
3
te
4
t
i
dt
=
s
−
1
24
e
−
4
t
−
1
6
te
−
2
t
−
1
12
e
−
2
t
−
1
12
e
2
t
+
1
6
te
4
t
−
1
24
e
4
t
i
and
X
p
=
ΦU
=
s
−
te
t
−
1
4
e
t
−
1
8
e
−
t
−
1
8
e
t
i
.
9.
From
X

=
L
32
−
2
−
1
e
X
+
L
2
1
e
e
−
t
we obtain
X
c
=
c
1
L
1
−
1
e
e
t
+
c
2
3L
1
−
1
e
te
t
+
L
0
1
/
2
e
e
t

.
Then
Φ
=
L
e
t
te
t
−
e
t
1
2
e
t
−
te
t
e
and
Φ
−
1
=
L
e
−
t
−
2
te
−
t
−
2
te
−
t
2
e
−
t
2
e
−
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
2
e
−
2
t
−
6
te
−
2
t
6
e
−
2
t
e
dt
=
L
1
2
e
−
2
t
+3
te
−
2
t
−
3
e
−
2
t
e
and
X
p
=
ΦU
=
L
1
/
2
−
2
e
e
−
t
.
10.
From
X

=
L
32
−
2
−
1
e
X
+
L
1
1
e
we obtain
X
c
=
c
1
L
1
−
1
e
e
t
+
c
2
3L
1
−
1
e
te
t
+
L
0
1
/
2
e
e
t

.
Then
Φ
=
L
e
t
te
t
−
e
t
1
2
e
t
−
te
t
e
and
Φ
−
1
=
L
e
−
t
−
2
te
−
t
−
2
te
−
t
2
e
−
t
2
e
−
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
e
−
t
−
4
te
−
t
2
e
−
t
e
dt
=
L
3
e
−
t
+4
te
−
t
−
2
e
−
t
e
and
X
p
=
ΦU
=
L
3
−
5
e
.
11.
From
X

=
L
0
−
1
10
e
X
+
L
sec
t
0
e
512

513. Exercises 10.4
we obtain
X
c
=
c
1
L
cos
t
sin
t
e
+
c
2
L
sin
t
−
cos
t
e
.
Then
Φ
=
L
cos
t
sin
t
sin
t
−
cos
t
e
and
Φ
−
1
=
L
cos
t
sin
t
sin
t
−
cos
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
1
tan
t
e
dt
=
L
t
ln
|
sec
t
|
e
and
X
p
=
ΦU
=
L
t
cos
t
+ sin
t
ln
|
sec
t
|
t
sin
t
−
cos
t
ln
|
sec
t
|
e
.
12.
From
X

=
L
1
−
1
11
e
X
+
L
3
3
e
e
t
we obtain
X
c
=
c
1
L
−
sin
t
cos
t
e
e
t
+
c
2
L
cos
t
sin
t
e
e
t
.
Then
Φ
=
L
−
sin
t
cos
t
cos
t
sin
t
e
e
t
and
Φ
−
1
=
L
−
sin
t
cos
t
cos
t
sin
t
e
e
−
t
so that
U
=
+
Φ
−
1
F
dt
=
+
L
−
3 sin
t
+ 3 cos
t
3 cos
t
+ 3 sin
t
e
dt
=
L
3 cos
t
+ 3 sin
t
3 sin
t
−
3 cos
t
e
and
X
p
=
ΦU
=
L
−
3
3
e
e
t
.
13.
From
X

=
L
1
−
1
11
e
X
+
L
cos
t
sin
t
e
e
t
we obtain
X
c
=
c
1
L
−
sin
t
cos
t
e
e
t
+
c
2
L
cos
t
sin
t
e
e
t
.
Then
Φ
=
L
−
sin
t
cos
t
cos
t
sin
t
e
e
t
and
Φ
−
1
=
L
−
sin
t
cos
t
cos
t
sin
t
e
e
−
t
so that
U
=
+
Φ
−
1
F
dt
=
+
L
0
1
e
dt
=
L
0
t
e
and
X
p
=
ΦU
=
L
cos
t
sin
t
e
te
t
.
14.
From
X

=
L
2
−
2
8
−
6
e
X
+
L
1
3
e
1
t
e
−
2
t
we obtain
X
c
=
c
1
L
1
2
e
e
−
2
t
+
c
2
3L
1
2
e
te
−
2
t
+
L
1
/
2
1
/
2
e
e
−
2
t

.
Then
Φ
=
L
12
t
+1
24
t
+1
e
e
−
2
t
and
Φ
−
1
=
L
−
(4
t
+1) 2
t
+1
2
−
1
e
e
2
t
513

514. Exercises 10.4
so that
U
=
+
Φ
−
1
F
dt
=
+
L
2
t
+2ln
t
−
ln
t
e
dt
and
X
p
=
ΦU
=
L
2
t
+ln
t
−
2
t
ln
t
4
t
+3ln
t
−
4
t
ln
t
e
e
−
2
t
.
15.
From
X

=
L
01
−
10
e
X
+
L
0
sec
t
tan
t
e
we obtain
X
c
=
c
1
L
cos
t
−
sin
t
e
+
c
2
L
sin
t
cos
t
e
.
Then
Φ
=
L
cos
t
sin
t
−
sin
t
cos
t
e
t
and
Φ
−
1
=
L
cos
t
−
sin
t
sin
t
cos
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
−
tan
2
t
tan
t
e
dt
=
L
t
−
tan
t
ln
|
sec
t
|
e
and
X
p
=
ΦU
=
L
cos
t
−
sin
t
e
t
+
L
−
sin
t
sin
t
tan
t
e
+
L
sin
t
cos
t
e
ln
|
sec
t
|
.
16.
From
X

=
L
01
−
10
e
X
+
L
1
cot
t
e
we obtain
X
c
=
c
1
L
cos
t
−
sin
t
e
+
c
2
L
sin
t
cos
t
e
.
Then
Φ
=
L
cos
t
sin
t
−
sin
t
cos
t
e
and
Φ
−
1
=
L
cos
t
−
sin
t
sin
t
cos
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
0
csc
t
e
dt
=
L
0
ln
|
csc
t
−
cot
t
|
e
and
X
p
=
ΦU
=
L
sin
t
ln
|
csc
t
−
cot
t
|
cos
t
ln
|
csc
t
−
cot
t
|
e
.
17.
From
X

=
L
12
−
1
/
21
e
X
+
L
csc
t
sec
t
e
e
t
we obtain
X
c
=
c
1
L
2 sin
t
cos
t
e
e
t
+
c
2
L
2 cos
t
−
sin
t
e
e
t
.
Then
Φ
=
L
2 sin
t
2 cos
t
cos
t
−
sin
t
e
e
t
and
Φ
−
1
=
L
1
2
sin
t
cos
t
1
2
cos
t
−
sin
t
e
e
−
t
so that
U
=
+
Φ
−
1
F
dt
=
+
L
3
2
1
2
cos
t
−
tan
t
e
dt
=
L
3
2
t
1
2
ln
|
sin
t
|−
ln
|
sec
t
|
e
and
X
p
=
ΦU
=
L
3 sin
t
3
2
cos
t
e
te
t
+
L
cos
t
−
1
2
sin
t
e
ln
|
sin
t
|
+
L
−
2 cos
t
sin
t
e
ln
|
sec
t
|
.
514

515. Exercises 10.4
18.
From
X

=
L
1
−
2
1
−
1
e
X
+
L
tan
t
1
e
we obtain
X
c
=
c
1
L
cos
t
−
sin
t
cos
t
e
+
c
2
L
cos
t
+ sin
t
sin
t
e
.
Then
Φ
=
L
cos
t
−
sin
t
cos
t
+ sin
t
cos
t
sin
t
e
and
Φ
−
1
=
L
−
sin
t
cos
t
+ sin
t
cos
t
sin
t
−
cos
t
e
so that
U
=
+
Φ
−
1
F
dt
=
+
L
2 cos
t
+ sin
t
−
sec
t
2 sin
t
−
cos
t
e
dt
=
L
2 sin
t
−
cos
t
−
ln
|
sec
t
+ tan
t
|
−
2 cos
t
−
sin
t
e
and
X
p
=
ΦU
=
L
3 sin
t
cos
t
−
cos
2
t
−
2 sin
2
t
+ (sin
t
−
cos
t
)ln
|
sec
t
+ tan
t
|
sin
2
t
−
cos
2
t
−
cos
t
(ln
|
sec
t
+ tan
t
|
)
e
.
19.
From
X

=



110
110
003



X
+



e
t
e
2
t
te
3
t



we obtain
X
c
=
c
1



1
−
1
0



+
c
2



1
1
0



e
2
t
+
c
3



0
0
1



e
3
t
.
Then
Φ
=



1
e
2
t
0
−
1
e
2
t
0
00
e
3
t



and
Φ
−
1
=



1
2
−
1
2
0
1
2
e
−
2
t
1
2
e
−
2
t
0
00
e
−
3
t



so that
U
=
+
Φ
−
1
F
dt
=
+



1
2
e
t
−
1
2
e
2
t
1
2
e
−
t
+
1
2
t



dt
=



1
2
e
t
−
1
4
e
2
t
−
1
2
e
−
t
+
1
2
t
1
2
t
2



and
X
p
=
ΦU
=



−
1
4
e
2
t
+
1
2
te
2
t
−
e
t
+
1
4
e
2
t
+
1
2
te
2
t
1
2
t
2
e
3
t



.
20.
From
X

=



3
−
1
−
1
11
−
1
1
−
11



X
+



0
t
2
e
t



we obtain
X
c
=
c
1



1
1
1



e
t
+
c
2



1
1
0



e
2
t
+
c
3



1
0
1



e
2
t
.
Then
Φ
=



e
t
e
2
t
e
2
t
e
t
e
2
t
0
e
t
0
e
2
t



and
Φ
−
1
=



−
e
−
t
e
−
t
e
−
t
e
−
2
t
0
−
e
−
2
t
e
−
2
t
−
e
−
2
t
0



515

516. Exercises 10.4
so that
U
=
+
Φ
−
1
F
dt
=
+



te
−
t
+2
−
2
e
−
t
−
te
−
2
t



dt
=



−
te
−
t
−
e
−
t
+2
t
2
e
−
t
1
2
te
−
2
t
+
1
4
e
−
2
t



and
X
p
=
ΦU
=



−
1
/
2
−
1
−
1
/
2



t
+



−
3
/
4
−
1
−
3
/
4



+



2
2
0



e
t
+



2
2
2



te
t
.
21.
From
X

=
L
3
−
1
−
13
e
X
+
L
4
e
2
t
4
e
4
t
e
we obtain
Φ
=
L
−
e
4
t
e
2
t
e
4
t
e
2
t
e
,
Φ
−
1
=
s
−
1
2
e
−
4
t
1
2
e
−
4
t
1
2
e
−
2
t
1
2
e
−
2
t
i
,
X
=
ΦΦ
−
1
(0)
X
(0) +
Φ
+
t
0
Φ
−
1
F
ds
=
Φ
·
L
0
1
e
+
Φ
·
L
e
−
2
t
+2
t
−
1
e
2
t
+2
t
−
1
e
and
=
L
2
2
e
te
2
t
+
L
−
1
1
e
e
2
t
+
L
−
2
2
e
te
4
t
+
L
2
0
e
e
4
t
.
22.
From
X

=
L
1
−
1
1
−
1
e
X
+
L
1
/t
1
/t
e
we obtain
Φ
=
L
11+
t
1
t
e
,
Φ
−
1
=
L
−
t
1+
t
1
−
1
e
,
and
X
=
ΦΦ
−
1
(1)
X
(1) +
Φ
+
t
1
Φ
−
1
F
ds
=
Φ
·
L
−
4
3
e
+
Φ
·
L
ln
t
0
e
=
L
3
3
e
t
−
L
1
4
e
+
L
1
1
e
ln
t.
23.
λ
1
=
−
1,
λ
2
=
−
2,
K
1
=

1
3

,
K
2
=

2
7

,
P
=

12
37

,
P
−
1
=

7
−
2
−
31

,
P
−
1
F
=

34
−
14

;
Y

=

−
10
0
−
2

Y
+

34
−
14

y
1
=34+
c
1
e
−
t
,y
2
=
−
7+
c
2
e
−
2
t
X
=
PY
=

12
37

34 +
c
1
e
−
t
−
7+
c
2
e
−
2
t

=

20 +
c
1
e
−
t
+2
c
2
e
−
2
t
53 + 3
c
1
e
−
t
+7
c
2
e
−
2
t

=
c
1

1
3

e
−
t
+
c
2

2
7

e
−
2
t
+

20
53

24.
λ
1
=
−
1,
λ
2
=4,
K
1
=

3
−
2

,
K
2
=

1
1

,
P
=

31
−
21

,
P
−
1
=
1
5

1
−
1
23

,
P
−
1
F
=

0
e
t

;
Y

=

−
10
04

Y
+

0
e
t

y
1
=
c
1
e
−
t
,y
2
=
−
1
3
e
t
+
c
2
e
4
t
X
=
PY
=

31
−
21

c
1
e
−
t
−
1
3
e
t
+
c
2
e
4
t

=

−
1
3
e
t
+3
c
1
e
−
t
+
c
2
e
4
t
−
1
3
e
t
−
2
c
1
e
−
t
+
c
2
e
4
t

=
c
1

3
−
2

e
−
t
+
c
2

1
1

e
4
t
−
1
3

1
1

e
t
516

517. Exercises 10.4
25.
λ
1
=0,
λ
2
= 10,
K
1
=

1
−
1

,
K
2
=

1
1

,
P
=

11
−
11

,
P
−
1
=
1
2

1
−
1
11

,
P
−
1
F
=

t
−
4
t
+4

;
Y

=

00
010

Y
+

t
−
4
t
+4

y
1
=
1
2
t
2
−
4
t
+
c
1
,y
2
=
−
1
10
t
−
41
100
+
c
2
e
10
t
X
=
PY
=

11
−
11

c
1
2
t
2
−
4
t
+
c
1
−
1
10
t
−
41
100
+
c
2
e
10
t
o
=

1
2
t
2
−
41
10
t
−
41
100
+
c
1
+
c
2
e
10
t
−
1
2
t
2
+
39
10
t
−
41
100
−
c
1
+
c
2
e
10
t

=
c
1

1
−
1

+
c
2

1
1

e
10
t
+
1
2

1
−
1

t
2
+
1
10

−
41
39

t
−
41
100

1
1

26.
λ
1
=
−
1,
λ
2
=1,
K
1
=

1
−
1

,
K
2
=

1
1

,
P
=

11
−
11

,
P
−
1
=
1
2

1
−
1
11

,
P
−
1
F
=

2
−
4
e
−
2
t
2+4
e
−
2
t

;
Y

=

−
10
01

Y
+

2
−
4
e
−
2
t
2+4
e
−
2
t

y
1
=2+4
e
−
2
t
+
c
1
e
−
t
,y
2
=
−
2
−
4
3
e
−
2
t
+
c
2
e
t
X
=
PY
=

11
−
11

2+4
e
−
2
t
+
c
1
e
−
t
−
2
−
4
3
e
−
2
t
+
c
2
e
t

=

8
3
e
−
2
t
+
c
1
e
−
t
+
c
2
e
t
−
4
−
16
3
e
−
2
t
−
c
1
e
−
t
+
c
2
e
t

=
c
1

1
−
1

e
−
t
+
c
2

1
1

e
t
+
8
3

1
−
2

e
−
2
t
+

0
−
4

27.
Let
I
=
L
i
1
i
2
e
so that
I

=
L
−
11 3
3
−
3
e
I
+
L
100 sin
t
0
e
and
X
c
=
c
1
L
1
3
e
e
−
2
t
+
c
2
L
3
−
1
e
e
−
12
t
.
Then
Φ
=
L
e
−
2
t
3
e
−
12
t
3
e
−
2
t
−
e
−
12
t
e
,
Φ
−
1
=
s
1
10
e
2
t
3
10
e
2
t
3
10
e
12
t
−
1
10
e
12
t
i
,
U
=
+
Φ
−
1
F
dt
=
+
L
10
e
2
t
sin
t
30
e
12
t
sin
t
e
dt
=
L
2
e
2
t
(2 sin
t
−
cos
t
)
6
29
e
12
t
(12 sin
t
−
cos
t
)
e
,
and
I
p
=
ΦU
=
s
332
29
sin
t
−
76
29
cos
t
276
29
sin
t
−
168
29
cos
t
i
so that
I
=
c
1
L
1
3
e
e
−
2
t
+
c
2
L
3
−
1
e
e
−
12
t
+
I
p
.
If
I
(0) =
L
0
0
e
then
c
1
= 2 and
c
2
=
6
29
.
28.
For the system
d
dt

i
1
q

=
c
−
20
3
−
50
3
1
3
−
5
3
o

i
1
q

+

2
0

517

518. Exercises 10.4
we find
λ
1
=
−
5,
λ
2
=
−
10
3
,
K
1
=

10
−
1

,
K
2
=

5
−
1

,
X
p
=
c
1
5
1
25
o
and so
X
=

i
1
q

=
c
1

10
−
1

e
−
5
t
+
c
2

5
−
1

e
−
10
t/
3
+
1
25

5
1

.
29.
Let
I
=

i
2
i
3

so that
I

=

−
2
−
2
−
2
−
5

I
+

60
60

and
X
c
=
c
1

2
−
1

e
−
t
+
c
2

1
2

e
−
6
t
.
If
X
p
=

a
1
b
1

then
X
p
=

30
0

so that
X
=
c
1

2
−
1

e
−
t
+
c
2

1
2

e
−
6
t
+

30
0

.For
I
(0) =

0
0

we find
c
1
=
−
12 and
c
2
=
−
6. Thus
i
1
(
t
)=
i
2
(
t
)+
i
3
(
t
)=
−
12
e
−
t
−
18
e
−
6
t
+30
.
30. (a)
The eigenvalues are 0, 1, 3, and 4, with corresponding eigenvectors





−
6
−
4
1
2





,





2
1
0
0





,





3
1
2
1





,
and





−
1
1
0
0





.
(b) Φ
=





−
62
e
t
3
e
3
t
−
e
4
t
−
4
e
t
e
3
t
e
4
t
102
e
3
t
0
20
e
3
t
0





,
Φ
−
1
=





00
−
1
3
2
3
1
3
e
−
t
1
3
e
−
t
−
2
e
−
t
8
3
e
−
t
00
2
3
e
−
3
t
−
1
3
e
−
3
t
−
1
3
e
−
4
t
2
3
e
−
4
t
0
1
3
e
−
4
t





(c) Φ
−
1
(
t
)
F
(
t
)=





2
3
−
1
3
e
2
t
1
3
e
−
2
t
+
8
3
e
−
t
−
2
e
t
+
1
3
t
−
1
3
e
−
3
t
+
2
3
e
−
t
2
3
e
−
5
t
+
1
3
e
−
4
t
−
1
3
te
−
3
t





,
+
Φ
−
1
(
t
)
F
(
t
)
dt
=





−
1
6
e
2
t
+
2
3
t
−
1
6
e
−
2
t
−
8
3
e
−
t
−
2
e
t
+
1
6
t
2
1
9
e
−
3
t
−
2
3
e
−
t
−
2
15
e
−
5
t
−
1
12
e
−
4
t
+
1
27
e
−
3
t
+
1
9
te
−
3
t





,
Φ
(
t
)
+
Φ
−
1
(
t
)
F
(
t
)
dt
=





−
5
e
2
t
−
1
5
e
−
t
−
1
27
e
t
−
1
9
te
t
+
1
3
t
2
e
t
−
4
t
−
59
12
−
2
e
2
t
−
3
10
e
−
t
+
1
27
e
t
+
1
9
te
t
+
1
6
t
2
e
t
−
8
3
t
−
95
36
−
3
2
e
2
t
+
2
3
t
+
2
9
−
e
2
t
+
4
3
t
−
1
9





,
Φ
(
t
)
C
=





−
6
c
1
+2
c
2
e
t
+3
c
3
e
3
t
−
c
4
e
4
t
−
4
c
1
+
c
2
e
t
+
c
3
e
3
t
+
c
4
e
4
t
c
1
+2
c
3
e
3
t
2
c
1
+
c
3
e
3
t





,
518

519. Exercises 10.5
Φ
(
t
)
C
+
Φ
(
t
)
+
Φ
−
1
(
t
)
F
(
t
)
dt
=





−
6
c
1
+2
c
2
e
t
+3
c
3
e
3
t
−
c
4
e
4
t
−
4
c
1
+
c
2
e
t
+
c
3
e
3
t
+
c
4
e
4
t
c
1
+2
c
3
e
3
t
2
c
1
+
c
3
e
3
t





+





−
5
e
2
t
−
1
5
e
−
t
−
1
27
e
t
−
1
9
te
t
+
1
3
t
2
e
t
−
4
t
−
59
12
−
2
e
2
t
−
3
10
e
−
t
+
1
27
e
t
+
1
9
te
t
+
1
6
t
2
e
t
−
8
3
t
−
95
36
−
3
2
e
2
t
+
2
3
t
+
2
9
−
e
2
t
+
4
3
t
−
1
9





(d) X
(
t
)=
c
1





−
6
−
4
1
2





+
c
2





2
1
0
0





+
c
3





3
1
2
1





+
c
4





−
1
1
0
0





+





−
5
e
2
t
−
1
5
e
−
t
−
1
27
e
t
−
1
9
te
t
+
1
3
t
2
e
t
−
4
t
−
59
12
−
2
e
2
t
−
3
10
e
−
t
+
1
27
e
t
+
1
9
te
t
+
1
6
t
2
e
t
−
8
3
t
−
95
36
−
3
2
e
2
t
+
2
3
t
+
2
9
−
e
2
t
+
4
3
t
−
1
9





Exercises 10.5
1.
For
A
=
A
10
02
D
we have
A
2
=
A
10
02
DA
10
02
D
=
A
10
04
D
,
A
3
=
AA
2
=
A
10
02
DA
10
04
D
=
A
10
08
D
,
A
4
=
AA
3
=
A
10
02
DA
10
08
D
=
A
10
016
D
,
and so on. In general
A
k
=
A
10
02
k
D
for
k
=1
,
2
,
3
,....
Thus
e
A
t
=
I
+
A
1!
t
+
A
2
2!
t
2
+
A
3
3!
t
3
+
···
=
A
10
01
D
+
1
1!
A
10
02
D
t
+
1
2!
A
10
04
D
t
2
+
1
3!
A
10
08
D
t
3
+
···
=



1+
t
+
t
2
2!
+
t
3
3!
+
···
0
01+
t
+
(2
t
)
2
2!
+
(2
t
)
3
3!
+
···



=
A
e
t
0
0
e
2
t
D
and
e
−
A
t
=
A
e
−
t
0
0
e
−
2
t
D
.
519

520. Exercises 10.5
2.
For
A
=
A
01
10
D
we have
A
2
=
A
01
10
DA
01
10
D
=
A
10
01
D
=
I
A
3
=
AA
2
=
A
01
10
D
I
=
A
01
10
D
=
A
A
4
=(
A
2
)
2
=
I
A
5
=
AA
4
=
AI
=
A
and so on. In general
A
k
=
;
A
,k
=1,3,5,
...
I
,k
=2,4,6,
...
.
Thus
e
A
t
=
I
+
A
1!
t
+
A
2
2!
t
2
+
A
3
3!
t
3
+
···
=
I
+
A
t
+
1
2!
I
t
2
+
1
3!
A
t
3
+
···
=
I
A
1+
1
2!
t
2
+
1
4!
t
4
+
···
D
+
A
A
t
+
1
3!
t
3
+
1
5!
t
5
+
···
D
=
I
cosh
t
+
A
sinh
t
=
A
cosh
t
sinh
t
sinh
t
cosh
t
D
and
e
−
A
t
=
A
cosh(
−
t
) sinh(
−
t
)
sinh(
−
t
) cosh(
−
t
)
D
=
A
cosh
t
−
sinh
t
−
sinh
t
cosh
t
D
.
3.
For
A
=



111
111
−
2
−
2
−
2



we have
A
2
=



111
111
−
2
−
2
−
2






111
111
−
2
−
2
−
2



=



000
000
000



.
Thus,
A
3
=
A
4
=
A
5
=
···
=
0
and
e
A
t
=
I
+
A
t
=



100
010
001



+



ttt
ttt
−
2
t
−
2
t
−
2
t



=



t
+1
tt
tt
+1
t
−
2
t
−
2
t
−
2
t
+1



.
4.
For
A
=



000
300
510



we have
520

521. Exercises 10.5
A
2
=



000
300
510






000
300
510



=



000
000
300



A
3
=
AA
2
=



000
300
510






000
000
300



=



000
000
000



.
Thus,
A
4
=
A
5
=
A
6
=
···
=
0
and
e
A
t
=
I
+
A
t
+
1
2
A
2
t
2
=



100
010
001



+



000
3
t
00
5
tt
0



+



000
000
3
2
t
2
00



=



100
3
t
10
3
2
t
2
+5
tt
1



.
5.
Using the result of Problem 1
X
=
A
e
t
0
0
e
2
t
DA
c
1
c
2
D
=
c
1
A
e
t
0
D
+
c
2
A
0
e
t
D
.
6.
Using the result of Problem 2
X
=
A
cosh
t
sinh
t
sinh
t
cosh
t
DA
c
1
c
2
D
=
c
1
A
cosh
t
sinh
t
D
+
c
2
A
sinh
t
cosh
t
D
.
7.
Using the result of Problem 3
X
=



t
+1
tt
tt
+1
t
−
2
t
−
2
t
−
2
t
+1






c
1
c
2
c
3



=
c
1



t
+1
t
−
2
t



+
c
2



t
t
+1
−
2
t



+
c
3



t
t
−
2
t
+1



.
8.
Using the result of Problem 4
X
=



100
3
t
10
3
2
t
2
+5
tt
1






c
1
c
2
c
3



=
c
1



1
3
t
3
2
t
2
+5
t



+
c
2



0
1
t



+
c
3



0
0
1



.
9.
To solve
X

=
A
10
02
D
X
+
A
3
−
1
D
we identify
t
0
=0,
F
(
s
)=
A
3
−
1
D
, and use the results of Problem 1 and equation (6) in the text.
X
(
t
)=
e
A
t
C
+
e
A
t
+
t
t
0
e
−
A
s
F
(
s
)
ds
=
A
e
t
0
0
e
2
t
DA
c
1
c
2
D
+
A
e
t
0
0
e
2
t
D
+
t
0
A
e
−
s
0
0
e
−
2
s
DA
3
−
1
D
ds
=
A
c
1
e
t
c
2
e
2
t
D
+
A
e
t
0
0
e
2
t
D
+
t
0
A
3
e
−
s
−
e
−
2
s
D
ds
521

522. Exercises 10.5
=
A
c
1
e
t
c
2
e
2
t
D
+
A
e
t
0
0
e
2
t
DA
−
3
e
−
s
1
2
e
−
2
s
D
(
(
(
(
t
0
=
A
c
1
e
t
c
2
e
2
t
D
+
A
e
t
0
0
e
2
t
DA
−
3
e
−
t
−
3
1
2
e
−
2
t
−
1
2
D
=
A
c
1
e
t
c
2
e
2
t
D
+
A
−
3
−
3
e
t
1
2
−
1
2
e
2
t
D
=
c
3
A
1
0
D
e
t
+
c
4
A
0
1
D
e
2
t
+
A
−
3
1
2
D
.
10.
To solve
X

=
A
10
02
D
X
+
A
t
e
4
t
D
we identify
t
0
=0,
F
(
s
)=
A
t
e
4
t
D
, and use the results of Problem 1 and equation (6) in the text.
X
(
t
)=
e
A
t
C
+
e
A
t
+
t
t
0
e
−
A
s
F
(
s
)
ds
=
A
e
t
0
0
e
2
t
DA
c
1
c
2
D
+
A
e
t
0
0
e
2
t
D
+
t
0
A
e
−
s
0
0
e
−
2
s
DA
s
e
4
s
D
ds
=
A
c
1
e
t
c
2
e
2
t
D
+
A
e
t
0
0
e
2
t
D
+
t
0
A
se
−
s
e
2
s
D
ds
=
A
c
1
e
t
c
2
e
2
t
D
+
A
e
t
0
0
e
2
t
DA
−
se
−
s
−
e
−
s
1
2
e
2
s
D
(
(
(
(
t
0
=
A
c
1
e
t
c
2
e
2
t
D
+
A
e
t
0
0
e
2
t
DA
−
te
−
t
−
e
−
t
+1
1
2
e
2
t
−
1
2
D
=
A
c
1
e
t
c
2
e
2
t
D
+
A
−
t
−
1+
e
t
1
2
e
4
t
−
1
2
e
2
t
D
=
c
3
A
1
0
D
e
t
+
c
4
A
0
1
D
e
2
t
+
A
−
t
−
1
1
2
e
4
t
D
.
11.
To solve
X

=
A
01
10
D
X
+
A
1
1
D
we identify
t
0
=0,
F
(
s
)=
A
1
1
D
, and use the results of Problem 2 and equation (6) in the text.
X
(
t
)=
e
A
t
C
+
e
A
t
+
t
t
0
e
−
A
s
F
(
s
)
ds
=
A
cosh
t
sinh
t
sinh
t
cosh
t
DA
c
1
c
2
D
+
A
cosh
t
sinh
t
sinh
t
cosh
t
D
+
t
0
A
cosh
s
−
sinh
s
−
sinh
s
cosh
s
DA
1
1
D
ds
=
A
c
1
cosh
t
+
c
2
sinh
t
c
1
sinh
t
+
c
2
cosh
t
D
+
A
cosh
t
sinh
t
sinh
t
cosh
t
D
+
t
0
A
cosh
s
−
sinh
s
−
sinh
s
cosh
s
D
ds
=
A
c
1
cosh
t
+
c
2
sinh
t
c
1
sinh
t
+
c
2
cosh
t
D
+
A
cosh
t
sinh
t
sinh
t
cosh
t
DA
sinh
s
−
cosh
s
−
cosh
s
+ sinh
s
D
(
(
(
(
t
0
522

523. Exercises 10.5
=
A
c
1
cosh
t
+
c
2
sinh
t
c
1
sinh
t
+
c
2
cosh
t
D
+
A
cosh
t
sinh
t
sinh
t
cosh
t
DA
sinh
t
−
cosh
t
−
cosh
t
+ sinh
t
D
=
A
c
1
cosh
t
+
c
2
sinh
t
c
1
sinh
t
+
c
2
cosh
t
D
+
A
sinh
2
t
−
cosh
2
t
sinh
2
t
−
cosh
2
t
D
=
c
1
A
cosh
t
sinh
t
D
+
c
2
A
sinh
t
cosh
t
D
−
A
1
1
D
.
12.
To solve
X

=
A
01
10
D
X
+
A
cosh
t
sinh
t
D
we identify
t
0
=0,
F
(
s
)=
A
cosh
t
sinh
t
D
, and use the results of Problem 2 and equation (6) in the text.
X
(
t
)=
e
A
t
C
+
e
A
t
+
t
t
0
e
−
A
s
F
(
s
)
ds
=
A
cosh
t
sinh
t
sinh
t
cosh
t
DA
c
1
c
2
D
+
A
cosh
t
sinh
t
sinh
t
cosh
t
D
+
t
0
A
cosh
s
−
sinh
s
−
sinh
s
cosh
s
DA
cosh
s
sinh
s
D
ds
=
A
c
1
cosh
t
+
c
2
sinh
t
c
1
sinh
t
+
c
2
cosh
t
D
+
A
cosh
t
sinh
t
sinh
t
cosh
t
D
+
t
0
A
1
0
D
ds
=
A
c
1
cosh
t
+
c
2
sinh
t
c
1
sinh
t
+
c
2
cosh
t
D
+
A
cosh
t
sinh
t
sinh
t
cosh
t
DA
s
0
D
(
(
(
(
t
0
=
A
c
1
cosh
t
+
c
2
sinh
t
c
1
sinh
t
+
c
2
cosh
t
D
+
A
cosh
t
sinh
t
sinh
t
cosh
t
DA
t
0
D
=
A
c
1
cosh
t
+
c
2
sinh
t
c
1
sinh
t
+
c
2
cosh
t
D
+
A
t
cosh
t
t
sinh
t
D
=
c
1
A
cosh
t
sinh
t
D
+
c
2
A
sinh
t
cosh
t
D
+
t
A
cosh
t
sinh
t
D
.
13.
We have
X
(0) =
c
1



1
0
0



+
c
2



0
1
0



+
c
3



0
0
1



=



c
1
c
2
c
3



=



1
−
4
6



.
Thus, the solution of the initial-value problem is
X
=



t
+1
t
−
2
t



−
4



t
t
+1
−
2
t



+6



t
t
−
2
t
+1



.
14.
We have
X
(0) =
c
3
A
1
0
D
+
c
4
A
0
1
D
+
A
−
3
1
2
D
=
A
c
3
−
3
c
4
+
1
2
D
=
A
4
3
D
.
Thus,
c
3
= 7 and
c
4
=
5
2
,so
X
=7
A
1
0
D
e
t
+
5
2
A
0
1
D
e
2
t
+
A
−
3
1
2
D
.
15.
From
s
I
−
A
=
A
s
−
4
−
3
4
s
+4
D
we find
(
s
I
−
A
)
−
1
=



3
/
2
s
−
2
−
1
/
2
s
+2
3
/
4
s
−
2
−
3
/
4
s
+2
−
1
s
−
2
+
1
s
+2
−
1
/
2
s
−
2
+
3
/
2
s
+2



523

524. Exercises 10.5
and
e
A
t
=
s
3
2
e
2
t
−
1
2
e
−
2
t
3
4
e
2
t
−
3
4
e
−
2
t
−
e
2
t
+
e
−
2
t
−
1
2
e
2
t
+
3
2
e
−
2
t
i
.
The general solution of the system is then
X
=
e
A
t
C
=
s
3
2
e
2
t
−
1
2
e
−
2
t
3
4
e
2
t
−
3
4
e
−
2
t
−
e
2
t
+
e
−
2
t
−
1
2
e
2
t
+
3
2
e
−
2
t
i
A
c
1
c
2
D
=
c
1
A
3
/
2
−
1
D
e
2
t
+
c
1
A
−
1
/
2
1
D
e
−
2
t
+
c
2
A
3
/
4
−
1
/
2
D
e
2
t
+
c
2
A
−
3
/
4
3
/
2
D
e
−
2
t
=
)
1
2
c
1
+
1
4
c
2
S
A
3
−
2
D
e
2
t
+
)
−
1
2
c
1
−
3
4
c
2
S
A
1
−
2
D
e
−
2
t
=
c
3
A
3
−
2
D
e
2
t
+
c
4
A
1
−
2
D
e
−
2
t
.
16.
From
s
I
−
A
=
A
s
−
42
−
1
s
−
1
D
we find
(
s
I
−
A
)
−
1
=



2
s
−
3
−
1
s
−
2
−
2
s
−
3
+
2
s
−
2
1
s
−
3
−
1
s
−
2
−
1
s
−
3
+
2
s
−
2



and
e
A
t
=
A
2
e
3
t
−
e
2
t
−
2
e
3
t
+2
e
2
t
e
3
t
−
e
2
t
−
e
3
t
+2
e
2
t
D
.
The general solution of the system is then
X
=
e
A
t
C
=
A
2
e
3
t
−
e
2
t
−
2
e
3
t
+2
e
2
t
e
3
t
−
e
2
t
−
e
3
t
+2
e
2
t
DA
c
1
c
2
D
=
c
1
A
2
1
D
e
3
t
+
c
1
A
−
1
−
1
D
e
2
t
+
c
2
A
−
2
−
1
D
e
3
t
+
c
2
A
2
2
D
e
2
t
=(
c
1
−
c
2
)
A
2
1
D
e
3
t
+(
−
c
1
+2
c
2
)
A
1
1
D
e
2
t
=
c
3
A
2
1
D
e
3
t
+
c
4
A
1
1
D
e
2
t
.
17.
From
s
I
−
A
=
A
s
−
59
−
1
s
+1
D
we find
(
s
I
−
A
)
−
1
=



1
s
−
2
+
3
(
s
−
2)
2
−
9
(
s
−
2)
2
1
(
s
−
2)
2
1
s
−
2
−
3
(
s
−
2)
2



and
e
A
t
=
A
e
2
t
+3
te
2
t
−
9
te
2
t
te
2
t
e
2
t
−
3
te
2
t
D
.
The general solution of the system is then
524

525. Exercises 10.5
X
=
e
A
t
C
=
A
e
2
t
+3
te
2
t
−
9
te
2
t
te
2
t
e
2
t
−
3
te
2
t
DA
c
1
c
2
D
=
c
1
A
1
0
D
e
2
t
+
c
1
A
3
1
D
te
2
t
+
c
2
A
0
1
D
e
2
t
+
c
2
A
−
9
−
3
D
te
2
t
=
c
1
A
1+3
t
t
D
e
2
t
+
c
2
A
−
9
t
1
−
3
t
D
e
2
t
.
18.
From
s
I
−
A
=
A
s
−
1
2
s
+2
D
we find
(
s
I
−
A
)
−
1
=



s
+1+1
(
s
+1)
2
+1
1
(
s
+1)
2
+1
−
2
(
s
+1)
2
+1
s
+1+1
(
s
+1)
2
+1



and
e
A
t
=
A
e
−
t
cos
t
+
e
−
t
sin
te
−
t
sin
t
−
2
e
−
t
sin
te
−
t
cos
t
−
e
−
t
sin
t
D
.
The general solution of the system is then
X
=
e
A
t
C
=
A
e
−
t
cos
t
+
e
−
t
sin
te
−
t
sin
t
−
2
e
−
t
sin
te
−
t
cos
t
−
e
−
t
sin
t
DA
c
1
c
2
D
=
c
1
A
1
0
D
e
−
t
cos
t
+
c
1
A
1
−
2
D
e
−
t
sin
t
+
c
2
A
0
1
D
e
−
t
cos
t
+
c
2
A
1
−
1
D
e
−
t
sin
t
=
c
1
A
cos
t
+ sin
t
−
2 sin
t
D
e
−
t
+
c
2
A
sin
t
cos
t
−
sin
t
D
e
−
t
.
19.
The eigenvalues are
λ
1
= 1 and
λ
2
= 6. This leads to the system
e
t
=
b
0
+
b
1
e
6
t
=
b
0
+6
b
1
,
which has the solution
b
0
=
6
5
e
t
−
1
5
e
6
t
and
b
1
=
−
1
5
e
t
+
1
5
e
6
t
. Then
e
A
t
=
b
0
I
+
b
1
A
=
s
4
5
e
t
+
1
5
e
6
t
2
5
e
t
−
2
5
e
6
t
2
5
e
t
−
2
5
e
6
t
1
5
e
t
+
4
5
e
6
t
i
.
The general solution of the system is then
X
=
e
A
t
C
=
s
4
5
e
t
+
1
5
e
6
t
2
5
e
t
−
2
5
e
6
t
2
5
e
t
−
2
5
e
6
t
1
5
e
t
+
4
5
e
6
t
i
A
c
1
c
2
D
=
c
1
A
4
/
5
2
/
5
D
e
t
+
c
1
A
1
/
5
−
2
/
5
D
e
6
t
+
c
2
A
2
/
5
1
/
5
D
e
t
+
c
2
A
−
2
/
5
4
/
5
D
e
6
t
=
)
2
5
c
1
+
1
5
c
2
S
A
2
1
D
e
t
+
)
1
5
c
1
−
2
5
c
2
S
A
1
−
2
D
e
6
t
=
c
3
A
2
1
D
e
t
+
c
4
A
1
−
2
D
e
6
t
.
525

526. Exercises 10.5
20.
The eigenvalues are
λ
1
= 2 and
λ
2
= 3. This leads to the system
e
2
t
=
b
0
+2
b
1
e
3
t
=
b
0
+3
b
1
,
which has the solution
b
0
=3
e
2
t
−
2
e
3
t
and
b
1
=
−
e
2
t
+
e
3
t
. Then
e
A
t
=
b
0
I
+
b
1
A
=
A
2
e
2
t
−
e
3
t
−
2
e
2
t
+2
e
3
t
e
2
t
−
e
3
t
−
e
2
t
+2
e
3
t
D
.
The general solution of the system is then
X
=
e
A
t
C
=
A
2
e
2
t
−
e
3
t
−
2
e
2
t
+2
e
3
t
e
2
t
−
e
3
t
−
e
2
t
+2
e
3
t
DA
c
1
c
2
D
=
c
1
A
2
1
D
e
2
t
+
c
1
A
−
1
−
1
D
e
3
t
+
c
2
A
−
2
−
1
D
e
2
t
+
c
2
A
2
2
D
e
3
t
=(
c
1
−
c
2
)
A
2
1
D
e
2
t
+(
−
c
1
+2
c
2
)
A
1
1
D
e
3
t
=
c
3
A
2
1
D
e
2
t
+
c
4
A
1
1
D
e
3
t
.
21.
The eigenvalues are
λ
1
=
−
1 and
λ
2
= 3. This leads to the system
e
−
t
=
b
0
−
b
1
e
3
t
=
b
0
+3
b
1
,
which has the solution
b
0
=
3
4
e
−
t
+
1
4
e
3
t
and
b
1
=
−
1
4
e
−
t
+
1
4
e
3
t
. Then
e
A
t
=
b
0
I
+
b
1
A
=
A
e
3
t
−
2
e
−
t
+2
e
3
t
0
e
−
t
D
.
The general solution of the system is then
X
=
e
A
t
C
=
s
e
3
t
−
2
e
−
t
+2
e
3
t
0
e
−
t
i
A
c
1
c
2
D
=
c
1
A
1
0
D
e
3
t
+
c
2
A
−
2
1
D
e
−
t
+
c
2
A
2
0
D
e
3
t
=
c
2
A
−
2
1
D
e
−
t
+(
c
1
+2
c
2
)
A
1
0
D
e
3
t
=
c
3
A
−
2
1
D
e
−
t
+
c
4
A
1
0
D
e
3
t
.
22.
The eigenvalues are
λ
1
=
1
4
and
λ
2
=
1
2
. This leads to the system
e
t/
4
=
b
0
+
1
4
b
1
e
t/
2
=
b
0
+
1
2
b
1
,
526

527. Exercises 10.5
which has the solution
b
0
=2
e
t/
4
+
e
t/
2
and
b
1
=
−
4
e
t/
4
+4
e
t/
2
. Then
e
A
t
=
b
0
I
+
b
1
A
=
A
−
2
e
t/
4
+3
e
t/
2
6
e
t/
4
−
6
e
t/
2
−
e
t/
4
+
e
t/
2
3
e
t/
4
−
2
e
t/
2
D
.
The general solution of the system is then
X
=
e
A
t
C
=
A
−
2
e
t/
4
+3
e
t/
2
6
e
t/
4
−
6
e
t/
2
−
e
t/
4
+
e
t/
2
3
e
t/
4
−
2
e
t/
2
DA
c
1
c
2
D
=
c
1
A
−
2
−
1
D
e
t/
4
+
c
1
A
3
1
D
e
t/
2
+
c
2
A
6
3
D
e
t/
4
+
c
2
A
−
6
−
2
D
e
t/
2
=(
−
c
1
+3
c
2
)
A
2
1
D
e
t/
4
+(
c
1
−
2
c
2
)
A
3
1
D
e
t/
2
=
c
3
A
2
1
D
e
t/
4
+
c
4
A
3
1
D
e
t/
2
.
23.
From equation (2) in the text we have
e
D
t
=
I
+
t
D
+
t
2
2!
D
2
+
t
3
3!
D
3
+
···
so that
P
e
D
t
P
−
1
=
PP
−
1
+
t
(
PDP
−
1
)+
t
2
2!
(
PD
2
P
−
1
)+
t
3
3!
(
PD
3
P
−
1
)+
···
.
But
PP
−
1
+
I
,
PDP
−
1
=
A
and
PD
n
P
−
1
=
A
n
(see Problem 37, Exercises 8
.
12). Thus,
P
e
D
t
P
−
1
=
I
+
t
A
+
t
2
2!
A
2
+
t
3
3!
A
3
+
···
=
e
A
t
.
24.
From equation (2) in the text
e
D
t
=





10
···
0
01
···
0
.
.
.
.
.
.
.
.
.
.
.
.
00
···
1





+





λ
1
0
···
0
0
λ
2
···
0
.
.
.
.
.
.
.
.
.
.
.
.
00
···
λ
n





+
1
2!
t
2






λ
2
1
0
···
0
0
λ
2
2
···
0
.
.
.
.
.
.
.
.
.
.
.
.
00
···
λ
2
n






+
1
3!
t
3






λ
3
1
0
···
0
0
λ
3
2
···
0
.
.
.
.
.
.
.
.
.
.
.
.
00
···
λ
3
n






+
···
=






1+
λ
1
t
+
1
2!
(
λ
1
t
)
2
+
···
0
···
0
01+
λ
2
t
+
1
2!
(
λ
2
t
)
2
+
··· ···
0
.
.
.
.
.
.
.
.
.
.
.
.
00
···
1+
λ
n
t
+
1
2!
(
λ
n
t
)
2
+
···






=





e
λ
1
t
0
···
0
0
e
λ
2
t
···
0
.
.
.
.
.
.
.
.
.
.
.
.
00
···
e
λ
n
t





527

528. Exercises 10.5
25.
From Problems 23 and 24 and equation (1) in the text
X
=
e
A
t
C
=
P
e
D
t
P
−
1
C
=

e
3
t
e
5
t
e
3
t
3
e
5
t

e
3
t
0
0
e
5
t

c
3
2
e
−
3
t
−
1
2
e
−
3
t
−
1
2
e
−
5
t
1
2
e
−
5
t
o

c
1
c
2

=
c
3
2
e
3
t
−
1
2
e
5
t
−
1
2
e
3
t
+
1
2
e
5
t
3
2
e
3
t
−
3
2
e
5
t
−
1
2
e
3
t
+
3
2
e
5
t
o

c
1
c
2

.
26.
From Problems 23 and 24 and equation (1) in the text
X
=
e
A
t
C
=
P
e
D
t
P
−
1
C
=

−
e
t
e
3
t
e
t
e
3
t

e
t
0
0
e
3
t

c
−
1
2
e
−
t
1
2
e
−
t
1
2
e
3
t
1
2
e
−
3
t
o

c
1
c
2

=
c
1
2
e
t
+
1
2
e
9
t
−
1
2
e
t
+
1
2
e
3
t
−
1
2
e
t
+
1
2
e
9
t
1
2
e
t
+
1
2
e
3
t
o

c
1
c
2

.
27. (a)
The following commands can be used in
Mathematica
:
A=
{{
4,2
}
,
{
3,3
}}
;
c=
{
c1,c2
}
;
m=MatrixExp[A t];
sol=Expand[m.c]
Collect[sol,
{
c1,c2
}
] // MatrixForm
The output gives
x
(
t
)=
c
1
A
2
5
e
t
+
3
5
e
6
t
D
+
c
2
A
−
2
5
e
t
+
2
5
e
6
t
D
y
(
t
)=
c
1
A
−
3
5
e
t
+
3
5
e
6
t
D
+
c
2
A
3
5
e
t
+
2
5
e
6
t
D
.
The eigenvalues are 1 and 6 with corresponding eigenvectors
A
−
2
3
D
and
A
1
1
D
,
so the solution of the system is
X
(
t
)=
b
1
A
−
2
3
D
e
t
+
b
2
A
1
1
D
e
6
t
or
x
(
t
)=
−
2
b
1
e
t
+
b
2
e
6
t
y
(
t
)=3
b
1
e
t
+
b
2
e
6
t
.
If we replace
b
1
with
−
1
5
c
1
+
1
5
c
2
and
b
2
with
3
5
c
1
+
2
5
c
2
, we obtain the solution found using the matrix
exponential.
(b)
x
(
t
)=
c
1
e
−
2
t
cos
t
−
(
c
1
+
c
2
)
e
−
2
t
sin
t
y
(
t
)=
c
2
e
−
2
t
cos
t
+(2
c
1
+
c
2
)
e
−
2
t
sin
t
28.
x
(
t
)=
c
1
(3
e
−
2
t
−
2
e
−
t
)+
c
3
(
−
6
e
−
2
t
+6
e
−
t
)
y
(
t
)=
c
2
(4
e
−
2
t
−
3
e
−
t
)+
c
4
(4
e
−
2
t
−
4
e
−
t
)
z
(
t
)=
c
1
(
e
−
2
t
−
e
−
t
)+
c
3
(
−
2
e
−
2
t
+3
e
−
t
)
w
(
t
)=
c
2
(
−
3
e
−
2
t
+3
e
−
t
)+
c
4
(
−
3
e
−
2
t
+4
e
−
t
)
528

529. Chapter 10 Review Exercises
Chapter 10 Review Exercises
1.
If
X
=
k
L
4
5
e
, then
X

=
0
and
k
L
14
2
−
1
eL
4
5
e
−
L
8
1
e
=
k
L
24
3
e
−
L
8
1
e
=
L
0
0
e
.
We see that
k
=
1
3
.
2.
Solving for
c
1
and
c
2
we find
c
1
=
−
3
4
and
c
2
=
1
4
.
3.
Since



466
132
−
1
−
4
−
3






3
1
−
1



=



12
4
−
4



=4



3
1
−
1



,
we see that
λ
= 4 is an eigenvalue with eigenvector
K
3
. The corresponding solution is
X
=
K
3
e
4
t
.
4.
The other eigenvalue is
λ
2
=1
−
2
i
with corresponding eigenvector
K
2
=
L
1
−
i
e
. The general solution is
X
(
t
)=
c
1
L
cos 2
t
−
sin 2
t
e
e
t
+
c
2
L
sin 2
t
cos 2
t
e
e
t
.
5.
We have det(
A
−
λ
I
)=(
λ
−
1)
2
= 0 and
K
=
L
1
−
1
e
. A solution to (
A
−
λ
I
)
P
=
K
is
P
=
L
0
1
e
so that
X
=
c
1
L
1
−
1
e
e
t
+
c
2
3L
1
−
1
e
te
t
+
L
0
1
e
e
t

.
6.
We have det(
A
−
λ
I
)=(
λ
+ 6)(
λ
+ 2) = 0 so that
X
=
c
1
L
1
−
1
e
e
−
6
t
+
c
2
L
1
1
e
e
−
2
t
.
7.
We have det(
A
−
λ
I
)=
λ
2
−
2
λ
+5=0. For
λ
=1+2
i
we obtain
K
1
=
L
1
i
e
and
X
1
=
L
1
i
e
e
(1+2
i
)
t
=
L
cos 2
t
−
sin 2
t
e
e
t
+
i
L
sin 2
t
cos 2
t
e
e
t
.
Then
X
=
c
1
L
cos 2
t
−
sin 2
t
e
e
t
+
c
2
L
sin 2
t
cos 2
t
e
e
t
.
8.
We have det(
A
−
λ
I
)=
λ
2
−
2
λ
+2=0. For
λ
=1+
i
we obtain
K
1
=
L
3
−
i
2
e
and
X
1
=
L
3
−
i
2
e
e
(1+
i
)
t
=
L
3 cos
t
+ sin
t
2 cos
t
e
e
t
+
i
L
−
cos
t
+ 3 sin
t
2 sin
t
e
e
t
.
Then
X
=
c
1
L
3 cos
t
+ sin
t
2 cos
t
e
e
t
+
c
2
L
−
cos
t
+ 3 sin
t
2 sin
t
e
e
t
.
529

530. Chapter 10 Review Exercises
9.
We have det(
A
−
λ
I
)=
−
(
λ
−
2)(
λ
−
4)(
λ
+ 3) = 0 so that
X
=
c
1



−
2
3
1



e
2
t
+
c
2



0
1
1



e
4
t
+
c
3



7
12
−
16



e
−
3
t
.
10.
We have det(
A
−
λ
I
)=
−
(
s
+2)(
s
2
−
2
s
+3) = 0. The eigenvalues are
λ
1
=
−
2,
λ
2
=1+
√
2
i
, and
λ
2
=1
−
√
2
i
,
with eigenvectors
K
1
=



−
7
5
4



,
K
2
=



1
√
2
i/
2
1



,
and
K
3
=



1
−
√
2
i/
2
1



.
Thus
X
=
c
1



−
7
5
4



e
−
2
t
+
c
2






1
0
1



cos
√
2
t
−



0
√
2
/
2
0



sin
√
2
t



e
t
+
c
3






0
√
2
/
2
0



cos
√
2
t
+



1
0
1



sin
√
2
t



e
t
=
c
1



−
7
5
4



e
−
2
t
+
c
2



cos
√
2
t
−
√
2 sin
√
2
t/
2
cos
√
2
t



e
t
+
c
3



sin
√
2
t
√
2 cos
√
2
t/
2
sin
√
2
t



e
t
.
11.
We have
X
c
=
c
1
L
1
0
e
e
2
t
+
c
2
L
4
1
e
e
4
t
.
Then
Φ
=
L
e
2
t
4
e
4
t
0
e
4
t
e
,
Φ
−
1
=
L
e
−
2
t
−
4
e
−
2
t
0
e
−
4
t
e
,
and
U
=
+
Φ
−
1
F
dt
=
+
L
2
e
−
2
t
−
64
te
−
2
t
16
te
−
4
t
e
dt
=
L
15
e
−
2
t
+32
te
−
2
t
−
e
−
4
t
−
4
te
−
4
t
e
,
so that
X
p
=
ΦU
=
L
11+16
t
−
1
−
4
t
e
.
12.
We have
X
c
=
c
1
L
2 cos
t
−
sin
t
e
e
t
+
c
2
L
2 sin
t
cos
t
e
e
t
.
Then
Φ
=
L
2 cos
t
2 sin
t
−
sin
t
cos
t
e
e
t
,
Φ
−
1
=
s
1
2
cos
t
−
sin
t
1
2
sin
t
cos
t
i
e
−
t
,
and
U
=
+
Φ
−
1
F
dt
=
+
L
cos
t
−
sec
t
sin
t
e
dt
=
L
sin
t
−
ln
|
sec
t
+ tan
t
|
−
cos
t
e
,
530

531. Chapter 10 Review Exercises
so that
X
p
=
ΦU
=
L
−
2 cos
t
ln
|
sec
t
+ tan
t
|
−
1 + sin
t
ln
|
sec
t
+ tan
t
|
e
13.
We have
X
c
=
c
1
L
cos
t
+ sin
t
2 cos
t
e
+
c
2
L
sin
t
−
cos
t
2 sin
t
e
.
Then
Φ
=
L
cos
t
+ sin
t
sin
t
−
cos
t
2 cos
t
2 sin
t
e
,
Φ
−
1
=
s
sin
t
1
2
cos
t
−
1
2
sin
t
−
cos
t
1
2
cos
t
+
1
2
sin
t
i
,
and
U
=
+
Φ
−
1
F
dt
=
+
s
1
2
sin
t
−
1
2
cos
t
+
1
2
csc
t
−
1
2
sin
t
−
1
2
cos
t
+
1
2
csc
t
i
dt
=
s
−
1
2
cos
t
−
1
2
sin
t
+
1
2
ln
|
csc
t
−
cot
t
|
1
2
cos
t
−
1
2
sin
t
+
1
2
ln
|
csc
t
−
cot
t
|
i
,
so that
X
p
=
ΦU
=
L
−
1
−
1
e
+
L
sin
t
sin
t
+ cos
t
e
ln
|
csc
t
−
cot
t
|
.
14.
We have
X
c
=
c
1
L
1
−
1
e
e
2
t
+
c
2
3L
1
−
1
e
te
2
t
+
L
1
0
e
e
2
t

.
Then
Φ
=
L
e
2
t
te
2
t
+
e
2
t
−
e
2
t
−
te
2
t
e
,
Φ
−
1
=
L
−
te
−
2
t
−
te
−
2
t
−
e
−
2
t
e
−
2
t
e
−
2
t
e
,
and
U
=
+
Φ
−
1
F
dt
=
+
L
t
−
1
−
1
e
dt
=
L
1
2
t
2
−
t
−
t
e
,
so that
X
p
=
ΦU
=
L
−
1
/
2
1
/
2
e
t
2
e
2
t
+
L
−
2
1
e
te
2
t
.
15. (a)
Letting
K
=



k
1
k
2
k
3



we note that (
A
−
2
I
)
K
=
0
implies that 3
k
1
+3
k
2
+3
k
3
=0,so
k
1
=
−
(
k
2
+
k
3
). Choosing
k
2
=0,
k
3
=1
and then
k
2
=1,
k
3
= 0 we get
K
1
=



−
1
0
1



and
K
2
=



−
1
1
0



,
respectively. Thus,
X
1
=



−
1
0
1



e
2
t
and
X
2
=



−
1
1
0



e
2
t
531

532. Chapter 10 Review Exercises
are two solutions.
(b)
From det(
A
−
λ
I
)=
λ
2
(3
−
λ
) = 0 we see that
λ
1
= 3, and 0 is an eigenvalue of multiplicity two. Letting
K
=



k
1
k
2
k
3



,
as in part (
a
), we note that (
A
−
0
I
)
K
=
AK
=
0
implies that
k
1
+
k
2
+
k
3
=0,so
k
1
=
−
(
k
2
+
k
3
).
Choosing
k
2
=0,
k
3
= 1, and then
k
2
=1,
k
3
= 0 we get
K
2
=



−
1
0
1



and
K
3
=



−
1
1
0



,
respectively. Since the eigenvector corresponding to
λ
1
=3is
K
1



1
1
1



,
the general solution of the system is
X
=
c
1



1
1
1



e
3
t
+
c
2



−
1
0
1



+
c
3



−
1
1
0



.
16.
For
X
=
L
c
1
c
2
e
e
t
we have
X

=
X
=
IX
.
532

533. 11
Systems of Nonlinear Differential Equations
Exercises 11.1
1.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
9 sin
x.
If (
x, y
) is a critical point,
y
= 0 and
−
9 sin
x
= 0. Therefore
x
=
±
nπ
and so the critical points are (
±
nπ,
0)
for
n
=0,1,2,
...
.
2.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
2
x
−
y
2
.
If (
x, y
) is a critical point, then
y
= 0 and so
−
2
x
−
y
2
=
−
2
x
= 0. Therefore (0
,
0) is the sole critical point.
3.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
x
2
−
y
(1
−
x
3
)
.
If (
x, y
) is a critical point,
y
= 0 and so
x
2
−
y
(1
−
x
3
)=
x
2
= 0. Therefore (0
,
0) is the sole critical point.
4.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
4
x
1+
x
2
−
2
y.
If (
x, y
) is a critical point,
y
= 0 and so
−
4
x
1+
x
2
−
2(0) = 0. Therefore
x
= 0 and so (0
,
0) is the sole critical
point.
5.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
x
+
5
3
.
If (
x, y
) is a critical point,
y
= 0 and
−
x
+
5
3
= 0. Hence
x
(
−
1+
5
2
) = 0 and so
x
=0,
y
1
65
,
−
y
1
65
. The
critical points are (0
,
0), (
y
1
65 .
0) and (
−
y
1
65 .
0).
6.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
x
+
5
|
x
|
.
If (
x, y
) is a critical point,
y
= 0 and
−
x
+
5
|
x
|
=
x
(
−
1+
5
|
x
|
) = 0. Hence
x
=0,1
65
,
−
1
65
. The critical points
are (0
,
0), (1
65.
0) and (
−
1
65.
0).
7.
From
x
+
xy
= 0 we have
x
(1+
y
) = 0. Therefore
x
=0or
y
=
−
1. If
x
= 0, then, substituting into
−
y
−
xy
=0,
we obtain
y
= 0, Likewise, if
y
=
−
1, 1 +
x
=0or
x
=
−
1. We may conclude that (0
,
0) and (
−
1
,
−
1) are
critical points of the system.
8.
From
y
2
−
x
= 0 we have
x
=
y
2
. Substituting into
x
2
−
y
= 0, we obtain
y
4
−
y
=0or
y
(
y
3
−
1) = 0. It follows
that
y
= 0, 1 and so (0
,
0) and (1
,
1) are the critical points of the system.
9.
From
x
−
y
= 0 we have
y
=
x
. Substituting into 3
x
2
−
4
y
= 0 we obtain 3
x
2
−
4
x
=
x
(3
x
−
4) = 0. It follows
that (0
,
0) and (4
/
3
,
4
/
3) are the critical points of the system.
10.
From
x
3
−
y
=0wehave
y
=
x
3
. Substituting into
x
−
y
3
= 0 we obtain
x
−
x
9
=0or
x
(1
−
x
8
). Therefore
x
=0,1,
−
1 and so the critical points of the system are (0
,
0), (1
,
1), and (
−
1
,
−
1).
533

534. Exercises 11.1
11.
From
x
(10
−
x
−
1
2
y
) = 0 we obtain
x
=0or
x
+
1
2
y
= 10. Likewise
y
(16
−
y
−
x
) = 0 implies that
y
=0or
x
+
y
= 16. We therefore have four cases. If
x
=0,
y
=0or
y
= 16. If
x
+
1
2
y
= 10, we may conclude that
y
(
−
1
2
y
+6) = 0 and so
y
= 0, 12. Therefore the critical points of the system are (0
,
0), (0
,
16), (10
,
0), and
(4
,
12).
12.
Adding the two equations we obtain 10
−
15
y
y
+5
= 0. It follows that
y
= 10, and from
−
2
x
+
y
+10 = 0 we
may conclude that
x
= 10. Therefore (10
,
10) is the sole critical point of the system.
13.
From
x
2
e
y
= 0 we have
x
= 0. Since
e
x
−
1=
e
0
−
1 = 0, the second equation is satisfied for an arbitrary value
of
y
. Therefore any point of the form (0
,y
) is a critical point.
14.
From sin
y
=0wehave
y
=
±
nπ
. From
e
x
−
y
= 1, we may conclude that
x
−
y
=0or
x
=
y
. The critical points
of the system are therefore (
±
nπ,
±
nπ
) for
n
=0,1,2,
...
.
15.
From
x
(1
−
x
2
−
3
y
2
)=0wehave
x
=0or
x
2
+3
y
2
=1. If
x
= 0, then substituting into
y
(3
−
x
2
−
3
y
2
) gives
y
(3
−
3
y
2
) = 0. Therefore
y
=0,1,
−
1. Likewise
x
2
=1
−
3
y
2
yields 2
y
= 0 so that
y
= 0 and
x
2
=1
−
3(0)
2
=1.
The critical points of the system are therefore (0
,
0), (0
,
1), (0
,
−
1), (1
,
0), and (
−
1
,
0).
16.
From
−
x
(4
−
y
2
) = 0 we obtain
x
=0,
y
=2,or
y
=
−
2. If
x
= 0, then substituting into 4
y
(1
−
x
2
) yields
y
= 0. Likewise
y
= 2 gives 8(1
−
x
2
)=0or
x
=1,
−
1. Finally
y
=
−
2 yields
−
8(1
−
x
2
)=0or
x
=1,
−
1.
The critical points of the system are therefore (0
,
0), (1
,
2), (
−
1
,
2), (1
,
−
2), and (
−
1
,
−
2).
17. (a)
From Exercises 10
.
2, Problem 1,
x
=
c
1
e
5
t
−
c
2
e
−
t
and
y
=2
c
1
e
5
t
+
c
2
e
−
t
.
(b)
From
X
(0) = (2
,
−
1) it follows that
c
1
= 0 and
c
2
= 2. Therefore
x
=
−
2
e
−
t
and
y
=2
e
−
t
.
(c)
18. (a)
From Exercises 10
.
2, Problem 6,
x
=
c
1
+2
c
2
e
−
5
t
and
y
=3
c
1
+
c
2
e
−
5
t
.
(b)
From
X
(0) = (3
,
4) it follows that
c
1
=
c
2
= 1. Therefore
x
=1+2
e
−
5
t
and
y
=3+
e
−
5
t
.
(c)
19. (a)
From Exercises 10
.
2, Problem 37,
x
=
c
1
(4 cos 3
t
−
3 sin 3
t
)+
c
2
(4 sin 3
t
+3 cos 3
t
) and
y
=
c
1
(5 cos 3
t
)+
c
2
(5 sin 3
t
). All solutions are one periodic with
p
=2
π/
3.
(b)
From
X
(0) = (4
,
5) it follows that
c
1
= 1 and
c
2
= 0. Therefore
x
= 4 cos 3
t
−
3 sin 3
t
and
y
= 5 cos 3
t
.
534

535. Exercises 11.1
(c)
20. (a)
From Exercises 10
.
2, Problem 34,
x
=
c
1
(sin
t
−
cos
t
)+
c
2
(
−
cos
t
−
sin
t
) and
y
=2
c
1
cos
t
+2
c
2
sin
t
. All
solutions are periodic with
p
=2
π
.
(b)
From
X
(0) = (
−
2
,
2) it follows that
c
1
=
c
2
= 1. Therefore
x
=
−
2 cos
t
and
y
= 2 cos
t
+2 sin
t
.
(c)
21. (a)
From Exercises 10
.
2, Problem 35,
x
=
c
1
(sin
t
−
cos
t
)
e
4
t
+
c
2
(
−
sin
t
−
cos
t
)
e
4
t
and
y
=2
c
1
(cos
t
)
e
4
t
+
2
c
2
(sin
t
)
e
4
t
. Because of the presence of
e
4
t
, there are no periodic solutions.
(b)
From
X
(0) = (
−
1
,
2) it follows that
c
1
= 1 and
c
2
= 0. Therefore
x
= (sin
t
−
cos
t
)
e
4
t
and
y
= 2(cos
t
)
e
4
t
.
(c)
22. (a)
From Exercises 10
.
2, Problem 38,
x
=
c
1
e
−
t
(2 cos 2
t
−
2 sin 2
t
)+
c
2
e
−
t
(2 cos 2
t
+2 sin 2
t
) and
y
=
c
1
e
−
t
cos 2
t
+
c
2
e
−
t
sin 2
t
. Because of the presence of
e
−
t
, there are no periodic solutions.
(b)
From
X
(0) = (2
,
1) it follows that
c
1
= 1 and
c
2
= 0. Therefore
x
=
e
−
t
(2 cos 2
t
−
2 sin 2
t
) and
y
=
e
−
t
cos 2
t
.
(c)
535

536. Exercises 11.1
23.
Switching to polar coordinates,
dr
dt
=
1
r
,
x
dx
dt
+
y
dy
dt
.
=
1
r
(
−
xy
−
x
2
r
4
+
xy
−
y
2
r
4
)=
−
r
5
dθ
dt
=
1
r
2
,
−
y
dx
dt
+
x
dy
dt
.
=
1
r
2
(
y
2
+
xyr
4
+
x
2
−
xyr
4
)=1
.
If we use separation of variables on
dr
dt
=
−
r
5
we obtain
r
=
,
1
4
t
+
c
1
.
1
/
4
and
θ
=
t
+
c
2
.
Since
X
(0)=(4
,
0),
r
= 4 and
θ
= 0 when
t
= 0. It follows that
c
2
= 0 and
c
1
=
1
256
. The final solution may
be written as
r
=
4
4
√
1024
t
+1
,θ
=
t
and so the solution spirals toward the origin as
t
increases.
24.
Switching to polar coordinates,
dr
dt
=
1
r
,
x
dx
dt
+
y
dy
dt
.
=
1
r
(
xy
−
x
2
r
2
−
xy
+
y
2
r
2
)=
r
3
dθ
dt
=
1
r
2
,
−
y
dx
dt
+
x
dy
dt
.
=
1
r
2
(
−
y
2
−
xyr
2
−
x
2
+
xyr
2
)=
−
1
.
If we use separation of variables, it follows that
r
=
1
√
−
2
t
+
c
1
and
θ
=
−
t
+
c
2
.
Since
X
(0) = (4
,
0),
r
= 4 and
θ
= 0 when
t
= 0. It follows that
c
2
= 0 and
c
1
=
1
16
. The final solution may be
written as
r
=
4
√
1
−
32
t
,θ
=
−
t.
Note that
r
→∞
as
t
→
n
1
32
π
−
. Because 0
≤
t
≤
1
32
, the curve is not a spiral.
25.
Switching to polar coordinates,
dr
dt
=
1
r
,
x
dx
dt
+
y
dy
dt
.
=
1
r
[
−
xy
+
x
2
(1
−
r
2
)+
xy
+
y
2
(1
−
r
2
)] =
r
(1
−
r
2
)
dθ
dt
=
1
r
2
,
−
y
dx
dt
+
x
dy
dt
.
=
1
r
2
[
y
2
−
xy
(1
−
r
2
)+
x
2
+
xy
(1
−
r
2
)] = 1
.
Now
dr
dt
=
r
−
r
3
or
dr
dt
−
r
=
−
r
3
is a Bernoulli differential equation. Following the procedure in Section 2.5
of the text, we let
w
=
r
−
2
so that
w
x
=
−
2
r
−
3
dr
dt
. Therefore
w
x
+2
w
= 2, a linear first order differential
equation. It follows that
w
=1+
c
1
e
−
2
t
and so
r
2
=
1
1+
c
1
e
−
2
t
. The general solution may be written as
r
=
1
√
1+
c
1
e
−
2
t
,θ
=
t
+
c
2
.
If
X
(0)=(1
,
0),
r
= 1 and
θ
= 0 when
t
= 0. Therefore
c
1
=0=
c
2
and so
x
=
r
cos
t
= cos
t
and
y
=
r
sin
t
= sin
t
. This solution generates the circle
r
=1. If
X
(0) = (2
,
0),
r
= 2 and
θ
= 0 when
t
=0.
536

537. Exercises 11.1
Therefore
c
1
=
−
3
/
4,
c
2
= 0 and so
r
=
1
5
1
−
3
4
e
−
2
t
,θ
=
t.
This solution spirals toward the circle
r
=1as
t
increases.
26.
Switching to polar coordinates,
dr
dt
=
1
r
,
x
dx
dt
+
y
dy
dt
.
=
1
r
/
xy
−
x
2
r
(4
−
r
2
)
−
xy
−
y
2
r
(4
−
r
2
)

=
r
2
−
4
dθ
dt
=
1
r
2
,
−
y
dx
dt
+
x
dy
dt
.
=
1
r
2

−
y
2
+
xy
r
(4
−
r
2
)
−
x
2
−
xy
r
(4
−
r
2
)

=
−
1
.
From Example 3, Section 2
.
2,
r
=2
1+
c
1
e
4
t
1
−
c
1
e
4
t
and
θ
=
−
t
+
c
2
.
If
X
(0)=(1
,
0),
r
= 1 and
θ
= 0 when
t
= 0. It follows that
c
2
= 0 and
c
1
=
−
1
3
. Therefore
r
=2
1
−
1
3
e
4
t
1+
1
3
e
4
t
and
θ
=
−
t.
Note that
r
= 0 when
e
4
t
=3or
t
=
ln 3
4
and
r
→−
2as
t
→∞
. The solution therefore approaches the circle
r
=2. If
X
(0)=(2
,
0), it follows that
c
1
=
c
2
= 0. Therefore
r
= 2 and
θ
=
−
t
so that the solution generates
the circle
r
= 2 traversed in the clockwise direction. Note also that the original system is not defined at (0
,
0)
but the corresponding polar system is defined for
r
= 0. If the Runge-Kutta method is applied to the original
system, the solution corresponding to
X
(0) = (1
,
0) will stall at the origin.
27.
The system has no critical points, so there are no periodic solutions.
28.
From
x
(6
y
−
1) = 0 and
y
(2
−
8
x
) = 0 we see that (0
,
0) and (1
/
4
,
1
/
6) are
critical points. From the graph we see that there are periodic solutions around
(1
/
4
,
1
/
6).
29.
The only critical point is (0
,
0). There appears to be a single periodic solution
around (0
,
0).
30.
The system has no critical points, so there are no periodic solutions.
31.
If
X
(
t
)=(
x
(
t
)
,y
(
t
)) is a solution,
d
dt
f
(
x
(
t
)
,y
(
t
)) =
∂f
∂x
dx
dt
+
∂f
∂y
dy
dt
=
QP
−
PQ
=0
,
using the chain rule. Therefore
f
(
x
(
t
)
,y
(
t
)) =
c
for some constant
c
, and the solution lies on a level curve of
f
.
537

538. Exercises 11.2
Exercises 11.2
1. (a)
If
X
(0) =
X
0
lies on the line
y
=2
x
, then
X
(
t
) approaches (0
,
0) along this line. For all other initial
conditions,
X
(
t
) approaches (0
,
0) from the direction determined by the line
y
=
−
x/
2.
(b)
2. (a)
If
X
(0) =
X
0
lies on the line
y
=
−
x
, then
X
(
t
) becomes unbounded along this line. For all other initial
conditions,
X
(
t
) becomes unbounded and
y
=
−
3
x/
2 serves as an asymptote.
(b)
3. (a)
All solutions are unstable spirals which become unbounded as
t
increases.
(b)
4. (a)
All solutions are spirals which approach the origin.
(b)
5. (a)
All solutions approach (0
,
0) from the direction specified by the line
y
=
x
.
538

539. Exercises 11.2
(b)
6. (a)
All solutions become unbounded and
y
=
x/
2 serves as the asymptote.
(b)
7. (a)
If
X
(0) =
X
0
lies on the line
y
=3
x
, then
X
(
t
) approaches (0
,
0) along this line. For all other initial
conditions,
X
(
t
) becomes unbounded and
y
=
x
serves as the asymptote.
(b)
8. (a)
The solutions are ellipses which encircle the origin.
(b)
9.
Since ∆ =
−
41
<
0, we may conclude from Figure 11
.
18 that (0
,
0) is a saddle point.
10.
Since ∆ = 29 and
τ
=
−
12,
τ
2
−
4∆
>
0 and so from Figure 11
.
18, (0
,
0) is a stable node.
11.
Since ∆ =
−
19
<
0, we may conclude from Figure 11
.
18 that (0
,
0) is a saddle point.
12.
Since ∆ = 1 and
τ
=
−
1,
τ
2
−
4∆ =
−
3 and so from Figure 11
.
18, (0
,
0) is a stable spiral point.
13.
Since ∆ = 1 and
τ
=
−
2,
τ
2
−
4∆ = 0 and so from Figure 11
.
18, (0
,
0) is a degenerate stable node.
14.
Since ∆ = 1 and
τ
=2,
τ
2
−
4∆ = 0 and so from Figure 11
.
18, (0
,
0) is a degenerate unstable node.
539

540. Exercises 11.2
15.
Since ∆ = 0
.
01 and
τ
=
−
0
.
03,
τ
2
−
4∆
<
0 and so from Figure 11
.
18, (0
,
0) is a stable spiral point.
16.
Since ∆ = 0
.
0016 and
τ
=0
.
08,
τ
2
−
4∆ = 0 and so from Figure 11
.
18, (0
,
0) is a degenerate unstable node.
17.
∆=1
−
µ
2
,
τ
= 0, and so we need ∆ = 1
−
µ
2
>
0 for (0
,
0) to be a center. Therefore
|
µ
|
<
1.
18.
Note that ∆ = 1 and
τ
=
µ
. Therefore we need both
τ
=
µ<
0 and
τ
2
−
4∆ =
µ
2
−
4
<
0 for (0
,
0) to be a
stable spiral point. These two conditions may be written as
−
2
<µ<
0.
19.
Note that ∆ =
µ
+1 and
τ
=
µ
+1 and so
τ
2
−
4∆ = (
µ
+1)
2
−
4(
µ
+1)=(
µ
+1)(
µ
−
3). It follows that
τ
2
−
4∆
<
0 if and only if
−
1
<µ<
3. We may conclude that (0
,
0) will be a saddle point when
µ<
−
1.
Likewise (0
,
0) will be an unstable spiral point when
τ
=
µ
+1
>
0 and
τ
2
−
4∆
<
0. This condition reduces to
−
1
<µ<
3.
20.
τ
=2
α
,∆=
α
2
+
β
2
>
0, and
τ
2
−
4∆ =
−
4
β<
0. If
α<
0, (0
,
0) is a stable spiral point. If
α>
0, (0
,
0) is an
unstable spiral point. Therefore (0
,
0) cannot be a node or saddle point.
21. AX
1
+
F
=
0
implies that
AX
1
=
−
F
or
X
1
=
−
A
−
1
F
. Since
X
p
(
t
)=
−
A
−
1
F
is a particular solution, it
follows from Theorem 10
.
6 that
X
(
t
)=
X
c
(
t
)+
X
1
is the general solution to
X
x
=
AX
+
F
.If
τ<
0 and ∆
>
0
then
X
c
(
t
) approaches (0
,
0) by Theorem 11
.
1(a). It follows that
X
(
t
) approaches
X
1
as
t
→∞
.
22.
If
bc <
1, ∆ =
ad
ˆ
x
ˆ
y
(1
−
bc
)
>
0 and
τ
2
−
4∆ = (
a
ˆ
x
−
d
ˆ
y
)
2
+4
abcd
ˆ
x
ˆ
y>
0. Therefore (0
,
0) is a stable node.
23. (a)
The critical point is
X
1
=(
−
3
,
4).
(b)
From the graph,
X
1
appears to be an unstable node or a saddle point.
(c)
Since ∆ =
−
1, (0
,
0) is a saddle point.
24. (a)
The critical point is
X
1
=(
−
1
,
−
2).
(b)
From the graph,
X
1
appears to be a stable node or a degenerate stable node.
(c)
Since
τ
=
−
16, ∆ = 64, and
τ
2
−
4∆ = 0, (0
,
0) is a degenerate stable node.
25. (a)
The critical point is
X
1
=(0
.
5
,
2).
540

541. Exercises 11.3
(b)
From the graph,
X
1
appears to be an unstable spiral point.
(c)
Since
τ
=0
.
2, ∆ = 0
.
03, and
τ
2
−
4∆ =
−
0
.
08, (0
,
0) is an unstable spiral point.
26. (a)
The critical point is
X
1
=(1
,
1).
(b)
From the graph,
X
1
appears to be a center.
(c)
Since
τ
= 0 and ∆ = 1, (0
,
0) is a center.
Exercises 11.3
1.
Switching to polar coordinates,
dr
dt
=
1
r
,
x
dx
dt
+
y
dy
dt
.
=
1
r
(
αx
2
−
βxy
+
xy
2
+
βxy
+
αy
2
−
xy
2
)=
1
r
αr
2
=
αr.
Therefore
r
=
ce
αt
and so
r
→
0 if and only if
α<
0.
2.
The differential equation
dr
dt
=
αr
(5
−
r
) is a logistic differential equation. [See Section 2
.
8, (4) and (5).] It
follows that
r
=
5
1+
c
1
e
−
5
αt
and
θ
=
−
t
+
c
2
.
If
α>
0,
r
→
5as
t
→
+
∞
and so the critical point (0
,
0) is unstable. If
α<
0,
r
→
0as
t
→
+
∞
and so (0
,
0)
is asymptotically stable.
3.
The critical points are
x
= 0 and
x
=
n
+1. Since
g
x
(
x
)=
k
(
n
+1)
−
2
kx
,
g
x
(0) =
k
(
n
+1)
>
0 and
g
x
(
n
+1) =
−
k
(
n
+1)
<
0. Therefore
x
= 0 is unstable while
x
=
n
+1 is asymptotically stable. See
Theorem 11
.
2.
4.
Note that
x
=
k
is the only critical point since ln(
x/k
) is not defined at
x
= 0. Since
g
x
(
x
)=
−
k
−
k
ln(
x/k
),
g
x
(
k
)=
−
k<
0. Therefore
x
=
k
is an asymptotically stable critical point by Theorem 11
.
2.
5.
The only critical point is
T
=
T
0
. Since
g
x
(
T
)=
k
,
g
x
(
T
0
)=
k>
0. Therefore
T
=
T
0
is unstable by
Theorem 11
.
2.
6.
The only critical point is
v
=
mg/k
.Now
g
(
v
)=
g
−
(
k/m
)
v
and so
g
x
(
v
)=
−
k/m <
0. Therefore
v
=
mg/k
is
an asymptotically stable critical point by Theorem 11
.
2.
541

542. Exercises 11.3
7.
Critical points occur at
x
=
α
,
β
. Since
g
x
(
x
)=
k
(
−
α
−
β
+2
x
),
g
x
(
α
)=
k
(
α
−
β
) and
g
x
(
β
)=
k
(
β
−
α
). Since
α>β
,
g
x
(
α
)
>
0 and so
x
=
α
is unstable. Likewise
x
=
β
is asymptotically stable.
8.
Critical points occur at
x
=
α
,
β
,
γ
. Since
g
x
(
x
)=
k
(
α
−
x
)(
−
β
−
γ
−
2
x
)+
k
(
β
−
x
)(
γ
−
x
)(
−
1)
,
g
x
(
α
)=
−
k
(
β
−
α
)(
γ
−
α
)
<
0 since
α>β>γ
. Therefore
x
=
α
is asymptotically stable. Similarly
g
x
(
β
)
>
0
and
g
x
(
γ
)
<
0. Therefore
x
=
β
is unstable while
x
=
γ
is asymptotically stable.
9.
Critical points occur at
P
=
a/b
,
c
but not at
P
= 0. Since
g
x
(
P
)=(
a
−
bP
)+(
P
−
c
)(
−
b
),
g
x
(
a/b
)=(
a/b
−
c
)(
−
b
)=
−
a
+
bc
and
g
x
(
c
)=
a
−
bc.
Since
a<bc
,
−
a
+
bc >
0 and
a
−
bc <
0. Therefore
P
=
a/b
is unstable while
P
=
c
is asymptotically stable.
10.
Since
A>
0, the only critical point is
A
=
K
2
. Since
g
x
(
A
)=
1
2
kKA
−
1
/
2
−
k
,
g
x
(
K
2
)=
−
k/
2
<
0. Therefore
A
=
K
2
is asymptotically stable.
11.
The sole critical point is (1
/
2
,
1) and
g
x
(
X
)=
,
−
2
y
−
2
x
2
y
2
x
−
1
.
.
Computing
g
x
((1
/
2
,
1)) we find that
τ
=
−
2 and ∆ = 2 so that
τ
2
−
4∆ =
−
4
<
0. Therefore (1
/
2
,
1) is a stable
spiral point.
12.
Critical points are (1
,
0) and (
−
1
,
0), and
g
x
(
X
)=
,
2
x
−
2
y
02
.
.
At
X
=(1
,
0),
τ
= 4, ∆ = 4, and so
τ
2
−
4∆ = 0. We may conclude that (1
,
0) is unstable but we are unable
to classify this critical point any further. At
X
=(
−
1
,
0), ∆ =
−
4
<
0 and so (
−
1
,
0) is a saddle point.
13.
y
x
=2
xy
−
y
=
y
(2
x
−
1). Therefore if (
x, y
) is a critical point, either
x
=1
/
2or
y
= 0. The case
x
=1
/
2
and
y
−
x
2
+2 = 0 implies that (
x, y
)=(1
/
2
,
−
7
/
4). The case
y
= 0 leads to the critical points (
√
2
,
0) and
(
−
√
2
,
0). We next use the Jacobian matrix
g
x
(
X
)=
,
−
2
x
1
2
y
2
x
−
1
.
to classify these three critical points. For
X
=(
√
2
,
0) or (
−
√
2
,
0),
τ
=
−
1 and ∆
<
0. Therefore both critical
points are saddle points. For
X
=(1
/
2
,
−
7
/
4),
τ
=
−
1,∆=7
/
2 and so
τ
2
−
4∆ =
−
13
<
0. Therefore
(1
/
2
,
−
7
/
4) is a stable spiral point.
14.
y
x
=
−
y
+
xy
=
y
(
−
1+
x
). Therefore if (
x, y
) is a critical point, either
y
=0or
x
= 1. The case
y
= 0 and
2
x
−
y
2
= 0 implies that (
x, y
)=(0
,
0). The case
x
= 1 leads to the critical points (1
,
√
2 ) and (1
,
−
√
2). We
next use the Jacobian matrix
g
x
(
X
)=
,
2
−
2
y
yx
−
1
.
to classify these critical points. For
X
=(0
,
0), ∆ =
−
2
<
0 and so (0
,
0) is a saddle point. For either (1
,
√
2)
or (1
,
−
√
2),
τ
=2,∆=4,andso
τ
2
−
4∆ =
−
12. Therefore (1
,
√
2 ) and (1
,
−
√
2 ) are unstable spiral points.
15.
Since
x
2
−
y
2
=0,
y
2
=
x
2
and so
x
2
−
3
x
+2=(
x
−
1)(
x
−
2) = 0. It follows that the critical points are (1
,
1),
(1
,
−
1), (2
,
2), and (2
,
−
2). We next use the Jacobian
g
x
(
X
)=
,
−
32
y
2
x
−
2
y
.
542

543. Exercises 11.3
to classify these four critical points. For
X
=(1
,
1),
τ
=
−
5, ∆ = 2, and so
τ
2
−
4∆ = 17
>
0. Therefore (1
,
1) is
a stable node. For
X
=(1
,
−
1), ∆ =
−
2
<
0 and so (1
,
−
1) is a saddle point. For
X
=(2
,
2), ∆ =
−
4
<
0 and
so we have another saddle point. Finally, if
X
=(2
,
−
2),
τ
=1,∆=4,andso
τ
2
−
4∆ =
−
15
<
0. Therefore
(2
,
−
2) is an unstable spiral point.
16.
From
y
2
−
x
2
=0,
y
=
x
or
y
=
−
x
. The case
y
=
x
leads to (4
,
4) and (
−
1
,
1) but the case
y
=
−
x
leads to
x
2
−
3
x
+4 = 0 which has no real solutions. Therefore (4
,
4) and (
−
1
,
1) are the only critical points. We next
use the Jacobian matrix
g
x
(
X
)=
,
yx
−
3
−
2
x
2
y
.
to classify these two critical points. For
X
=(4
,
4),
τ
= 12, ∆ = 40, and so
τ
2
−
4∆
<
0. Therefore (4
,
4) is
an unstable spiral point. For
X
=(
−
1
,
1),
τ
=
−
3, ∆ = 10, and so
x
2
−
4∆
<
0. It follows that (
−
1
,
−
1) is a
stable spiral piont.
17.
Since
x
x
=
−
2
xy
= 0, either
x
=0or
y
=0. If
x
=0,
y
(1
−
y
2
) = 0 and so (0
,
0), (0
,
1), and (0
,
−
1) are critical
points. The case
y
= 0 leads to
x
= 0. We next use the Jacobian matrix
g
x
(
X
)=
,
−
2
y
−
2
x
−
1+
y
1+
x
−
3
y
2
.
to classify these three critical points. For
X
=(0
,
0),
τ
= 1 and ∆ = 0 and so the test is inconclusive. For
X
=(0
,
1),
τ
=
−
4, ∆ = 4 and so
τ
2
−
4∆ = 0. We can conclude that (0
,
1) is a stable critical point but we
are unable to classify this critical point further in this borderline case. For
X
=(0
,
−
1), ∆ =
−
4
<
0 and so
(0
,
−
1) is a saddle point.
18.
We found that (0
,
0), (0
,
1), (0
,
−
1), (1
,
0) and (
−
1
,
0) were the critical points in Exercise 15, Section 11
.
1. The
Jacobian is
g
x
(
X
)=
,
1
−
3
x
2
−
3
y
2
−
6
xy
−
2
xy
3
−
x
2
−
9
y
2
.
.
For
X
=(0
,
0),
τ
=4,∆=3andso
τ
2
−
4∆=4
>
0. Therefore (0
,
0) is an unstable node. Both (0
,
1) and
(0
,
−
1) give
τ
=
−
8, ∆ = 12, and
τ
2
−
4∆=16
>
0. These two critical points are therefore stable nodes. For
X
=(1
,
0) or (
−
1
,
0), ∆ =
−
4
<
0 and so saddle points occur.
19.
We found the critical points (0
,
0), (10
,
0), (0
,
16) and (4
,
12) in Exercise 11, Section 11
.
1. Since the Jacobian is
g
x
(
X
)=
,
10
−
2
x
−
1
2
y
−
1
2
x
−
y
16
−
2
y
−
x
.
we may classify the critical points as follows:
X
τ
∆
τ
2
−
4∆ Conclusion
(0
,
0) 26 160 36 unstable node
(10
,
0)
−
4
−
60 – saddle point
(0
,
16)
−
14
−
32 – saddle point
(4
,
12)
−
16 24 160 stable node
20.
We found the sole critical point (10
,
10) in Exercise 12, Section 11
.
1. The Jacobian is
g
x
(
X
)=


−
21
2
−
1
−
15
(
y
+5)
2


,
g
x
((10
,
10)) has trace
τ
=
−
46
/
15, ∆ = 2
/
15, and
τ
2
−
4∆
>
0. Therefore (0
,
0) is a stable node.
543

544. Exercises 11.3
21.
The corresponding plane autonomous system is
θ
x
=
y, y
x
= (cos
θ
−
1
2
) sin
θ.
Since
|
θ
|
<π
, it follows that critical points are (0
,
0), (
π/
3
,
0) and (
−
π/
3
,
0). The Jacobian matrix is
g
x
(
X
)=
,
01
cos 2
θ
−
1
2
cos
θ
0
.
and so at (0
,
0),
τ
=0and∆=
−
1
/
2. Therefore (0
,
0) is a saddle point. For
X
=(
±
π/
3
,
0),
τ
= 0 and ∆ = 3
/
4.
It is not possible to classify either critical point in this borderline case.
22.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
x
+
,
1
2
−
3
y
2
.
y
−
x
2
.
If (
x, y
) is a critical point,
y
= 0 and so
−
x
−
x
2
=
−
x
(1 +
x
) = 0. Therefore (0
,
0) and (
−
1
,
0) are the only two
critical points. We next use the Jacobian matrix
g
x
(
X
)=
,
01
−
1
−
2
x
1
2
−
9
y
2
.
to classify these critical points. For
X
=(0
,
0),
τ
=1
/
2, ∆ = 1, and
τ
2
−
4∆
<
0. Therefore (0
,
0) is an unstable
spiral point. For
X
=(
−
1
,
0),
τ
=1
/
2, ∆ =
−
1 and so (
−
1
,
0) is a saddle point.
23.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
x
2
−
y
(1
−
x
3
)
and the only critical point is (0
,
0). Since the Jacobian matrix is
g
x
(
X
)=
,
01
2
x
+3
x
2
yx
3
−
1
.
,
τ
=
−
1 and ∆ = 0, and we are unable to classify the critical point in this borderline case.
24.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
4
x
1+
x
2
−
2
y
and the only critical point is (0
,
0). Since the Jacobian matrix is
g
x
(
X
)=


01
−
4
1
−
x
2
(1 +
x
2
)
2
−
2


,
τ
=
−
2, ∆ = 4,
τ
2
−
4∆ =
−
12, and so (0
,
0) is a stable spiral point.
25.
In Exercise 5, Section 11
.
1, we showed that (0
,
0), (
y
1
65 .
0) and (
−
y
1
65 .
0) are the critical points. We will
use the Jacobian matrix
g
x
(
X
)=
,
01
−
1+3
5
2
0
.
to classify these three critical points. For
X
=(0
,
0),
τ
= 0 and ∆ = 1 and we are unable to classify this critical
point. For (
±
y
1
65 .
0),
τ
= 0 and ∆ =
−
2 and so both of these critical points are saddle points.
26.
In Exercise 6, Section 11
.
1, we showed that (0
,
0), (1
65.
0), and (
−
1
65.
0) are the critical points. Since
D
x
x
|
x
|
=
2
|
x
|
, the Jacobian matrix is
g
x
(
X
)=
,
01
2
5
|
x
|−
10
.
.
544

545. Exercises 11.3
For
X
=(0
,
0),
τ
= 0, ∆ = 1 and we are unable to classify this critical point. For (
±
1
65.
0),
τ
=0,∆=
−
1,
and so both of these critical points are saddle points.
27.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
(
β
+
α
2
y
2
)
x
1+
α
2
x
2
and the Jacobian matrix is
g
x
(
X
)=


01
(
β
+
αy
2
)(
α
2
x
2
−
1)
(1 +
α
2
x
2
)
2
−
2
α
2
yx
1+
α
2
x
2


.
For
X
=(0
,
0),
τ
= 0 and ∆ =
β
. Since
β<
0, we may conclude that (0
,
0) is a saddle point.
28.
From
x
x
=
−
αx
+
xy
=
x
(
−
α
+
y
) = 0, either
x
=0or
y
=
α
.If
x
= 0, then 1
−
βy
= 0 and so
y
=1
/β
. The
case
y
=
α
implies that 1
−
βα
−
x
2
=0or
x
2
=1
−
αβ
. Since
αβ >
1, this equation has no real solutions. It
follows that (0
,
1
/β
) is the unique critical point. Since the Jacobian matrix is
g
x
(
X
)=
,
−
α
+
yx
−
2
x
−
β
.
,
τ
=
−
α
−
β
+
1
β
=
−
β
+
1
−
αβ
β
<
0 and ∆ =
αβ
−
1
>
0. Therefore (0
,
1
/β
) is a stable critical point.
29. (a)
The graphs of
−
x
+
y
−
x
3
= 0 and
−
x
−
y
+
y
2
= 0 are shown in the figure.
The Jacobian matrix is
g
x
(
X
)=
,
−
1
−
3
x
2
1
−
1
−
1+2
y
.
.
For
X
=(0
,
0),
τ
=
−
2, ∆ = 2,
τ
2
−
4∆ =
−
4, and so (0
,
0) is a stable spiral
point.
(b)
For
X
1
,∆=
−
6
.
07
<
0 and so a saddle point occurs at
X
1
.
30. (a)
The corresponding plane autonomous system is
x
x
=
y, y
x
=
5
(
y
−
1
3
y
3
)
−
x
and so the only critical point is (0
,
0). Since the Jacobian matrix is
g
x
(
X
)=
,
01
−
1
5
(1
−
y
2
)
.
,
τ
=
5
, ∆ = 1, and so
τ
2
−
4∆ =
5
2
−
4 at the critical point (0
,
0).
(b)
When
τ
=
5N
0, (0
,
0) is an unstable critical point.
(c)
When
5g
0 and
τ
2
−
4∆ =
5
2
−
4
<
0, (0
,
0) is a stable spiral point. These two requirements can be written
as
−
2
g5g
0.
(d)
When
5
=0,
x
xx
+
x
= 0 and so
x
=
c
1
cos
t
+
c
2
sin
t
. Therefore all solutions are periodic (with period 2
π
)
and so (0
,
0) is a center.
31.
dy
dx
=
y
x
x
x
=
−
2
x
3
y
may be solved by separating variables. It follows that
y
2
+
x
4
=
c
.If
X
(0)=(
x
0
,
0) where
x
0
>
0, then
c
=
x
4
0
so that
y
2
=
x
4
0
−
x
4
. Therefore if
−
x
0
<x<x
0
,
y
2
>
0 and so there are two values of
y
corresponding to each value of
x
. Therefore the solution
X
(
t
) with
X
(0) = (
x
0
,
0) is periodic and so (0
,
0) is a
center.
545

546. Exercises 11.3
32.
dy
dx
=
y
x
x
x
=
x
2
−
2
x
y
may be solved by separating variables. It follows that
y
2
2
=
x
3
3
−
x
2
+
c
and since
X
(0) = (
x
(0)
,x
x
(0)) = (1
,
0),
c
=
2
3
. Therefore
y
2
2
=
x
3
−
3
x
2
+2
3
=
(
x
−
1)(
x
2
−
2
x
−
2)
3
.
But (
x
−
1)(
x
2
−
2
x
−
2)
>
0 for 1
−
√
3
<x<
1 and so each
x
in this interval has 2 corresponding values of
y
.
therefore
X
(
t
) is a periodic solution.
33. (a)
x
x
=2
xy
= 0 implies that either
x
=0or
y
=0. If
x
= 0, then from 1
−
x
2
+
y
2
=0,
y
2
=
−
1 and there are
no real solutions. If
y
=0,1
−
x
2
= 0 and so (1
,
0) and (
−
1
,
0) are critical points. The Jacobian matrix is
g
x
(
X
)=
,
2
y
2
x
−
2
x
2
y
.
and so
τ
= 0 and ∆ = 4 at either
X
=(1
,
0) or (
−
1
,
0). We obtain no information about these critical
points in this borderline case.
(b)
dy
dx
=
y
x
x
x
=
1
−
x
2
+
y
2
2
xy
or 2
xy
dy
dx
=1
−
x
2
+
y
2
. Letting
µ
=
y
2
x
, it follows
that
dµ
dx
=
1
x
2
−
1 and so
µ
=
−
1
x
−
x
+2
c
. Therefore
y
2
x
=
−
1
x
−
x
+2
c
which can be put in the form
(
x
−
c
)
2
+
y
2
=
c
2
−
1
.
The solution curves are shown and so both (1
,
0) and (
−
1
,
0) are centers.
34. (a)
dy
dx
=
y
x
x
x
=
−
x
−
y
2
y
=
−
x
y
−
y
and so
dy
dx
+
y
=
−
xy
−
1
.
(b)
Let
w
=
y
1
−
n
=
y
2
. It follows that
dw
dx
+2
w
=
−
2
x
, a linear first order differential equation whose solution
is
y
2
=
w
=
ce
−
2
x
+
,
1
2
−
x
.
.
Since
x
(0) =
1
2
and
y
(0) =
x
x
(0) = 0, 0 =
c
and so
y
2
=
1
2
−
x,
a parabola with vertex at (1
/
2
,
0). Therefore the solution
X
(
t
) with
X
(0) = (1
/
2
,
0) is not periodic.
35.
dy
dx
=
y
x
x
x
=
x
3
−
x
y
and so
y
2
2
=
x
4
4
−
x
2
2
+
c
or
y
2
=
x
4
2
−
x
2
+
c
1
. Since
x
(0) = 0 and
y
(0) =
x
x
(0) =
v
0
,it
follows that
c
1
=
v
2
0
and so
y
2
=
1
2
x
4
−
x
2
+
v
2
0
=
(
x
2
−
1)
2
+2
v
2
0
−
1
2
.
The
x
-intercepts on this graph satisfy
x
2
=1
±
5
1
−
2
v
2
0
and so we must require that 1
−
2
v
2
0
≥
0 (or
|
v
0
|≤
1
2
√
2 ) for real solutions to exist. If
x
2
0
=1
−
y
1
−
2
v
2
0
and
−
x
0
<x<x
0
, then (
x
2
−
1)
2
+2
v
2
0
−
1
>
0 and so there are two corresponding values of
y
. Therefore
X
(
t
)
with
X
(0) = (0
,v
0
) is periodic provided that
|
v
0
|≤
1
2
√
2.
546

547. Exercises 11.3
36.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
5
2
−
x
+1
and so the critical points must satisfy
y
= 0 and
x
=
1
±
√
1
−
4
5
2
5
.
Therefore we must require that
5
≤
1
4
for real solutions to exist. We will use the Jacobian matrix
g
x
(
X
)=
,
01
2
5
−
10
.
to attempt to classify ((1
±
√
1
−
4
5
)
/
2
5.
0) when
5
≤
1
/
4. Note that
τ
= 0 and ∆ =
∓
√
1
−
4
5
.For
X
= ((1 +
√
1
−
4
5
)
/
2
5.
0) and
5g
1
/
4, ∆
<
0 and so a saddle point occurs. For
X
= ((1
−
√
1
−
4
5
)
/
2
5.
0)
∆
≥
0 and we are not able to classify this critical point using linearization.
37.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
α
L
x
−
β
L
x
3
−
R
L
y
where
x
=
q
and
y
=
q
x
.If
X
=(
x, y
) is a critical point,
y
= 0 and
−
αx
−
βx
3
=
−
x
(
α
+
βx
2
)=0. If
β>
0,
α
+
βx
2
= 0 has no real solutions and so (0
,
0) is the only critical point. Since
g
x
(
X
)=
w
01
−
α
−
3
βx
2
L
−
R
L
f
,
τ
=
−
R/L <
0 and ∆ =
α/L >
0. Therefore (0
,
0) is a stable critical point. If
β<
0, (0
,
0) and (
±
ˆ
x,
0), where
ˆ
x
2
=
−
α/β
are critical points. At
X
(
±
ˆ
x,
0),
τ
=
−
R/L <
0 and ∆ =
−
2
α/L <
0. Therefore both critical
points are saddles.
38.
If we let
dx/dt
=
y
, then
dy/dt
=
−
x
3
−
x
. From this we obtain the first-order differential equation
dy
dx
=
dy/dt
dx/dt
=
−
x
3
+
x
y
.
Separating variables and integrating we obtain
∂
ydy
=
−
∂
(
x
3
+
x
)
dx
and
1
2
y
2
=
−
1
4
x
4
−
1
2
x
2
+
c
1
.
Completing the square we can write the solution as
y
2
=
−
1
2
(
x
2
+1)
2
+
c
2
.If
X
(0) = (
x
0
,
0), then
c
2
=
1
2
(
x
2
0
+1)
2
and so
y
2
=
−
1
2
(
x
2
+1)
2
+
1
2
(
x
2
0
+1)
2
=
x
4
0
+2
x
2
0
+1
−
x
4
−
2
x
2
−
1
2
=
(
x
2
0
+
x
2
)(
x
2
0
−
x
2
)+2(
x
2
0
−
x
2
)
2
=
(
x
2
0
+
x
2
+2)(
x
2
0
−
x
2
)
2
.
Note that
y
= 0 when
x
=
−
x
0
. In addition, the right-hand side is positive for
−
x
0
<x<x
0
, and so
there are two corresponding values of
y
for each
x
between
−
x
0
and
x
0
. The solution
X
=
X
(
t
) that satisfies
X
(0) = (
x
0
,
0) is therefore periodic, and so (0
,
0) is a center.
39. (a)
Letting
x
=
θ
and
y
=
x
x
we obtain the system
x
x
=
y
and
y
x
=1
/
2
−
sin
x
. Since sin
π/
6 = sin 5
π/
6=1
/
2
we see that (
π/
6
,
0) and (5
π/
6
,
0) are critical points of the system.
547

548. Exercises 11.3
(b)
The Jacobian matrix is
g
x
(
X
)=
,
01
−
cos
x
0
.
and so
A
1
=
g
x
=((
π/
6
,
0)) =
,
01
−
√
3
/
20
.
and
A
2
=
g
x
= ((5
π/
6
,
0)) =
,
01
√
3
/
20
.
.
Since det
A
1
>
0 and the trace of
A
1
is 0, no conclusion can be drawn regarding the critical point (
π/
6
,
0).
Since det
A
2
<
0, we see that (5
π/
6
,
0) is a saddle point.
(c)
From the system in part (a) we obtain the first-order differential equation
dy
dx
=
1
/
2
−
sin
x
y
.
Separating variables and integrating we obtain
∂
ydy
=
∂
,
1
2
−
sin
x
.
dx
and
1
2
y
2
=
1
2
x
+cos
x
+
c
1
or
y
2
=
x
+2 cos
x
+
c
2
.
For
x
0
near
π/
6, if
X
(0) = (
x
0
,
0) then
c
2
=
−
x
0
−
2 cos
x
0
and
y
2
=
x
+2 cos
x
−
x
0
−
2 cos
x
0
. Thus, there
are two values of
y
for each
x
in a sufficiently small interval around
π/
6. Therefore (
π/
6
,
0) is a center.
40. (a)
Writing the system as
x
x
=
x
(
x
3
−
2
y
3
) and
y
x
=
y
(2
x
3
−
y
3
) we see that (0
,
0) is a critical point. Setting
x
3
−
2
y
3
= 0 we have
x
3
=2
y
3
and 2
x
3
−
y
3
=4
y
3
−
y
3
=3
y
3
. Thus, (0
,
0) is the only critical point of the
system.
(b)
From the system we obtain the first-order differential equation
dy
dx
=
2
x
3
y
−
y
4
x
4
−
2
xy
3
or
(2
x
3
y
−
y
4
)
dx
+(2
xy
3
−
x
4
)
dy
=0
which is homogeneous. If we let
y
=
ux
it follows that
(2
x
4
u
−
x
4
u
4
)
dx
+(2
x
4
u
3
−
x
4
)(
udx
+
xdu
)=0
x
4
u
(1 +
u
3
)
dx
+
x
5
(2
u
3
−
1)
du
=0
1
x
dx
+
2
u
3
−
1
u
(
u
3
+1)
du
=0
1
x
dx
+
,
1
u
+1
−
1
u
+
2
u
−
1
u
2
−
u
+1
.
du
=0
.
Integrating gives
ln
|
x
|
+ln
|
u
+1
|−
ln
|
u
|
+ln
|
u
2
−
u
+1
|
=
c
1
548

549. Exercises 11.4
or
x
,
u
+1
u
.
(
u
2
−
u
+1) =
c
2
x
,
y
+
x
y
.,
y
2
x
2
−
y
x
+1
.
=
c
2
(
xy
+
x
2
)(
y
2
−
xy
+
x
2
)=
c
2
x
2
y
xy
3
+
x
4
=
c
2
x
2
y
x
3
+
y
2
=3
c
3
xy.
(c)
We see from the graph that (0
,
0) is unstable. It is not possible to classify
the critical point as a node, saddle, center, or spiral point.
Exercises 11.4
1.
We are given that
x
(0) =
θ
(0) =
π
3
and
y
(0) =
θ
x
(0) =
w
0
. Since
y
2
=
2
g
l
cos
x
+
c
,
w
2
0
=
2
g
l
cos
π
3
+
c
=
g
l
+
c
and so
c
=
w
2
0
−
g
l
. Therefore
y
2
=
2
g
l
,
cos
x
−
1
2
+
l
2
g
w
2
0
.
and the
x
-intercepts occur where cos
x
=
1
2
−
l
2
g
w
2
0
and so
1
2
−
l
2
g
w
2
0
must be greater than
−
1 for solutions to
exist. This condition is equivalent to
|
w
0
|
<
Q
3
g
l
.
2. (a)
Since
y
2
=
2
g
l
cos
x
+
c
,
x
(0) =
θ
(0) =
θ
0
and
y
(0) =
θ
x
(0) = 0,
c
=
−
2
g
l
cos
θ
0
and so
y
2
=
2
g
l
(cos
θ
−
cos
θ
0
).
When
θ
=
−
θ
0
,
y
2
=
2
g
l
(cos(
−
θ
0
)
−
cos(
θ
0
)) = 0. Therefore
y
=
dθ
dt
= 0 when
θ
=
θ
0
.
(b)
Since
y
=
dθ
dt
and
θ
is decreasing between the time when
θ
=
θ
0
,
t
= 0, and
θ
=
−
θ
0
, that is,
t
=
T
,
dθ
dt
=
−
Q
2
g
l
y
cos
θ
−
cos
θ
0
.
Therefore
dt
dθ
=
−
P
l
2
g
1
√
cos
θ
−
cos
θ
0
and so
T
=
−
P
l
2
g
∂
θ
=
−
θ
0
θ
=
θ
0
1
√
cos
θ
−
cos
θ
0
dθ
=
P
l
2
g
∂
θ
0
−
θ
0
1
√
cos
θ
−
cos
θ
0
dθ.
3.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
g
f
x
(
x
)
1+[
f
x
(
x
)]
2
−
β
m
y
549

550. Exercises 11.4
and
∂
∂x
,
−
g
f
x
(
x
)
1+[
f
x
(
x
)]
2
−
β
m
y
.
=
−
g
(1 +[
f
x
(
x
)]
2
)
f
xx
(
x
)
−
f
x
(
x
)2
f
x
(
x
)
f
xx
(
x
)
(1 +[
f
x
(
x
)]
2
)
2
.
If
X
1
=(
x
1
,y
1
) is a critical point,
y
1
= 0 and
f
x
(
x
1
) = 0. The Jacobian at this critical point is therefore
g
x
(
X
1
)=
,
01
−
gf
xx
(
x
1
)
−
β
m
.
.
4.
When
β
= 0 the Jacobian matrix is
,
01
−
gf
xx
(
x
1
)0
.
which has complex eigenvalues
λ
=
±
y
gf
xx
(
x
1
)
i
. The approximating linear system with
x
x
(0) = 0 has solution
x
(
t
)=
x
(0) cos
y
gf
xx
(
x
1
)
t
and period 2
π/
y
gf
xx
(
x
1
) . Therefore
p
≈
2
π/
y
gf
xx
(
x
1
) for the actual solution.
5. (a)
If
f
(
x
)=
x
2
2
,
f
x
(
x
)=
x
and so
dy
dx
=
y
x
x
x
=
−
g
x
1+
x
2
1
y
.
We may separate variables to show that
y
2
=
−
g
ln(1 +
x
2
)+
c
. But
x
(0) =
x
0
and
y
(0) =
x
x
(0) =
v
0
.
Therefore
c
=
v
2
0
+
g
ln(1 +
x
2
0
) and so
y
2
=
v
2
0
−
g
ln
,
1+
x
2
1+
x
2
0
.
.
Now
v
2
0
−
g
ln
,
1+
x
2
1+
x
2
0
.
≥
0 if and only if
x
2
≤
e
v
2
0
/g
(1 +
x
2
0
)
−
1
.
Therefore, if
|
x
|≤
[
e
v
2
0
/g
(1 +
x
2
0
)
−
1]
1
/
2
, there are two values of
y
for a given value of
x
and so the solution
is periodic.
(b)
Since
z
=
x
2
2
, the maximum height occurs at the largest value of
x
on the cycle. From (a),
x
max
=
[
e
v
2
0
/g
(1 +
x
2
0
)
−
1]
1
/
2
and so
z
max
=
x
2
max
2
=
1
2
[
e
v
2
0
/g
(1 +
x
2
0
)
−
1]
.
6. (a)
If
f
(
x
) = cosh
x
,
f
x
(
x
) = sinh
x
and [
f
x
(
x
)]
2
+1 = sinh
2
x
+1 = cosh
2
x
. Therefore
dy
dx
=
y
x
x
x
=
−
g
sinh
x
cosh
2
x
1
y
.
We may separate variables to show that
y
2
=
2
g
cosh
x
+
c
. But
x
(0) =
x
0
and
y
(0) =
x
x
(0) =
v
0
. Therefore
c
=
v
2
0
−
2
g
cosh
x
0
and so
y
2
=
2
g
cosh
x
−
2
g
cosh
x
0
+
v
2
0
.
Now
2
g
cosh
x
−
2
g
cosh
x
0
+
v
2
0
≥
0 if and only if cosh
x
≤
2
g
cosh
x
0
2
g
−
v
2
0
cosh
x
0
550

551. x(t)
y(t)
t
x,
y
20 40 60 80 100
5
10
Exercises 11.4
and the solution to this inequality is an interval [
−
a, a
]. Therefore each
x
in (
−
a, a
) has two corresponding
values of
y
and so the solution is periodic.
(b)
Since
z
= cosh
x
, the maximum height occurs at the largest value of
x
on the cycle. From (a),
x
max
=
a
where cosh
a
=
2
g
cosh
x
0
2
g
−
v
2
0
cosh
x
0
. Therefore
z
max
=
2
g
cosh
x
0
2
g
−
v
2
0
cosh
x
0
.
7.
If
x
m
<x
1
<x
n
, then
F
(
x
1
)
>F
(
x
m
)=
F
(
x
n
). Letting
x
=
x
1
,
G
(
y
)=
c
0
F
(
x
1
)
=
F
(
x
m
)
G
(
a/b
)
F
(
x
1
)
<G
(
a/b
)
.
Therefore from Figure 11
.
35(b),
G
(
y
)=
c
0
F
(
x
1
)
has two solutions
y
1
and
y
2
that satisfy
y
1
< a/b < y
2
.
8.
From (i) when
y
=
a/b
,
x
n
is taken on at some time
t
. From (iii) if
x>x
n
there is no corresponding value of
y
. Therefore the maximum number of predators is
x
n
and
x
n
occurs when
y
=
a/b
.
9. (a)
In the Lotka-Volterra Model the average number of predators is
d/c
and the average number of prey is
a/b
.
But
x
y
=
−
ax
+
bxy
−
α
1
x
=
−
(
a
+
α
1
)
x
+
bxy
y
y
=
−
cxy
+
dy
−
α
2
y
=
−
cxy
+(
d
−
α
2
)
y
and so the new critical point in the first quadrant is (
d/c
−
α
2
/c, a/b
+
α
1
/b
).
(b)
The average number of predators
d/c
−
α
2
/c
has decreased while the average number of prey
a/b
+
α
1
/b
has
increased. The fishery science model is consistent with Volterra’s principle.
10. (a)
Solving
x
(
−
0
.
1+0
.
02
y
)=0
y
(0
.
2
−
0
.
025
x
)=0
in the first quadrant we obtain the critical point
(8
,
5). The graphs are plotted using
x
(0)=7
and
y
(0) = 4.
(b)
The graph in part (a) was obtained using
NDSolve
in
Mathematica
. We see that the period is around
40. Since
x
(0) = 7, we use the
FindRoot
equation solver in
Mathematica
to approximate the solution of
x
(
t
) = 7 for
t
near 40. From this we see that the period is more closely approximated by
t
=44
.
65.
11.
Solving
x
(20
−
0
.
4
x
−
0
.
3
y
)=0
y
(10
−
0
.
1
y
−
0
.
3
x
)=0
we see that critical points are (0
,
0), (0
,
100), (50
,
0), and (20
,
40). The Jacobian matrix is
g
y
(
X
)=
t
0
.
08(20
−
0
.
8
x
−
0
.
3
y
)
−
0
.
024
x
−
0
.
018
y
0
.
06(10
−
0
.
2
y
−
0
.
3
x
)
/
and so
A
1
=
g
y
((0
,
0)) =
t
1
.
60
00
.
6
/
A
3
=
g
y
((50
,
0)) =
t
−
1
.
6
−
1
.
2
0
−
0
.
3
/
A
2
=
g
y
((0
,
100)) =
t
−
0
.
80
−
1
.
8
−
0
.
6
/
A
4
=
g
y
((20
,
40)) =
t
−
0
.
64
−
0
.
48
−
0
.
72
−
0
.
24
/
.
551

552. Exercises 11.4
Since det(
A
1
)=∆
1
=0
.
96
>
0,
τ
=2
.
2
>
0, and
τ
2
1
−
4∆
1
=1
>
0, we see that (0
,
0) is an unstable node.
Since det(
A
2
)=∆
2
=0
.
48
>
0,
τ
=
−
1
.
4
<
0, and
τ
2
2
−
4∆
2
=0
.
04
>
0, we see that (0
,
100) is a stable node.
Since det(
A
3
)=∆
3
=0
.
48
>
0,
τ
=
−
1
.
9
<
0, and
τ
2
3
−
4∆
3
=1
.
69
>
0, we see that (50
,
0) is a stable node.
Since det(
A
4
)=
−
0
.
192
<
0 we see that (20
,
40) is a saddle point.
12.
∆=
r
1
r
2
,
τ
=
r
1
+
r
2
and
τ
2
−
4∆ = (
r
1
+
r
2
)
2
−
4
r
1
r
2
=(
r
1
−
r
2
)
2
. Therefore when
r
1

=
r
2
,(0
,
0) is an
unstable node.
13.
For
X
=(
K
1
,
0),
τ
=
−
r
1
+
r
2
,
1
−
K
1
K
2
α
21
.
and ∆ =
−
r
1
r
2
,
1
−
K
1
K
2
α
21
.
. If we let
c
=1
−
K
1
K
2
α
21
,
τ
2
−
4∆ = (
cr
2
+
r
1
)
2
>
0. Now if
k
1
>
K
2
α
21
,
c<
0 and so
τ<
0, ∆
>
0. Therefore (
K
1
,
0) is a stable node. If
K
1
<
K
2
α
21
,
c>
0 and so ∆
<
0. In this case (
K
1
,
0) is a saddle point.
14.
(ˆ
x,
ˆ
y
) is a stable node if and only if
K
1
α
12
>K
2
and
K
2
α
21
>K
1
. [See Figure 11
.
38(a) in the text.] From Problem
12, (0
.
0) is an unstable node and from Problem 13, since
K
1
<
K
2
α
21
,(
K
1
,
0) is a saddle point. Finally, when
K
2
<
K
1
α
12
,(0
,K
2
) is a saddle point. This is Problem 12 with the roles of 1 and 2 interchanged. Therefore
(0
,
0), (
K
1
,
0), and (0
,K
2
) are unstable.
15.
K
1
α
12
<K
2
<K
1
α
21
and so
α
12
α
21
>
1. Therefore ∆ = (1
−
α
12
α
21
)ˆ
x
ˆ
y
r
1
r
2
K
1
K
2
<
0 and so (ˆ
x,
ˆ
y
) is a saddle
point.
16. (a)
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
g
l
sin
x
−
β
ml
y
and so critical points must satisfy both
y
= 0 and sin
x
= 0. Therefore (
±
nπ,
0) are critical points.
(b)
The Jacobian matrix
w
01
−
g
l
cos
x
−
β
ml
f
has trace
τ
=
−
β
ml
and determinant ∆ =
g
l
>
0at(0
,
0). Therefore
τ
2
−
4∆ =
β
2
m
2
l
2
−
4
g
l
=
β
2
−
4
glm
2
m
2
l
2
.
We may conclude that (0
,
0) is a stable spiral point provided
β
2
−
4
glm
2
<
0or
β<
2
m
√
gl
.
17. (a)
The corresponding plane autonomous system is
x
=
y, y
x
=
−
β
m
y
|
y
|−
k
m
x
and so a critical point must satisfy both
y
= 0 and
x
= 0. Therefore (0
,
0) is the unique critical point.
(b)
The Jacobian matrix is
w
01
−
k
m
−
β
m
2
|
y
|
f
552

553. Exercises 11.4
and so
τ
=0and∆=
k
m
>
0. Therefore (0
,
0) is a center, stable spiral point, or an unstable spiral point.
Physical considerations suggest that (0
,
0) must be asymptotically stable and so (0
,
0) must be a stable
spiral point.
18. (a)
The magnitude of the frictional force between the bead and the wire is
µ
(
mg
cos
θ
) for some
µ>
0. The
component of this frictional force in the
x
-direction is
(
µmg
cos
θ
) cos
θ
=
µmg
cos
2
θ.
But
cos
θ
=
1
y
1+[
f
x
(
x
)]
2
and so
µmg
cos
2
θ
=
µmg
1+[
f
x
(
x
)]
2
.
It follows from Newton’s Second Law that
mx
xx
=
−
mg
f
x
(
x
)
1+[
f
x
(
x
)]
2
−
βx
x
+
mg
µ
1+[
f
x
(
x
)]
2
and so
x
xx
=
g
µ
−
f
x
(
x
)
1+[
f
x
(
x
)]
2
−
β
m
x
x
.
(b)
A critical point (
x, y
) must satisfy
y
= 0 and
f
x
(
x
)=
µ
. Therefore critical points occur at (
x
1
,
0) where
f
x
(
x
1
)=
µ
. The Jacobian matrix of the plane autonomous system is
g
x
(
X
)=


01
g
(1 +[
f
x
(
x
)]
2
)(
−
f
xx
(
x
))
−
(
µ
−
f
x
(
x
))2
f
x
(
x
)
f
xx
(
x
)
(1 +[
f
x
(
x
)]
2
)
2
−
β
m


and so at a critical point
X
1
,
g
x
(
X
)=


01
−
gf
xx
(
x
1
)
1+
µ
2
−
β
m


.
Therefore
τ
=
−
β
m
<
0 and ∆ =
gf
xx
(
x
1
)
1+
µ
2
. When
f
xx
(
x
1
)
<
0, ∆
<
0 and so a saddle point occurs. When
f
xx
(
x
1
)
>
0 and
τ
2
−
4∆ =
β
2
m
2
−
4
g
f
xx
(
x
1
)
1+
µ
2
<
0
,
(
x
1
,
0) is a stable spiral point. This condition may also be written as
β
2
<
4
gm
2
f
xx
(
x
1
)
1+
µ
2
.
19.
dy
dx
=
y
x
x
x
=
−
f
(
x
)
y
and so using separation of variables,
y
2
2
=
−
∂
x
0
f
(
µ
)
dµ
+
c
or
y
2
+2
F
(
x
)=
c
.Wemay
conclude that for a given value of
x
there are at most two corresponding values of
y
.If(0
,
0) were a stable spiral
point there would exist an
x
with more than two corresponding values of
y
. Note that the condition
f
(0) = 0
is required for (0
,
0) to be a critical point of the corresponding plane autonomous system
x
x
=
y
,
y
x
=
−
f
(
x
).
20. (a)
x
x
=
x
(
−
a
+
by
) = 0 implies that
x
=0or
y
=
a/b
.If
x
= 0, then, from
−
cxy
+
r
K
y
(
K
−
y
)=0
,
y
=0or
K
. Therefore (0
,
0) and (0
,K
) are critical points. If ˆ
y
=
a/b
, then
ˆ
y

−
cx
+
r
K
(
K
−
ˆ
y
)

=0
.
553

554. Exercises 11.4
The corresponding value of
x
,
x
=ˆ
x
, therefore satisfies the equation
c
ˆ
x
=
r
K
(
K
−
ˆ
y
).
(b)
The Jacobian matrix is
g
x
(
X
)=
w
−
a
+
by bx
−
cy
−
cx
+
r
K
(
K
−
2
y
)
f
and so at
X
1
=(0
,
0), ∆ =
−
ar <
0. For
X
1
=(0
,K
), ∆ =
n
(
Kb
−
a
)=
−
rb
<
K
−
a
b
τ
. Since we are given
that
K>
a
b
,∆
<
0 in this case. Therefore (0
,
0) and (0
,K
) are each saddle points. For
X
1
=(ˆ
x,
ˆ
y
) where
ˆ
y
=
a
b
and
c
ˆ
x
=
r
K
(
K
−
ˆ
y
), we may write the Jacobian matrix as
g
x
((ˆ
x,
ˆ
y
)) =
w
0
b
ˆ
x
−
c
ˆ
y
−
r
K
ˆ
y
f
and so
τ
=
−
r
K
ˆ
y<
0 and ∆ =
bc
ˆ
x
ˆ
y>
0. Therefore (ˆ
x,
ˆ
y
) is a stable critical point and so it is either a
stable node (perhaps degenerate) or a stable spiral point.
(c)
Write
τ
2
−
4∆ =
r
2
K
2
ˆ
y
2
−
4
bc
ˆ
x
ˆ
y
=ˆ
y
/
r
2
K
2
ˆ
y
−
4
bc
ˆ
x

=ˆ
y
/
r
2
K
2
ˆ
y
−
4
b
r
K
(
K
−
ˆ
y
)

using
c
ˆ
x
=
r
K
(
K
−
ˆ
y
)=
r
K
ˆ
y
c<
r
K
+4
b
τ
ˆ
y
−
4
bK

.
Therefore
τ
2
−
4∆
<
0 if and only if
ˆ
y<
4
bK
r
K
+4
b
=
4
bK
2
r
+4
bK
.
Note that
4
bK
2
r
+4
bK
=
4
bK
r
+4
bK
·
K
≈
K
where
K
is large, and ˆ
y
=
a
b
<K
. Therefore
τ
2
−
4∆
<
0 when
K
is large and a stable spiral point will
result.
21.
The equation
x
x
=
α
y
1+
y
x
−
x
=
x
,
αy
1+
y
−
1
.
=0
implies that
x
=0or
y
=
1
α
−
1
. When
α>
0, ˆ
y
=
1
α
−
1
>
0. If
x
= 0, then from the differential equation for
y
x
,
y
=
β
. On the other hand, if ˆ
y
=
1
α
−
1
,
ˆ
y
1+ˆ
y
=
1
α
and so
1
α
ˆ
x
−
1
α
−
1
+
β
= 0. It follows that
ˆ
x
=
α
,
β
−
1
α
−
1
.
=
α
α
−
1
[(
α
−
1)
β
−
1]
and if
β
(
α
−
1)
>
1, ˆ
x>
0. Therefore (ˆ
x,
ˆ
y
) is the unique critical point in the first quadrant. The Jacobian
matrix is
g
x
(
X
)=



α
y
y
+1
−
1
αx
(1 +
y
)
2
−
y
1+
y
−
x
(1 +
y
)
2
−
1



554

555. Exercises 11.5
and for
X
=(ˆ
x,
ˆ
y
), the Jacobian can be written in the form
g
x
((ˆ
x,
ˆ
y
)) =



0
(
α
−
1)
2
α
ˆ
x
−
1
α
−
(
α
−
1)
2
α
2
−
1



.
It follows that
τ
=
−
/
(
α
−
1)
2
α
2
ˆ
x
+1

<
0
,
∆=
(
α
−
1)
2
α
2
ˆ
x
and so
τ
=
−
(∆ +1). Therefore
τ
2
−
4∆ = (∆ +1)
2
−
4∆ = (∆
−
1)
2
>
0. Therefore (ˆ
x,
ˆ
y
) is a stable node.
22.
Letting
y
=
x
x
we obtain the plane autonomous system
x
x
=
y
y
x
=
−
8
x
+6
x
3
−
x
5
.
Solving
x
5
−
6
x
3
+8
x
=
x
(
x
2
−
4)(
x
2
−
2) = 0 we see that critical points
are (0
,
0), (0
,
−
2), (0
,
2), (0
,
−
√
2 ), and (0
,
√
2 ). The Jacobian matrix
is
g
x
(
X
)=
,
01
−
8+18
x
2
−
5
x
4
0
.
and we see that det(
g
x
(
X
)) = 5
x
4
−
18
x
2
+8 and the trace of
g
x
(
X
)is
0. Since det(
g
x
((
±
√
2
,
0))) =
−
8
<
0, (
±
√
2
,
0) are saddle points. For the other critical points the determinant
is positive and linearization discloses no information. The graph of the phase plane suggests that (0
,
0) and
(
±
2
,
0) are centers.
Exercises 11.5
1.
y
x
=
x
−
y
= 0 implies that
y
=
x
and so 2 +
xy
=2+
x
2
>
0. Therefore the system has no critical points, and
so, by the corollary to Theorem 11
.
4, there are no periodic solutions.
2.
x
x
=2
x
−
xy
=
x
(2
−
y
) = 0 implies that
x
=0or
y
=2. If
x
= 0, then from
−
1
−
x
2
+2
x
−
y
2
=0,
y
2
=
−
1
and there are no real solutions. If
y
=2,
−
x
2
+2
x
−
5 = 0 which has no real solutions. Therefore the system
has no critical points and so, by the corollary to Theorem 11
.
4, there are no periodic solutions.
3.
For
P
=
−
x
+
y
2
and
Q
=
x
−
y
,
∂P
∂x
+
∂Q
∂y
=
−
2
<
0. Therefore there are no periodic solutions by
Theorem 11
.
5.
4.
For
P
=
xy
2
−
x
2
y
and
Q
=
x
2
y
−
1,
∂P
∂x
+
∂Q
∂y
=
y
2
−
2
xy
+
x
2
=(
y
−
x
)
2
. Therefore
∂P
∂x
+
∂Q
∂y
does not
change signs and so there are no periodic solutions by Theorem 11
.
5.
5.
For
P
=
−
µx
−
y
and
Q
=
x
+
y
3
,
∂P
∂x
+
∂Q
∂y
=
−
µ
+9
y
2
>
0 since
µ<
0. Therefore there are no periodic
solutions by Theorem 11
.
5.
6.
From
y
x
=
xy
−
y
=
y
(
x
−
1) = 0 either
y
=0or
x
=1. If
y
= 0, then from 2
x
+
y
2
=0,
x
= 0. Likewise
x
=1
implies that 2 +
y
2
= 0, which has no real solutions. Therefore (0
,
0) is the only critical point. But
g
x
((0
,
0))
has determinant ∆ =
−
2. The single critical point is a saddle point and so, by the corollary to Theorem 11
.
4,
there are no periodic solutions.
555

556. Exercises 11.5
7.
The corresponding plane autonomous system is
x
x
=
y
,
y
x
=2
x
−
y
4
. Therefore (0
,
0) is the only critical point.
But
g
x
((0
,
0)) has determinant ∆ =
−
2
<
0. The single critical point is a saddle point and so, by the corollary
to Theorem 11
.
4, there are no periodic solutions.
8.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
x
+
,
1
2
+3
y
2
.
y
−
x
2
and so
∂P
∂x
+
∂Q
∂y
=0+
1
2
+9
y
2
>
0. Therefore there are no periodic solutions by Theorem 11
.
5.
9.
For
δ
(
x, y
)=
e
ax
+
by
,
∂
∂x
(
δP
)+
∂
∂y
(
δQ
) can be simplified to
e
ax
+
by
[
−
bx
2
−
2
ax
+
axy
+(2
b
+1)
y
]
.
Setting
a
= 0 and
b
=
−
1
/
2,
∂
∂
(
δP
)+
∂
∂y
(
δQ
)=
1
2
x
2
e
−
1
2
y
which does not change signs. Therefore by Theorem 11
.
6 there are no periodic solutions.
10.
For
δ
(
x, y
)=
ax
2
+
by
2
,
∂
∂x
(
δP
)+
∂
∂y
(
δQ
) can be simplified to
−
5
ax
4
−
3
bx
2
y
2
+10(
a
−
b
)
x
2
y.
Setting
a
=
b
=1,
∂
∂x
(
δP
)+
∂
∂y
(
δQ
)=
−
5
x
4
−
3
x
2
y
2
which does not change signs. Therefore by Theorem 11
.
6 there are no periodic solutions.
11.
For
P
=
x
(1
−
x
2
−
3
y
2
) and
Q
=
y
(3
−
x
2
−
3
y
2
),
∂P
∂x
+
∂Q
∂y
= 4(1
−
x
2
−
3
y
2
) and so
∂P
∂x
+
∂Q
∂y
>
0
for
x
2
+3
y
2
<
1. Therefore there are no periodic solutions in the elliptical region
x
2
+3
y
2
<
1.
12.
The corresponding plane autonomous system is
x
x
=
y
,
y
x
=
g
(
x, y
) and so
∂P
∂x
+
∂Q
∂y
=
∂g
∂y
=
∂g
∂x
x

=0
in the region
R
. Therefore
∂P
∂x
+
∂Q
∂y
cannot change signs and so there are no periodic solutions by
Theorem 11
.
5.
13.
For
δ
(
x, y
)=
1
xy
,
δP
=
−
a
y
+
b
,
δQ
=
−
c
+
r
K
1
x
(
K
−
y
) and so
∂
(
δP
)
∂x
+
∂
(
δQ
)
∂y
=
−
r
Kx
<
0
in the first quadrant. By Theorem 11
.
6, there are no periodic solutions in the first quadrant.
14.
If
n
=(
−
2
x,
−
2
y
),
V
·
n
=2
xy
+2
x
2
e
x
+
y
−
2
xy
+2
y
2
e
x
+
y
=2(
x
2
+
y
2
)
e
x
+
y
≥
0
.
Therefore any circular region of the form
x
2
+
y
2
≤
r
2
is an invariant region by Theorem 11
.
7.
15.
If
n
=(
−
2
x,
−
2
y
),
V
·
n
=2
x
2
−
4
xy
+2
y
2
+2
y
4
=2(
x
−
y
)
2
+2
y
4
≥
0
.
556

557. Exercises 11.5
Therefore
x
2
+
y
2
≤
r
serves as an invariant region for any
r>
0 by Theorem 11
.
7.
16. n
=
−∇
t
=(
−
6
x
5
,
−
6
y
). Since the corresponding plane autonomous system is
x
x
=
y
,
y
x
=
−
y
−
y
3
−
x
5
,
n
·
V
=
−
6
x
5
y
+6
y
2
+6
y
4
+6
x
5
y
=6
y
2
+6
y
4
.
Therefore
n
·
V
≥
0 and so by Theorem 11
.
7, the region
x
6
+3
y
2
≤
1 serves as an invariant region.
17.
We showed in Example 8 that
1
16
≤
x
2
+
y
2
≤
1 is an invariant region for the plane autonomous system. If
the only critical point is (0
,
0), this critical point lies outside the invariant region and so Theorem 11
.
8(ii) is
applicable. There is at least one periodic solution in
R
.
18.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
y
(1
−
3
x
2
−
2
y
2
)
−
x
and it is easy to see that (0
,
0) is the only critical point. If
n
=(
−
2
x,
−
2
y
) then
V
·
n
=
−
2
xy
−
2
y
2
(1
−
3
x
2
−
2
y
2
)+2
xy
=
−
2
y
2
(1
−
2
r
2
−
x
2
)
.
If
r
=
1
2
√
2, 2
r
2
= 1 and so
V
·
n
=2
x
2
y
2
≥
0. Therefore
1
4
≤
x
2
+
y
2
≤
1
2
serves as an invariant region. By
Theorem 11
.
8(ii) there is at least one periodic solution.
19.
If
r<
1 and
n
=(
−
2
x,
−
2
y
) then
V
·
n
=
−
2
xy
+2
xy
+2
y
2
(1
−
x
2
)=2
y
2
(1
−
x
2
)
≥
0
since
x
2
<
1. Therefore
x
2
+
y
2
≤
r
2
serves as an invariant region. Now (0
,
0) is the only critical point and,
since
τ
=
−
1 and ∆ = 1,
τ
2
−
4∆
<
0. Therefore (0
,
0) is a stable spiral point and so, by Theorem 11
.
9(ii),
lim
t
→∞
X
(
t
)=(0
,
0).
20.
Since
∂P
∂x
+
∂Q
∂y
=
−
1
−
3
y
2
<
0, there are no periodic solutions. If
n
=(
−
2
x,
−
2
y
),
V
·
n
=
−
2
xy
+2
x
2
+2
xy
+2
y
4
=2(
x
2
+
y
4
)
≥
0
.
Therefore the circular region
x
2
+
y
2
≤
r
2
serves as an invariant region for any
r>
0. If (
x, y
) is a critical point.
y
−
x
=0or
y
=
x
. From
−
x
−
y
3
= 0 we have
−
y
(1 +
y
2
) = 0. Therefore
y
= 0 and so (0
,
0) is the only critical
point. It is easy to check that
τ
=
−
1,∆=1,
τ
2
−
4∆ =
−
3 and so (0
,
0) is a stable spiral point. By Theorem
11
.
9(ii), (0
,
0) is globally stable. For any initial condition, lim
t
→∞
X
(
t
)=(0
,
0).
21. (a)
∂P
∂x
+
∂Q
∂y
=2
xy
−
1
−
x
2
≤
2
x
−
1
−
x
2
=
−
(
x
−
1)
2
≤
0. Therefore there are no periodic solutions.
(b)
If (
x, y
) is a critical point,
x
2
y
=
1
2
and so from
x
x
=
x
2
y
−
x
+1,
1
2
−
x
+1 = 0. Therefore
x
=3
/
2 and so
y
=2
/
9. For this critical point,
τ
=
−
31
/
12
<
0,∆=9
/
4
>
0, and
τ
2
−
4∆
<
0. Therefore (3
/
2
,
2
/
9) is a
stable spiral point and so, from Theorem 11
.
9(ii), lim
x
→∞
X
(
t
)=(3
/
2
,
2
/
9).
22. (a)
From
x
,
2
y
y
+2
−
1
.
= 0, either
x
=0or
y
= 2. For the case
x
= 0, from
y
,
1
−
2
x
y
+2
−
y
8
.
=0,
y
<
1
−
y
8
τ
= 0. Therefore (0
,
0) and (0
,
8) are critical points. If
y
=2,1
−
2
x
4
−
1
4
= 0 and so
x
=3
/
2.
Therefore (3
/
2
,
2) is the additional critical point. We may classify these critical points as follows:
X
τ
∆
τ
2
−
4∆ Conclusion
(0
,
0) –
−
1 – saddle point
(0
,
8) –
−
3
5
– saddle point
(
3
2
,
2)
1
8
3
8
−
95
64
unstable spiral point
557

558. Exercises 11.5
(b)
By Theorem 11
.
8(i), since there is a unique unstable critical point inside the invariant region, there is at
least one periodic solution.
Chapter 11 Review Exercises
1.
True
2.
True
3.
A center or a saddle point
4.
Complex with negative real parts
5.
False; there are initial conditions for which lim
t
→∞
X
(
t
)=(0
,
0).
6.
True
7.
False; this is a borderline case. See Figure 11
.
25 in the text.
8.
False; see Figure 11
.
29 in the text.
9.
True
10.
False; we also need to have no critical points on the boundary of
R
.
11.
Switching to polar coordinates,
dr
dt
=
1
r
,
x
dx
dt
+
y
dy
dt
.
=
1
r
(
−
xy
−
x
2
r
3
+
xy
−
y
2
r
3
)=
−
r
4
dθ
dt
=
1
r
2
,
−
y
dx
dt
+
x
dy
dt
.
=
1
r
2
(
y
2
+
xyr
3
+
x
2
−
xyr
3
)=1
.
Using separation of variables it follows that
r
=
1
3
√
3
t
+
c
1
and
θ
=
t
+
c
2
. Since
X
(0) = (1
,
0),
r
= 1 and
θ
=0.
It follows that
c
1
=1,
c
2
= 0, and so
r
=
1
3
√
3
t
+1
,θ
=
t.
As
t
→∞
,
r
→
0 and the solution spirals toward the origin.
12. (a)
If
X
(0) =
X
0
lies on the line
y
=
−
2
x
, then
X
(
t
) approaches (0
,
0) along this line. For all other initial
conditions,
X
(
t
) approaches (0
,
0) from the direction determined by the line
y
=
x
.
(b)
If
X
(0) =
X
0
lies on the line
y
=
−
x
, then
X
(
t
) approaches (0
,
0) along this line. For all other initial
conditions,
X
(
t
) becomes unbounded and
y
=2
x
serves as an asymptote.
13. (a)
τ
=0,∆=11
>
0 and so (0
,
0) is a center.
(b)
τ
=
−
2, ∆ = 1,
τ
2
−
4∆ = 0 and so (0
,
0) is a degenerate stable node.
14.
From
x
x
=
x
(1 +
y
−
3
x
) = 0, either
x
=0or1+
y
−
3
x
=0. If
x
= 0, then, from
y
(4
−
2
x
−
y
) = 0 we obtain
y
(4
−
y
) = 0. It follows that (0
,
0) and (0
,
4) are critical points. If 1 +
y
−
3
x
= 0, then
y
(5
−
5
x
) = 0. Therefore
(1
/
3
,
0) and (1
,
2) are the remaining critical points. We will use the Jacobian matrix
g
x
(
X
)=
/
1+
y
−
6
xx
−
2
y
4
−
2
x
−
2
y

558

559. Chapter 11 Review Exercises
to classify these four critical points. The results are as follows:
X
τ
∆
τ
2
−
4∆ Conclusion
(0
,
0) 5 4 9 unstable node
(0
,
4) –
−
20 – saddle point
(
1
3
,
0) –
−
10
3
– saddle point
(1
,
2)
−
510
−
15 stable spiral point
If
δ
(
x, y
)=
1
xy
,
δP
=
1
y
+1
−
3
x
y
and
δQ
=
4
x
−
2
−
y
x
. It follows that
∂
∂x
(
δP
)+
∂
∂y
(
δQ
)=
−
3
y
−
1
x
<
0
in quadrant one. Therefore there are no periodic solutions in the first quadrant.
15.
The corresponding plane autonomous system is
x
x
=
y
,
y
x
=
µ
(1
−
x
2
)
−
x
and so the Jacobian at the critical
point (0
,
0) is
g
x
((0
,
0)) =
/
01
−
1
µ

.
Therefore
τ
=
µ
, ∆ = 1 and
τ
2
−
4∆ =
µ
2
−
4. Now
µ
2
−
4
<
0 if and only if
−
2
<µ<
2. We may therefore
conclude that (0
,
0) is a stable node for
µ<
−
2, a stable spiral point for
−
2
<µ<
0, an unstable spiral point
for 0
<µ<
2, and an unstable node for
µ>
2.
16.
Critical points occur at
x
=
±
1. Since
g
x
(
x
)=
−
1
2
e
−
x/
2
(
x
2
−
4
x
−
1)
,
g
x
(1)
>
0 and
g
x
(
−
1)
<
0. Therefore
x
= 1 is unstable and
x
=
−
1 is asymptotically stable.
17.
dy
dx
=
y
x
x
x
=
−
2
x
y
y
2
+1
y
. We may separate variables to show that
y
y
2
+1=
−
x
2
+
c
. But
x
(0) =
x
0
and
y
(0) =
x
x
(0) = 0. It follows that
c
=1+
x
2
0
so that
y
2
=(1+
x
2
0
−
x
2
)
2
−
1
.
Note that 1 +
x
2
0
−
x
2
>
1 for
−
x
0
<x<x
0
and
y
= 0 for
x
=
±
x
0
. Each
x
with
−
x
0
<x<x
0
has two
corresponding values of
y
and so the solution
X
(
t
) with
X
(0) = (
x
0
,
0) is periodic.
18.
The corresponding plane autonomous system is
x
x
=
y, y
x
=
−
β
m
y
−
k
m
(
s
+
x
)
3
+
g
and so the Jacobian is
g
x
(
X
)=
/
01
−
3
k
m
(
s
+
x
)
2
−
β
m

.
For
X
=(0
,
0),
τ
=
−
β
m
<
0, ∆ =
3
k
m
s
2
>
0. Therefore
τ
2
−
4∆ =
β
2
m
2
−
12
k
m
s
2
=
1
m
2
(
β
2
−
12
kms
2
)
.
Therefore (0
,
0) is a stable node if
β
2
>
12
kms
2
and a stable spiral point provided
β
2
<
12
kms
2
, where
ks
3
=
mg
.
559

560. Chapter 11 Review Exercises
19.
For
P
=4
x
+2
y
−
2
x
2
and
Q
=4
x
−
3
y
+4
xy
.
∂P
∂x
+
∂Q
∂y
=4
−
4
x
−
3+4
x
=1
>
0
.
Therefore there are no periodic solutions by Theorem 11
.
5.
20.
If (
x, y
) is a critical point,
P
=
5
+
y
−
x
(
x
2
+
y
2
)=0
Q
=
−
x
+
51
−
y
(
x
2
+
y
2
)=0
.
But
yP
−
xQ
=
y
2
+
x
2
. Therefore
x
2
+
y
2
= 0 and so
x
=
y
= 0. It follows that (0
,
0) is the only critical point
and
g
x
((0
,
0)) =
/
5
1
−
1
5

so that
τ
=2
5
,∆=
5
2
+1
>
0, and so
τ
2
−
4∆ =
−
4. Note that if
n
=(
−
2
x,
−
2
y
),
n
·
V
=2
r
2
(
r
2
−
5
). If
5N
0,
then
τ>
0 and so (0
,
0) is an unstable spiral point. If
r
=2
5
,
n
·
V
≥
0 and so
x
2
+
y
2
≤
4
5
2
is an invariant
region. It follows from Theorem 11
.
8(i) that there is at least one periodic solution in this region. If
5g
0,
τ<
0
and so (0
,
0) is a stable spiral point. Note that
n
·
V
=2
r
2
(
r
2
−
5
)
≥
0 for any
r
and so
x
2
+
y
2
≤
r
2
is an
invariant region. It follows from Theorem 11
.
9(ii) that lim
t
→
+
∞
X
(
t
)=(0
,
0) for any choice of initial position
X
0
.
21. (a)
If
x
=
θ
and
y
=
x
x
=
θ
x
, the corresponding plane autonomous system is
x
x
=
y, y
x
=
ω
2
sin
x
cos
x
−
g
l
sin
x
−
β
ml
y.
Therefore
∂P
∂x
+
∂Q
∂y
=
−
β
ml
<
0 and so there are no periodic solutions.
(b)
If (
x, y
) is a critical point,
y
= 0 and so sin
x
(
ω
2
cos
x
−
g/l
) = 0. Either sin
x
= 0 (in which case
x
=0)
of cos
x
=
g/ω
2
l
. But if
ω
2
<g/l
,
g/ω
2
l>
1 and so the latter equation has no real solutions. Therefore
(0
,
0) is the only critical point if
ω
2
<g/l
. The Jacobian matrix is
g
x
(
X
)=
/
01
ω
2
cos 2
x
−
g
l
cos
x
−
β
ml

and so
τ
=
−
β/ml <
0 and ∆ =
g/l
−
ω
2
>
0 for
X
=(0
,
0). It follows that (0
,
0) is asymptotically stable
and so after a small displacement, the pendulum will return to
θ
=0,
θ
x
=0.
(c)
If
ω
2
>g/l
, cos
x
=
g/ω
2
l
will have two solutions
x
=
±
ˆ
x
that satisfy
−
π<x<π
. Therefore (
±
ˆ
x,
0)
are two additional critical points. If
X
1
=(0
,
0), ∆ =
g/l
−
ω
2
<
0 and so (0
,
0) is a saddle point. If
X
1
=(
±
ˆ
x,
0),
τ
=
−
β/ml <
0 and
∆=
g
l
cos ˆ
x
−
ω
2
cos 2ˆ
x
=
g
2
ω
2
l
2
−
ω
2
,
2
g
2
ω
4
l
2
−
1
.
=
ω
2
−
g
2
ω
2
l
2
>
0
.
Therefore (ˆ
x,
0) and (
−
ˆ
x,
0) are each stable. When
θ
(0) =
θ
0
,
θ
x
(0) = 0 and
θ
0
is small we expect the
pendulum to reach one of these two stable equilibrium positions.
(d)
In (
b
), (0
,
0) is a stable spiral point provided
τ
2
−
4∆ =
β
2
m
2
l
2
−
4
<
g
l
−
ω
2
τ
<
0
.
This condition is equivalent to
β<
2
ml
y
g/l
−
ω
2
.In(
c
), (
±
ˆ
x,
0) are stable spiral points provided that
τ
2
−
4∆ =
β
2
m
2
l
2
−
4
,
ω
2
−
g
2
ω
2
l
2
.
<
0
.
560

561. Chapter 11 Review Exercises
This condition is equivalent to
β<
2
ml
y
ω
2
−
g
2
/
(
ω
2
l
2
).
22.
The corresponding plane autonomous system is
x
x
=
y, y
x
=2
ky
−
cy
3
−
ω
2
x
and so (0
,
0) is the only critical point. Since
g
x
(
X
)=
/
01
−
ω
2
2
k
−
3
cy
2

.
τ
=2
k
and ∆ =
ω
2
>
0. Since
k>
0, (0
,
0) is an unstabale critical point. Assuming that a Type I invariant
region exists that contains (0
,
0) in its interior, we may apply Theorem 11
.
8(i) to conclude that there is at least
one periodic solution.
561

562. 12
Orthogonal Functions and Fourier Series
Exercises 12.1
1.
592
5
2
−
2
xx
2
dx
=
1
4
x
4
9
9
9
9
2
−
2
=0
2.
5
1
−
1
x
3
(
x
2
+1)
dx
=
1
6
x
6
9
9
9
9
1
−
1
+
1
4
x
4
9
9
9
9
1
−
1
=0
3.
5
2
0
e
x
(
xe
−
x
−
e
−
x
)
dx
=
5
2
0
(
x
−
1)
dx
=
2
1
2
x
2
−
x
=
9
9
9
9
2
0
=0
4.
5
π
0
cos
x
sin
2
xdx
=
1
3
sin
3
x
9
9
9
9
π
0
=0
5.
5
π/
2
−
π/
2
x
cos 2
xdx
=
1
2
2
1
2
cos 2
x
+
x
sin 2
x
=
9
9
9
9
π/
2
−
π/
2
=0
6.
5
5
π/
4
π/
4
e
x
sin
xdx
=
2
1
2
e
x
sin
x
−
1
2
e
x
cos
x
=
9
9
9
9
5
π/
4
π/
4
=0
7.
For
m
5
=
n
5
π/
2
0
sin(2
n
+1)
x
sin(2
m
+1)
xdx
=
1
2
5
π/
2
0
[cos 2(
n
−
m
)
x
−
cos 2(
n
+
m
+1)
x
]
dx
=
1
4(
n
−
m
)
sin 2(
n
−
m
)
x
9
9
9
9
π/
2
0
−
1
4(
n
+
m
+1)
sin 2(
n
+
m
+1)
x
9
9
9
9
π/
2
0
=0
.
For
m
=
n
5
π/
2
0
sin
2
(2
n
+
1
)
xdx
=
5
π/
2
0
2
1
2
−
1
2
cos 2(2
n
+1)
x
=
dx
=
1
2
x
9
9
9
9
π/
2
0
−
1
4(2
n
+1)
sin 2(2
n
+1)
x
9
9
9
9
π/
2
0
=
π
4
so that
9
sin(2
n
+1)
x
9
=
1
2
√
π.
562

563. Exercises 12.1
8.
For
m
5
=
n
5
π/
2
0
cos(2
n
+1)
x
cos(2
m
+1)
xdx
=
1
2
5
π/
2
0
[cos 2(
n
−
m
)
x
+ cos 2(
n
+
m
+1)
x
]
dx
=
1
4(
n
−
m
)
sin 2(
n
−
m
)
x
9
9
9
9
π/
2
0
+
1
4(
n
+
m
+1)
sin 2(
n
+
m
+1)
x
9
9
9
9
π/
2
0
=0
.
For
m
=
n
5
π/
2
0
cos
2
(2
n
+1)
xdx
=
5
π/
2
0
2
1
2
+
1
2
cos 2(2
n
+1)
x
=
dx
=
1
2
x
9
9
9
9
π/
2
0
+
1
4(2
n
+1)
sin 2(2
n
+1)
x
9
9
9
9
π/
2
0
=
π
4
so that
9
cos(2
n
+1)
x
9
=
1
2
√
π.
9.
For
m
5
=
n
5
π
0
sin
nx
sin
mx dx
=
1
2
5
π
0
[cos(
n
−
m
)
x
−
cos(
n
+
m
)
x
]
dx
=
1
2(
n
−
m
)
sin(
n
−
m
)
x
9
9
9
9
π
0
−
1
2(
n
+
m
)
sin 2(
n
+
m
)
x
9
9
9
9
π
0
=0
.
For
m
=
n
5
π
0
sin
2
nx dx
=
5
π
0
1
1
2
−
1
2
cos 2
nx
4
dx
=
1
2
x
9
9
9
9
π
0
−
1
4
n
sin 2
nx
9
9
9
9
π
0
=
π
2
so that
9
sin
nx
9
=
0
π
2
.
10.
For
m
5
=
n
5
p
0
sin
nπ
p
x
sin
mπ
p
xdx
=
1
2
5
p
0
2
cos
(
n
−
m
)
π
p
x
−
cos
(
n
+
m
)
π
p
x
=
dx
=
p
2(
n
−
m
)
π
sin
(
n
−
m
)
π
p
x
9
9
9
9
p
0
−
p
2(
n
+
m
)
π
sin
(
n
+
m
)
π
p
x
9
9
9
9
p
0
=0
.
For
m
=
n
5
p
0
sin
2
nπ
p
xdx
=
5
p
0
1
1
2
−
1
2
cos
2
nπ
p
x
4
dx
=
1
2
x
9
9
9
9
p
0
−
p
4
nπ
sin
2
nπ
p
x
9
9
9
9
p
0
=
p
2
so that




sin
nπ
p
x




=
0
p
2
.
563

564. Exercises 12.1
11.
For
m
5
=
n
5
p
0
cos
nπ
p
x
cos
mπ
p
xdx
=
1
2
5
p
0
2
cos
(
n
−
m
)
π
p
x
+ cos
(
n
+
m
)
π
p
x
=
dx
=
p
2(
n
−
m
)
π
sin
(
n
−
m
)
π
p
x
9
9
9
9
p
0
+
p
2(
n
+
m
)
π
sin
(
n
+
m
)
π
p
x
9
9
9
9
p
0
=0
.
For
m
=
n
5
p
0
cos
2
nπ
p
xdx
=
5
p
0
2
1
2
+
1
2
cos
2
nπ
p
x
=
dx
=
1
2
x
9
9
9
9
p
0
+
p
4
nπ
sin
2
nπ
p
x
9
9
9
9
p
0
=
p
2
.
Also
5
p
0
1
·
cos
nπ
p
xdx
=
p
nπ
sin
nπ
p
x
9
9
9
9
p
0
= 0 and
5
p
0
1
2
dx
=
p
so that
9
1
9
=
√
p
and




cos
nπ
p
x




=
0
p
2
.
12.
For
m
5
=
n
, we use Problems 11 and 10:
5
p
−
p
cos
nπ
p
x
cos
mπ
p
xdx
=2
5
p
0
cos
nπ
p
x
cos
mπ
p
xdx
=0
5
p
−
p
sin
nπ
p
x
sin
mπ
p
xdx
=2
5
p
0
sin
nπ
p
x
sin
mπ
p
xdx
=0
.
Also
5
p
−
p
sin
nπ
p
x
cos
mπ
p
xdx
=
1
2
5
p
−
p
2
sin
(
n
−
m
)
π
p
x
+ sin
(
n
+
m
)
π
p
x
=
dx
=0
,
5
p
−
p
1
·
cos
nπ
p
xdx
=
p
nπ
sin
nπ
p
x
9
9
9
9
p
−
p
=0
,
5
p
−
p
1
·
sin
nπ
p
xdx
=
−
p
nπ
cos
nπ
p
x
9
9
9
9
p
−
p
=0
,
and
5
p
−
p
sin
nπ
p
x
cos
nπ
p
xdx
=
5
p
−
p
1
2
sin
2
nπ
p
xdx
=
−
p
4
nπ
cos
2
nπ
p
x
9
9
9
9
p
−
p
=0
.
For
m
=
n
5
p
−
p
cos
2
nπ
p
xdx
=
5
p
−
p
2
1
2
+
1
2
cos
2
nπ
p
x
=
dx
=
p,
5
p
−
p
sin
2
nπ
p
xdx
=
5
p
−
p
2
1
2
−
1
2
cos
2
nπ
p
x
=
dx
=
p,
and
5
p
−
p
1
2
dx
=2
p
so that
9
1
9
=

2
p,




cos
nπ
p
x




=
√
p,
and




sin
nπ
p
x




=
√
p.
564

565. Exercises 12.1
13.
Since
5
∞
−∞
e
−
x
2
·
1
·
2
xdx
=
−
e
−
x
2
9
9
9
9
0
−∞
−
e
−
x
2
9
9
9
9
∞
0
=0
,
5
∞
−∞
e
−
x
2
·
1
·
(4
x
2
−
2)
dx
=2
5
∞
−∞
x

2
xe
−
x
2

dx
−
2
5
∞
−∞
e
−
x
2
dx
=2

−
xe
−
x
2
9
9
9
9
∞
−∞
+
5
∞
−∞
e
−
x
2
dx
o
−
2
5
∞
−∞
e
−
x
2
dx
=2

−
xe
−
x
2
9
9
9
9
0
−∞
−
xe
−
x
2
9
9
9
9
∞
0
o
=0
,
and
5
∞
−∞
e
−
x
2
·
2
x
·
(4
x
2
−
2)
dx
=4
5
∞
−∞
x
2

2
xe
−
x
2

dx
−
4
5
∞
−∞
xe
−
x
2
dx
=4

−
x
2
e
−
x
2
9
9
9
9
∞
−∞
+2
5
∞
−∞
xe
−
x
2
dx
o
−
4
5
∞
−∞
xe
−
x
2
dx
=4

−
x
2
e
−
x
2
9
9
9
9
0
−∞
−
x
2
e
−
x
2
9
9
9
9
∞
0
o
+2
5
∞
−∞
2
xe
−
x
2
dx
=0
,
the functions are orthogonal.
14.
Since
5
∞
0
e
−
x
2
·
1(1
−
x
)
dx
=(
x
−
1)
e
−
x
9
9
9
9
∞
0
−
5
∞
0
e
−
x
dx
=0
,
5
∞
0
e
−
x
·
1
·
2
1
2
x
2
−
2
x
+1
=
dx
=
2
2
x
−
1
−
1
2
x
2
=
e
−
x
9
9
9
9
∞
0
+
5
∞
0
e
−
x
(
x
−
2)
dx
=1+(2
−
x
)
e
−
x
9
9
9
9
∞
0
+
5
∞
0
e
−
x
dx
=0
,
and
5
∞
0
e
−
x
·
(1
−
x
)
2
1
2
x
2
−
2
x
+1
=
dx
=
5
∞
0
e
−
x
2
−
1
2
x
3
+
5
2
x
2
−
3
x
+1
=
dx
=
e
−
x
2
1
2
x
3
−
5
2
x
2
+3
x
−
1
=
9
9
9
9
∞
0
+
5
∞
0
e
−
x
2
−
3
2
x
2
+5
x
−
3
=
dx
=1+
e
−
x
2
3
2
x
2
−
5
x
+3
=
9
9
9
9
∞
0
+
5
∞
0
e
−
x
(5
−
3
x
)
dx
=1
−
3+
e
−
x
(3
x
−
5)
9
9
9
9
∞
0
−
3
5
∞
0
e
−
x
dx
=0
,
the functions are orthogonal.
15.
By orthogonality
s
b
a
φ
0
(
x
)
φ
n
(
x
)
dx
= 0 for
n
=1,2,3,
...
; that is,
s
b
a
φ
n
(
x
)
dx
= 0 for
n
=1,2,3,
...
.
565

566. Exercises 12.1
16.
Using the facts that
φ
0
and
φ
1
are orthogonal to
φ
n
for
n>
1, we have
5
b
a
(
αx
+
β
)
φ
n
(
x
)
dx
=
α
5
b
a
xφ
n
(
x
)
dx
+
β
5
b
a
1
·
φ
n
(
x
)
dx
=
α
5
b
a
φ
1
(
x
)
φ
n
(
x
)
dx
+
β
5
b
a
φ
0
(
x
)
φ
n
(
x
)
dx
=
α
·
0+
β
·
0=0
for
n
=2,3,4,
...
.
17.
Using the fact that
φ
n
and
φ
m
are orthogonal for
n
5
=
m
we have
9
φ
m
(
x
)+
φ
n
(
x
)
9
2
=
5
b
a
[
φ
m
(
x
)+
φ
n
(
x
)]
2
dx
5
b
a
i
φ
2
m
(
x
)+2
φ
m
(
x
)
φ
n
(
x
)+
φ
2
n
(
x
)
n
dx
=
5
b
a
φ
2
m
(
x
)
dx
+2
5
b
a
φ
m
(
x
)
φ
n
(
x
)
dx
+
5
b
a
φ
2
n
(
x
)
dx
=
9
φ
m
(
x
)
9
2
+
9
φ
n
(
x
)
9
2
.
18.
Setting
0=
5
2
−
2
f
3
(
x
)
f
1
(
x
)
dx
=
5
2
−
2
3
x
2
+
c
1
x
3
+
c
2
x
4
F
dx
=
16
3
+
64
5
c
2
and
0=
5
2
−
2
f
3
(
x
)
f
2
(
x
)
dx
=
5
2
−
2
3
x
3
+
c
1
x
4
+
c
2
x
5
F
dx
=
64
5
c
1
we obtain
c
1
= 0 and
c
2
=
−
5
/
12.
19.
Since sin
nx
is an odd function on [
−
π, π
],
(1
,
sin
nx
)=
5
π
−
π
sin
nx dx
=0
and
f
(
x
) = 1 is orthogonal to every member of
{
sin
nx
}
.Thus
{
sin
nx
}
is not complete.
20.
(
f
1
+
f
2
,f
3
)=
5
b
a
[
f
1
(
x
)+
f
2
(
x
)]
f
3
(
x
)
dx
=
5
b
a
f
1
(
x
)
f
3
(
x
)
dx
+
5
b
a
f
2
(
x
)
f
3
(
x
)
dx
=(
f
1
,f
3
)+(
f
2
,f
3
)
21. (a)
The fundamental period is 2
π/
2
π
=1.
(b)
The fundamental period is 2
π/
(4
/L
)=
1
2
πL
.
(c)
The fundamental period of sin
x
+ sin 2
x
is 2
π
.
(d)
The fundamental period of sin 2
x
+ cos 4
x
is 2
π/
2=
π
.
(e)
The fundamental period of sin 3
x
+ cos 4
x
is 2
π
since the smallest integer multiples of 2
π/
3 and 2
π/
4=
π/
2
that are equal are 3 and 4, respectively.
(f)
The fundamental period of
f
(
x
)is2
π/
(
nπ/p
)=2
p/n
.
566

567. Exercises 12.2
Exercises 12.2
1.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
π
0
1
dx
=1
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nπ
π
xdx
=
1
π
5
π
0
cos
nx dx
=0
b
n
=
1
π
5
π
−
π
f
(
x
) sin
nπ
π
xdx
=
1
π
5
π
0
sin
nx dx
=
1
nπ
(1
−
cos
nπ
)=
1
nπ
[1
−
(
−
1)
n
]
f
(
x
)=
1
2
+
1
π
∞
r
n
=1
1
−
(
−
1)
n
n
sin
nx
2.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
0
−
π
−
1
dx
+
1
π
5
π
0
2
dx
=1
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nx dx
=
1
π
5
0
−
π
−
cos
nx dx
+
1
π
5
π
0
2 cos
nx dx
=0
b
n
=
1
π
5
π
−
π
f
(
x
) sin
nx dx
=
1
π
5
0
−
π
−
sin
nx dx
+
1
π
5
π
0
2 sin
nx dx
=
3
nπ
[1
−
(
−
1)
n
]
f
(
x
)=
1
2
+
3
π
∞
r
n
=1
1
−
(
−
1)
n
n
sin
nx
3.
a
0
=
5
1
−
1
f
(
x
)
dx
=
5
0
−
1
1
dx
+
5
1
0
xdx
=
3
2
a
n
=
5
1
−
1
f
(
x
) cos
nπx dx
=
5
0
−
1
cos
nπx dx
+
5
1
0
x
cos
nπx dx
=
1
n
2
π
2
[(
−
1)
n
−
1]
b
n
=
5
1
−
1
f
(
x
) sin
nπx dx
=
5
0
−
1
sin
nπx dx
+
5
1
0
x
sin
nπx dx
=
−
1
nπ
f
(
x
)=
3
4
+
∞
r
n
=1
1
(
−
1)
n
−
1
n
2
π
2
cos
nπx
−
1
nπ
sin
nπx
4
4.
a
0
=
5
1
−
1
f
(
x
)
dx
=
5
1
0
xdx
=
1
2
a
n
=
5
1
−
1
f
(
x
) cos
nπx dx
=
5
1
0
x
cos
nπx dx
=
1
n
2
π
2
[(
−
1)
n
−
1]
b
n
=
5
1
−
1
f
(
x
) sin
nπx dx
=
5
1
0
x
sin
nπx dx
=
(
−
1)
n
+1
nπ
f
(
x
)=
3
4
+
∞
r
n
=1
1
(
−
1)
n
−
1
n
2
π
2
cos
nπx
+
(
−
1)
n
+1
nπ
sin
nπx
4
5.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
π
0
x
2
dx
=
1
3
π
2
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nx dx
=
1
π
5
π
0
x
2
cos
nx dx
=
1
π
2
x
2
π
sin
nx
9
9
9
9
π
0
−
2
n
5
π
0
x
sin
nx dx
=
=
2(
−
1)
n
n
2
b
n
=
1
π
5
π
0
x
2
sin
nx dx
=
1
π
2
−
x
2
n
cos
nx
9
9
9
9
π
0
+
2
n
5
π
0
x
cos
nx dx
=
=
π
n
(
−
1)
n
+1
+
2
n
3
π
[(
−
1)
n
−
1]
567

568. Exercises 12.2
f
(
x
)=
π
2
6
+
∞
r
n
=1
1
2(
−
1)
n
n
2
cos
nx
+
2
π
n
(
−
1)
n
+1
+
2[(
−
1)
n
−
1]
n
3
π
=
sin
nx
4
6.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
0
−
π
π
2
dx
+
1
π
5
π
0
3
π
2
−
x
2
F
dx
=
5
3
π
2
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nx dx
=
1
π
5
0
−
π
π
2
cos
nx dx
+
1
π
5
π
0
3
π
2
−
x
2
F
cos
nx dx
=
1
π
2
π
2
−
x
2
n
sin
nx
9
9
9
9
π
0
+
2
n
5
π
0
x
sin
nx dx
=
=
2
n
2
(
−
1)
n
+1
b
n
=
1
π
5
π
−
π
f
(
x
) sin
nx dx
=
1
π
5
0
−
π
π
2
sin
nx dx
+
1
π
5
π
0
3
π
2
−
x
2
F
sin
nx dx
=
π
n
[(
−
1)
n
−
1] +
1
π
2
x
2
−
π
2
n
cos
nx
9
9
9
9
π
0
−
2
n
5
π
0
x
cos
nx dx
=
=
π
n
(
−
1)
n
+
2
n
3
π
[1
−
(
−
1)
n
]
f
(
x
)=
5
π
2
6
+
∞
r
n
=1
1
2
n
2
(
−
1)
n
+1
cos
nx
+
2
π
n
(
−
1)
n
+
2[1
−
(
−
1
n
]
n
3
π
=
sin
nx
4
7.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
π
−
π
(
x
+
π
)
dx
=2
π
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nx dx
=
1
π
5
π
−
π
(
x
+
π
) cos
nx dx
=0
b
n
=
1
π
5
π
−
π
f
(
x
) sin
nx dx
=
2
n
(
−
1)
n
+1
f
(
x
)=
π
+
∞
r
n
=1
2
n
(
−
1)
n
+1
sin
nx
8.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
π
−
π
(3
−
2
x
)
dx
=6
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nx dx
=
1
π
5
π
−
π
(3
−
2
x
) cos
nx dx
=0
b
n
=
1
π
5
π
−
π
(3
−
2
x
) sin
nx dx
=
4
n
(
−
1)
n
f
(
x
)=3+4
∞
r
n
=1
(
−
1)
n
n
sin
nx
9.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
π
0
sin
xdx
=
2
π
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nx dx
=
1
π
5
π
0
sin
x
cos
nx dx
=
1
2
π
5
π
0
[sin(
n
+1)
x
+ sin(1
−
n
)
x
]
dx
=
1+(
−
1)
n
π
(1
−
n
2
)
for
n
=2
,
3
,
4
,...
a
1
=
1
2
π
5
π
0
sin 2
xdx
=0
b
n
=
1
π
5
π
−
π
f
(
x
) sin
nx dx
=
1
π
5
π
0
sin
x
sin
nx dx
=
1
2
π
5
π
0
[cos(1
−
n
)
x
−
cos(1 +
n
)
x
]
dx
= 0 for
n
=2
,
3
,
4
,...
568

569. Exercises 12.2
b
1
=
1
2
π
5
π
0
(1
−
cos 2
x
)
dx
=
1
2
f
(
x
)=
1
π
+
1
2
sin
x
+
∞
r
n
=2
1+(
−
1)
n
π
(1
−
n
2
)
cos
nx
10.
a
0
=
2
π
5
π/
2
−
π/
2
f
(
x
)
dx
=
2
π
5
π/
2
0
cos
xdx
=
2
π
a
n
=
2
π
5
π/
2
−
π/
2
f
(
x
) cos 2
nx dx
=
2
π
5
π/
2
0
cos
x
cos 2
nx dx
=
1
π
5
π/
2
0
[cos(2
n
−
1)
x
+ cos(2
n
+1)
x
]
dx
=
2(
−
1)
n
+1
π
(4
n
2
−
1)
b
n
=
2
π
5
π/
2
−
π/
2
f
(
x
) sin 2
nx dx
=
2
π
5
π/
2
0
cos
x
sin 2
nx dx
=
1
π
5
π/
2
0
[sin(2
n
−
1)
x
+ sin(2
n
+1)
x
]
dx
=
4
n
π
(4
n
2
−
1)
f
(
x
)=
1
π
+
∞
r
n
=1
1
2(
−
1)
n
+1
π
(4
n
2
−
1)
cos 2
nx
+
4
n
π
(4
n
2
−
1)
sin 2
nx
4
11.
a
0
=
1
2
5
2
−
2
f
(
x
)
dx
=
1
2
2
5
0
−
1
−
2
dx
+
5
1
0
1
dx
=
=
−
1
2
a
n
=
1
2
5
2
−
2
f
(
x
) cos
nπ
2
xdx
=
1
2
2
5
0
−
1
[
−
2 cos
nπ
2
x
]
dx
+
5
1
0
cos
nπ
2
xdx
=
=
−
1
nπ
sin
nπ
2
b
n
=
1
2
5
2
−
2
f
(
x
) sin
nπ
2
xdx
=
1
2
2
5
0
−
1
[
−
2 sin
nπ
2
x
]
dx
+
5
1
0
sin
nπ
2
xdx
=
=
3
nπ

1
−
cos
nπ
2

f
(
x
)=
−
1
4
+
∞
r
n
=1
1
−
1
nπ
sin
nπ
2
cos
nπ
2
x
+
3
nπ

1
−
cos
nπ
2

sin
nπ
2
x
4
12.
a
0
=
1
2
5
2
−
2
f
(
x
)
dx
=
1
2
2
5
1
0
xdx
+
5
2
1
1
dx
=
=
3
4
a
n
=
1
2
5
2
−
2
f
(
x
) cos
nπ
2
xdx
=
1
2
2
5
1
0
x
cos
nπ
2
xdx
+
5
2
1
cos
nπ
2
xdx
=
=
2
n
2
π
2

cos
nπ
2
−
1

b
n
=
1
2
5
2
−
2
f
(
x
) sin
nπ
2
xdx
=
1
2
2
5
1
0
x
sin
nπ
2
xdx
+
5
2
1
sin
nπ
2
xdx
=
=
2
n
2
π
2

sin
nπ
2
+
nπ
2
(
−
1)
n
+1

f
(
x
)=
3
8
+
∞
r
n
=1
1
2
n
2
π
2

cos
nπ
2
−
1

cos
nπ
2
x
+
2
n
2
π
2

sin
nπ
2
+
nπ
2
(
−
1)
n
+1

sin
nπ
2
x
4
13.
a
0
=
1
5
5
5
−
5
f
(
x
)
dx
=
1
5
2
5
0
−
5
1
dx
+
5
5
0
(1 +
x
)
dx
=
=
9
2
a
n
=
1
5
5
5
−
5
f
(
x
) cos
nπ
5
xdx
=
1
5
2
5
0
−
5
cos
nπ
5
xdx
+
5
5
0
(1 +
x
) cos
nπ
5
xdx
=
=
5
n
2
π
2
[(
−
1)
n
−
1]
b
n
=
1
5
5
5
−
5
f
(
x
) sin
nπ
5
xdx
=
1
5
2
5
0
−
5
sin
nπ
5
xdx
+
5
5
0
(1 +
x
) cos
nπ
5
xdx
=
=
5
nπ
(
−
1)
n
+1
f
(
x
)=
9
4
+
∞
r
n
=1
1
5
n
2
π
2
[(
−
1)
n
−
1] cos
nπ
5
x
+
5
nπ
(
−
1)
n
+1
sin
nπ
5
x
4
569

570. Exercises 12.2
14.
a
0
=
1
2
5
2
−
2
f
(
x
)
dx
=
1
2
2
5
0
−
2
(2 +
x
)
dx
+
5
2
0
2
dx
=
=3
a
n
=
1
2
5
2
−
2
f
(
x
) cos
nπ
2
xdx
=
1
2
2
5
0
−
2
(2 +
x
) cos
nπ
2
xdx
+
5
2
0
2 cos
nπ
2
xdx
=
=
2
n
2
π
2
[1
−
(
−
1)
n
]
b
n
=
1
2
5
2
−
2
f
(
x
) sin
nπ
2
xdx
=
1
2
2
5
0
−
2
(2 +
x
) sin
nπ
2
xdx
+
5
2
0
2 sin
nπ
2
xdx
=
=
2
nπ
(
−
1)
n
+1
f
(
x
)=
3
2
+
∞
r
n
=1
1
2
n
2
π
2
[1
−
(
−
1)
n
] cos
nπ
2
x
+
2
nπ
(
−
1)
n
+1
sin
nπ
2
x
4
15.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
π
−
π
e
x
dx
=
1
π
(
e
π
−
e
−
π
)
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nx dx
=
(
−
1)
n
(
e
π
−
e
−
π
)
π
(1 +
n
2
)
b
n
=
1
π
5
π
−
π
f
(
x
) sin
nx dx
=
1
π
5
π
−
π
e
x
sin
nx dx
=
(
−
1)
n
n
(
e
−
π
−
e
π
)
π
(1 +
n
2
)
f
(
x
)=
e
π
−
e
−
π
2
π
+
∞
r
n
=1
1
(
−
1)
n
(
e
π
−
e
−
π
)
π
(1 +
n
2
)
cos
nx
+
(
−
1)
n
n
(
e
−
π
−
e
π
)
π
(1 +
n
2
)
sin
nx
4
16.
a
0
=
1
π
5
π
−
π
f
(
x
)
dx
=
1
π
5
π
0
(
e
x
−
1)
dx
=
1
π
(
e
π
−
π
−
1)
a
n
=
1
π
5
π
−
π
f
(
x
) cos
nx dx
=
1
π
5
π
0
(
e
x
−
1) cos
nx dx
=
[
e
π
(
−
1)
n
−
1]
π
(1 +
n
2
)
b
n
=
1
π
5
π
−
π
f
(
x
) sin
nx dx
=
1
π
5
π
0
(
e
x
−
1) sin
nx dx
=
1
π
2
ne
π
(
−
1)
n
+1
1+
n
2
+
n
1+
n
2
+
(
−
1)
n
n
−
1
n
=
f
(
x
)=
e
π
−
π
−
1
2
π
+
∞
r
n
=1
1
e
π
(
−
1)
n
−
1
π
(1 +
n
2
)
cos
nx
+
2
n
1+
n
2
i
e
π
(
−
1)
n
+1
+1
n
+
(
−
1)
n
−
1
n
=
sin
nx
4
17.
The function in Problem 5 is discontinuous at
x
=
π
, so the corresponding Fourier series converges to
π
2
/
2at
x
=
π
. That is,
π
2
2
=
π
2
6
+
∞
r
n
=1
1
2(
−
1)
n
n
2
cos
nπ
+
2
π
n
(
−
1)
n
+1
+
2[(
−
1)
n
−
1]
n
3
π
=
sin
nπ
4
=
π
2
6
+
∞
r
n
=1
2(
−
1)
n
n
2
(
−
1)
n
=
π
2
6
+
∞
r
n
=1
2
n
2
=
π
2
6
+2
2
1+
1
2
2
+
1
3
2
+
···
=
and
π
2
6
=
1
2
2
π
2
2
−
π
2
6
=
=1+
1
2
2
+
1
3
2
+
···
.
At
x
= 0 the series converges to 0 and
0=
π
2
6
+
∞
r
n
=1
2(
−
1)
n
n
2
=
π
2
6
+2
2
−
1+
1
2
2
−
1
3
2
+
1
4
2
−···
=
so
π
2
12
=1
−
1
2
2
+
1
3
2
−
1
4
2
+
···
.
18.
From Problem 17
π
2
8
=
1
2
2
π
2
6
+
π
2
12
=
=
1
2
2
2+
2
3
2
+
2
5
2
+
···
=
=1+
1
3
2
+
1
5
2
+
···
.
570

571. Exercises 12.2
19.
The function in Problem 7 is continuous at
x
=
π/
2so
3
π
2
=
f

π
2

=
π
+
∞
r
n
=1
2
n
(
−
1)
n
+1
sin
nπ
2
=
π
+2
2
1
−
1
3
+
1
5
−
1
7
+
···
=
and
π
4
=1
−
1
3
+
1
5
−
1
7
+
···
.
20.
The function in Problem 9 is continuous at
x
=
π/
2so
1=
f

π
2

=
1
π
+
1
2
+
∞
r
n
=2
1+(
−
1)
n
π
(1
−
n
2
)
cos
nπ
2
=
1
π
+
1
2
+
2
3
π
−
2
3
·
5
π
+
2
5
·
7
π
−···
and
π
=1+
π
2
+
2
3
−
2
3
·
5
+
2
5
·
7
−···
or
π
4
=
1
2
+
1
1
·
3
−
1
3
·
5
+
1
5
·
7
−···
.
21.
Writing
f
(
x
)=
a
0
2
+
a
1
cos
π
p
x
+
···
+
a
n
cos
nπ
p
x
+
···
+
b
1
sin
π
p
x
+
···
+
b
n
sin
nπ
p
x
+
···
we see that
f
2
(
x
) consists exclusively of squared terms of the form
a
2
0
4
,a
2
n
cos
2
nπ
p
x, b
2
n
sin
2
nπ
p
x
and cross-product terms, with
m
5
=
n
, of the form
a
0
a
n
cos
nπ
p
x, a
0
b
n
sin
nπ
p
x,
2
a
m
a
n
cos
mπ
p
x
cos
nπ
p
x,
2
a
m
b
n
cos
mπ
p
x
sin
nπ
p
x,
2
b
m
b
n
sin
mπ
p
x
sin
nπ
p
x.
The integral of each cross-product term taken over the interval (
−
p, p
) is zero by orthogonality. For the squared
terms we have
a
2
0
4
5
p
−
p
dx
=
a
2
0
p
2
,a
2
n
5
p
−
p
cos
2
nπ
p
xdx
=
a
2
n
p, b
2
n
5
p
−
p
sin
2
nπ
p
xdx
=
b
2
n
p.
Thus
RMS
(
f
)=
t
h
h
a
1
4
a
2
0
+
1
2
∞
r
n
=1
(
a
2
n
+
b
2
n
)
.
571

572. Exercises 12.3
Exercises 12.3
1.
Since
f
(
−
x
) = sin(
−
3
x
)=
−
sin 3
x
=
−
f
(
x
),
f
(
x
) is an odd function.
2.
Since
f
(
−
x
)=
−
x
cos(
−
x
)=
−
x
cos
x
=
−
f
(
x
),
f
(
x
) is an odd function.
3.
Since
f
(
−
x
)=(
−
x
)
2
−
x
=
x
2
−
x
,
f
(
x
) is neither even nor odd.
4.
Since
f
(
−
x
)=(
−
x
)
3
+4
x
=
−
(
x
3
−
4
x
)=
−
f
(
x
),
f
(
x
) is an odd function.
5.
Since
f
(
−
x
)=
e
|−
x
|
=
e
|
x
|
=
f
(
x
),
f
(
x
) is an even function.
6.
Since
f
(
−
x
)=
e
−
x
−
e
x
=
−
f
(
x
),
f
(
x
) is an odd function.
7.
For 0
<x<
1,
f
(
−
x
)=(
−
x
)
2
=
x
2
=
−
f
(
x
),
f
(
x
) is an odd function.
8.
For 0
≤
x<
2,
f
(
−
x
)=
−
x
+5=
f
(
x
),
f
(
x
) is an even function.
9.
Since
f
(
x
) is not defined for
x<
0, it is neither even nor odd.
10.
Since
f
(
−
x
)=
9
9
(
−
x
)
5
9
9
=
9
9
x
5
9
9
=
f
(
x
),
f
(
x
) is an even function.
11.
Since
f
(
x
) is an odd function, we expand in a sine series:
b
n
=
2
π
5
π
0
1
·
sin
nx dx
=
2
nπ
[1
−
(
−
1)
n
]
.
Thus
f
(
x
)=
∞
r
n
=1
2
nπ
[1
−
(
−
1)
n
] sin
nx.
12.
Since
f
(
x
) is an even function, we expand in a cosine series:
a
0
=
5
2
1
1
dx
=1
a
n
=
5
2
1
cos
nπ
2
xdx
=
−
2
nπ
sin
nπ
2
.
Thus
f
(
x
)=
1
2
+
∞
r
n
=1
−
2
nπ
sin
nπ
2
cos
nπ
2
x.
13.
Since
f
(
x
) is an even function, we expand in a cosine series:
a
0
=
2
π
5
π
0
xdx
=
π
a
n
=
2
π
5
π
0
x
cos
nx dx
=
2
n
2
π
[(
−
1)
n
−
1]
.
Thus
f
(
x
)=
π
2
+
∞
r
n
=1
2
n
2
π
[(
−
1)
n
−
1] cos
nx.
14.
Since
f
(
x
) is an odd function, we expand in a sine series:
b
n
=
2
π
5
π
0
x
sin
nx dx
=
2
n
(
−
1)
n
+1
.
Thus
f
(
x
)=
∞
r
n
=1
2
n
(
−
1)
n
+1
sin
nx.
572

573. Exercises 12.3
15.
Since
f
(
x
) is an even function, we expand in a cosine series:
a
0
=2
5
1
0
x
2
dx
=
2
3
a
n
=2
5
1
0
x
2
cos
nπx dx
=2

x
2
nπ
sin
nπx
9
9
9
9
1
0
−
2
nπ
5
1
0
x
sin
nπx dx
o
=
4
n
2
π
2
(
−
1)
n
.
Thus
f
(
x
)=
1
3
+
∞
r
n
=1
4
n
2
π
2
(
−
1)
n
cos
nπx.
16.
Since
f
(
x
) is an odd function, we expand in a sine series:
b
n
=2
5
1
0
x
2
sin
nπx dx
=2

−
x
2
nπ
cos
nπx
9
9
9
9
1
0
+
2
nπ
5
1
0
x
cos
nπx dx
o
=
2(
−
1)
n
+1
nπ
+
4
n
3
π
3
[(
−
1)
n
−
1]
.
Thus
f
(
x
)=
∞
r
n
=1
2
2(
−
1)
n
+1
nπ
+
4
n
3
π
3
[(
−
1)
n
−
1]
=
sin
nπx.
17.
Since
f
(
x
) is an even function, we expand in a cosine series:
a
0
=
2
π
5
π
0
(
π
2
−
x
2
)
dx
=
4
3
π
2
a
n
=
2
π
5
π
0
(
π
2
−
x
2
) cos
nx dx
=
2
π
2
π
2
−
x
2
n
sin
nx
9
9
9
9
π
0
+
2
n
5
π
0
x
sin
nx dx
=
=
4
n
2
(
−
1)
n
+1
.
Thus
f
(
x
)=
2
3
π
2
+
∞
r
n
=1
4
n
2
(
−
1)
n
+1
cos
nx dx.
18.
Since
f
(
x
) is an odd function, we expand in a sine series:
b
n
=
2
π
5
π
0
x
3
sin
nx dx
=
2
π
2
−
x
3
n
cos
nx
9
9
9
9
π
0
+
3
n
5
π
0
x
2
cos
nx dx
=
=
2
π
2
n
(
−
1)
n
+1
−
12
n
2
π
5
π
0
x
sin
nx dx
=
2
π
2
n
(
−
1)
n
+1
−
12
n
2
π
2
−
x
n
cos
nx
9
9
9
9
π
0
+
1
n
5
π
0
cos
nx dx
=
=
2
π
2
n
(
−
1)
n
+1
+
12
n
3
(
−
1)
n
.
Thus
f
(
x
)=
∞
r
n
=1
2
2
π
2
n
(
−
1)
n
+1
+
12
n
3
(
−
1)
n
=
sin
nx.
19.
Since
f
(
x
) is an odd function, we expand in a sine series:
b
n
=
2
π
5
π
0
(
x
+ 1) sin
nx dx
=
2(
π
+1)
nπ
(
−
1)
n
+1
+
2
nπ
.
Thus
f
(
x
)=
∞
r
n
=1
2
2(
π
+1)
nπ
(
−
1)
n
+1
+
2
nπ
=
sin
nx.
573

574. Exercises 12.3
20.
Since
f
(
x
) is an odd function, we expand in a sine series:
b
n
=2
5
1
0
(
x
−
1) sin
nπx dx
=2
1
5
1
0
x
sin
nπx dx
−
5
1
0
sin
nπx dx
4
=2
1
1
n
2
π
2
sin
nπx
−
x
nπ
cos
nπx
+
1
nπ
cos
nπx
4
1
0
=
−
2
nπ
.
Thus
f
(
x
)=
−
∞
r
n
=1
2
nπ
sin
nπx.
21.
Since
f
(
x
) is an even function, we expand in a cosine series:
a
0
=
5
1
0
xdx
+
5
2
1
1
dx
=
3
2
a
n
=
5
1
0
x
cos
nπ
2
xdx
+
5
2
1
cos
nπ
2
xdx
=
4
n
2
π
2

cos
nπ
2
−
1

.
Thus
f
(
x
)=
3
4
+
∞
r
n
=1
4
n
2
π
2

cos
nπ
2
−
1

cos
nπ
2
x.
22.
Since
f
(
x
) is an odd function, we expand in a sine series:
b
n
=
1
π
5
π
0
x
sin
n
2
xdx
+
5
2
π
π
π
sin
n
2
xdx
=
4
n
2
π
sin
nπ
2
+
2
n
(
−
1)
n
+1
.
Thus
f
(
x
)=
∞
r
n
=1
2
4
n
2
π
sin
nπ
2
+
2
n
(
−
1)
n
+1
=
sin
n
2
x.
23.
Since
f
(
x
) is an even function, we expand in a cosine series:
a
0
=
2
π
5
π
0
sin
xdx
=
4
π
a
n
=
2
π
5
π
0
sin
x
cos
nx dx
=
1
π
5
π
0
[sin(
n
+1)
x
+ sin(1
−
n
)
x
]
dx
=
2
π
(1
−
n
2
)
[1+(
−
1)
n
] for
n
=2
,
3
,
4
,...
a
1
=
1
π
5
π
0
sin 2
xdx
=0
.
Thus
f
(
x
)=
2
π
+
∞
r
n
=2
2[1+(
−
1)
n
]
π
(1
−
n
2
)
cos
nx.
24.
Since
f
(
x
) is an even function, we expand in a cosine series. [See the solution of Problem 10 in Exercise 12
.
2
for the computation of the integrals.]
a
0
=
2
π/
2
5
π/
2
0
cos
xdx
=
4
π
a
n
=
2
π/
2
5
π/
2
0
cos
x
cos
nπ
π/
2
xdx
=
4(
−
1)
n
+1
π
(4
n
2
−
1)
574

575. Exercises 12.3
Thus
f
(
x
)=
2
π
+
∞
r
n
=1
4(
−
1)
n
+1
π
(4
n
2
−
1)
cos 2
nx.
25.
a
0
=2
5
1
/
2
0
1
dx
=1
a
n
=2
5
1
/
2
0
1
·
cos
nπx dx
=
2
nπ
sin
nπ
2
b
n
=2
5
1
/
2
0
1
·
sin
nπx dx
=
2
nπ

1
−
cos
nπ
2

f
(
x
)=
1
2
+
∞
r
n
=1
2
nπ
sin
nπ
2
cos
nπx
f
(
x
)=
∞
r
n
=1
2
nπ

1
−
cos
nπ
2

sin
nπx
26.
a
0
=2
5
1
1
/
2
1
dx
=1
a
n
=2
5
1
1
/
2
1
·
cos
nπx dx
=
−
2
nπ
sin
nπ
2
b
n
=2
5
1
1
/
2
1
·
sin
nπx dx
=
2
nπ

cos
nπ
2
+(
−
1)
n
+1

f
(
x
)=
1
2
+
∞
r
n
=1
2
−
2
nπ
sin
nπ
2
cos
nπx
=
f
(
x
)=
∞
r
n
=1
2
nπ

cos
nπ
2
+(
−
1)
n
+1

sin
nπx
27.
a
0
=
4
π
5
π/
2
0
cos
xdx
=
4
π
a
n
=
4
π
5
π/
2
0
cos
x
cos 2
nx dx
=
2
π
5
π/
2
0
[cos(2
n
+1)
x
+ cos(2
n
−
1)
x
]
dx
=
4(
−
1)
n
π
(1
−
4
n
2
)
b
n
=
4
π
5
π/
2
0
cos
x
sin 2
nx dx
=
2
π
5
π/
2
0
[sin(2
n
+1)
x
+ sin(2
n
−
1)
x
]
dx
=
8
n
π
(4
n
2
−
1)
f
(
x
)=
2
π
+
∞
r
n
=1
4(
−
1)
n
π
(1
−
4
n
2
)
cos 2
nx
f
(
x
)=
∞
r
n
=1
8
n
π
(4
n
2
−
1)
sin 2
nx
28.
a
0
=
2
π
5
π
0
sin
xdx
=
4
π
a
n
=
2
π
5
π
0
sin
x
cos
nx dx
=
1
π
5
π
0
[sin(
n
+1)
x
−
sin(
n
−
1)
x
]
dx
=
2[(
−
1)
n
+1]
π
(1
−
n
2
)
for
n
=2
,
3
,
4
,...
b
n
=
2
π
5
π
0
sin
x
sin
nx dx
=
1
π
5
π
0
[cos(
n
−
1)
x
−
cos(
n
+1)
x
]
dx
= 0 for
n
=2
,
3
,
4
,...
a
1
=
1
π
5
π
0
sin 2
xdx
=0
575

576. Exercises 12.3
b
1
=
2
π
5
π
0
sin
2
xdx
=1
f
(
x
) = sin
x
f
(
x
)=
2
π
+
2
π
∞
r
n
=2
(
−
1)
n
+1
1
−
n
2
cos
nx
29.
a
0
=
2
π

5
π/
2
0
xdx
+
5
π
π/
2
(
π
−
x
)
dx
o
=
π
2
a
n
=
2
π

5
π/
2
0
x
cos
nx dx
+
5
π
π/
2
(
π
−
x
) cos
nx dx
o
=
2
n
2
π

2 cos
nπ
2
+(
−
1)
n
+1
−
1

b
n
=
2
π

5
π/
2
0
x
sin
nx dx
+
5
π
π/
2
(
π
−
x
) sin
nx dx
o
=
4
n
2
π
sin
nπ
2
f
(
x
)=
π
4
+
∞
r
n
=1
2
n
2
π

2 cos
nπ
2
+(
−
1)
n
+1
−
1

cos
nx
f
(
x
)=
∞
r
n
=1
4
n
2
π
sin
nπ
2
sin
nx
30.
a
0
=
1
π
5
2
π
π
(
x
−
π
)
dx
=
π
2
a
n
=
1
π
5
2
π
π
(
x
−
π
) cos
n
2
xdx
=
4
n
2
π
[
(
−
1)
n
−
cos
nπ
2
]
b
n
=
1
π
5
2
π
π
(
x
−
π
) sin
n
2
xdx
=
2
n
(
−
1)
n
+1
−
4
n
2
π
sin
nπ
2
f
(
x
)=
π
4
+
∞
r
n
=1
4
n
2
π
[
(
−
1)
n
−
cos
nπ
2
]
cos
n
2
x
f
(
x
)=
∞
r
n
=1
2
2
n
(
−
1)
n
+1
−
4
n
2
π
sin
nπ
2
=
sin
n
2
x
31.
a
0
=
5
1
0
xdx
+
5
2
1
1
dx
=
3
2
a
n
=
5
1
0
x
cos
nπ
2
xdx
=
4
n
2
π
2

cos
nπ
2
−
1

b
n
=
5
1
0
x
sin
nπ
2
xdx
+
5
2
1
1
·
sin
nπ
2
xdx
=
4
n
2
π
2
sin
nπ
2
+
2
nπ
(
−
1)
n
+1
f
(
x
)=
3
4
+
∞
r
n
=1
4
n
2
π
2

cos
nπ
2
−
1

cos
nπ
2
x
f
(
x
)=
∞
r
n
=1
2
4
n
2
π
2
sin
nπ
2
+
2
nπ
(
−
1)
n
+1
4
sin
nπ
2
x
576

577. Exercises 12.3
32.
a
0
=
5
1
0
1
dx
+
5
2
1
(2
−
x
)
dx
=
3
2
a
n
=
5
1
0
1
·
cos
nπ
2
xdx
+
5
2
1
(2
−
x
) cos
nπ
2
xdx
=
4
n
2
π
2

cos
nπ
2
+(
−
1)
n
+1

b
n
=
5
1
0
1
·
sin
nπ
2
xdx
+
5
2
1
(2
−
x
) sin
nπ
2
xdx
=
2
nπ
+
4
n
2
π
2
sin
nπ
2
f
(
x
)=
3
4
+
∞
r
n
=1
4
n
2
π
2

cos
nπ
2
+(
−
1)
n
+1

cos
nπ
2
x
f
(
x
)=
∞
r
n
=1
2
2
nπ
+
4
n
2
π
2
sin
nπ
2
=
sin
nπ
2
x
33.
a
0
=2
5
1
0
(
x
2
+
x
)
dx
=
5
3
a
n
=2
5
1
0
(
x
2
+
x
) cos
nπx dx
=
2(
x
2
+
x
)
nπ
sin
nπx
9
9
9
9
1
0
−
2
nπ
5
1
0
(2
x
+ 1) sin
nπx dx
=
2
n
2
π
2
[3(
−
1)
n
−
1]
b
n
=2
5
1
0
(
x
2
+
x
) sin
nπx dx
=
−
2(
x
2
+
x
)
nπ
cos
nπx
9
9
9
9
1
0
+
2
nπ
5
1
0
(2
x
+ 1) cos
nπx dx
=
4
nπ
(
−
1)
n
+1
+
4
n
3
π
3
[(
−
1)
n
−
1]
f
(
x
)=
5
6
+
∞
r
n
=1
2
n
2
π
2
[3(
−
1)
n
−
1] cos
nπx
f
(
x
)=
∞
r
n
=1
2
4
nπ
(
−
1)
n
+1
+
4
n
3
π
3
[(
−
1)
n
−
1]
=
sin
nπx
34.
a
0
=
5
2
0
(2
x
−
x
2
)
dx
=
4
3
a
n
=
5
2
0
(2
x
−
x
2
) cos
nπ
2
xdx
=
8
n
2
π
2
[(
−
1)
n
+1
−
1]
b
n
=
5
2
0
(2
x
−
x
2
) sin
nπ
2
xdx
=
16
n
3
π
3
[1
−
(
−
1)
n
]
f
(
x
)=
2
3
+
∞
r
n
=1
8
n
2
π
2
[(
−
1)
n
+1
−
1] cos
nπ
2
x
f
(
x
)=
∞
r
n
=1
16
n
3
π
3
[1
−
(
−
1)
n
] sin
nπ
2
x
35.
a
0
=
1
π
5
2
π
0
x
2
dx
=
8
3
π
2
a
n
=
1
π
5
2
π
0
x
2
cos
nx dx
=
4
n
2
b
n
=
1
π
5
2
π
0
x
2
sin
nx dx
=
−
4
π
n
f
(
x
)=
4
3
π
2
+
∞
r
n
=1
2
4
n
2
cos
nx
−
4
π
n
sin
nx
=
577

578. Exercises 12.3
36.
a
0
=
2
π
5
π
0
xdx
=
π
a
n
=
2
π
5
π
0
x
cos 2
nx dx
=0
b
n
=
2
π
5
π
0
x
sin 2
nx dx
=
−
1
n
f
(
x
)=
π
2
+
∞
r
n
=1
2
−
1
n
sin 2
nx
=
37.
a
0
=2
5
1
0
(
x
+1)
dx
=3
a
n
=2
5
1
0
(
x
+ 1) cos 2
nπx dx
=0
b
n
=2
5
1
0
(
x
+ 1) sin 2
nπx dx
=
−
1
nπ
f
(
x
)=
3
2
−
∞
r
n
=1
1
nπ
sin 2
nπx
38.
a
0
=2
5
2
0
(2
−
x
)
dx
=2
a
n
=2
5
2
0
(2
−
x
) cos
nπx dx
=0
b
n
=2
5
2
0
(2
−
x
) sin
nπx dx
=
2
nπ
f
(
x
)=1+
∞
r
n
=1
2
nπ
sin
nπx
39.
We have
b
n
=
2
π
5
π
0
5 sin
nt dt
=
10
nπ
[1
−
(
−
1)
n
]
so that
f
(
t
)=
∞
r
n
=1
10[1
−
(
−
1)
n
]
nπ
sin
nt.
Substituting the assumption
x
p
(
t
)=
∞
r
n
=1
B
n
sin
nt
into the differential equation then gives
x
22
p
+10
x
p
=
∞
r
n
=1
B
n
(10
−
n
2
) sin
nt
=
∞
r
n
=1
10[1
−
(
−
1)
n
]
nπ
sin
nt
and so
B
n
=
10[1
−
(
−
1)
n
]
nπ
(10
−
n
2
)
.Thus
x
p
(
t
)=
10
π
∞
r
n
=1
1
−
(
−
1)
n
n
(10
−
n
2
)
sin
nt.
40.
We have
b
n
=
2
π
5
1
0
(1
−
t
) sin
nπt dt
=
2
nπ
578

579. Exercises 12.3
so that
f
(
t
)=
∞
r
n
=1
2
nπ
sin
nπt.
Substituting the assumption
x
p
(
t
)=
∞
r
n
=1
B
n
sin
nπt
into the differential equation then gives
x
22
p
+10
x
p
=
∞
r
n
=1
B
n
(10
−
n
2
π
2
) sin
nπt
=
∞
r
n
=1
2
nπ
sin
nπt
and so
B
n
=
2
nπ
(10
−
n
2
π
2
)
.Thus
x
p
(
t
)=
2
π
∞
r
n
=1
1
n
(10
−
n
2
π
2
)
sin
nπt.
41.
We have
a
0
=
2
π
5
π
0
(2
πt
−
t
2
)
dt
=
4
3
π
2
a
n
=
2
π
5
π
0
(2
πt
−
t
2
) cos
nt dt
=
−
4
n
2
so that
f
(
t
)=
2
π
2
3
+
∞
r
n
=1
(
−
4)
n
2
cos
nt.
Substituting the assumption
x
p
(
t
)=
A
0
2
+
∞
r
n
=1
A
n
cos
nt
into the differential equation then gives
1
4
x
22
p
+12
x
p
=6
A
0
+
∞
r
n
=1
A
n
2
−
1
4
n
2
+12
=
cos
nt
=
2
π
2
3
+
∞
r
n
=1
(
−
4)
n
2
cos
nt
and
A
0
=
π
2
9
,
A
n
=
16
n
2
(
n
2
−
48)
.Thus
x
p
(
t
)=
π
2
18
+16
∞
r
n
=1
1
n
2
(
n
2
−
48)
cos
nt.
42.
We have
a
0
=
2
(1
/
2)
5
1
/
2
0
tdt
=
1
2
a
n
=
2
(1
/
2)
5
1
/
2
0
t
cos 2
nπt dt
=
1
n
2
π
2
[(
−
1)
n
−
1]
so that
f
(
t
)=
1
4
+
∞
r
n
=1
(
−
1)
n
−
1
n
2
π
2
cos 2
nπt.
Substituting the assumption
x
p
(
t
)=
A
0
2
+
∞
r
n
=1
A
n
cos 2
nπt
579

580. 20 40 60 80
t
-4
-2
2
4
x
Exercises 12.3
into the differential equation then gives
1
4
x
22
p
+12
x
p
=6
A
0
+
∞
r
n
=1
A
n
(12
−
n
2
π
2
) cos 2
nπt
=
1
4
+
∞
r
n
=1
(
−
1)
n
−
1
n
2
π
2
cos 2
nπt
and
A
0
=
1
24
,
A
n
=
(
−
1)
n
−
1
n
2
π
2
(12
−
n
2
π
2
)
.Thus
x
p
(
t
)=
1
48
+
1
π
2
∞
r
n
=1
(
−
1)
n
−
1
n
2
(12
−
n
2
π
2
)
cos 2
nπt.
43. (a)
The general solution is
x
(
t
)=
c
1
cos
√
10
t
+
c
2
sin
√
10
t
+
x
p
(
t
), where
x
p
(
t
)=
10
π
∞
r
n
=1
1
−
(
−
1)
n
n
(10
−
n
2
)
sin
nt.
The initial condition
x
(0) = 0 implies
c
1
+
x
p
(0) = 0. Since
x
p
(0) = 0, we have
c
1
= 0 and
x
(
t
)=
c
2
sin
√
10
t
+
x
p
(
t
). Then
x
2
(
t
)=
c
2
√
10 cos
√
10
t
+
x
2
p
(
t
) and
x
2
(0) = 0 implies
c
2
√
10 +
10
π
∞
r
n
=1
1
−
(
−
1)
n
10
−
n
2
cos 0 = 0
.
Thus
c
2
=
−
√
10
π
∞
r
n
=1
1
−
(
−
1)
n
10
−
n
2
and
x
(
t
)=
10
π
∞
r
n
=1
1
−
(
−
1)
n
10
−
n
2
1
1
n
sin
nt
−
1
√
10
sin
√
10
t
4
.
(b)
The graph is plotted using eight nonzero terms in the series expansion of
x
(
t
).
44. (a)
The general solution is
x
(
t
)=
c
1
cos 4
√
3
t
+
c
2
sin 4
√
3
t
+
x
p
(
t
), where
x
p
(
t
)=
π
2
18
+16
∞
r
n
=1
1
n
2
(
n
2
−
48)
cos
nt.
The initial condition
x
(0) = 0 implies
c
1
+
x
p
(0) = 1 or
c
1
=1
−
x
p
(0) = 1
−
π
2
18
−
16
∞
r
n
=1
1
n
2
(
n
2
−
48)
.
Now
x
2
(
t
)=
−
4
√
3
c
1
sin 4
√
3
t
+4
c
2
cos 4
√
3
t
+
x
2
p
(
t
), so
x
2
(0) = 0 implies 4
c
2
+
x
2
p
(0) = 0. Since
x
2
p
(0) = 0, we
have
c
2
= 0 and
580

581. 2 4 6 8101214
t
-1
-0.5
0.5
1
1.5
x
Exercises 12.3
x
(
t
)=

1
−
π
2
18
−
16
∞
r
n
=1
1
n
2
(
n
2
−
48)
o
cos 4
√
3
t
+
π
2
18
+16
∞
r
n
=1
1
n
2
(
n
2
−
48)
cos
nt
=
π
2
18
+
2
1
−
π
2
18
=
cos 4
√
3
t
+16
∞
r
n
=1
1
n
2
(
n
2
−
48)
i
cos
nt
−
cos 4
√
3
t
n
.
(b)
The graph is plotted using five nonzero terms in the series expansion of
x
(
t
).
45.
We have
b
n
=
2
L
5
L
0
w
0
x
L
sin
nπ
L
xdx
=
2
w
0
nπ
(
−
1)
n
+1
so that
w
(
x
)=
∞
r
n
=1
2
w
0
nπ
(
−
1)
n
+1
sin
nπ
L
x.
If we assume
y
(
x
)=
∞
r
n
=1
B
n
sin
nπ
L
x
then
y
(4)
=
∞
r
n
=1
n
4
π
4
L
4
B
n
sin
nπ
L
x
and so the differential equation
EIy
(4)
=
w
(
x
) gives
B
n
=
2
w
0
(
−
1)
n
+1
L
4
EIn
5
π
5
.
Thus
y
(
x
)=
2
w
0
L
4
EIπ
5
∞
r
n
=1
(
−
1)
n
+1
n
5
sin
nπ
L
x.
46.
We have
b
n
=
2
L
5
2
L/
3
L/
3
w
0
sin
nπ
L
xdx
=
2
w
0
nπ
1
cos
nπ
3
−
cos
2
nπ
3
4
so that
w
(
x
)=
∞
r
n
=1
2
w
0
nπ
1
cos
nπ
3
−
cos
2
nπ
3
4
sin
nπ
L
x.
If we assume
y
(
x
)=
∞
r
n
=1
B
n
sin
nπ
L
x
then
y
(4)
=
∞
r
n
=1
n
4
π
4
L
4
B
n
sin
nπ
L
x
581

582. 2 4 6 8 10 12 14
x
-3
-2
-1
1
2
3
fHxL
2 4 6 8 10
x
-0.5
-0.25
0.25
0.5
0.75
1
1.25
1.5
fHxL
Exercises 12.3
and so the differential equation
EIy
(4)
=
w
(
x
) gives
B
n
=2
w
0
L
4
cos
nπ
3
−
cos
2
nπ
3
EIn
5
π
5
.
Thus
y
(
x
)=
2
w
0
L
4
EIπ
5
∞
r
n
=1
cos
nπ
3
−
cos
2
nπ
3
n
5
sin
nπ
L
x.
47.
The graph is obtained by summing the series from
n
= 1 to 20. It appears that
f
(
x
)=
A
x,
0
<x<π
−
π, π<x<
2
π
.
48.
The graph is obtained by summing the series from
n
= 1 to 10. It appears that
f
(
x
)=
A
1
−
x,
0
<x<
1
0
,
1
<x<
2.
49. (a)
If
f
and
g
are even and
h
(
x
)=
f
(
x
)
g
(
x
) then
h
(
−
x
)=
f
(
−
x
)
g
(
−
x
)=
f
(
x
)
g
(
x
)=
h
(
x
)
and
h
is even.
(c)
If
f
is even and
g
is odd and
h
(
x
)=
f
(
x
)
g
(
x
) then
h
(
−
x
)=
f
(
−
x
)
g
(
−
x
)=
f
(
x
)[
−
g
(
x
)] =
−
h
(
x
)
and
h
is odd.
(d)
Let
h
(
x
)=
f
(
x
)
±
g
(
x
) where
f
and
g
are even. Then
h
(
−
x
)=
f
(
−
x
)
±
g
(
−
x
)=
f
(
x
)
±
g
(
x
)
,
and
h
is an even function.
582

583. Exercises 12.4
(f)
If
f
is even then
1
a
−
a
f
(
x
)
dx
=
−
1
0
a
f
(
−
u
)
du
+
1
a
0
f
(
x
)
dx
=
1
a
0
f
(
u
)
du
+
1
a
0
f
(
x
)
dx
=2
1
a
0
f
(
x
)
dx.
(g)
If
f
is odd then
1
a
−
a
f
(
x
)
dx
=
−
1
0
−
a
f
(
−
x
)
dx
+
1
a
0
f
(
x
)
dx
=
1
0
a
f
(
u
)
du
+
1
a
0
f
(
x
)
dx
=
−
1
a
0
f
(
u
)
du
+
1
a
0
f
(
x
)
dx
=0
.
50.
The answers are not unique. The function in Problem 47 could also be
f
(
x
)=



x,
0
<x<π
1
,x
=
π
−
π, π < x <
2
π.
In Problem 48 the function could also be
f
(
x
)=









0
,
−
2
<x<
−
1
x
+1
,
−
1
<x<
0
−
x
+1
,
0
<x<
1
0
,
1
<x<
2.
Exercises 12.4
In this section we make use of the following identities due to Euler’s formula:
e
inπ
=
e
−
inπ
=(
−
1)
n
,e
−
2
inπ
=1
,e
−
inπ/
2
=(
−
i
)
n
.
1.
Identifying
p
= 2 we have
c
n
=
1
4
1
2
−
2
f
(
x
)
e
−
inπx/
2
dx
=
1
4
4
1
0
−
2
(
−
1)
e
−
inπx/
2
dx
+
1
2
0
e
−
inπx/
2
dx
5
=
i
2
nπ
b
−
1+
e
inπ
+
e
−
inπ
−
1
c
=
i
2
nπ
[
−
1+(
−
1)
n
+(
−
1)
n
−
1] =
1
−
(
−
1)
n
nπi
and
c
0
=
1
4
1
2
−
2
f
(
x
)
dx
=0
.
Thus
f
(
x
)=
∞
f
n
=
−∞
n
.
=0
1
−
(
−
1)
n
inπ
e
inπx/
2
.
583

584. Exercises 12.4
2.
Identifying 2
p
=2or
p
=1wehave
c
n
=
1
2
1
2
0
f
(
x
)
e
−
inπx
dx
=
1
2
1
2
1
e
−
inπx
dx
=
−
1
2
inπ
e
−
inπx
.
.
.
2
1
=
−
1
2
inπ
d
e
−
2
inπ
−
e
−
inπ
e
=
−
1
2
inπ
[1
−
(
−
1)
n
]=
i
2
nπ
[1
−
(
−
1)
n
]
and
c
0
=
1
2
1
2
0
f
(
x
)
dx
=
1
2
1
2
1
dx
=
1
2
.
Thus
f
(
x
)=
1
2
+
i
2
π
∞
f
n
=
−∞
n
.
=0
1
−
(
−
1)
n
n
e
inπx
.
3.
Identifying
p
=1
/
2wehave
c
n
=
1
1
/
2
−
1
/
2
f
(
x
)
e
−
2
inπx
dx
=
1
1
/
4
0
e
−
2
inπx
dx
=
−
1
2
inπ
e
−
2
inπx
.
.
.
.
1
/
4
0
=
−
1
2
inπ
b
e
−
inπ/
2
−
1
c
=
−
1
2
inπ
[(
−
i
)
n
−
1] =
i
2
nπ
[(
−
i
)
n
−
1]
and
c
0
=
1
1
/
4
0
dx
=
1
4
.
Thus
f
(
x
)=
1
4
+
i
2
π
∞
f
n
=
−∞
n
.
=0
(
−
i
)
n
−
1
n
e
2
inπx
.
4.
Identifying
p
=
π
we have
c
n
=
1
2
π
1
π
−
π
f
(
x
)
e
−
inx/π
dx
=
1
2
π
1
π
0
xe
−
inx/π
dx
=
1
2
2
π
n
2
+
ix
n
3
e
−
inx/π
.
.
.
.
π
0
=
π
(1 +
in
)
2
n
2
e
−
in
−
π
2
n
2
and
c
0
=
1
2
π
1
π
0
xdx
=
π
4
.
Thus
f
(
x
)=
π
4
+
π
2
∞
f
n
=
−∞
n
.
=0
1
n
2
g
(1 +
in
)
e
−
in
−
1
F
e
inx
.
5.
Identifying 2
p
=2
π
or
p
=
π
we have
c
n
=
1
2
π
1
2
π
0
f
(
x
)
e
−
inx
dx
=
1
2
π
1
2
π
0
xe
−
inx
dx
=
1
2
π
2
1
n
2
+
ix
n
3
e
−
inx
.
.
.
.
2
π
0
=
1+2
inπ
2
n
2
π
−
1
2
n
2
π
=
i
n
584

585. -
5p
2
-
3p
2
-
p
2
p
2
3p
2
5p
2
frequency
0.2
0.4
0.6
cHnL
n-5-4 -3-2-1 0 1 2 3 4 5
c
n
0.1273 0.0000 0.2122 0.0000 0.6366 0.0000 0.6366 0.0000 0.2122 0.0000 0.1273
-10 p-6 p-2 p 2p 6p 10 p
frequency
0.05
0.1
0.15
0.2
0.25
cHnL
n-5-4 -3-2-1 0 1 2 3 4 5
c
n
0.0450 0.0000 0.0750 0.1592 0.2251 0.2500 0.2251 0.1592 0.0750 0.0000 0.0450
Exercises 12.4
and
c
0
=
1
2
π
1
2
π
0
xdx
=
π.
Thus
f
(
x
)=
π
+
∞
f
n
=
−∞
n
.
=0
i
n
e
inx
.
6.
Identifying
p
= 1 we have
c
n
=
1
2
1
1
−
1
f
(
x
)
e
−
inπx
dx
=
1
2
4
1
0
−
1
e
x
e
−
inπx
dx
+
1
1
0
e
−
x
e
−
inπx
dx
5
=
1
2
4
−
1
1
−
inπ
e
(1
−
inπ
)
x
.
.
.
0
−
1
−
1
1+
inπ
e
−
(1+
inπ
)
x
.
.
.
1
0
5
=
e
−
(
−
1)
n
e
(1
−
inπ
)
+
1
−
e
−
1
(
−
1)
n
1+
inπ
=
2[
e
−
(
−
1)
n
]
e
(1 +
n
2
π
2
)
.
Thus
f
(
x
)=
∞
f
n
=
−∞
2[
e
−
(
−
1)
n
]
e
(1 +
n
2
π
2
)
e
inπx
.
7.
The fundamental period is
T
=4,so
ω
=2
π/
4=
π/
2
and the values of
nω
are 0,
±
π/
2,
±
π
,
±
3
π/
2,
...
. From
Problem 1,
c
0
= 0 and
|
c
n
|
=(1
−
(
−
1)
n
)
/nπ
. The table
shows some values of
n
with corresponding values of
|
c
n
|
.
The graph is a portion of the frequency spectrum.
8.
The fundamental period is
T
=1,so
ω
=2
π
and the
values of
nω
are 0,
±
2
π
,
±
4
π
,
±
6
π
,
...
. From Problem 3,
c
0
=
1
4
and
|
c
n
|
=
|
(
−
i
)
n
−
1
|
/
2
nπ
,or
c
1
=
c
−
1
=
√
2
/
2
π
,
c
2
=
c
−
2
=1
/
2
π
,
c
3
=
c
−
3
=
√
2
/
6
π
,
c
4
=
c
−
4
=0,
c
5
=
c
−
5
=
√
2
/
10
π
,
c
6
=
c
−
6
=1
/
6
π
,
c
7
=
c
−
7
=
√
2
/
14
π
,
c
8
=
c
−
8
=0,
...
. The table shows some values of
n
with
corresponding values of
|
c
n
|
. The graph is a portion of the
frequency spectrum.
585

586. -6 -4 -2 2 4 6
frequency
0.2
0.4
0.6
cHnL
-2 p -p p 2p
x
1
2
3
4
f@xD
n-5-4 -3-2-1 0 1 2 3 4 5
c
n
0.0198 0.0759 0.2380 0.4265 0.5784 0.6366 0.5784 0.4265 0.2380 0.0759 0.0198
-10 -5 5 10
frequency
0.1
0.2
0.3
cHnL
-2 p -p p 2p
x
0.2
0.4
0.6
0.8
1
f@xD
Exercises 12.4
9.
Identifying 2
p
=
π
or
p
=
π/
2, and using sin
x
=
(
e
ix
−
e
−
ix
)
/
2
i
, we have
c
n
=
1
π
5
π
0
f
(
x
)
e
−
2
inx/π
dx
=
1
π
5
π
0
(sin
x
)
e
−
2
inx/π
dx
=
1
π
5
π
0
1
2
i
(
e
ix
−
e
−
ix
)
e
−
2
inx/π
dx
=
1
2
πi
5
π
0

e
(1
−
2
n/π
)
ix
−
e
−
(1+2
n/π
)
ix

dx
=
1
2
πi
1
1
i
(1
−
2
n/π
)
e
(1
−
2
n/π
)
ix
+
1
i
(1+2
n/π
)
e
−
(1+2
n/π
)
ix
4
π
0
=
π
(1 +
e
−
2
in
)
π
2
−
4
n
2
.
The fundamental period is
T
=
π
,so
ω
=2
π/π
= 2 and the values of
nω
are 0,
±
2,
±
4,
±
6,
...
. Values of
|
c
n
|
for
n
=0,
±
1,
±
2,
±
3,
±
4, and
±
5 are shown in the table. The bottom graph is a portion of the frequency
spectrum.
10.
Identifying 2
p
=
π
or
p
=
π/
2, and using cos
x
=
(
e
ix
−
e
−
ix
)
/
2, we have
c
n
=
1
π
5
π
0
f
(
x
)
e
−
2
inx/π
dx
=
1
π
5
π/
2
0
(cos
x
)
e
−
2
inx/π
dx
=
1
π
5
π/
2
0
1
2
(
e
ix
−
e
−
ix
)
e
−
2
inx/π
dx
=
1
2
π
5
π/
2
0

e
(1
−
2
n/π
)
ix
−
e
−
(1+2
n/π
)
ix

dx
=
1
2
π
1
1
i
(1
−
2
n/π
)
e
(1
−
2
n/π
)
ix
+
1
i
(1+2
n/π
)
e
−
(1+2
n/π
)
ix
4
π/
2
0
=
2
ne
−
in
+
iπ
π
2
−
4
n
2
.
586

587. n-5-4 -3-2-1 0 1 2 3 4 5
c
n
0.1447 0.1954 0.2437 0.2833 0.3093 0.3183 0.3093 0.2833 0.2437 0.1954 0.1447
Exercises 12.5
The fundamental period is
T
=
π
,so
ω
=2
π/π
= 2 and the values of
nω
are 0,
±
2,
±
4,
±
6,
...
. Values of
|
c
n
|
for
n
=0,
±
1,
±
2,
±
3,
±
4, and
±
5 are shown in the table. The bottom graph is a portion of the frequency
spectrum.
11. (a)
Adding
c
n
=
1
2
(
a
n
−
ib
n
) and
c
−
n
=
1
2
(
a
n
+
ib
n
) we get
c
n
+
c
−
n
=
a
n
. Subtracting, we get
c
n
−
c
−
n
=
−
ib
n
.
Multiplying both sides by
i
we obtain
i
(
c
n
−
c
−
n
)=
b
n
.
(b)
From
a
n
=
c
n
+
c
−
n
=(
−
1)
n
sinh
π
π
1
1
−
in
n
2
+1
+
1+
in
n
2
+1
4
=
2(
−
1)
n
sinh
π
π
(
n
2
+1)
,n
=0
,
1
,
2
,...
and
b
n
=
i
(
c
n
−
c
−
n
)=
i
(
−
1)
n
sinh
π
π
1
1
−
in
n
2
+1
−
1+
in
n
2
+1
4
=
i
(
−
1)
n
sinh
π
π
1
−
2
in
n
2
+1
4
=
2(
−
1)
n
n
sinh
π
π
(
n
2
+1)
,
the Fourier series of
f
is
f
(
x
)=
sinh
π
π
+
2 sinh
π
π
∞
r
n
=1
1
(
−
1)
n
n
2
+1
cos
nx
+
n
(
−
1)
n
n
2
+1
sin
nx
4
.
12.
From Problem 11 and the fact that
f
is odd,
c
n
+
c
−
n
=
a
n
=0,so
c
−
n
=
−
c
n
. Then
b
n
=
i
(
c
n
−
c
−
n
)=2
ic
n
.
From Problem 1,
b
n
=2
i
[1
−
(
−
1)
n
]
/nπi
= 2[1
−
(
−
1)
n
]
/nπ
, and the Fourier sine series of
f
is
f
(
x
)=
∞
r
i
=1
2[1
−
(
−
1)
n
nπ
sin
nπx
2
.
Exercises 12.5
1.
For
λ
≤
0 the only solution of the boundary-value problem is
y
=0. For
λ>
0 we have
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx.
Now
y
2
(
x
)=
−
c
1
√
λ
sin
√
λx
+
c
2
√
λ
cos
√
λx
and
y
2
(0) = 0 implies
c
2
=0,so
y
(1) +
y
2
(1) =
c
1
(cos
√
λ
−
√
λ
sin
√
λ
) = 0 or cot
√
λ
=
√
λ.
The eigenvalues are
λ
n
=
x
2
n
where
x
1
,
x
2
,
x
3
,
...
are the consecutive positive solutions of cot
√
λ
=
√
λ
. The
corresponding eigenfunctions are cos
√
λ
n
x
= cos
x
n
x
for
n
=1,2,3,
...
. Using a CAS we find that the
first four eigenvalues are 0
.
7402, 11
.
7349, 41
.
4388, and 90
.
8082 with corresponding eigenfunctions cos 0
.
8603
x
,
cos 3
.
4256
x
, cos 6
.
4373
x
, and cos 9
.
5293
x
.
2.
For
λ<
0 the only solution of the boundary-value problem is
y
=0. For
λ
= 0 we have
y
=
c
1
x
+
c
2
.Now
y
2
=
c
1
and the boundary conditions both imply
c
1
+
c
2
=0. Thus,
λ
= 0 is an eigenvalue with corresponding
eigenfunction
y
0
=
x
−
1.
587

588. Exercises 12.5
For
λ>
0wehave
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx
and
y
2
=
−
c
1
√
λ
sin
√
λx
+
c
2
√
λ
cos
√
λx.
The boundary conditions imply
c
1
+
c
2
√
λ
=0
c
1
cos
√
λ
+
c
2
sin
√
λ
=0
which gives
−
c
2
√
λ
cos
√
λ
+
c
2
sin
√
λ
= 0 or tan
√
λ
=
√
λ.
The eigenvalues are
λ
n
=
x
2
n
where
x
1
,
x
2
,
x
3
,
...
are the consecutive positive solutions of tan
√
λ
=
√
λ
. The
corresponding eigenfunctions are
√
λ
n
cos
√
λ
n
x
−
sin
√
λ
n
x
(obtained by taking
c
2
=
−
1 in the first equation
of the system.) Using a CAS we find that the first four positive eigenvalues are 20
.
1907, 59
.
6795, 118
.
9000,
and 197
.
858 with corresponding eigenfunctions 4
.
4934 cos 4
.
4934
x
−
sin 4
.
4934
x
,7
.
7253 cos 7
.
7253
x
−
sin 7
.
7253
x
,
10
.
9041 cos 10
.
9041
x
−
sin 10
.
9041
x
, and 14
.
0662 cos 14
.
0662
x
−
sin 14
.
0662
x
.
3.
For
λ
= 0 the solution of
y
22
=0is
y
=
c
1
x
+
c
2
. The condition
y
2
(0) = 0 implies
c
1
=0,so
λ
=0is
an eigenvalue with corresponding eigenfunction 1. For
λ<
0 we have
y
=
c
1
cosh
√
−
λx
+
c
2
sin
√
−
λx
and
y
2
=
c
1
√
−
λ
sinh
√
−
λx
+
c
2
√
−
λ
cosh
√
−
λx
. The condition
y
2
(0) = 0 implies
c
2
= 0 and so
y
=
c
1
cosh
√
−
λx
.
Now the condition
y
2
(
L
) = 0 implies
c
1
=0. Thus
y
= 0 and there are no negative eigenvalues. For
λ>
0we
have
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx
and
y
2
=
−
c
1
√
λ
sin
√
λx
+
c
2
√
λ
cos
√
λx
. The condition
y
2
(0) = 0 implies
c
2
= 0 and so
y
=
c
1
cos
√
λx
. Now the condition
y
2
(
L
) = 0 implies
−
c
1
√
λ
sin
√
λL
=0. For
c
1
5
= 0 this
condition will hold when
√
λL
=
nπ
or
λ
=
n
2
π
2
/L
2
, where
n
=1,2,3,
...
. These are the positive eigenvalues
with corresponding eigenfunctions cos(
nπ/L
)
x
,
n
=1,2,3,
...
.
4.
For
λ<
0wehave
y
=
c
1
cosh
√
−
λx
+
c
2
sinh
√
−
λx
y
2
=
c
1
√
−
λ
sinh
√
−
λx
+
c
2
√
−
λ
cosh
√
−
λx.
Using the fact that cosh
x
is an even function and sinh
x
isoddwehave
y
(
−
L
)=
c
1
cosh(
−
√
−
λL
)+
c
2
sinh(
−
√
−
λL
)
=
c
1
cosh
√
−
λL
−
c
2
sinh
√
−
λL
and
y
2
(
−
L
)=
c
1
√
−
λ
sinh(
−
√
−
λL
)+
c
2
√
−
λ
cosh(
−
√
−
λL
)
=
−
c
1
√
−
λ
sinh
√
−
λL
+
c
2
√
−
λ
cosh
√
−
λL.
The boundary conditions imply
c
1
cosh
√
−
λL
−
c
2
sinh
√
−
λL
=
c
1
cosh
√
−
λL
+
c
2
sinh
√
−
λL
or
2
c
2
sinh
√
−
λL
=0
and
−
c
1
√
−
λ
sinh
√
−
λL
+
c
2
√
−
λ
cosh
√
−
λL
=
c
1
√
−
λ
sinh
√
−
λL
+
c
2
√
−
λ
cosh
√
−
λL
or
2
c
1
√
−
λ
sinh
√
−
λL
=0
.
588

589. Exercises 12.5
Since
√
−
λL
5
=0,
c
1
=
c
2
= 0 and the only solution of the boundary-value problem in this case is
y
=0.
For
λ
=0wehave
y
=
c
1
x
+
c
2
y
2
=
c
1
.
From
y
(
−
L
)=
y
(
L
) we obtain
−
c
1
L
+
c
2
=
c
1
L
+
c
2
.
Then
c
1
= 0 and
y
= 1 is an eigenfunction corresponding to the eigenvalue
λ
=0.
For
λ>
0wehave
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx
y
2
=
−
c
1
√
λ
sin
√
λx
+
c
2
√
λ
cos
√
λx.
The first boundary condition implies
c
1
cos
√
λL
−
c
2
sin
√
λL
=
c
1
cos
√
λL
+
c
2
sin
√
λL
or
2
c
2
sin
√
λL
=0
.
Thus, if
c
1
= 0 and
c
2
5
=0,
√
λL
=
nπ
or
λ
=
n
2
π
2
L
2
,n
=1
,
2
,
3
,... .
The corresponding eigenfunctions are sin
nπ
L
x
, for
n
=1,2,3,
...
. Similarly, the second boundary condition
implies
2
c
1
√
λ
sin
√
λL
=0
.
If
c
2
= 0 and
c
1
5
=0,
√
λL
=
nπ
or
λ
=
n
2
π
2
L
2
,n
=1
,
2
,
3
,...,
and the corresponding eigenfunctions are cos
nπ
L
x
, for
n
=1,2,3,
...
.
5.
The eigenfunctions are cos
√
λ
n
x
where cot
√
λ
n
=
√
λ
n
.Thus
9
cos

λ
n
x
9
2
=
5
1
0
cos
2

λ
n
xdx
=
1
2
5
1
0

1 + cos 2

λ
n
x

dx
=
1
2
2
x
+
1
2
√
λ
n
sin 2

λ
n
x
=
9
9
9
9
1
0
=
1
2
2
1+
1
2
√
λ
n
sin 2

λ
n
=
=
1
2
1
1+
1
2
√
λ
n

2 sin

λ
n
cos

λ
n

4
=
1
2
1
1+
1
√
λ
n
sin

λ
n
cot

λ
n
sin

λ
n
4
=
1
2
1
1+
1
√
λ
n

sin

λ
n


λ
n

sin

λ
n

4
=
1
2

1 + sin
2

λ
n

.
589

590. Exercises 12.5
6.
The eigenfunctions are sin
√
λ
n
x
where tan
√
λ
n
=
−
λ
n
.Thus
9
sin

λ
n
x
9
2
=
5
1
0
sin
2

λ
n
xdx
=
1
2
5
1
0

1
−
cos 2

λ
n
x

dx
=
1
2
2
x
−
1
2
√
λ
n
sin 2

λ
n
x
=
9
9
9
9
1
0
=
1
2
2
1
−
1
2
√
λ
n
sin 2

λ
n
=
=
1
2
1
1
−
1
2
√
λ
n

2 sin

λ
n
cos

λ
n

4
=
1
2
1
1
−
1
√
λ
n
tan

λ
n
cos

λ
n
cos

λ
n
4
=
1
2
1
1
−
1
√
λ
n

−

λ
n
cos
2

λ
n

4
=
1
2

1 + cos
2

λ
n

.
7. (a)
If
λ
≤
0 the initial conditions imply
y
=0. For
λ>
0 the general solution of the Cauchy-Euler differential
equation is
y
=
c
1
cos(
√
λ
ln
x
)+
c
2
sin(
√
λ
ln
x
). The condition
y
(1) = 0 implies
c
1
= 0, so that
y
=
c
2
sin(
√
λ
ln
x
). The condition
y
(5) = 0 implies
√
λ
ln 5 =
nπ
,
n
=1,2,3,
...
. Thus, the eigenvalues are
n
2
π
2
/
(ln 5)
2
for
n
=1,2,3,
...
, with corresponding eigenfunctions sin[(
nπ/
ln 5) ln
x
].
(b)
The self-adjoint form is
d
dx
[
xy
2
]+
λ
x
y
=0
.
(c)
An orthogonality relation is
5
5
1
sin

mπ
ln 5
ln
x

sin

nπ
ln 5
ln
x

dx
=0
.
8. (a)
The roots of the auxiliary equation
m
2
+
m
+
λ
= 0 are
1
2
(
−
1
±
√
1
−
4
λ
). When
λ
= 0 the general solution
of the differential equation is
c
1
+
c
2
e
−
x
. The initial conditions imply
c
1
+
c
2
= 0 and
c
1
+
c
2
e
−
2
= 0. Since
the determinant of the coefficients is not 0, the only solution of this homogeneous system is
c
1
=
c
2
=0,in
which case
y
= 0. Similarly, if 0
<λ<
1
4
, the general solution is
y
=
c
1
e
1
2
(
−
1+
√
1
−
4
λ
)
x
+
c
2
e
1
2
(
−
1
−
√
1
−
4
λ
)
x
.
In this case the initial conditions again imply
c
1
=
c
2
= 0, and so
y
= 0. Now, for
λ>
1
4
, the general
solution of the differential equation is
y
=
c
1
e
−
x/
2
cos
√
4
λ
−
1
x
+
c
2
e
−
x/
2
sin
√
4
λ
−
1
x.
The condition
y
(0) = 0 implies
c
1
=0so
y
=
c
2
e
−
x/
2
sin
√
4
λ
−
1
x
. From
y
(2) =
c
2
e
−
1
sin 2
√
4
λ
−
1=0
we see that the eigenvalues are determined by 2
√
4
λ
−
1=
nπ
for
n
=1,2,3,
...
. Thus, the eigenvalues
are
n
2
π
2
/
4
2
+1
/
4 for
n
=1,2,3,
...
, with corresponding eigenfunctions
e
−
x/
2
sin
nπ
2
x
.
(b)
The self-adjoint form is
d
dx
[
e
x
y
2
]+
λe
x
y
=0
.
(c)
An orthogonality relation is
5
2
0
e
x

e
−
x/
2
sin
mπ
2
x

e
−
x/
2
cos
nπ
2
x

dx
=
5
2
0

sin
mπ
2
x

cos
nπ
2
x

dx
=0
.
590

591. Exercises 12.5
9. (a)
An orthogonality relation is
5
1
0
cos
x
m
x
cos
x
n
x
=0
where
x
m
5
=
x
n
are positive solutions of cot
x
=
x
.
(b)
Referring to Problem 1 we use a CAS to compute
5
1
0
(cos 0
.
8603
x
)(cos 3
.
4256
x
)
dx
=
−
1
.
8771
×
10
−
6
.
10. (a)
An orthogonality relation is
5
1
0
(
x
m
cos
x
m
x
−
sin
x
m
x
)(
x
n
cos
x
n
x
−
sin
x
n
x
)
dx
=0
where
x
m
5
=
x
n
are positive solutions of tan
x
=
x
.
(b)
Referring to Problem 2 we use a CAS to compute
5
1
0
(4
.
4934 cos 4
.
4934
x
−
sin 4
.
4934
x
)(7
.
7253 cos 7
.
7253
x
−
sin 7
.
7253
x
)
dx
=
−
2
.
5650
×
10
−
4
.
11.
To obtain the self-adjoint form we note that an integrating factor is (1
/x
)
e
s
(1
−
x
)
dx/x
=
e
−
x
. Thus, the
differential equation is
xe
−
x
y
22
+(1
−
x
)
e
−
x
y
2
+
ne
−
x
y
=0
and the self-adjoint form is
d
dx
i
xe
−
x
y
2
n
+
ne
−
x
y
=0
.
Identifying the weight function
p
(
x
)=
e
−
x
and noting that since
r
(
x
)=
xe
−
x
,
r
(0) = 0 and lim
x
→∞
r
(
x
)=0,
we have the orthogonality relation
5
∞
0
e
−
x
L
n
(
x
)
L
m
(
x
)
dx
=0
,m
5
=
n.
12.
To obtain the self-adjoint form we note that an integrating factor is
e
s
−
2
xdx
=
e
−
x
2
. Thus, the differential
equation is
e
−
x
2
y
22
−
2
xe
−
x
2
y
2
+2
ne
−
x
2
y
=0
and the self-adjoint form is
d
dx
[
e
−
x
2
y
2
]
+2
ne
−
x
2
y
=0
.
Identifying the weight function
p
(
x
)=2
e
−
x
2
and noting that since
r
(
x
)=
e
−
x
2
, lim
x
→−∞
r
(
x
) = lim
x
→∞
r
(
x
)=
0, we have the orthogonality relation
5
∞
∞
2
e
−
x
2
H
n
(
x
)
H
m
(
x
)
dx
=0
,m
5
=
n.
13. (a)
The differential equation is
(1 +
x
2
)
y
22
+2
xy
2
+
λ
1+
x
2
y
=0
.
591

592. Exercises 12.5
Letting
x
= tan
θ
we have
θ
= tan
−
1
x
and
dy
dx
=
dy
dθ
dθ
dx
=
1
1+
x
2
dy
dθ
d
2
y
dx
2
=
d
dx
1
1
1+
x
2
dy
dθ
4
=
1
1+
x
2
2
d
2
y
dθ
2
dθ
dx
=
−
2
x
(1 +
x
2
)
2
dy
dθ
=
1
(1 +
x
2
)
2
d
2
y
dθ
2
−
2
x
(1 +
x
2
)
2
dy
dθ
.
The differential equation can then be written in terms of
y
(
θ
)as
(1 +
x
2
)
1
1
(1 +
x
2
)
2
d
2
y
dθ
2
−
2
x
(1 +
x
2
)
2
dy
dθ
4
+2
x
1
1
1+
x
2
dy
dθ
4
+
λ
1+
x
2
y
=
1
1+
x
2
d
2
y
dθ
2
+
λ
1+
x
2
y
=0
or
d
2
y
dθ
2
+
λy
=0
.
The boundary conditions become
y
(0) =
y
(
π/
4)=0. For
λ
≤
0 the only solution of the boundary-
value problem is
y
=0. For
λ>
0 the general solution of the differential equation is
y
=
c
1
cos
√
λθ
+
c
2
sin
√
λθ
. The condition
y
(0) = 0 implies
c
1
=0so
y
=
c
2
sin
√
λθ
. Now the condition
y
(
π/
4) = 0 implies
c
2
sin
√
λπ/
4 = 0. For
c
2
5
= 0 this condition will hold when
√
λπ/
4=
nπ
or
λ
=16
n
2
, where
n
=
1, 2, 3,
...
. These are the eigenvalues with corresponding eigenfunctions sin 4
nθ
= sin(4
n
tan
−
1
x
), for
n
=1,2,3,
...
.
(b)
An orthogonality relation is
5
1
0
1
x
2
+1
sin(4
m
tan
−
1
x
) sin(4
n
tan
−
1
x
)
dx
=0
.
14. (a)
This is the parametric Bessel equation with
ν
= 1. The general solution is
y
=
c
1
J
1
(
λx
)+
c
2
Y
1
(
λx
)
.
Since
Y
is unbounded at 0 we must have
c
2
= 0, so that
y
=
c
1
J
1
(
λx
). The condition
J
1
(3
λ
) = 0 defines the
eigenvalues
λ
1
,
λ
2
,
λ
3
,
...
. (When
λ
= 0 the differential equation is Cauchy-Euler and the only solution
satisfying the boundary condition is
y
=0,so
λ
= 0 is not an eigenvalue.)
(b)
From Table 5
.
2 in the text we see that eigenvalues are determined by 3
λ
1
=3
.
832, 3
λ
2
=7
.
016, 3
λ
3
=10
.
173,
and 3
λ
4
=13
.
323. The first four eigenvalues are thus
λ
1
=1
.
2773,
λ
2
=2
.
3387,
λ
3
=3
.
391, and
λ
4
=4
.
441.
592

593. Exercises 12.6
Exercises 12.6
1.
Identifying
b
= 3, the first four eigenvalues are
λ
1
=
3
.
832
3
≈
1
.
277
λ
2
=
7
.
016
3
≈
2
.
339
λ
3
=
10
.
173
3
=3
.
391
λ
4
=
13
.
323
3
≈
4
.
441
.
2.
We first note from Case III in the text that 0 is an eigenvalue. Now, since
J
2
0
(2
λ
) = 0 is equivalent to
J
1
(2
λ
)=0,
the next three eigenvalues are
λ
2
=
3
.
832
2
=1
.
916
λ
3
=
7
.
016
2
=3
.
508
λ
4
=
10
.
173
2
=5
.
087
.
3.
The boundary condition indicates that we use (15) and (16) of Section 12.6. With
b
= 2 we obtain
c
i
=
2
4
J
2
1
(2
λ
i
)
i
2
0
xJ
0
(
λ
i
x
)
dx
t
=
λ
i
xdt
=
λ
i
dx
=
1
2
J
2
1
(2
λ
i
)
·
1
λ
2
i
i
2
λ
i
0
tJ
0
(
t
)
dt
=
1
2
λ
2
i
J
2
1
(2
λ
i
)
i
2
λ
i
0
d
dt
[
tJ
1
(
t
)]
dt
[From (4) in the text]
=
1
2
λ
2
i
J
2
1
(2
λ
i
)
tJ
1
(
t
)
x
x
x
x
2
λ
i
0
=
1
λ
i
J
1
(2
λ
i
)
.
Thus
f
(
x
)=
∞
r
i
=1
1
λ
i
J
1
(2
λ
i
)
J
0
(
λ
i
x
)
.
4.
The boundary condition indicates that we use (19) and (20) of Section 12.6. With
b
= 2 we obtain
c
1
=
2
4
i
2
0
xdx
=
2
4
x
2
2
x
x
x
x
2
0
=1
,
c
i
=
2
4
J
2
0
(2
λ
i
)
i
2
0
xJ
0
(
λ
i
x
)
dx
t
=
λ
i
xdt
=
λ
i
dx
=
1
2
J
2
0
(2
λ
i
)
·
1
λ
2
i
i
2
λ
i
0
tJ
0
(
t
)
dt
593

594. Exercises 12.6
=
1
2
λ
2
i
J
2
0
(2
λ
i
)
i
2
λ
i
0
d
dt
[
tJ
1
(
t
)]
dt
[From (4) in the text]
=
1
2
λ
2
i
J
2
0
(2
λ
i
)
tJ
1
(
t
)
x
x
x
x
2
λ
i
0
=
J
1
(2
λ
i
)
λ
i
J
2
0
(2
λ
i
)
.
Now since
J
2
0
(2
λ
i
) = 0 is equivalent to
J
1
(2
λ
i
) = 0 we conclude
c
i
= 0 for
i
=2,3,4,
...
. Thus the expansion
of
f
on 0
<x<
2 consists of a series with one nontrivial term:
f
(
x
)=
c
1
=1
.
5.
The boundary condition indicates that we use (17) and (18) of Section 12.6. With
b
= 2 and
h
= 1 we obtain
c
i
=
2
λ
2
i
(4
λ
2
i
+1)
J
2
0
(2
λ
i
)
i
2
0
xJ
0
(
λ
i
x
)
dx
t
=
λ
i
xdt
=
λ
i
dx
=
2
λ
2
i
(4
λ
2
i
+1)
J
2
0
(2
λ
i
)
·
1
λ
2
i
i
2
λ
i
0
tJ
0
(
t
)
dt
=
2
(4
λ
2
i
+1)
J
2
0
(2
λ
i
)
i
2
λ
i
0
d
dt
[
tJ
1
(
t
)]
dt
[From (4) in the text]
=
2
(4
λ
2
i
+1)
J
2
0
(2
λ
i
)
tJ
1
(
t
)
x
x
x
x
2
λ
i
0
=
4
λ
i
J
1
(2
λ
i
)
(4
λ
2
i
+1)
J
2
0
(2
λ
i
)
.
Thus
f
(
x
)=4
∞
r
i
=1
λ
i
J
1
(2
λ
i
)
(4
λ
2
i
+1)
J
2
0
(2
λ
i
)
J
0
(
λ
i
x
)
.
6.
Writing the boundary condition in the form
2
J
0
(2
λ
)+2
λJ
2
0
(2
λ
)=0
we identify
b
= 2 and
h
= 2. Using (17) and (18) of Section 12.6 we obtain
c
i
=
2
λ
2
i
(4
λ
2
i
+4)
J
2
0
(2
λ
i
)
i
2
0
xJ
0
(
λ
i
x
)
dx
t
=
λ
i
xdt
=
λ
i
dx
=
λ
2
i
2(
λ
2
i
+1)
J
2
0
(2
λ
i
)
·
1
λ
2
i
i
2
λ
i
0
tJ
0
(
t
)
dt
=
1
2(
λ
2
i
+1)
J
2
0
(2
λ
i
)
i
2
λ
i
0
d
dt
[
tJ
1
(
t
)]
dt
[From (4) in the text]
=
1
2(
λ
2
i
+1)
J
2
0
(2
λ
i
)
tJ
1
(
t
)
x
x
x
x
2
λ
i
0
=
λ
i
J
1
(2
λ
i
)
(
λ
2
i
+1)
J
2
0
(2
λ
i
)
.
594

595. Exercises 12.6
Thus
f
(
x
)=
∞
r
i
=1
λ
i
J
1
(2
λ
i
)
(
λ
2
i
+1)
J
2
0
(2
λ
i
)
J
0
(
λ
i
x
)
.
7.
The boundary condition indicates that we use (17) and (18) of Section 12.6. With
n
=1,
b
= 4, and
h
=3we
obtain
c
i
=
2
λ
2
i
(16
λ
2
i
−
1+9)
J
2
1
(4
λ
i
)
i
4
0
xJ
1
(
λ
i
x
)5
xdx
t
=
λ
i
xdt
=
λ
i
dx
=
5
λ
2
i
4(2
λ
2
i
+1)
J
2
1
(4
λ
i
)
·
1
λ
3
i
i
4
λ
i
0
t
2
J
1
(
t
)
dt
=
5
4
λ
i
(2
λ
2
i
+1)
J
2
1
(4
λ
i
)
i
4
λ
i
0
d
dt
[
t
2
J
2
(
t
)]
dt
[From (4) in the text]
=
5
4
λ
i
(2
λ
2
i
+1)
J
2
1
(4
λ
i
)
t
2
J
2
(
t
)
x
x
x
x
4
λ
i
0
=
20
λ
i
J
2
(4
λ
i
)
(2
λ
2
i
+1)
J
2
1
(4
λ
i
)
.
Thus
f
(
x
)=20
∞
r
i
=1
λ
i
J
2
(4
λ
i
)
(2
λ
2
i
+1)
J
2
1
(4
λ
i
)
J
1
(
λ
i
x
)
.
8.
The boundary condition indicates that we use (15) and (16) of Section 12.6. With
n
= 2 and
b
= 1 we obtain
c
1
=
2
J
2
3
(
λ
i
)
i
1
0
xJ
2
(
λ
i
x
)
x
2
dx
t
=
λ
i
xdt
=
λ
i
dx
=
2
J
2
3
(
λ
i
)
·
1
λ
4
i
i
λ
i
0
t
3
J
2
(
t
)
dt
=
2
λ
4
i
J
2
3
(
λ
i
)
i
λ
i
0
d
dt
[
t
3
J
3
(
t
)]
dt
[From (4) in the text]
=
2
λ
4
i
J
2
3
(
λ
i
)
t
3
J
3
(
t
)
x
x
x
x
λ
i
0
=
2
λ
i
J
3
(
λ
i
)
.
Thus
f
(
x
)=2
∞
r
i
=1
1
λ
i
J
3
(
λ
i
)
J
2
(
λ
i
x
)
.
9.
The boundary condition indicates that we use (19) and (20) of Section 12
.
6. With
b
= 3 we obtain
c
1
=
2
9
i
3
0
xx
2
dx
=
2
9
x
4
4
x
x
x
x
3
0
=
9
2
,
c
i
=
2
9
J
2
0
(3
λ
i
)
i
3
0
xJ
0
(
λ
i
x
)
x
2
dx
t
=
λ
i
xdt
=
λ
i
dx
595

596. Exercises 12.6
=
2
9
J
2
0
(3
λ
i
)
·
1
λ
4
i
i
3
λ
i
0
t
3
J
0
(
t
)
dt
=
2
9
λ
4
i
J
2
0
(3
λ
i
)
i
3
λ
i
0
t
2
d
dt
[
tJ
1
(
t
)]
dt
u
=
t
2
dv
=
d
dt
[
tJ
1
(
t
)]
dt
du
=2
tdt v
=
tJ
1
(
t
)
=
2
9
λ
4
i
J
2
0
(3
λ
i
)

t
3
J
1
(
t
)
x
x
x
x
3
λ
i
0
−
2
i
3
λ
i
0
t
2
J
1
(
t
)
dt
o
With
n
= 0 in equation (5) in Section 12.6 in the text we have
J
2
0
(
x
)=
−
J
1
(
x
), so the boundary condition
J
2
0
(3
λ
i
) = 0 implies
J
1
(3
λ
i
) = 0. Then
c
i
=
2
9
λ
4
i
J
2
0
(3
λ
i
)

−
2
i
3
λ
i
0
d
dt
i
t
2
J
2
(
t
)
n
dt
o
=
2
9
λ
4
i
J
2
0
(3
λ
i
)

−
2
t
2
J
2
(
t
)
x
x
x
x
3
λ
i
0
o
=
2
9
λ
4
i
J
2
0
(3
λ
i
)
i
−
18
λ
2
i
J
2
(3
λ
i
)
n
=
−
4
J
2
(3
λ
i
)
λ
2
i
J
2
0
(3
λ
i
)
.
Thus
f
(
x
)=
9
2
−
4
∞
r
i
=1
J
2
(3
λ
i
)
λ
2
i
J
2
0
(3
λ
i
)
J
0
(
λ
i
x
)
.
10.
The boundary condition indicates that we use (15) and (16) of Section 12
.
6. With
b
= 1 it follows that
c
i
=
2
J
2
1
(
λ
i
)
i
1
0
x
3
1
−
x
2
F
J
0
(
λ
i
x
)
dx
=
2
J
2
1
(
λ
i
)
1
i
1
0
xJ
0
(
λ
i
x
)
dx
−
i
1
0
x
3
J
0
(
λ
i
x
)
dx
4
t
=
λ
i
xdt
=
λ
i
dx
=
2
J
2
1
(
λ
i
)
e
1
λ
2
i
i
λ
i
0
tJ
0
(
t
)
dt
−
1
λ
4
i
i
λ
i
0
t
3
J
0
(
t
)
dt

=
2
J
2
1
(
λ
i
)
e
1
λ
2
i
i
λ
i
0
d
dt
(
tJ
1
(
t
))
dt
−
1
λ
4
i
i
λ
i
0
t
2
d
dt
[
tJ
1
(
t
)]
dt

u
=
t
2
dv
=
d
dt
[
tJ
1
(
t
)]
dt
du
=2
tdt v
=
tJ
1
(
t
)
=
2
J
2
1
(
λ
i
)
e
1
λ
2
i
tJ
1
(
t
)
x
x
x
x
λ
i
0
−
1
λ
4
i

t
3
J
1
(
t
)
x
x
x
x
λ
i
0
−
2
i
λ
i
0
t
2
J
1
(
t
)
dt
ou
=
2
J
2
1
(
λ
i
)
e
J
1
(
λ
i
)
λ
i
−
J
1
(
λ
i
)
λ
i
+
2
λ
4
i
i
λ
i
0
d
dt
i
t
2
J
2
(
t
)
n
dt

=
2
J
2
1
(
λ
i
)
e
2
λ
4
i
t
2
J
2
(
t
)
x
x
x
x
λ
i
0

=
4
J
2
(
λ
i
)
λ
2
i
J
2
1
(
λ
i
)
.
596

597. 5 10 15 20 25 30
x
-4
-2
2
4
y
12345
x
5
10
15
20
S1
12345
x
5
10
15
20
S2
12345
x
5
10
15
20
S3
12345
x
5
10
15
20
S4
12345
x
5
10
15
20
S5
1234
x
5
10
15
20
S10
10 20 30 40 50
x
-10
-5
5
10
15
20
S10
Exercises 12.6
Thus
f
(
x
)=4
∞
r
i
=1
J
2
(
λ
i
)
λ
2
i
J
2
1
(
λ
i
)
j
0
(
λ
i
x
)
.
11. (a)
(b)
Using
FindRoot
in
Mathematica
we find the roots
x
1
=2
.
9496,
x
2
=5
.
84113,
x
3
=8
.
87273,
x
4
=11
.
9561,
and
x
5
=15
.
0624.
(c)
Dividing the roots in part (b) by 4 we find the eigenvalues
λ
1
=0
.
7374,
λ
2
=1
.
46028,
λ
3
=2
.
21818,
λ
4
=2
.
98904, and
λ
5
=3
.
76559.
(d)
The next five eigenvalues are
λ
6
=4
.
54508,
λ
7
=5
.
32626,
λ
8
=6
.
1085,
λ
9
=6
.
89145, and
λ
10
=7
.
6749.
12. (a)
From Problem 7, the coefficients of the Fourier-Bessel series are
c
1
=
20
λ
i
J
2
(4
λ
i
)
(2
λ
2
i
+1)
J
2
1
(4
λ
i
)
.
Using a CAS we find
c
1
=26
.
7896,
c
2
=
−
12
.
4624,
c
3
=7
.
1404,
c
4
=
−
4
.
68705, and
c
5
=3
.
35619.
(b)
(c)
13.
Since
f
is expanded as a series of Bessel functions,
J
1
(
λ
i
x
) and
J
1
is an odd function, the series should represent
an odd function.
597

598. -2 -1 1 2
x
0.5
1
1.5
2
y
-1-0.5 0.51
x
1
2
3
4
5
y
-1 -0.5 0.5 1
x
0.2
0.4
0.6
0.8
1
fHxL
Exercises 12.6
14. (a)
Since
J
0
is an even function, a series expansion of a function de-
fined on (0
,
2) would converge to the even extension of the function
on (
−
2
,
0).
(b)
In Section 5
.
3 we saw that
J
2
2
(
x
)=2
J
2
(
x
)
/x
−
J
3
(
x
). Since
J
2
is even and
J
3
is odd
we see that
J
2
2
(
−
x
)=2
J
2
(
−
x
)
/
(
−
x
)
−
J
3
(
−
x
)
=
−
2
J
2
(
x
)
/x
+
J
3
(
x
)=
−
J
2
2
(
x
)
,
so that
J
2
2
is an odd function. Now, if
f
(
x
)=3
J
2
(
x
)+2
xJ
2
2
(
x
), we see that
f
(
−
x
)=3
J
2
(
−
x
)
−
2
xJ
2
2
(
−
x
)
=3
J
2
(
x
)+2
xJ
2
2
(
x
)=
f
(
x
)
,
so that
f
is an even function. Thus, a series expansion of a function defined on (0
,
1)
would converge to the even extension of the function on (
−
1
,
0).
15.
We compute
c
0
=
1
2
5
1
0
xP
0
(
x
)
dx
=
1
2
5
1
0
xdx
=
1
4
c
1
=
3
2
5
1
0
xP
1
(
x
)
dx
=
3
2
5
1
0
x
2
dx
=
1
2
c
2
=
5
2
5
1
0
xP
2
(
x
)
dx
=
5
2
5
1
0
1
2
(3
x
3
−
x
)
dx
=
5
16
c
3
=
7
2
5
1
0
xP
3
(
x
)
dx
=
7
2
5
1
0
1
2
(5
x
4
−
3
x
2
)
dx
=0
c
4
=
9
2
5
1
0
xP
4
(
x
)
dx
=
9
2
5
1
0
1
8
(35
x
5
−
30
x
3
+3
x
)
dx
=
−
3
32
c
5
=
11
2
5
1
0
xP
5
(
x
)
dx
=
11
2
5
1
0
1
8
(63
x
6
−
70
x
4
+15
x
2
)
dx
=0
c
6
=
13
2
5
1
0
xP
6
(
x
)
dx
=
13
2
5
1
0
1
16
(231
x
7
−
315
x
5
+ 105
x
3
−
5
x
)
dx
=
13
256
.
Thus
f
(
x
)=
1
4
P
0
(
x
)+
1
2
P
1
(
x
)+
5
16
P
2
(
x
)
−
3
32
P
4
(
x
)+
13
256
P
6
(
x
)+
···
.
598

599. -1 -0.5 0.5 1
x
0.5
1
1.5
2
2.5
3
fHxL
Exercises 12.6
16.
We compute
c
0
=
1
2
5
1
−
1
e
x
P
0
(
x
)
dx
=
1
2
5
1
−
1
e
x
dx
=
1
2
(
e
−
e
−
1
)
c
1
=
3
2
5
1
−
1
e
x
P
1
(
x
)
dx
=
3
2
5
1
−
1
xe
x
dx
=3
e
−
1
c
2
=
5
2
5
1
−
1
e
x
P
2
(
x
)
dx
=
5
2
5
1
−
1
1
2
(3
x
2
e
x
−
e
x
)
dx
=
5
2
(
e
−
7
e
−
1
)
c
3
=
7
2
5
1
−
1
e
x
P
3
(
x
)
dx
=
7
2
5
1
−
1
1
2
(5
x
3
e
x
−
3
xe
x
)
dx
=
7
2
(
−
5
e
+37
e
−
1
)
c
4
=
9
2
5
1
−
1
e
x
P
4
(
x
)
dx
=
9
2
5
1
−
1
1
8
(35
x
4
e
x
−
30
x
2
e
x
+3
e
x
)
dx
=
9
2
(36
e
−
266
e
−
1
)
.
Thus
f
(
x
)=
1
2
(
e
−
e
−
1
)
P
0
(
x
)+3
e
−
1
P
1
(
x
)+
5
2
(
e
−
7
e
−
1
)
P
2
(
x
)
+
7
2
(
−
5
e
+37
e
−
1
)
P
3
(
x
)+
9
2
(36
e
−
266
e
−
1
)
P
4
(
x
)+
···
.
17.
Using cos
2
θ
=
1
2
(cos 2
θ
+1)wehave
P
2
(cos
θ
)=
1
2
(3 cos
2
θ
−
1) =
3
2
cos
2
θ
−
1
2
=
3
4
(cos 2
θ
+1)
−
1
2
=
3
4
cos 2
θ
+
1
4
=
1
4
(3 cos 2
θ
+1)
.
18.
From Problem 17 we have
P
2
(cos 2
θ
)=
1
4
(3 cos 2
θ
+1)
or
cos 2
θ
=
4
3
P
2
(cos
θ
)
−
1
3
.
Then, using
P
0
(cos
θ
)=1,
F
(
θ
)=1
−
cos 2
θ
=1
−
1
4
3
P
2
(cos
θ
)
−
1
3
4
=
4
3
−
4
3
P
2
(cos
θ
)=
4
3
P
0
(cos
θ
)
−
4
3
P
2
(cos
θ
)
.
19.
If
f
is an even function on (
−
1
,
1) then
5
1
−
1
f
(
x
)
P
2
n
(
x
)
dx
=2
5
1
0
f
(
x
)
P
2
n
(
x
)
dx
and
5
1
−
1
f
(
x
)
P
2
n
+1
(
x
)
dx
=0
.
Thus
c
2
n
=
2(2
n
)+1
2
5
1
−
1
f
(
x
)
P
2
n
(
x
)
dx
=
4
n
+1
2
2
2
5
1
0
f
(
x
)
P
2
n
(
x
)
dx
=
=(4
n
+1)
5
1
0
f
(
x
)
P
2
n
(
x
)
dx,
599

600. -1-0.5 0.5 1
x
0.2
0.4
0.6
0.8
1
S
-1-0.5 0.5 1
x
-1
-0.5
0.5
1
S
Exercises 12.6
c
2
n
+1
= 0, and
f
(
x
)=
∞
r
n
=0
c
2
n
P
2
n
(
x
)
.
20.
If
f
is an odd function on (
−
1
,
1) then
5
1
−
1
f
(
x
)
P
2
n
(
x
)
dx
=0
and
5
1
−
1
f
(
x
)
P
2
n
+1
(
x
)
dx
=2
5
1
0
f
(
x
)
P
2
n
+1
(
x
)
dx.
Thus
c
2
n
+1
=
2(2
n
+1)+1
2
5
1
−
1
f
(
x
)
P
2
n
+1
(
x
)
dx
=
4
n
+3
2
2
2
5
1
0
f
(
x
)
P
2
n
+1
(
x
)
dx
=
=(4
n
+1)
5
1
0
f
(
x
)
P
2
n
+1
(
x
)
dx,
c
2
n
= 0, and
f
(
x
)=
∞
r
n
=0
c
2
n
+1
P
2
n
+1
(
x
)
.
21.
From (26) in Problem 19 in the text we find
c
0
=
5
1
0
xP
0
(
x
)
dx
=
5
1
0
xdx
=
1
2
,
c
2
=5
5
1
0
xP
2
(
x
)
dx
=5
5
1
0
1
2
(3
x
3
−
x
)
dx
=
5
8
,
c
4
=9
5
1
0
xP
4
(
x
)
dx
=9
5
1
0
1
8
(35
x
5
−
30
x
3
+3
x
)
dx
=
−
3
16
,
and
c
6
=13
5
1
0
xP
6
(
x
)
dx
=13
5
1
0
1
16
(231
x
7
−
315
x
5
+ 105
x
3
−
5
x
)
dx
=
13
128
.
Hence, from (25) in the text,
f
(
x
)=
1
2
P
0
(
x
)+
5
8
P
2
(
x
)
−
3
16
P
4
(
x
)+
13
128
P
6
+
···
.
On the interval
−
1
<x<
1 this series represents the function
f
(
x
)=
|
x
|
.
22.
From (28) in Problem 20 in the text we find
c
1
=3
5
1
0
P
1
(
x
)
dx
=3
5
1
0
xdx
=
3
2
,
c
3
=7
5
1
0
P
3
(
x
)
dx
=7
5
1
0
1
2
3
5
x
3
−
3
x
F
dx
=
−
7
8
,
c
5
=11
5
1
0
P
5
(
x
)
dx
=11
5
1
0
1
8
3
63
x
5
−
70
x
3
+15
x
F
dx
=
11
16
and
c
7
=15
5
1
0
P
7
(
x
)
dx
=15
5
1
0
1
16
3
429
x
7
−
693
x
5
+ 315
x
3
−
35
x
F
dx
=
−
75
128
.
600

601. Chapter 12 Review Exercises
Hence, from (27) in the text,
f
(
x
)=
3
2
P
1
(
x
)
−
7
8
P
3
(
x
)+
11
16
P
5
(
x
)
−
75
128
P
7
(
x
)+
···
.
On the interval
−
1
<x<
1 this series represents the odd function
f
(
x
)=
o
−
1
,
−
1
<x<
0
1
,
0
<x<
1.
Chapter 12 Review Exercises
1.
True, since
)
π
−
π
(
x
2
−
1)
x
5
dx
=0
2.
Even, since if
f
and
g
are odd then
h
(
−
x
)=
f
(
−
x
)
g
(
−
x
)=
−
f
(
x
)[
−
g
(
x
)] =
f
(
x
)
g
(
x
)=
h
(
x
)
3.
Cosine, since
f
is even
4.
True
5.
False; the Sturm-Liouville problem,
d
dx
[
r
(
x
)
y
2
]+
λp
(
x
)
y
=0
,y
2
(
a
)=0
,y
2
(
b
)=0
,
on the interval [
a, b
], has eigenvalue
λ
=0.
6.
3
/
2, the average of 3 and 0
7.
The Fourier series will converge to 1, the cosine series to 1, and the sine series to 0 at
x
= 0. Respectively, this
is because the rule (
x
2
+ 1) defining
f
(
x
) determines a continuous function on (
−
3
,
3), the even extension of
f
to (
−
3
,
0) is continuous at 0, and the odd extension of
f
to (
−
3
,
0) approaches
−
1as
x
approaches 0 from the
left.
8.
cos 5
x
, since the general solution is
y
=
c
1
cos
√
λx
+
c
2
sin
√
λx
and
y
2
(0) = 0 implies
c
2
=0.
9.
Since the coefficient of
y
in the differential equation is
n
2
, the weight function is the integrating factor
1
a
(
x
)
e
)
(
b/a
)
dx
=
1
1
−
x
2
e
)
−
x
1
−
x
2
dx
=
1
1
−
x
2
e
1
2
ln(1
−
x
2
)
=
√
1
−
x
2
1
−
x
2
=
1
√
1
−
x
2
on the interval [
−
1
,
1]. The orthogonality relation is
1
1
−
1
1
√
1
−
x
2
T
m
(
x
)
T
n
(
x
)
dx
=0
,m
1
=
n.
10.
Since
P
n
(
x
) is orthogonal to
P
0
(
x
) = 1 for
n>
0,
1
1
−
1
P
n
(
x
)
dx
=
1
1
−
1
P
0
(
x
)
P
n
(
x
)
dx
=0
.
11.
We know from a half-angle formula in trigonometry that cos
2
x
=
1
2
+
1
2
cos 2
x
, which is a cosine series.
12. (a)
For
m
1
=
n
1
L
0
sin
(2
n
+1)
π
2
L
x
sin
(2
m
+1)
π
2
L
xdx
=
1
2
1
L
0
2
cos
n
−
m
L
πx
−
cos
n
+
m
+
π
L
πx
3
dx
=0
.
(b)
From
1
L
0
sin
2
(2
n
+1)
π
2
L
xdx
=
1
L
0
2
1
2
−
1
2
cos
(2
n
+1)
π
2
L
x
3
dx
=
L
2
601

602. Chapter 12 Review Exercises
we see that




sin
(2
n
+1)
π
2
L
x




=
0
L
2
.
13.
Since
A
0
=
5
0
−
1
(
−
2
x
)
dx
=1
,
A
n
=
5
0
−
1
(
−
2
x
) cos
nπx dx
=
2
n
2
π
2
[(
−
1)
n
−
1]
,
and
B
n
=
5
0
−
1
(
−
2
x
) sin
nπx dx
=
4
nπ
(
−
1)
n
for
n
=1,2,3,
...
we have
f
(
x
)=
1
2
+
∞
r
n
=1
2
2
n
2
π
2
[(
−
1)
n
−
1] cos
nπx
+
4
nπ
(
−
1)
n
sin
nπx
=
.
14.
Since
A
0
=
5
1
−
1
(2
x
2
−
1)
dx
=
−
2
3
,
A
n
=
5
1
−
1
(2
x
2
−
1) cos
nπx dx
=
8
n
2
π
2
(
−
1)
n
,
and
B
n
=
5
1
−
1
(2
x
2
−
1) sin
nπx dx
=0
for
n
=1,2,3,
...
we have
f
(
x
)=
−
1
3
+
∞
r
n
=1
8
n
2
π
2
(
−
1)
n
cos
nπx.
15.
Since
A
0
=2
5
1
0
e
−
x
dx
and
A
n
=2
5
1
−
1
e
−
x
cos
nπx dx
=
2
1+
n
2
π
2
[(1
−
(
−
1)
n
e
−
1
]
for
n
=1,2,3,
...
we have
f
(
x
)=1
−
e
−
1
+2
∞
r
n
=1
1
−
(
−
1)
n
e
−
1
1+
n
2
π
2
cos
nπx.
Since
B
n
=2
5
1
0
e
−
x
sin
nπx dx
=
2
nπ
1+
n
2
π
2
[(1
−
(
−
1)
n
e
−
1
]
for
n
=1,2,3,
...
we have
f
(
x
)=
∞
r
n
=1
2
nπ
1+
n
2
π
2
[(1
−
(
−
1)
n
e
−
1
] sin
nπx.
602

603. -1-0.5 0.51
x
0.5
1
1.5
2
fHxL
-1-0.5 0.51
x
-1
-0.5
0.5
1
y
-1-0.5 0.51
x
-1
-0.5
0.5
1
y
-1-0.5 0.51
x
-1
-0.5
0.5
1
y
Chapter 12 Review Exercises
16.
f
(
x
)=
|
x
|−
1
f
(
x
)=2
x
2
−
1
f
(
x
)=
e
−|
x
|
f
(
x
)=



e
−
x
,
0
<x<
1
0
,x
=0
−
e
x
,
−
1
<x<
0
17.
For
λ>
0 a general solution of the given differential equation is
y
=
c
1
cos(3
√
λ
ln
x
)+
c
2
sin(3
√
λ
ln
x
)
and
y
2
=
−
3
c
1
√
λ
x
sin(3
√
λ
ln
x
)+
3
c
2
√
λ
x
cos(3
√
λ
ln
x
)
.
Since ln 1 = 0, the boundary condition
y
2
(1) = 0 implies
c
2
= 0. Therefore
y
=
c
1
cos(3
√
λ
ln
x
)
.
Using ln
e
= 1 we find that
y
(
e
) = 0 implies
c
1
cos 3
√
λ
= 0 or 3
√
λ
=
2
n
−
1
2
π
, for
n
=1,2,3,
...
. The
eigenvalues are
λ
=(2
n
−
1)
2
π
2
/
36 with corresponding eigenfunctions cos
2
2
n
−
1
2
π
ln
x
=
for
n
=1,2,3,
...
.
18.
To obtain the self-adjoint form of the differential equation in Problem 17 we note that an integrating factor is
(1
/x
2
)
e
s
dx/x
=1
/x
. Thus the weight function is 9
/x
and an orthogonality relation is
i
e
1
9
x
cos
2
2
n
−
1
2
π
ln
x
=
cos
2
2
m
−
1
2
π
ln
x
=
dx
=0
,m
i
=
n.
19.
The boundary condition indicates that we use (15) and (16) of Section 12.6. With
b
= 4 we obtain
c
i
=
2
16
J
2
1
(4
λ
i
)
i
4
0
xJ
0
(
λ
i
x
)
f
(
x
)
dx
=
1
8
J
2
1
(4
λ
i
)
i
2
0
xJ
0
(
λ
i
x
)
dx
t
=
λ
i
xdt
=
λ
i
dx
=
1
8
J
2
1
(4
λ
i
)
·
1
λ
2
i
i
2
λ
i
0
tJ
0
(
t
)
dt
=
1
8
J
2
1
(4
λ
i
)
i
2
λ
i
0
d
dt
[
tJ
1
(
t
)]
dt
[From (4) in 12.6 in the text]
=
1
8
J
2
1
(4
λ
i
)
tJ
1
(
t
)
x
x
x
x
2
λ
i
0
=
J
1
(2
λ
i
)
4
λ
i
J
2
1
(4
λ
i
)
.
Thus
f
(
x
)=
1
4
∞
r
i
=1
J
1
(2
λ
i
)
λ
i
J
2
1
(4
λ
i
)
J
0
(
λ
i
x
)
.
20.
Since
f
(
x
)=
x
4
is a polynomial in
x
, an expansion of
f
in polynomials in
x
must terminate with the term
having the same degree as
f
. Using the fact that
x
4
P
1
(
x
) and
x
4
P
3
(
x
) are odd functions, we see immediately
603

604. Chapter 12 Review Exercises
that
c
1
=
c
3
=0. Now
c
0
=
1
2
5
1
−
1
x
4
P
0
(
x
)
dx
=
1
2
5
1
−
1
x
4
dx
=
1
5
c
2
=
5
2
5
1
−
1
x
4
P
2
(
x
)
dx
=
5
2
5
1
−
1
1
2
(3
x
6
−
x
4
)
dx
=
4
7
c
4
=
9
2
5
1
−
1
x
4
P
4
(
x
)
dx
=
9
2
5
1
−
1
1
8
(35
x
8
−
30
x
6
+3
x
4
)
dx
=
8
35
.
Thus
f
(
x
)=
1
5
P
0
(
x
)+
4
7
P
2
(
x
)+
8
35
P
4
(
x
)
.
604

605. 13
Boundary-Value Problems in
Rectangular Coordinates
Exercises 13.1
1.
If
u
=
XY
then
u
x
=
X

Y,
u
y
=
XY

,
X

Y
=
XY

,
and
X

X
=
Y

Y
=
±
λ
2
.
Then
X

∓
λ
2
X
= 0 and
Y

∓
λ
2
Y
=0
so that
X
=
A
1
e
±
λ
2
x
,
Y
=
A
2
e
±
λ
2
y
,
and
u
=
XY
=
c
1
e
c
2
(
x
+
y
)
.
2.
If
u
=
XY
then
u
x
=
X

Y,
u
y
=
XY

,
X

Y
=
−
3
XY

,
and
X

−
3
X
=
Y

Y
=
±
λ
2
.
Then
X

±
3
λ
2
X
= 0 and
Y

∓
λ
2
Y
=0
so that
X
=
A
1
e
∓
3
λ
2
x
,
Y
=
A
2
e
±
λ
2
y
,
and
u
=
XY
=
c
1
e
c
2
(
y
−
3
x
)
.
3.
If
u
=
XY
then
u
x
=
X

Y,
u
y
=
XY

,
X

Y
=
X
(
Y
−
Y

)
,
and
605

606. Exercises 13.1
X

X
=
Y
−
Y

Y
=
±
λ
2
.
Then
X

∓
λ
2
X
= 0 and
Y

−
(1
∓
λ
2
)
Y
=0
so that
X
=
A
1
e
±
λ
2
x
,
Y
=
A
2
e
(1
∓
λ
2
)
y
,
and
u
=
XY
=
c
1
e
y
+
c
2
(
x
−
y
)
.
4.
If
u
=
XY
then
u
x
=
X

Y,
u
y
=
XY

,
X

Y
=
X
(
Y
+
Y

)
,
and
X

X
=
Y
+
Y

Y
=
±
λ
2
.
Then
X

∓
λ
2
X
= 0 and
Y

−
(
−
1
±
λ
2
)
Y
=0
so that
X
=
A
1
e
±
λ
2
x
,
Y
=
A
2
e
(
−
1
±
λ
2
)
y
,
and
u
=
XY
=
c
1
e
−
y
+
c
2
(
x
+
y
)
.
5.
If
u
=
XY
then
u
x
=
X

Y,
u
y
=
XY

,
xX

Y
=
yXY

,
and
xX

X
=
yY

Y
=
±
λ
2
.
Then
X
∓
1
x
λ
2
X
= 0 and
Y

∓
1
y
λ
2
Y
=0
so that
X
=
A
1
x
±
λ
2
,
Y
=
A
2
y
±
λ
2
,
and
u
=
XY
=
c
1
(
xy
)
c
2
.
606

607. Exercises 13.1
6.
If
u
=
XY
then
u
x
=
X

Y,
u
y
=
XY

,
yX

Y
=
xXY

,
and
X

xX
=
Y

−
yY
=
±
λ
2
.
Then
X
∓
λ
2
xX
= 0 and
Y

±
λ
2
yY
=0
so that
X
=
A
1
e
±
λ
2
x
2
/
2
,
Y
=
A
2
e
∓
λ
2
y
2
/
2
,
and
u
=
XY
=
c
1
e
c
2
(
x
2
−
y
2
)
.
7.
If
u
=
XY
then
u
xx
=
X

Y, u
yy
=
XY

,u
yx
=
X

Y

,
and
X

Y
+
X

Y

+
XY

=0
,
which is not separable.
8.
If
u
=
XY
then
u
yx
=
X

Y

,
yX

Y

+
XY
=0
,
and
X

−
X
=
Y
yY

=
±
λ
2
.
Then
X

∓
λ
2
X
= 0 and
±
λ
2
yY

−
Y
=0
so that
X
=
A
1
e
∓
λ
2
x
,
Y
=
A
2
y
±
1
/λ
2
,
and
u
=
XY
=
c
1
e
−
c
2
x
y
1
/c
2
.
9.
If
u
=
XT
then
u
t
=
XT

,
u
xx
=
X

T,
kX

T
−
XT
=
XT

,
and we choose
T

T
=
kX

−
X
X
=
−
1
±
kλ
2
607

608. Exercises 13.1
so that
T

−
(
−
1
±
kλ
2
)
T
= 0 and
X

−
(
±
λ
2
)
X
=0
.
For
λ
2
>
0 we obtain
X
=
A
1
cosh
λx
+
A
2
sinh
λx
and
T
=
A
3
e
(
−
1+
kλ
2
)
t
so that
u
=
XT
=
e
(
−
1+
kλ
2
)
t
(
c
1
cosh
λx
+
c
2
sinh
λx
)
.
For
−
λ
2
<
0 we obtain
X
=
A
1
cos
λx
+
A
2
sin
λx
and
T
=
A
3
e
(
−
1
−
kλ
2
)
t
so that
u
=
XT
=
e
(
−
1
−
kλ
2
)
t
(
c
3
cos
λx
+
c
4
sin
λx
)
.
If
λ
2
= 0 then
X

= 0 and
T

+
T
=0
,
and we obtain
X
=
A
1
x
+
A
2
and
T
=
A
3
e
−
t
.
In this case
u
=
XT
=
e
−
t
(
c
5
x
+
c
6
)
10.
If
u
=
XT
then
u
t
=
XT

,
u
xx
=
X

T,
kX

T
=
XT

,
and
X

X
=
T

kT
=
±
λ
2
so that
X

∓
λ
2
X
= 0 and
T

∓
λ
2
kT
=0
.
For
λ
2
>
0 we obtain
X
=
A
1
cos
λx
+
A
2
sin
λx
and
T
=
A
3
e
λ
2
kt
,
so that
u
=
XT
=
e
−
c
2
3
kt
(
c
1
cos
c
3
x
+
c
2
sin
c
3
x
)
.
For
−
λ
2
<
0 we obtain
X
=
A
1
e
−
λx
+
A
2
e
λx
,
T
=
A
3
e
λ
2
kt
,
and
u
=
XT
=
e
c
2
3
kt
I
c
1
e
−
c
3
x
+
c
2
e
c
3
x
f
.
For
λ
2
= 0 we obtain
T
=
A
3
,X
=
A
1
x
+
A
2
,
and
u
=
XT
=
c
1
x
+
c
2
.
608

609. Exercises 13.1
11.
If
u
=
XT
then
u
xx
=
X

T,
u
tt
=
XT

,
a
2
X

T
=
XT

,
and
X

X
=
T

a
2
T
=
±
λ
2
so that
X

∓
λ
2
X
= 0 and
T

∓
a
2
λ
2
T
=0
.
For
λ
2
>
0 we obtain
X
=
A
1
cos
λx
+
A
2
sin
λx,
T
=
A
3
cos
aλt
+
A
4
sin
aλt,
and
u
=
XT
=(
A
1
cos
λx
+
A
2
sin
λx
)(
A
3
cos
aλt
+
A
4
sin
aλt
)
.
For
−
λ
2
<
0 we obtain
X
=
A
1
e
λx
+
A
2
e
−
λx
,
T
=
A
3
e
aλt
+
A
4
e
−
aλt
,
and
u
=
XT
=
I
A
1
e
λx
+
A
2
e
−
λx
fI
A
3
e
aλt
+
A
4
e
−
aλt
f
.
For
λ
2
= 0 we obtain
X
=
A
1
x
+
A
2
,
T
=
A
3
t
+
A
4
,
and
u
=
XT
=(
A
1
x
+
A
2
)(
A
3
t
+
A
4
)
.
12.
If
u
=
XT
then
u
t
=
XT

,
u
tt
=
XT

,
u
xx
=
X

T,
a
2
X

T
=
XT

+2
kXT

,
and
X

X
=
T

+2
kT

a
2
T
=
±
λ
2
so that
X

∓
λ
2
X
= 0 and
T

+2
kT

∓
a
2
λ
2
T
=0
.
For
λ
2
>
0 we obtain
X
=
A
1
e
λx
+
A
2
e
−
λx
,
T
=
A
3
e
(
−
k
+
√
k
2
+
a
2
λ
2
)
t
+
A
4
e
(
−
k
−
√
k
2
+
a
2
λ
2
)
t
,
and
609

610. Exercises 13.1
u
=
XT
=
I
A
1
e
λx
+
A
2
e
−
λx
f
=
A
3
e
(
−
k
+
√
k
2
+
a
2
λ
2
)
t
+
A
4
e
(
−
k
−
√
k
2
+
a
2
λ
2
)
t
t
.
For
−
λ
2
<
0 we obtain
X
=
A
1
cos
λx
+
A
2
sin
λx.
If
k
2
−
a
2
λ
2
>
0 then
T
=
A
3
e
(
−
k
+
√
k
2
−
a
2
λ
2
)
t
+
A
4
e
(
−
k
−
√
k
2
−
a
2
λ
2
)
t
.
If
k
2
−
a
2
λ
2
<
0 then
T
=
e
−
kt
=
A
3
cos
h
a
2
λ
2
−
k
2
t
+
A
4
sin
h
a
2
λ
2
−
k
2
t
t
.
If
k
2
−
a
2
λ
2
= 0 then
T
=
A
3
e
−
kt
+
A
4
te
−
kt
so that
u
=
XT
=(
A
1
cos
λx
+
A
2
sin
λx
)
=
A
3
e
(
−
k
+
√
k
2
−
a
2
λ
2
)
t
+
A
4
e
(
−
k
−
√
k
2
−
a
2
λ
2
)
t
t
=(
A
1
cos
λx
+
A
2
sin
λx
)
e
−
kt
=
A
3
cos
h
a
2
λ
2
−
k
2
t
+
A
4
sin
h
a
2
λ
2
−
k
2
t
t
=
e
A
1
cos
k
a
x
+
A
2
sin
k
a
x
n
I
A
3
e
−
kt
+
A
4
te
−
kt
f
.
For
λ
2
= 0 we obtain
XA
1
x
+
A
2
,
T
=
A
3
+
A
4
e
−
2
kt
,
and
u
=
XT
=(
A
1
x
+
A
2
)(
A
3
+
A
4
e
−
2
kt
)
.
13.
If
u
=
XY
then
u
xx
=
X

Y,
u
yy
=
XY

,
X

Y
+
xY

=0
,
and
X

−
X
=
Y

Y
=
±
λ
2
so that
X

±
λ
2
X
= 0 and
Y

∓
λ
2
Y
=0
.
For
λ
2
>
0 we obtain
X
=
A
1
cos
λx
+
A
2
sin
λx,
Y
=
A
3
e
λy
+
A
4
e
−
λy
,
and
u
=
XY
=(
A
1
cos
λx
+
A
2
sin
λx
)(
A
3
e
λy
+
A
4
e
−
λy
)
.
For
−
λ
2
<
0 we obtain
X
=
A
1
e
λx
+
A
2
e
−
λx
,
Y
=
A
3
cos
λy
+
A
4
sin
λy,
and
610

611. Exercises 13.1
u
=
XY
=(
A
1
e
λx
+
A
2
e
−
λx
)(
A
3
cos
λy
+
A
4
sin
λy
)
.
For
λ
2
= 0 we obtain
u
=
XY
=(
A
1
x
+
A
2
)(
A
3
y
+
A
4
)
.
14.
If
u
=
XY
then
u
xx
=
X

Y,
u
yy
=
XY

,
x
2
X

Y
+
xY

=0
,
and
x
2
X

−
X
=
Y

Y
=
±
λ
2
so that
x
2
X

±
λ
2
X
= 0 and
Y

∓
λ
2
Y
=0
.
For
λ
2
>
0 we obtain
Y
=
A
3
e
λy
+
A
4
e
−
λy
.
If 1
−
4
λ
2
>
0 then
X
=
A
1
x
(1
/
2+
√
1
−
4
λ
2
/
2)
+
A
2
x
(1
/
2
−
√
1
−
4
λ
2
/
2)
.
If 1
−
4
λ
2
<
0 then
X
=
x
1
/
2
e
A
1
cos
1
2
h
4
λ
2
−
1ln
x
+
A
2
sin
1
2
h
4
λ
2
−
1ln
x
n
.
If 1
−
4
λ
2
= 0 then
X
=
A
1
x
1
/
2
+
A
2
x
1
/
2
ln
x
so that
u
=
XY
=
=
A
1
x
(1
/
2+
√
1
−
4
λ
2
/
2)
+
A
2
x
(1
/
2
−
√
1
−
4
λ
2
/
2)
t
I
A
3
e
λy
+
A
4
e
−
λy
f
=
x
1
/
2
e
A
1
cos
1
2
h
4
λ
2
−
1ln
x
+
A
2
sin
1
2
h
4
λ
2
−
1ln
x
n
=
A
3
e
y/
2
+
A
4
e
−
y/
2
t
=
=
A
1
x
1
/
2
+
A
2
x
1
/
2
ln
x
t=
A
3
e
y/
2
+
A
4
e
−
y/
2
t
.
For
−
λ
2
<
0 we obtain
X
=
A
1
e
(1
/
2+
√
1+4
λ
2
/
2)
x
+
A
2
e
(1
/
2
−
√
1+4
λ
2
/
2)
x
,
Y
=
A
3
cos
λy
+
A
4
sin
λy,
and
u
=
XY
=
=
A
1
e
(1
/
2+
√
1+4
λ
2
/
2)
x
+
A
2
e
(1
/
2
−
√
1+4
λ
2
/
2)
x
t
(
A
3
cos
λy
+
A
4
sin
λy
)
.
For
λ
2
= 0 we obtain
X
=
A
1
x
+
A
2
,
Y
=
A
3
y
+
A
4
,
and
u
=
XY
=(
A
1
x
+
A
2
)(
A
3
y
+
A
4
)
.
611

612. Exercises 13.1
15.
If
u
=
XY
then
u
xx
=
X

Y,
u
yy
=
XY

,
X

Y
+
XY

=
XY,
and
X

X
=
Y
−
Y

Y
=
±
λ
2
so that
X

∓
λ
2
X
= 0 and
Y

+(
±
λ
2
−
1)
Y
=0
.
For
λ
2
>
0 we obtain
X
=
A
1
e
λx
+
A
2
e
−
λx
.
If
λ
2
−
1
>
0 then
Y
=
A
3
cos
h
λ
2
−
1
y
+
A
4
sin
h
λ
2
−
1
y.
If
λ
2
−
1
<
0 then
Y
=
A
3
e
√
1
−
λ
2
y
+
A
4
e
−
√
1
−
λ
2
y
.
If
λ
2
−
1 = 0 then
Y
=
A
3
y
+
A
4
so that
u
=
XY
=
I
A
1
e
λx
+
A
2
e
−
λx
f
=
A
3
cos
h
λ
2
−
1
y
+
A
4
sin
h
λ
2
−
1
y
t
,
=
I
A
1
e
λx
+
A
2
e
−
λx
f
=
A
3
e
√
1
−
λ
2
y
+
A
4
e
−
√
1
−
λ
2
y
t
=(
A
1
e
x
+
A
2
e
−
x
)(
A
3
y
+
A
4
)
.
For
−
λ
2
<
0 we obtain
X
=
A
1
cos
λx
+
A
2
sin
λx,
Y
=
A
3
e
√
1+
λ
2
y
+
A
4
e
−
√
1+
λ
2
y
,
and
u
=
XY
=(
A
1
cos
λx
+
A
2
sin
λx
)
=
A
3
e
√
1+
λ
2
y
+
A
4
e
−
√
1+
λ
2
y
t
.
For
λ
2
= 0 we obtain
X
=
A
1
x
+
A
2
,
Y
=
A
3
e
y
+
A
4
e
−
y
,
and
u
=
XY
=(
A
1
x
+
A
2
)(
A
3
e
y
+
A
4
e
−
y
)
.
16.
If
u
=
XT
then
u
tt
=
XT

,u
xx
=
X

T,
and
a
2
X

T
−
g
=
XT

,
which is not separable.
17.
Identifying
A
=
B
=
C
= 1, we compute
B
2
−
4
AC
=
−
3
<
0. The equation is elliptic.
18.
Identifying
A
=3,
B
= 5, and
C
= 1, we compute
B
2
−
4
AC
=13
>
0. The equation is hyperbolic.
19.
Identifying
A
=1,
B
= 6, and
C
= 9, we compute
B
2
−
4
AC
= 0. The equation is parabolic.
20.
Identifying
A
=1,
B
=
−
1, and
C
=
−
3, we compute
B
2
−
4
AC
=13
>
0. The equation is hyperbolic.
612

613. Exercises 13.1
21.
Identifying
A
=1,
B
=
−
9, and
C
= 0, we compute
B
2
−
4
AC
=81
>
0. The equation is hyperbolic.
22.
Identifying
A
=0,
B
= 1, and
C
= 0, we compute
B
2
−
4
AC
=1
>
0. The equation is hyperbolic.
23.
Identifying
A
=1,
B
= 2, and
C
= 1, we compute
B
2
−
4
AC
= 0. The equation is parabolic.
24.
Identifying
A
=1,
B
= 0, and
C
= 1, we compute
B
2
−
4
AC
=
−
4
<
0. The equation is elliptic.
25.
Identifying
A
=
a
2
,
B
= 0, and
C
=
−
1, we compute
B
2
−
4
AC
=4
a
2
>
0. The equation is hyperbolic.
26.
Identifying
A
=
k>
0,
B
= 0, and
C
= 0, we compute
B
2
−
4
AC
=
−
4
k<
0. The equation is elliptic.
27.
If
u
=
RT
then
u
r
=
R

T,
u
rr
=
R

T,
u
t
=
RT

,
RT

=
k
e
R

T
+
1
r
R

T
n
,
and
r
2
R

+
rR

r
2
R
=
T

kT
=
±
λ
2
.
If we use
−
λ
2
<
0 then
r
2
R

+
rR

+
λ
2
r
2
R
= 0 and
T

∓
λ
2
kT
=0
so that
R
=
A
2
J
0
(
λr
)+
A
3
Y
0
(
λr
)
,
T
=
A
1
e
−
kλ
2
t
,
and
u
=
RT
=
e
−
kλ
2
t
[
c
1
J
0
(
λr
)+
c
2
Y
0
(
λr
)]
28.
If
u
=
R
Θ then
u
r
=
R

Θ
,
u
rr
=
R

Θ
,
u
t
=
R
Θ

,
u
tt
=
R
Θ

,
R

Θ+
1
r
R

Θ+
1
r
2
R
Θ

=0
,
and
r
2
R

+
rR

R
=
−
Θ

Θ
=
λ
2
.
Then
r
2
R

+
rR

−
λ
2
R
= 0 and Θ

+
λ
2
Θ=0
so that
R
=
c
3
r
λ
+
c
4
r
−
λ
Θ
θ
=
c
1
cos
λθ
+
c
2
sin
λθ,
and
u
=
R
Θ=(
c
1
cos
λθ
+
c
2
sin
λθ
)(
c
3
r
λ
+
c
4
r
−
λ
)
.
613

614. Exercises 13.1
29.
For
u
=
A
1
e
λ
2
y
cosh 2
λx
+
B
1
e
λ
2
y
sinh 2
λx
we compute
∂
2
u
∂x
2
=4
λ
2
A
1
e
λ
2
y
cosh 2
λx
+4
λ
2
B
1
e
λ
2
y
sinh 2
λx
and
∂u
∂y
=
λ
2
A
1
e
λ
2
y
cosh 2
λx
+
λ
2
B
1
e
λ
2
y
sinh 2
λx.
Then
∂
2
u/∂x
2
=4
∂u/∂y
.
For
u
=
A
2
e
−
λ
2
y
cos 2
λx
+
B
2
e
−
λ
2
y
sin 2
λx
we compute
∂
2
u
∂x
2
=
−
4
λ
2
A
2
e
−
λ
2
y
cos 2
λx
−
4
λ
2
B
2
e
−
λ
2
y
sin 2
λx
and
∂u
∂y
=
−
λ
2
A
2
cos 2
λx
−
λ
2
B
2
sin 2
λx.
Then
∂
2
u/∂x
2
=4
∂u/∂y
.
For
u
=
A
3
x
+
B
3
we compute
∂
2
u/∂x
2
=
∂u/∂y
= 0. Then
∂
2
u/∂x
2
=4
∂u/∂y
.
30.
We identify
A
=
xy
+1,
B
=
x
+2
y
, and
C
= 1. Then
B
2
−
4
AC
=
x
2
+4
y
2
−
4. The equation
x
2
+4
y
2
=4
defines an ellipse. The partial differential equation is hyperbolic outside the ellipse, parabolic on the ellipse,
and elliptic inside the ellipse.
Exercises 13.2
1.
k
∂
2
u
∂x
2
=
∂u
∂t
,
0
<x<L, t>
0
u
(0
,t
)=0
,
∂u
∂x




x
=
L
=0
,t>
0
u
(
x,
0) =
f
(
x
)
,
0
<x<L
2.
k
∂
2
u
∂x
2
=
∂u
∂t
,
0
<x<L, t>
0
u
(0
,t
)=
u
0
,u
(
L, t
)=
u
1
,t>
0
u
(
x,
0) = 0
,
0
<x<L
3.
k
∂
2
u
∂x
2
=
∂u
∂t
,
0
<x<L, t>
0
u
(0
,t
) = 100
,
∂u
∂x




x
=
L
=
−
hu
(
L, t
)
,t>
0
u
(
x,
0) =
f
(
x
)
,
0
<x<L
4.
k
∂
2
u
∂x
2
+
h
(
u
−
50) =
∂u
∂t
,
0
<x<L, t>
0
∂u
∂x




x
=0
=0
,
∂u
∂x




x
=
L
t>
0
u
(
x,
0) = 100
,
0
<x<L
5.
a
2
∂
2
u
∂x
2
=
∂
2
u
∂t
2
,
0
<x<L, t>
0
u
(0
,t
)=0
,u
(
L, t
)=0
,t>
0
u
(
x,
0) =
x
(
L
−
x
)
,
∂u
∂x




t
=0
=0
,
0
<x<L
614

615. Exercises 13.3
6.
a
2
∂
2
u
∂x
2
=
∂
2
u
∂t
2
,
0
<x<L, t>
0
u
(0
,t
)=0
,u
(
L, t
)=0
,t>
0
u
(
x,
0) = 0
,
∂u
∂x




t
=0
= sin
πx
L
,
0
<x<L
7.
a
2
∂
2
u
∂x
2
−
2
β
∂u
∂t
=
∂
2
u
∂t
2
,
0
<x<L, t>
0
u
(0
,t
)=0
,u
(
L, t
) = sin
πt, t >
0
u
(
x,
0) =
f
(
x
)
,
∂u
∂t




t
=0
=0
,
0
<x<L
8.
a
2
∂
2
u
∂x
2
+
Ax
=
∂
2
u
∂t
2
,
0
<x<L, t>
0
,A
a constant
u
(0
,t
)=0
,u
(
L, t
)=0
,t>
0
u
(
x,
0) = 0
,
∂u
∂x




t
=0
=0
,
0
<x<L
9.
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0
,
0
<x<
4
,
0
<y<
2
∂u
∂x




x
=0
=0
,u
(4
,y
)=
f
(
y
)
,
0
<y<
2
∂u
∂y




y
=0
=0
,u
(
x,
2) = 0
,
0
<x<
4
10.
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0
,
0
<x<π, y>
0
u
(0
,y
)=
e
−
y
,u
(
π, y
)=

100
,
0
<y
≤
1
0
,y>
1
u
(
x,
0) =
f
(
x
)
,
0
<x<π
Exercises 13.3
1.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X
kk
+
λ
2
X
=0
,
X
(0) = 0
,
X
(
L
)=0
,
and
T
k
+
kλ
2
T
=0
.
Then
X
=
c
1
sin
nπ
L
x
and
T
=
c
2
e
−
kn
2
π
2
L
2
t
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
e
−
kn
2
π
2
L
2
t
sin
nπ
L
x.
615

616. Exercises 13.3
Imposing
u
(
x,
0) =
∞

n
=1
A
n
sin
nπ
L
x
gives
A
n
=
2
L

L/
2
0
sin
nπ
L
xdx
=
2
nπ
L
1
−
cos
nπ
2
t
for
n
=1,2,3,
...
so that
u
(
x, t
)=
2
π
∞

n
=1
1
−
cos
nπ
2
n
e
−
kn
2
π
2
L
2
t
sin
nπ
L
x.
2.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X
kk
+
λ
2
X
=0
,
X
(0) = 0
,
X
(
L
)=0
,
and
T
k
+
kλ
2
T
=0
.
Then
X
=
c
1
sin
nπ
L
x
and
T
=
c
2
e
−
kn
2
π
2
L
2
t
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
e
−
kn
2
π
2
L
2
t
sin
nπ
L
x.
Imposing
u
(
x,
0) =
∞

n
=1
A
n
sin
nπ
L
x
gives
A
n
=
2
L

L
0
x
(
L
−
x
) sin
nπ
L
xdx
=
4
L
2
n
3
π
3
[1
−
(
−
1)
n
]
for
n
=1,2,3,
...
so that
u
(
x, t
)=
4
L
2
π
3
∞

n
=1
1
−
(
−
1)
n
n
3
e
−
kn
2
π
2
L
2
t
sin
nπ
L
x.
3.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X
kk
+
λ
2
X
=0
,
X
k
(0) = 0
,
X
k
(
L
)=0
,
and
T
k
+
kλ
2
T
=0
.
Then
X
=
c
1
cos
nπ
L
x
and
T
=
c
2
e
−
kn
2
π
2
L
2
t
for
n
=0,1,2,
...
so that
u
=
∞

n
=0
A
n
e
−
kn
2
π
2
L
2
t
cos
nπ
L
x.
616

617. Exercises 13.3
Imposing
u
(
x,
0) =
f
(
x
)=
∞

n
=0
A
n
cos
nπ
L
x
gives
u
(
x, t
)=
1
L

L
0
f
(
x
)
dx
+
2
L
∞

n
=1


L
0
f
(
x
) cos
nπ
L
xdx
o
e
−
kn
2
π
2
L
2
t
cos
nπ
L
x.
4.
If
L
= 2 and
f
(
x
)is
x
for 0
<x<
1 and
f
(
x
) is 0 for 1
<x<
2 then
u
(
x, t
)=
1
4
+4
∞

n
=1
3
1
2
nπ
sin
nπ
2
+
1
n
2
π
2
L
cos
nπ
2
−
1
t
(
e
−
kn
2
π
2
4
t
cos
nπ
2
x.
5.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X
kk
+
λ
2
X
=0
,
X
k
(0) = 0
,
X
k
(
L
)=0
,
and
T
k
+(
h
+
kλ
2
)
T
=0
.
Then
X
=
c
1
cos
nπ
L
x
and
T
=
c
2
e
−
n
h
+
kn
2
π
2
L
2
π
t
for
n
=0,1,2,
...
so that
u
=
∞

n
=0
A
n
e
−
n
h
+
kn
2
π
2
L
2
π
t
cos
nπ
L
x.
Imposing
u
(
x,
0) =
f
(
x
)=
∞

n
=0
A
n
cos
nπ
L
x
gives
u
(
x, t
)=
e
−
ht
L

L
0
f
(
x
)
dx
+
2
L
∞

n
=1


L
0
f
(
x
) cos
nπ
L
xdx
o
e
−
n
h
+
kn
2
π
2
L
2
π
t
cos
nπ
L
x.
6.
In Problem 5 we instead find that
X
(0) = 0 and
X
(
L
) = 0 so that
X
=
c
1
sin
nπ
L
x
and
u
=
2
L
∞

n
=1


L
0
f
(
x
) sin
nπ
L
xdx
o
e
−
n
h
+
kn
2
π
2
L
2
π
t
sin
nπ
L
x.
617

618. 1 2 3 4 5 6
t
20
40
60
80
100
u
x=3p 4
x=5p 6
x=11p 12
x=p
0
20
40
60
80
100
x
0
50
100
150
200
t
0
10
20
30
40
uHx,tL
0
20
40
60
Exercises 13.3
7.
8. (a)
The solution is
u
(
x, t
)=
∞

n
=1
A
n
e
−
k
(
n
2
π
2
/
100
2
)
t
sin
nπ
100
x,
where
A
n
=
2
100
3

50
0
0
.
8
x
sin
nπ
100
xdx
+

100
50
0
.
8(100
−
x
) sin
nπ
100
xdx
(
=
320
n
2
π
2
sin
nπ
x
.
Thus,
u
(
x, t
)=
320
π
2
∞

n
=1
1
n
2
=
sin
nπ
2
t
e
−
k
(
n
2
π
2
/
100
2
)
t
sin
nπ
100
x.
(b)
Since
A
n
= 0 for
n
even, the first five nonzero terms correspond to
n
= 1, 3, 5, 7, 9. In this case
sin(
nπ/
2) = sin(2
p
−
1)
/
2=(
−
1)
p
+1
for
p
=1,2,3,4,5,and
u
(
x, t
)=
320
π
2
∞

p
=1
(
−
1)
p
+1
(2
p
−
1)
2
e
(
−
1
.
6352(2
p
−
1)
2
π
2
/
100
2
)
t
sin
(2
p
−
1)
π
100
x.
Exercises 13.4
1.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
L
)=0
,
and
T

+
λ
2
a
2
T
=0
.
618

619. Exercises 13.4
Then
X
=
c
1
sin
nπ
L
x
and
T
=
c
2
cos
nπa
L
t
+
c
3
sin
nπa
L
t
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
=
A
n
cos
nπa
L
t
+
B
n
sin
nπa
L
t
t
sin
nπ
L
x.
Imposing
u
(
x,
0) =
1
4
x
(
L
−
x
)=
∞

n
=1
A
n
sin
nπ
L
x
and
u
t
(
x,
0) = 0 =
∞

n
=1
B
n
nπa
L
sin
nπ
L
x
gives
A
n
=
L
n
3
π
3
[1
−
(
−
1)
n
] and
B
n
=0
for
n
=1,2,3,
...
so that
u
(
x, t
)=
L
2
π
3
∞

n
=1
1
−
(
−
1)
n
n
3
cos
nπa
L
t
sin
nπ
L
x.
2.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
L
)=0
,
and
T

+
λ
2
a
2
T
=0
.
Then
X
=
c
1
sin
nπ
L
x
and
T
=
c
2
cos
nπa
L
t
+
c
3
sin
nπa
L
t
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
=
A
n
cos
nπa
L
t
+
B
n
sin
nπa
L
t
t
sin
nπ
L
x.
Imposing
u
(
x,
0) = 0 =
∞

n
=1
A
n
sin
nπ
L
x
and
u
t
(
x,
0) =
x
(
L
−
x
)=
∞

n
=1
B
n
nπa
L
sin
nπ
L
x
gives
A
n
= 0 and
B
n
nπa
L
=
4
L
2
n
3
π
3
[1
−
(
−
1)
n
]
for
n
=1,2,3,
...
so that
u
(
x, t
)=
4
L
3
aπ
4
∞

n
=1
1
−
(
−
1)
n
n
4
sin
nπa
L
t
sin
nπ
L
x.
619

620. Exercises 13.4
3.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
L
)=0
,
and
T

+
λ
2
a
2
T
=0
.
Then
X
=
c
1
sin
nπ
L
x
and
T
=
c
2
cos
nπa
L
t
+
c
3
sin
nπa
L
t
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
=
A
n
cos
nπa
L
t
+
B
n
sin
nπa
L
t
t
sin
nπ
L
x.
Imposing
u
(
x,
0) =
∞

n
=1
A
n
sin
nπ
L
x
gives
A
n
=
2
L


L/
3
0
3
L
x
sin
nπ
L
xdx
+

2
L/
3
L/
3
sin
nπ
L
xdx
+

L
2
L/
3
e
3
−
3
L
x
n
sin
nπ
L
xdx
o
so that
A
1
=
6
√
3
π
2
,
A
2
=
A
3
=
A
4
=0
,
A
5
=
−
6
√
3
5
2
π
2
,
A
6
=0
,
a
7
=
6
√
3
7
2
π
2
... .
Imposing
u
t
(
x,
0) = 0 =
∞

n
=1
B
n
nπa
L
sin
nπ
L
x
gives
B
n
= 0 for
n
=1,2,3,
...
so that
u
(
x, t
)=
6
√
3
π
2
e
cos
πa
L
t
sin
π
L
x
−
1
5
2
cos
5
πa
L
t
sin
5
π
L
x
+
1
7
2
cos
7
πa
L
t
sin
7
π
L
x
−···
n
.
4.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
π
)=0
,
and
T

+
λ
2
a
2
T
=0
.
Then
X
=
c
1
sin
nx
and
T
=
c
2
cos
nat
+
c
3
sin
nat
620

621. Exercises 13.4
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
(
A
n
cos
nat
+
B
n
sin
nat
) sin
nx.
Imposing
u
(
x,
0) =
1
6
x
(
π
2
−
x
2
)=
∞

n
=1
A
n
sin
nx
and
u
t
(
x,
0) = 0
gives
B
n
= 0 and
A
n
=
2
n
3
(
−
1)
n
+1
for
n
=1,2,3,
...
so that
u
(
x, t
)=2
∞

n
=1
(
−
1)
n
+1
n
3
cos
nat
sin
nx.
5.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
π
)=0
,
and
T

+
λ
2
a
2
T
=0
.
Then
X
=
c
1
sin
nx
and
T
=
c
2
cos
nat
+
c
3
sin
nat
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
(
A
n
cos
nat
+
B
n
sin
nat
) sin
nx.
Imposing
u
(
x,
0) = 0 =
∞

n
=1
A
n
sin
nx
and
u
t
(
x,
0) = sin
x
=
∞

n
=1
B
n
na
sin
nx
gives
A
n
=0
,B
1
=
1
a
2
,
and
B
n
=0
for
n
=2,3,4,
...
so that
u
(
x, t
)=
1
a
sin
at
sin
x.
6.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(1) = 0
,
and
T

+
λ
2
a
2
T
=0
.
Then
X
=
c
1
sin
nπx
and
T
=
c
2
cos
nπat
+
c
3
sin
nπat
621

622. Exercises 13.4
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
(
A
n
cos
nπat
+
B
n
sin
nπat
) sin
nπx.
Imposing
u
(
x,
0) = 0
.
01 sin 3
πx
=
∞

n
=1
A
n
sin
nπx
and
u
t
(
x,
0) = 0 =
∞

n
=1
B
n
nπa
sin
nπx
gives
B
n
= 0 for
n
=1,2,3,
...
,
A
3
=0
.
01, and
A
n
= 0 for
n
=1,2,4,5,6,
...
so that
u
(
x, t
)=0
.
01 sin 3
πx
cos 3
πat.
7.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
L
)=0
,
and
T

+
λ
2
a
2
T
=0
.
Then
X
=
c
1
sin
nπ
L
x
and
T
=
c
2
cos
nπa
L
t
+
c
3
sin
nπa
L
t
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
=
A
n
cos
nπa
L
t
+
B
n
sin
nπa
L
t
t
sin
nπ
L
x.
Imposing
u
(
x,
0) =
∞

n
=1
A
n
sin
nπ
L
x
and
u
t
(
x,
0) = 0 =
∞

n
=1
B
n
nπa
L
sin
nπ
L
x
gives
B
n
= 0 and
A
n
=
8
h
n
2
π
2
sin
nπ
2
for
n
=1,2,3,
...
so that
u
(
x, t
)=
8
h
π
2
∞

n
=1
1
n
2
sin
nπ
2
cos
nπa
L
t
sin
nπ
L
x.
8.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X

(0) = 0
,
X

(
L
)=0
,
and
T

+
a
2
λ
2
T
=0
.
622

623. Exercises 13.4
Then
X
=
c
1
cos
nπ
L
x
and
T
=
c
2
cos
nπa
L
t
for
n
=1,2,3,
...
. Since
λ
= 0 is also an eigenvalue with eigenfunction
X
(
x
)=1wehave
u
=
A
0
+
∞

n
=1
A
n
cos
nπa
L
t
cos
nπ
L
x.
Imposing
u
(
x,
0) =
x
=
∞

n
=0
A
n
cos
nπ
L
x
gives
A
0
=
1
L

L
0
xdx
=
L
2
and
A
n
=
2
L

L
0
x
cos
nπ
L
xdx
=
2
L
n
2
π
2
[(
−
1)
n
−
1]
for
n
=1,2,3,
...
, so that
u
(
x, t
)=
L
2
+
2
L
π
2
∞

n
=1
(
−
1)
n
−
1
n
2
cos
nπa
L
t
cos
nπ
L
x.
9.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
π
)=0
,
and
T

+2
βT

+
λ
2
T
=0
.
Then
X
=
c
1
sin
nx
and
T
=
e
−
βt
=
c
2
cos
h
n
2
−
β
2
t
+
c
3
sin
h
n
2
−
β
2
t
t
so that
u
=
∞

n
=1
e
−
βt
=
A
n
cos
h
n
2
−
β
2
t
+
B
n
sin
h
n
2
−
β
2
t
t
sin
nx.
Imposing
u
(
x,
0) =
f
(
x
)=
∞

n
=1
A
n
sin
nx
and
u
t
(
x,
0) = 0 =
∞

n
=1
=
B
n
h
n
2
−
β
2
−
βA
n
t
sin
nx
gives
u
(
x, t
)=
e
−
βt
∞

n
=1
A
n

cos
h
n
2
−
β
2
t
+
β
h
n
2
−
β
2
sin
h
n
2
−
β
2
t
o
sin
nx,
where
A
n
=
2
π

π
0
f
(
x
) sin
nx dx.
10.
Using
u
=
XT
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0,
X
(0) = 0,
X
(
π
) = 0 and
623

624. Exercises 13.4
T

+(1+
λ
2
)
T
=0,
T

(0) = 0. Then
X
=
c
2
sin
nx
and
T
=
c
3
cos
√
n
2
+1
t
for
n
=1,2,3,
...
so that
u
=
∞

n
1
B
n
cos
h
n
2
+1
t
sin
nx.
Imposing
u
(
x,
0) =
∞

n
=1
B
n
sin
nx
gives
B
n
=
2
π

π/
2
0
x
sin
nxdx
+
2
π

π
π/
2
(
π
−
x
) sin
nxdx
=
4
πn
2
sin
nπ
2
=
)
0
,n
even
4
πn
2
(
−
1)
(
n
+3)
/
2
,n
=2
k
−
1,
k
=1,2,3,
...
.
Thus with
n
=2
k
−
1,
u
(
x, t
)=
4
π
∞

n
=1
sin
nπ
2
n
2
cos
h
n
2
+1
t
sin
nx
=
4
π
∞

k
=1
(
−
1)
k
+1
(2
k
−
1)
2
cos
h
(2
k
−
1)
2
+1
t
sin(2
k
−
1)
x.
11.
Separatingvariables in the partial differential equation gives
X
(4)
X
=
−
T

a
2
T
=
λ
4
so that
X
(4)
−
λ
4
X
=0
T

+
a
2
λ
4
T
=0
and
X
=
c
1
cosh
λx
+
c
2
sinh
λx
+
c
3
cos
λx
+
c
4
sin
λx
T
=
c
5
cos
aλ
2
t
+
c
6
sin
aλ
2
t.
The boundary conditions translate into
X
(0) =
X
(
L
) = 0 and
X

(0) =
X

(
L
) = 0. From
X
(0) =
X

(0) = 0
we find
c
1
=
c
3
= 0. From
X
(
L
)=
c
2
sinh
λL
+
c
4
sin
λL
=0
X

(
L
)=
λ
2
c
2
sinh
λL
−
λ
2
c
4
sin
λL
=0
we see by subtraction that
c
4
sin
λL
= 0. This equation yields the eigenvalues
λ
=
nπL
for
n
=1,2,3,
...
.
The correspondingeigenfunctions are
X
=
c
4
sin
nπ
L
x.
Thus
u
(
x, t
)=
∞

n
=1
e
A
n
cos
n
2
π
2
L
2
at
+
B
n
sin
n
2
π
2
L
2
at
n
sin
nπ
L
x.
From
u
(
x,
0) =
f
(
x
)=
∞

n
=1
A
n
sin
nπ
L
x
we obtain
A
n
=
2
L

L
0
f
(
x
) sin
nπ
L
x dx.
From
∂u
∂t
=
∞

n
=1
e
−
A
n
n
2
π
2
a
L
2
sin
n
2
π
2
L
2
at
+
B
n
n
2
π
2
a
L
2
cos
n
2
π
2
L
2
at
n
sin
nπ
L
x
624

625. Exercises 13.4
and
∂u
∂t




t
=0
=
g
(
x
)=
∞

n
=1
B
n
n
2
π
2
a
L
2
sin
nπ
L
x
we obtain
B
n
n
2
π
2
a
L
2
=
2
L

L
0
g
(
x
) sin
nπ
L
xdx
and
B
n
=
2
L
n
2
π
2
a

L
0
g
(
x
) sin
nπ
L
x dx.
12. (a)
Using
X
=
c
1
cosh
λx
+
c
2
sinh
λx
+
c
3
cos
λx
+
c
4
sin
λx
and
X
(0) = 0,
X

(0) = 0 we find, in turn,
c
3
=
−
c
1
and
c
4
=
−
c
2
. The conditions
X
(
L
) = 0 and
X

(
L
)=0
then yield the system of equations for
c
1
and
c
2
:
c
1
(cosh
λL
−
cos
λL
)+
c
2
(sinh
λL
−
sin
λL
)=0
c
1
(
λ
sinh
λL
+
λ
sin
λL
)+
c
2
(
λ
cosh
λL
−
λ
cos
λL
)=0
.
In order that this system have nontrivial solutions the determinant of the coefficients must be zero:
λ
(cosh
λL
−
cos
λL
)
2
−
λ
(sinh
2
λL
−
sin
2
λL
)=0
.
λ
= 0 is not an eigenvalue since this leads to
X
= 0. Thus the last equation simplifies to cosh
λL
cos
λL
=1
or cosh
x
cos
x
= 1, where
x
=
λL
.
(b)
The equation cosh
x
cos
x
= 1 is the same as cos
x
= sech
x
. The figure
indicates that the equation has an infinite number of roots.
(c)
Usinga CAS we find the first four positive roots to be
x
1
=4
.
7300,
x
2
=7
.
8532,
x
3
=10
.
9956, and
x
4
=14
.
1372. Thus the first four eigenvalues are
λ
1
=
x
1
/L
=4
.
7300
/L
,
λ
2
=
x
2
/L
=7
.
8532
/L
,
λ
3
=
x
3
/L
=10
.
9956
/L
, and
λ
4
=14
.
1372
/L
.
13.
From (5) in the text we have
u
(
x, t
)=
∞

n
=1
=
A
n
cos
nπa
L
t
+
B
n
sin
nπa
L
t
t
sin
nπ
L
x.
Since
u
t
(
x,
0) =
g
(
x
)=0wehave
B
n
= 0 and
u
(
x, t
)=
∞

n
=1
A
n
cos
nπa
L
t
sin
nπ
L
x
=
∞

n
=1
A
n
1
2
1
sin
=
nπ
L
x
+
nπa
L
t
t
+ sin
=
nπ
L
x
−
nπa
L
t
t+
=
1
2
∞

n
=1
A
n
1
sin
nπ
L
(
x
+
at
) + sin
nπ
L
(
x
−
at
)
+
.
From
u
(
x,
0) =
f
(
x
)=
∞

n
=1
A
n
sin
nπ
L
x
625

626. Exercises 13.4
we identify
f
(
x
+
at
)=
∞

n
=1
A
n
sin
nπ
L
(
x
+
at
)
and
f
(
x
−
at
)=
∞

n
=1
A
n
sin
nπ
L
(
x
−
at
)
,
so that
u
(
x, t
)=
1
2
[
f
(
x
+
at
)+
f
(
x
−
at
)]
.
14. (a)
We note that
ξ
x
=
η
x
=1,
ξ
t
=
a
, and
η
t
=
−
a
. Then
∂u
∂x
=
∂u
∂ξ
∂ξ
∂x
+
∂u
∂η
∂η
∂x
=
u
ξ
+
u
η
and
∂
2
u
∂x
2
=
∂
∂x
(
u
ξ
+
u
η
)=
∂u
ξ
∂ξ
∂ξ
∂x
+
∂u
ξ
∂η
∂η
∂x
+
∂u
η
∂ξ
∂ξ
∂x
+
∂u
η
∂η
∂η
∂x
=
u
ξξ
+2
u
ξη
+
u
ηη
.
Similarly
∂
2
u
∂t
2
=
a
2
(
u
ξξ
−
2
u
ξη
+
u
ηη
)
.
Thus
a
2
∂
2
u
∂x
2
=
∂
2
u
∂t
2
becomes
∂
2
u
∂ξ∂η
=0
.
(b)
Integrating
∂
2
u
∂ξ∂η
=
∂
∂η
u
ξ
=0
we obtain

∂
∂η
u
ξ
dη
=

0
dη
u
ξ
=
f
(
ξ
)
.
Integrating this result with respect to
ξ
we obtain

∂u
∂ξ
dξ
=

f
(
ξ
)
dξ
u
=
F
(
ξ
)+
G
(
η
)
.
Since
ξ
=
x
+
at
and
η
=
x
−
at
, we then have
u
=
F
(
ξ
)+
G
(
η
)=
F
(
x
+
at
)+
G
(
x
−
at
)
.
Next, we have
u
(
x, t
)=
F
(
x
+
at
)+
G
(
x
−
at
)
u
(
x,
0) =
F
(
x
)+
G
(
x
)=
f
(
x
)
u
t
(
x,
0) =
aF

(
x
)
−
aG

(
x
)=
g
(
x
)
Integrating the last equation with respect to
x
gives
F
(
x
)
−
G
(
x
)=
1
a

x
x
0
g
(
s
)
ds
+
c
1
.
626

627. Exercises 13.4
Substituting
G
(
x
)=
f
(
x
)
−
F
(
x
) we obtain
F
(
x
)=
1
2
f
(
x
)+
1
2
a

x
x
0
g
(
s
)
ds
+
C
where
c
=
c
1
/
2. Thus
G
(
x
)=
1
2
f
(
x
)
−
1
2
a

x
x
0
g
(
s
)
ds
−
c.
(c)
From the expressions for
F
and
G
,
F
(
x
+
at
)=
1
2
f
(
x
+
at
)+
1
2
a

x
+
at
x
0
g
(
s
)
ds
+
c
G
(
x
−
at
)=
1
2
f
(
x
−
at
)
−
1
2
a

x
−
at
x
0
g
(
s
)
ds
−
c.
Thus,
u
(
x, t
)=
F
(
x
+
at
)+
G
(
x
−
at
)=
1
2
[
f
(
x
+
at
)+
f
(
x
−
at
)] +
1
2
a

x
+
at
x
−
at
g
(
s
)
ds.
Here we have used
−

x
−
at
x
0
g
(
s
)
ds
=

x
0
x
−
at
g
(
s
)
ds
.
15.
u
(
x, t
)=
1
2
[sin(
x
+
at
) + sin(
x
−
at
)] +
1
2
a

x
+
at
x
−
at
ds
=
1
2
[sin
x
cos
at
+ cos
x
sin
at
+ sin
x
cos
at
−
cos
x
sin
at
]+
1
2
a
s



x
+
at
x
−
at
= sin
x
cos
at
+
t
16.
u
(
x, t
)=
1
2
sin(
x
+
at
) + sin(
x
−
at
)] +
1
2
a

x
+
at
x
−
at
cos
sds
= sin
x
cos
at
+
1
2
a
[sin(
x
+
at
)
−
sin(
x
−
at
)] = sin
x
cos
at
+
1
a
cos
x
sin
at
17.
u
(
x, t
)=0+
1
2
a

x
+
at
x
−
at
sin 2
sds
=
1
2
a
3
−
cos(2
x
+2
at
) + cos(2
x
−
2
at
)
2
(
=
1
4
a
[
−
cos 2
x
cos 2
at
+ sin 2
x
sin 2
at
+ cos 2
x
cos 2
at
+ sin 2
x
sin 2
at
]=
1
2
a
sin 2
x
sin 2
at
18.
19. (a)
(b)
Since
g
(
x
) = 0, d’Alembert’s solution with
a
=1is
u
(
x, t
)=
1
2
[
f
(
x
+
t
)+
f
(
x
−
t
)]
.
627

628. -6-4-20 2 4 6
1
t= 0
-6-4-20 2 4 6
1
t= 0.2
-6-4-20 2 4 6
1
t= 0.4
-6-4-20 2 4 6
1
t=
0.6
-6-4-20 2 4 6
1
t=
0.8
-6-4-20 2 4 6
1
t= 1.
-6-4-20 2 4 6
1
t= 2.
-6-4-20 2 4 6
1
t= 3.
-6-4-20 2 4 6
1
t= 4.
-6-4-20 2 4 6
1
t= 5.
-6-4-20 2 4 6
1
t= 0
-6-4-20 2 4 6
1
t= 0.2
-6-4-20 2 4 6
1
t= 0.4
-6-4-20 2 4 6
1
t=
0.6
-6-4-20 2 4 6
1
t=
0.8
-6-4-20 2 4 6
1
t= 1.
Exercises 13.4
Sample plots are shown below.
(c)
The single peaked wave disolves into two peaks moving outward.
20. (a)
With
a
= 1, d’Alembert’s solution is
u
(
x, t
)=
1
2

x
+
t
x
−
t
g
(
s
)
ds
where
g
(
s
)=

1
,
|
s
|≤
0
.
1
0
,
|
s
|
>
0
.
1
.
Sample plots are shown below.
628

629. -6-4-20 2 4 6
1
t= 2.
-6-4-20 2 4 6
1
t= 3.
-6-4-20 2 4 6
1
t= 4.
-6-4-20 2 4 6
1
t= 5.
0.511.522.53
x
-1
-0.5
0.5
1
u
0.511.522.53
x
-1
-0.5
0.5
1
u
0.511.522.53
x
-1
-0.5
0.5
1
u
0.511.522.53
x
-1
-0.5
0.5
1
u
Exercises 13.5
(b)
Some frames of the movie are shown in part (a), The string has a roughly rectangular shape with the base
on the
x
-axis increasingin length.
21. (a)
and
(b)
With the given parameters, the solution is
u
(
x, t
)=
8
π
2
∞

n
=1
1
n
2
sin
nπ
2
cos
nt
sin
nx.
For
n
even, sin(
nπ/
2) = 0, so the first six nonzero terms correspond to
n
= 1, 3, 5, 7, 9, 11. In this case
sin(
nπ/
2) = sin(2
p
−
1)
/
2=(
−
1)
p
+1
for
p
=1,2,3,4,5,6,and
u
(
x, t
)=
8
π
2
∞

p
=1
(
−
1)
p
+1
(2
p
−
1)
2
cos(2
p
−
1)
t
sin(2
p
−
1)
x.
Frames of the movie correspondingto
t
=0
.
5, 1, 1
.
5, and 2 are shown.
Exercises 13.5
1.
Using
u
=
XY
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
a
)=0
,
and
Y

−
λ
2
Y
=0
,
Y
(0) = 0
.
Then
X
=
c
1
sin
nπ
a
x
and
Y
=
c
2
sinh
nπ
a
y
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
sin
nπ
a
x
sinh
nπ
a
y.
629

630. Exercises 13.5
Imposing
u
(
x, b
)=
f
(
x
)=
∞

n
=1
A
n
sinh
nπb
a
sin
nπ
a
x
gives
A
n
sinh
nπb
a
=
2
a

a
0
f
(
x
) sin
nπ
a
xdx
so that
u
(
x, y
)=
∞

n
=1
A
n
sin
nπ
a
x
sinh
nπ
a
y
where
A
n
=
2
a
csch
nπb
a

a
0
f
(
x
) sin
nπ
a
x dx.
2.
Using
u
=
XY
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
a
)=0
,
and
Y

−
λ
2
Y
=0
,
Y

(0) = 0
.
Then
X
=
c
1
sin
nπ
a
x
and
Y
=
c
2
cosh
nπ
a
y
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
sin
nπ
a
x
cosh
nπ
a
y.
Imposing
u
(
x, b
)=
f
(
x
)=
∞

n
=1
A
n
cosh
nπb
a
sin
nπ
a
x
gives
A
n
cosh
nπb
a
=
2
a

a
0
f
(
x
) sin
nπ
a
xdx
so that
u
(
x, y
)=
∞

n
=1
A
n
sin
nπ
a
x
cosh
nπ
a
y
where
A
n
=
2
a
sech
nπb
a

a
0
f
(
x
) sin
nπ
a
x dx.
3.
Using
u
=
XY
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
a
)=0
,
and
630

631. Exercises 13.5
Y

−
λ
2
Y
=0
,
Y
(
b
)=0
.
Then
X
=
c
1
sin
nπ
a
x
and
Y
=
c
2
cosh
nπ
a
y
−
c
2
cosh
nπb
a
sinh
nπb
a
sinh
nπ
a
y
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
sin
nπ
a
x

cosh
nπ
a
y
−
cosh
nπb
a
sinh
nπb
a
sinh
nπ
a
y
o
.
Imposing
u
(
x,
0) =
f
(
x
)=
∞

n
=1
A
n
sin
nπ
a
x
gives
A
n
=
2
a

a
0
f
(
x
) sin
nπ
a
xdx
so that
u
(
x, y
)=
2
a
∞

n
=1
e

a
0
f
(
x
) sin
nπ
a
xdx
n
sin
nπ
a
x

cosh
nπ
a
y
−
cosh
nπb
a
sinh
nπb
a
sinh
nπ
a
y
o
.
4.
Using
u
=
XY
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X

(0) = 0
,
X

(
a
)=0
,
and
Y

−
λ
2
Y
=0
,
Y
(
b
)=0
.
Then
X
=
c
1
cos
nπ
a
x
for
n
=0,1,2,
...
and
Y
=
c
2
(
y
−
b
)or
Y
=
c
2
cosh
nπ
a
y
−
c
2
cosh
nπb
a
sinh
nπb
a
sinh
nπ
a
y
so that
u
=
A
0
(
y
−
b
)+
∞

n
=1
A
n
cos
nπ
a
x

cosh
nπ
a
y
−
cosh
nπb
a
sinh
nπb
a
sinh
nπ
a
y
o
.
Imposing
u
(
x,
0) =
x
=
−
A
0
b
+
∞

n
=1
A
n
cos
nπ
a
x
gives
−
A
0
b
=
1
a

a
0
xdx
=
1
2
a
and
A
n
=
2
a

a
0
x
cos
nπ
a
xdx
=
2
a
n
2
π
2
[(
−
1)
n
−
1]
631

632. Exercises 13.5
so that
u
(
x, y
)=
a
2
b
(
b
−
y
)+
2
a
π
2
∞

n
=1
(
−
1)
n
−
1
n
2
cos
nπ
a
x

cosh
nπ
a
y
−
cosh
nπb
a
sinh
nπb
a
sinh
nπ
a
y
o
.
5.
Using
u
=
XY
and
λ
2
as a separation constant leads to
X

+
−
λ
2
X
=0
,
X
(0) = 0
,
and
Y

+
λ
2
Y
=0
,
Y

(0) = 0
,
Y

(1) = 0
.
Then
Y
=
c
1
cos
nπy
for
n
=0,1,2,
...
and
X
=
c
2
x
or
X
=
c
2
sinh
nπx
for
n
=1,2,3,
...
so that
u
=
A
0
x
+
∞

n
=1
A
n
sinh
nπx
cos
nπy.
Imposing
u
(1
,y
)=1
−
y
=
A
0
+
∞

n
=1
A
n
sinh
nπx
cos
nπy
gives
A
0
=

1
0
(1
−
y
)
dy
and
A
n
sinh
nπ
=2

1
0
(1
−
y
) cos
nπy
=
2[1
−
(
−
1)
n
]
n
2
π
2
sinh
nπ
for
n
=1,2,3,
...
so that
u
(
x, y
)=
1
2
x
+
2
π
2
∞

n
=1
1
−
(
−
1)
n
n
2
sinh
nπ
sinh
nπx
cos
nπy.
6.
Using
u
=
XY
and
λ
2
as a separation constant leads to
X

−
λ
2
X
=0
,
X

(1) = 0
,
and
Y

+
λ
2
Y
=0
,
Y

(0) = 0
,
Y

(
π
)=0
.
Then
Y
=
c
1
cos
ny
632

633. Exercises 13.5
for
n
=0,1,2,
...
and
X
=
c
2
cosh
nx
−
c
2
sinh
n
cosh
n
sinh
nx
for
n
=0,1,2,
...
so that
u
=
A
0
+
∞

n
=1
A
n
e
cosh
nx
−
sinh
n
cosh
n
sinh
nx
n
cos
ny.
Imposing
u
(0
,y
)=
g
(
y
)=
A
0
+
∞

n
=1
A
n
cos
ny
gives
A
0
=
1
π

π
0
g
(
y
)
dy
and
A
n
=
2
π

π
0
g
(
y
) cos
nydy
for
n
=1,2,3,
...
so that
u
(
x, y
)=
1
π

π
0
g
(
y
)
dy
+
∞

n
=1
e
2
π

π
0
g
(
y
) cos
nydy
ne
cosh
nx
−
sinh
n
cosh
n
sinh
nx
n
cos
ny.
7.
Using
u
=
XY
and
λ
2
as a separation constant leads to
X

−
λ
2
X
=0
,
X

(0) =
X
(0)
,
and
Y

+
λ
2
Y
=0
,
Y
(0) = 0
,
Y
(
π
)=0
.
Then
Y
=
c
1
sin
ny
and
X
=
c
2
(
n
cosh
nx
+ sinh
nx
)
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
(
n
cosh
nx
+ sinh
nx
) sin
ny.
Imposing
u
(
π, y
)=1=
∞

n
=1
A
n
(
n
cosh
nπ
+ sinh
nπ
) sin
ny
gives
A
n
(
n
cosh
nπ
+ sinh
nπ
)=
2
π

π
0
sin
nydy
=
2[1
−
(
−
1)
n
]
nπ
for
n
=1,2,3,
...
so that
u
(
x, y
)=
2
π
∞

n
=1
1
−
(
−
1)
n
n
n
cosh
nx
+ sinh
nx
n
cosh
nπ
+ sinh
nπ
sin
ny.
8.
Using
u
=
XY
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(1) = 0
and
633

634. Exercises 13.5
Y

−
λ
2
Y
=0
,
Y

(0) =
Y
(0)
.
Then
X
=
c
1
sin
nπx
and
Y
=
c
2
(
n
cosh
nπy
+ sinh
nπy
)
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
(
n
cosh
nπy
+ sinh
nπy
) sin
nπx.
Imposing
u
(
x,
1) =
f
(
x
)=
∞

n
=1
A
n
(
n
cosh
nπ
+ sinh
nπ
) sin
nπx
gives
A
n
(
n
cosh
nπ
+ sinh
nπ
)=
2
π

π
0
f
(
x
) sin
nπx dx
for
n
=1,2,3,
...
so that
u
(
x, y
)=
∞

n
=1
A
n
(
n
cosh
nπy
+ sinh
nπy
) sin
nπx
where
A
n
=
2
nπ
cosh
nπ
+
π
sinh
nπ

1
0
f
(
x
) sin
nπx dx.
9.
This boundary-value problem has the form of Problem 1 in this section, with
a
=
b
=1,
f
(
x
) = 100, and
g
(
x
) = 200. The solution, then, is
u
(
x, y
)=
∞

n
=1
(
A
n
cosh
nπy
+
B
n
sinh
nπy
) sin
nπx,
where
A
n
=2

1
0
100 sin
nπx dx
= 200
e
1
−
(
−
1)
n
nπ
n
and
B
n
=
1
sinh
nπ
3
2

1
0
200 sin
nπx dx
−
A
n
cosh
nπ
(
=
1
sinh
nπ
3
400
e
1
−
(
−
1)
n
nπ
n
−
200
e
1
−
(
−
1)
n
nπ
n
cosh
nπ
(
= 200
3
1
−
(
−
1)
n
nπ
(
[2 csch
nπ
−
coth
nπ
]
.
10.
This boundary-value problem has the form of Problem 2 in this section, with
a
= 1 and
b
= 1. Thus, the
solution has the form
u
(
x, y
)=
∞

n
=1
(
A
n
cosh
nπx
+
B
n
sinh
nπx
) sin
nπy.
The boundary condition
u
(0
,y
)=10
y
implies
10
y
=
∞

n
=1
A
n
sin
nπy
634

635. Exercises 13.5
and
A
n
=
2
1

1
0
10
y
sin
nπy dy
=
20
nπ
(
−
1)
n
+1
.
The boundary condition
u
x
(1
,y
)=
−
1 implies
−
1=
∞

n
=1
(
nπA
n
sinh
nπ
+
nπB
n
cosh
nπ
) sin
nπy
and
nπA
n
sinh
nπ
+
nπB
n
cosh
nπ
=
2
1

1
0
(
−
sin
nπy
)
dy
A
n
sinh
nπ
+
B
n
cos
nπ
=
−
2
nπ
w
1
−
(
−
1)
n
i
B
n
=
2
nπ
w
(
−
1)
n
−
1
i
sech
nπ
−
20
nπ
(
−
1)
n
+1
tanh
nπ.
11.
Using
u
=
XY
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(
π
)=0
,
and
Y

−
λ
2
Y
=0
.
Then the boundedness of
u
as
y
→∞
gives
Y
=
c
1
e
−
ny
and
X
=
c
2
sin
nx
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
e
−
ny
sin
nx.
Imposing
u
(
x,
0) =
f
(
x
)=
∞

n
=1
A
n
sin
nx
gives
A
n
=
2
π

π
0
f
(
x
) sin
nxdx
so that
u
(
x, y
)=
∞

n
=1
e
2
π

π
0
f
(
x
) sin
nxdx
n
e
−
ny
sin
nx.
12.
Using
u
=
XY
and
−
λ
2
as a separation constant leads to
X

+
λ
2
X
=0
,
X

(0) = 0
,
X

(
π
)=0
,
and
Y

−
λ
2
Y
=0
.
By the boundedness of
u
as
y
→∞
we obtain
Y
=
c
1
e
−
ny
for
n
=1,2,3,
...
or
Y
=
c
1
and
X
=
c
2
cos
nx
for
n
=0,1,2,
...
so that
u
=
A
0
+
∞

n
=1
A
n
e
−
ny
cos
nx.
635

636. Exercises 13.5
Imposing
u
(
x,
0) =
f
(
x
)=
A
0
+
∞

n
=1
A
n
cos
nx
gives
A
0
=
1
π

π
0
f
(
x
)
dx
and
A
n
=
2
π

π
0
f
(
x
) cos
nxdx
so that
u
(
x, y
)=
1
π

π
0
f
(
x
)
dx
+
∞

n
=1
e
2
π

π
0
f
(
x
) cos
nxdx
n
e
−
ny
cos
nx.
13.
Since the boundary conditions at
y
= 0 and
y
=
b
are functions of
x
we choose to separate Laplace’s equation
as
X

X
=
−
Y

Y
=
−
λ
2
so that
X

+
λ
2
X
=0
Y

−
λ
2
Y
=0
and
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
Y
(
y
)=
c
3
cosh
λy
+
c
2
sinh
λy.
Now
X
(0) = 0 gives
c
1
= 0 and
X
(
a
) = 0 implies sin
λa
=0or
λ
=
nπ/a
for
n
=1,2,3,
...
.Thus
u
n
(
x, y
)=
XY
=
=
A
n
cosh
nπ
a
y
+
B
n
sinh
nπ
a
y
t
sin
nπ
a
x
and
u
(
x, y
)=
∞

n
=1
=
A
n
cosh
nπ
a
y
+
B
n
sinh
nπ
a
y
t
sin
nπ
a
x.
(
1
)
At
y
= 0 we then have
f
(
x
)=
∞

n
=1
A
n
sin
nπ
a
x
and consequently
A
n
=
2
a

a
0
f
(
x
) sin
nπ
a
x dx.
(
2
)
At
y
=
b
,
g
(
y
)=
∞

n
=1
=
A
n
cosh
nπ
a
b
+
B
n
sinh
nπ
b
a
t
sin
nπ
a
x
indicates that the entire expression in the parentheses is given by
A
n
cosh
nπ
a
b
+
B
n
sinh
nπ
a
b
=
2
a

a
0
g
(
x
) sin
nπ
a
x dx.
We can now solve for
B
n
:
B
n
sinh
nπ
a
b
=
2
a

a
0
g
(
x
) sin
nπ
a
xdx
−
A
n
cosh
nπ
a
b
B
n
=
1
sinh
nπ
a
b
e
2
a

a
0
g
(
x
) sin
nπ
a
xdx
−
A
n
cosh
nπ
a
b
n
.
(
3
)
A solution to the given boundary-value problem consists of the series (1) with coefficients
A
n
and
B
n
given in
(2) and (3), respectively.
636

637. Exercises 13.5
14.
Since the boundary conditions at
x
= 0 and
x
=
a
are functions of
y
we choose to separate Laplace’s equation
as
X

X
=
−
Y

Y
=
λ
2
so that
X

−
λ
2
X
=0
Y

+
λ
2
Y
=0
and
X
(
x
)=
c
1
cosh
λx
+
c
2
sinh
λx
Y
(
y
)=
c
3
cos
λy
+
c
2
sin
λy.
Now
Y
(0) = 0 gives
c
3
= 0 and
Y
(
b
) = 0 implies sin
λb
=0or
λ
=
nπ/b
for
n
=1,2,3,
...
.Thus
u
n
(
x, y
)=
XY
=
=
A
n
cosh
nπ
b
x
+
B
n
sinh
nπ
b
x
t
sin
nπ
b
y
and
u
(
x, y
)=
∞

n
=1
=
A
n
cosh
nπ
b
x
+
B
n
sinh
nπ
b
x
t
sin
nπ
b
y.
(
4
)
At
x
= 0 we then have
F
(
y
)=
∞

n
=1
A
n
sin
nπ
b
y
and consequently
A
n
=
2
b

b
0
F
(
y
) sin
nπ
b
ydy.
(
5
)
At
x
=
a
,
G
(
y
)=
∞

n
=1
=
A
n
cosh
nπ
b
a
+
B
n
sinh
nπ
b
a
t
sin
nπ
b
y
indicates that the entire expression in the parentheses is given by
A
n
cosh
nπ
b
a
+
B
n
sinh
nπ
b
a
=
2
b

b
0
G
(
y
) sin
nπ
b
ydy.
We can now solve for
B
n
:
B
n
sinh
nπ
b
a
=
2
b

b
0
G
(
y
) sin
nπ
b
ydy
−
A
n
cosh
nπ
b
a
B
n
=
1
sinh
nπ
b
a

2
b

b
0
G
(
y
) sin
nπ
b
ydy
−
A
n
cosh
nπ
b
a
o
.
(
6
)
A solution to the given boundary-value problem consists of the series (4) with coefficients
A
n
and
B
n
given in
(5) and (6), respectively.
637

638. Exercises 13.5
In Problems 15 and 16 we refer to the discussion in the text under the heading
Superposition Principle
.
15.
We identify
a
=
b
=
π
,
f
(
x
)=0,
g
(
x
)=1,
F
(
y
) = 1, and
G
(
y
) = 1. Then
A
n
= 0 and
u
1
(
x, y
)=
∞

n
=1
B
n
sinh
ny
sin
nx
where
B
n
=
2
π
sinh
nπ

π
0
sin
nxdx
=
2[1
−
(
−
1)
n
]
nπ
sinh
nπ
.
Next
u
2
(
x, y
)=
∞

n
=1
(
A
n
cosh
nx
+
B
n
sinh
nx
) sin
ny
where
A
n
=
2
π

π
0
sin
nydy
=
2[1
−
(
−
1)
n
]
nπ
and
B
n
=
1
sinh
nπ
e
2
π

π
0
sin
nydy
−
A
n
cosh
nπ
n
=
1
sinh
nπ
e
2[1
−
(
−
1)
n
]
nπ
−
2[1
−
(
−
1)
n
]
nπ
cosh
nπ
n
=
2[1
−
(
−
1)
n
]
nπ
sinh
nπ
(1
−
cosh
nπ
)
.
Now
A
n
cosh
nx
+
B
n
sinh
nx
=
2[1
−
(
−
1)
n
]
nπ
3
cosh
nx
+
sinh
nx
sinh
nπ
(1
−
cosh
nπ
)
(
=
2[1
−
(
−
1)
n
]
nπ
sinh
nπ
[cosh
nx
sinh
nπ
+ sinh
nx
−
sinh
nx
cosh
nπ
]
=
2[1
−
(
−
1)
n
]
nπ
sinh
nπ
[sinh
nx
+ sinh
n
(
π
−
x
)]
and
u
(
x, y
)=
u
1
+
u
2
=
2
π
∞

n
=1
1
−
(
−
1)
n
n
sinh
nπ
sinh
ny
sin
nx
+
2
π
∞

n
=1
[1
−
(
−
1)
n
][sinh
nx
+ sinh
n
(
π
−
x
)]
n
sinh
nπ
sin
ny.
16.
We identify
a
=
b
=2,
f
(
x
)=0,
g
(
x
)=

x,
0
<x<
1
2
−
x,
1
<x<
2
,
F
(
y
)=0,and
G
(
y
)=
y
(2
−
y
). Then
A
n
=0
and
u
1
(
x, y
)=
∞

n
=1
B
n
sinh
nπ
2
y
sin
nπ
2
x
where
B
n
=
1
sinh
nπ

2
0
g
(
x
) sin
nπ
2
xdx
=
1
sinh
nπ
e

1
0
x
sin
nπ
2
xdx
+

2
1
(2
−
x
) sin
nπ
2
xdx
n
=
8 sin
nπ
2
n
2
π
2
sinh
nπ
.
638

639. 0.511.522.53
x
0.5
1
1.5
2
2.5
3
y
u=0
u=0 u=0
u=0
u=0
u=0
u=1
u=1
u=x u=2-x
u=yH2-yL
u=yH2-yL
0
1
2
3
x
0
1
2
3
y
0
100
200
uHx,yL
0
1
2
3
x
Exercises 13.5
Next, since
A
n
=0in
u
2
, we have
u
2
(
x, y
)=
∞

n
=1
B
n
sinh
nπ
2
x
sin
nπ
2
where
B
n
=
1
sinh
nπ

b
0
y
(2
−
y
) sin
nπ
2
ydy
=
16[1
−
(
−
1)
n
]
n
3
π
3
sinh
nπ
.
Thus
u
(
x, y
)=
u
1
+
u
2
=
8
π
2
∞

n
=1
sin
nπ
2
n
2
sinh
nπ
sinh
nπ
2
y
sin
nπ
2
x
+
16
π
3
∞

n
=1
[1
−
(
−
1)
n
]
n
3
sinh
nπ
sinh
nπ
2
x
sin
nπ
2
y.
17.
From the figure showing the boundary conditions we see that the maxi-
mum value of the temperature is 1 at (1
,
2) and (2
,
1).
18. (a)
(b)
The maximum value occurs at (
π/
2
,π
) and is
f
(
π/
2) = 25
π
2
.
(c)
The coefficients are
A
n
=
2
π
csch
nπ

π
0
100
x
(
π
−
x
) sin
nxdx
=
200 csch
nπ
π
3
200
n
3
=
1
−
(
−
1)
n
t
(
=
400
n
3
π
1
1
−
(
−
1)
n
+
csch
nπ.
639

640. 0 0.20.40.60.8 1
0
0.2
0.4
0.6
0.8
1
170
140
110
80
80
60
60
30
30
0
0.2
0.4
0.6
0.8
1
x
0
0.2
0.4
0.6
0.8
1
y
0
50
100
150
200
uHx,yL
0
0.2
0.4
0.6
0.8
x
0 0.20.40.60.8 1
0
0.2
0.4
0.6
0.8
1
2
1
0.5
0.2
0
-0.05
Exercises 13.5
19. (a)
(b)
20.
Exercises 13.6
1.
Using
v
(
x, t
)=
u
(
x, t
)
−
100 we wish to solve
kv
xx
=
v
t
subject to
v
(0
,t
)=0,
v
(1
,t
)=0,and
v
(
x,
0) =
−
100.
Let
v
=
XT
and use
−
λ
2
as a separation constant so that
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(1) = 0
,
and
T

+
λ
2
kT
=0
.
640

641. Exercises 13.6
Then
X
=
c
1
sin
nπx
and
T
=
c
2
e
−
kn
2
π
2
t
for
n
=1,2,3,
...
so that
v
=
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx.
Imposing
v
(
x,
0) =
−
100 =
∞

n
=1
A
n
sin
nπx
gives
A
n
=2

1
0
(
−
100) sin
nπx dx
=
−
200
nπ
[1
−
(
−
1)
n
]
so that
u
(
x, t
)=
v
(
x, t
) + 100 = 100 +
200
π
∞

n
=1
(
−
1)
n
−
1
n
e
−
kn
2
π
2
t
sin
nπx.
2.
Letting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) and proceedingas in Example 1 in the text we find
ψ
(
x
)=
u
0
−
u
0
x
. Then
v
(
x, t
)=
u
(
x, t
)+
u
0
x
−
u
0
and we wish to solve
kv
xx
=
v
t
subject to
v
(0
,t
)=0,
v
(1
,t
) = 0, and
v
(
x,
0) =
f
(
x
)+
u
0
x
−
u
0
. Let
v
=
XT
and use
−
λ
2
as a separation constant so that
X

+
λ
2
X
=0
,
X
(0) = 0
,
X
(1) = 0
,
and
T

+
λ
2
kT
=0
.
Then
X
=
c
1
sin
nπx
and
T
=
c
2
e
−
kn
2
π
2
t
for
n
=1,2,3,
...
so that
v
=
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx.
Imposing
v
(
x,
0) =
f
(
x
)+
u
0
x
−
u
0
=
∞

n
=1
A
n
sin
nπx
gives
A
n
=2

1
0
I
f
(
x
)+
u
0
x
−
u
0
f
sin
nπx dx
so that
u
(
x, t
)=
v
(
x, t
)+
u
0
−
u
0
x
=
u
0
−
u
0
x
+
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx.
3.
If we let
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
), then we obtain as in Example 1 in the text
kψ

+
r
=0
or
ψ
(
x
)=
−
r
2
k
x
2
+
c
1
x
+
c
2
.
641

642. Exercises 13.6
The boundary conditions become
u
(0
,t
)=
v
(0
,t
)+
ψ
(0) =
u
0
u
(1
,t
)=
v
(1
,t
)+
ψ
(1) =
u
0
.
Letting
ψ
(0) =
ψ
(1) =
u
0
we obtain homogeneous boundary conditions in
v
:
v
(0
,t
) = 0 and
v
(1
,t
)=0
.
Now
ψ
(0) =
ψ
(1) =
u
0
implies
c
2
=
u
0
and
c
1
=
r/
2
k
.Thus
ψ
(
x
)=
−
r
2
k
x
2
+
r
2
k
x
+
u
0
=
u
0
−
r
2
k
x
(
x
−
1)
.
To determine
v
(
x, t
) we solve
k
∂
2
v
∂x
2
=
∂v
dt
,
0
<x<
1
,t>
0
v
(0
,t
)=0
,v
(1
,t
)=0
,
v
(
x,
0) =
r
2
k
x
(
x
−
1)
−
u
0
.
Separatingvariables, we find
v
(
x, t
)=
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx,
where
A
n
=2

1
0
1
r
2
k
x
(
x
−
1)
−
u
0
+
sin
nπx dx
=2
1
u
0
nπ
+
r
kn
3
π
3
+
[(
−
1)
n
−
1]
.
(
1
)
Hence, a solution of the original problem is
u
(
x, t
)=
ψ
(
x
)+
v
(
x, t
)
=
u
0
−
r
2
k
x
(
x
−
1) +
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx,
where
A
n
is defined in (1).
4.
If we let
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
), then we obtain as in Example 1 in the text
kψ

+
r
=0
or
ψ
(
x
)=
−
r
2
k
x
2
+
c
1
x
+
c
2
.
The boundary conditions become
u
(0
,t
)=
v
(0
,t
)+
ψ
(0) =
u
0
u
(1
,t
)=
v
(1
,t
)+
ψ
(1) =
u
1
.
Letting
ψ
(0) =
u
0
and
ψ
(1) =
u
1
we obtain homogeneous boundary conditions in
v
:
v
(0
,t
) = 0 and
v
(1
,t
)=0
.
Now
ψ
(0) =
u
0
and
ψ
(1) =
u
1
imply
c
2
=
u
0
and
c
1
=
u
1
−
u
0
+
r/
2
k
.Thus
ψ
(
x
)=
−
r
2
k
x
2
+
=
u
1
−
u
0
+
r
2
k
t
x
+
u
0
.
642

643. Exercises 13.6
To determine
v
(
x, t
) we solve
k
∂
2
v
∂x
2
=
∂v
dt
,
0
<x<
1
,t>
0
v
(0
,t
)=0
,v
(1
,t
)=0
,
v
(
x,
0) =
f
(
x
)
−
ψ
(
x
)
.
Separatingvariables, we find
v
(
x, t
)=
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx,
where
A
n
=2

1
0
[
f
(
x
)
−
ψ
(
x
)] sin
nπx dx.
(
2
)
Hence, a solution of the original problem is
u
(
x, t
)=
ψ
(
x
)+
v
(
x, t
)
=
−
r
2
k
x
2
+
=
u
1
−
u
0
+
r
2
k
t
x
+
u
0
+
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx,
where
A
n
is defined in (2).
5.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation gives
k
∂
2
v
∂x
2
+
kψ

+
Ae
−
βx
=
∂v
∂t
.
This equation will be homogeneous provided
ψ
satisfies
kψ

+
Ae
−
βx
=0
.
The general solution of this differential equation is
ψ
(
x
)=
−
A
β
2
k
e
−
βx
+
c
1
x
+
c
2
.
From
ψ
(0) = 0 and
ψ
(1) = 0 we find
c
1
=
A
β
2
k
(
e
−
β
−
1) and
c
2
=
A
β
2
k
.
Hence
ψ
(
x
)=
−
A
β
2
k
e
−
βx
+
A
β
2
k
(
e
−
β
−
1)
x
+
A
β
2
k
=
A
β
2
k
w
1
−
e
−
βx
+(
e
−
β
−
1)
x
i
.
Now the new problem is
k
∂
2
v
∂x
2
=
∂v
∂t
,
0
<x<
1
,t>
0
,
v
(0
,t
)=0
,v
(1
,t
)=0
,t>
0
,
v
(
x,
0) =
f
(
x
)
−
ψ
(
x
)
,
0
<x<
1
.
Identifyingthis as the heat equation solved in Section 13
.
3 in the text with
L
= 1 we obtain
v
(
x, t
)=
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx
643

644. Exercises 13.6
where
A
n
=2

1
0
[
f
(
x
)
−
ψ
(
x
)] sin
nπx dx.
Thus
u
(
x, t
)=
A
β
2
k
w
1
−
e
−
βx
+(
e
−
β
−
1)
x
i
+
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx.
6.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation gives
k
∂
2
v
∂x
2
+
kψ

−
hv
−
hψ
=
∂v
∂t
.
This equation will be homogeneous provided
ψ
satisfies
kψ

−
hψ
=0
.
Since
k
and
h
are positive, the general solution of this latter equation is
ψ
(
x
)=
c
1
cosh
c
h
k
x
+
c
2
sinh
c
h
k
x.
From
ψ
(0) = 0 and
ψ
(
π
)=
u
0
we find
c
1
= 0 and
c
2
=
u
0
/
sinh
h
h/k π
. Hence
ψ
(
x
)=
u
0
sinh
h
h/k x
sinh
h
h/k π
.
Now the new problem is
k
∂
2
v
∂x
2
−
hv
=
∂v
∂t
,
0
<x<π, t>
0
v
(0
,t
)=0
,v
(
π, t
)=0
,t>
0
v
(
x,
0) =
−
ψ
(
x
)
,
0
<x<π.
If we let
v
=
XT
then
X

X
=
T

+
hT
kT
=
−
λ
2
gives the separated differential equations
X

+
λ
2
X
= 0 and
T

+
I
h
+
kλ
2
f
T
=0
.
The respective solutions are
X
(
x
)=
c
3
cos
λx
+
c
4
sin
λx
T
(
t
)=
c
5
e
−
(
h
+
kλ
2
)
t
.
From
X
(0) = 0 we get
c
3
= 0 and from
X
(
π
)=0wefind
λ
=
n
for
n
=1,2,3,
...
. Consequently, it follows
that
v
(
x, t
)=
∞

n
=1
A
n
e
−
(
h
+
kn
2
)
t
sin
nx
where
A
n
=
−
2
π

π
0
ψ
(
x
) sin
nx dx.
Hence a solution of the original problem is
u
(
x, t
)=
u
0
sinh
h
h/k x
sinh
h
h/k π
+
e
−
ht
∞

n
=1
A
n
e
−
kn
2
t
sin
nx
644

645. Exercises 13.6
where
A
n
=
−
2
π

π
0
u
0
sinh
h
h/k x
sinh
h
h/k π
sin
nx dx.
Usingthe exponential definition of the hyperbolic sine and integration by parts we find
A
n
=
2
u
0
nk
(
−
1)
n
π
(
h
+
kn
2
)
.
7.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation gives
k
∂
2
v
∂x
2
+
kψ

−
hv
−
hψ
+
hu
0
=
∂v
∂t
.
This equation will be homogeneous provided
ψ
satisfies
kψ

−
hψ
+
hu
0
=0 or
kψ

−
hψ
=
−
hu
0
.
This second-order, linear, non-homogeneous differential equation has solution
ψ
(
x
)=
c
1
cosh
c
h
k
x
+
c
2
sinh
c
h
k
x
+
u
0
,
where we assume
h>
0 and
k>
0. From
ψ
(0) =
u
0
and
ψ
(1) = 0 we find
c
1
= 0 and
c
2
=
−
u
0
/
sinh
h
h/k
. Thus, the steady-state solution is
ψ
(
x
)=
−
u
0
sinh
p
h
k
sinh
c
h
k
x
+
u
0
=
u
0


1
−
sinh
p
h
k
x
sinh
p
h
k


.
8.
The partial differential equation is
k
∂
2
u
∂x
2
−
hu
=
∂u
∂t
.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) gives
k
∂
2
v
∂x
2
+
kψ

−
hv
−
hψ
=
∂v
∂t
.
This equation will be homogeneous provided
ψ
satisfies
kψ

−
hψ
=0
.
Assuming
h>
0 and
k>
0, we have
ψ
=
c
1
e
√
h/k x
+
c
2
e
−
√
h/k x
,
where we have used the exponential form of the solution since the rod is infinite. Now, in order that the
steady-state temperature
ψ
(
x
) be bounded as
x
→∞
, we require
c
1
= 0. Then
ψ
(
x
)=
c
2
e
−
√
h/k x
and
ψ
(0) =
u
0
implies
c
2
=
u
0
.Thus
ψ
(
x
)=
u
0
e
−
√
h/k x
.
9.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation gives
a
2
∂
2
v
∂x
2
+
a
2
ψ

+
Ax
=
∂
2
v
∂t
2
.
This equation will be homogeneous provided
ψ
satisfies
a
2
ψ

+
Ax
=0
.
645

646. Exercises 13.6
The general solution of this differential equation is
ψ
(
x
)=
−
A
6
a
2
x
3
+
c
1
x
+
c
2
.
From
ψ
(0) = 0 we obtain
c
2
= 0, and from
ψ
(1) = 0 we obtain
c
1
=
A/
6
a
2
. Hence
ψ
(
x
)=
A
6
a
2
(
x
−
x
3
)
.
Now the new problem is
a
2
∂
2
v
∂x
2
=
∂
2
v
∂t
2
v
(0
,t
)=0
,v
(1
,t
)=0
,t>
0
,
v
(
x,
0) =
−
ψ
(
x
)
,v
t
(
x,
0) = 0
,
0
<x<
1
.
Identifyingthis as the
wave
equation solved in Section 13
.
4 in the text with
L
=1,
f
(
x
)=
−
ψ
(
x
), and
g
(
x
)=0
we obtain
v
(
x, t
)=
∞

n
=1
A
n
cos
nπat
sin
nπx
where
A
n
=2

1
0
[
−
ψ
(
x
)] sin
nπx dx
=
A
3
a
2

1
0
(
x
3
−
x
) sin
nπx dx
=
2
A
(
−
1)
n
a
2
π
3
n
3
.
Thus
u
(
x, t
)=
A
6
a
2
(
x
−
x
3
)+
2
A
a
2
π
3
∞

n
=1
(
−
1)
n
n
3
cos
nπat
sin
nπx.
10.
We solve
a
2
∂
2
u
∂x
2
−
g
=
∂
2
u
∂t
2
,
0
<x<
1
,t>
0
u
(0
,t
)=0
,u
(1
,t
)=0
,t>
0
u
(
x,
0) = 0
,
∂u
∂t




t
=0
=0
,
0
<x<
1
.
The partial differential equation is nonhomogeneous. The substitution
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) yields a homoge-
neous partial differential equation provided
ψ
satisfies
a
2
ψ

−
g
=0
.
By integrating twice we find
ψ
(
x
)=
g
2
a
2
x
2
+
c
1
x
+
c
2
.
The imposed conditions
ψ
(0) = 0 and
ψ
(1) = 0 then lead to
c
2
= 0 and
c
1
=
−
g/
2
a
2
. Hence
ψ
(
x
)=
g
2
a
2
I
x
2
−
x
f
.
The new problem is now
a
2
∂
2
v
∂x
2
=
∂
2
v
∂t
2
,
0
<x<
1
,t>
0
v
(0
,t
)=0
,v
(1
,t
)=0
v
(
x,
0) =
g
2
a
2
I
x
−
x
2
f
,
∂v
∂t




t
=0
=0
.
646

647. Exercises 13.6
Substituting
v
=
XT
we find in the usual manner
X

+
λ
2
X
=0
T

+
a
2
λ
2
T
=0
with solutions
X
(
x
)=
c
3
cos
λx
+
c
4
sin
λx
T
(
t
)=
c
5
cos
aλt
+
c
6
sin
aλt.
The conditions
X
(0) = 0 and
X
(1) = 0 imply in turn that
c
3
= 0 and
λ
=
nπ
for
n
=1,2,3,
...
. The condition
T

(0) = 0 implies
c
6
= 0. Hence, by the superposition principle
v
(
x, t
)=
∞

n
=1
A
n
cos(
anπt
) sin(
nπx
)
.
At
t
=0,
g
2
a
2
I
x
−
x
2
f
=
∞

n
=1
A
n
sin
nπx
and so
A
n
=
g
a
2

1
0
I
x
−
x
2
f
sin
nπx dx
=
2
g
a
2
n
3
π
3
[1
−
(
−
1)
n
]
.
Thus the solution to the original problem is
u
(
x, t
)=
ψ
(
x
)+
v
(
x, t
)=
g
2
a
2
I
x
2
−
x
f
+
2
g
a
2
π
3
∞

n
=1
1
−
(
−
1)
n
n
3
cos(
anπt
) sin(
nπx
)
.
11.
Substituting
u
(
x, y
)=
v
(
x, y
)+
ψ
(
y
) into Laplace’s equation we obtain
∂
2
v
∂x
2
+
∂
2
v
∂y
2
+
ψ

(
y
)=0
.
This equation will be homogeneous provided
ψ
satisfies
ψ
(
y
)=
c
1
y
+
c
2
. Considering
u
(
x,
0) =
v
(
x,
0) +
ψ
(0) =
u
1
u
(
x,
1) =
v
(
x,
1) +
ψ
(1) =
u
0
u
(0
,y
)=
v
(0
,y
)+
ψ
(
y
)=0
we require that
ψ
(0) =
u
1
,
ψ
1
=
u
0
and
v
(0
,y
)=
−
ψ
(
y
). Then
c
1
=
u
0
−
u
1
and
c
2
=
u
1
. The new
boundary-value problem is
∂
2
v
∂x
2
+
∂
2
v
∂y
2
=0
v
(
x,
0) = 0
,v
(
x,
1)=0
,
v
(0
,y
)=
−
ψ
(
y
)
,
0
<y<
1
,
where
v
(
x, y
) is bounded at
x
→∞
. This problem is similar to Problem 11 in Section 13
.
5. The solution is
v
(
x, y
)=
∞

n
=1
e
2

1
0
[
−
ψ
(
y
) sin
nπy
]
dy
n
e
−
nπx
sin
nπy
=2
∞

n
=1
3
(
u
1
−
u
0
)

1
0
y
sin
nπy dy
−
u
1

1
0
sin
nπy dy
(
e
−
nπx
sin
nπy
=
2
π
∞

n
=1
u
0
(
−
1)
n
−
u
1
n
e
−
nπx
sin
nπy.
647

648. Exercises 13.6
Thus
u
(
x, y
)=
v
(
x, y
)+
ψ
(
y
)
=(
u
0
−
u
1
)
y
+
u
1
+
2
π
∞

n
=1
u
0
(
−
1)
n
−
u
1
n
e
−
nπx
sin
nπy.
12.
Substituting
u
(
x, y
)=
v
(
x, y
)+
ψ
(
x
) into Poisson’s equation we obtain
∂
2
v
∂x
2
+
ψ

(
x
)+
h
+
∂
2
v
∂y
2
=0
.
The equation will be homogeneous provided
ψ
satisfies
ψ

(
x
)+
h
=0or
ψ
(
x
)=
−
h
2
x
2
+
c
1
x
+
c
2
. From
ψ
(0) = 0
we obtain
c
2
= 0. From
ψ
(
π
) = 1 we obtain
c
1
=
1
π
+
hπ
2
.
Then
ψ
(
x
)=
e
1
π
+
hπ
2
n
x
−
h
2
x
2
.
The new boundary-value problem is
∂
2
v
∂x
2
+
∂
2
v
∂y
2
=0
v
(0
,y
)=0
,v
(
π, y
)=0
,
v
(
x,
0) =
−
ψ
(
x
)
,
0
<x<π.
This is Problem 11 in Section 13
.
5. The solution is
v
(
x, y
)=
∞

n
=1
A
n
e
−
ny
sin
nx
where
A
n
=
2
π

π
0
[
−
ψ
(
x
) sin
nx
]
dx
=
2(
−
1)
n
m
e
1
π
+
hπ
2
n
−
h
(
−
1)
n
e
π
n
+
2
n
2
n
.
Thus
u
(
x, y
)=
v
(
x, y
)+
ψ
(
x
)=
e
1
π
+
hπ
2
n
x
−
h
2
x
2
+
∞

n
=1
A
n
e
−
ny
sin
nx.
Exercises 13.7
1.
Referringto Example 1 in the text we have
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
and
T
(
t
)=
c
3
e
−
kλ
2
t
.
From
X

(0) = 0 (since the left end of the rod is insulated), we find
c
2
= 0. Then
X
(
x
)=
c
1
cos
λx
and the
other boundary condition
X

(1) =
−
hX
(1) implies
−
λ
sin
λ
+
h
cos
λ
= 0 or cot
λ
=
λ
h
.
648

649. Exercises 13.7
Denotingthe consecutive positive roots of this latter equation by
λ
n
for
n
=1,2,3,
...
,wehave
u
(
x, t
)=
∞

n
=1
A
n
e
−
kλ
2
n
t
cos
λ
n
x.
From the initial condition
u
(
x,
0) = 1 we obtain
1=
∞

n
=1
A
n
cos
λ
n
x
and
A
n
=
F
1
0
cos
λ
n
xdx
F
1
0
cos
2
λ
n
xdx
=
sin
λ
n
/λ
n
1
2
1
1+
1
2
λ
n
sin 2
λ
n
+
=
2 sin
λ
n
λ
n
1
1+
1
λ
n
sin
λ
n
cos
λ
n
+
=
2 sin
λ
n
λ
n
1
1+
1
hλ
n
sin
λ
n
(
λ
n
sin
λ
n
)
+
=
2
h
sin
λ
n
λ
n
[
h
+ sin
2
λ
n
]
.
The solution is
u
(
x, t
)=2
h
∞

n
=1
sin
λ
n
λ
n
(
h
+ sin
2
λ
n
)
e
−
kλ
2
n
t
cos
λ
n
x.
2.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation gives
k
∂
2
v
∂x
2
+
kψ
kk
=
∂v
∂t
.
This equation will be homogeneous if
ψ
kk
(
x
)=0or
ψ
(
x
)=
c
1
x
+
c
2
. The boundary condition
u
(0
,t
) = 0 implies
ψ
(0) = 0 which implies
c
2
=0. Thus
ψ
(
x
)=
c
1
x
. Usingthe second boundary condition we obtain
−
e
∂v
∂x
+
ψ
k
n




x
=1
=
−
h
[
v
(1
,t
)+
ψ
(1)
−
u
0
]
,
which will be homogeneous when
−
ψ
k
(1) =
−
hψ
(1) +
hu
0
.
Since
ψ
(1) =
ψ
k
(1) =
c
1
we have
−
c
1
=
−
hc
1
+
hu
0
and
c
1
=
hu
0
/
(
h
−
1). Thus
ψ
(
x
)=
hu
0
h
−
1
x.
The new boundary-value problem is
k
∂
2
v
∂x
2
=
∂v
∂t
,
0
<x<
1
,t>
0
v
(0
,t
)=0
,
∂v
∂x




x
=1
=
−
hv
(1
,t
)
,h>
0
,t>
0
v
(
x,
0) =
f
(
x
)
−
hu
0
h
−
1
x,
0
<x<
1
.
Referringto Example 1 in the text we see that
v
(
x, t
)=
∞

n
=1
A
n
e
−
kλ
2
n
t
sin
λ
n
x
649

650. Exercises 13.7
and
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
)=
hu
0
h
−
1
x
+
∞

n
=1
A
n
e
−
kλ
2
n
t
sin
λ
n
x
where
f
(
x
)
−
hu
0
h
−
1
x
=
∞

n
=1
A
n
sin
λ
n
x
and
λ
n
is a solution of
λ
n
cos
λ
n
=
−
h
sin
λ
n
. The coefficients are
A
n
=
F
1
0
[
f
(
x
)
−
hu
0
x/
(
h
−
1)] sin
λ
n
xdx
F
1
0
sin
2
λ
n
xdx
=
F
1
0
[
f
(
x
)
−
hu
0
x/
(
h
−
1)] sin
λ
n
xdx
1
2
1
1
−
1
2
λ
n
sin 2
λ
n
+
=
2
F
1
0
[
f
(
x
)
−
hu
0
x/
(
h
−
1)] sin
λ
n
xdx
1
−
1
λ
n
sin
λ
n
cos
λ
n
=
2
F
1
0
[
f
(
x
)
−
hu
0
x/
(
h
−
1)] sin
λ
n
xdx
1
−
1
hλ
n
(
h
sin
λ
n
) cos
λ
n
=
2
F
1
0
[
f
(
x
)
−
hu
0
x/
(
h
−
1)] sin
λ
n
xdx
1
−
1
hλ
n
(
−
λ
n
cos
λ
n
) cos
λ
n
=
2
h
h
+ cos
2
λ
n

1
0
3
f
(
x
)
−
hu
0
h
−
1
x
(
sin
λ
n
x dx.
3.
Separatingvariables in Laplace’s equation gives
X
kk
+
λ
2
X
=0
Y
kk
−
λ
2
Y
=0
and
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
Y
(
y
)=
c
3
cosh
λy
+
c
4
sinh
λy.
From
u
(0
,y
) = 0 we obtain
X
(0) = 0 and
c
1
= 0. From
u
x
(
a, y
)=
−
hu
(
a, y
) we obtain
X
k
(
a
)=
−
hX
(
a
) and
λ
cos
λa
=
−
h
sin
λa
or tan
λa
=
−
λ
h
.
Let
λ
n
, where
n
=1,2,3,
...
, be the consecutive positive roots of this equation. From
u
(
x,
0) = 0 we obtain
Y
(0) = 0 and
c
3
=0. Thus
u
(
x, y
)=
∞

n
=1
A
n
sinh
λ
n
y
sin
λ
n
x.
Now
f
(
x
)=
∞

n
=1
A
n
sinh
λ
n
b
sin
λ
n
x
and
A
n
sinh
λ
n
b
=
F
a
0
f
(
x
) sin
λ
n
xdx
F
a
0
sin
2
λ
n
xdx
.
650

651. Exercises 13.7
Since

a
0
sin
2
λ
n
xdx
=
1
2
3
a
−
1
2
λ
n
sin 2
λ
n
a
(
=
1
2
3
a
−
1
λ
n
sin
λ
n
a
cos
λ
n
a
(
=
1
2
3
a
−
1
hλ
n
(
h
sin
λ
n
a
) cos
λ
n
a
(
=
1
2
3
a
−
1
hλ
n
(
−
λ
n
cos
λ
n
a
) cos
λ
n
a
(
=
1
2
h
w
ah
+ cos
2
λ
n
a
i
,
we have
A
n
=
2
h
sinh
λ
n
b
[
ah
+ cos
2
λ
n
a
]

a
0
f
(
x
) sin
λ
n
x dx.
4.
Letting
u
(
x, y
)=
X
(
x
)
Y
(
y
) and separatingvariables gives
X
kk
Y
+
XY
kk
=0
.
The boundary conditions
∂u
∂y




y
=0
= 0 and
∂u
∂y




y
=1
=
−
hu
(
x,
1)
correspond to
X
(
x
)
Y
k
(0) = 0 and
X
(
x
)
Y
k
(1) =
−
hX
(
x
)
Y
(1)
or
Y
k
(0) = 0 and
Y
k
(1) =
−
hY
(1)
.
Since these homogeneous boundary conditions are in terms of
Y
, we separate the differential equation as
X
kk
X
=
−
Y
kk
Y
=
λ
2
.
Then
Y
kk
+
λ
2
Y
=0
and
X
kk
−
λ
2
X
=0
have solutions
Y
(
y
)=
c
1
cos
λy
+
c
2
sin
λy
and
X
(
x
)=
c
3
e
−
λx
+
c
4
e
λx
.
We use exponential functions in the solution of
X
(
x
) since cosh
λx
and sinh
λx
are both unbounded as
x
→∞
.
Now,
Y
k
(0) = 0 implies
c
2
=0,so
Y
(
y
)=
c
1
cos
λy
. Since
Y
k
(
y
)=
−
c
1
λ
sin
λy
, the boundary condition
Y
k
(1) =
−
hY
(1) implies
−
c
1
λ
sin
λ
=
−
hc
1
cos
λ
or cot
λ
=
λ
h
.
Consideration of the graphs of
f
(
λ
) = cot
λ
and
g
(
λ
)=
λ/h
show that cos
λ
=
λh
has an infinite number of roots.
The consecutive positive roots
λ
n
for
n
=1,2,3,
...
, are the eigenvalues of the problem. The corresponding
eigenfunctions are
Y
n
(
y
)=
c
1
cos
λ
n
y
. The condition lim
x
→∞
u
(
x, y
) = 0 is equivalent to lim
x
→∞
X
(
x
)=0. Thus
c
4
= 0 and
X
(
x
)=
c
3
e
−
λx
. Therefore
u
n
(
x, y
)=
X
n
(
x
)
Y
n
(
x
)=
A
n
e
−
λ
n
x
cos
λ
n
y
651

652. Exercises 13.7
and by the superposition principle
u
(
x, y
)=
∞

n
=1
A
n
e
−
λ
n
x
cos
λ
n
y.
[It is easily shown that there are no eigenvalues corresponding to
λ
= 0.] Finally, the condition
u
(0
,y
)=
u
0
implies
u
0
=
∞

n
=1
A
n
cos
λ
n
y.
This is not a Fourier cosine series since the coefficients
λ
n
of
y
are not integer multiples of
π/p
, where
p
=1in
this problem. The functions cos
λ
n
y
are however orthogonal since they are eigenfunctions of the Sturm-Lionville
problem
Y
kk
+
λ
2
Y
=0
,
Y
k
(0) = 0
Y
k
(1) +
hY
(1) = 0
,
with weight function
p
(
x
) = 1. Thus we find
A
n
=
F
1
0
u
0
cos
λ
n
ydy
F
1
0
cos
2
λ
n
ydy
.
Now

1
0
u
0
cos
λ
n
ydy
=
u
0
λ
n
sin
λ
n
y




1
0
=
u
0
λ
n
sin
λ
n
and

1
0
cos
2
λ
n
ydy
=
1
2

1
0
(1 + cos 2
λ
n
y
)
dy
=
1
2
3
y
+
1
2
λ
n
sin 2
λ
n
y
(
1
0
=
1
2
3
1+
1
2
λ
n
sin 2
λ
n
(
=
1
2
3
1+
1
λ
n
sin
λ
n
cos
λ
n
(
.
Since cot
λ
=
λ/h
,
cos
λ
λ
=
sin
λ
h
and

1
0
cos
2
λ
n
ydy
=
1
2
3
1+
sin
2
λ
n
h
(
.
Then
A
n
=
u
0
λ
n
sin
λ
n
1
2
w
1+
1
h
sin
2
λ
n
i
=
2
hu
0
sin
λ
n
λ
n
n
h
+ sin
2
λ
n
π
and
u
(
x, y
)=2
hu
0
∞

n
=1
sin
λ
n
λ
n
n
1 + sin
2
λ
n
π
e
−
λ
n
x
cos
λ
n
y
where
λ
n
for
n
=1,2,3,
...
are the consecutive positive roots of cot
λ
=
λ/h
.
5.
The boundary-value problem is
k
∂
2
u
∂x
2
=
∂u
∂t
,
0
<x<L, t>
0
,
u
(0
,t
)=0
,
∂u
∂x




x
=
L
=0
,t>
0
,
652

653. Exercises 13.7
u
(
x,
0) =
f
(
x
)
,
0
<x<L.
Separation of variables leads to
X

+
λ
2
X
=0
T

+
kλ
2
T
=0
and
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
T
(
t
)=
c
3
e
−
kλ
2
t
.
From
X
(0) = 0 we find
c
1
= 0. From
X

(
L
) = 0 we obtain cos
λL
= 0 and
λ
=
π
(2
n
−
1)
2
L
,n
=1
,
2
,
3
,... .
Thus
u
(
x, t
)=
∞

n
=1
A
n
e
−
k
(2
n
−
1)
2
π
2
t/
4
L
2
sin
e
2
n
−
1
2
L
n
πx
where
A
n
=
F
L
0
f
(
x
) sin
I
2
n
−
1
2
L
f
πx dx
F
L
0
sin
2
I
2
n
−
1
2
L
f
πx dx
=
2
L

L
0
f
(
x
) sin
e
2
n
−
1
2
L
n
πx dx.
6.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation gives
a
2
∂
2
v
∂x
2
+
ψ

(
x
)=
∂
2
v
∂t
2
.
This equation will be homogeneous if
ψ

(
x
)=0or
ψ
(
x
)=
c
1
x
+
c
2
. The boundary condition
u
(0
,t
) = 0 implies
ψ
(0) = 0 which implies
c
2
=0. Thus
ψ
(
x
)=
c
1
x
. Usingthe second boundary condition, we obtain
E
e
∂v
∂x
+
ψ

n




x
=
L
=
F
0
,
which will be homogeneous when
Eψ

(
L
)=
F
0
.
Since
ψ

(
x
)=
c
1
we conclude that
c
1
=
F
0
/E
and
ψ
(
x
)=
F
0
E
x.
The new boundary-value problem is
a
2
∂
2
v
∂x
2
=
∂
2
v
∂t
2
,
0
<x<L, t>
0
v
(0
,t
)=0
,
∂v
∂x




x
=
L
=0
,t>
0
,
v
(
x,
0) =
−
F
0
E
x,
∂v
∂t




t
=0
=0
,
0
<x<L.
Referringto Example 2 in the text we see that
v
(
x, t
)=
∞

n
=1
A
n
cos
a
e
2
n
−
1
2
L
n
πt
sin
e
2
n
−
1
2
L
n
πx
where
−
F
0
E
x
=
∞

n
=1
A
n
sin
e
2
n
−
1
2
L
n
πx
653

654. Exercises 13.7
and
A
n
=
−
F
0
F
L
0
x
sin
I
2
n
−
1
2
L
f
πx dx
E
F
L
0
sin
2
I
2
n
−
1
2
L
f
πx dx
=
8
F
0
L
(
−
1)
n
Eπ
2
(2
n
−
1)
2
.
Thus
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
)
=
F
0
E
x
+
8
F
0
L
Eπ
2
∞

n
=1
(
−
1)
n
(2
n
−
1)
2
cos
a
e
2
n
−
1
2
L
n
πt
sin
e
2
n
−
1
2
L
n
πx.
7.
Separation of variables leads to
Y

+
λ
2
Y
=0
X

−
λ
2
X
=0
and
Y
(
y
)=
c
1
cos
λy
+
c
2
sin
λy
X
(
x
)=
c
3
cosh
λx
+
c
4
sinh
λx.
From
Y
(0) = 0 we find
c
1
= 0. From
Y

(1) = 0 we obtain cos
λ
= 0 and
λ
=
π
(2
n
−
1)
2
,n
=1
,
2
,
3
,... .
Thus
Y
(
y
)=
c
2
sin
e
2
n
−
1
2
n
πy.
From
X

(0) = 0 we find
c
4
= 0. Then
u
(
x, y
)=
∞

n
=1
A
n
cosh
e
2
n
−
1
2
n
πx
sin
e
2
n
−
1
2
n
πy
where
u
0
=
u
(1
,y
)=
∞

n
=1
A
n
cosh
e
2
n
−
1
2
n
π
sin
e
2
n
−
1
2
n
πy
and
A
n
cosh
e
2
n
−
1
2
n
π
=
F
1
0
u
0
sin
I
2
n
−
1
2
f
πy dy
F
1
0
sin
2
I
2
n
−
1
2
f
πy dy
=
4
u
0
(2
n
−
1)
π
.
Thus
u
(
x, y
)=
4
u
0
π
∞

n
=1
1
(2
n
−
1) cosh
I
2
n
−
1
2
f
π
cosh
e
2
n
−
1
2
n
πx
sin
e
2
n
−
1
2
n
πy.
8.
The boundary-value problem is
k
∂
2
u
∂x
2
=
∂u
∂t
,
0
<x<
1
,t>
0
∂u
∂x




x
=0
=
hu
(0
,t
)
,
∂u
∂x




x
=1
=
−
hu
(1
,t
)
,h>
0
,t>
0
,
u
(
x,
0) =
f
(
x
)
,
0
<x<
1
.
Referringto Example 1 in the text we have
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
and
T
(
t
)=
c
3
e
−
kλ
2
t
.
654

655. Exercises 13.7
Applyingthe boundary conditions, we obtain
X
k
(0) =
hX
(0)
X
k
(1) =
−
hX
(1)
or
λc
2
=
hc
1
−
λc
1
sin
λ
+
λc
2
cos
λ
=
−
hc
1
cos
λ
−
hc
2
sin
λ.
Choosing
c
1
=
λ
and
c
2
=
h
(to satisfy the first equation above) we obtain
−
λ
2
sin
λ
+
hλ
cos
λ
=
−
hλ
cos
λ
−
h
2
sin
λ
2
hλ
cos
λ
=(
λ
2
−
h
2
) sin
λ.
The eigenvalues
λ
n
are the consecutive positive roots of
tan
λ
=
2
hλ
λ
2
−
h
2
.
Then
u
(
x, t
)=
∞

n
=1
A
n
e
−
kλ
2
n
t
(
λ
n
cos
λ
n
x
+
h
sin
λ
n
x
)
where
f
(
x
)=
u
(
x,
0) =
∞

n
=1
A
n
(
λ
n
cos
λ
n
x
+
h
sin
λ
n
x
)
and
A
n
=
F
1
0
f
(
x
)(
λ
n
cos
λ
n
x
+
h
sin
λ
n
x
)
dx
F
1
0
(
λ
n
cos
λ
n
x
+
h
sin
λ
n
x
)
2
dx
=
2
λ
2
n
+2
h
+
h
2

1
0
f
(
x
)(
λ
n
cos
λ
n
x
+
h
sin
λ
n
x
)
dx.
[Note: the evaluation and simplification of the integral in the denominator requires the use of the relationship
(
λ
2
−
h
2
) sin
λ
=2
hλ
cos
λ
.]
9. (a)
Using
u
=
XT
and separation constant
λ
4
we find
X
(4)
−
λ
4
X
=0
and
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
+
c
3
cosh
λx
+
c
4
sinh
λx.
Since
u
=
XT
the boundary conditions become
X
(0) = 0
,X
k
(0) = 0
,X
kk
(1) = 0
,X
kkk
(1) = 0
.
Now
X
(0) = 0 implies
c
1
+
c
3
= 0, while
X
k
(0) = 0 implies
c
2
+
c
4
=0. Thus
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
−
c
1
cosh
λx
−
c
2
sinh
λx.
The boundary condition
X
kk
(1) = 0 implies
−
c
1
cos
λ
−
c
2
sin
λ
−
c
1
cosh
λ
−
c
2
sinh
λ
=0
while the boundary condition
X
kkk
(1) = 0 implies
c
1
sin
λ
−
c
2
cos
λ
−
c
1
sinh
λ
−
c
2
cosh
λ
=0
.
655

656. 0 1 2 3 4 5
10
20
30
40
50
Exercises 13.7
We then have the system of two equations in two unknowns
(cos
λ
+ cosh
λ
)
c
1
+ (sin
λ
+ sinh
λ
)
c
2
=0
(sin
λ
−
sinh
λ
)
c
1
−
(cos
λ
+ cosh
λ
)
c
2
=0
.
This homogeneous system will have nontrivial solutions for
c
1
and
c
2
provided




cos
λ
+ cosh
λ
sin
λ
+ sinh
λ
sin
λ
−
sinh
λ
−
cos
λ
−
cosh
λ




=0
or
−
2
−
2 cos
λ
cosh
λ
=0
.
Thus, the eigenvalues are determined by the equation cos
λ
cosh
λ
=
−
1.
(b)
Usinga computer to graph cosh
λ
and
−
1
/
cos
λ
=
−
sec
λ
we
see that the first two positive eigenvalues occur near 1
.
9 and 4
.
7.
ApplyingNewton’s method with these initial values we find that
the eigenvalues are
λ
1
=1
.
8751 and
λ
2
=4
.
6941.
10. (a)
In this case the boundary conditions are
u
(0
,t
)=0
,
∂u
∂x



x
=0
=0
u
(1
,t
)=0
,
∂u
∂x



x
=1
=0
.
Separatingvariables leads to
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
+
c
3
cosh
λx
+
c
4
sinh
λx
subject to
X
(0) = 0
,X

(0) = 0
,X
(1) = 0
,
and
X

(1) = 0
.
Now
X
(0) = 0 implies
c
1
+
c
3
= 0 while
X

(0) = 0 implies
c
2
+
c
4
=0. Thus
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
−
c
1
cosh
λx
−
c
2
sinh
λx.
The boundary condition
X
(1) = 0 implies
c
1
cos
λ
+
c
2
sin
λ
−
c
1
cosh
λ
−
c
2
sinh
λ
=0
while the boundary condition
X

(1) = 0 implies
−
c
1
sin
λ
+
c
2
cos
λ
−
c
1
sinh
λ
−
c
2
cosh
λ
=0
.
We then have the system of two equations in two unknowns
(cos
λ
−
cosh
λ
)
c
1
+ (sin
λ
−
sinh
λ
)
c
2
=0
−
(sin
λ
+ sinh
λ
)
c
1
+ (cos
λ
−
cosh
λ
)
c
2
=0
.
This homogeneous system will have nontrivial solutions for
c
1
and
c
2
provided




cos
λ
−
cosh
λ
sin
λ
−
sinh
λ
−
sin
λ
−
sinh
λ
cos
λ
−
cosh
λ




=0
656

657. 0 2 4 6 810
20
40
60
80
100
Exercises 13.8
or
2
−
2 cos
λ
cosh
λ
=0
.
Thus, the eigenvalues are determined by the equation cos
λ
cosh
λ
=1.
(b)
Usinga computer to graph cosh
λ
and 1
/
cos
λ
= sec
λ
we see that
the first two positive eigenvalues occur near the vertical asymp-
totes of sec
λ
,at3
π/
2 and 5
π/
2. ApplyingNewton’s method with
these initial values we find that the eigenvalues are
λ
1
=4
.
7300
and
λ
2
=7
.
8532.
Exercises 13.8
1.
This boundary-value problem was solved in Example 1 in the text. Identifying
b
=
c
=
π
and
f
(
x, y
)=
u
0
we
have
u
(
x, y, t
)=
∞

m
=1
∞

n
=1
A
mn
e
−
k
(
m
2
+
n
2
)
t
sin
mx
sin
ny
where
A
mn
=
4
π
2

π
0

π
0
u
0
sin
mx
sin
nydxdy
=
4
u
0
π
2

π
0
sin
mx dx

π
0
sin
nydy
=
4
u
0
mnπ
2
[1
−
(
−
1)
m
][1
−
(
−
1)
n
]
.
2.
This boundary-value problem was solved in Example 1 in the text. Identifying
b
=
c
= 1 and
f
(
x, y
)=
xy
(
x
−
1)(
y
−
1) we have
u
(
x, y, t
)=
∞

m
=1
∞

n
=1
A
mn
e
−
kπ
(
m
2
+
n
2
)
t
sin
mπx
sin
nπy
where
A
mn
=4

1
0

1
0
xy
(
x
−
1)(
y
−
1) sin
mπx
sin
nπy dx dy
=4
e

1
0
x
(
x
−
1) sin
mπx dx
ne

1
0
y
(
y
−
1) sin
nπy dy
n
=4
e
2
m
3
π
3
w
(
−
1)
m
−
1
i
ne
2
n
3
π
3
w
(
−
1)
n
−
1
i
n
=
16
m
3
n
3
π
6
w
(
−
1)
m
−
1
iw
(
−
1)
n
−
1
i
.
In Problems 3 and 4 we need to solve the partial differential equation
a
2
e
∂
2
u
∂x
2
+
∂
2
u
∂y
2
n
=
∂
2
u
∂t
2
.
657

658. Exercises 13.8
To separate this equation we try
u
(
x, y, t
)=
X
(
x
)
Y
(
y
)
T
(
t
):
a
2
(
X

YT
+
XY

T
)=
XY T

X

X
=
−
Y

Y
+
T

a
2
T
=
−
λ
2
.
Then
X

+
λ
2
X
=0
(1)
Y

Y
=
T

a
2
T
+
λ
2
=
−
µ
2
Y

+
µ
2
Y
=0
(2)
T

+
a
2
I
λ
2
+
µ
2
f
T
=0
.
(3)
The general solutions of equations (1), (2), and (3) are, respectively,
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
Y
(
y
)=
c
3
cos
µy
+
c
4
sin
µy
T
(
t
)=
c
5
cos
a
h
λ
2
+
µ
2
t
+
c
6
sin
a
h
λ
2
+
µ
2
t.
3.
The conditions
X
(0) = 0 and
Y
(0) = 0 give
c
1
= 0 and
c
3
= 0. The conditions
X
(
π
) = 0 and
Y
(
π
) = 0 yield
two sets of eigenvalues:
λ
=
m, m
=1
,
2
,
3
,...
and
µ
=
n,n
=1
,
2
,
3
,... .
A product solution of the partial differential equation that satisfies the boundary conditions is
u
mn
(
x, y, t
)=
=
A
mn
cos
a
h
m
2
+
n
2
t
+
B
mn
sin
a
h
m
2
+
n
2
t
t
sin
mx
sin
ny.
To satisfy the initial conditions we use the superposition principle:
u
(
x, y, t
)=
∞

m
=1
∞

n
=1
=
A
mn
cos
a
h
m
2
+
n
2
t
+
B
mn
sin
a
h
m
2
+
n
2
t
t
sin
mx
sin
ny.
The initial condition
u
t
(
x, y,
0) = 0 implies
B
mn
= 0 and
u
(
x, y, t
)=
∞

m
=1
∞

n
=1
A
mn
cos
a
h
m
2
+
n
2
t
sin
mx
sin
ny.
At
t
= 0 we have
xy
(
x
−
π
)(
y
−
π
)=
∞

m
=1
∞

n
=1
A
mn
sin
mx
sin
ny.
Using(11) and (12) in the text, it follows that
A
mn
=
4
π
2

π
0

π
0
xy
(
x
−
π
)(
y
−
π
) sin
mx
sin
nydxdy
=
4
π
2

π
0
x
(
x
−
π
) sin
mx dx

π
0
y
(
y
−
π
) sin
nydy
=
16
m
3
n
3
π
2
[(
−
1)
m
−
1][(
−
1)
n
−
1]
.
658

659. Exercises 13.8
4.
The conditions
X
(0) = 0 and
Y
(0) = 0 give
c
1
= 0 and
c
3
= 0. The conditions
X
(
b
) = 0 and
Y
(
c
) = 0 yield
two sets of eigenvalues
λ
=
mπ/b, m
=1
,
2
,
3
,...
and
µ
=
nπ/c, n
=1
,
2
,
3
,....
A product solution of the partial differential that satisfies the boundary conditions is
u
mn
(
x, y, t
)=(
A
mn
cos
aω
mn
t
+
B
mn
sin
aω
mn
t
) sin
=
mπ
b
x
t
sin
=
nπ
c
y
t
,
where
ω
mn
=
h
(
mπ/b
)
2
+(
nπ/c
)
2
. To satisfy the initial conditions we use the superposition principle:
u
(
x, y, t
)=
∞

m
=1
∞

n
=1
(
A
mn
cos
aω
mn
t
+
B
mn
sin
aω
mn
t
) sin
=
mπ
b
x
t
sin
=
nπ
c
y
t
.
At
t
= 0 we have
f
(
x, y
)=
∞

m
=1
∞

n
=1
A
mn
sin
=
mπ
b
x
t
sin
=
nπ
c
y
t
and
g
(
x, y
)=
∞

m
=1
∞

n
=1
B
mn
aω
mn
sin
=
mπ
b
x
t
sin
=
nπ
c
y
t
.
Using(11) and (12) in the text, it follows that
A
mn
=
4
bc

c
0

b
0
f
(
x, y
) sin
=
mπ
b
x
t
sin
=
nπ
c
y
t
dx dy
B
mn
=
4
abcω
mn

c
0

b
0
g
(
x, y
) sin
=
mπ
b
x
t
sin
=
nπ
c
y
t
dx dy.
To separate Laplace’s equation in three dimensions we try
u
(
x, y, z
)=
X
(
x
)
Y
(
y
)
Z
(
z
):
X

YZ
+
XY

Z
+
XY Z

=0
X

X
=
−
Y

Y
−
Z

Z
=
−
λ
2
.
Then
X

+
λ
2
X
=0
(4)
Y

Y
=
−
Z

Z
+
λ
2
=
−
µ
2
Y

+
µ
2
Y
=0
(5)
Z

−
(
λ
2
+
µ
2
)
Z
=0
.
(6)
The general solutions of equations (4), (5), and (6) are, respectively
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
Y
(
y
)=
c
3
cos
µy
+
c
4
sin
µy
Z
(
z
)=
c
5
cosh
h
λ
2
+
µ
2
z
+
c
6
sinh
h
λ
2
+
µ
2
z.
5.
The boundary and initial conditions are
u
(0
,y,z
)=0
,u
(
a, y, z
)=0
u
(
x,
0
,z
)=0
,u
(
x, b, z
)=0
u
(
x, y,
0) = 0
,u
(
x, y, c
)=
f
(
x, y
)
.
659

660. Exercises 13.8
The conditions
X
(0) =
Y
(0) =
Z
(0) = 0 give
c
1
=
c
3
=
c
5
= 0. The conditions
X
(
a
) = 0 and
Y
(
b
) = 0 yield
two sets of eigenvalues:
λ
=
mπ
a
,m
=1
,
2
,
3
,...
and
µ
=
nπ
b
,n
=1
,
2
,
3
,... .
By the superposition principle
u
(
x, y, t
)=
∞

m
=1
∞

n
=1
A
mn
sinh
ω
mn
z
sin
mπ
a
x
sin
nπ
b
y
where
ω
2
mn
=
m
2
π
2
a
2
+
n
2
π
2
b
2
and
A
mn
=
4
ab
sinh
ω
mn
c

b
0

a
0
f
(
x, y
) sin
mπ
a
x
sin
nπ
b
ydxdy.
6.
The boundary and initial conditions are
u
(0
,y,z
)=0
,
u
(
x,
0
,z
)=0
,
u
(
x, y,
0) =
f
(
x, y
)
,
u
(
a, y, z
)=0
,
u
(
x, b, z
)=0
,
u
(
x, y, c
)=0
.
The conditions
X
(0) =
Y
(0) = 0 give
c
1
=
c
3
= 0. The conditions
X
(
a
)=
Y
(
b
) = 0 yield two sets of eigenvalues:
λ
=
mπ
a
,m
=1
,
2
,
3
,...
and
µ
=
nπ
b
,n
=1
,
2
,
3
,... .
Let
ω
2
mn
=
m
2
π
2
a
2
+
n
2
π
2
b
2
.
Then the boundary condition
Z
(
c
) = 0 gives
c
5
cosh
cω
mn
+
c
6
sinh
cω
mn
=0
from which we obtain
Z
(
z
)=
c
5
e
cosh
w
mn
z
−
cosh
cω
mn
sinh
cω
mn
sinh
ωz
n
=
c
5
sinh
cω
mn
(sinh
cω
mn
cosh
ω
mn
z
−
cosh
cω
mn
sinh
ω
mn
z
)
=
c
mn
sinh
ω
mn
(
c
−
z
)
.
By the superposition principle
u
(
x, y, t
)=
∞

m
=1
∞

n
=1
A
mn
sinh
ω
mn
(
c
−
z
) sin
mπ
a
x
sin
nπ
b
y
where
A
mn
=
4
ab
sinh
cω
mn

b
0

a
0
f
(
x, y
) sin
mπ
a
x
sin
nπ
b
ydxdy.
7.
The boundary and initial conditions are
u
(0
,y,z
)=0
,
u
(
x,
0
,z
)=0
,
u
(
x, y,
0) =
−
u
0
,
u
(1
,y,z
)=0
,
u
(
x,
1
,z
)=0
,
u
(
x, y,
1) =
u
0
.
660

661. Chapter 13 Review Exercises
Applyingthe superposition principle to the solutions in Problems 5 and 6, with
a
=
b
=
c
= 1 and
f
(
x, y
)=
u
0
in Problem 5 and
f
(
x, y
)=
−
u
0
in Problem 6, we get
u
(
x, y, t
)=
∞

m
=1
∞

n
=1
A
mn
[sinh
ω
mn
z
−
sinh
ω
m
(1
−
z
)] sin
mπx
sin
nπy
where
ω
2
mn
=(
m
2
+
n
2
)
π
2
and
A
mn
=
4
sinh
ω
mn

1
0

1
0
u
0
sin
mπx
sin
nπy dx dy
=
4
u
0
sinh
ω
mn
e

1
0
sin
mπx dx
ne

1
0
sin
nπy dy
n
=
4
u
0
sinh
ω
mn
e
1
mπ
w
1
−
(
−
1)
n
i
ne
1
nπ
w
1
−
(
−
1)
n
i
n
=
4
u
0
mnπ
2
sinh
ω
mn
w
1
−
(
−
1)
m
iw
1
−
(
−
1)
n
i
.
Chapter 13 Review Exercises
1.
Let
u
=
XY
so that
u
yx
=
X

Y

,
X

Y

=
XY,
and
X

X
=
Y
Y

=
±
λ
2
.
Then
X

∓
λ
2
X
= 0 and
Y
∓
λ
2
Y

=0
.
For
λ
2
>
0 we obtain
X
=
c
1
e
λ
2
x
and
Y
=
c
2
e
y/λ
2
.
For
−
λ
2
<
0 we obtain
X
=
c
1
e
−
λ
2
x
and
Y
=
c
2
e
−
y/λ
2
.
For
λ
2
= 0 we obtain
X
=
c
1
and
Y
= 0. The general form is
u
=
XY
=
A
1
e
A
2
x
+
y/A
2
.
2.
If
u
=
XY
then
u
x
=
X

Y,
u
y
=
XY

,
u
xx
=
X

Y,
u
yy
=
XY

,
and
X

Y
+
XY

+2
X

Y
+2
XY

=0
661

662. Chapter 13 Review Exercises
so that
(
X

+2
X

)
Y
+
X
(
Y

+2
Y

)=0
,
X

+2
X

−
X
=
Y

+2
Y

Y
=
±
λ
2
,
X

+2
X

±
λ
2
X
=0
,
and
Y

+2
Y

∓
λ
2
Y
=0
.
Using
λ
2
as a separation constant we obtain
Y
=
c
1
e
(
−
1+
√
1+
λ
2
)
y
+
c
2
e
(
−
1
−
√
1+
λ
2
)
y
.
If 1
−
λ
2
<
0 then
X
=
e
−
x
=
c
3
cos
h
λ
2
−
1
x
+
c
4
sin
h
λ
2
−
1
x
t
.
If 1
−
λ
2
>
0 then
X
=
c
3
e
(
−
1+
√
1
−
λ
2
)
x
+
c
4
e
(
−
1
−
√
1
−
λ
2
)
x
.
If 1
−
λ
2
= 0 then
X
=
c
3
e
−
x
+
c
4
xe
−
x
so that
u
=
XY
=
=
A
1
e
(
−
1+
√
1+
A
5
)
y
+
A
2
e
(
−
1
−
√
1+
A
5
)
y
t
e
−
x
=
A
3
cos
h
A
5
−
1
x
+
A
4
sin
h
A
5
−
1
x
t
=
=
A
1
e
(
−
1+
√
1+
A
5
)
y
+
A
2
e
(
−
1
−
√
1+
A
5
)
y
t=
A
3
e
(
−
1+
√
1
−
A
5
)
x
+
A
4
e
(
−
1
−
√
1
−
A
5
)
x
t
=
=
A
1
e
(
−
1+
√
2)
y
+
A
2
e
(
−
1
−
√
2)
y
t
I
A
3
e
−
x
+
A
4
xe
−
x
f
.
Using
−
λ
2
we obtain the same three solutions except that
x
and
y
are interchanged. Using
λ
2
= 0 we obtain
u
=
XY
=
I
A
1
+
A
2
e
−
2
x
fI
A
3
+
A
4
e
−
2
y
f
.
3.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation we obtain
k
∂
2
v
∂x
2
+
kψ

(
x
)=
∂v
∂t
.
This equation will be homogeneous provided
ψ
satisfies
kψ

=0 or
ψ
=
c
1
x
+
c
2
.
Considering
u
(0
,t
)=
v
(0
,t
)+
ψ
(0) =
u
0
we set
ψ
(0) =
u
0
so that
ψ
(
x
)=
c
1
x
+
u
0
.Now
−
∂u
∂x




x
=
π
=
−
∂v
∂x




x
=
π
−
ψ

(
x
)=
v
(
π, t
)+
ψ
(
π
)
−
u
1
is equivalent to
∂v
∂x




x
=
π
+
v
(
π, t
)=
u
1
−
ψ

(
x
)
−
ψ
(
π
)=
u
1
−
c
1
−
(
c
1
π
+
u
0
)
,
which will be homogeneous when
u
1
−
c
1
−
c
1
π
−
u
0
=0 or
c
1
=
u
1
−
u
0
1+
π
.
662

663. Chapter 13 Review Exercises
The steady-state solution is
ψ
(
x
)=
e
u
1
−
u
0
1+
π
n
x
+
u
0
.
4.
The solution of the problem represents the heat of a thin rod of length
π
. The left boundary
x
= 0 is kept at
constant temperature
u
0
for
t>
0. Heat is lost from the right end of the rod by being in contact with a medium
that is held at constant temperature
u
1
.
5.
The boundary-value problem is
a
2
∂
2
u
∂x
2
=
∂
2
u
∂t
2
,
0
<x<
1
,t>
0
,
u
(0
,t
)=0
,u
=(1
,t
)=0
,t>
0
,
u
(
x,
0) = 0
,
∂u
∂t




t
=0
=
g
(
x
)
,
0
<x<
1
.
From Section 13.4 in the text we see that
A
n
=0,
B
n
=
2
nπa

1
0
g
(
x
) sin
nπx dx
=
2
nπa

3
/
4
1
/
4
h
sin
nπx dx
=
2
h
nπa
e
−
1
nπ
cos
nπx
n




3
/
4
1
/
4
=
2
h
n
2
π
2
a
e
cos
nπ
4
−
cos
3
nπ
4
n
and
u
(
x, t
)=
∞

n
=1
B
n
sin
nπat
sin
nπx.
6.
The boundary-value problem is
∂
2
u
∂x
2
+
x
2
=
∂
2
u
∂t
2
,
0
<x<
1
,t>
0
,
u
(0
,t
)=1
,u
(1
,t
)=0
,t>
0
,
u
(
x,
0) =
f
(
x
)
,u
t
(
x,
0) = 0
,
0
<x<
1
.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation gives
∂
2
v
∂x
2
+
ψ

(
x
)+
x
2
=
∂
2
v
∂t
2
.
This equation will be homogeneous provided
ψ

(
x
)+
x
2
=0or
ψ
(
x
)=
−
1
12
x
4
+
c
1
x
+
c
2
.
From
ψ
(0) = 1 and
ψ
(1) = 0 we obtain
c
1
=
−
11
/
12 and
c
2
= 1. The new problem is
∂
2
v
∂x
2
=
∂
2
v
∂t
2
,
0
<x<
1
,t>
0
,
v
(0
,t
)=0
,v
(1
,t
)=0
,t>
0
,
v
(
x,
0) =
f
(
x
)
−
ψ
(
x
)
,v
t
(
x,
0) = 0
,
0
<x<
1
.
From Section 13.4 in the text we see that
B
n
=0,
A
n
=2

1
0
[
f
(
x
)
−
ψ
(
x
)] sin
nπx dx
=2

1
0
3
f
(
x
)+
1
12
x
4
+
11
12
x
−
1
(
sin
nπx dx,
663

664. Chapter 13 Review Exercises
and
v
(
x, t
)=
∞

n
=1
A
n
cos
nπt
sin
nπx.
Thus
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
)=
−
1
12
x
4
−
11
12
x
+1+
∞

n
=1
A
n
cos
nπt
sin
nπx.
7.
Using
u
=
XY
and
λ
2
as a separation constant leads to
X

−
λ
2
X
=0
,
X
(0) = 0
,
and
Y

+
λ
2
Y
=0
,
Y
(0) = 0
,
Y
(
π
)=0
.
Then
Y
=
c
1
sin
ny
and
X
=
c
2
sinh
nx
for
n
=1,2,3,
...
so that
u
=
∞

n
=1
A
n
sinh
nx
sin
ny.
Imposing
u
(
π, y
) = 50 =
∞

n
=1
A
n
sinh
nπ
sin
ny
gives
A
n
sinh
nπ
=
100
nπ
1
−
(
−
1)
n
sinh
nπ
so that
u
(
x, y
)=
100
π
∞

n
=1
1
−
(
−
1)
n
n
sinh
nπ
sinh
nx
sin
ny.
8.
Using
u
=
XY
and
λ
2
as a separation constant leads to
X

−
λ
2
X
=0
,
and
Y

+
λ
2
Y
=0
,
Y

(0) = 0
,
Y

(
π
)=0
.
Then
Y
=
c
1
cos
ny
for
n
=0,1,2,
...
and
X
=
c
2
or
X
=
c
2
e
−
nx
for
n
=1,2,3,
...
(since
u
must be bounded as
x
→∞
) so that
u
=
A
0
+
∞

n
=1
A
n
e
−
nx
cos
ny.
664

665. Chapter 13 Review Exercises
Imposing
u
(0
,y
) = 50 =
A
0
+
∞

n
=1
A
n
cos
ny
gives
A
0
=
1
π

π
0
50
dy
=50
and
A
n
=
2
π

π
0
50 cos
nydy
=0
for
n
=1,2,3,
...
so that
u
(
x, y
)=50
.
9.
Using
u
=
XY
and
λ
2
as a separation constant leads to
X

−
λ
2
X
=0
,
and
Y

+
λ
2
Y
=0
,
Y
(0) = 0
,
Y
(
π
)=0
.
Then
Y
=
c
1
sin
ny
and
X
=
c
2
e
−
nx
for
n
=1,2,3,
...
(since
u
must be bounded as
x
→∞
) so that
u
=
∞

n
=1
A
n
e
−
nx
sin
ny.
Imposing
u
(0
,y
) = 50 =
∞

n
=1
A
n
sin
ny
gives
A
n
=
2
π

π
0
50 sin
nydy
=
100
nπ
[1
−
(
−
1)
n
]
so that
u
(
x, y
)=
∞

n
=1
100
nπ
[1
−
(
−
1)
n
]
e
−
nx
sin
ny.
10.
The boundary-value problem is
k
∂
2
u
∂x
2
=
∂u
∂t
,
−
L<x<L, t>
0
,
u
(
−
L, t
)=0
,u
(
L, t
)=0
,t>
0
,
u
(
x,
0) =
u
0
,
−
L<x<L.
Referringto Section 13
.
3 in the text we have
X
(
x
)=
c
1
cos
λx
+
c
2
sin
λx
and
T
(
t
)=
c
3
e
−
kλ
2
t
.
665

666. Chapter 13 Review Exercises
Usingthe boundary conditions
u
(
−
L,
0) =
X
(
−
L
)
T
(0) = 0 and
u
(
L,
0) =
X
(
L
)
T
(0) = 0 we obtain
c
1
cos(
−
λL
)+
c
2
sin(
−
λL
)=0
c
1
cos
λL
+
c
2
sin
λL
=0
or
c
1
cos
λL
−
c
2
sin
λL
=0
c
1
cos
λL
+
c
2
sin
λL
=0
.
Adding, we find cos
λL
= 0 which gives the eigenvalues
λ
=
2
n
−
1
2
L
π, n
=1
,
2
,
3
,... .
Thus
u
(
x, t
)=
∞

n
=1
A
n
e
−
(
2
n
−
1
2
L
π
)
2
kt
cos
e
2
n
−
1
2
L
n
πx.
From
u
(
x,
0) =
u
0
=
∞

n
=1
A
n
cos
e
2
n
−
1
2
L
n
πx
we find
A
n
=
2
F
L
0
u
0
cos
n
2
n
−
1
2
L
π
πx dx
2
F
L
0
cos
2
n
2
n
−
1
2
L
π
πx dx
=
u
0
(
−
1)
n
+1
2
L/π
(2
n
−
1)
L/
2
=
4
u
0
(
−
1)
n
+1
π
(2
n
−
1)
.
11.
The coefficients of the series
u
(
x,
0) =
∞

n
=1
B
n
sin
nx
are
B
n
=
2
π

π
0
sin
x
sin
nxdx
=
2
π

π
0
1
2
[cos(1
−
n
)
x
−
cos(1 +
n
)
x
]
dx
=
1
π
3
sin(1
−
n
)
x
1
−
n




π
0
−
sin(1 +
n
)
x
1+
n




π
0
(
= 0 for
n

=1
.
For
n
=1,
B
1
=
2
π

π
0
sin
2
xdx
=
1
π

π
0
(1
−
cos 2
x
)
dx
=1
.
Thus
u
(
x, t
)=
∞

n
=1
B
n
e
−
n
2
t
sin
nx
reduces to
u
(
x, t
)=
e
−
t
sin
x
for
n
=1.
12.
Substituting
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
) into the partial differential equation gives
k
∂
2
v
∂x
2
+
kψ

+ sin 2
πx
=
∂v
∂t
.
This equation will be homogeneous provided
ψ
satisfies
kψ

+ sin 2
πx
=0
.
The general solution of this equation is
ψ
(
x
)=
1
4
kπ
2
sin 2
πx
+
c
1
x
+
c
2
.
666

667. Chapter 13 Review Exercises
From
ψ
(0) =
ψ
(1) = 0 we find that
c
1
=
c
2
= 0 and
ψ
(
x
)=
1
4
kπ
2
sin 2
πx.
Now the new problem is
k
∂
2
v
∂x
2
=
∂v
∂t
,
0
<x<
1
,t>
0
v
(0
,t
)=0
,v
(1
,t
)=0
,t>
0
v
(
x,
0) = sin
πx
−
ψ
(
x
)
,
0
<x<
1
.
If we let
v
=
XT
then
X

X
=
T

kT
=
−
λ
2
gives the separated differential equations
X

+
λ
2
X
= 0 and
T

+
kλ
2
T
=0
.
The respective solutions are
X
(
x
)=
c
3
cos
λx
+
c
4
sin
λx
T
(
t
)=
c
5
e
−
kλ
2
t
.
From
X
(0) = 0 we get
c
3
= 0 and from
X
(1) = 0 we find
λ
=
nπ
for
n
=1,2,3,
...
. Consequently, it follows
that
v
(
x, t
)=
∞

n
=1
A
n
e
−
kn
2
π
2
t
sin
nπx
where
v
(
x,
0) = sin
πx
−
1
4
kπ
2
sin 2
πx
=0
implies
A
n
=2

1
0
e
sin
πx
−
1
4
kπ
2
sin 2
πx
n
sin
nπx dx.
By orthogonality
A
n
= 0 for
n
=3,4,5,
...
, and only
A
1
and
A
2
can be nonzero. We have
A
1
=2
3

1
0
sin
2
πx dx
−
1
4
kπ
2

1
0
sin 2
πx
sin
πx dx
(
=2

1
0
1
2
(1
−
cos 2
πx
)
dx
=1
and
A
2
=2
3

1
0
sin
πx
sin 2
πx dx
−
1
4
kπ
2

1
0
sin
2
2
πx dx
(
=
−
1
2
kπ
2

1
0
1
2
(1
−
cos 4
πx
)
dx
=
−
1
4
kπ
2
.
Therefore
v
(
x, t
)=
A
1
e
−
kπ
2
t
sin
πx
+
A
2
e
−
k
4
π
2
t
sin 2
πx
=
e
−
kπ
2
t
sin
πx
−
1
4
kπ
2
e
−
4
kπ
2
t
sin 2
πx
and
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
)=
e
−
kπ
2
t
sin
πx
+
1
4
kπ
2
(1
−
e
−
4
kπ
2
t
) sin 2
πx.
13.
Using
u
=
XT
and
−
λ
2
as a separation constant we find
X

+2
X

+
λ
2
X
= 0 and
T

+2
T

+(1+
λ
2
)
T
=0
.
667

668. Chapter 13 Review Exercises
Thus for
λ>
1
X
=
c
1
e
−
x
cos
h
λ
2
−
1
x
+
c
2
e
−
x
sin
h
λ
2
−
1
x
T
=
c
3
e
−
t
cos
λt
+
c
4
e
−
t
sin
λt.
For 0
≤
λ
≤
1 we only obtain
X
= 0. Now the boundary conditions
X
(0) = 0 and
X
(
π
) = 0 give, in turn,
c
1
= 0 and
√
λ
2
−
1
π
=
nπ
or
λ
2
=
n
2
+1,
n
=1,2,3,
...
. The correspondingsolutions are
X
=
c
2
e
−
x
sin
nx
.
The initial condition
T

(0) = 0 implies
c
3
=
λc
4
and so
T
=
c
4
e
−
t
1
h
n
2
+ 1 cos
h
n
2
+1
t
+ sin
h
n
2
+1
t
+
.
Using
u
=
XT
and the superposition principle, a formal series solution is
u
(
x, t
)=
e
−
(
x
+
t
)
∞

n
=1
A
n
1
h
n
2
+ 1 cos
h
n
2
+1
t
+ sin
h
n
2
+1
t
+
sin
nx.
14.
Letting
c
=
XT
and separatingvariables we obtain
kX

−
hX

X
=
T

T
or
X

−
αX

X
=
T

kT
=
−
λ
2
where
α
=
h/k
. This leads to the separated differential equations
X

−
αX

+
λ
2
X
= 0 and
T

+
kλ
2
T
=0
.
The solution of the second equation is
T
(
t
)=
c
3
e
−
kλ
2
t
.
For the first equation we have
m
=
1
2
(
α
±
√
α
2
−
4
λ
2
), and we consider three cases usingthe boundary conditions
X
(0) =
X
(1) = 0:
α
2
>
4
λ
2
The solution is
X
=
c
1
e
m
1
x
+
c
2
e
m
2
x
, where the boundary conditions imply
c
1
=
c
2
=0,so
X
=0.
(Note in this case that if
λ
= 0, the solution is
X
=
c
1
+
c
2
e
αx
and the boundary conditions again imply
c
1
=
c
2
=0,so
X
= 0.)
α
2
=4
λ
2
The solution is
X
=
c
1
e
m
1
x
+
c
2
xe
m
1
x
, where the boundary conditions imply
c
1
=
c
2
=0,so
X
=0.
α
2
<
4
λ
2
The solution is
X
(
x
)=
c
1
e
αx/
2
cos
√
4
λ
2
−
α
2
2
x
+
c
2
e
αx/
2
sin
√
4
λ
2
−
α
2
2
x.
From
X
(0) = 0 we see that
c
1
= 0. From
X
(1) = 0 we find
1
2
h
4
λ
2
−
α
2
=
nπ
or
λ
2
=
1
4
(4
n
2
π
2
+
α
2
)
.
Thus
X
(
x
)=
c
2
e
αx/
2
sin
nπx,
and
c
(
x, t
)=
∞

n
=1
A
n
e
αx/
2
e
−
k
(4
n
2
π
2
+
α
2
)
t/
4
sin 2
πx.
The initial condition
c
(
x,
0) =
c
0
implies
c
0
=
∞

n
=1
A
n
e
α/
2
sin
nπx.
(
1
)
668

669. Chapter 13 Review Exercises
From the self-adjoint form
d
dx
[
e
−
αx
X

]+
λ
2
e
−
αx
X
=0
the eigenfunctions are orthogonal on [0
,
1] with weight function
e
−
αx
. That is

1
0
e
−
αx
(
e
αx/
2
sin
nπx
)(
e
αx/
2
sin
mπx
)
dx
=0
,n

=
m.
Multiplying(1) by
e
−
αx
e
αx/
2
sin
mπx
and integrating we obtain

1
0
c
0
e
−
αx
e
αx/
2
sin
mπx dx
=
∞

n
=1
A
n
e
−
αx
e
αx/
2
(sin
mπx
)
e
αx/
2
sin
nπx dx
c
0

1
0
e
−
αx/
2
sin
nπx dx
=
A
n

1
0
sin
2
nπx dx
=
1
2
A
n
and
A
n
=2
c
0

1
0
e
−
αx/
2
sin
nπx dx
=
4
c
0
[2
e
α/
2
nπ
−
2
nπ
(
−
1)
n
]
e
α/
2
(
α
2
+4
n
2
π
2
)
=
8
nπc
0
[
e
α/
2
−
(
−
1)
n
]
e
α/
2
(
α
2
+4
n
2
π
2
)
.
669

670. 14
Boundary-Value Problems in Other
Coordinate Systems
Exercises 14.1
1.
We have
A
0
=
1
2
π
W
π
0
u
0
dθ
=
u
0
2
A
n
=
1
π
W
π
0
u
0
cos
nθ dθ
=0
B
n
=
1
π
W
π
0
u
0
sin
nθ dθ
=
u
0
nπ
[1
−
(
−
1)
n
]
and so
u
(
r, θ
)=
u
0
2
+
u
0
π
∞
e
n
=1
1
−
(
−
1)
n
n
r
n
sin
nθ.
2.
We have
A
0
=
1
2
π
W
π
0
θdθ
+
1
2
π
W
2
π
π
(
π
−
θ
)
dθ
=0
A
n
=
1
π
W
π
0
θ
cos
nθ dθ
+
1
π
W
2
π
π
(
π
−
θ
) cos
nθ dθ
=
2
n
2
π
[(
−
1)
n
−
1]
B
n
=
1
π
W
π
0
θ
sin
nθ dθ
+
1
π
W
2
π
π
(
π
−
θ
) sin
nθ dθ
=
1
n
[1
−
(
−
1)
n
]
and so
u
(
r, θ
)=
∞
e
n
=1
r
n
h
(
−
1)
n
−
1
n
2
π
cos
nθ
+
1
−
(
−
1)
n
n
sin
nθ
a
.
3.
We have
A
0
=
1
2
π
W
2
π
0
(2
πθ
−
θ
2
)
dθ
=
2
π
2
3
A
n
=
1
π
W
2
π
0
(2
πθ
−
θ
2
) cos
nθ dθ
=
−
4
n
2
B
n
=
1
π
W
2
π
0
(2
πθ
−
θ
2
) sin
nθ dθ
=0
and so
u
(
r, θ
)=
2
π
2
3
−
4
∞
e
n
=1
r
n
n
2
cos
nθ.
4.
We have
A
0
=
1
2
π
W
2
π
0
θdθ
=
π
A
n
=
1
π
W
2
π
0
θ
cos
nθ dθ
=0
B
n
=
1
π
W
2
π
0
θ
sin
nθ dθ
=
−
2
n
670

671. Exercises 14.1
and so
u
(
r, θ
)=
π
−
2
∞
e
n
=1
r
n
n
sin
nθ.
5.
As in Example 1 in the text we have
R
(
r
)=
c
3
r
n
+
c
4
r
−
n
.In order that the solution be bounded as
r
→∞
we
must define
c
3
= 0.Hence
u
(
r, θ
)=
A
0
+
∞
e
n
=1
r
−
n
(
A
n
cos
nθ
+
B
n
sin
nθ
)
A
0
=
1
2
π
W
2
π
0
f
(
θ
)
dθ
where
A
n
=
c
n
π
W
2
π
0
f
(
θ
) cos
nθ dθ
B
n
=
c
n
π
W
2
π
0
f
(
θ
) sin
nθ dθ.
6.
We solve
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
=0
,
0
<θ<
π
2
,
0
<r<c,
u
(
c, θ
)=
f
(
θ
)
,
0
<θ<
π
2
,
u
(
r,
0) = 0
,u
(
r, π/
2) = 0
,
0
<r<c.
Proceeding as in Example 1 in the text we obtain the separated differential equations
r
2
R
ee
+
rR
e
−
λ
2
R
=0
Θ
ee
+
λ
2
Θ=0
with solutions
Θ(
θ
)=
c
1
cos
λθ
+
c
2
sin
λθ
R
(
r
)=
c
3
r
λ
+
c
4
r
−
λ
.
Since we want
R
(
r
) to be bounded as
r
→
0 we require
c
4
= 0.Applying the boundary conditions Θ(0) = 0
and Θ(
π/
2) = 0 we find that
c
1
= 0 and
λ
=2
n
for
n
=1,2,3,
...
.Therefore
u
(
r, θ
)=
∞
e
n
=1
A
n
r
2
n
sin 2
nθ.
From
u
(
c, θ
)=
f
(
θ
)=
∞
e
n
=1
A
n
c
n
sin 2
nθ
we find
A
n
=
4
πc
2
n
W
π/
2
0
f
(
θ
) sin 2
nθ dθ.
7.
Referring to the solution of Problem 6 above we have
Θ(
θ
)=
c
1
cos
λθ
+
c
2
sin
λθ
R
(
r
)=
c
3
r
n
.
Applying the boundary conditions Θ
e
(0) = 0 and Θ
e
(
π/
2) = 0 we find that
c
2
= 0 and
λ
=2
n
for
671

672. Exercises 14.1
n
=0,1,2,
...
.Therefore
u
(
r, θ
)=
A
0
+
∞
e
n
=1
A
n
r
2
n
cos 2
nθ.
From
u
(
c, θ
)=
v
1
,
0
<θ<π/
4
0
,π/
4
<θ<π/
2
=
A
0
+
∞
e
n
=1
A
n
c
2
n
cos 2
nθ
we find
A
0
=
1
π/
2
W
π/
4
0
dθ
=
1
2
and
c
2
n
A
n
=
2
π/
2
W
π/
4
0
cos 2
nθ dθ
=
2
nπ
sin
nπ
2
.
Thus
u
(
r, θ
)=
1
2
+
2
π
∞
e
n
=1
1
n
sin
nπ
2
=
r
c
1
2
n
cos 2
nθ.
8.
We solve
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
=0
,
0
<θ<π/
4
,r>
0
u
(
r,
0) = 0
,r>
0
u
(
r, π/
4) = 30
,r>
0
.
Proceeding as in Example 1 in the text we find the separated ordinary differential equations to be
r
2
R
ee
+
rR
e
−
λ
2
R
=0
Θ
ee
+
λ
2
Θ=0
.
The corresponding general solutions are
R
(
r
)=
c
1
r
λ
+
c
2
r
−
λ
Θ(
θ
)=
c
3
cos
λθ
+
c
4
sin
λθ.
The condition Θ(0) = 0 implies
c
3
= 0 so that Θ =
c
4
sin
λθ
.Now, in order that the temperature be bounded
as
r
→∞
we define
c
1
= 0.Similarly, in order that the temperature be bounded as
r
→
0 we are forced to
define
c
2
= 0.Thus
R
(
r
) = 0 and so no nontrivial solution exists for
λ>
0.For
λ
= 0 the separated differential
equations are
r
2
R
ee
+
rR
e
= 0 and Θ
ee
=0
.
Solutions of these latter equations are
R
(
r
)=
c
1
+
c
2
ln
r
and Θ(
θ
)=
c
2
θ
+
c
3
.
Θ(0) = 0 still implies
c
3
= 0, whereas boundedness as
r
→
0 demands
c
2
= 0.Thus, a product solution is
u
=
c
1
c
2
θ
=
Aθ.
From
u
(
r, π/
4) = 0 we obtain
A
= 120
/π
.Thus, a solution to the problem is
u
(
r, θ
)=
120
π
θ.
672

673. Exercises 14.1
9.
Proceeding as in Example 1 in the text and again using the periodicity of
u
(
r, θ
), we have
Θ(
θ
)=
c
1
cos
λθ
+
c
2
sin
λθ
where
λ
=
n
for
n
=0,1,2,
...
.Then
R
(
r
)=
c
3
r
n
+
c
4
r
−
n
.
[We do not have
c
4
= 0 in this case since 0
<a
≤
r
.] Since
u
(
b, θ
)=0wehave
u
(
r, θ
)=
A
0
ln
r
b
+
∞
e
n
=1
h2
b
r

n
−
=
r
b
1
n
a
[
A
n
cos
nθ
+
B
n
sin
nθ
]
.
From
u
(
a, θ
)=
f
(
θ
)=
A
0
ln
a
b
+
∞
e
n
=1
h2
b
a

n
−
=
a
b
1
n
a
[
A
n
cos
nθ
+
B
n
sin
nθ
]
we find
A
0
ln
a
b
=
1
2
π
W
2
π
0
f
(
θ
)
dθ,
h2
b
a

n
−
=
a
b
1
n
a
A
n
=
1
π
W
2
π
0
f
(
θ
) cos
nθ dθ,
and
h2
b
a

n
−
=
a
b
1
n
a
B
n
=
1
π
W
2
π
0
f
(
θ
) sin
nθ dθ.
10.
Substituting
u
(
r, θ
)=
v
(
r, θ
)+
ψ
(
r
) into the partial differential equation we obtain
∂
2
v
∂r
2
+
ψ
ee
(
r
)+
1
r
h
∂v
∂r
+
ψ
e
(
r
)
a
+
1
r
2
∂
2
v
∂θ
2
=0
.
This equation will be homogeneous provided
ψ
ee
(
r
)+
1
r
ψ
e
(
r
)=0 or
r
2
ψ
ee
(
r
)+
rψ
e
(
r
)=0
.
The general solution of this Cauchy-Euler differential equation is
ψ
(
r
)=
c
1
+
c
2
ln
r.
From
u
0
=
u
(
a, θ
)=
v
(
a, θ
)+
ψ
(
a
) and
u
1
=
u
(
b, θ
)=
v
(
b, θ
)+
ψ
(
b
)
we see that in order for the boundary values
v
(
a, θ
) and
v
(
b, θ
) to be 0 we need
ψ
(
a
)=
u
0
and
ψ
(
b
)=
u
1
.From
this we have
ψ
(
a
)=
c
1
+
c
2
ln
a
=
u
0
ψ
(
b
)=
c
1
+
c
2
ln
b
=
u
1
.
Solving for
c
1
and
c
2
we obtain
c
1
=
u
1
ln
a
−
u
0
ln
b
ln
a/b
and
c
2
=
u
0
−
u
1
ln
a/b
.
Then
ψ
(
r
)=
u
1
ln
a
−
u
0
ln
b
ln
a/b
+
u
0
−
u
1
ln
a/b
ln
r
=
u
0
ln
r/b
−
u
1
ln
r/a
ln
a/b
.
673

674. Exercises 14.1
From Problem 9 with
f
(
θ
) = 0 we see that the solution of
∂
2
v
∂r
2
+
1
r
∂v
∂r
+
1
r
2
∂
2
v
∂θ
2
=0
,
0
<θ<
2
π, a < r < b,
v
(
a, θ
)=0
,v
(
b, θ
)=0
,
0
<θ<
2
π
is
v
(
r, θ
) = 0.Thus the steady-state temperature of the ring is
u
(
r, θ
)=
v
(
r, θ
)+
ψ
(
r
)=
u
0
ln
r/b
−
u
1
ln
r/a
ln
a/b
.
11.
We solve
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
=0
,
0
<θ<π, a<r<b,
u
(
a, θ
)=
θ
(
π
−
θ
)
,u
(
b, θ
)=0
,
0
<θ<π,
u
(
r,
0) = 0
,u
(
r, π
)=0
,a<r<b.
Proceeding as in Example 1 in the text we obtain the separated differential equations
r
2
R
ee
+
rR
e
−
λ
2
R
=0
Θ
ee
+
λ
2
Θ=0
with solutions
Θ(
θ
)=
c
1
cos
λθ
+
c
2
sin
λθ
R
(
r
)=
c
3
r
λ
+
c
4
r
−
λ
.
Applying the boundary conditions Θ(0) = 0 and Θ(
π
) = 0 we find that
c
1
= 0 and
λ
=
n
for
n
=1,2,3,
...
.The boundary condition
R
(
b
) = 0 gives
c
3
b
n
+
c
4
b
−
n
= 0 and
c
4
=
−
c
3
b
2
n
.
Then
R
(
r
)=
c
3

r
n
−
b
2
n
r
n

=
c
3

r
2
n
−
b
2
n
r
n

and
u
(
r, θ
)=
∞
e
n
=1
A
n

r
2
n
−
b
2
n
r
n

sin
nθ.
From
u
(
a, θ
)=
θ
(
π
−
θ
)=
∞
e
n
=1
A
n

a
2
n
−
b
2
n
a
n

sin
nθ
we find
A
n

a
2
n
−
b
2
n
a
n

=
2
π
W
π
0
(
θπ
−
θ
2
) sin
nθ dθ
=
4
n
3
π
[1
−
(
−
1)
n
]
.
Thus
u
(
r, θ
)=
4
π
∞
e
n
=1
1
−
(
−
1)
n
n
3
r
2
n
−
b
2
n
a
2
n
−
b
2
n
=
a
r
1
n
sin
nθ.
12.
Letting
u
(
r, θ
)=
v
(
r, θ
)+
ψ
(
θ
) we obtain
ψ
ee
(
θ
) = 0 and so
ψ
(
θ
)=
c
1
θ
+
c
2
.From
ψ
(0) = 0 and
ψ
(
π
)=
u
0
we find, in turn,
c
2
= 0 and
c
1
=
u
0
/π
.Therefore
ψ
(
θ
)=
u
0
π
θ
.
Now
u
(1
,θ
)=
v
(1
,θ
)+
ψ
(
θ
) so that
v
(1
,θ
)=
u
0
−
u
0
π
θ
.From
v
(
r, θ
)=
∞
e
n
=1
A
n
r
n
sin
nθ
and
v
(1
,θ
)=
∞
e
n
=1
A
n
sin
nθ
674

675. Exercises 14.1
we obtain
A
n
=
2
π
W
π
0
=
u
0
−
u
0
π
θ
1
sin
nθ dθ
=
2
u
0
πn
.
Thus
u
(
r, θ
)=
u
0
π
θ
+
2
u
0
π
∞
e
n
=1
r
n
n
sin
nθ.
13.
We solve
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
=0
,
0
<θ<π,
0
<r<
2
,
u
(2
,θ
)=
v
u
0
,
0
<θ<π/
2
0
,π/
2
<θ<π
∂u
∂θ




θ
=0
=0
,
∂u
∂θ




θ
=
π
=0
,
0
<r<
2
.
Proceeding as in Example 1 in the text we obtain the separated differential equations
r
2
R
ee
+
rR
e
−
λ
2
R
=0
Θ
ee
+
λ
2
Θ=0
with solutions
Θ(
θ
)=
c
1
cos
λθ
+
c
2
sin
λθ
R
(
r
)=
c
3
r
λ
+
c
4
r
−
λ
.
Applying the boundary conditions Θ
e
(0) = 0 and Θ
e
(
π
) = 0 we find that
c
2
= 0 and
λ
=
n
for
n
=0,1,2,
...
.Since we want
R
(
r
) to be bounded as
r
→
0 we require
c
4
= 0.Thus
u
(
r, θ
)=
A
0
+
∞
e
n
=1
A
n
r
n
cos
nθ.
From
u
(2
,θ
)=
v
u
0
,
0
<θ<π/
2
0
,π/
2
<θ<π
=
A
0
+
∞
e
n
=1
A
n
2
n
cos
nθ
we find
A
0
=
1
2
2
π
W
π/
2
0
u
0
dθ
=
u
0
2
and
2
n
A
n
=
2
u
0
π
W
π/
2
0
cos
nθ dθ
=
2
u
0
π
sin
nπ/
2
n
.
Therefore
u
(
r, θ
)=
u
0
2
+
2
u
0
π
∞
e
n
=1
1
n
=
sin
nπ
2
1=
r
2
1
n
cos
nθ.
14.
Let
u
1
be the solution of the boundary-value problem
∂
2
u
1
∂r
2
+
1
r
∂u
1
∂r
+
1
r
2
∂
2
u
1
∂θ
2
=0
,
0
<θ<π, a<r<b
u
1
(
a, θ
)=
f
(
θ
)
,
0
<θ<π
u
1
(
b, θ
)=0
,
0
<θ<π,
675

676. 1 2 3 4 5 6
q
20
40
60
80
100
u
r=0.9
r=0.7
r=0.5
r=0.3
r=0.1
Exercises 14.1
and let
u
2
be the solution of the boundary-value problem
∂
2
u
2
∂r
2
+
1
r
∂u
2
∂r
+
1
r
2
∂
2
u
1
∂θ
2
=0
,
0
<θ<π, a<r<b
u
2
(
a, θ
)=0
,
0
<θ<π
u
2
(
b, θ
)=
g
(
θ
)
,
0
<θ<π.
Each of these problems can be solved using the methods shown in Problem 9 of this section.Now if
u
(
r, θ
)=
u
1
(
t, θ
)+
u
2
(
r, θ
), then
u
(
a, θ
)=
u
1
(
a, θ
)+
u
2
(
a, θ
)=
f
(
θ
)
u
(
b, θ
)=
u
1
(
b, θ
)+
u
2
(
b, θ
)=
g
(
θ
)
and
u
(
r, θ
) will be the steady-state temperature of the circular ring with boundary conditions
u
(
a, θ
)=
f
(
θ
)
and
u
(
b, θ
)=
g
(
θ
).
15. (a)
From Problem 1 in this section, with
u
0
= 100,
u
(
r, θ
) = 50 +
100
π
∞
e
n
=1
1
−
(
−
1)
n
n
r
n
sin
nθ.
(b)
(c)
Using a partial sum including the term with
r
99
we find
u
(0
.
9
,
1
.
3)
≈
96
.
5268
u
(0
.
9
,
2
π
−
1
.
3)
≈
3
.
4731
u
(0
.
7
,
2)
≈
87
.
871
u
(0
.
7
,
2
π
−
2)
≈
12
.
129
u
(0
.
5
,
3
.
5)
≈
36
.
0744
u
(0
.
5
,
2
π
−
3
.
5)
≈
63
.
9256
u
(0
.
3
,
4)
≈
35
.
2674
u
(0
.
3
,
2
π
−
4)
≈
64
.
7326
u
(0
.
1
,
5
.
5)
≈
45
.
4934
u
(0
.
1
,
2
π
−
5
.
5)
≈
54
.
5066
(d)
At the center of the plate
u
(0
,
0) = 50.From the graphs in part (b) we observe that the solution curves
are symmetric about the point (
π,
50).In part (c) we observe that the horizontal pairs add up to 100, and
hence average 50.This is consistent with the observation about part (b), so it is appropriate to say the
average temperature in the plate is 50
◦
.
676

677. Exercises 14.2
Exercises 14.2
1.
Referring to the solution of Example 1 in the text we have
R
(
r
)=
c
1
J
0
(
λ
n
r
) and
T
(
t
)=
c
3
cos
aλ
n
t
+
c
4
sin
aλ
n
t
where the eigenvalues
λ
n
are the positive roots of
J
0
(
λc
) = 0.Now, the initial condition
u
(
r,
0) =
R
(
r
)
T
(0) = 0
implies
c
3
= 0.Thus
u
(
r, t
)=
∞
e
n
=1
A
n
sin
aλ
n
tJ
0
(
λ
n
r
)
∂u
∂t
=
∞
e
n
=1
aλ
n
A
n
cos
aλ
n
tJ
0
(
λ
n
r
)
.
From
∂u
∂t




t
=0
=1=
∞
e
n
=1
aλ
n
A
n
J
0
λ
n
r
we find
aλ
n
A
n
=
2
c
2
J
2
1
(
λ
n
c
)
W
c
0
rJ
0
(
λ
n
r
)
dr
x
=
λ
n
r, dx
=
λ
n
dr
=
2
c
2
J
2
1
(
λ
n
c
)
W
λ
n
c
0
1
λ
2
n
xJ
0
(
x
)
dx
=
2
c
2
J
2
1
(
λ
n
c
)
W
λ
n
c
0
1
λ
2
n
d
dx
[
xJ
1
(
x
)]
dx
see (4) of Section 11
.
5 in text
=
2
c
2
λ
2
n
J
2
1
(
λ
n
c
)

xJ
1
(
x
)





λ
n
c
0
=
2
cλ
n
J
1
(
λ
n
c
)
.
Then
A
n
=
2
acλ
2
n
J
1
(
λ
n
c
)
and
u
(
r, t
)=
2
ac
∞
e
n
=1
J
0
(
λ
n
r
)
λ
2
n
J
1
(
λ
n
c
)
sin
aλ
n
t.
2.
From Example 1 in the text we have
B
n
= 0 and
A
n
=
2
J
2
1
(
λ
n
)
W
1
0
r
(1
−
r
2
)
J
0
(
λ
n
r
)
dr.
From Problem 10, Exercises 12
.
6 we obtained
A
n
=
4
J
2
(
λ
n
)
λ
2
n
J
2
1
(
λ
n
)
.Thus
u
(
r, t
)=4
∞
e
n
=1
J
2
(
λ
n
)
J
2
1
(
λ
n
)
cos
aλ
n
tJ
0
(
λ
n
r
)
.
3.
Referring to Example 2 in the text we have
R
(
r
)=
c
1
J
0
(
λr
)+
c
2
Y
0
(
λr
)
Z
(
z
)=
c
3
cosh
λz
+
c
4
sinh
λz
where
c
2
= 0 and
J
0
(2
λ
) = 0 defines the positive eigenvalues
λ
n
.From
Z
(4) = 0 we obtain
c
3
cosh 4
λ
n
+
c
4
sinh 4
λ
n
=0 or
c
4
=
−
c
3
cosh 4
λ
n
sinh 4
λ
n
.
677

678. Exercises 14.2
Then
Z
(
z
)=
c
3
h
cosh
λ
n
z
−
cosh 4
λ
n
sinh 4
λ
n
sinh
λ
n
z
a
=
c
3
sinh 4
λ
n
cosh
λ
n
z
−
cosh 4
λ
n
sinh
λ
n
z
sinh 4
λ
n
=
c
3
sinh
λ
n
(4
−
z
)
sinh 4
λ
n
and
u
(
r, z
)=
∞
e
n
=1
A
n
sinh
λ
n
(4
−
z
)
sinh 4
λ
n
J
0
(
λ
n
r
)
.
From
u
(
r,
0) =
u
0
=
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
we obtain
A
n
=
2
u
0
4
J
2
1
(2
λ
n
)
W
2
0
rJ
0
(
λ
n
r
)
dr
=
u
0
λ
n
J
1
(2
λ
n
)
.
Thus the temperature in the cylinder is
u
(
r, z
)=
u
0
∞
e
n
=1
sinh
λ
n
(4
−
z
)
J
0
(
λ
n
r
)
λ
n
sinh 4
λ
n
J
1
(2
λ
n
)
.
4. (a)
The boundary condition
u
r
(2
,z
) = 0 implies
R
e
(2) = 0 or
J
e
0
(2
λ
) = 0.Thus
λ
= 0 is also an eigenvalue
and the separated equations are in this case
rR
ee
+
R
e
= 0 and
z
ee
= 0.The solutions of these equations are
then
R
(
r
)=
c
1
+
c
2
ln
r, Z
(
z
)=
c
3
z
+
c
4
.
Now
Z
(0) = 0 yields
c
4
= 0 and the implicit condition that the temperature is bounded as
r
→
0 demands
that we define
c
2
= 0.Using the superposition principle then gives
u
(
r, z
)=
A
1
z
+
∞
e
n
=2
A
n
sinh
λ
n
zJ
0
(
λ
n
r
)
.
(1)
At
z
= 4 we obtain
f
(
r
)=4
A
1
+
∞
e
n
=2
A
n
sinh 4
λ
n
J
0
(
λ
n
r
)
.
Thus from (19) and (20) of Section 12
.
6 in the text we can write with
b
=2,
A
1
=
1
8
W
2
0
rf
(
r
)
dr
(2)
A
n
=
1
2 sinh 4
λ
n
J
2
0
(2
λ
n
)
W
2
0
rf
(
r
)
J
0
(
λ
n
r
)
dr.
(3)
A solution of the problem consists of the series (1) with coefficients
A
1
and
A
n
defined in (2) and (3),
respectively.
(b)
When
f
(
r
)=
u
0
we get
A
1
=
u
0
/
4 and
A
n
=
u
0
J
1
(2
λ
n
)
λ
n
sinh 4
λ
n
J
2
0
(2
λ
n
)
=0
since
J
e
0
(2
λ
) = 0 is equivalent to
J
1
(2
λ
) = 0.A solution of the problem is then
u
(
r, z
)=
u
0
4
z
.
678

679. Exercises 14.2
5.
Letting
u
(
r, t
)=
R
(
r
)
T
(
t
) and separating variables we obtain
R
ee
+
1
r
R
e
R
=
T
e
kT
=
µ
and
R
ee
+
1
r
R
e
−
µR
=0
,T
e
−
µkT
=0
.
From the second equation we find
T
(
t
)=
e
µkt
.If
µ>
0,
T
(
t
) increases without bound as
t
→∞
.Thus we
assume
µ
=
−
λ
2
≤
0.Now
R
ee
+
1
r
R
e
+
λ
2
R
=0
is a parametric Bessel equation with solution
R
(
r
)=
c
1
J
0
(
λr
)+
c
2
Y
0
(
λr
)
.
Since
Y
0
is unbounded as
r
→
0 we take
c
2
= 0.Then
R
(
r
)=
c
1
J
0
(
λr
) and the boundary condition
u
(
c, t
)=
R
(
c
)
T
(
t
) = 0 implies
J
0
(
λc
) = 0.This latter equation defines the positive eigenvalues
λ
n
.Thus
u
(
r, t
)=
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
e
−
λ
2
n
kt
.
From
u
(
r,
0) =
f
(
r
)=
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
we find
A
n
=
2
c
2
J
2
1
(
λ
n
c
)
W
c
0
rJ
0
(
λ
n
r
)
f
(
r
)
dr, n
=1
,
2
,
3
,... .
6.
If the edge
r
=
c
is insulated we have the boundary condition
u
r
(
c, t
) = 0.Referring to the solution of
Problem 5 above we have
R
e
(
c
)=
λc
1
J
e
0
(
λc
)=0
which defines an eigenvalue
λ
= 0 and positive eigenvalues
λ
n
.Thus
u
(
r, t
)=
A
0
+
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
e
−
λ
2
n
kt
.
From
u
(
r,
0) =
f
(
r
)=
A
0
+
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
we find
A
0
=
2
c
2
W
c
0
rf
(
r
)
dr
A
n
=
2
c
2
J
2
0
(
λ
n
c
)
W
c
0
rJ
0
(
λ
n
r
)
f
(
r
)
dr.
7.
Referring to Problem 5 above we have
T
(
t
)=
e
−
λ
2
kt
and
R
(
r
)=
c
1
J
0
(
λr
).The boundary condition
hu
(1
,t
)+
u
r
(1
,t
) = 0 implies
hJ
0
(
λ
)+
λJ
e
0
(
λ
) = 0 which defines positive eigenvalues
λ
n
.Now
u
(
r, t
)=
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
e
−
λ
2
n
kt
where
A
n
=
2
λ
2
n
(
λ
2
n
+
h
2
)
J
2
0
(
λ
n
)
W
1
0
rJ
0
(
λ
n
r
)
f
(
r
)
dr.
679

680. Exercises 14.2
8.
We solve
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
∂
2
u
∂z
2
=0
,
0
<r<
1
,z>
0
∂u
∂r




r
=1
=
−
hu
(1
,z
)
,z>
0
u
(
r,
0) =
u
0
,
0
<r<
1
.
assuming
u
=
RZ
we get
R
ee
+
1
r
R
e
R
=
−
Z
ee
Z
=
−
λ
2
and so
rR
ee
+
R
e
+
λ
2
rR
= 0 and
Z
ee
−
λ
2
Z
=0
.
Therefore
R
(
r
)=
c
1
J
0
(
λr
)+
c
2
Y
0
(
λr
) and
Z
(
z
)=
c
3
e
−
λz
+
c
4
e
λz
.
We use the exponential form of the solution of
Z
ee
−
λ
2
Z
= 0 since the domain of the variable
z
is a semi-infinite
interval.As usual we define
c
2
= 0 since the temperature is surely bounded as
r
→
0.Hence
R
(
r
)=
c
1
J
0
(
λr
).
Now the boundary-condition
u
r
(1
,z
)+
hu
(1
,z
) = 0 is equivalent to
λJ
e
0
(
λ
)+
hJ
0
(
λ
)=0
.
(4)
The eigenvalues
λ
n
are the positive roots of (4).Finally, we must now define
c
4
= 0 since the temperature is
also expected to be bounded as
z
→∞
.A product solution of the partial differential equation that satisfies the
first boundary condition is given by
u
n
(
r, z
)=
A
n
e
−
λ
n
z
J
0
(
λ
n
r
)
.
Therefore
u
(
r, z
)=
∞
e
n
=1
A
n
e
−
λ
n
z
J
0
(
λ
n
r
)
is another formal solution.At
z
= 0 we have
u
0
=
A
n
J
0
(
λ
n
r
).In view of (4) we use (17) and (18) of Section
12
.
6 in the text with the identification
b
=1:
A
n
=
2
λ
2
n
(
λ
2
n
+
h
2
)
J
2
0
(
λ
n
)
W
1
0
rJ
0
(
λ
n
r
)
u
0
dr
=
2
λ
2
n
u
0
(
λ
2
n
+
h
2
)
J
2
0
(
λ
n
)
λ
2
n
tJ
1
(
t
)




λ
n
0
=
2
λ
n
u
0
J
1
(
λ
n
)
(
λ
2
n
+
h
2
)
J
2
0
(
λ
n
)
.
(5)
Since
J
e
0
=
−
J
1
[see (5) of Section 12
.
6] it follows from (4) that
λ
n
J
1
(
λ
n
)=
hJ
0
(
λ
n
).Thus (5) simplifies to
A
n
=
2
u
0
h
(
λ
2
n
+
h
2
)
J
2
0
(
λ
n
)
.
A solution to the boundary-value problem is then
u
(
r, z
)=2
u
0
h
∞
e
n
=1
e
−
λ
n
z
(
λ
2
n
+
h
2
)
J
0
(
λ
n
)
J
0
(
λ
n
r
)
.
9.
Substituting
u
(
r, t
)=
v
(
r, t
)+
ψ
(
r
) into the partial differential equation gives
∂
2
v
∂r
2
+
1
r
∂v
∂r
+
ψ
ee
+
1
r
ψ
e
=
∂v
∂t
.
680

681. Exercises 14.2
This equation will be homogeneous provided
ψ
ee
+
1
r
ψ
e
=0or
ψ
(
r
)=
c
1
ln
r
+
c
2
.Since ln
r
is unbounded as
r
→
0 we take
c
1
= 0.Then
ψ
(
r
)=
c
2
and using
u
(2
,t
)=
v
(2
,t
)+
ψ
(2) = 100 we set
c
2
=
ψ
(
r
) = 100.Referring
to Problem 5 above, the solution of the boundary-value problem
∂
2
v
∂r
2
+
1
r
∂v
∂r
=
∂v
∂t
,
0
<r<
2
,t>
0
,
v
(2
,t
)=0
,t>
0
,
v
(
r,
0) =
u
(
r,
0)
−
ψ
(
r
)
is
v
(
r, t
)=
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
e
−
λ
2
n
t
where
A
n
=
2
2
2
J
2
1
(2
λ
n
)
W
2
0
rJ
0
(
λ
n
r
)[
u
(
r,
0)
−
ψ
(
r
)]
dr
=
1
2
J
2
1
(2
λ
n
)
i
W
1
0
rJ
0
(
λ
n
r
)[200
−
100]
dr
+
W
2
1
rJ
0
(
λ
n
r
)[100
−
100]
dr
n
=
50
J
2
1
(2
λ
n
)
W
1
0
rJ
0
(
λ
n
r
)
dr
x
=
λ
n
r, dx
=
λ
n
dr
=
50
J
2
1
(2
λ
n
)
W
λ
n
0
1
λ
2
n
xJ
0
(
x
)
dx
=
50
λ
2
n
J
2
1
(2
λ
n
)
W
λ
n
0
d
dx
[
xJ
1
(
x
)]
dx
see (4) of Section 12.6 in text
=
50
λ
2
n
J
2
1
(2
λ
n
)

xJ
1
(
x
)





λ
n
0
=
50
J
1
(
λ
n
)
λ
n
J
2
1
(2
λ
n
)
.
Thus
u
(
r, t
)=
v
(
r, t
)+
ψ
(
r
) = 100 + 50
∞
e
n
=1
J
1
(
λ
n
)
J
0
(
λ
n
r
)
λ
n
J
2
1
(2
λ
n
)
e
−
λ
2
t
.
10.
Letting
u
(
r, t
)=
u
(
r, t
)+
ψ
(
r
) we obtain
rψ
ee
+
ψ
e
=
−
βr
.The general solution of this nonhomogeneous equation
is found with the aid of variation of parameters:
ψ
=
c
1
+
c
2
ln
r
−
β
r
2
4
.In order that this solution be bounded
as
r
→
0 we define
c
2
= 0.Using
ψ
(1) = 0 then gives
c
1
=
β
4
and so
ψ
(
r
)=
β
4
(1
−
r
2
).Using
v
=
RT
we find
that a solution of
∂
2
v
∂r
2
+
1
r
∂v
∂r
=
∂v
∂t
,
0
<r<
1
,t>
0
v
(1
,t
)=0
,t>
0
v
(
r,
0) =
−
ψ
(
r
)
,
0
<r<
1
is
v
(
r, t
)=
∞
e
n
=1
A
n
e
−
λ
2
n
t
J
0
(
λ
n
r
)
where
A
n
=
−
β
4
2
J
2
1
(
λ
n
)
W
1
0
r
(1
−
r
2
)
J
0
(
λ
n
r
)
dr
681

682. Exercises 14.2
and the eigenvalues are defined by
J
0
(
λ
) = 0.From the result of Problem 10, Exercises 12
.
6 (see also
Problem 2 of this exercise set) we get
A
n
=
−
βJ
2
(
λ
n
)
λ
2
n
J
2
1
(
λ
n
)
.
Thus from
u
=
v
+
ψ
(
r
) it follows that
u
(
r, t
)=
β
4
(1
−
r
2
)
−
β
∞
e
n
=1
J
2
(
λ
n
)
λ
2
n
J
2
1
(
λ
n
)
e
−
λ
2
n
t
J
0
(
λ
n
r
)
.
11. (a)
Writing the partial differential equation in the form
g

x
∂
2
u
∂x
2
+
∂u
∂x

=
∂
2
u
∂t
2
and separating variables we obtain
xX
ee
+
X
e
X
=
T
ee
gT
=
−
λ
2
.
Then
xX
ee
+
X
e
+
λ
2
X
= 0 and
T
ee
+
gλ
2
T
=0
.
Letting
x
=
τ
2
/
4 in the first equation we obtain
dx/dτ
=
τ/
2or
dτ/dx
=2
τ
.Then
dX
dx
=
dX
dτ
dτ
dx
=
2
τ
dX
dτ
and
d
2
X
dx
2
=
d
dx

2
τ
dX
dτ

=
2
τ
d
dx

dX
dτ

+
dX
dτ
d
dx

2
τ

=
2
τ
d
dτ

dX
dτ

dτ
dx
+
dX
dτ
d
dτ

2
τ

dτ
dx
=
4
τ
2
d
2
X
dτ
2
−
4
τ
3
dX
dτ
.
Thus
xX
ee
+
X
e
+
λ
2
X
=
τ
2
4

4
τ
2
d
2
X
dτ
2
−
4
τ
3
dX
dτ

+
2
τ
dX
dτ
+
λ
2
X
=
d
2
X
dτ
2
+
1
τ
dX
dτ
+
λ
2
X
=0
.
This is a parametric Bessel equation with solution
X
(
τ
)=
c
1
J
0
(
λτ
)+
c
2
Y
0
(
λτ
)
.
(b)
To insure a finite solution at
x
= 0 (and thus
τ
= 0) we set
c
2
= 0.The condition
u
(
L, t
)=
X
(
L
)
T
(
t
)=0
implies
X


x
=
L
=
X


τ
=2
√
L
=
c
1
J
0
(2
λ
√
L
) = 0, which defines positive eigenvalues
λ
n
.The solution of
T
ee
+
gλ
2
T
=0is
T
(
t
)=
c
3
cos
λ
n
√
gt
+
c
4
sin
λ
n
√
gt.
The boundary condition
u
t
(
x,
0) =
X
(
x
)
T
e
(0) = 0 implies
c
4
= 0.Thus
u
(
τ,t
)=
∞
e
n
=1
A
n
cos
λ
n
√
gtJ
0
(
λ
n
τ
)
.
From
u
(
τ,
0) =
f
(
τ
2
/
4) =
∞
e
n
=1
A
n
J
0
(
λ
n
τ
)
682

683. 2 4 6 8 10
r
-1
-0.5
0.5
1
u
t=0
t=4
t=10
t=12
t=20
Exercises 14.2
we find
A
n
=
2
(2
√
L
)
2
J
2
1
(2
λ
n
√
L
)
W
2
√
L
0
τJ
0
(
λ
n
τ
)
f
(
τ
2
/
4)
dτ
v
=
τ/
2
,dv
=
dτ/
2
=
1
2
LJ
2
1
(2
λ
n
√
L
)
W
√
L
0
2
vJ
0
(2
λ
n
v
)
f
(
v
2
)2
dv
=
2
LJ
2
1
(2
λ
n
√
L
)
W
√
L
0
vJ
0
(2
λ
n
v
)
f
(
v
2
)
dv.
12. (a)
We need to solve
J
0
(10
λ
) = 0. From the Table 5.2 in Section 5.3 in the text we see that
λ
1
=0
.
2405,
λ
2
=0
.
5520, and
λ
3
=0
.
8654.Using a CAS to compute
A
1
we find:
A
1
=
2
10
2
J
2
1
(10
λ
1
)
W
10
0
rJ
0
(
λ
1
r
)(1
−
r/
10)
dr
=0
.
7845
.
In a similar fashion we compute
A
2
=0
.
0687 and
A
3
=0
.
0531.Then, from Example 1 in the text with
a
= 1 and
g
(
r
) = 0 [so that
B
n
=0]wehave
u
(
r, t
)
≈
3
e
n
=1
A
n
(cos
λ
n
t
)
J
0
(
λ
n
r
)
=0
.
7845 cos(0
.
2405
t
)
J
0
(0
.
2405
r
)+0
.
0687 cos(0
.
5520
t
)
J
0
(0
.
5520
r
)
+0
.
0531 cos(0
.
8654
t
)
J
0
(0
.
8654
r
)
.
(b)
13.
Because of the nonhomogeneous boundary condition
u
(
c, t
) = 200 we use the substitution
u
(
r, t
)=
v
(
r, t
)+
ψ
(
r
).
This gives
k

∂
2
v
∂r
2
+
1
r
∂v
∂r
+
ψ
ee
+
1
r
ψ
e

=
∂v
∂t
.
This equation will be homogeneous provided
ψ
ee
+(1
/r
)
ψ
e
=0or
ψ
(
r
)=
c
1
ln
r
+
c
2
.Since ln
r
is unbounded as
r
→
0 we take
c
1
= 0.Then
ψ
(
r
)=
c
2
and using
u
(
c, t
)=
v
(
c, t
)+
c
2
= 200 we set
c
2
= 200, giving
v
(
c, t
)=0.
Referring to Problem 5 in this section, the solution of the boundary-value problem
k

∂
2
v
∂r
2
+
1
r
∂v
∂r

=
∂v
∂t
,
0
<r<c, t>
0
v
(
c, t
)=0
,t>
0
v
(
r,
0) =
−
200
,
0
<r<c
683

684. 200400600800
t
50
100
150
200
u
r=0
r=5
Exercises 14.2
is
v
(
r, t
)=
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
e
−
λ
2
n
kt,
where the eigenvalues
λ
n
are defined by
J
0
(
λc
) = 0, and
A
n
=
2
c
2
J
2
1
(
λ
n
c
)
W
c
0
rJ
0
(
λ
n
r
)(
−
200)
dr
=
−
400
c
2
J
2
1
(
λ
n
c
)
W
c
0
rJ
0
(
λ
n
r
)
dr.
Taking
c
= 10 and
k
=0
.
1wehave
u
(
r, t
)=
v
(
r, t
) + 200 = 200 +
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
e
−
0
.
1
λn
2
t
,
where the eigenvalues
λ
n
are defined by
J
0
(10
λ
) = 0 and
A
n
=
−
4
J
2
1
(10
λ
n
)
W
10
0
rJ
0
(
λ
n
r
)
dr.
Using a CAS we find that the first five eigenvalues are
λ
1
=0
.
240483,
λ
2
=0
.
552008,
λ
3
=0
.
865373,
λ
4
=1
.
17915, and
λ
5
=1
.
49309.From
the graph we see that
u
(5
,t
)
≈
100 when
t
= 132,
u
(0
,t
)
≈
100 when
t
= 198,
u
(5
,t
)
≈
200 when
t
= 808, and
u
(0
,t
)
≈
200 when
t
= 885.
14. (a)
The boundary-value problem is
a
2

∂
2
u
∂r
2
+
1
r
∂u
∂r

=
∂
2
u
∂t
2
,
0
<r<
1
,t>
0
u
(1
,t
)=0
,t>
0
u
(
r,
0) = 0
,
∂u
∂t



t
=0
=
v
−
v
0
,
0
≤
r<b
0
,b
≤
r<
1
,
0
<r<
1
,
and the solution is
u
(
r, t
)=
∞
e
n
=1
(
A
n
cos
aλ
n
t
+
B
n
sin
aλ
n
t
)
J
0
(
λ
n
r
)
,
where the eigenvalues
λ
n
are defined by
J
0
(
λ
)=0and
A
n
= 0 since
f
(
r
) = 0.The coefficients
B
n
are
given by
B
n
=
2
aλ
n
J
2
1
(
λ
n
)
W
c
0
rJ
0
(
λ
n
r
)
g
(
r
)
dr
=
−
2
v
0
aλ
n
J
2
1
(
λ
n
)
W
b
0
rJ
0
(
λ
n
r
)
dr
let
x
=
λ
n
r
=
−
2
v
0
aλ
n
J
2
1
(
λ
n
)
W
λ
n
b
0
x
λ
n
J
0
(
x
)
1
λ
n
dx
=
−
2
v
0
aλ
3
n
J
2
1
(
λ
n
)
W
λ
n
b
0
xJ
0
(
x
)
dx
=
−
2
v
0
aλ
3
n
J
2
1
(
λ
n
)

xJ
1
(
x
)




λ
n
b
0
=
−
2
v
0
aλ
3
n
J
1
(
λ
n
)

λ
n
bJ
1
(
λ
n
b
)

=
−
2
v
0
b
aλ
2
n
J
1
(
λ
n
b
)
J
2
1
(
λ
n
)
.
Thus,
u
(
r, t
)=
−
2
v
0
b
a
∞
e
n
=1
1
λ
2
n
J
1
(
λ
n
b
)
J
2
1
(
λ
n
)
sin(
aλ
n
t
)
J
0
(
λ
n
r
)
.
684

685. -1-0.5 0.5 1
r
-0.1
-0.075
-0.05
-0.025
0.025
0.05
0.075
0.1
u
-1-0.5 0.5 1
r
-0.1
-0.075
-0.05
-0.025
0.025
0.05
0.075
0.1
u
-1-0.5 0.5 1
r
-0.1
-0.075
-0.05
-0.025
0.025
0.05
0.075
0.1
u
-1-0.5 0.5 1
r
-0.1
-0.075
-0.05
-0.025
0.025
0.05
0.075
0.1
u
-1-0.5 0.5 1
r
-0.1
-0.075
-0.05
-0.025
0.025
0.05
0.075
0.1
u
-1-0.5 0.5 1
r
-0.1
-0.075
-0.05
-0.025
0.025
0.05
0.075
0.1
u
Exercises 14.2
(b)
The standing wave
u
n
(
r, t
) is given by
u
n
(
r, t
)=
B
n
sin(
aλ
n
t
)
J
0
(
λ
n
r
), which has frequency
f
n
=
aλ
n
/
2
π
,
where
λ
n
is the
n
th positive zero of
J
0
(
x
).The fundamental frequency is
f
1
=
aλ
1
/
2
π
.The next two
frequencies are
f
2
=
aλ
2
2
π
=
λ
2
λ
1

aλ
1
2
π

=
5
.
520
2
.
405
f
1
=2
.
295
f
1
and
f
3
=
aλ
3
2
π
=
λ
3
λ
1

aλ
1
2
π

=
8
.
654
2
.
405
f
1
=3
.
598
f
1
.
(c)
With
a
=1,
b
=
1
4
, and
v
0
= 1, the solution becomes
u
(
r, t
)=
−
1
2
∞
e
n
=1
1
λ
2
n
J
1
(
λ
n
/
4)
J
1
(
λ
n
)
sin(
λ
n
t
)
J
0
(
λ
n
r
)
.
The graphs of
S
5
(
r, t
) for
t
= 1, 2, 3, 4, 5, 6 are shown below.
(d)
Three frames from the movie are shown.
15. (a)
First we see that
R
ee
Θ+
1
r
R
e
Θ+
1
r
2
R
Θ
ee
R
Θ
=
T
ee
a
2
T
=
−
λ
2
.
This gives
T
ee
+
a
2
λ
2
T
= 0.Then from
R
ee
+
1
r
R
e
+
λ
2
R
−
R/r
2
=
Θ
ee
Θ
=
−
ν
2
685

686. Exercises 14.2
we get Θ
ee
+
ν
2
Θ = 0 and
r
2
R
ee
+
rR
e
+(
λ
2
r
2
−
ν
2
)
R
=0
.
(b)
The general solutions of the differential equations in part (a) are
T
=
c
1
cos
aλt
+
c
2
sin
aλt
Θ=
c
3
cos
νθ
+
c
4
cos
νθ
R
=
c
5
J
ν
(
λr
)+
c
6
Y
ν
(
λr
)
.
(c)
Implicitly we expect
u
(
r, θ, t
)=
u
(
r, θ
+2
π, t
) and so Θ must be 2
π
-periodic.Therefore
ν
=
n
,
n
=0,1,
2,
...
.The corresponding eigenfunctions are 1, cos
θ
, cos 2
θ
,
...
, sin
θ
, sin 2
θ
,
...
.Arguing that
u
(
r, θ, t
)
is bounded as
r
→
0 we then define
c
6
= 0 and so
R
=
c
3
J
n
(
λr
).But
R
(
c
) = 0 gives
J
n
(
λc
) = 0; this
equation defines the eigenvalues
λ
n
.For each
n
,
λ
ni
=
x
ni
/c
,
i
=1,2,3,
...
.
(d)
u
(
r, θ, t
)=
n
e
i
=1
(
A
0
i
cos
aλ
0
i
t
+
B
0
i
sin
aλ
0
i
t
)
J
0
(
λ
0
i
r
)
+
∞
e
n
=1
∞
e
i
=1
[
(
A
ni
cos
aλ
ni
t
+
B
ni
sin
aλ
ni
t
) cos
nθ
+(
C
ni
cos
aλ
ni
t
+
D
ni
sin
aλ
ni
t
) sin
nθ
(
J
n
(
λ
ni
r
)
Exercises 14.3
1.
To compute
A
n
=
2
n
+1
2
c
n
W
π
0
f
(
θ
)
P
n
(cos
θ
) sin
θdθ
we substitute
x
= cos
θ
and
dx
=
−
sin
θdθ
.Then
A
n
=
2
n
+1
2
c
n
W
−
1
1
F
(
x
)
P
n
(
x
)(
−
dx
)=
2
n
+1
2
c
n
W
1
−
1
F
(
x
)
P
n
(
x
)
dx
where
F
(
x
)=
v
0
,
−
1
<x<
0
50
,
0
<x<
1
=50
v
0
,
−
1
<x<
0
1
,
0
<x<
1
.
The coefficients
A
n
are computed in Example 3 of Section 12
.
6.Thus
u
(
r, θ
)=
∞
e
n
=0
A
n
r
n
P
n
(cos
θ
)
=50
h
1
2
P
0
(cos
θ
)+
3
4

r
c

P
1
(cos
θ
)
−
7
16
=
r
c
1
3
P
3
(cos
θ
)+
11
32
=
r
c
1
5
P
5
(cos
θ
)+
···
a
.
2.
In the solution of the Cauchy-Euler equation,
R
(
r
)=
c
1
r
n
+
c
2
r
−
(
n
+1)
,
we define
c
1
= 0 since we expect the potential
u
to be bounded as
r
→∞
.Hence
u
n
(
r, θ
)=
A
n
r
−
(
n
+1)
P
n
(cos
θ
)
u
(
r, θ
)=
∞
e
n
=0
A
n
r
−
(
n
+1)
P
n
(cos
θ
)
.
686

687. Exercises 14.3
When
r
=
c
we have
f
(
θ
)=
∞
e
n
=0
A
n
c
−
(
n
+1)
P
n
(cos
θ
)
so that
A
n
=
c
n
+1
(2
n
+1)
2
W
π
0
f
(
θ
)
P
n
(cos
θ
) sin
θdθ.
The solution of the problem is then
u
(
r, θ
)=
∞
e
n
=0

2
n
+1
2
W
π
0
f
(
θ
)
P
n
(cos
θ
) sin
θdθ

=
c
r
1
n
+1
P
n
(cos
θ
)
.
3.
The coefficients are given by
A
n
=
2
n
+1
2
c
n
W
π
0
cos
θP
n
(cos
θ
) sin
θdθ
=
2
n
+1
2
c
n
W
π
0
P
1
(cos
θ
)
P
n
(cos
θ
) sin
θdθ
x
= cos
θ, dx
=
−
sin
θdθ
=
2
n
+1
2
c
n
W
1
−
1
P
1
(
x
)
P
n
(
x
)
dx.
Since
P
n
(
x
) and
P
m
(
x
) are orthogonal for
m
1
=
n
,
A
n
= 0 for
n
1
= 1 and
A
1
=
2(1) + 1
2
c
1
W
1
−
1
P
1
(
x
)
P
1
(
x
)
dx
=
3
2
c
W
1
−
1
x
2
dx
=
1
c
.
Thus
u
(
r, θ
)=
r
c
P
1
(cos
θ
)=
r
c
cos
θ.
4.
The coefficients are given by
A
n
=
2
n
+1
2
c
n
W
π
0
(1
−
cos 2
θ
)
P
n
(cos
θ
) sin
θdθ.
These were computed in Problem 18 of Section 12
.
6.Thus
u
(
r, θ
)=
4
3
P
0
(cos
θ
)
−
4
3
=
r
c
1
2
P
2
(cos
θ
)
.
5.
Referring to Example 1 in the text we have
Θ=
P
n
(cos
θ
) and
R
=
c
1
r
n
+
c
2
r
−
(
n
+1)
.
Since
u
(
b, θ
)=
R
(
b
)Θ(
θ
)=0,
c
1
b
n
+
c
2
b
−
(
n
+1)
=0 or
c
1
=
−
c
2
b
−
2
n
−
1
,
and
R
(
r
)=
−
c
2
b
−
2
n
−
1
r
n
+
c
2
r
−
(
n
+1)
=
c
2

b
2
n
+1
−
r
2
n
+1
b
2
n
+1
r
n
+1

.
Then
u
(
r, θ
)=
∞
e
n
=0
A
n
b
2
n
+1
−
r
2
n
+1
b
2
n
+1
r
n
+1
P
n
(cos
θ
)
where
b
2
n
+1
−
a
2
n
+1
b
2
n
+1
a
n
+1
A
n
=
2
n
+1
2
W
π
0
f
(
θ
)
P
n
(cos
θ
) sin
θdθ.
687

688. Exercises 14.3
6.
Referring to Example 1 in the text we have
R
(
r
)=
c
1
r
n
and Θ(
θ
)=
P
n
(cos
θ
)
.
Now Θ(
π/
2) = 0 implies that
n
is odd, so
u
(
r, θ
)=
∞
e
n
=0
A
2
n
+1
r
2
n
+1
P
2
n
+1
(cos
θ
)
.
From
u
(
c, θ
)=
f
(
θ
)=
∞
e
n
=0
A
2
n
+1
c
2
n
+1
P
2
n
+1
(cos
θ
)
we see that
A
2
n
+1
c
2
n
+1
=(4
n
+3)
W
π/
2
0
f
(
θ
) sin
θP
2
n
+1
(cos
θ
)
dθ.
Thus
u
(
r, θ
)=
∞
e
n
=0
A
2
n
+1
r
2
n
+1
P
2
n
+1
(cos
θ
)
where
A
2
n
+1
=
4
n
+3
c
2
n
+1
W
π/
2
0
f
(
θ
) sin
θP
2
n
+1
(cos
θ
)
dθ.
7.
Referring to Example 1 in the text we have
r
2
R
ee
+2
rR
e
−
λ
2
R
=0
sin
θ
Θ
ee
+ cos
θ
Θ
e
+
λ
sin
θ
Θ=0
.
Substituting
x
= cos
θ
,0
≤
θ
≤
π/
2, the latter equation becomes
(1
−
x
2
)
d
2
Θ
dx
2
−
2
x
d
Θ
dx
+
λ
2
Θ=0
,
0
≤
x
≤
1
.
Taking the solutions of this equation to be the Legendre polynomials
P
n
(
x
) corresponding to
λ
2
=
n
(
n
+ 1) for
n
=1,2,3,
...
,wehaveΘ=
P
n
(cos
θ
).Since
∂u
∂θ




θ
=
π/
2
=Θ
e
(
π/
2)
R
(
r
)=0
we have
Θ
e
(
π/
2) =
−
(sin
π/
2)
P
e
n
(cos
π/
2) =
−
P
e
n
(0) = 0
.
As noted in the hint,
P
e
n
(0) = 0 only if
n
is even.Thus Θ =
P
n
(cos
θ
),
n
=0,2,4,
...
.As in Example 1,
R
(
r
)=
c
1
r
n
.Hence
u
(
r, θ
)=
∞
e
n
=0
A
2
n
r
2
n
P
2
n
(cos
θ
)
.
At
r
=
c
,
f
(
θ
)=
∞
e
n
=0
A
2
n
c
2
n
P
2
n
(cos
θ
)
.
Using Problem 19 in Section 12
.
6, we obtain
c
2
n
A
2
n
=(4
n
+1)
W
0
π/
2
f
(
θ
)
P
2
n
(cos
θ
)(
−
sin
θ
)
dθ
and
688

689. Exercises 14.3
A
2
n
=
4
n
+1
c
2
n
W
π/
2
0
f
(
θ
) sin
θP
2
n
(cos
θ
)
dθ.
8.
Referring to Example 1 in the text we have
R
(
r
)=
c
1
r
n
+
c
2
r
−
(
n
−
1)
and Θ(
θ
)=
P
n
(cos
θ
)
.
Since we expect
u
(
r, θ
) to be bounded as
r
→∞
, we define
c
1
= 0.Also Θ(
π/
2) = 0 implies that
n
is odd, so
u
(
r, θ
)=
∞
e
n
=0
A
2
n
+1
r
−
2(
n
+1)
P
2
n
+1
(cos
θ
)
.
From
u
(
c, θ
)=
f
(
θ
)=
∞
e
n
=0
A
2
n
+1
c
−
2(
n
+1)
P
2
n
+1
(cos
θ
)
we see that
A
2
n
+1
c
−
2(
n
+1)
=(4
n
+3)
W
π/
2
0
f
(
θ
) sin
θP
2
n
+1
(cos
θ
)
dθ.
Thus
u
(
r, θ
)=
∞
e
n
=0
A
2
n
+1
r
−
2(
n
+1)
P
2
n
+1
(cos
θ
)
where
A
2
n
+1
=(4
n
+3)
c
2(
n
+1)
W
π/
2
0
f
(
θ
) sin
θP
2
n
+1
(cos
θ
)
dθ.
9.
Checking the hint, we find
1
r
∂
2
∂r
2
(
ru
)=
1
r
∂
∂r
h
r
∂u
∂r
+
u
a
=
1
r
h
r
∂
2
u
∂r
2
+
∂u
∂r
+
∂u
∂r
a
=
∂
2
u
∂r
2
+
2
r
∂u
∂r
.
The partial differential equation then becomes
∂
2
∂r
2
(
ru
)=
r
∂u
∂t
.
Now, letting
ru
(
r, t
)=
v
(
r, t
)+
ψ
(
r
), since the boundary condition is nonhomogeneous, we obtain
∂
2
∂r
2
[
v
(
r, t
)+
ψ
(
r
)] =
r
∂
∂t
h
1
r
v
(
r, t
)+
ψ
(
r
)
a
or
∂
2
v
∂r
2
+
ψ
ee
(
r
)=
∂v
∂t
.
This differential equation will be homogeneous if
ψ
ee
(
r
)=0or
ψ
(
r
)=
c
1
r
+
c
2
.Now
u
(
r, t
)=
1
r
v
(
r, t
)+
1
r
ψ
(
r
) and
1
r
ψ
(
r
)=
c
1
+
c
2
r
.
Since we want
u
(
r, t
) to be bounded as
r
approaches 0, we require
c
2
= 0.Then
ψ
(
r
)=
c
1
r
.When
r
=1
u
(1
,t
)=
v
(1
,t
)+
ψ
(1) =
v
(1
,t
)+
c
1
= 100
,
and we will have the homogeneous boundary condition
v
(1
,t
) = 0 when
c
1
= 100.Consequently,
ψ
(
r
) = 100
r
.
The initial condition
u
(
r,
0) =
1
r
v
(
r,
0) +
1
r
ψ
(
r
)=
1
r
v
(
r,
0) + 100 = 0
689

690. Exercises 14.3
implies
v
(
r,
0) =
−
100
r
.We are thus led to solve the new boundary-value problem
∂
2
v
∂r
2
=
∂v
∂t
,
0
<r<
1
,t>
0
,
v
(1
,t
)=0
,
lim
r
→
0
1
r
v
(
r, t
)
<
∞
,
v
(
r,
0) =
−
100
r.
Letting
v
(
r, t
)=
R
(
r
)
T
(
t
) and separating variables leads to
R
ee
+
λ
2
R
= 0 and
T
e
+
λ
2
T
=0
with solutions
R
(
r
)=
c
3
cos
λr
+
c
4
sin
λr
and
T
(
t
)=
c
5
e
−
λ
2
t
.
The boundary conditions are equivalent to
R
(1) = 0 and lim
r
→
0
1
r
R
(
r
)
<
∞
.Since
lim
r
→
0
1
r
R
(
r
) = lim
r
→
0
c
3
cos
λr
r
+ lim
r
→
0
c
4
sin
λr
r
= lim
r
→
0
c
3
cos
λr
r
+
c
4
λ<
∞
we must have
c
3
= 0.Then
R
(
r
)=
c
4
sin
λr
, and
R
(1) = 0 implies
λ
=
nπ
for
n
=1,2,3,
...
.Thus
v
n
(
r, t
)=
A
n
e
−
n
2
π
2
t
sin
nπr
for
n
=1,2,3,
...
.Using the condition lim
r
→
0
1
r
R
(
r
)
<
∞
it is easily shown that there are no eigenvalues for
λ
= 0, nor does setting the common constant to +
λ
2
when separating variables lead to any solutions.Now, by
the superposition principle,
v
(
r, t
)=
∞
e
n
=1
A
n
e
−
n
2
π
2
t
sin
nπr.
The initial condition
v
(
r,
0) =
−
100
r
implies
−
100
r
=
∞
e
n
=1
A
n
sin
nπr.
This is a Fourier sine series and so
A
n
=2
W
1
0
(
−
100
r
sin
nπr
)
dr
=
−
200
i
−
r
nπ
cos
nπr




1
0
+
W
1
0
1
nπ
cos
nπr dr
n
=
−
200
i
−
cos
nπ
nπ
+
1
n
2
π
2
sin
nπr




1
0
n
=
−
200
h
−
(
−
1)
n
nπ
a
=
(
−
1)
n
200
nπ
.
A solution of the problem is thus
u
(
r, t
)=
1
r
v
(
r, t
)+
1
r
ψ
(
r
)=
1
r
∞
e
n
=1
(
−
1)
n
20
nπ
e
−
n
2
π
2
t
sin
nπr
+
1
r
(100
r
)
=
200
πr
∞
e
n
=1
(
−
1)
n
n
e
−
n
2
π
2
t
sin
nπr
+ 100
.
10.
Referring to Problem 9 we have
∂
2
v
∂r
2
+
ψ
ee
(
r
)=
∂v
∂t
690

691. Exercises 14.3
where
ψ
(
r
)=
c
1
r
.Since
u
(
r, t
)=
1
r
v
(
r, t
)+
1
r
ψ
(
r
)=
1
r
v
(
r, t
)+
c
1
we have
∂u
∂r
=
1
r
v
r
(
r, t
)
−
1
r
2
v
(
r, t
)
.
When
r
=1,
∂u
∂r




r
=1
=
v
r
(1
,t
)
−
v
(1
,t
)
and
∂u
∂r




r
=1
+
hu
(1
,t
)=
v
r
(1
,t
)
−
v
(1
,t
)+
h
[
v
(1
,t
)+
ψ
(1)] =
v
r
(1
,t
)+(
h
−
1)
v
(1
,t
)+
hc
1
.
Thus the boundary condition
∂u
∂r




r
=1
+
hu
(1
,t
)=
hu
1
will be homogeneous when
hc
1
=
hu
1
or
c
1
=
u
1
.Consequently
ψ
(
r
)=
u
1
r
.The initial condition
u
(
r,
0) =
1
r
v
(
r,
0) +
1
r
ψ
(
r
)=
1
r
v
(
r,
0) +
u
1
=
u
0
implies
v
(
r,
0) = (
u
0
−
u
1
)
r
.We are thus led to solve the new boundary-value problem
∂
2
v
∂r
2
=
∂v
∂t
,
0
<r<
1
,t>
0
,
v
r
(1
,t
)+(
h
−
1)
v
(1
,t
)=0
,t>
0
,
lim
r
→
0
1
r
v
(
r, t
)
<
∞
,
v
(
r,
0)=(
u
0
−
u
1
)
r.
Separating variables as in Problem 9 leads to
R
(
r
)=
c
3
cos
λr
+
c
4
sin
λr
and
T
(
t
)=
c
5
e
−
λ
2
t
.
The boundary conditions are equivalent to
R
e
(1)+(
h
−
1)
R
(1) = 0 and lim
r
→
0
1
r
R
(
r
)
<
∞
.
As in Problem 6 we use the second condition to determine that
c
3
= 0 and
R
(
r
)=
c
4
sin
λr
.Then
R
e
(1)+(
h
−
1)
R
(1) =
c
4
λ
cos
λ
+
c
4
(
h
−
1) sin
λ
=0
and the eigenvalues
λ
n
are the consecutive nonnegative roots of tan
λ
=
λ/
(1
−
h
).Now
v
(
r, t
)=
∞
e
n
=1
A
n
e
−
λ
2
n
t
sin
λ
n
r.
From
v
(
r,
0) = (
u
0
−
u
1
)
r
=
∞
e
n
=1
A
n
sin
λ
n
r
we obtain
A
n
=
W
1
0
(
u
0
−
u
1
)
r
sin
λ
n
rdr
W
1
0
sin
2
λ
n
rdr
.
691

692. Exercises 14.3
We compute the integrals
W
1
0
r
sin
λ
n
rdr
=

1
λ
2
n
sin
λ
n
r
−
1
λ
n
cos
λ
n
r





1
0
=
1
λ
2
n
sin
λ
n
−
1
λ
n
cos
λ
n
and
W
1
0
sin
2
λ
n
rdr
=

1
2
r
−
1
4
λ
n
sin 2
λ
n
r





1
0
=
1
2
−
1
4
λ
n
sin 2
λ
n
.
Using
λ
n
cos
λ
n
=
−
(
h
−
1) sin
λ
n
we then have
A
n
=(
u
0
−
u
1
)
1
λ
2
n
sin
λ
n
−
1
λ
n
cos
λ
n
1
2
−
1
4
λ
n
sin 2
λ
n
=(
u
0
−
u
1
)
4 sin
λ
n
−
4
λ
n
cos
λ
n
2
λ
2
n
−
λ
n
2 sin
λ
n
cos
λ
n
=2(
u
0
−
u
1
)
sin
λ
n
+(
h
−
1) sin
λ
n
λ
2
n
+(
h
−
1) sin
λ
n
sin
λ
n
=2(
u
0
−
u
1
)
h
sin
λ
n
λ
2
n
+(
h
−
1) sin
2
λ
n
.
Therefore
u
(
r, t
)=
1
r
v
(
r, t
)+
1
r
ψ
(
r
)=
u
1
+2(
u
0
−
u
1
)
h
∞
e
n
=1
sin
λ
n
sin
λ
n
r
r
[
λ
2
n
+(
h
−
1) sin
2
λ
n
]
e
−
λ
2
n
t
.
11.
We write the differential equation in the form
a
2
1
r
∂
2
∂r
2
(
ru
)=
∂
2
u
∂t
2
or
a
2
∂
2
∂r
2
(
ru
)=
r
∂
2
u
∂t
2
,
and then let
v
(
r, t
)=
ru
(
r, t
).The new boundary-value problem is
a
2
∂
2
v
∂r
2
=
∂
2
v
∂t
2
,
0
<r<c, t>
0
v
(
c, t
)=0
,t>
0
v
(
r,
0) =
rf
(
r
)
,
∂v
∂t




t
=0
=
rg
(
r
)
.
Letting
v
(
r, t
)=
R
(
r
)
T
(
t
) and separating variables we obtain
R
ee
+
λ
2
R
=0
T
ee
+
a
2
λ
2
T
=0
and
R
(
r
)=
c
1
cos
λr
+
c
2
sin
λr
T
(
t
)=
c
3
cos
aλt
+
c
4
sin
aλt.
Since
u
(
r, t
)=
v
(
r, t
)
/r
, in order to insure boundedness at
r
= 0 we define
c
1
= 0.Then
R
(
r
)=
c
2
sin
λr
and
the condition
R
(
c
) = 0 implies
λ
=
nπ/c
.Thus
v
(
r, t
)=
∞
e
n
=1
=
A
n
cos
nπa
c
t
+
B
n
sin
nπa
c
t
1
sin
nπ
c
r.
From
v
(
r,
0) =
rf
(
r
)=
∞
e
n
=1
A
n
sin
nπ
c
r
we see that
A
n
=
2
c
W
c
0
rf
(
r
) sin
nπ
c
rdr.
692

693. Exercises 14.3
From
∂v
∂t




t
=0
=
rg
(
r
)=
∞
e
n
=1
=
B
n
nπa
c
1
sin
nπ
c
r
we see that
B
n
=
c
nπa
·
2
c
W
c
0
rg
(
r
) sin
nπ
c
rdr
=
2
nπa
W
c
0
rg
(
r
) sin
nπ
c
rdr.
12.
Proceeding as in Example 1 we obtain
Θ(
θ
)=
P
n
(cos
θ
) and
R
(
r
)=
c
1
r
n
+
c
2
r
−
(
n
+1)
so that
u
(
r, θ
)=
∞
e
n
=0
(
A
n
r
n
+
B
n
r
−
(
n
+1)
)
P
n
(cos
θ
)
.
To satisfy lim
r
→∞
u
(
r, θ
)=
−
Er
cos
θ
we must have
A
n
= 0 for
n
=2,3,4,
...
.Then
lim
r
→∞
u
(
r, θ
)=
−
Er
cos
θ
=
A
0
·
1+
A
1
r
cos
θ,
so
A
0
= 0 and
A
1
=
−
E
.Thus
u
(
r, θ
)=
−
Er
cos
θ
+
∞
e
n
=0
B
n
r
−
(
n
+1)
P
n
(cos
θ
)
.
Now
u
(
c, θ
)=0=
−
Ec
cos
θ
+
∞
e
n
=0
B
n
c
−
(
n
+1)
P
n
(cos
θ
)
so
∞
e
n
=0
B
n
c
−
(
n
+1)
P
n
(cos
θ
)=
Ec
cos
θ
and
B
n
c
−
(
n
+1)
=
2
n
+1
2
W
π
0
Ec
cos
θP
n
(cos
θ
) sin
θdθ.
Now cos
θ
=
P
1
(cos
θ
) so, for
n
1
=1,
W
π
0
cos
θP
n
(cos
θ
) sin
θdθ
=0
by orthogonality.Thus
B
n
= 0 for
n
1
= 1 and
B
1
=
3
2
Ec
3
W
π
0
cos
2
θ
sin
θdθ
=
Ec
3
.
Therefore,
u
(
r, θ
)=
−
Er
cos
θ
+
Ec
3
r
−
2
cos
θ.
693

694. Chapter 14 Review Exercises
Chapter 14 Review Exercises
1.
We have
A
0
=
1
2
π
W
π
0
u
0
dθ
+
1
2
π
W
2
π
π
(
−
u
0
)
dθ
=0
A
n
=
1
c
n
π
W
π
0
u
0
cos
nθ dθ
+
1
c
n
π
W
2
π
π
(
−
u
0
) cos
nθ dθ
=0
B
n
=
1
c
n
π
W
π
0
u
0
sin
nθ dθ
+
1
c
n
π
W
2
π
π
(
−
u
0
) sin
nθ dθ
=
2
u
0
c
n
nπ
[1
−
(
−
1)
n
]
and so
u
(
r, θ
)=
2
u
0
π
∞
e
n
=1
1
−
(
−
1)
n
n
=
r
c
1
n
sin
nθ.
2.
We have
A
0
=
1
2
π
W
π/
2
0
dθ
+
1
2
π
W
2
π
3
π/
2
dθ
=
1
2
A
n
=
1
c
n
π
W
π/
2
0
cos
nθ dθ
+
1
c
n
π
W
2
π
3
π/
2
cos
nθ dθ
=
1
c
n
nπ
h
sin
nπ
2
−
sin
3
nπ
2
a
B
n
=
1
c
n
π
W
π/
2
0
sin
nθ dθ
+
1
c
n
π
W
2
π
3
π/
2
sin
nθ dθ
=
1
c
n
nπ
h
cos
3
nπ
2
−
cos
nπ
2
a
and so
u
(
r, θ
)=
1
2
+
1
π
∞
e
n
=1
=
r
c
1
n
h
sin
nπ
2
−
sin
3
nπ
2
n
cos
nθ
+
cos
3
nπ
2
−
cos
nπ
2
n
sin
nθ
a
.
3.
The conditions Θ(0) = 0 and Θ(
π
) = 0 applied to Θ =
c
1
cos
λθ
+
c
2
sin
λθ
give
c
1
= 0 and
λ
=
n
,
n
=1,2,3,
...
, respectively.Thus we have the Fourier sine-series coefficients
A
n
=
2
π
W
π
0
u
0
(
πθ
−
θ
2
) sin
nθ dθ
=
4
u
0
n
3
π
[1
−
(
−
1)
n
]
.
Thus
u
(
r, θ
)=
4
u
0
π
∞
e
n
=1
1
−
(
−
1)
n
n
3
r
n
sin
nθ.
4.
In this case
A
n
=
2
π
W
π
0
sin
θ
sin
nθ dθ
=
1
π
W
π
0
[cos(1
−
n
)
θ
−
cos(1 +
n
)
θ
]
dθ
=0
,n
1
=1
.
For
n
=1,
A
1
=
2
π
W
π
0
sin
2
θdθ
=
1
π
W
π
0
(1
−
cos 2
θ
)
dθ
=1
.
Thus
u
(
r, θ
)=
∞
e
n
=1
A
n
r
n
sin
nθ
reduces to
u
(
r, θ
)=
r
sin
θ.
694

695. Chapter 14 Review Exercises
5.
We solve
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
=0
,
0
<θ<
π
4
,
1
2
<r<
1
,
u
(
r,
0) = 0
,u
(
r, π/
4) = 0
,
1
2
<r<
1
,
u
(1
/
2
,θ
)=
u
0
,u
r
(1
,θ
)=0
,
0
<θ<
π
4
.
Proceeding as in Example 1 in Section 14
.
1 in the text we obtain the separated equations
r
2
R
ee
+
rR
e
−
λ
2
R
=0
Θ
ee
+
λ
2
Θ=0
with solutions
Θ(
θ
)=
c
1
cos
λθ
+
c
2
sin
λθ
R
(
r
)=
c
3
r
λ
+
c
4
r
−
λ
.
Applying the boundary conditions Θ(0) = 0 and Θ(
π/
4) = 0 gives
c
1
= 0 and
λ
=4
n
for
n
=1,2,3,
...
.From
R
r
(1) = 0 we obtain
c
3
=
c
4
.Therefore
u
(
r, θ
)=
∞
e
n
=1
A
n

r
4
n
+
r
−
4
n

sin 4
nθ.
From
u
(1
/
2
,θ
)=
u
0
=
∞
e
n
=1
A
n

1
2
4
n
+
1
2
−
4
n

sin 4
nθ
we find
A
n

1
2
4
n
+
1
2
−
4
n

=
2
π/
4
W
π/
4
0
u
0
sin 4
nθ dθ
=
2
u
0
nπ
[1
−
(
−
1)
n
]
or
A
n
=
2
u
0
nπ
(2
4
n
+2
−
4
n
)
[1
−
(
−
1)
n
]
.
Thus the steady-state temperature in the plate is
u
(
r, θ
)=
2
u
0
π
∞
e
n
=1
[
r
4
n
+
r
−
4
n
][1
−
(
−
1)
n
]
n
[2
4
n
+2
−
4
n
]
sin 4
nθ.
6.
We solve
∂
2
u
∂r
2
+
1
r
∂u
∂r
+
1
r
2
∂
2
u
∂θ
2
=0
,r>
1
,
0
<θ<π,
u
(
r,
0) = 0
,u
(
r, π
)=0
,r>
1
,
u
(1
,θ
)=
f
(
θ
)
,
0
<θ<π.
Separating variables we obtain
Θ(
θ
)=
c
1
cos
λθ
+
c
2
sin
λθ
R
(
r
)=
c
3
r
λ
+
c
4
r
−
λ
.
Applying the boundary conditions Θ(0) = 0, and Θ(
π
) = 0 gives
c
1
= 0 and
λ
=
n
for
n
=1,2,3,
...
.
Assuming
f
(
θ
) to be bounded, we expect the solution
u
(
r, θ
) to also be bounded as
r
→∞
.This requires that
c
3
= 0.Therefore
u
(
r, θ
)=
∞
e
n
=1
A
n
r
−
n
sin
nθ.
695

696. Chapter 14 Review Exercises
From
u
(1
,θ
)=
f
(
θ
)=
∞
e
n
=1
A
n
sin
nθ
we obtain
A
n
=
2
π
W
π
0
f
(
θ
) sin
nθ dθ.
7.
Letting
u
(
r, t
)=
R
(
r
)
T
(
t
) and separating variables we obtain
R
ee
+
1
r
R
e
−
hR
R
=
T
e
T
=
µ
so
R
ee
+
1
r
R
e
−
(
µ
+
h
)
R
= 0 and
T
e
−
µT
=0
.
From the second equation we find
T
(
t
)=
c
1
e
µt
.If
µ>
0,
T
(
t
) increases without bound as
t
→∞
.Thus we
assume
µ
≤
0.Since
h>
0 we can take
µ
=
−
λ
2
−
h
.Then
R
ee
+
1
r
R
e
+
λ
2
R
=0
is a parametric Bessel equation with solution
R
(
r
)=
c
1
J
0
(
λr
)+
c
2
Y
0
(
λr
)
.
Since
Y
0
is unbounded as
r
→
0 we take
c
2
= 0.Then
R
(
r
)=
c
1
J
0
(
λr
) and the boundary condition
u
(1
,t
)=
R
(1)
T
(
t
) = 0 implies
J
0
(
λ
) = 0.This latter equation defines the positive eigenvalues
λ
n
.Thus
u
(
r, t
)=
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
e
(
−
λ
2
n
−
h
)
t
.
From
u
(
r,
0) = 1 =
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
we find
A
n
=
2
J
2
1
(
λ
n
)
W
1
0
rJ
0
(
λ
n
r
)
dr
x
=
λ
n
r, dx
=
λ
n
,dr
=
2
J
2
1
(
λ
n
)
W
λ
n
0
1
λ
2
n
xJ
0
(
x
)
dx.
From recurrence relation (4) in Section 12
.
6 of the text we have
xJ
0
(
x
)=
d
dx
[
xJ
1
(
x
)]
.
Then
A
n
=
2
λ
2
n
J
2
1
(
λ
n
)
W
λ
n
0
d
dx
[
xJ
1
(
x
)]
dx
=
2
λ
2
n
J
2
1
(
λ
n
)

xJ
1
(
x
)





λ
n
0
=
2
λ
1
J
1
(
λ
n
)
λ
2
n
J
2
1
(
λ
n
)
=
2
λ
n
J
1
(
λ
n
)
and
u
(
r, t
)=2
e
−
ht
∞
e
n
=1
J
0
(
λ
n
r
)
λ
n
J
1
(
λ
n
)
e
−
λ
2
n
t
8.
Proceeding in the usual manner we find
u
(
r, t
)=
∞
e
n
=1
A
n
cos
aλ
n
tJ
0
(
λ
n
r
)
696

697. Chapter 14 Review Exercises
where the eigenvalues are defined by
J
0
(
λ
) = 0.Thus the initial condition gives
u
0
J
0
(
x
k
r
)=
∞
e
n
=1
A
n
J
0
(
λ
n
r
)
and so
A
n
=
2
J
2
1
(
λ
n
)
W
1
0
r
(
u
0
J
0
(
x
k
r
))
J
0
(
λ
n
r
)
dr.
But
J
0
(
λ
) = 0 implies that the eigenvalues are the positive zeros of
J
0
, that is,
λ
n
=
x
n
,
n
=1,2,3,
...
.
Therefore
A
n
=
2
u
0
J
2
1
(
λ
n
)
W
1
0
rJ
0
(
λ
k
r
)
J
0
(
λ
n
r
)
dr
=0
,n
1
=
k
by orthogonality.For
n
=
k
,
A
k
=
2
u
0
J
2
1
(
λ
k
)
W
1
0
rJ
2
0
(
λ
k
)
dr
=
u
0
by (6) of Section 12
.
6.Thus the solution
u
(
r, t
) reduces to one term when
n
=
k
, and
u
(
r, t
)=
u
0
cos
aλ
k
tJ
0
(
λ
k
r
)=
u
0
cos
ax
k
tJ
0
(
x
k
r
)
.
9.
Referring to Example 2 in Section 14
.
2 we have
R
(
r
)=
c
1
J
0
(
λr
)+
c
2
Y
0
(
λr
)
Z
(
z
)=
c
3
cosh
λz
+
c
4
sinh
λz
where
c
2
= 0 and
J
0
(2
λ
) = 0 defines the positive eigenvalues
λ
n
.From
Z
e
(0) = 0 we obtain
c
4
= 0.Then
u
(
r, z
)=
∞
e
n
=1
A
n
cosh
λ
n
zJ
0
(
λ
n
r
)
.
From
u
(
r,
4)=50=
∞
e
n
=1
A
n
cosh 4
λ
n
J
0
(
λ
n
r
)
we obtain (as in Example 1 of Section 14
.
1)
A
n
cosh 4
λ
n
=
2(50)
4
J
2
1
(2
λ
n
)
W
2
0
rJ
0
(
λ
n
r
)
dr
=
50
λ
n
J
1
(2
λ
n
)
.
Thus the temperature in the cylinder is
u
(
r, z
)=50
∞
e
n
=1
cosh
λ
n
zJ
0
(
λ
n
r
)
λ
n
cosh 4
λ
n
J
1
(2
λ
n
)
.
10.
Using
u
=
RZ
and
−
λ
2
as a separation constant leads to
r
2
R
ee
+
rR
e
+
λ
2
r
2
R
=0
,R
e
(1) = 0
,
and
Z
ee
−
λ
2
Z
=0
.
Thus
R
(
r
)=
c
1
J
0
(
λr
)+
c
2
Y
0
(
λr
)
Z
(
z
)=
c
3
cosh
λz
+
c
4
sinh
λz
for
λ>
0.Arguing that
u
(
r, z
) is bounded as
r
→
0 we defined
c
2
= 0.Since the eigenvalues are defined by
J
e
0
(
λ
) = 0 we know that
λ
= 0 is an eigenvalue.The solutions are then
R
(
r
)=
c
1
+
c
2
ln
r
and
Z
(
z
)=
c
3
z
+
c
4
697

698. Chapter 14 Review Exercises
where
c
2
= 0.Thus a formal solution is
u
(
r, z
)=
A
0
z
+
B
0
+
∞
e
n
=1
(
A
n
sinh
λ
n
z
+
B
n
cosh
λ
n
z
)
J
0
(
λ
n
r
)
.
Finally, the specified conditions
z
= 0 and
z
= 1 give, in turn,
B
0
=2
W
1
0
rf
(
r
)
dr
B
n
=
2
J
2
0
(
λ
n
)
W
1
0
rf
(
r
)
J
0
(
λ
n
r
)
dr
A
0
=
−
B
0
+2
W
1
0
rg
(
r
)
dr
A
n
=
1
sinh
λ
n
h
−
B
n
cosh
λ
n
+
2
J
2
0
(
λ
n
)
W
1
0
rg
(
r
)
J
0
(
λ
n
r
)
dr
a
.
11.
Referring to Example 1 in Section 14
.
3 of the text we have
u
(
r, θ
)=
∞
e
n
=0
A
n
r
n
P
n
(cos
θ
)
.
For
x
= cos
θ
u
(1
,θ
)=
v
100 0
<θ<π/
2
−
100
π/
2
<θ<π
= 100
v
−
1
,
−
1
<x<
0
1
,
0
<x<
1
=
g
(
x
)
.
From Problem 22 in Exercise 12
.
6wehave
u
(
r, θ
) = 100
h
3
2
rP
1
(cos
θ
)
−
7
8
r
3
P
3
(cos
θ
)+
11
16
r
5
P
5
(cos
θ
)+
···
a
.
12.
Since
1
r
∂
2
∂r
2
(
ru
)=
1
r
∂
∂r
h
r
∂u
∂r
+
u
a
=
1
r
h
r
∂
2
u
∂r
2
+
∂u
∂r
+
∂u
∂r
a
=
∂
2
u
∂r
2
+
2
r
∂u
r
the differential equation becomes
1
r
∂
2
∂r
2
(
ru
)=
∂
2
u
∂t
2
or
∂
2
∂r
2
(
ru
)=
r
∂
2
u
∂t
2
.
Letting
v
(
r, t
)=
ru
(
r, t
) we obtain the boundary-value problem
∂
2
v
∂r
2
=
∂
2
v
∂t
2
,
0
<r<
1
,t>
0
∂v
∂r




r
=1
−
v
(1
,t
)=0
,t>
0
v
(
r,
0) =
rf
(
r
)
,
∂v
∂t




t
=0
=
rg
(
r
)
,
0
<r<
1
.
If we separate variables using
v
(
r, t
)=
R
(
r
)
T
(
t
) then we obtain
R
(
r
)=
c
1
cos
λr
+
c
2
sin
λr
T
(
t
)=
c
3
cos
λt
+
c
4
sin
λt.
Since
u
(
r, t
)=
v
(
r, t
)
/r
, in order to insure boundedness at
r
= 0 we define
c
1
= 0.Then
R
(
r
)=
c
2
sin
λr
.Now
the boundary condition
R
e
(1)
−
R
(1) = 0 implies
λ
cos
λ
−
sin
λ
= 0.Thus, the eigenvalues
λ
n
are the positive
solutions of tan
λ
=
λ
.We now have
v
n
(
r, t
)=(
A
n
cos
λ
n
t
+
B
n
sin
λ
n
t
) sin
λ
n
r.
698

699. Chapter 14 Review Exercises
For the eigenvalue
λ
=0,
R
(
r
)=
c
1
r
+
c
2
and
T
(
t
)=
c
3
t
+
c
4
,
and boundedness at
r
= 0 implies
c
2
= 0.We then take
v
0
(
r, t
)=
A
0
tr
+
B
0
r
so that
v
(
r, t
)=
A
0
tr
+
B
0
r
+
∞
e
n
=1
(
a
n
cos
λ
n
t
+
B
n
sin
λ
n
t
) sin
λ
n
r.
Now
v
(
r,
0) =
rf
(
r
)=
B
0
r
+
∞
e
n
=1
A
n
sin
λ
n
r.
Since
{
r,
sin
λ
n
r
}
is an orthogonal set on [0
,
1],
W
1
0
r
sin
λ
n
rdr
= 0 and
W
1
0
sin
λ
n
r
sin
λ
n
rdr
=0
for
m
1
=
n
.Therefore
W
1
0
r
2
f
(
r
)
dr
=
B
0
W
1
0
r
2
dr
=
1
3
B
0
and
B
0
=3
W
1
0
r
2
f
(
r
)
dr.
Also
W
1
0
rf
(
r
) sin
λ
n
rdr
=
A
n
W
1
0
sin
2
λ
n
rdr
and
A
n
=
)
1
0
rf
(
r
) sin
λ
n
rdr
)
1
0
sin
2
λ
n
rdr
.
Now
W
1
0
sin
2
λ
n
rdr
=
1
2
W
1
0
(1
−
cos 2
λ
n
r
)
dr
=
1
2
h
1
−
sin 2
λ
n
2
λ
n
a
=
1
2
[1
−
cos
2
λ
n
]
.
Since tan
λ
n
=
λ
n
,
1+
λ
2
n
= 1 + tan
2
λ
n
= sec
2
λ
n
=
1
cos
2
λ
n
and
cos
2
λ
n
=
1
1+
λ
2
n
.
Then
W
1
0
sin
2
λ
n
rdr
=
1
2
h
1
−
1
1+
λ
2
n
a
=
λ
2
n
2(1 +
λ
2
n
)
and
A
n
=
2(1 +
λ
2
n
)
λ
2
n
W
1
0
rf
(
r
) sin
λ
n
rdr.
Similarly, setting
∂v
∂t




t
=0
=
rg
(
r
)=
A
0
r
+
∞
e
n
=1
B
n
λ
n
sin
λ
n
r
699

700. Chapter 14 Review Exercises
we obtain
A
0
=3
W
1
0
r
2
g
(
r
)
dr
and
B
n
=
2(1 +
λ
2
n
)
λ
3
n
W
1
0
rg
(
r
) sin
λ
n
rdr.
Therefore, since
v
(
r, t
)=
ru
(
r, t
) we have
u
(
r, t
)=
A
0
t
+
B
0
+
∞
e
n
=1
(
A
n
cos
λ
n
t
+
B
n
sin
λ
n
t
)
sin
λ
n
r
r
,
where the
λ
n
are solutions of tan
λ
=
λ
and
A
0
=3
W
1
0
r
2
g
(
r
)
dr
B
0
=3
W
1
0
r
2
f
(
r
)
dr
A
n
=
2(1 +
λ
2
n
)
λ
2
n
W
1
0
rf
(
r
) sin
λ
n
rdr
B
n
=
2(1 +
λ
2
n
)
λ
3
n
W
1
0
rg
(
r
) sin
λ
n
rdr
for
n
=1,2,3,
...
.
13.
We note that the differential equation can be expressed in the form
d
dx
[
xu
e
]=
−
λ
2
xu.
Thus
u
n
d
dx
[
xu
e
m
]=
−
λ
2
m
xu
m
u
n
and
u
m
d
dx
[
xu
e
n
]=
−
λ
2
n
xu
n
u
m
.
Subtracting we obtain
u
n
d
dx
[
xu
e
m
]
−
u
m
d
dx
[
xu
e
n
]=(
λ
2
n
−
λ
2
m
)
xu
m
u
n
and
W
b
a
u
n
d
dx
[
xu
e
m
]
dx
−
W
b
a
u
m
d
dx
[
xu
e
n
]=(
λ
2
n
−
λ
2
m
)
W
b
a
xu
m
u
n
dx.
Using integration by parts this becomes
u
n
xu
e
m



b
a
−
W
b
a
xu
e
m
u
e
n
dx
−
u
m
xu
e
n



b
a
+
W
b
a
xu
e
n
u
e
m
dx
=
b
[
u
n
(
b
)
u
e
m
(
b
)
−
u
m
(
b
)
u
e
n
(
b
)]
−
a
[
u
n
(
a
)
u
e
m
(
a
)
−
u
m
(
a
)
u
e
n
(
a
)]
=(
λ
2
n
−
λ
2
m
)
W
b
a
xu
m
u
n
dx.
Since
u
(
x
)=
Y
0
(
λa
)
J
0
(
λx
)
−
J
0
(
λa
)
Y
0
(
λx
)
700

701. Chapter 14 Review Exercises
we have
u
n
(
b
)=
Y
0
(
λ
n
a
)
J
0
(
λ
n
b
)
−
J
0
(
λ
n
a
)
Y
(
λ
n
b
)=0
by the definition of the
λ
n
.Similarly
u
m
(
b
) = 0.Also
u
n
(
a
)=
Y
0
(
λa
)
J
0
(
λ
n
a
)
−
J
0
(
λ
n
a
)
Y
0
(
λ
n
a
)=0
and
u
m
(
a
) = 0.Therefore
W
b
a
xu
m
u
n
dx
=
1
λ
2
n
−
λ
2
m
(
b
[
u
n
(
b
)
u
e
m
(
b
)
−
u
m
(
b
)
u
e
n
(
b
)]
−
a
[
u
n
(
a
)
u
e
m
(
a
)
−
u
m
(
a
)
u
e
n
(
a
)]) = 0
and the
u
n
(
x
) are orthogonal with respect to the weight function
x
.
14.
Letting
u
(
r, t
)=
R
(
r
)
T
(
t
) and separating variables we obtain
rR
ee
+
R
e
+
λ
2
rR
=0
T
e
+
λ
2
T
=0
,
with solutions
R
(
r
)=
c
1
J
0
(
λr
)+
c
2
Y
0
(
λr
)
T
(
t
)=
c
3
e
−
λ
2
t
.
Now the boundary conditions imply
R
(
a
)=0=
c
1
J
0
(
λa
)+
c
2
Y
0
(
λa
)
R
(
b
)=0=
c
1
J
0
(
λb
)+
c
2
Y
0
(
λb
)
so that
c
2
=
−
c
1
J
0
(
λa
)
Y
0
(
λa
)
and
c
1
J
0
(
λb
)
−
c
1
J
0
(
λa
)
Y
(
λa
)
Y
0
(
λb
)=0
or
Y
0
(
λa
)
J
0
(
λb
)
−
J
0
(
λa
)
Y
0
(
λb
)=0
.
This equation defines
λ
n
for
n
=1,2,3,
...
.Now
R
(
r
)=
c
1
J
0
(
λr
)
−
c
1
J
0
(
λa
)
Y
0
(
λa
)
Y
0
(
λr
)=
c
1
Y
0
(
λa
)
[
Y
0
(
λa
)
J
0
(
λr
)
−
J
0
(
λa
)
Y
0
(
λr
)
(
and
u
n
(
r, t
)=
A
n
[
Y
0
(
λ
n
a
)
J
0
(
λ
n
r
)
−
J
0
(
λ
n
a
)
Y
0
(
λ
n
r
)
(
e
−
λ
2
n
t
=
A
n
u
n
(
r
)
e
−
λ
2
n
t
.
Thus
u
(
r, t
)=
∞
e
n
=1
A
n
u
n
(
r
)
e
−
λ
2
n
t
.
From the initial condition
u
(
r,
0) =
f
(
r
)=
∞
e
n
=1
A
n
u
n
(
r
)
we obtain
A
n
=
)
b
a
rf
(
r
)
u
n
(
r
)
dr
)
b
a
ru
2
n
(
r
)
dr
.
701

702. 15
Integral Transform Method
Exercises 15.1
1. (a)
The result follows by letting
τ
=
u
2
or
u
=
√
τ
in erf(
√
t
)=
2
√
π
T
√
t
0
e
−
u
2
du
.
(b)
Using
{
t
−
1
/
2
}
=
√
π
s
1
/
2
and the first translation theorem, it follows from the convolution theorem that
h
erf(
√
t
)
e
=
1
√
π
r
T
t
0
e
−
τ
√
τ
dτ
s
=
1
√
π
{
1
}
h
t
−
1
/
2
e
−
t
e
=
1
√
π
1
s
h
t
−
1
/
2
e
u
u
u
u
s
→
s
+1
=
1
√
π
1
s
√
π
√
s
+1
=
1
s
√
s
+1
.
2.
Since erfc(
√
t
)=1
−
erf(
√
t
)wehave
h
erfc(
√
t
)
e
=
{
1
}−
h
erf(
√
t
)
e
=
1
s
−
1
s
√
s
+1
=
1
s
l
1
−
1
√
s
+1

.
3.
By the first translation theorem,
h
e
t
erf(
√
t
)
e
=
h
erf(
√
t
)
e
u
u
u
u
s
→
s
−
1
=
1
s
√
s
+1
u
u
u
u
s
→
s
−
1
=
1
√
s
(
s
−
1)
.
4.
By the first translation theorem and the result of Problem 2,
h
e
t
erfc(
√
t
)
e
=
h
erfc(
√
t
)
e
u
u
u
u
s
→
s
−
1
=

1
s
−
1
s
√
s
+1

u
u
u
u
s
→
s
−
1
=
1
s
−
1
−
1
√
s
(
s
−
1)
=
√
s
−
1
√
s
(
s
−
1)
=
√
s
−
1
√
s
(
√
s
+1)(
√
s
−
1)
=
1
√
s
(
√
s
+1)
.
5.
From table entry 3 and the first translation theorem we have

e
−
Gt/C
erf

x
2
y
RC
t
in
=

e
−
Gt/C
g
1
−
erfc

x
2
y
RC
t
i=n
=
h
e
−
Gt/C
e
−

e
−
Gt/C
erfc

x
2
y
RC
t
in
=
1
s
+
G/C
−
e
−
x
√
RC
√
s
s
u
u
u
u
s
→
s
+
G/C
=
1
s
+
G/C
−
e
−
x
√
RC
√
s
+
G/C
s
+
G/C
=
C
Cs
+
G
(
1
−
e
x
√
RCs
+
RG
)
.
702

703. Exercises 15.1
6.
We first compute
sinh
a
√
s
s
sinh
√
s
=
e
a
√
s
−
e
−
a
√
s
s
(
e
√
s
−
e
−
√
s
)
=
e
(
a
−
1)
√
s
−
e
−
(
a
+1)
√
s
s
(1
−
e
−
2
√
s
)
=
e
(
a
−
1)
√
s
s
2
1+
e
−
2
√
s
+
e
−
4
√
s
+
···
.
−
e
−
(
a
+1)
√
s
s
2
1+
e
−
2
√
s
+
e
−
4
√
s
+
···
.
=
g
e
−
(1
−
a
)
√
s
s
+
e
−
(3
−
a
)
√
s
s
+
e
−
(5
−
a
)
√
s
s
+
···
=
−
g
e
−
(1+
a
)
√
s
s
+
e
−
(3+
a
)
√
s
s
+
e
−
(5+
a
)
√
s
s
+
···
=
=
∞
U
n
=0
g
e
−
(2
n
+1
−
a
)
√
s
s
−
e
−
(2
n
+1+
a
)
√
s
s
=
.
Then
r
sinh
a
√
s
s
sinh
√
s
s
=
∞
U
n
=0
gw
e
−
(2
n
+1
−
a
)
√
s
s
n
−

−
e
−
(2
n
+1+
a
)
√
s
s
n=
=
∞
U
n
=0
l
erfc

2
n
+1
−
a
2
√
t

−
erfc

2
n
+1+
a
2
√
t

=
∞
U
n
=0
fl
1
−
erf

2
n
+1
−
a
2
√
t

−
l
1
−
erf

2
n
+1+
a
2
√
t
oto
=
∞
U
n
=0
l
erf

2
n
+1+
a
2
√
t

−
erf

2
n
+1
−
a
2
√
t

.
7.
Taking the Laplace transform of both sides of the equation we obtain
{
y
(
t
)
}
=
{
1
}−
r
T
t
0
y
(
τ
)
√
t
−
τ
dτ
s
Y
(
s
)=
1
s
−
Y
(
s
)
√
π
√
s
√
s
+
√
π
√
s
Y
(
s
)=
1
s
Y
(
s
)=
1
√
s
(
√
s
+
√
π
)
.
Thus
y
(
t
)=
r
1
√
s
(
√
s
+
√
π
)
s
=
e
πt
erfc(
√
πt
)
.
By entry 5 in the table
8.
Using entries 3 and 5 in the table, we have

−
e
ab
e
b
2
t
erfc

b
√
t
+
a
2
√
t

+erfc

a
2
√
t

n
=
−
r
e
ab
e
b
2
t
erfc

b
√
t
+
a
2
√
t
os
+
r
a
2
√
t
s
=
−
e
−
a
√
s
√
s
(
√
s
+
b
)
+
e
−
a
√
s
s
703

704. -10 -5 5 10
x
-2
-1
1
2
y
erfHxL
erfcHxL
Exercises 15.1
=
e
−
a
√
s
l
1
s
−
1
√
s
(
√
s
+
b
)

=
e
−
a
√
s
l
1
s
−
√
s
s
(
√
s
+
b
)

=
e
−
a
√
s
l
√
s
+
b
−
√
s
s
(
√
s
+
b
)

=
be
−
a
√
s
s
(
√
s
+
b
)
.
9.
T
b
a
e
−
u
2
du
=
T
0
a
e
−
u
2
du
+
T
b
0
e
−
u
2
du
=
T
b
0
e
−
u
2
du
−
T
a
0
e
−
u
2
du
=
√
π
2
erf(
b
)
−
√
π
2
erf(
a
)=
√
π
2
[erf(
b
)
−
erf(
a
)]
10.
Since
f
(
x
)=
e
−
x
2
is an even function,
T
a
−
a
e
−
u
2
du
=2
T
a
0
e
−
u
2
du.
Therefore,
T
a
−
a
e
−
u
2
du
=
√
π
erf(
a
)
.
11.
The function erf(
x
) is symmetric with respect to the origin,
while erfc(
x
) appears to be symmetric with respect to the point
(0
,
1). From the graph it appears that lim
x
→−∞
erf(
x
)=
−
1
and lim
x
→−∞
erfc(
x
)=2.
Exercises 15.2
1.
The boundary-value problem is
a
2
∂
2
u
∂x
2
=
∂
2
u
∂t
2
,
0
<x<L, t>
0
,
u
(0
,t
)=0
,u
(
L, t
)=0
,t>
0
,
u
(
x,
0) =
A
sin
π
L
x,
∂u
∂t
u
u
u
u
t
=0
=0
.
Transforming the partial differential equation gives
d
2
U
dx
2
−
(
s
a
)
2
U
=
−
s
a
2
A
sin
π
L
x.
Using undetermined coefficients we obtain
U
(
x, s
)=
c
1
cosh
s
a
x
+
c
2
sinh
s
a
x
+
As
s
2
+
a
2
π
2
/L
2
sin
π
L
x.
The transformed boundary conditions,
U
(0
,s
)=0,
U
(
L, s
) = 0 give in turn
c
1
= 0 and
c
2
= 0. Therefore
U
(
x, s
)=
As
s
2
+
a
2
π
2
/L
2
sin
π
L
x
704

705. Exercises 15.2
and
u
(
x, t
)=
A
r
s
s
2
+
a
2
π
2
/L
2
s
sin
π
L
x
=
A
cos
aπ
L
t
sin
π
L
x.
2.
The transformed equation is
d
2
U
dx
2
−
s
2
U
=
−
2 sin
πx
−
4 sin 3
πx
and so
U
(
x, s
)=
c
1
cosh
sx
+
c
2
sinh
sx
+
2
s
2
+
π
2
sin
πx
+
4
s
2
+9
π
2
sin 3
πx.
The transformed boundary conditions,
U
(0
,s
) = 0 and
U
(1
,s
) = 0 give
c
1
= 0 and
c
2
=0. Thus
U
(
x, s
)=
2
s
2
+
π
2
sin
πx
+
4
s
2
+9
π
2
sin 3
πx
and
u
(
x, t
)=2
r
1
s
2
+
π
2
s
sin
πx
+4
r
1
s
2
+9
π
2
s
sin 3
πx
=
2
π
sin
πt
sin
πx
+
4
3
π
sin 3
πt
sin 3
πx.
3.
The solution of
a
2
d
2
U
dx
2
−
s
2
U
=0
is in this case
U
(
x, s
)=
c
1
e
−
(
x/a
)
s
+
c
2
e
(
x/a
)
s
.
Since lim
x
→∞
u
(
x, t
)=0wehavelim
x
→∞
U
(
x, s
)=0. Thus
c
2
= 0 and
U
(
x, s
)=
c
1
e
−
(
x/a
)
s
.
If
{
u
(0
,t
)
}
=
{
f
(
t
)
}
=
F
(
s
) then
U
(0
,s
)=
F
(
s
). From this we have
c
1
=
F
(
s
) and
U
(
x, s
)=
F
(
s
)
e
−
(
x/a
)
s
.
Hence, by the second translation theorem,
u
(
x, t
)=
f
(
t
−
x
a
)(
t
−
x
a
)
.
4.
Expressing
f
(
t
) in the form (sin
πt
)[1
−
(
t
−
1)] and using the result of Problem 3 we find
u
(
x, t
)=
f
(
t
−
x
a
)(
t
−
x
a
)
= sin
π
(
t
−
x
a
)2
1
−
(
t
−
x
a
−
1
). (
t
−
x
a
)
= sin
π
(
t
−
x
a
)2 (
t
−
x
a
)
−
(
t
−
x
a
)(
t
−
x
a
−
1
).
= sin
π
(
t
−
x
a
)2 (
t
−
x
a
)
−
(
t
−
x
a
−
1
).
Now
(
t
−
x
a
)
−
(
t
−
x
a
−
1
)
=





0
,
0
≤
t < x/a
1
, x/a
≤
t
≤
x/a
+1
0
, t > x/a
+1
=
r
0
,x<a
(
t
−
1) or
x>at
1
,a
(
t
−
1)
≤
x
≤
at
705

706. Exercises 15.2
so
u
(
x, t
)=
r
0
,x<a
(
t
−
1) or
x>at
sin
π
(
t
−
x/a
)
,a
(
t
−
1)
≤
x
≤
at
.
The graph is shown for
t>
1.
5.
We use
U
(
x, s
)=
c
1
e
−
(
x/a
)
s
−
g
s
3
.
Now
{
u
(0
,t
)
}
=
U
(0
,s
)=
Aω
s
2
+
ω
2
and so
U
(0
,s
)=
c
1
−
g
s
3
=
Aω
s
2
+
ω
2
or
c
1
=
g
s
3
+
Aω
s
2
+
ω
2
.
Therefore
U
(
x, s
)=
Aω
s
2
+
ω
2
e
−
(
x/a
)
s
+
g
s
3
e
−
(
x/a
)
s
−
g
s
3
and
u
(
x, t
)=
A
r
ωe
−
(
x/a
)
s
s
2
+
ω
2
s
+
g
r
e
−
(
x/a
)
s
s
3
s
−
g
r
1
s
3
s
=
A
sin
ω
(
t
−
x
a
)(
t
−
x
a
)
+
1
2
g
(
t
−
x
a
)
2
(
t
−
x
a
)
−
1
2
gt
2
.
6.
Transforming the partial differential equation gives
d
2
U
dx
2
−
s
2
U
=
−
ω
s
2
+
ω
2
sin
πx.
Using undetermined coefficients we obtain
U
(
x, s
)=
c
1
cosh
sx
+
c
2
sinh
sx
+
ω
(
s
2
+
π
2
)(
s
2
+
ω
2
)
sin
πx.
The transformed boundary conditions
U
(0
,s
)=0and
U
(1
,s
) = 0 give, in turn,
c
1
= 0 and
c
2
= 0. Therefore
U
(
x, s
)=
ω
(
s
2
+
π
2
)(
s
2
+
ω
2
)
sin
πx
and
u
(
x, t
)=
ω
sin
πx
r
1
(
s
2
+
π
2
)(
s
2
+
ω
2
)
s
=
ω
ω
2
−
π
2
sin
πx
r
1
π
π
s
2
+
π
2
−
1
ω
ω
s
2
+
ω
2
s
=
ω
π
(
ω
2
−
π
2
)
sin
πt
sin
πx
−
1
ω
2
−
π
2
sin
ωt
sin
πx.
7.
We use
U
(
x, s
)=
c
1
cosh
s
a
x
+
2
sinh
s
a
x.
706

707. Exercises 15.2
Now
U
(0
,s
) = 0 implies
c
1
=0,so
U
(
x, s
)=
c
2
sinh(
s/a
)
x
. The condition
E
dU
dx
u
u
u
u
x
=
L
=
F
0
then yields
c
2
=
F
0
a/Es
cosh(
s/a
)
L
and so
U
(
x, s
)=
aF
0
Es
sinh(
s/a
)
x
cosh(
s/a
)
L
=
aF
0
Es
e
(
s/a
)
x
−
e
−
(
s/a
)
x
e
(
s/a
)
L
+
e
−
(
s/a
)
L
=
aF
0
Es
e
(
s/a
)(
x
−
L
)
−
e
−
(
s/a
)(
x
+
L
)
1+
e
−
2
sL/a
=
aF
0
E
l
e
−
(
s/a
)(
L
−
x
)
s
−
e
−
(
s/a
)(3
L
−
x
)
s
+
e
−
(
s/a
)(5
L
−
x
)
s
−···

−
aF
0
E
l
e
−
(
s/a
)(
L
+
x
)
s
−
e
−
(
s/a
)(3
L
+
x
)
s
+
e
−
(
s/a
)(5
L
+
x
)
s
−···

=
aF
0
E
∞
U
n
=0
(
−
1)
n
l
e
−
(
s/a
)(2
nL
+
L
−
x
)
s
−
e
−
(
s/a
)(2
nL
+
L
+
x
)
s

and
u
(
x, t
)=
aF
0
E
∞
U
n
=0
(
−
1)
n
lr
e
−
(
s/a
)(2
nL
+
L
−
x
)
s
s
−
r
e
−
(
s/a
)(2
nL
+
L
+
x
)
s
st
=
aF
0
E
∞
U
n
=0
(
−
1)
n
g

t
−
2
nL
+
L
−
x
a

t
−
2
nL
+
L
−
x
a

−

t
−
2
nL
+
L
+
x
a

t
−
2
nL
+
L
+
x
a

=
.
8.
We use
U
(
x, s
)=
c
1
e
−
(
x/a
)
s
+
c
2
e
(
x/a
)
s
−
v
0
s
2
.
Now lim
x
→∞
dU
dx
= 0 implies
c
2
= 0, and
U
(0
,s
) = 0 then gives
c
1
=
v
0
/s
2
. Hence
U
(
x, s
)=
v
0
s
2
e
−
(
x/a
)
s
−
v
0
s
2
and
u
(
x, t
)=
v
0
(
t
−
x
a
)(
t
−
x
a
)
−
v
0
t.
9.
Transforming the partial differential equation gives
d
2
U
dx
2
−
s
2
U
=
−
sxe
−
x
.
Using undetermined coefficients we obtain
U
(
x, s
)=
c
1
e
−
sx
+
c
2
e
sx
−
2
s
(
s
2
−
1)
2
e
−
x
+
s
s
2
−
1
xe
−
x
.
The transformed boundary conditions lim
x
→∞
U
(
x, s
)=0and
U
(0
,s
) = 0 give, in turn,
c
2
= 0 and
c
1
=2
s/
(
s
2
−
1)
2
. Therefore
U
(
x, s
)=
2
s
(
s
2
−
1)
2
e
−
sx
−
2
s
(
s
2
−
1)
2
e
−
x
+
s
s
2
−
1
xe
−
x
.
From entries (13) and (26) in the table we obtain
u
(
x, t
)=
r
2
s
(
s
2
−
1)
2
e
−
sx
−
2
s
(
s
2
−
1)
2
e
−
x
+
s
s
2
−
1
xe
−
x
s
=2(
t
−
x
) sinh(
t
−
x
)(
t
−
x
)
−
te
−
x
sinh
t
+
xe
−
x
cosh
t.
707

708. 0
2
4
6
8
10
x
5
10
15
t
0
25
50
75
100
uHx,tL
0
2
4
6
8
500100015002000
t
20
40
60
80
100
uH10,tL
Exercises 15.2
10.
We use
U
(
x, s
)=
c
1
e
−
xs
+
c
2
e
xs
+
s
s
2
−
1
e
−
x
.
Now lim
x
→∞
u
(
x, t
) = 0 implies lim
x
→∞
U
(
x, s
) = 0, so we define
c
2
= 0. Then
U
(
x, s
)=
c
1
e
−
xs
+
s
s
2
−
1
e
−
x
.
Finally,
U
(0
,s
)=1
/s
gives
c
1
=1
/s
−
s/
(
s
2
−
1). Thus
U
(
x, s
)=
1
s
−
s
s
2
−
1
e
−
xs
+
s
s
2
−
1
e
−
x
and
u
(
x, t
)=
−
r
s
s
2
−
1
e
−
(
x/a
)
s
s
+
r
s
s
2
−
1
s
e
−
x
=
−
cosh
(
t
−
x
a
)(
t
−
x
a
)
+
e
−
x
cosh
t.
11. (a)
We use
U
(
x, s
)=
c
1
e
−
√
s/k x
+
c
2
e
√
s/k x
.
Now lim
x
→∞
u
(
x, t
) = 0 implies lim
x
→∞
U
(
x, s
) = 0, so we define
c
2
= 0. Then
U
(
x, s
)=
c
1
e
−
√
s/k x
.
Finally, from
U
(0
,s
)=
u
0
/s
we obtain
c
1
=
u
0
/s
.Thus
U
(
x, s
)=
u
0
e
−
√
s/k x
s
and
u
(
x, t
)=
u
0

e
−
√
s/k x
s
n
=
u
0

e
−
(
x/
√
k
)
√
s
s
n
=
u
0
erfc

x
2
√
kt

.
Since erfc(0) = 1,
lim
t
→∞
u
(
x, t
) = lim
t
→∞
u
0
erfc(
x/
2
√
kt
)=
u
0
.
(b)
12. (a)
Transforming the partial differential equation and using the initial condition gives
k
d
2
U
dx
2
−
sU
=0
.
Since the domain of the variable
x
is an infinite interval we write the general solution of this differential
equation as
U
(
x, s
)=
c
1
e
−
√
s/k x
+
c
2
e
−
√
s/k x
.
708

709. 0
2
4
6
8
10
x
5
10
15
t
0
10
20
30
40
uHx,tL
0
2
4
6
8
500100015002000
t
100
200
300
400
500
uH10,tL
Exercises 15.2
Transforming the boundary conditions gives
U
r
(0
,s
)=
−
A/s
and lim
x
→∞
U
(
x, s
) = 0. Hence we find
c
2
=0
and
c
1
=
A
√
k/s
√
s
. From
U
(
x, s
)=
A
√
k
e
−
√
s/k x
s
√
s
we see that
u
(
x, t
)=
A
√
k

e
−
√
s/k x
s
√
s
n
.
With the identification
a
=
x/
√
k
it follows from entry 49 of the table in Appendix III that
u
(
x, t
)=
A
√
k

2
y
t
π
e
−
x
2
/
4
kt
−
x
√
k
(
x/
2
√
kt
)
n
=2
A
y
kt
π
e
−
x
2
/
4
kt
−
Ax
erfc
(
x/
2
√
kt
)
.
Since erfc(0) = 1,
lim
t
→∞
u
(
x, t
) = lim
t
→∞

2
A
y
kt
π
e
−
x
2
/
4
kt
−
Ax
erfc
x
2
√
kt

=
∞
.
(b)
13.
We use
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
+
u
1
s
.
The condition lim
x
→∞
u
(
x, t
)=
u
1
implies lim
x
→∞
U
(
x, s
)=
u
1
/s
, so we define
c
2
= 0. Then
U
(
x, s
)=
c
1
e
−
√
sx
+
u
1
s
.
From
U
(0
,s
)=
u
0
/s
we obtain
c
1
=(
u
0
−
u
1
)
/s
.Thus
U
(
x, s
)=(
u
0
−
u
1
)
e
−
√
sx
s
+
u
1
s
and
u
(
x, t
)=(
u
0
−
u
1
)

e
−
x
√
s
s
n
+
u
1
r
1
s
s
=(
u
0
−
u
1
) erfc

x
2
√
t

+
u
1
.
14.
We use
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
+
u
1
x
s
.
The condition lim
x
→∞
u
(
x, t
)
/x
=
u
1
implies lim
x
→∞
U
(
x, s
)
/x
=
u
1
/s
, so we define
c
2
= 0. Then
U
(
x, s
)=
c
1
e
−
√
sx
+
u
1
x
s
.
From
U
(0
,s
)=
u
0
/s
we obtain
c
1
=
u
0
/s
. Hence
U
(
x, s
)=
u
0
e
−
√
sx
s
+
u
1
x
s
709

710. Exercises 15.2
and
u
(
x, t
)=
u
0

e
−
x
√
s
s
n
+
u
1
x
r
1
s
s
=
u
0
erfc

x
2
√
t

+
u
1
x.
15.
We use
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
+
u
0
s
.
The condition lim
x
→∞
u
(
x, t
)=
u
0
implies lim
x
→∞
U
(
x, s
)=
u
0
/s
, so we define
c
2
= 0. Then
U
(
x, s
)=
c
1
e
−
√
sx
+
u
0
s
.
The transform of the remaining boundary conditions gives
dU
dx
u
u
u
u
x
=0
=
U
(0
,s
)
.
This condition yields
c
1
=
−
u
0
/s
(
√
s
+1). Thus
U
(
x, s
)=
−
u
0
e
−
√
sx
s
(
√
s
+1)
+
u
0
s
and
u
(
x, t
)=
−
u
0

e
−
x
√
s
s
(
√
s
+1)
n
+
u
0
r
1
s
s
=
u
0
e
x
+
t
erfc

√
t
+
x
2
√
t

−
u
0
erfc

x
2
√
t

+
u
0
By (5) in the table in 15.1.
16.
We use
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
.
The condition lim
x
→∞
u
(
x, t
) = 0 implies lim
x
→∞
U
(
x, s
) = 0, so we define
c
2
= 0. Hence
U
(
x, s
)=
c
1
e
−
√
sx
.
The remaining boundary condition transforms into
dU
dx
u
u
u
u
x
=0
=
U
(0
,s
)
−
50
s
.
This condition gives
c
1
=50
/s
(
√
s
+1). Therefore
U
(
x, s
)=50
e
−
√
sx
s
(
√
s
+1)
and
u
(
x, t
)=50

e
−
x
√
s
s
(
√
s
+1)
n
=
−
50
e
x
+
t
erfc

√
t
+
x
2
√
t

+50 erfc

x
2
√
t

.
17.
We use
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
.
The condition lim
x
→∞
u
(
x, t
) = 0 implies lim
x
→∞
U
(
x, s
) = 0, so we define
c
2
= 0. Hence
U
(
x, s
)=
c
1
e
−
√
sx
.
The transform of
u
(0
,t
)=
f
(
t
)is
U
(0
,s
)=
F
(
s
). Therefore
U
(
x, s
)=
F
(
s
)
e
−
√
sx
710

711. Exercises 15.2
and
u
(
x, t
)=
h
F
(
s
)
e
−
x
√
s
e
=
x
2
√
π
T
t
0
f
(
t
−
τ
)
e
−
x
2
/
4
τ
τ
3
/
2
dτ.
18.
We use
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
.
The condition lim
x
→∞
u
(
x, t
) = 0 implies lim
x
→∞
U
(
x, s
) = 0, so we define
c
2
= 0. Then
U
(
x, s
)=
c
1
e
−
√
sx
.
The transform of the remaining boundary condition gives
dU
dx
u
u
u
u
x
=0
=
−
F
(
s
)
where
F
(
s
)=
{
f
(
t
)
}
. This condition yields
c
1
=
F
(
s
)
/
√
s
.Thus
U
(
x, s
)=
F
(
s
)
e
−
√
sx
√
s
.
Using entry (44) of the table and the convolution theorem we obtain
u
(
x, t
)=

F
(
s
)
·
e
−
√
sx
√
s
n
=
1
√
π
T
t
0
f
(
τ
)
e
−
x
2
/
4(
t
−
τ
)
√
t
−
τ
dτ.
19.
Transforming the partial differential equation gives
d
2
U
dx
2
−
sU
=
−
60
.
Using undetermied coefficients we obtain
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
+
60
s
.
The condition lim
x
→∞
u
(
x, t
) = 60 implies lim
x
→∞
U
(
x, s
)=60
/s
, so we define
c
2
= 0. The transform of the
remaining boundary condition gives
U
(0
,s
)=
60
s
+
40
s
e
−
2
s
.
This condition yields
c
1
=
40
s
e
−
2
s
.Thus
U
(
x, s
)=
60
s
+40
e
−
2
s
e
−
√
sx
s
.
Using entry (46) of the table in Appendix III and the second translation theorem we obtain
u
(
x, t
)=

60
s
+40
e
−
2
s
e
−
√
sx
s
n
= 60 +40 erfc

x
2
√
t
−
2

(
t
−
2)
.
20.
The solution of the transformed equation
d
2
U
dx
2
−
sU
=
−
100
by undetermined coefficients is
U
(
x, s
)=
c
1
e
√
sx
+
c
2
e
−
√
sx
+
100
s
.
From the fact that lim
x
→∞
U
(
x, s
) = 100
/s
we see that
c
1
=0. Thus
U
(
x, s
)=
c
2
e
−
√
sx
+
100
s
.
(
1
)
711

712. Exercises 15.2
Now the transform of the boundary condition at
x
=0is
U
(0
,s
)=20
l
1
s
−
1
s
e
−
s

.
It follows from (1) that
20
s
−
20
s
e
−
s
=
c
2
+
100
s
or
c
2
=
−
80
s
−
20
s
e
−
s
and so
U
(
x, s
)=

−
80
s
−
20
s
e
−
s

e
−
√
sx
+
100
s
=
100
s
−
80
s
e
−
√
sx
−
20
s
e
−
√
sx
e
−
s
.
Thus
u
(
x, t
) = 100
r
1
s
s
−
80

e
−
√
sx
s
n
−
20

e
−
√
sx
s
e
−
s
n
= 100
−
80 erfc
(
x/
2
√
t
)
−
20 erfc
,
x/
2
√
t
−
1
c
(
t
−
1)
.
21.
Transforming the partial differential equation gives
d
2
U
dx
2
−
sU
=0
and so
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
.
The condition lim
x
→−∞
u
(
x, t
) = 0 implies lim
x
→−∞
U
(
x, s
) = 0, so we define
c
1
= 0. The transform of the
remaining boundary condition gives
dU
dx
u
u
u
u
x
=1
=
100
s
−
U
(1
,s
)
.
This condition yields
c
2
√
se
√
s
=
100
s
−
c
2
e
√
s
from which it follows that
c
2
=
100
s
(
√
s
+1)
e
−
√
s
.
Thus
U
(
x, s
) = 100
e
−
(1
−
x
)
√
s
s
(
√
s
+1)
.
Using entry (49) of the table in Appendix III we obtain
u
(
x, t
) = 100

e
−
(1
−
x
)
√
s
s
(
√
s
+1)
n
= 100
l
−
e
1
−
x
+
t
erfc

√
t
+
1
−
x
√
t

+erfc

1
−
x
2
√
t

.
22.
Transforming the partial differential equation gives
k
d
2
U
dx
2
−
sU
=
−
r
s
.
Using undetermined coefficients we obtian
U
(
x, s
)=
c
1
e
−
√
s/k x
+
c
2
e
√
s/k x
+
r
s
2
.
712

713. Exercises 15.2
The condition lim
x
→∞
∂u
∂x
= 0 implies lim
x
→∞
dU
dx
= 0, so we define
c
2
= 0. The transform of the remaining boundary
condition gives
U
(0
,s
) = 0. This condition yields
c
1
=
−
r/s
2
.Thus
U
(
x, s
)=
r
g
1
s
2
−
e
−
√
s/k x
s
2
=
.
Using entries (3) and (46) of the table in Appendix III and the convolution theorem we obtain
u
(
x, t
)=
r

1
s
2
−
1
s
·
e
−
√
s/k x
s
n
=
rt
−
r
T
t
0
erfc

x
2
√
kτ

dτ.
23.
The solution of
d
2
U
dx
2
−
sU
=
−
u
0
−
u
0
sin
π
L
x
is
U
(
x, s
)=
c
1
cosh(
√
sx
)+
c
2
sinh(
√
sx
)+
u
0
s
+
u
0
s
+
π
2
/L
2
sin
π
L
x.
The transformed boundary conditions
U
(0
,s
)=
u
0
/s
and
U
(
L, s
)=
u
0
/s
give, in turn,
c
1
= 0 and
c
2
=0.
Therefore
U
(
x, s
)=
u
0
s
+
u
0
s
+
π
2
/L
2
sin
π
L
x
and
u
(
x, t
)=
u
0
r
1
s
s
+
u
0
r
1
s
+
π
2
/L
2
s
sin
π
L
x
=
u
0
+
u
0
e
−
π
2
t/L
2
sin
π
L
x.
24.
The transform of the partial differential equation is
k
d
2
U
dx
2
−
hU
+
h
u
m
s
=
sU
−
u
0
or
k
d
2
U
dx
2
−
(
h
+
s
)
U
=
−
h
u
m
s
−
u
0
.
By undetermined coefficients we find
U
(
x, s
)=
c
1
e
√
(
h
+
s
)
/k x
+
c
2
e
−
√
(
h
+
s
)
/k x
+
hu
m
+
u
0
s
s
(
s
+
h
)
.
The transformed boundary conditions are
U
r
(0
,s
)=0and
U
r
(
L, s
) = 0. These conditions imply
c
1
= 0 and
c
2
= 0. By partial fractions we then get
U
(
x, s
)=
hu
m
+
u
0
s
s
(
s
+
h
)
=
u
m
s
−
u
m
s
+
h
+
u
0
s
+
h
.
Therefore,
u
(
x, t
)=
u
m
r
1
s
s
−
u
m
r
1
s
+
h
s
+
u
0
r
1
s
+
h
s
=
u
m
−
u
m
e
−
ht
+
u
0
e
−
ht
.
25.
We use
U
(
x, s
)=
c
1
cosh
y
s
k
x
+
c
2
sinh
y
s
k
x
+
u
0
s
.
The transformed boundary conditions
dU
dx
u
u
u
u
x
=0
= 0 and
U
(1
,s
) = 0 give, in turn,
c
2
= 0 and
713

714. Exercises 15.2
c
1
=
−
u
0
/s
cosh
v
s/k
. Therefore
U
(
x, s
)=
u
0
s
−
u
0
cosh
v
s/k x
s
cosh
v
s/k
=
u
0
s
−
u
0
e
√
s/k
x
+
e
−
√
s/k x
s
(
e
√
s/k
+
e
−
√
s/k
)
=
u
0
s
−
u
0
e
√
s/k
(
x
−
1)
+
e
−
√
s/k
(
x
+1)
s
(1 +
e
−
2
√
s/k
)
=
u
0
s
−
u
0
g
e
−
√
s/k
(1
−
x
)
s
−
e
−
√
s/k
(3
−
x
)
s
+
e
−
√
s/k
(5
−
x
)
s
−···
=
−
u
0
g
e
−
√
s/k
(1+
x
)
s
−
e
−
√
s/k
(3+
x
)
s
+
e
−
√
s/k
(5+
x
)
s
−···
=
=
u
0
s
−
u
0
∞
U
n
=0
(
−
1)
n
g
e
−
(2
n
+1
−
x
)
√
s/
√
k
s
+
e
−
(2
n
+1+
x
)
√
s/
√
k
s
=
and
u
(
x, t
)=
u
0
r
1
s
s
−
u
0
∞
U
n
=0
(
−
1)
n
gw
e
−
(2
n
+1
−
x
)
√
s/
√
k
s
n
−

e
−
(2
n
+1+
x
)
√
s/
√
k
s
n=
=
u
0
−
u
0
∞
U
n
=0
(
−
1)
n
l
erfc

2
n
+1
−
x
2
√
kt

−
erfc

2
n
+1+
x
2
√
kt

.
26.
We use
c
(
x, s
)=
c
1
cosh
y
s
D
x
+
c
2
sinh
y
s
D
x.
The transform of the two boundary conditions are
c
(0
,s
)=
c
0
/s
and
c
(1
,s
)=
c
0
/s
. From these conditions we
obtain
c
1
=
c
0
/s
and
c
2
=
c
0
(1
−
cosh
v
s/D
)
/s
sinh
v
s/D .
Therefore
c
(
x, s
)=
c
0
g
cosh
v
s/D x
s
+
(1
−
cosh
v
s/D
)
s
sinh
v
s/D
sinh
v
s/D x
=
=
c
0
g
sinh
v
s/D
(1
−
x
)
s
sinh
v
s/D
+
sin
v
s/D x
s
sinh
v
s/D
=
=
c
0
g
e
√
s/D
(1
−
x
)
−
e
−
√
s/D
(1
−
x
)
s
(
e
√
s/D
−
e
−
√
s/D
)
+
e
√
s/D x
−
e
−
√
s/D x
s
(
e
√
s/D
−
e
−
√
s/D
)
=
=
c
0
g
e
−
√
s/D x
−
e
−
√
s/D
(2
−
x
)
s
(1
−
e
−
2
√
s/D
)
+
e
√
s/D
(
x
−
1)
−
e
−
√
s/D
(
x
+1)
s
(1
−
e
−
2
√
s/D
)
=
=
c
0
(
e
−
√
s/D x
−
e
−
√
s/D
(2
−
x
)
)
s
(
1+
e
−
2
√
s/D
+
e
−
4
√
s/D
+
···
)
+
c
0
(
e
√
s/D
(
x
−
1)
−
e
−
√
s/D
(
x
+1)
s
(
1+
e
−
2
√
s/D
+
e
−
4
√
s/D
+
···
)
714

715. Exercises 15.2
=
c
0
∞
U
n
=0
g
e
−
(2
n
+
x
)
√
s/D
s
−
e
−
(2
n
+2
−
x
)
√
s/D
s
=
+
c
0
∞
U
n
=0
g
e
−
(2
n
+1
−
x
)
√
s/D
s
−
e
−
(2
n
+1+
x
)
√
s/D
s
=
and
c
(
x, t
)=
c
0
∞
U
n
=0





e
−
(2
n
+
x
)
√
D
√
s
s



−



e
−
(2
n
+2
−
x
)
√
D
√
s
s





+
c
0
∞
U
n
=0





e
−
(2
n
+1
−
x
)
√
D
√
s
s



−



e
−
(2
n
+1+
x
)
√
D
√
s
s





=
c
0
∞
U
n
=0
l
erfc

2
n
+
x
2
√
Dt

−
erfc

2
n
+2
−
x
2
√
Dt

+
c
0
∞
U
n
=0
l
erfc

2
n
+1
−
x
2
√
Dt

−
erfc

2
n
+1+
x
2
√
Dt

.
Now using erfc(
x
)=1
−
erf(
x
) we get
c
(
x, t
)=
c
0
∞
U
n
=0
l
erf

2
n
+2
−
x
2
√
Dt

−
erf

2
n
+
x
2
√
Dt

+
c
0
∞
U
n
=0
l
erf

2
n
+1+
x
2
√
Dt

−
erf

2
n
+1
−
x
2
√
Dt

.
27.
We use
U
(
x, s
)=
c
1
e
−
√
RCs
+
RG x
+
c
2
e
√
RCs
+
RG
+
Cu
0
Cs
+
G
.
The condition lim
x
→∞
∂u/∂x
= 0 implies lim
x
→∞
dU/dx
= 0, so we define
c
2
= 0. Applying
U
(0
,s
)=0to
U
(
x, s
)=
c
1
e
−
√
RCsRG x
+
Cu
0
Cs
+
G
gives
c
1
=
−
Cu
0
/
(
Cs
+
G
). Therefore
U
(
x, s
)=
−
Cu
0
e
−
√
RCs
+
RG x
Cs
+
G
+
Cu
0
Cs
+
G
and
u
(
x, t
)=
u
0
r
1
s
+
G/C
s
−
u
0

e
−
x
√
RC
√
s
+
G/C
s
+
G/C
n
=
u
0
e
−
Gt/C
−
u
0
e
−
Gt/C
erfc

x
√
RC
2
√
t
i
=
u
0
e
−
Gt/C
g
1
−
erfc

x
2
y
RC
t
i=
=
u
0
e
−
Gt/C
erf

x
2
y
RC
t
i
.
715

716. x
cHx,tL
t=0.1
t=0.5
t=1
t=2
t=5
Exercises 15.2
28.
We use
U
(
x, s
)=
c
1
e
−
√
s
+
hx
+
c
2
e
√
s
+
hx
.
The condition lim
x
→∞
u
(
x, t
) = 0 implies lim
x
→∞
U
(
x, s
) = 0, so we take
c
2
= 0. Therefore
U
(
x, s
)=
c
1
e
−
√
s
+
hx
.
The Laplace transform of
u
(0
,t
)=
u
0
is
U
(0
,s
)=
u
0
/s
and so
U
(
x, s
)=
u
0
e
−
√
s
+
hx
s
and
u
(
x, t
)=
u
0

e
−
√
s
+
hx
s
n
=
u
0
r
1
s
e
−
√
s
+
hx
s
.
From the first translation theorem,
h
e
−
√
s
+
hx
e
=
e
−
ht
{
e
−
x
√
s
}
=
e
−
ht
x
2
√
πt
3
e
−
x
2
/
4
t
.
Thus, from the convolution theorem we obtain
u
(
x, s
)=
u
0
x
2
√
π
T
t
0
e
−
hτ
−
x
2
/
4
τ
τ
3
/
2
dτ.
29. (a)
Letting
C
(
x, s
)=
{
c
(
x, t
)
}
we obtain
d
2
C
dx
2
−
s
k
C
= 0 subject to
dC
dx
u
u
u
u
x
=0
=
−
A.
The solution of this initial-value problem is
C
(
x, s
)=
A
√
k
e
−
(
x/
√
k
)
√
s
√
s
,
so that
c
(
x, t
)=
A
y
k
πt
e
−
x
2
/
4
kt
.
(b)
(c)
T
∞
0
c
(
x, t
)
dx
=
Ak
erf
x
2
√
kt
u
u
u
u
∞
0
=
Ak
(1
−
0) =
Ak
30. (a)
We use
U
(
x, s
)=
c
1
e
−
(
s/a
)
x
+
c
2
e
(
s/a
)
x
+
v
2
0
F
0
(
a
2
−
v
2
0
)
s
2
e
−
(
s/v
0
)
x
.
The condition lim
x
→∞
u
(
x, t
) = 0 implies lim
x
→∞
U
(
x, s
) = 0, so we must define
c
2
= 0. Consequently
U
(
x, s
)=
c
1
e
−
(
s/a
)
x
+
v
2
0
F
0
(
a
2
−
v
2
0
)
s
2
e
−
(
s/v
0
)
x
.
716

717. Exercises 15.3
The remaining boundary condition transforms into
U
(0
,s
) = 0. From this we find
c
1
=
−
v
2
0
F
0
/
(
a
2
−
v
2
0
)
s
2
.
Therefore, by the second translation theorem
U
(
x, s
)=
−
v
2
0
F
0
(
a
2
−
v
2
0
)
s
2
e
−
(
s/a
)
x
+
v
2
0
F
0
(
a
2
−
v
2
0
)
s
2
e
−
(
s/v
0
)
x
and
u
(
x, t
)=
v
2
0
F
0
a
2
−
v
2
0
lr
e
−
(
x/v
0
)
s
s
2
s
−
r
e
−
(
x/a
)
s
s
2
st
=
v
2
0
F
0
a
2
−
v
2
0
lf
t
−
x
v
0

t
−
x
v
0

−
(
t
−
x
a
)(
t
−
x
a
)

.
(b)
In the case when
v
0
=
a
the solution of the transformed equation is
U
(
x, s
)=
c
1
e
−
(
s/a
)
x
+
c
2
e
(
s/a
)
x
−
F
0
2
as
xe
−
(
s/a
)
x
.
The usual analysis then leads to
c
1
= 0 and
c
2
= 0. Therefore
U
(
x, s
)=
−
F
0
2
as
xe
−
(
s/a
)
x
and
u
(
x, t
)=
−
xF
0
2
a
r
e
−
(
x/a
)
s
s
s
=
−
xF
0
2
a
(
t
−
x
a
)
.
Exercises 15.3
1.
From formulas (5) and (6) in the text,
A
(
α
)=
T
0
−
1
(
−
1) cos
αx dx
+
T
1
0
(2) cos
αx dx
=
−
sin
α
α
+2
sin
α
α
=
sin
α
α
and
B
(
α
)=
T
0
−
1
(
−
1) sin
αx dx
+
T
1
0
(2) sin
αx dx
=
1
−
cos
α
α
−
2
cos
α
−
1
α
=
3(1
−
cos
α
)
α
.
Hence
f
(
x
)=
1
π
T
∞
0
sin
α
cos
αx
+3(1
−
cos
α
) sin
αx
α
dα.
2.
From formulas (5) and (6) in the text,
A
(
α
)=
T
2
π
π
4 cos
αx dx
=4
sin 2
πα
−
sin
πα
α
and
B
(
α
)=
T
2
π
π
4 sin
αx dx
=4
cos
πα
−
cos 2
πα
α
.
717

718. Exercises 15.3
Hence
f
(
x
)=
4
π
T
∞
0
(sin 2
πα
−
sin
πα
) cos
αx
+(cos
πα
−
cos 2
πα
) sin
αx
α
dα
=
4
π
T
∞
0
sin 2
πα
cos
αx
−
cos 2
πα
sin
αx
−
sin
πα
cos
αx
+cos
πα
sin
αx
α
dα
=
4
π
T
∞
0
sin
α
(2
π
−
x
)
−
sin
α
(
π
−
x
)
α
dα.
3.
From formulas (5) and (6) in the text,
A
(
α
)=
T
3
0
x
cos
αx dx
=
x
sin
αx
α
u
u
u
u
3
0
−
1
α
T
3
0
sin
αx dx
=
3 sin 3
α
α
+
cos
αx
α
2
u
u
u
u
3
0
=
3
α
sin 3
α
+cos 3
α
−
1
α
2
and
B
(
α
)=
T
3
0
x
sin
αx dx
=
−
x
cos
αx
α
u
u
u
u
3
0
+
1
α
T
3
0
cos
αx dx
=
−
3 cos 3
α
α
+
sin
αx
α
2
u
u
u
u
3
0
=
sin 3
α
−
3
α
cos 3
α
α
2
.
Hence
f
(
x
)=
1
π
T
∞
0
(3
α
sin 3
α
+cos 3
α
−
1) cos
αx
+(sin 3
α
−
3
α
cos 3
α
) sin
αx
α
2
dα
=
1
π
T
∞
0
3
α
(sin 3
α
cos
αx
−
cos 3
α
sin
αx
) +cos 3
α
cos
αx
+sin 3
α
sin
αx
−
cos
αx
α
2
dα
=
1
π
T
∞
0
3
α
sin
α
(3
−
x
) +cos
α
(3
−
x
)
−
cos
αx
α
2
dα.
4.
From formulas (5) and (6) in the text,
A
(
α
)=
T
∞
−∞
f
(
x
) cos
αx dx
=
T
0
−∞
0
·
cos
αx dx
+
T
π
0
sin
x
cos
αx dx
+
T
∞
π
0
·
cos
αx dx
=
1
2
T
π
0
[sin(1 +
α
)
x
+sin(1
−
α
)
x
]
dx
=
1
2
l
−
cos(1 +
α
)
x
1+
α
−
cos(1
−
α
)
x
1
−
α

π
0
=
−
1
2
l
cos(1 +
α
)
π
−
1
1+
α
+
cos(1
−
α
)
π
−
1
1
−
α

=
−
1
2
l
cos(1 +
α
)
π
−
α
cos(1 +
α
)
π
+cos(1
−
α
)
π
+
α
cos(1
−
α
)
π
−
2
1
−
α
2

=
1 +cos
απ
1
−
α
2
,
718

719. Exercises 15.3
and
B
(
α
)=
T
π
0
sin
x
sin
αx dx
=
1
2
T
π
0
[cos(1
−
α
)
x
−
cos(1 +
α
)]
dx
=
1
2
l
sin(1
−
α
)
π
1
−
α
−
sin(1 +
α
)
π
1+
α

=
sin
απ
1
−
α
2
.
Hence
f
(
x
)=
1
π
T
∞
0
cos
αx
+cos
αx
cos
απ
+sin
αx
sin
απ
1
−
α
2
dα
=
1
π
T
∞
0
cos
αx
+cos
α
(
x
−
π
)
1
−
α
2
dα.
5.
From formula (5) in the text,
A
(
α
)=
T
∞
0
e
−
x
cos
αx dx.
Recall
{
cos
kt
}
=
s/
(
s
2
+
k
2
). If we set
s
= 1 and
k
=
α
we obtain
A
(
α
)=
1
1+
α
2
.
Now
B
(
α
)=
T
∞
0
e
−
x
sin
αx dx.
Recall
{
sin
kt
}
=
k/
(
s
2
+
k
2
). If we set
s
= 1 and
k
=
α
we obtain
B
(
α
)=
α
1+
α
2
.
Hence
f
(
x
)=
1
π
T
∞
0
cos
αx
+
α
sin
αx
1+
α
2
dα.
6.
From formulas (5) and (6) in the text,
A
(
α
)=
T
1
−
1
e
x
cos
αx dx
=
e
(cos
α
+
α
sin
α
)
−
e
−
1
(cos
α
−
α
sin
α
)
1+
α
2
=
2(sinh 1) cos
α
−
2
α
(cosh 1) sin
α
1+
α
2
and
B
(
α
)=
T
1
−
1
e
x
sin
αx dx
=
e
(sin
α
−
α
cos
α
)
−
e
−
1
(
−
sin
α
−
α
cos
α
)
1+
α
2
=
2(cosh 1) sin
α
−
2
α
(sinh 1) cos
α
1+
α
2
.
Hence
f
(
x
)=
1
π
T
∞
0
[
A
(
α
) cos
αx
+
B
(
α
) sin
αx
]
dα.
7.
The function is odd. Thus from formula (11) in the text
B
(
α
)=5
T
1
0
sin
αx dx
=
5(1
−
cos
α
)
α
.
719

720. Exercises 15.3
Hence from formula (10) in the text,
f
(
x
)=
10
π
T
∞
0
(1
−
cos
α
) sin
αx
α
dα.
8.
The function is even. Thus from formula (9) in the text
A
(
α
)=
π
T
2
1
cos
αx dx
=
π

sin 2
α
−
sin
α
α

.
Hence from formula (8) in the text,
f
(
x
)=2
T
∞
0
(sin 2
α
−
sin
α
) cos
αx
α
dα.
9.
The function is even. Thus from formula (9) in the text
A
(
α
)=
T
π
0
x
cos
αx dx
=
x
sin
αx
α
u
u
u
u
π
0
−
1
α
T
π
0
sin
αx dx
=
πα
sin
πα
α
+
1
α
2
cos
αx
u
u
u
u
π
0
=
πα
sin
πα
+cos
πα
−
1
α
2
.
Hence from formula (8) in the text
f
(
x
)=
2
π
T
∞
0
(
πα
sin
πα
+cos
πα
−
1) cos
αx
α
2
dα.
10.
The function is odd. Thus from formula (11) in the text
B
(
α
)=
T
π
0
x
sin
αx dx
=
−
x
cos
αx
α
u
u
u
u
π
0
+
1
α
T
π
0
cos
αx dx
=
−
π
cos
πα
α
+
1
α
2
sin
αx
u
u
u
u
π
0
=
−
πα
cos
πα
+sin
πα
α
2
.
Hence from formula (10) in the text,
f
(
x
)=
2
π
T
∞
0
(
−
πα
cos
πα
+sin
πα
) sin
αx
α
2
dα.
11.
The function is odd. Thus from formula (11) in the text
B
(
α
)=
T
∞
0
(
e
−
x
sin
x
) sin
αx dx
=
1
2
T
∞
0
e
−
x
[cos(1
−
α
)
x
−
cos(1 +
α
)
x
]
dx
=
1
2
T
∞
0
e
−
x
cos(1
−
α
)
xdx
−
1
2
T
∞
0
e
−
x
cos(1 +
α
)
x, dx.
Now recall
{
cos
kt
}
=
T
∞
0
e
−
st
cos
kt dt
=
s/
(
s
2
+
k
2
)
.
If we set
s
= 1, and in turn,
k
=1
−
α
and then
k
=1+
α
, we obtain
B
(
α
)=
1
2
1
1+(1
−
α
)
2
−
1
2
1
1 +(1 +
α
)
2
=
1
2
(1 +
α
)
2
−
(1
−
α
)
2
[1 +(1
−
α
)
2
][1 +(1 +
α
)
2
]
.
Simplifying the last expression gives
B
(
α
)=
2
α
4+
α
4
.
720

721. Exercises 15.3
Hence from formula (10) in the text
f
(
x
)=
4
π
T
∞
0
α
sin
αx
4+
α
4
dα.
12.
The function is odd. Thus from formula (11) in the text
B
(
α
)=
T
∞
0
xe
−
x
sin
αx dx.
Now recall
{
t
sin
kt
}
=
−
d
ds
{
sin
kt
}
=2
ks/
(
s
2
+
k
2
)
2
.
If we set
s
= 1 and
k
=
α
we obtain
B
(
α
)=
2
α
(1 +
α
2
)
2
.
Hence from formula (10) in the text
f
(
x
)=
4
π
T
∞
0
α
sin
αx
(1 +
α
2
)
2
dα.
13.
For the cosine integral,
A
(
α
)=
T
∞
0
e
−
kx
cos
αx dx
=
k
k
2
+
α
2
.
Hence
f
(
x
)=
2
π
T
∞
0
k
cos
αx
k
2
+
α
2
=
2
k
π
T
∞
0
cos
αx
k
2
+
α
2
dα.
For the sine integral,
B
(
α
)=
T
∞
0
e
−
kx
sin
αx dx
=
α
k
2
+
α
2
.
Hence
f
(
x
)=
2
π
T
∞
0
α
sin
αx
k
2
+
α
2
dα.
14.
From Problem 13 the cosine and sine integral representations of
e
−
kx
,k>
0, are respectively,
e
−
kx
=
2
k
π
T
∞
0
cos
αx
k
2
+
α
2
dα
and
e
−
kx
=
2
π
T
∞
0
α
sin
αx
k
2
+
α
2
dα.
Hence, the cosine integral representation of
f
(
x
)=
e
−
x
−
e
−
3
x
is
e
−
x
−
e
−
3
x
=
2
π
T
∞
0
cos
αx
1+
α
2
dα
−
2(3)
π
T
∞
0
cos
αx
9+
α
2
dα
=
4
π
T
∞
0
3
−
α
2
(1 +
α
2
)(9+
α
2
)
cos
αx dα.
The sine integral representation of
f
is
e
−
x
−
e
−
3
x
=
2
π
T
∞
0
α
sin
αx
1+
α
2
dα
−
2
π
T
∞
0
α
sin
αx
9+
α
2
dα
=
16
π
T
∞
0
α
sin
αx
(1 +
α
2
)(9+
α
2
)
dα.
15.
For the cosine integral,
A
(
α
)=
T
∞
0
xe
−
2
x
cos
αx dx.
But we know
{
t
cos
kt
}
=
−
d
ds
s
(
s
2
+
k
2
)
=
(
s
2
−
k
2
)
(
s
2
+
k
2
)
2
.
If we set
s
= 2 and
k
=
α
we obtain
A
(
α
)=
4
−
α
2
(4 +
α
2
)
2
.
721

722. Exercises 15.3
Hence
f
(
x
)=
2
π
T
∞
0
(4
−
α
2
) cos
αx
(4 +
α
2
)
2
dα.
For the sine integral,
B
(
α
)=
T
∞
0
xe
−
2
x
sin
αx dx.
From Problem 12, we know
{
t
sin
kt
}
=
2
ks
(
s
2
+
k
2
)
2
.
If we set
s
= 2 and
k
=
α
we obtain
B
(
α
)=
4
α
(4 +
α
2
)
2
.
Hence
f
(
x
)=
8
π
T
∞
0
α
sin
αx
(4 +
α
2
)
2
dα.
16.
For the cosine integral,
A
(
α
)=
T
∞
0
e
−
x
cos
x
cos
αx dx
=
1
2
T
∞
0
e
−
x
[cos(1 +
α
)
x
+cos(1
−
α
)
x
]
dx
=
1
2
1
1 +(1 +
α
)
2
+
1
2
1
1+(1
−
α
)
2
=
1
2
1+(1
−
α
)
2
+1 +(1 +
α
)
2
[1 +(1 +
α
)
2
][1 +(1
−
α
)
2
]
=
2+
α
2
4+
α
4
.
Hence
f
(
x
)=
2
π
T
∞
0
(2 +
α
2
) cos
αx
4+
α
4
dα.
For the sine integral,
B
(
α
)=
T
∞
0
e
−
x
cos
x
sin
αx dx
=
1
2
T
∞
0
e
−
x
[sin(1 +
α
)
x
−
sin(1
−
α
)
x
]
dx
=
1
2
1+
α
1 +(1 +
α
)
2
−
1
2
1
−
α
1+(1
−
α
)
2
=
1
2
l
(1 +
α
)[1 +(1
−
α
)
2
]
−
(1
−
α
)[1 +(1 +
α
)
2
]
[1 +(1 +
α
)
2
][1 +(1
−
α
)
2
]

=
α
3
4+
α
4
.
Hence
f
(
x
)=
2
π
T
∞
0
α
3
sin
αx
4+
α
4
dα.
722

723. Exercises 15.4
17.
By formula (8) in the text
f
(
x
)=2
π
T
∞
0
e
−
α
cos
αx dα
=
2
π
1
1+
x
2
,x>
0
.
18.
From the formula for sine integral of
f
(
x
)wehave
f
(
x
)=
2
π
T
∞
0

T
∞
0
f
(
x
) sin
αx dx

sin
αx dx
=
2
π
l
T
1
0
1
·
sin
αx dα
+
T
∞
1
0
·
sin
αx dα

=
2
π
(
−
cos
αx
)
x
u
u
u
u
1
0
=
2
π
1
−
cos
x
x
.
19. (a)
From formula (7) in the text with
x
=2,wehave
1
2
=
2
π
T
∞
0
sin
α
cos
α
α
dα
=
1
π
T
∞
0
sin 2
α
α
dα.
If we let
α
=
x
we obtain
T
∞
0
sin 2
x
x
dx
=
π
2
.
(b)
If we now let 2
x
=
kt
where
k>
0, then
dx
=(
k/
2)
dt
and the integral in part (a) becomes
T
∞
0
sin
kt
kt/
2
(
k/
2)
dt
=
T
∞
0
sin
kt
t
dt
=
π
2
.
20.
With
f
(
x
)=
e
−|
x
|
, formula (16) in the text is
C
(
α
)=
T
∞
−∞
e
−|
x
|
e
iαx
dx
=
T
∞
−∞
e
−|
x
|
cos
αx dx
+
i
T
∞
−∞
e
−|
x
|
sin
αx dx.
The imaginary part in the last line is zero since the integrand is an odd function of
x
. Therefore,
C
(
α
)=
T
∞
−∞
e
−|
x
|
cos
αx dx
=2
T
∞
0
e
−
x
cos
αx dx
=
2
1+
α
2
and so from formula (15) in the text,
f
(
x
)=
1
π
T
∞
−∞
cos
αx
1+
α
2
dα
=
2
π
T
∞
0
cos
αx
1+
α
2
dα.
This is the same result obtained from formulas (8) and (9) in the text.
Exercises 15.4
For the boundary-value problems in this section it is sometimes useful to note that the identities
e
iα
= cos
α
+
i
sin
α
and
e
−
iα
= cos
α
−
i
sin
α
imply
e
iα
+
e
−
iα
= 2 cos
α
and
e
iα
−
e
−
iα
=2
i
sin
α.
723

724. Exercises 15.4
1.
Using the Fourier transform, the partial differential equation becomes
dU
dt
+
kα
2
U
= 0 and so
U
(
α, t
)=
ce
−
kα
2
t
.
Now
{
u
(
x,
0)
}
=
U
(
α,
0) =
h
e
−|
x
|
e
.
We have
h
e
−|
x
|
e
=
T
∞
−∞
e
−|
x
|
e
iαx
dx
=
T
∞
−∞
e
−|
x
|
(cos
αx
+
i
sin
αx
)
dx
=
T
∞
−∞
e
−|
x
|
cos
αx dx.
The integral
T
∞
−∞
e
−|
x
|
sin
αx dx
=0
since the integrand is an odd function of
x
. Continuing we obtain
h
e
−|
x
|
e
=2
T
∞
0
e
−
x
cos
αx dx
=
2
1+
α
2
.
But
U
(
α,
0) =
c
=
2
1+
α
2
gives
U
(
α, t
)=
2
e
−
kα
2
t
1+
α
2
and so
u
(
x, t
)=
2
2
π
T
∞
−∞
e
−
kα
2
t
e
−
iαx
1+
α
2
dα
=
1
π
T
∞
−∞
e
−
kα
2
t
1+
α
2
(cos
αx
−
i
sin
αx
)
dα
=
1
π
T
∞
−∞
e
−
kα
2
t
cos
αx
1+
α
2
dα
=
2
π
T
∞
0
e
−
kα
2
t
cos
αx
1+
α
2
dα.
2.
Since the domain of
x
is (
−∞
,
∞
) we transform the differential equation using the Fourier transform:
−
kα
2
U
(
α, t
)=
du
dt
du
dt
+
kα
2
U
(
α, t
)=0
U
(
α, t
)=
ce
−
kα
2
t
.
(
1
)
The transform of the initial condition is
{
u
(
x,
0)
}
=
T
∞
−∞
u
(
x,
0)
e
iαx
dx
=
T
0
−
1
(
−
100
e
iαx
)
dx
+
T
1
0
100
e
iαx
dx
=
−
100
1
−
e
−
iα
iα
+100
e
iα
−
1
iα
= 100
e
iα
+
e
−
iα
−
2
iα
= 100
2 cos
α
−
2
iα
= 200
cos
α
−
1
iα
.
Thus
U
(
α,
0) = 200
cos
α
−
1
iα
,
and since
c
=
U
(
α,
0) in (1) we have
U
(
α, t
) = 200
cos
α
−
1
iα
e
−
kα
2
t
.
724

725. Exercises 15.4
Applying the inverse Fourier transform we obtain
u
(
x, t
)=
−
1
{
U
(
α, t
)
}
=
1
2
π
T
∞
−∞
200
cos
α
−
1
iα
e
−
kα
2
t
e
−
iαx
dα
=
100
π
T
∞
−∞
200
cos
α
−
1
iα
e
−
kα
2
t
(cos
αx
−
i
sin
αx
)
dα
=
100
π
T
∞
−∞
cos
αx
(cos
α
−
1)
iα
e
−
kα
2
t
&
'( )
odd function
dα
−
100
π
T
∞
−∞
sin
αx
(cos
α
−
1)
α
e
−
kα
2
t
&
'( )
even function
dα
=
200
π
T
∞
0
sin
αx
(1
−
cos
α
)
α
e
−
kα
2
t
dα.
3.
Using the Fourier transform, the partial differential equation equation becomes
dU
dt
+
kα
2
U
= 0 and so
U
(
α, t
)=
ce
−
kα
2
t
.
Now
{
u
(
x,
0)
}
=
U
(
α,
0) =
√
πe
−
α
2
/
4
by the given result. This gives
c
=
√
πe
−
α
2
/
4
and so
U
(
α, t
)=
√
πe
−
(
1
4
+
kt
)
α
2
.
Using the given Fourier transform again we obtain
u
(
x, t
)=
√
π
−
1
{
e
−
(1+4
kt
)
α
2
/
4
}
=
1
√
1+4
kt
e
−
x
2
/
(1+4
kt
)
.
4. (a)
We use
U
(
α, t
)=
ce
−
kα
2
t
. The Fourier transform of the boundary condition is
U
(
α,
0) =
F
(
α
). This gives
c
=
F
(
α
) and so
U
(
α, t
)=
F
(
α
)
e
−
kα
2
t
. By the convolution theorem and the given result, we obtain
u
(
x, t
)=
−
1
{
F
(
α
)
·
e
−
kα
2
t
}
=
1
2
√
kπt
T
∞
−∞
f
(
τ
)
e
−
(
x
−
τ
)
2
/
4
kt
dτ.
(b)
Using the definition of
f
and the solution is part (a) we obtain
u
(
x, t
)=
u
0
2
√
kπt
T
1
−
1
e
−
(
x
−
τ
)
2
/
4
kt
dτ.
If
u
=
x
−
τ
2
√
kt
, then
dτ
=
−
2
√
kt du
and the integral becomes
u
(
x, t
)=
u
0
√
π
T
(
x
+1)
/
2
√
πt
(
x
−
1)
/
2
√
kt
e
−
u
2
du.
Using the result in Problem 9, Exercises 15
.
1, we have
u
(
x, t
)=
u
0
2
l
erf

x
+1
2
√
kt

−
erf

x
−
1
2
√
kt

.
725

726. -4
-2
0
2
4
x
2
4
6
t
0
20
40
uHx,tL
4
-2
0
2
x -4 -2 2 4
t
20
40
60
80
100
u
t=0.05
t=15
Exercises 15.4
5.
Since erf(0) = 0 and lim
x
→∞
erf(
x
) = 1, we have
lim
t
→∞
u
(
x, t
) = 50[erf(0)
−
erf(0)] = 0
and
lim
x
→∞
u
(
x, t
) = 50[erf(
∞
)
−
erf(
∞
)] = 50[1
−
1] = 0
.
6. (a)
Using the Fourier sine transform, the partial differential equation becomes
dU
dt
+
kα
2
U
=
kαu
0
.
The general solution of this linear equation is
U
(
α, t
)=
ce
−
kα
2
t
+
u
0
α
.
But
U
(
α,
0) = 0 implies
c
=
−
u
0
/α
and so
U
(
α, t
)=
u
0
1
−
e
−
kα
2
t
α
and
u
(
x, t
)=
2
u
0
π
T
∞
0
1
−
e
−
kα
2
t
α
sin
αx dα.
(b)
The solution of part (a) can be written
u
(
x, t
)=
2
u
0
π
T
∞
0
sin
αx
α
dα
−
2
u
0
π
T
∞
0
sin
αx
α
e
−
kα
2
t
dα.
Using
T
∞
0
sin
αx
α
dα
=
π/
2 the last line becomes
u
(
x, t
)=
u
0
−
2
u
0
π
T
∞
0
sin
αx
α
e
−
kα
2
t
dα.
7.
Using the Fourier sine transform we find
U
(
α, t
)=
ce
−
kα
2
t
.
Now
S
{
u
(
x,
0)
}
=
U
(
α,
0) =
T
1
0
sin
αx dx
=
1
−
cos
α
α
.
726

727. Exercises 15.4
From this we find
c
=(1
−
cos
α
)
/α
and so
U
(
α, t
)=
1
−
cos
α
α
e
−
kα
2
t
and
u
(
x, t
)=
2
π
x
∞
0
1
−
cos
α
α
e
−
kα
2
t
sin
αx dα.
8.
Since the domain of
x
is (0
,
∞
) and the condition at
x
= 0 involves
∂u/∂x
we use the Fourier cosine transform:
−
kα
2
U
(
α, t
)
−
ku
x
(0
,t
)=
dU
dt
dU
dt
+
kα
2
U
(
α, t
)=
kA
U
(
α, t
)=
ce
−
kα
2
t
+
A
α
2
.
Since
{
u
(
x,
0)
}
=
U
(
α,
0)=0
we find
c
=
−
A/α
2
, so that
U
(
α, t
)=
A
1
−
e
−
kα
2
t
α
2
.
Applying the inverse Fourier cosine transform we obtain
u
(
x, t
)=
C
−
1
{
U
(
α, t
)
}
=
2
A
π
x
∞
0
1
−
e
−
kα
2
t
α
2
cos
αx dα.
9.
Using the Fourier cosine transform we find
U
(
α, t
)=
ce
−
kα
2
t
.
Now
C
{
u
(
x,
0)
}
=
x
1
0
cos
αx dx
=
sin
α
α
=
U
(
α,
0)
.
From this we obtain
c
= (sin
α
)
/α
and so
U
(
α, t
)=
sin
α
α
e
−
kα
2
t
and
u
(
x, t
)=
2
π
x
∞
0
sin
α
α
e
−
kα
2
t
cos
αx dα.
10.
Using the Fourier sine transform we find
U
(
α, t
)=
ce
−
kα
2
t
+
1
α
.
Now
S
{
u
(
x,
0)
}
=
S
*
e
−
x
+
=
x
∞
0
e
−
x
sin
αx dx
=
α
1+
α
2
=
U
(
α,
0)
.
From this we obtain
c
=
α/
(1 +
α
2
)
−
1
/α
. Therefore
U
(
α, t
)=

α
1+
α
2
−
1
α

e
−
kα
2
t
+
1
α
=
1
α
−
e
−
kα
2
t
α
(1 +
α
2
)
and
u
(
x, t
)=
2
π
x
∞
0

1
α
−
e
−
kα
2
t
α
(1 +
α
2
)
i
sin
αx dα.
727

728. Exercises 15.4
11. (a)
Using the Fourier transform we obtain
U
(
α, t
)=
c
1
cos
αat
+
c
2
sin
αat.
If we write
{
u
(
x,
0)
}
=
{
f
(
x
)
}
=
F
(
α
)
and
{
u
t
(
x,
0)
}
=
{
g
(
x
)
}
=
G
(
α
)
we first obtain
c
1
=
F
(
α
) from
U
(
α,
0) =
F
(
α
) and then
c
2
=
G
(
α
)
/αa
from
dU
dt
u
u
u
u
t
=0
=
G
(
α
). Thus
U
(
α, t
)=
F
(
α
) cos
αat
+
G
(
α
)
αa
sin
αat
and
u
(
x, t
)=
1
2
π
T
∞
−∞

F
(
α
) cos
αat
+
G
(
α
)
αa
sin
αat

e
−
iαx
dα.
(b)
If
g
(
x
) = 0 then
c
2
= 0 and
u
(
x, t
)=
1
2
π
T
∞
−∞
F
(
α
) cos
αate
−
iαx
dα
=
1
2
π
T
∞
−∞
F
(
α
)

e
αati
+
e
−
αati
2

e
−
iαx
dα
=
1
2
l
1
2
π
T
∞
−∞
F
(
α
)
e
−
i
(
x
−
at
)
α
dα
+
1
2
π
T
∞
−∞
F
(
α
)
e
−
i
(
x
+
at
)
α
dα

=
1
2
[
f
(
x
−
at
)+
f
(
x
+
at
)]
.
12.
Using the Fourier sine transform we obtain
U
(
α, t
)=
c
1
cos
αat
+
c
2
sin
αat.
Now
S
{
u
(
x,
0)
}
=
*
xe
−
x
+
=
T
∞
0
xe
−
x
sin
αx dx
=
2
α
(1 +
α
2
)
2
=
U
(
α,
0)
.
Also,
S
{
u
t
(
x,
0)
}
=
dU
dt
u
u
u
u
t
=0
=0
.
This last condition gives
c
2
= 0. Then
U
(
α,
0) = 2
α/
(1 +
α
2
)
2
yields
c
1
=2
α/
(1 +
α
2
)
2
. Therefore
U
(
α, t
)=
2
α
(1 +
α
2
)
2
cos
αat
and
u
(
x, t
)=
4
π
T
∞
0
α
cos
αat
(1 +
α
2
)
2
sin
αx dα.
13.
Using the Fourier cosine transform we obtain
U
(
x, α
)=
c
1
cosh
αx
+
c
2
sinh
αx.
728

729. Exercises 15.4
Now the Fourier cosine transforms of
u
(0
,y
)=
e
−
y
and
u
(
π, y
) = 0 are, respectively,
U
(0
,α
)=1
/
(1 +
α
2
) and
U
(
π, α
) = 0. The first of these conditions gives
c
1
=1
/
(1 +
α
2
). The second condition gives
c
2
=
−
cosh
απ
(1 +
α
2
) sinh
απ
.
Hence
U
(
x, α
)=
cosh
αx
1+
α
2
−
cosh
απ
sinh
αx
(1 +
α
2
) sinh
απ
=
sinh
απ
cosh
απ
−
cosh
απ
sinh
αx
(1 +
α
2
) sinh
απ
=
sinh
α
(
π
−
x
)
(1 +
α
2
) sinh
απ
and
u
(
x, t
)=
2
π
T
∞
0
sinh
α
(
π
−
x
)
(1 +
α
2
) sinh
απ
cos
αy dα.
14.
Since the boundary condition at
y
= 0 now involves
u
(
x,
0) rather than
u
r
(
x,
0), we use the Fourier sine
transform. The transform of the partial differential equation is then
d
2
U
dx
2
−
α
2
U
+
αu
(
x,
0) = 0 or
d
2
U
dx
2
−
α
2
U
=
−
α.
The solution of this differential equation is
U
(
x, α
)=
c
1
cosh
αx
+
c
2
sinh
αx
+
1
α
.
The transforms of the boundary conditions at
x
= 0 and
x
=
π
in turn imply that
c
1
=1
/α
and
c
2
=
cosh
απ
α
sinh
απ
−
1
α
sinh
απ
+
α
(1 +
α
2
) sinh
απ
.
Hence
U
(
α, x
)=
1
α
−
cosh
αx
α
+
cosh
απ
α
sinh
απ
sinh
αx
−
sinh
αx
α
sinh
απ
+
α
sinh
αx
(1 +
α
2
) sinh
απ
=
1
α
−
sinh
α
(
π
−
x
)
α
sinh
απ
−
sinh
αx
α
(1 +
α
2
) sinh
απ
.
Taking the inverse transform it follows that
u
(
x, y
)=
2
π
T
∞
0

1
α
−
sinh
α
(
π
−
x
)
α
sinh
απ
−
sinh
αx
α
(1 +
α
2
) sinh
απ

sin
αy dα.
15.
Using the Fourier cosine transform with respect to
x
gives
U
(
α, y
)=
c
1
e
−
αy
+
c
2
e
αy
.
Since we expect
u
(
x, y
) to be bounded as
y
→∞
we define
c
2
=0. Thus
U
(
α, y
)=
c
1
e
−
αy
.
Now
C
{
u
(
x,
0)
}
=
T
1
0
50 cos
αx dx
=50
sin
α
α
and so
U
(
α, y
)=50
sin
α
α
e
−
αy
and
u
(
x, y
)=
100
π
T
∞
0
sin
α
α
e
−
αy
cos
αx dα.
729

730. Exercises 15.4
16.
The boundary condition
u
(0
,y
) = 0 indicates that we now use the Fourier sine transform. We still have
U
(
α, y
)=
c
1
e
−
αy
, but
S
{
u
(
x,
0)
}
=
T
1
0
50 sin
αx dx
= 50(1
−
cos
α
)
/α
=
U
(
α,
0)
.
This gives
c
1
= 50(1
−
cos
α
)
/α
and so
U
(
α, y
)=50
1
−
cos
α
α
e
−
αy
and
u
(
x, y
)=
100
π
T
∞
0
1
−
cos
α
α
e
−
αy
sin
αx dα.
17.
We use the Fourier sine transform with respect to
x
to obtain
U
(
α, y
)=
c
1
cosh
αy
+
c
2
sinh
αy.
The transforms of
u
(
x,
0) =
f
(
x
) and
u
(
x,
2) = 0 give, in turn,
U
(
α,
0) =
F
(
α
) and
U
(
α,
2) = 0. The first
condition gives
c
1
=
F
(
α
) and the second condition then yields
c
2
=
−
F
(
α
) cosh 2
α
sinh 2
α
.
Hence
U
(
α, y
)=
F
(
α
) cosh
αy
−
F
(
α
) cosh 2
α
sinh
αy
sinh 2
α
=
F
(
α
)
sinh 2
α
cosh
αy
−
cosh 2
α
sinh
αy
sinh 2
α
=
F
(
α
)
sinh
α
(2
−
y
)
sinh 2
α
and
u
(
x, y
)=
2
π
T
∞
0
F
(
α
)
sinh
α
(2
−
y
)
sinh 2
α
sin
αx dα.
18.
The domain of
y
and the boundary condition at
y
= 0 suggest that we use a Fourier cosine transform. The
transformed equation is
d
2
U
dx
2
−
α
2
U
−
u
y
(
x,
0) = 0 or
d
2
U
dx
2
−
α
2
U
=0
.
Because the domain of the variable
x
is a finite interval we choose to write the general solution of the latter
equation as
U
(
x, α
)=
c
1
cosh
αx
+
c
2
sinh
αx.
Now
U
(0
,α
)=
F
(
α
), where
F
(
α
) is the Fourier cosine transform of
f
(
y
), and
U
r
(
π, α
) = 0 imply
c
1
=
F
(
α
)
and
c
2
=
−
F
(
α
) sinh
απ/
cosh
απ
.Thus
U
(
x, α
)=
F
(
α
) cosh
αx
−
F
(
α
)
sinh
απ
cosh
απ
sinh
αx
=
F
(
α
)
cosh
α
(
π
−
x
)
cosh
απ
.
Using the inverse transform we find that a solution to the problem is
u
(
x, y
)=
2
π
T
∞
0
F
(
α
)
cosh
α
(
π
−
x
)
cosh
απ
cos
αy dα.
730

731. Exercises 15.4
19.
We solve two boundary-value problems:
Using the Fourier sine transform with respect to
y
gives
u
1
(
x, y
)=
2
π
T
∞
0
αe
−
αx
1+
α
2
sin
αy dα.
The Fourier sine transform with respect to
x
yields the solution to the second problem:
u
2
(
x, y
)=
2
π
T
∞
0
αe
−
αy
1+
α
2
sin
αx dα.
We define the solution of the original problem to be
u
(
x, y
)=
u
1
(
x, y
)+
u
2
(
x, y
)=
2
π
T
∞
0
α
1+
α
2
,
e
−
αx
sin
αy
+
e
−
αy
sin
αx
-
dα.
20.
We solve the three boundary-value problems:
Using separation of variables we find the solution of the first problem is
u
1
(
x, y
)=
∞
U
n
=1
A
n
e
−
ny
sin
nx
where
A
n
=
2
π
T
π
0
f
(
x
) sin
nx dx.
Using the Fourier sine transform with respect to
y
gives the solution of the second problem:
u
2
(
x, y
)=
200
π
T
∞
0
(1
−
cos
α
) sinh
α
(
π
−
x
)
α
sinh
απ
sin
αy dα.
Also, the Fourier sine transform with respect to
y
gives the solution of the third problem:
u
3
(
x, y
)=
2
π
T
∞
0
α
sinh
αx
(1 +
α
2
) sinh
απ
sin
αy dα.
The solution of the original problem is
u
(
x, y
)=
u
1
(
x, y
)+
u
2
(
x, y
)+
u
3
(
x, y
)
.
21.
Using the Fourier transform with respect to
x
gives
U
(
α, y
)=
c
1
cosh
αy
+
c
2
sinh
αy.
The transform of the boundary condition
∂u
∂y
u
u
u
u
y
=0
=0 is
dU
dy
u
u
u
u
y
=0
= 0. This condition gives
c
2
= 0. Hence
U
(
α, y
)=
c
1
cosh
αy.
Now by the given information the transform of the boundary condition
u
(
x,
1) =
e
−
x
2
is
731

732. Exercises 15.4
U
(
α,
1) =
√
πe
−
α
2
/
4
. This condition then gives
c
1
=
√
πe
−
α
2
/
4
cosh
α
. Therefore
U
(
α, y
)=
√
π
e
−
α
2
/
4
cosh
αy
cosh
α
and
U
(
x, y
)=
1
2
√
π
T
∞
−∞
e
−
α
2
/
4
cosh
αy
cosh
α
e
−
iαx
dα
=
1
2
√
π
T
∞
−∞
e
−
α
2
/
4
cosh
αy
cosh
α
cos
αx dα
=
1
√
π
T
∞
0
e
−
α
2
/
4
cosh
αy
cosh
α
cos
αx dα.
22.
Entries 42 and 43 of the table in Appendix III imply
T
∞
0
e
−
st
sin
at
t
dt
= arctan
a
s
and
T
∞
0
e
−
st
sin
at
cos
bt
t
dt
=
1
2
arctan
a
+
b
s
+
1
2
arctan
a
−
b
s
.
Identifying
α
=
t, x
=
a
, and
y
=
s
, the solution of Problem 16 is
u
(
x, y
)=
100
π
T
∞
0
1
−
cos
α
α
e
−
αy
sin
αx dα
=
100
π
l
T
∞
0
sin
αx
α
e
−
αy
dα
−
T
∞
0
sin
αx
cos
α
α
e
−
αy
dα

=
100
π
l
arctan
x
y
−
1
2
arctan
x
+1
y
−
1
2
arctan
x
−
1
y

.
Exercises 15.5
1.
We show that
1
4
F
4
F
4
=
I
:
1
4
F
4
F
4
=
1
4





1111
1
−
i
−
1
i
1
−
11
−
1
1
i
−
1
−
i










1111
1
i
−
1
−
i
1
−
11
−
1
1
−
i
−
1
i





=
1
4





4000
0400
0040
0004





=
I.
Thus
F
−
1
4
=
1
4
F
4
.
2.
We have
T
∞
−∞
f
(
x
)
δ
a
(
x
−
a
)
dx
=
1
2
*
T
a
+
a
a
−
a
f
(
x
)
dx
=
1
2
*
f
(
c
)(2
*
)=
f
(
c
)
by the mean value theorem for integrals.
3.
By the sifting property,
{
δ
(
x
)
}
=
T
∞
−∞
δ
(
x
)
e
iαx
dx
=
e
iα
0
=1
.
732

733. Exercises 15.5
4.
We already know that
f
∗
δ
=
δ
∗
f
. Then, by the sifting property,
(
f
∗
δ
)(
x
)=
T
∞
∞
f
(
τ
)
δ
(
x
−
τ
)
dτ
=
T
∞
−∞
f
(
τ
)
δ
(
τ
−
x
)
dτ
=
f
(
x
)
.
5.
Using integration by parts with
u
=
f
(
x
) and
dv
=
δ
r
(
x
−
a
) we find
T
∞
−∞
f
(
x
)
δ
r
(
x
−
a
)
dx
=
−
T
∞
−∞
f
r
(
x
)
δ
(
x
−
a
)
dx
=
−
f
r
(
a
)
by the sifting property.
6.
Using a CAS we find
{
g
(
x
)
}
=
1
2
[sign(
A
−
α
) +sign(
A
+
α
)]
where sign(
t
)=1if
t>
0 and sign
t
=
−
1if
t<
0. Thus
{
g
(
x
)
}
=
r
1
,
−
A<α<A
0
,
elsewhere.
7.
Using
ω
8
=
√
2
2
+
i
√
2
2
ω
2
8
=
i
ω
3
8
=
−
√
2
2
+
i
√
2
2
ω
4
8
=
−
1
ω
5
8
=
−
√
2
2
−
i
√
2
2
ω
6
8
=
−
i
ω
7
8
=
√
2
2
−
i
√
2
2
ω
8
8
=1
we have
F
8=



















11111111
1
√
2
2
+
i
√
2
2
i
−
√
2
2
+
i
√
2
2
−
1
−
√
2
2
−
i
√
2
2
−
i
√
2
2
−
i
√
2
2
1
i
−
1
−
i
1
i
−
1
−
i
1
−
√
2
2
+
i
√
2
2
−
i
√
2
2
+
i
√
2
2
−
1
√
2
2
−
i
√
2
2
i
−
√
2
2
−
i
√
2
2
1
−
11
−
11
−
11
−
1
1
−
√
2
2
−
i
√
2
2
i
√
2
2
−
i
√
2
2
−
1
√
2
2
+
i
√
2
2
−
i
−
√
2
2
+
i
√
2
2
1
−
i
−
1
i
1
−
i
−
1
i
1
√
2
2
−
i
√
2
2
−
i
−
√
2
2
−
i
√
2
2
−
1
−
√
2
2
+
i
√
2
2
i
√
2
2
+
i
√
2
2



















.
In factored form
F
8=

I
4
D
4
I
4
−
D
4

F
4
0
0
F
4

P,
where
I
4
is the 4
×
4 identity matrix.
D
4
=





100 0
0
√
2
/
2+
i
√
2
/
20 0
00
i
0
000
−
√
2
/
2+
i
√
2
/
2





,
and
P
is the 8
×
8 matrix with 1 in positions (1
,
1), (2
,
3), (3
,
5), (4
,
7), (5
,
2), (6
,
4), (7
,
6), and (8
,
8).
733

734. -1-0.5 0.5 1
x
-1
-0.5
0.5
1
y
-2 -1 1 2
0.2
0.4
0.6
0.8
Exercises 15.5
8.
The 8th roots of unity,
ω
1
8
,
ω
2
8
,
...
,
ω
8
8
are shown in the solution
of Problem 7 above. The points in the complex plane are equally
spaced on the perimeter of the unit circle.
9.
The Fourier transform of
g
(
x
) = (sin 2
x
)
/πx
is
G
(
α
)=
r
1
,
−
2
<α<
2
0
,
elsewhere.
This implies that (
f
∗
g
)(
x
)=
−
1
{
F
(
α
)
G
(
α
)
}
is band-limited.
The graph of
F
(
α
)
G
(
α
), which is identical to the graph of (
f
∗
g
), is shown.
10.
For
N
=6,
F
6
=











11 1 11 1
11
/
2+
√
3
i/
2
−
1
/
2+
√
3
i/
2
−
1
−
1
/
2
−
√
3
i/
21
/
2
−
√
3
i/
2
1
−
1
/
2+
√
3
i/
2
−
1
/
2
−
√
3
i/
21
−
1
/
2+
√
3
i/
2
−
1
/
2
−
√
3
i/
2
1
−
11
−
11
−
1
1
−
1
/
2
−
√
3
i/
2
−
1
/
2+
√
3
i/
21
−
1
/
2
−
√
3
i/
2
−
1
/
2+
√
3
i/
2
11
/
2
−
√
3
i/
2
−
1
/
2
−
√
3
i/
2
−
1
−
1
/
2+
√
3
i/
21
/
2+
√
3
i/
2











.
If, for example,
f
=(2
,
0
,
1
,
6
,
2
,
3), then
c
=
1
6
F
6
f
=










7
/
3
−
2
/
3+
√
3
i/
3
5
/
6
−
√
3
i/
6
−
2
/
3
5
/
6
−
√
3
i/
6
−
2
/
3
−
i/
√
3










.
Chapter 5 Review Exercises
1.
The partial differential equation and the boundary conditions indicate that the Fourier cosine transform is
appropriate for the problem. We find in this case
u
(
x, y
)=
2
π
T
∞
0
sinh
αy
α
(1 +
α
2
) cosh
απ
cos
αx dα.
2.
We use the Laplace transform and undetermined coefficients to obtain
U
(
x, s
)=
c
1
cosh
√
sx
+
c
2
sinh
√
sx
+
50
s
+4
π
2
sin 2
πx.
The transformed boundary conditions
U
(0
,s
)=0and
U
(1
,s
) = 0 give, in turn,
c
1
= 0 and
c
2
= 0. Hence
U
(
x, s
)=
50
s
+4
π
2
sin 2
πx
734

735. Chapter 5 Review Exercises
and
u
(
x, t
) = 50 sin 2
πx
r
1
s
+4
π
2
s
=50
e
−
4
π
2
t
sin 2
πx.
3.
The Laplace transform gives
U
(
x, s
)=
c
1
e
−
√
s
+
hx
+
c
2
e
√
s
+
hx
+
u
0
s
+
h
.
The condition lim
x
→∞
∂u/∂x
= 0 implies lim
x
→∞
dU/dx
= 0 and so we define
c
2
=0. Thus
U
(
x, s
)=
c
1
e
−
√
s
+
hx
+
u
0
s
+
h
.
The condition
U
(0
,s
) = 0 then gives
c
1
=
−
u
0
/
(
s
+
h
) and so
U
(
x, s
)=
u
0
s
+
h
−
u
0
e
−
√
s
+
hx
s
+
h
.
With the help of the first translation theorem we then obtain
u
(
x, t
)=
u
0
r
1
s
+
h
s
−
u
0

e
−
√
s
+
hx
s
+
h
n
=
u
0
e
−
ht
−
u
0
e
−
ht
erfc

x
2
√
t

=
u
0
e
−
ht
l
1
−
erfc

x
2
√
t

=
u
0
e
−
ht
erf

x
2
√
t

.
4.
Using the Fourier transform and the result
*
e
−|
x
|
+
=1
/
(1 +
α
2
)wefind
u
(
x, t
)=
1
2
π
T
∞
−∞
1
−
e
−
α
2
t
α
2
(1 +
α
2
)
e
−
iαx
dα
=
1
2
π
T
∞
−∞
1
−
e
−
α
2
t
α
2
(1 +
α
2
)
cos
αx dα
=
1
π
T
∞
0
1
−
e
−
α
2
t
α
2
(1 +
α
2
)
cos
αx dα.
5.
The Laplace transform gives
U
(
x, s
)=
c
1
e
−
√
sx
+
c
2
e
√
sx
.
The condition lim
x
→∞
u
(
x, t
) = 0 implies lim
x
→∞
U
(
x, s
) = 0 and so we define
c
2
=0. Thus
U
(
x, s
)=
c
1
e
−
√
sx
.
The transform of the remaining boundary condition is
U
(0
,s
)=1
/s
2
. This gives
c
1
=1
/s
2
. Hence
U
(
x, s
)=
e
−
√
sx
s
2
and
u
(
x, t
)=

1
s
e
−
√
sx
s
n
.
Using
r
1
s
s
= 1 and

e
−
√
sx
s
n
= erfc

x
2
√
t

,
it follows from the convolution theorem that
u
(
x, t
)=
T
t
0
erfc

x
2
√
τ

dτ.
6.
The Laplace transform and undetermined coefficients gives
U
(
x, s
)=
c
1
cosh
sx
+
c
2
sinh
sx
+
s
−
1
s
2
+
π
2
sin
πx.
735

736. Chapter 5 Review Exercises
The conditions
U
(0
,s
) = 0 and
U
(1
,s
) = 0 give, in turn,
c
1
= 0 and
c
2
=0. Thus
U
(
x, s
)=
s
−
1
s
2
+
π
2
sin
πx
and
u
(
x, t
) = sin
πx
r
s
s
2
+
π
2
s
−
1
π
sin
πx
r
π
s
2
+
π
2
s
= (sin
πx
) cos
πt
−
1
π
(sin
πx
) sin
πt.
7.
The Fourier transform gives the solution
u
(
x, t
)=
u
0
2
π
T
∞
−∞

e
iαπ
−
1
iα

e
−
iαx
e
−
kα
2
t
dα
=
u
0
2
π
T
∞
−∞
e
iα
(
π
−
x
)
−
e
−
iαx
iα
e
−
kα
2
t
dα
=
u
0
2
π
T
∞
−∞
cos
α
(
π
−
x
)+
i
sin
α
(
π
−
x
)
−
cos
αx
+
i
sin
αx
iα
e
−
kα
2
t
dα.
Since the imaginary part of the integrand of the last integral is an odd function of
α
, we obtain
u
(
x, t
)=
u
0
2
π
T
∞
−∞
sin
α
(
π
−
x
) +sin
αx
α
e
−
kα
2
t
dα.
8.
Using the Fourier cosine transform we obtain
U
(
x, α
)=
c
1
cosh
αx
+
c
2
sinh
αx.
The condition
U
(0
,α
) = 0 gives
c
1
=0. Thus
U
(
x, α
)=
c
2
sinh
αx.
Now
C
{
u
(
π, y
)
}
=
T
2
1
cos
αy dy
=
sin 2
α
−
sin
α
α
=
U
(
π, α
)
.
This last condition gives
c
2
= (sin 2
α
−
sin
α
)
/α
sinh
απ
. Hence
U
(
x, α
)=
sin 2
α
−
sin
α
α
sinh
απ
sinh
αx
and
u
(
x, y
)=
2
π
T
∞
0
sin 2
α
−
sin
α
α
sinh
απ
sinh
αx
cos
αy dα.
9.
We solve the two problems
∂
2
u
1
∂x
2
+
∂
2
u
1
∂y
2
=0
,x>
0
,y>
0
,
u
1
(0
,y
)=0
,y>
0
,
u
1
(
x,
0) =
r
100
,
0
<x<
1
0
,x>
1
and
∂
2
u
2
∂x
2
+
∂
2
u
2
∂y
2
=0
,x>
0
,y>
0
,
u
2
(0
,y
)=
r
50
,
0
<y<
1
0
,y>
1
u
2
(
x,
0) = 0
.
736

737. Chapter 5 Review Exercises
Using the Fourier sine transform with respect to
x
we find
u
1
(
x, y
)=
200
π
T
∞
0

1
−
cos
α
α

e
−
αy
sin
αx dα.
Using the Fourier sine transform with respect to
y
we find
u
2
(
x, y
)=
100
π
T
∞
0

1
−
cos
α
α

e
−
αx
sin
αy dα.
The solution of the problem is then
u
(
x, y
)=
u
1
(
x, y
)+
u
2
(
x, y
)
.
10.
The Laplace transform gives
U
(
x, s
)=
c
1
cosh
√
sx
+
c
2
sinh
√
sx
+
r
s
2
.
The condition
∂u
∂x
u
u
u
u
x
=0
= 0 transforms into
dU
dx
u
u
u
u
x
=0
= 0. This gives
c
2
= 0. The remaining condition
u
(1
,t
)=0
transforms into
U
(1
,s
) = 0. This condition then implies
c
1
=
−
r/s
2
cosh
√
s
. Hence
U
(
x, s
)=
r
s
2
−
r
cosh
√
sx
s
2
cosh
√
s
.
Using geometric series and the convolution theorem we obtain
u
(
x, t
)=
r
r
1
s
2
s
−
r
r
cosh
√
sx
s
2
cosh
√
s
s
=
rt
−
r
∞
U
n
=0
(
−
1)
n
l
T
t
0
erfc

2
n
+1
−
x
2
√
τ

dτ
+
T
t
0
erfc

2
n
+1+
x
2
√
τ

dτ

.
11.
The Fourier sine transform with respect to
x
and undetermined coefficients give
U
(
α, y
)=
c
1
cosh
αy
+
c
2
sinh
αy
+
A
α
.
The transforms of the boundary conditions are
dU
dy
u
u
u
u
y
=0
= 0 and
dU
dy
u
u
u
u
y
=
π
=
Bα
1+
α
2
.
The first of these conditions gives
c
2
= 0 and so
U
(
α, y
)=
c
1
cosh
αy
+
A
α
.
The second transformed boundary condition yields
c
1
=
B/
(1 +
α
2
) sinh
απ
. Therefore
U
(
α, y
)=
B
cosh
αy
(1 +
α
2
) sinh
απ
+
A
α
and
u
(
x, y
)=
2
π
T
∞
0

B
cosh
αy
(1 +
α
2
) sinh
απ
+
A
α

sin
αx dα.
12.
Using the Laplace transform gives
U
(
x, s
)=
c
1
cosh
√
sx
+
c
2
sinh
√
sx.
737

738. Chapter 5 Review Exercises
The condition
u
(0
,t
)=
u
0
transforms into
U
(0
,s
)=
u
0
/s
. This gives
c
1
=
u
0
/s
. The condition
u
(1
,t
)=
u
0
transforms into
U
(1
,s
)=
u
0
/s
. This implies that
c
2
=
u
0
(1
−
cosh
√
s
)
/s
sinh
√
s
. Hence
U
(
x, s
)=
u
0
s
cosh
√
sx
+
u
0
l
1
−
cosh
√
s
s
sinh
√
s

sinh
√
sx
=
u
0
l
sinh
√
s
cosh
√
sx
−
cosh sinh
√
s
sinh
√
sx
+sinh
√
sx
s
sinh
√
s

=
u
0
l
sinh
√
s
(1
−
x
) +sinh
√
sx
s
sinh
√
s

=
u
0
l
sinh
√
s
(1
−
x
)
s
sinh
√
s
+
sinh
√
sx
s
sinh
√
s

and
u
(
x, t
)=
u
0
lr
sinh
√
s
(1
−
x
)
s
sinh
√
s
s
+
r
sinh
√
sx
s
sinh
√
s
st
=
u
0
∞
U
n
=0
l
erf

2
n
+2
−
x
2
√
t

−
erf

2
n
+
x
2
√
t

+
u
0
∞
U
n
=0
l
erf

2
n
+1+
x
2
√
t

−
erf

2
n
+1
−
x
2
√
t

.
13.
Using the Fourier transform gives
U
(
α, t
)=
c
1
e
−
kα
2
t
.
Now
u
(
α,
0) =
T
∞
0
e
−
x
e
iαx
dx
=
e
(
iα
−
1)
x
iα
−
1
u
u
u
u
∞
0
=0
−
1
iα
−
1
=
1
1
−
iα
=
c
1
so
U
(
α, t
)=
1+
iα
1+
α
2
e
−
kα
2
t
and
u
(
x, t
)=
1
2
π
T
∞
−∞
1+
iα
1+
α
2
e
−
kα
2
t
e
−
iαx
dα.
Since
1+
iα
1+
α
2
(cos
αx
−
i
sin
αx
)=
cos
αx
+
α
sin
αx
1+
α
2
+
i
(
α
cos
αx
−
sin
αx
)
1+
α
2
and the integral of the product of the second term with
e
−
kα
2
t
is 0 (it is an odd function), we have
u
(
x, t
)=
1
2
π
T
∞
−∞
cos
αx
+
α
sin
αx
1+
α
2
e
−
kα
2
t
dα.
738

739. 16
Numerical Solutions of
Partial Differential Equations
Exercises 16.1
1.
The figure shows the values of
u
(
x, y
) along the boundary. We need to determine
u
11
and
u
21
. The system is
u
21
+2+0+0
−
4
u
11
=0
1+2+
u
11
+0
−
4
u
21
=0
or
−
4
u
11
+
u
21
=
−
2
u
11
−
4
u
21
=
−
3
.
Solving we obtain
u
11
=11
/
15 and
u
21
=14
/
15.
2.
The figure shows the values of
u
(
x, y
) along the boundary. We need to determine
u
11
,
u
21
, and
u
31
. By symmetry
u
11
=
u
31
and the system is
u
21
+0+0+100
−
4
u
11
=0
u
31
+0+
u
11
+100
−
4
u
21
=0
0+0+
u
21
+100
−
4
u
31
=0
or
−
4
u
11
+
u
21
=
−
100
2
u
11
−
4
u
21
=
−
100
.
Solving we obtain
u
11
=
u
31
= 250
/
7 and
u
21
= 300
/
7.
3.
The figure shows the values of
u
(
x, y
) along the boundary. We need to determine
u
11
,
u
21
,
u
12
, and
u
22
. By symmetry
u
11
=
u
21
and
u
12
=
u
22
. The system is
u
21
+
u
12
+0+0
−
4
u
11
=0
0+
u
22
+
u
11
+0
−
4
u
21
=0
u
22
+
√
3
/
2+0+
u
11
−
4
u
12
=0
0+
√
3
/
2+
u
12
+
u
21
−
4
u
22
=0
or
3
u
11
+
u
12
=0
u
11
−
3
u
12
=
−
√
3
2
.
Solving we obtain
u
11
=
u
21
=
√
3
/
16 and
u
12
=
u
22
=3
√
3
/
16.
4.
The figure shows the values of
u
(
x, y
) along the boundary. We need to determine
u
11
,
u
21
,
u
12
, and
u
22
. The system is
u
21
+
u
12
+8+0
−
4
u
11
=0
0+
u
22
+
u
11
+0
−
4
u
21
=0
u
22
+0+16+
u
11
−
4
u
12
=0
0+0+
u
12
+
u
21
−
4
u
22
=0
or
−
4
u
11
+
u
21
+
u
12
=
−
8
u
11
−
4
u
21
+
u
22
=0
u
11
−
4
u
12
+
u
22
=
−
16
u
21
+
u
12
−
4
u
22
=0
.
Solving we obtain
u
11
=11
/
3,
u
21
=4
/
3,
u
12
=16
/
3, and
u
22
=5
/
3.
739

740. Exercises 16.1
5.
The figure shows the values of
u
(
x, y
) along the boundary. For Gauss-Seidel the co-
efficients of the unknowns
u
11
,
u
21
,
u
31
,
u
12
,
u
22
,
u
32
,
u
13
,
u
23
,
u
33
are shown in the
matrix

















0
.
25 0
.
2500000
.
25 0
.
25 0
.
250000
0
.
25000
.
25000
.
25000
.
25 0
.
25 0 0
0
.
25 0
.
25 0
.
25 0
.
25 0
00
.
25 0
.
25 0 0 0
.
25
000
.
25 0 0 0
.
25 0
0000
.
25 0
.
25 0
.
25
00000
.
25 0
.
25 0

















The constant terms in the equations are 0, 0, 6
.
25, 0, 0, 12
.
5, 6
.
25, 12
.
5, 37
.
5. We use 25 as the initial guess
for each variable. Then
u
11
=6
.
25,
u
21
=
u
12
=12
.
5,
u
31
=
u
13
=18
.
75,
u
22
= 25,
u
32
=
u
23
=37
.
5, and
u
33
=56
.
25
6.
The coefficients of the unknowns are the same as shown above in Problem 5. The constant terms are 7
.
5, 5,
20, 10, 0, 15, 17
.
5, 5, 27
.
5. We use 32
.
5 as the initial guess for each variable. Then
u
11
=21
.
92,
u
21
=28
.
30,
u
31
=38
.
17,
u
12
=29
.
38,
u
22
=33
.
13,
u
32
=44
.
38,
u
13
=22
.
46,
u
23
=30
.
45, and
u
33
=46
.
21.
7. (a)
Using the difference approximations for
u
xx
and
u
yy
we obtain
u
xx
+
u
yy
=
1
h
2
(
u
i
+1
,j
+
u
i,j
+1
+
u
i
−
1
,j
+
u
i,j
−
1
−
4
u
ij
)=
f
(
x, y
)
so that
u
i
+1
,j
+
u
i,j
+1
+
u
i
−
1
,j
+
u
i,j
−
1
−
4
u
ij
=
h
2
f
(
x, y
)
.
(b)
By symmetry, as shown in the figure, we need only solve for
u
1
,
u
2
,
u
3
,
u
4
, and
u
5
. The difference equations are
u
2
+0+0+1
−
4
u
1
=
1
4
(
−
2)
u
3
+0+
u
1
+1
−
4
u
2
=
1
4
(
−
2)
u
4
+0+
u
2
+
u
5
−
4
u
3
=
1
4
(
−
2)
0+0+
u
3
+
u
3
−
4
u
4
=
1
4
(
−
2)
u
3
+
u
3
+1+1
−
4
u
5
=
1
4
(
−
2)
or
u
1
=0
.
25
u
2
+0
.
375
u
2
=0
.
25
u
1
+0
.
25
u
3
+0
.
375
u
3
=0
.
25
u
2
+0
.
25
u
4
+0
.
25
u
5
+0
.
125
u
4
=0
.
5
u
3
+0
.
125
u
5
=0
.
5
u
3
+0
.
625
.
Using Gauss-Seidel iteration we find
u
1
=0
.
5427,
u
2
=0
.
6707,
u
3
=0
.
6402,
u
4
=0
.
4451, and
u
5
=0
.
9451.
740

741. TIME
X=0.25 X=0.50 X=0.75 X=1.00 X=1.25 X=1.50 X=1.75
0.000 1.0000 1.0000 1.0000 1.0000 0.0000 0.0000 0.0000
0.025 0.6000 1.0000 1.0000 0.6000 0.4000 0.0000 0.0000
0.050 0.5200 0.8400 0.8400 0.6800 0.3200 0.1600 0.0000
0.075 0.4400 0.7120 0.7760 0.6000 0.4000 0.1600 0.0640
0.100 0.3728 0.6288 0.6800 0.5904 0.3840 0.2176 0.0768
0.125 0.3261 0.5469 0.6237 0.5437 0.4000 0.2278 0.1024
0.150 0.2840 0.4893 0.5610 0.5182 0.3886 0.2465 0.1116
0.175 0.2525 0.4358 0.5152 0.4835 0.3836 0.2494 0.1209
0.200 0.2248 0.3942 0.4708 0.4562 0.3699 0.2517 0.1239
0.225 0.2027 0.3571 0.4343 0.4275 0.3571 0.2479 0.1255
0.250 0.1834 0.3262 0.4007 0.4021 0.3416 0.2426 0.1242
0.275 0.1672 0.2989 0.3715 0.3773 0.3262 0.2348 0.1219
0.300 0.1530 0.2752 0.3448 0.3545 0.3101 0.2262 0.1183
0.325 0.1407 0.2541 0.3209 0.3329 0.2943 0.2166 0.1141
0.350 0.1298 0.2354 0.2990 0.3126 0.2787 0.2067 0.1095
0.375 0.1201 0.2186 0.2790 0.2936 0.2635 0.1966 0.1046
0.400 0.1115 0.2034 0.2607 0.2757 0.2488 0.1865 0.0996
0.425 0.1036 0.1895 0.2438 0.2589 0.2347 0.1766 0.0945
0.450 0.0965 0.1769 0.2281 0.2432 0.2211 0.1670 0.0896
0.475 0.0901 0.1652 0.2136 0.2283 0.2083 0.1577 0.0847
0.500 0.0841 0.1545 0.2002 0.2144 0.1961 0.1487 0.0800
(continued)
Exercises 16.2
8.
By symmetry, as shown in the figure, we need only solve for
u
1
,
u
2
,
u
3
,
u
4
, and
u
5
. The
difference equations are
u
2
+0+0+
u
3
−
4
u
1
=
−
1
0+0+
u
1
+
u
4
−
4
u
2
=
−
1
u
4
+
u
1
+0+
u
5
−
4
u
3
=
−
1
u
2
+
u
2
+
u
3
+
u
3
−
4
u
4
=
−
1
u
3
+
u
3
+0+0
−
4
u
5
=
−
1
or
u
1
=0
.
25
u
2
+0
.
25
u
3
+0
.
25
u
2
=0
.
25
u
1
+0
.
25
u
4
+0
.
25
u
3
=0
.
25
u
1
+0
.
25
u
4
+0
.
25
u
5
+0
.
25
u
4
=0
.
5
u
2
+0
.
5
u
3
+0
.
25
u
5
=0
.
5
u
3
+0
.
25
.
Using Gauss-Seidel iteration we find
u
1
=0
.
6157,
u
2
=0
.
6493,
u
3
=0
.
8134,
u
4
=0
.
9813, and
u
5
=0
.
6567.
Exercises 16.2
1.
We identify
c
=1,
a
=2,
T
=1,
n
= 8, and
m
= 40. Then
h
=2
/
8=0
.
25,
k
=1
/
40=0
.
025, and
λ
=2
/
5=0
.
4.
741

742. (continued)
TIME
X=0.25 X=0.50 X=0.75 X=1.00 X=1.25 X=1.50 X=1.75
0.525 0.0786 0.1446 0.1876 0.2014 0.1845 0.1402 0.0755
0.550 0.0736 0.1354 0.1759 0.1891 0.1735 0.1320 0.0712
0.575 0.0689 0.1269 0.1650 0.1776 0.1632 0.1243 0.0670
0.600 0.0645 0.1189 0.1548 0.1668 0.1534 0.1169 0.0631
0.625 0.0605 0.1115 0.1452 0.1566 0.1442 0.1100 0.0594
0.650 0.0567 0.1046 0.1363 0.1471 0.1355 0.1034 0.0559
0.675 0.0532 0.0981 0.1279 0.1381 0.1273 0.0972 0.0525
0.700 0.0499 0.0921 0.1201 0.1297 0.1196 0.0914 0.0494
0.725 0.0468 0.0864 0.1127 0.1218 0.1124 0.0859 0.0464
0.750 0.0439 0.0811 0.1058 0.1144 0.1056 0.0807 0.0436
0.775 0.0412 0.0761 0.0994 0.1074 0.0992 0.0758 0.0410
0.800 0.0387 0.0715 0.0933 0.1009 0.0931 0.0712 0.0385
0.825 0.0363 0.0671 0.0876 0.0948 0.0875 0.0669 0.0362
0.850 0.0341 0.0630 0.0823 0.0890 0.0822 0.0628 0.0340
0.875 0.0320 0.0591 0.0772 0.0836 0.0772 0.0590 0.0319
0.900 0.0301 0.0555 0.0725 0.0785 0.0725 0.0554 0.0300
0.925 0.0282 0.0521 0.0681 0.0737 0.0681 0.0521 0.0282
0.950 0.0265 0.0490 0.0640 0.0692 0.0639 0.0489 0.0265
0.975 0.0249 0.0460 0.0601 0.0650 0.0600 0.0459 0.0249
1.000 0.0234 0.0432 0.0564 0.0610 0.0564 0.0431 0.0233
(x,y) exact approx abs error
(0.25,0.1) 0.3794 0.3728 0.0066
(1,0.5) 0.1854 0.2144 0.0290
(1.5,0.8) 0.0623 0.0712 0.0089
Exercises 16.2
2.
742

743. TIME
X=0.25 X=0.50 X=0.75 X=1.00 X=1.25 X=1.50 X=1.75
0.000 1.0000 1.0000 1.0000 1.0000 0.0000 0.0000 0.0000
0.025 0.7074 0.9520 0.9566 0.7444 0.2545 0.0371 0.0053
0.050 0.5606 0.8499 0.8685 0.6633 0.3303 0.1034 0.0223
0.075 0.4684 0.7473 0.7836 0.6191 0.3614 0.1529 0.0462
0.100 0.4015 0.6577 0.7084 0.5837 0.3753 0.1871 0.0684
0.125 0.3492 0.5821 0.6428 0.5510 0.3797 0.2101 0.0861
0.150 0.3069 0.5187 0.5857 0.5199 0.3778 0.2247 0.0990
0.175 0.2721 0.4652 0.5359 0.4901 0.3716 0.2329 0.1078
0.200 0.2430 0.4198 0.4921 0.4617 0.3622 0.2362 0.1132
0.225 0.2186 0.3809 0.4533 0.4348 0.3507 0.2358 0.1160
0.250 0.1977 0.3473 0.4189 0.4093 0.3378 0.2327 0.1166
0.275 0.1798 0.3181 0.3881 0.3853 0.3240 0.2275 0.1157
0.300 0.1643 0.2924 0.3604 0.3626 0.3097 0.2208 0.1136
0.325 0.1507 0.2697 0.3353 0.3412 0.2953 0.2131 0.1107
0.350 0.1387 0.2495 0.3125 0.3211 0.2808 0.2047 0.1071
0.375 0.1281 0.2313 0.2916 0.3021 0.2666 0.1960 0.1032
0.400 0.1187 0.2150 0.2725 0.2843 0.2528 0.1871 0.0989
0.425 0.1102 0.2002 0.2549 0.2675 0.2393 0.1781 0.0946
0.450 0.1025 0.1867 0.2387 0.2517 0.2263 0.1692 0.0902
0.475 0.0955 0.1743 0.2236 0.2368 0.2139 0.1606 0.0858
0.500 0.0891 0.1630 0.2097 0.2228 0.2020 0.1521 0.0814
0.525 0.0833 0.1525 0.1967 0.2096 0.1906 0.1439 0.0772
0.550 0.0779 0.1429 0.1846 0.1973 0.1798 0.1361 0.0731
0.575 0.0729 0.1339 0.1734 0.1856 0.1696 0.1285 0.0691
0.600 0.0683 0.1256 0.1628 0.1746 0.1598 0.1214 0.0653
0.625 0.0641 0.1179 0.1530 0.1643 0.1506 0.1145 0.0617
0.650 0.0601 0.1106 0.1438 0.1546 0.1419 0.1080 0.0582
0.675 0.0564 0.1039 0.1351 0.1455 0.1336 0.1018 0.0549
0.700 0.0530 0.0976 0.1270 0.1369 0.1259 0.0959 0.0518
0.725 0.0497 0.0917 0.1194 0.1288 0.1185 0.0904 0.0488
0.750 0.0467 0.0862 0.1123 0.1212 0.1116 0.0852 0.0460
0.775 0.0439 0.0810 0.1056 0.1140 0.1050 0.0802 0.0433
0.800 0.0413 0.0762 0.0993 0.1073 0.0989 0.0755 0.0408
0.825 0.0388 0.0716 0.0934 0.1009 0.0931 0.0711 0.0384
0.850 0.0365 0.0674 0.0879 0.0950 0.0876 0.0669 0.0362
0.875 0.0343 0.0633 0.0827 0.0894 0.0824 0.0630 0.0341
0.900 0.0323 0.0596 0.0778 0.0841 0.0776 0.0593 0.0321
0.925 0.0303 0.0560 0.0732 0.0791 0.0730 0.0558 0.0302
0.950 0.0285 0.0527 0.0688 0.0744 0.0687 0.0526 0.0284
0.975 0.0268 0.0496 0.0647 0.0700 0.0647 0.0495 0.0268
1.000 0.0253 0.0466 0.0609 0.0659 0.0608 0.0465 0.0252
(x,y) exact approx abs error
(0.25,0.1) 0.3794 0.4015 0.0221
(1,0.5) 0.1854 0.2228 0.0374
(1.5,0.8) 0.0623 0.0755 0.0132
Exercises 16.2
3.
We identify
c
=1,
a
=2,
T
=1,
n
= 8, and
m
= 40. Then
h
=2
/
8=0
.
25,
k
=1
/
40=0
.
025, and
λ
=2
/
5=0
.
4.
743

744. TIME
X=0.25 X=0.50 X=0.75 X=1.00 X=1.25 X=1.50 X=1.75
0.00 1.00 1.00 1.00 1.00 0.00 0.00 0.00
0.05 0.20 1.00 1.00 0.20 0.80 0.00 0.00
0.10 0.68 0.36 0.36 1.32 -0.32 0.64 0.00
0.15 -0.12 0.62 1.13 -0.76 1.76 -0.64 0.51
0.20 0.56 0.44 -0.79 2.77 -2.18 2.20 -0.82
0.25 0.01 -0.44 3.04 -4.03 5.28 -3.72 2.25
0.30 -0.36 2.70 -5.41 9.07 -9.37 8.26 -4.33
0.35 2.38 -6.24 12.67 -17.26 19.49 -15.91 9.20
0.40 -6.42 15.78 -26.40 36.08 -38.23 32.50 -18.25
0.45 16.47 -35.72 57.33 -73.35 77.80 -64.68 36.94
0.50 -38.46 80.48 -121.66 152.12 -157.11 130.60 -73.91
0.55 87.46 -176.38 259.07 -314.28 320.44 -263.18 148.83
0.60 -193.58 383.05 -547.97 652.17 -654.23 533.32 -299.84
0.65 422.59 -823.07 1156.96 -1353.07 1340.93 -1083.25 606.56
0.70 -912.01 1757.48 -2435.09 2810.16 -2753.61 2207.94 -1230.53
0.75 1953.19 -3732.17 5115.16 -5837.05 5666.65 -4512.08 2504.67
0.80 -4157.65 7893.99 -10724.47 12127.68 -11679.29 9244.30 -5112.47
0.85 8809.78 -16642.09 22452.02 -25199.62 24105.16 -18979.99 10462.92
0.90 -18599.54 34994.69 -46944.58 52365.51 -49806.79 39042.46 -21461.75
0.95 39155.48 -73432.11 98054.91 -108820.40 103010.45 -80440.31 44111.02
1.00 -82238.97 153827.58 -204634.95 226144.53 -213214.84 165961.36 -90818.86
(x,y) exact approx abs error
(0.25,0.1) 0.3794 0.6800 0.3006
(1,0.5) 0.1854 152.1152 151.9298
(1.5,0.8) 0.0623 9244.3042 9244.2419
Exercises 16.2
4.
We identify
c
=1,
a
=2,
T
=1,
n
= 8, and
m
= 20. Then
h
=2
/
8=0
.
25,
h
=1
/
20=0
.
05, and
λ
=4
/
5=0
.
8.
In this case
λ
=0
.
8 is greater than 0
.
5 and the procedure is unstable.
744

745. TIME
X=0.25 X=0.50 X=0.75 X=1.00 X=1.25 X=1.50 X=1.75
0.00 1.0000 1.0000 1.0000 1.0000 0.0000 0.0000 0.0000
0.05 0.5265 0.8693 0.8852 0.6141 0.3783 0.0884 0.0197
0.10 0.3972 0.6551 0.7043 0.5883 0.3723 0.1955 0.0653
0.15 0.3042 0.5150 0.5844 0.5192 0.3812 0.2261 0.1010
0.20 0.2409 0.4171 0.4901 0.4620 0.3636 0.2385 0.1145
0.25 0.1962 0.3452 0.4174 0.4092 0.3391 0.2343 0.1178
0.30 0.1631 0.2908 0.3592 0.3624 0.3105 0.2220 0.1145
0.35 0.1379 0.2482 0.3115 0.3208 0.2813 0.2056 0.1077
0.40 0.1181 0.2141 0.2718 0.2840 0.2530 0.1876 0.0993
0.45 0.1020 0.1860 0.2381 0.2514 0.2265 0.1696 0.0904
0.50 0.0888 0.1625 0.2092 0.2226 0.2020 0.1523 0.0816
0.55 0.0776 0.1425 0.1842 0.1970 0.1798 0.1361 0.0732
0.60 0.0681 0.1253 0.1625 0.1744 0.1597 0.1214 0.0654
0.65 0.0599 0.1104 0.1435 0.1544 0.1418 0.1079 0.0582
0.70 0.0528 0.0974 0.1268 0.1366 0.1257 0.0959 0.0518
0.75 0.0466 0.0860 0.1121 0.1210 0.1114 0.0851 0.0460
0.80 0.0412 0.0760 0.0991 0.1071 0.0987 0.0754 0.0408
0.85 0.0364 0.0672 0.0877 0.0948 0.0874 0.0668 0.0361
0.90 0.0322 0.0594 0.0776 0.0839 0.0774 0.0592 0.0320
0.95 0.0285 0.0526 0.0687 0.0743 0.0686 0.0524 0.0284
1.00 0.0252 0.0465 0.0608 0.0657 0.0607 0.0464 0.0251
(x,y) exact approx abs error
(0.25,0.1) 0.3794 0.3972 0.0178
(1,0.5) 0.1854 0.2226 0.0372
(1.5,0.8) 0.0623 0.0754 0.0131
TIME
X=2 X=4 X=6 X=8 X=10 X=12 X=14 X=16 X=18
0 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000
1 28.7216 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 28.7216
2 27.5521 29.9455 30.0000 30.0000 30.0000 30.0000 30.0000 29.9455 27.5521
3 26.4800 29.8459 29.9977 30.0000 30.0000 30.0000 29.9977 29.8459 26.4800
4 25.4951 29.7089 29.9913 29.9999 30.0000 29.9999 29.9913 29.7089 25.4951
5 24.5882 29.5414 29.9796 29.9995 30.0000 29.9995 29.9796 29.5414 24.5882
6 23.7515 29.3490 29.9618 29.9987 30.0000 29.9987 29.9618 29.3490 23.7515
7 22.9779 29.1365 29.9373 29.9972 29.9998 29.9972 29.9373 29.1365 22.9779
8 22.2611 28.9082 29.9057 29.9948 29.9996 29.9948 29.9057 28.9082 22.2611
9 21.5958 28.6675 29.8670 29.9912 29.9992 29.9912 29.8670 28.6675 21.5958
10 20.9768 28.4172 29.8212 29.9862 29.9985 29.9862 29.8212 28.4172 20.9768
Exercises 16.2
5.
We identify
c
=1,
a
=2,
T
=1,
n
= 8, and
m
= 20. Then
h
=2
/
8=0
.
25,
k
=1
/
20 = 0
.
05, and
λ
=4
/
5=0
.
8.
6.
We identify
c
=15
/
88
≈
0
.
1705,
a
= 20,
T
=10,
n
= 10, and
m
= 10. Then
h
=2,
k
= 1, and
λ
=15
/
352
≈
0
.
0426.
745

746. TIME
X=5 X=10 X=15 X=20 X=25 X=30 X=35 X=40 X=45
0 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000
1 29.7955 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 30.0000 29.7955
2 29.5937 29.9986 30.0000 30.0000 30.0000 30.0000 30.0000 29.9986 29.5937
3 29.3947 29.9959 30.0000 30.0000 30.0000 30.0000 30.0000 29.9959 29.3947
4 29.1984 29.9918 30.0000 30.0000 30.0000 30.0000 30.0000 29.9918 29.1984
5 29.0047 29.9864 29.9999 30.0000 30.0000 30.0000 29.9999 29.9864 29.0047
6 28.8136 29.9798 29.9998 30.0000 30.0000 30.0000 29.9998 29.9798 28.8136
7 28.6251 29.9720 29.9997 30.0000 30.0000 30.0000 29.9997 29.9720 28.6251
8 28.4391 29.9630 29.9995 30.0000 30.0000 30.0000 29.9995 29.9630 28.4391
9 28.2556 29.9529 29.9992 30.0000 30.0000 30.0000 29.9992 29.9529 28.2556
10 28.0745 29.9416 29.9989 30.0000 30.0000 30.0000 29.9989 29.9416 28.0745
TIME
X=2 X=4 X=6 X=8 X=10 X=12 X=14 X=16 X=18
0 18.0000 32.0000 42.0000 48.0000 50.0000 48.0000 42.0000 32.0000 18.0000
1 16.1481 30.1481 40.1481 46.1481 48.1481 46.1481 40.1481 30.1481 16.1481
2 15.1536 28.2963 38.2963 44.2963 46.2963 44.2963 38.2963 28.2963 15.1536
3 14.2226 26.8414 36.4444 42.4444 44.4444 42.4444 36.4444 26.8414 14.2226
4 13.4801 25.4452 34.7764 40.5926 42.5926 40.5926 34.7764 25.4452 13.4801
5 12.7787 24.2258 33.1491 38.8258 40.7407 38.8258 33.1491 24.2258 12.7787
6 12.1622 23.0574 31.6460 37.0842 38.9677 37.0842 31.6460 23.0574 12.1622
7 11.5756 21.9895 30.1875 35.4385 37.2238 35.4385 30.1875 21.9895 11.5756
8 11.0378 20.9636 28.8232 33.8340 35.5707 33.8340 28.8232 20.9636 11.0378
9 10.5230 20.0070 27.5043 32.3182 33.9626 32.3182 27.5043 20.0070 10.5230
10 10.0420 19.0872 26.2620 30.8509 32.4400 30.8509 26.2620 19.0872 10.0420
TIME
X=10 X=20 X=30 X=40 X=50 X=60 X=70 X=80 X=90
0 8.0000 16.0000 24.0000 32.0000 40.0000 32.0000 24.0000 16.0000 8.0000
1 8.0000 16.0000 23.6075 31.3459 39.2151 31.6075 23.7384 15.8692 8.0000
2 8.0000 15.9936 23.2279 30.7068 38.4452 31.2151 23.4789 15.7384 7.9979
3 7.9999 15.9812 22.8606 30.0824 37.6900 30.8229 23.2214 15.6076 7.9937
4 7.9996 15.9631 22.5050 29.4724 36.9492 30.4312 22.9660 15.4769 7.9874
5 7.9990 15.9399 22.1606 28.8765 36.2228 30.0401 22.7125 15.3463 7.9793
6 7.9981 15.9118 21.8270 28.2945 35.5103 29.6500 22.4610 15.2158 7.9693
7 7.9967 15.8791 21.5037 27.7261 34.8117 29.2610 22.2112 15.0854 7.9575
8 7.9948 15.8422 21.1902 27.1709 34.1266 28.8733 21.9633 14.9553 7.9439
9 7.9924 15.8013 20.8861 26.6288 33.4548 28.4870 21.7172 14.8253 7.9287
10 7.9894 15.7568 20.5911 26.0995 32.7961 28.1024 21.4727 14.6956 7.9118
Exercises 16.2
7.
We identify
c
=15
/
88
≈
0
.
1705,
a
= 50,
T
=10,
n
= 10, and
m
= 10. Then
h
=5,
k
= 1, and
λ
=3
/
440
≈
0
.
0068.
8.
We identify
c
=50
/
27
≈
1
.
8519,
a
= 20,
T
=10,
n
= 10, and
m
= 10. Then
h
=2,
k
= 1, and
λ
=25
/
54
≈
0
.
4630.
9.
We identify
c
= 260
/
159
≈
1
.
6352,
a
= 100,
T
=10,
n
= 10, and
m
= 10. Then
h
=10,
k
= 1, and
λ
=13
/
795
≈
0
.
0164.
746

747. Exercises 16.2
10. (a)
With
n
=4wehave
h
=1
/
2 so that
λ
=1
/
100=0
.
01.
(b)
We observe that
α
= 2(1+ 1
/λ
) = 202 and
β
= 2(1
−
1
/λ
)=
−
198. The system of equations is
−
u
01
+
αu
11
−
u
21
=
u
20
−
βu
10
+
u
00
−
u
11
+
αu
21
−
u
31
=
u
30
−
βu
20
+
u
10
−
u
21
+
αu
31
−
u
41
=
u
40
−
βu
30
+
u
20
.
Now
u
00
=
u
01
=
u
40
=
u
41
= 0, so the system is
αu
11
−
u
21
=
u
20
−
βu
10
−
u
11
+
αu
21
−
u
31
=
u
30
−
βu
20
+
u
10
−
u
21
+
αu
31
=
−
βu
30
+
u
20
or
202
u
11
−
u
21
= sin
π
+ 198 sin
π
2
=198
−
u
11
+ 202
u
21
−
u
31
= sin
3
π
2
+ 198 sin
π
+ sin
π
2
=0
−
u
21
+ 202
u
31
= 198 sin
3
π
2
+ sin
π
=
−
198
.
(c)
The solution of this system is
u
11
≈
0
.
9802,
u
21
=0,
u
31
≈−
0
.
9802. The corresponding entries in
Table 16
.
3 in the text are 0
.
9768, 0, and
−
0
.
9768.
11. (a)
The differential equation is
k
∂
2
u
∂x
2
=
∂u
∂t
where
k
=
K/γρ
. If we let
u
(
x, t
)=
v
(
x, t
)+
ψ
(
x
), then
∂
2
u
∂x
2
=
∂
2
v
∂x
2
+
ψ
√√
and
∂u
∂t
=
∂v
∂t
.
Substituting into the differential equation gives
k
∂
2
v
∂x
2
+
kψ
√√
=
∂v
∂t
.
Requiring
kψ
√√
=0wehave
ψ
(
x
)=
c
1
x
+
c
2
. The boundary conditions become
u
(0
,t
)=
v
(0
,t
)+
ψ
(0) = 20 and
u
(20
,t
)=
v
(20
,t
)+
ψ
(20) = 30
.
Letting
ψ
(0) = 20 and
ψ
(20) = 30 we obtain the homogeneous boundary conditions in
v
:
v
(0
,t
)=
v
(20
,t
) = 0. Now
ψ
(0) = 20 and
ψ
(20) = 30 imply that
c
1
=1
/
2 and
c
2
= 20. The steady-state solution is
ψ
(
x
)=
1
2
x
+ 20.
(b)
To use the Crank-Nicholson method we identify
c
= 375
/
212
≈
1
.
7689,
a
= 20,
T
= 400,
n
= 5, and
m
= 40. Then
h
=4,
k
= 10, and
λ
= 1875
/
1696
≈
1
.
1055.
747

748. TIME
X=4.00 X=8.00 X=12.00 X=16.00
0.00 50.0000 50.0000 50.0000 50.0000
10.00 32.7433 44.2679 45.4228 38.2971
20.00 29.9946 36.2354 38.3148 35.8160
30.00 26.9487 32.1409 34.0874 32.9644
40.00 25.2691 29.2562 31.2704 31.2580
50.00 24.1178 27.4348 29.4296 30.1207
60.00 23.3821 26.2339 28.2356 29.3810
70.00 22.8995 25.4560 27.4554 28.8998
80.00 22.5861 24.9481 26.9482 28.5859
90.00 22.3817 24.6176 26.6175 28.3817
100.00 22.2486 24.4022 26.4023 28.2486
110.00 22.1619 24.2620 26.2620 28.1619
120.00 22.1055 24.1707 26.1707 28.1055
130.00 22.0687 24.1112 26.1112 28.0687
140.00 22.0447 24.0724 26.0724 28.0447
150.00 22.0291 24.0472 26.0472 28.0291
160.00 22.0190 24.0307 26.0307 28.0190
170.00 22.0124 24.0200 26.0200 28.0124
180.00 22.0081 24.0130 26.0130 28.0081
190.00 22.0052 24.0085 26.0085 28.0052
200.00 22.0034 24.0055 26.0055 28.0034
210.00 22.0022 24.0036 26.0036 28.0022
220.00 22.0015 24.0023 26.0023 28.0015
230.00 22.0009 24.0015 26.0015 28.0009
240.00 22.0006 24.0010 26.0010 28.0006
250.00 22.0004 24.0007 26.0007 28.0004
260.00 22.0003 24.0004 26.0004 28.0003
270.00 22.0002 24.0003 26.0003 28.0002
280.00 22.0001 24.0002 26.0002 28.0001
290.00 22.0001 24.0001 26.0001 28.0001
300.00 22.0000 24.0001 26.0001 28.0000
310.00 22.0000 24.0001 26.0001 28.0000
320.00 22.0000 24.0000 26.0000 28.0000
330.00 22.0000 24.0000 26.0000 28.0000
340.00 22.0000 24.0000 26.0000 28.0000
350.00 22.0000 24.0000 26.0000 28.0000
Exercises 16.2
We observe that the approximate steady-state temperatures agree exactly with the corresponding values of
ψ
(
x
).
12.
We identify
c
=1,
a
=1,
T
=1,
n
= 5, and
m
= 20. Then
h
=0
.
2,
k
=0
.
04, and
λ
= 1. The values below
were obtained using
Excel
, which carries more than 12 significant digits. In order to see evidence of instability
use 0
≤
t
≤
2.
748

749. TIME
X=0.2 X=0.4 X=0.6 X=0.8
TIME
X=0.2 X=0.4 X=0.6 X=0.8
0.00 0.5878 0.9511 0.9511 0.5878 1.04 0.0000 0.0000 0.0000 0.0000
0.04 0.3633 0.5878 0.5878 0.3633 1.08 0.0000 0.0000 0.0000 0.0000
0.08 0.2245 0.3633 0.3633 0.2245 1.12 0.0000 0.0000 0.0000 0.0000
0.12 0.1388 0.2245 0.2245 0.1388 1.16 0.0000 0.0000 0.0000 0.0000
0.16 0.0858 0.1388 0.1388 0.0858 1.20 -0.0001 0.0001 -0.0001 0.0001
0.20 0.0530 0.0858 0.0858 0.0530 1.24 0.0001 -0.0002 0.0002 -0.0001
0.24 0.0328 0.0530 0.0530 0.0328 1.28 -0.0004 0.0006 -0.0006 0.0004
0.28 0.0202 0.0328 0.0328 0.0202 1.32 0.0010 -0.0015 0.0015 -0.0010
0.32 0.0125 0.0202 0.0202 0.0125 1.36 -0.0025 0.0040 -0.0040 0.0025
0.36 0.0077 0.0125 0.0125 0.0077 1.40 0.0065 -0.0106 0.0106 -0.0065
0.40 0.0048 0.0077 0.0077 0.0048 1.44 -0.0171 0.0277 -0.0277 0.0171
0.44 0.0030 0.0048 0.0048 0.0030 1.48 0.0448 -0.0724 0.0724 -0.0448
0.48 0.0018 0.0030 0.0030 0.0018 1.52 -0.1172 0.1897 -0.1897 0.1172
0.52 0.0011 0.0018 0.0018 0.0011 1.56 0.3069 -0.4965 0.4965 -0.3069
0.56 0.0007 0.0011 0.0011 0.0007 1.60 -0.8034 1.2999 -1.2999 0.8034
0.60 0.0004 0.0007 0.0007 0.0004 1.64 2.1033 -3.4032 3.4032 -2.1033
0.64 0.0003 0.0004 0.0004 0.0003 1.68 -5.5064 8.9096 -8.9096 5.5064
0.68 0.0002 0.0003 0.0003 0.0002 1.72 14.416 -23.326 23.326 -14.416
0.72 0.0001 0.0002 0.0002 0.0001 1.76 -37.742 61.067 -61.067 37.742
0.76 0.0001 0.0001 0.0001 0.0001 1.80 98.809 -159.88 159.88 -98.809
0.80 0.0000 0.0001 0.0001 0.0000 1.84 -258.68 418.56 -418.56 258.685
0.84 0.0000 0.0000 0.0000 0.0000 1.88 677.24 -1095.8 1095.8 -677.245
0.88 0.0000 0.0000 0.0000 0.0000 1.92 -1773.1 2868.9 -2868.9 1773.1
0.92 0.0000 0.0000 0.0000 0.0000 1.96 4641.9 -7510.8 7510.8 -4641.9
0.96 0.0000 0.0000 0.0000 0.0000 2.00 -12153 19663 -19663 12153
1.00 0.0000 0.0000 0.0000 0.0000
TIME
X=0.25 X=0.5 X=0.75
0.00 0.1875 0.2500 0.1875
0.10 0.1775 0.2400 0.1775
0.20 0.1491 0.2100 0.1491
0.30 0.1066 0.1605 0.1066
0.40 0.0556 0.0938 0.0556
0.50 0.0019 0.0148 0.0019
0.60 -0.0501 -0.0682 -0.0501
0.70 -0.0970 -0.1455 -0.0970
0.80 -0.1361 -0.2072 -0.1361
0.90 -0.1648 -0.2462 -0.1648
1.00 -0.1802 -0.2591 -0.1802
Exercises 16.3
Exercises 16.3
1. (a)
Identifying
h
=1
/
4 and
k
=1
/
10 we see that
λ
=2
/
5.
749

750. TIME
X=0.4 X=0.8 X=1.2 X=1.6
0.00 0.0032 0.5273 0.5273 0.0032
0.10 0.0194 0.5109 0.5109 0.0194
0.20 0.0652 0.4638 0.4638 0.0652
0.30 0.1318 0.3918 0.3918 0.1318
0.40 0.2065 0.3035 0.3035 0.2065
0.50 0.2743 0.2092 0.2092 0.2743
0.60 0.3208 0.1190 0.1190 0.3208
0.70 0.3348 0.0413 0.0413 0.3348
0.80 0.3094 -0.0180 -0.0180 0.3094
0.90 0.2443 -0.0568 -0.0568 0.2443
1.00 0.1450 -0.0768 -0.0768 0.1450
TIME
X=0.1 X=0.2 X=0.3 X=0.4 X=0.5 X=0.6 X=0.7 X=0.8 X=0.9
0.00 0.0000 0.0000 0.0000 0.0000 0.0000 0.5000 0.5000 0.5000 0.5000
0.04 0.0000 0.0000 0.0000 0.0000 0.0800 0.4200 0.5000 0.5000 0.4200
0.08 0.0000 0.0000 0.0000 0.0256 0.2432 0.2568 0.4744 0.4744 0.2312
0.12 0.0000 0.0000 0.0082 0.1126 0.3411 0.1589 0.3792 0.3710 0.0462
0.16 0.0000 0.0026 0.0472 0.2394 0.3076 0.1898 0.2108 0.1663 -0.0496
0.20 0.0008 0.0187 0.1334 0.3264 0.2146 0.2651 0.0215 -0.0933 -0.0605
0.24 0.0071 0.0657 0.2447 0.3159 0.1735 0.2463 -0.1266 -0.3056 -0.0625
0.28 0.0299 0.1513 0.3215 0.2371 0.2013 0.0849 -0.2127 -0.3829 -0.1223
0.32 0.0819 0.2525 0.3168 0.1737 0.2033 -0.1345 -0.2580 -0.3223 -0.2264
0.36 0.1623 0.3197 0.2458 0.1657 0.0877 -0.2853 -0.2843 -0.2104 -0.2887
0.40 0.2412 0.3129 0.1727 0.1583 -0.1223 -0.3164 -0.2874 -0.1473 -0.2336
0.44 0.2657 0.2383 0.1399 0.0658 -0.3046 -0.2761 -0.2549 -0.1565 -0.0761
0.48 0.1965 0.1410 0.1149 -0.1216 -0.3593 -0.2381 -0.1977 -0.1715 0.0800
0.52 0.0466 0.0531 0.0225 -0.3093 -0.2992 -0.2260 -0.1451 -0.1144 0.1300
0.56 -0.1161 -0.0466 -0.1662 -0.3876 -0.2188 -0.2114 -0.1085 0.0111 0.0602
0.60 -0.2194 -0.2069 -0.3875 -0.3411 -0.1901 -0.1662 -0.0666 0.1140 -0.0446
0.64 -0.2485 -0.4290 -0.5362 -0.2611 -0.2021 -0.0969 0.0012 0.1084 -0.0843
0.68 -0.2559 -0.6276 -0.5625 -0.2503 -0.1993 -0.0298 0.0720 0.0068 -0.0354
0.72 -0.3003 -0.6865 -0.5097 -0.3230 -0.1585 0.0156 0.0893 -0.0874 0.0384
0.76 -0.3722 -0.5652 -0.4538 -0.4029 -0.1147 0.0289 0.0265 -0.0849 0.0596
0.80 -0.3867 -0.3464 -0.4172 -0.4068 -0.1172 -0.0046 -0.0712 -0.0005 0.0155
0.84 -0.2647 -0.1633 -0.3546 -0.3214 -0.1763 -0.0954 -0.1249 0.0665 -0.0386
0.88 -0.0254 -0.0738 -0.2202 -0.2002 -0.2559 -0.2215 -0.1079 0.0385 -0.0468
0.92 0.2064 -0.0157 -0.0325 -0.1032 -0.3067 -0.3223 -0.0804 -0.0636 -0.0127
0.96 0.3012 0.1081 0.1380 -0.0487 -0.2974 -0.3407 -0.1250 -0.1548 0.0092
1.00 0.2378 0.3032 0.2392 -0.0141 -0.2223 -0.2762 -0.2481 -0.1840 -0.0244
Exercises 16.3
(b)
Identifying
h
=2
/
5 and
k
=1
/
10 we see that
λ
=1
/
4.
(c)
Identifying
h
=1
/
10 and
k
=1
/
25 we see that
λ
=2
√
2
/
5.
2. (a)
In Section 13
.
4 the solution of the wave equation is shown to be
u
(
x, t
)=
∞
6
n
=1
(
A
n
cos
nπt
+
B
n
sin
nπt
) sin
nπx
where
A
n
=2

1
0
sin
πx
sin
nπx dx
=

1
,n
=1
0
,n
=2,3,4,
...
750

751. TIME
x=0.25 x=0.50 x=0.75
0.0 0.7071 1.0000 0.7071
0.1 0.6740 0.9531 0.6740
0.2 0.5777 0.8169 0.5777
0.3 0.4272 0.6042 0.4272
0.4 0.2367 0.3348 0.2367
0.5 0.0241 0.0340 0.0241
i,j approx exact error
1,1 0.6740 0.6725 0.0015
1,2 0.5777 0.5721 0.0056
1,3 0.4272 0.4156 0.0116
1,4 0.2367 0.2185 0.0182
1,5 0.0241 0.0000 0.0241
2,1 0.9531 0.9511 0.0021
2,2 0.8169 0.8090 0.0079
2,3 0.6042 0.5878 0.0164
2,4 0.3348 0.3090 0.0258
2,5 0.0340 0.0000 0.0340
3,1 0.6740 0.6725 0.0015
3,2 0.5777 0.5721 0.0056
3,3 0.4272 0.4156 0.0116
3,4 0.2367 0.2185 0.0182
3,5 0.0241 0.0000 0.0241
Exercises 16.3
and
B
n
=
2
nπ

1
0
0
dx
=0
.
Thus
u
(
x, t
) = cos
πt
sin
πx
.
(b)
We have
h
=1
/
4,
k
=0
.
5
/
5=0
.
1and
λ
=0
.
4. Now
u
0
,j
=
u
4
,j
=0or
j
=0,1,
...
, 5, and the initial
values of
u
are
u
1
,
0
=
u
(1
/
4
,
0) = sin
π/
4
≈
0
.
7071,
u
2
,
0
=
u
(1
/
2
,
0) = sin
π/
2=1,
u
3
,
0
=
u
(3
/
4
,
0) =
sin 3
π/
4
≈
0
.
7071. From equation (6) in the text we have
u
i,
1
=0
.
8(
u
i
+1
,
0
+
u
i
−
1
,
0
)+0
.
84
u
i,
0
+0
.
1(0)
.
Then
u
1
,
1
≈
0
.
6740,
u
2
,
1
=0
.
9531,
u
3
,
1
=0
.
6740. From equation (3) in the text we have for
j
=1,2,3,
...
u
i,j
+1
=0
.
16
u
i
+1
,j
+ 2(0
.
84)
u
i,j
+0
.
16
u
i
−
1
,j
−
u
i,j
−
1
.
The results of the calculations are given in the table.
(c)
751

752. TIME
X=0.2 X=0.4 X=0.6 X=0.8
0.00 0.5878 0.9511 0.9511 0.5878
0.05 0.5808 0.9397 0.9397 0.5808
0.10 0.5599 0.9059 0.9059 0.5599
0.15 0.5256 0.8505 0.8505 0.5256
0.20 0.4788 0.7748 0.7748 0.4788
0.25 0.4206 0.6806 0.6806 0.4206
0.30 0.3524 0.5701 0.5701 0.3524
0.35 0.2757 0.4460 0.4460 0.2757
0.40 0.1924 0.3113 0.3113 0.1924
0.45 0.1046 0.1692 0.1692 0.1046
0.50 0.0142 0.0230 0.0230 0.0142
TIME
X=0.2 X=0.4 X=0.6 X=0.8
0.00 0.5878 0.9511 0.9511 0.5878
0.03 0.5860 0.9482 0.9482 0.5860
0.05 0.5808 0.9397 0.9397 0.5808
0.08 0.5721 0.9256 0.9256 0.5721
0.10 0.5599 0.9060 0.9060 0.5599
0.13 0.5445 0.8809 0.8809 0.5445
0.15 0.5257 0.8507 0.8507 0.5257
0.18 0.5039 0.8153 0.8153 0.5039
0.20 0.4790 0.7750 0.7750 0.4790
0.23 0.4513 0.7302 0.7302 0.4513
0.25 0.4209 0.6810 0.6810 0.4209
0.28 0.3879 0.6277 0.6277 0.3879
0.30 0.3527 0.5706 0.5706 0.3527
0.33 0.3153 0.5102 0.5102 0.3153
0.35 0.2761 0.4467 0.4467 0.2761
0.38 0.2352 0.3806 0.3806 0.2352
0.40 0.1929 0.3122 0.3122 0.1929
0.43 0.1495 0.2419 0.2419 0.1495
0.45 0.1052 0.1701 0.1701 0.1052
0.48 0.0602 0.0974 0.0974 0.0602
0.50 0.0149 0.0241 0.0241 0.0149
Exercises 16.3
3. (a)
Identifying
h
=1
/
5 and
k
=0
.
5
/
10=0
.
05 we see that
λ
=0
.
25.
(b)
Identifying
h
=1
/
5 and
k
=0
.
5
/
20=0
.
025 we see that
λ
=0
.
125.
4.
We have
λ
= 1. The initial values of
n
are
u
1
,
0
=
u
(0
.
2
,
0) = 0
.
16,
u
2
,
0
=
u
(0
.
4) = 0
.
24,
u
3
,
0
=0
.
24, and
u
4
,
0
=0
.
16. From equation (6) in the text we have
u
i,
1
=
1
2
(
u
i
+1
,
0
+
u
i
−
1
,
0
)+0
u
i,
0
+
k
·
0=
1
2
(
u
i
+1
,
0
+
u
i
−
1
,
0
)
.
Then, using
u
0
,
0
=
u
5
,
0
= 0, we find
u
1
,
1
=0
.
12,
u
2
,
1
=0
.
2,
u
3
,
1
=0
.
2, and
u
4
,
1
=0
.
12.
752

753. TIME
X=10 X=20 X=30 X=40 X=50
0.00000 0.1000 0.2000 0.3000 0.2000 0.1000
0.20045 0.1000 0.2000 0.2750 0.2000 0.1000
0.40089 0.1000 0.1938 0.2125 0.1938 0.1000
0.60134 0.0984 0.1688 0.1406 0.1688 0.0984
0.80178 0.0898 0.1191 0.0828 0.1191 0.0898
1.00223 0.0661 0.0531 0.0432 0.0531 0.0661
1.20268 0.0226 -0.0121 0.0085 -0.0121 0.0226
1.40312 -0.0352 -0.0635 -0.0365 -0.0635 -0.0352
1.60357 -0.0913 -0.1011 -0.0950 -0.1011 -0.0913
1.80401 -0.1271 -0.1347 -0.1566 -0.1347 -0.1271
2.00446 -0.1329 -0.1719 -0.2072 -0.1719 -0.1329
2.20491 -0.1153 -0.2081 -0.2402 -0.2081 -0.1153
2.40535 -0.0920 -0.2292 -0.2571 -0.2292 -0.0920
2.60580 -0.0801 -0.2230 -0.2601 -0.2230 -0.0801
2.80624 -0.0838 -0.1903 -0.2445 -0.1903 -0.0838
3.00669 -0.0932 -0.1445 -0.2018 -0.1445 -0.0932
3.20713 -0.0921 -0.1003 -0.1305 -0.1003 -0.0921
3.40758 -0.0701 -0.0615 -0.0440 -0.0615 -0.0701
3.60803 -0.0284 -0.0205 0.0336 -0.0205 -0.0284
3.80847 0.0224 0.0321 0.0842 0.0321 0.0224
4.00892 0.0700 0.0953 0.1087 0.0953 0.0700
4.20936 0.1064 0.1555 0.1265 0.1555 0.1064
4.40981 0.1285 0.1962 0.1588 0.1962 0.1285
4.61026 0.1354 0.2106 0.2098 0.2106 0.1354
4.81070 0.1273 0.2060 0.2612 0.2060 0.1273
5.01115 0.1070 0.1955 0.2851 0.1955 0.1070
5.21159 0.0821 0.1853 0.2641 0.1853 0.0821
5.41204 0.0625 0.1689 0.2038 0.1689 0.0625
5.61249 0.0539 0.1347 0.1260 0.1347 0.0539
5.81293 0.0520 0.0781 0.0526 0.0781 0.0520
6.01338 0.0436 0.0086 -0.0080 0.0086 0.0436
6.21382 0.0156 -0.0564 -0.0604 -0.0564 0.0156
6.41427 -0.0343 -0.1043 -0.1107 -0.1043 -0.0343
6.61472 -0.0931 -0.1364 -0.1578 -0.1364 -0.0931
6.81516 -0.1395 -0.1630 -0.1942 -0.1630 -0.1395
7.01561 -0.1568 -0.1915 -0.2150 -0.1915 -0.1568
7.21605 -0.1436 -0.2173 -0.2240 -0.2173 -0.1436
7.41650 -0.1129 -0.2263 -0.2297 -0.2263 -0.1129
7.61695 -0.0824 -0.2078 -0.2336 -0.2078 -0.0824
7.81739 -0.0625 -0.1644 -0.2247 -0.1644 -0.0625
8.01784 -0.0526 -0.1106 -0.1856 -0.1106 -0.0526
8.21828 -0.0440 -0.0611 -0.1091 -0.0611 -0.0440
8.41873 -0.0287 -0.0192 -0.0085 -0.0192 -0.0287
8.61918 -0.0038 0.0229 0.0867 0.0229 -0.0038
8.81962 0.0287 0.0743 0.1500 0.0743 0.0287
9.02007 0.0654 0.1332 0.1755 0.1332 0.0654
9.22051 0.1027 0.1858 0.1799 0.1858 0.1027
9.42096 0.1352 0.2160 0.1872 0.2160 0.1352
9.62140 0.1540 0.2189 0.2089 0.2189 0.1540
9.82185 0.1506 0.2030 0.2356 0.2030 0.1506
10.02230 0.1226 0.1822 0.2461 0.1822 0.1226
Exercises 16.3
5.
We identify
c
= 24944
.
4,
k
=0
.
00020045 seconds = 0
.
20045 milliseconds, and
λ
=0
.
5. Time in the table is
expressed in milliseconds.
753

754. TIME
X=10 X=20 X=30 X=40 X=50
0.00000 0.2000 0.2667 0.2000 0.1333 0.0667
0.10022 0.1958 0.2625 0.2000 0.1333 0.0667
0.20045 0.1836 0.2503 0.1997 0.1333 0.0667
0.30067 0.1640 0.2307 0.1985 0.1333 0.0667
0.40089 0.1384 0.2050 0.1952 0.1332 0.0667
0.50111 0.1083 0.1744 0.1886 0.1328 0.0667
0.60134 0.0755 0.1407 0.1777 0.1318 0.0666
0.70156 0.0421 0.1052 0.1615 0.1295 0.0665
0.80178 0.0100 0.0692 0.1399 0.1253 0.0661
0.90201 -0.0190 0.0340 0.1129 0.1184 0.0654
1.00223 -0.0435 0.0004 0.0813 0.1077 0.0638
1.10245 -0.0626 -0.0309 0.0464 0.0927 0.0610
1.20268 -0.0758 -0.0593 0.0095 0.0728 0.0564
1.30290 -0.0832 -0.0845 -0.0278 0.0479 0.0493
1.40312 -0.0855 -0.1060 -0.0639 0.0184 0.0390
1.50334 -0.0837 -0.1237 -0.0974 -0.0150 0.0250
1.60357 -0.0792 -0.1371 -0.1275 -0.0511 0.0069
1.70379 -0.0734 -0.1464 -0.1533 -0.0882 -0.0152
1.80401 -0.0675 -0.1515 -0.1747 -0.1249 -0.0410
1.90424 -0.0627 -0.1528 -0.1915 -0.1595 -0.0694
2.00446 -0.0596 -0.1509 -0.2039 -0.1904 -0.0991
2.10468 -0.0585 -0.1467 -0.2122 -0.2165 -0.1283
2.20491 -0.0592 -0.1410 -0.2166 -0.2368 -0.1551
2.30513 -0.0614 -0.1349 -0.2175 -0.2507 -0.1772
2.40535 -0.0643 -0.1294 -0.2154 -0.2579 -0.1929
2.50557 -0.0672 -0.1251 -0.2105 -0.2585 -0.2005
2.60580 -0.0696 -0.1227 -0.2033 -0.2524 -0.1993
2.70602 -0.0709 -0.1219 -0.1942 -0.2399 -0.1889
2.80624 -0.0710 -0.1225 -0.1833 -0.2214 -0.1699
2.90647 -0.0699 -0.1236 -0.1711 -0.1972 -0.1435
3.00669 -0.0678 -0.1244 -0.1575 -0.1681 -0.1115
3.10691 -0.0649 -0.1237 -0.1425 -0.1348 -0.0761
3.20713 -0.0617 -0.1205 -0.1258 -0.0983 -0.0395
3.30736 -0.0583 -0.1139 -0.1071 -0.0598 -0.0042
3.40758 -0.0547 -0.1035 -0.0859 -0.0209 0.0279
3.50780 -0.0508 -0.0889 -0.0617 0.0171 0.0552
3.60803 -0.0460 -0.0702 -0.0343 0.0525 0.0767
3.70825 -0.0399 -0.0478 -0.0037 0.0840 0.0919
3.80847 -0.0318 -0.0221 0.0297 0.1106 0.1008
3.90870 -0.0211 0.0062 0.0648 0.1314 0.1041
4.00892 -0.0074 0.0365 0.1005 0.1464 0.1025
4.10914 0.0095 0.0680 0.1350 0.1558 0.0973
4.20936 0.0295 0.1000 0.1666 0.1602 0.0897
4.30959 0.0521 0.1318 0.1937 0.1606 0.0808
4.40981 0.0764 0.1625 0.2148 0.1581 0.0719
4.51003 0.1013 0.1911 0.2291 0.1538 0.0639
4.61026 0.1254 0.2164 0.2364 0.1485 0.0575
4.71048 0.1475 0.2373 0.2369 0.1431 0.0532
4.81070 0.1659 0.2526 0.2315 0.1379 0.0512
4.91093 0.1794 0.2611 0.2217 0.1331 0.0514
5.01115 0.1867 0.2620 0.2087 0.1288 0.0535
Exercises 16.3
6.
We identify
c
= 24944
.
4,
k
=0
.
00010022 seconds = 0
.
10022 milliseconds, and
λ
=0
.
25. Time in the table is
expressed in milliseconds.
754

755. TIME
X=0.0 X=0.2 X=0.4 X=0.6 X=0.8 X=1.0
0.00 0.0000 0.2000 0.4000 0.6000 0.8000 0.0000
0.01 0.0000 0.2000 0.4000 0.6000 0.5500 0.0000
0.02 0.0000 0.2000 0.4000 0.5375 0.4250 0.0000
0.03 0.0000 0.2000 0.3844 0.4750 0.3469 0.0000
0.04 0.0000 0.1961 0.3609 0.4203 0.2922 0.0000
0.05 0.0000 0.1883 0.3346 0.3734 0.2512 0.0000
Chapter 16 Review Exercises
Chapter 16 Review Exercises
1.
Using the figure we obtain the system
u
21
+0+0+0
−
4
u
11
=0
u
31
+0+
u
11
+0
−
4
u
21
=0
50+0+
u
21
+0
−
4
u
31
=0
.
By Gauss-Elimination then,



−
410
0
1
−
41
0
01
−
4
−
50



row
−
−
−
−
−
−→
operations



1
−
41
0
01
−
4
−
50
001
13
.
3928



.
The solution is
u
11
=0
.
8929,
u
21
=3
.
5714,
u
31
=13
.
3928.
2.
By symmetry we observe that
u
i,
1
=
u
i,
3
for
i
=1,2,
...
, 7. We then
use Gauss-Seidel iteration with an initial guess of 7
.
5 for all variables
to solve the system
u
11
=0
.
25
u
21
+0
.
25
u
12
u
21
=0
.
25
u
31
+0
.
25
u
22
+0
.
25
u
11
u
31
=0
.
25
u
41
+0
.
25
u
32
+0
.
25
u
21
u
41
=0
.
25
u
51
+0
.
25
u
42
+0
.
25
u
31
u
51
=0
.
25
u
61
+0
.
25
u
52
+0
.
25
u
41
u
61
=0
.
25
u
71
+0
.
25
u
62
+0
.
25
u
51
u
71
=12
.
5+0
.
25
u
72
+0
.
25
u
61
u
12
=0
.
25
u
22
+0
.
5
u
11
u
22
=0
.
25
u
32
+0
.
5
u
21
+0
.
25
u
12
u
32
=0
.
25
u
42
+0
.
5
u
31
+0
.
25
u
22
u
42
=0
.
25
u
52
+0
.
5
u
41
+0
.
25
u
32
u
52
=0
.
25
u
62
+0
.
5
u
51
+0
.
25
u
42
u
62
=0
.
25
u
72
+0
.
5
u
61
+0
.
25
u
52
u
72
=12
.
5+0
.
5
u
71
+0
.
25
u
62
.
After 30 iterations we obtain
u
11
=
u
13
=0
.
1765,
u
21
=
u
23
=0
.
4566,
u
31
=
u
33
=1
.
0051,
u
41
=
u
43
=2
.
1479,
u
51
=
u
53
=4
.
5766,
u
61
=
u
63
=9
.
8316,
u
71
=
u
73
=21
.
6051,
u
12
=0
.
2494,
u
22
=0
.
6447,
u
32
=1
.
4162,
u
42
=3
.
0097,
u
52
=6
.
3269,
u
62
=13
.
1447,
u
72
=26
.
5887.
3. (a)
755

756. TIME
X=0.0 X=0.2 X=0.4 X=0.6 X=0.8 X=1.0
0.00 0.0000 0.2000 0.4000 0.6000 0.8000 0.0000
0.01 0.0000 0.2000 0.4000 0.6000 0.8000 0.0000
0.02 0.0000 0.2000 0.4000 0.6000 0.5500 0.0000
0.03 0.0000 0.2000 0.4000 0.5375 0.4250 0.0000
0.04 0.0000 0.2000 0.3844 0.4750 0.3469 0.0000
0.05 0.0000 0.1961 0.3609 0.4203 0.2922 0.0000
Chapter 16 Review Exercises
(b)
(c)
The table in part (b) is the same as the table in part (a) shifted downward one row.
756

757. 17
Functions of a Complex Variable
Exercises 17.1
1.
3+3
i
2.
−
4
i
3.
i
8
=(
i
2
)
4
=(
−
1)
4
=1
4.
i
11
=
i
(
i
2
)
5
=
i
(
−
1)
5
=
−
i
5.
7
−
13
i
6.
−
3
−
9
i
7.
−
7+5
i
8.
−
7+8
i
9.
11
−
10
i
10.
3
4
+
2
3
i
11.
−
5+12
i
12.
−
2
−
2
i
13.
−
2
i
14.
i
1+
i
·
1
−
i
1
−
i
=
i
+1
2
=
1
2
+
1
2
i
15.
2
−
4
i
3+5
i
·
3
−
5
i
3
−
5
i
=
−
14
−
22
i
34
=
−
7
17
−
11
17
i
16.
10
−
5
i
6+2
i
·
6
−
2
i
6
−
2
i
=
50
−
50
i
40
=
5
4
−
5
4
i
17.
9+7
i
1+
i
·
1
−
i
1
−
i
=
16
−
2
i
2
=8
−
i
18.
3
−
i
11
−
2
i
·
11 + 2
i
11 + 2
i
=
35
−
5
i
125
=
7
25
−
1
25
i
19.
2
−
11
i
6
−
i
·
6+
i
6+
i
=
23
−
64
i
37
=
23
37
−
64
37
i
20.
4+3
i
3+4
i
·
3
−
4
i
3
−
4
i
=
24
−
7
i
25
=
24
25
−
7
25
i
21.
(1 +
i
)(10 + 10
i
) = 10(1 +
i
)
2
=20
i
22.
[(1 +
i
)(1
−
i
)]
2
(1
−
i
)=4
−
4
i
23.
20+23
i
+
1
2
−
i
·
2+
i
2+
i
=20+23
i
+
2
5
+
1
5
i
=
102
5
+
116
5
i
24.
(2+3
i
)(
−
i
)
2
=
−
2
−
3
i
25.
i
9+7
i
·
9
−
7
i
9
−
7
i
=
7+9
i
130
=
7
130
+
9
130
i
26.
1
6+8
i
·
6
−
8
i
6
−
8
i
=
6
−
8
i
84
=
1
14
−
2
21
i
27.
x
x
2
+
y
2
28.
x
2
−
y
2
29.
−
2
y
−
4
30.
0
31.
1
(
x
−
1)
2
+(
y
−
3)
2
32.
1
36
x
2
+16
y
2
33.
2
x
+2
yi
=
−
9+2
i
implies 2
x
=
−
9 and 2
y
= 2. Hence
z
=
−
9
2
+
i
.
34.
−
x
+3
yi
=
−
7+6
i
implies
−
x
−
7 and 3
y
= 6. Hence
z
=7+2
i
.
35.
x
2
−
y
2
+2
xyi
=0+
i
implies
x
2
−
y
2
= 0 and 2
xy
=1. Now
y
=
x
implies 2
x
2
= 1 and so
x
=
±
1
/
√
2 . The
choice
y
=
−
x
gives
−
2
x
2
= 1 which has no real solution. Hence
z
=
1
√
2
+
1
√
2
i
and
z
=
−
1
√
2
−
1
√
2
i
.
757

758. Exercises 17.1
36.
x
2
−
y
2
−
4
x
+(
−
2
xy
−
4
y
)
i
=0+0
i
implies
x
2
−
y
2
−
4
x
= 0 and
y
(
−
2
x
−
4) = 0. If
y
= 0 then
x
(
x
−
4) = 0
and so
z
= 0 and
z
=4. If
−
2
x
−
4=0or
x
=
−
2 then 12
−
y
2
=0or
y
=
±
2
√
3 . This gives
z
=
−
2+2
√
3
i
and
z
=
−
2
−
2
√
3
i
.
37.
|
10+8
i
|
=
√
164 and
|
11
−
6
i
|
=
√
157 . Hence 11
−
6
i
is closer to the origin.
38.
|
1
2
−
1
4
i
|
=
√
5
4
and
|
2
3
+
1
6
i
|
=
√
17
6
. Since
√
5
4
<
√
17
6
,
1
2
−
1
4
i
is closer to the origin.
39.
|
z
1
−
z
2
|
=
|
(
x
1
−
x
2
)+
i
(
y
1
−
y
2
)
|
=
1
(
x
1
−
x
2
)
2
+(
y
1
−
y
2
)
2
which is the distance formula in the plane.
40.
By the triangle inequality,
|
z
+6+8
i
|≤|
z
|
+
|
6+8
i
|
. On the circle,
|
z
|
= 2 and so
|
z
+6+8
i
|≤
2+
√
100 = 12.
Exercises 17.2
1.
2(cos 2
π
+
i
sin 2
π
)
2.
10(cos
π
+
i
sin
π
)
3.
3
.
cos
3
π
2
+
i
sin
3
π
2
2
4.
6
3
cos
π
2
+
i
sin
π
2
4
5.
√
2
3
cos
π
4
+
i
sin
π
4
4
6.
5
√
2
.
cos
7
π
4
+
i
sin
7
π
4
2
7.
2
.
cos
5
π
6
+
i
sin
5
π
6
2
8.
4
.
cos
4
π
3
+
i
sin
4
π
3
2
9.
3
√
2
2
.
cos
5
π
4
+
i
sin
5
π
4
2
10.
6
5
cos
3
−
π
6
4
+
i
sin
3
−
π
6
46
11.
z
=
−
5
√
3
2
−
5
2
i
12.
z
=
−
8+8
i
13.
z
=5
.
5433 + 2
.
2961
i
14.
z
=8
.
0902 + 5
.
8779
i
15.
z
1
z
2
=8

cos
.
π
8
+
3
π
8
2
+
i
sin
.
π
8
+
3
π
8
28
=8
i
;
z
1
z
2
=
1
2

cos
.
π
8
−
3
π
8
2
+
i
sin
.
π
8
−
3
π
8
28
=
√
2
4
−
√
2
4
i
16.
z
1
z
2
=
√
6
5
cos
3
π
4
+
π
12
4
+
i
sin
3
π
4
+
π
12
46
=
√
6
2
+
3
√
2
2
i
z
1
z
2
=
√
6
3
5
cos
3
π
4
−
π
12
4
+
i
sin
3
π
4
−
π
12
46
=
√
2
2
+
√
6
6
i
17.

3
√
2
.
cos
7
π
4
+
i
sin
7
π
4
28
5
10
3
cos
π
3
+
i
sin
π
3
46
=30
√
2

cos
.
7
π
4
+
π
3
2
+
i
sin
.
7
π
4
+
π
3
28
=40
.
9808 + 10
.
9808
i
18.
5
4
√
2
3
cos
π
4
+
i
sin
π
4
46

√
2
.
cos
3
π
4
+
i
sin
3
π
4
28
=8

cos
.
π
4
+
3
π
4
2
+
i
sin
.
π
4
+
3
π
4
28
=
−
8
19.
cos
3
π
2
+
i
sin
3
π
2
2
√
2

cos
7
π
4
+
i
sin
7
π
4

=
√
2
4

cos
.
3
π
2
−
7
π
4
2
+
i
sin
.
3
π
2
−
7
π
4
28
=
1
4
−
1
4
i
758

759. Exercises 17.2
20.
2
√
2
t
cos
π
3
+
i
sin
π
3
e
2

cos
2
π
3
+
i
sin
2
π
3

=
√
2

cos
z
π
3
−
2
π
3
i
+
i
sin
z
π
3
−
2
π
3
i.
=
√
2
2
−
√
6
2
i
21.
2
9

cos
9
π
3
+
i
sin
9
π
3

=
−
512
22.
(2
√
2)
5

cos
z
−
5
π
4
i
+
i
sin
z
−
5
π
4
i.
=
−
128 + 128
i
23.

√
2
2

10

cos
10
π
4
+
i
sin
10
π
4

=
1
32
i
24.
(2
√
2)
4

cos
8
π
3
+
i
sin
8
π
3

=
−
32+32
√
3
i
25.
cos
12
π
8
+
i
sin
12
π
8
=
−
i
26.
(
√
3)
6

cos
12
π
9
+
i
sin
12
π
9

=
−
27
2
−
27
√
3
2
i
27.
8
1
/
3
=2

cos
2
kπ
3
+
i
sin
2
kπ
3

,
k
=0,1,2
w
0
= 2[cos 0 +
i
sin 0] = 2;
w
1
=2

cos
2
π
3
+
i
sin
2
π
3

=
−
1+
√
3
i
w
2
=2

cos
4
π
3
+
i
sin
4
π
3

=
−
1
−
√
3
i
28.
(1)
1
/
8
= cos
kπ
4
+
i
sin
kπ
4
,
k
=0,1,2,
...
,7
w
0
= cos 0 +
i
sin 0 = 1;
w
1
= cos
π
4
+
i
sin
π
4
=
√
2
2
+
√
2
2
i
w
2
= cos
π
2
+
i
sin
π
2
=
i
;
w
3
= cos
3
π
4
+
i
sin
3
π
4
=
−
√
2
2
+
√
2
2
i
w
4
= cos
π
+
i
sin
π
=
−
1;
w
5
= cos
5
π
4
+
i
sin
5
π
4
=
−
√
2
2
−
√
2
2
i
w
6
= cos
3
π
2
+
i
sin
3
π
2
=
−
i
;
w
7
= cos
7
π
4
+
i
sin
7
π
4
=
√
2
2
−
√
2
2
i
29.
(
i
)
1
/
2
= cos
y
π
4
+
kπ
x
+
i
sin
y
π
4
+
kπ
x
,
k
=0,1
w
0
= cos
π
4
+
i
sin
π
4
=
√
2
2
+
√
2
2
i
w
1
= cos
5
π
4
+
i
sin
5
π
4
=
−
√
2
2
−
√
2
2
i
759

760. Exercises 17.2
30.
(
−
1+
i
)
1
/
3
=2
1
/
6

cos
z
π
4
+
2
kπ
3
i
+
i
sin
z
π
4
+
2
kπ
3
i.
,
k
=0,1,2
w
0
=2
1
/
6
t
cos
π
4
+
i
sin
π
4
e
=
1
3
√
2
+
1
3
√
2
i
=0
.
7937 + 0
.
7937
i
w
1
=2
1
/
6

cos
11
π
12
+
i
sin
11
π
12

=
−
1
.
0842 + 0
.
2905
i
w
2
=2
1
/
6

cos
19
π
12
+
i
sin
19
π
12

=0
.
2905
−
1
.
0842
i
31.
(
−
1+
√
3
i
)
1
/
2
=2
1
/
2
t
cos
y
π
3
+
kπ
x
+
i
sin
y
π
3
+
kπ
xe
,
k
=0,1
w
0
=2
1
/
2
t
cos
π
3
+
i
sin
π
3
e
=
√
2
2
+
√
6
2
i
w
2
=2
1
/
2

cos
4
π
3
+
i
sin
4
π
3

=
−
√
2
2
−
√
6
2
i
32.
(
−
1
−
√
3
i
)
1
/
4
=2
1
/
4

cos
z
π
3
+
kπ
2
i
+
i
sin
z
π
3
+
kπ
2
i.
,
k
=0,1,2,3
w
0
=2
1
/
4
t
cos
π
3
+
i
sin
π
3
e
=2
1
/
4

1
2
+
√
3
2
i
a
w
1
=2
1
/
4

cos
5
π
6
+
i
sin
5
π
6

=2
1
/
4

−
√
3
2
+
1
2
i
a
w
2
=2
1
/
4

cos
4
π
3
+
i
sin
4
π
3

=2
1
/
4

−
1
2
−
√
3
2
i
a
;
w
3
=2
1
/
4

cos
11
π
6
+
i
sin
11
π
6

=2
1
/
4

√
3
2
−
1
2
i
a
33.
The solutions are the four fourth roots of
−
1;
w
k
= cos
π
+2
kπ
4
+
i
sin
π
+2
kπ
4
,k
=0
,
1
,
2
,
3
.
We have
w
1
= cos
π
4
+
i
sin
π
4
=
√
2
2
+
√
2
2
i
w
2
= cos
3
π
4
+
i
sin
3
π
4
=
−
√
2
2
+
√
2
2
i
w
3
= cos
5
π
4
+
i
sin
5
π
4
=
−
√
2
2
−
√
2
2
i
w
4
= cos
7
π
4
+
i
sin
7
π
4
=
√
2
2
−
√
2
2
i.
34.
(
z
4
−
1)
2
= 0 is the same as (
z
−
i
)
2
(
z
+
i
)
2
(
z
−
1)
2
(
z
+1)
2
=0. Thus
z
1
=1,
z
2
=
−
1,
z
3
=
i
, and
z
4
=
−
i
are
roots of multiplicity two.
35.
y
cos
π
9
+
i
sin
π
9
x
12
t
2
y
cos
π
6
+
i
sin
π
6
xe
5
=2
5
z
cos
4
π
3
+
i
sin
4
π
3
iz
cos
5
π
6
+
i
sin
5
π
6
i
=32

cos
z
4
π
3
+
5
π
6
i
+
i
sin
z
4
π
3
+
5
π
6
i.
=32
z
cos
13
π
6
+
i
sin
13
π
6
i
=32
y
cos
π
6
+
i
sin
π
6
x
=16
√
3+16
i
36.

8
z
cos
3
π
8
+
i
sin
3
π
8
i.
3
t
2
y
cos
π
16
+
i
sin
π
16
xe
10
=
2
9
2
10

cos
z
9
π
8
−
10
π
16
i
+
i
z
9
π
8
−
10
π
16
i.
=
1
2
y
cos
π
2
+
i
sin
π
2
x
=
1
2
i
760

761. Exercises 17.3
37.
We have
(cos 2
θ
+
i
sin
θ
)
2
= cos 2
θ
+
i
sin 2
θ
Also
(cos
θ
+
i
sin
θ
)
2
= cos
2
θ
−
sin
2
θ
+ (2 sin
θ
cos
θ
)
i.
Equating real and imaginary parts gives
cos 2
θ
= cos
2
θ
−
sin
2
θ,
sin 2
θ
= 2 sin
θ
cos
θ.
38.
We have
(cos
θ
+
i
sin
θ
)
3
= cos 3
θ
+
i
sin 3
θ.
Also
(cos
θ
+
i
sin
θ
)
3
= cos
3
θ
+ 3 cos
2
θ
(
i
sin
θ
)+3cos
θ
(
i
sin
θ
)
2
+(
i
sin
θ
)
3
= cos
3
θ
−
3 cos
θ
sin
2
θ
+ (3 cos
2
θ
sin
θ
−
sin
3
θ
)
i.
Equating real and imaginary parts gives
cos 3
θ
= cos
3
θ
−
3 cos
θ
sin
2
θ,
sin 3
θ
= 3 cos
2
θ
sin
θ
−
sin
3
θ.
39. (a)
Arg(
z
1
)=
π
, Arg(
z
2
)=
π
2
, Arg(
z
1
z
2
)=
−
π
2
, Arg(
z
1
) + Arg(
z
2
)=
3
π
2
5
= Arg(
z
1
z
2
)
(b)
Arg(
z
1
/z
2
)=
−
π
2
, Arg(
z
1
)
−
Arg(
z
2
)=
π
−
π
2
=
π
2
5
= Arg(
z
1
/z
2
)
40. (a)
If we take arg(
z
1
)=
π
and arg(
z
2
)=
π/
2 then arg(
z
1
) + arg(
z
2
)=3
π/
2 is an argument of the product
z
1
z
2
=
−
5
i
. With these same arguments we see that arg(
z
1
)
−
arg(
z
2
)=
π/
2 is an argument of the quotient
z
1
/z
2
=
1
5
i
.
(b)
If we take arg(
z
1
)=
π
and arg(
z
2
)=
−
π/
2 then arg(
z
1
) + arg(
z
2
)=
π/
2 is an argument of the product
z
1
z
2
=5
i
. With these same arguments we see that arg(
z
1
)
−
arg(
z
2
)=3
π/
2 is an argument of the quotient
z
1
/z
2
=
−
1
5
i
.
Exercises 17.3
1. 2. 3.
4. 5. 6.
761

762. Exercises 17.3
7. 8. 9.
10. 11. 12.
13. 14. 15.
16. 17. 18.
19. 20. 21.
762

763. Exercises 17.4
22.
23.
The given equation is equivalent to (
x
+1)
2
+
y
2
=
x
2
+(
y
−
1)
2
. This simplifies to
y
=
−
x
which describes a
straight line through the origin.
24.
|
Re(
z
)
|
=
|
x
|
is the same as
√
x
2
and
|
z
|
=
1
x
2
+
y
2
. Since
y
2
≥
0 the inequality
√
x
2
≤
1
x
2
+
y
2
is true for
all complex numbers.
25.
The given equation simplifies to the equation
x
2
−
y
2
= 1 which is a hyperbola with center at the origin.
26.
Since
|
z
−
i
|
and
|
z
−
(
−
i
)
|
represent distances from the point (
x, y
)to
i
and
−
i
, respectively, the equation is
the distance formula definition of an ellipse with foci at (0
,
1) and (0
,
−
1).
Exercises 17.4
1.
Substituting
y
= 2 into
u
=
x
2
−
y
2
,
v
=2
xy
gives the parametric equations
u
=
x
2
−
4,
v
=4
x
.
Using
x
=
v/
4 the first equation gives
u
=
v
2
/
16
−
4. The graph is the parabola shown.
2.
Substituting
x
=
−
3into
u
=
x
2
−
y
2
,
v
=2
xy
gives the parametric equations
u
=9
−
y
2
,
v
=
−
6
y
. Using
y
=
−
v/
6 the first equation gives
u
=9
−
v
2
/
36. The graph is the parabola
shown.
3.
x
= 0 gives
u
=
−
y
2
,
v
= 0. Since
−
y
2
≤
0 for all real values of
y
, the image is the origin
and the negative
u
-axis.
4.
y
= 0 gives
u
=
x
2
,
v
= 0. Since
x
2
≥
0 for all real values of
x
, the image is the origin and
the positive
u
-axis.
5.
y
=
x
gives
u
=0,
v
=2
x
2
. Since
x
2
≥
0 for all real values of
x
, the image is the origin and the
positive
v
-axis.
763

764. Exercises 17.4
6.
y
=
−
x
gives
u
=0,
v
=
−
2
x
2
. Since
−
x
2
≤
0 for all real values of
x
, the image is the origin and the
negative
v
-axis.
7.
f
(
z
)=(6
x
−
5) +
i
(6
y
+9)
8.
f
(
z
)=(7
x
−
9
y
−
3) +
i
(7
y
−
9
x
+2)
9.
f
(
z
)=(
x
2
−
y
2
−
3
x
)+
i
(2
xy
−
3
y
+4)
10.
f
(
z
)=(3
x
2
−
3
y
2
+2
x
)+
i
(
−
6
xy
+2
y
)
11.
f
(
z
)=(
x
3
−
3
xy
2
−
4
x
)+
i
(3
x
2
y
−
y
3
−
4
y
)
12.
f
(
z
)=(
x
4
−
6
x
2
y
2
+
y
4
)+
i
(4
x
3
y
−
4
xy
3
)
13.
f
(
z
)=
z
x
+
x
x
2
+
y
2
i
+
i
z
y
−
y
x
2
+
y
2
i
14.
f
(
z
)=
x
2
+
y
2
+
x
(
x
+1)
2
+
y
2
+
i
y
(
x
+1)
2
+
y
2
15. (a)
f
(0+2
i
)=
−
4+
i
(b)
f
(2
−
i
)=3
−
9
i
(c)
f
(5+3
i
)=1+86
i
16. (a)
f
(1 +
i
)=3
−
2
i
(b)
f
(2
−
i
)=
7
2
+10
i
(c)
f
(1+4
i
)=3
−
32
i
17. (a)
f
(4
−
6
i
)=14
−
20
i
(b)
f
(
−
5+12
i
)=
−
13+43
i
(c)
f
(2
−
7
i
)=3
−
26
i
18. (a)
f
(0 +
π
4
i
)=
√
2
2
+
√
2
2
i
(b)
f
(
−
1
−
πi
)=
−
e
−
1
(c)
f
(3+
π
3
i
)=
1
2
e
3
+
√
3
2
e
3
i
19.
lim
z
→
i
(4
z
3
−
5
z
2
+4
z
+1
−
5
i
)=6
−
5
i
20.
lim
z
→
1
−
i
5
z
2
−
2
z
+2
z
+1
=
5(1
−
i
)
2
−
2(1
−
i
)+2
2
−
i
=
8
5
−
16
5
i
21.
lim
z
→
i
z
4
−
1
z
−
i
= lim
z
→
i
(
z
2
−
1)(
z
−
i
)(
z
+
i
)
z
−
i
=
−
4
i
22.
lim
z
→
1+
i
z
2
−
2
z
+2
z
2
−
2
i
= lim
z
→
1+
i
[
z
−
(1 +
i
)][
z
−
(1
−
i
)]
[
z
−
(1 +
i
)][
z
−
(
−
1
−
i
)]
=
1
2
+
1
2
i
23.
Along the
y
-axis, lim
z
→
0
x
+
iy
x
−
iy
= lim
y
→
0
iy
−
iy
=
−
1, whereas along the
x
-axis, lim
z
→
0
x
+
iy
x
−
iy
= lim
x
→
0
x
x
=1.
24.
Along the line
x
= 1, lim
z
→
1
x
+
y
−
1
z
−
1
= lim
y
→
0
y
iy
=
1
i
=
−
i
, whereas along the
x
-axis,
lim
z
→
1
x
+
y
−
1
z
−
1
= lim
x
→
1
x
−
1
x
−
1
=1.
25.
f
i
(
z
) = lim
∆
z
→
0
(
z
+∆
z
)
2
−
z
2
∆
z
= lim
∆
z
→
0
2
z
∆
z
+(∆
z
)
2
∆
z
= lim
∆
z
→
0
(2
z
+∆
z
)=2
z
26.
f
i
(
z
) = lim
∆
z
→
0
1
z
+∆
z
−
1
z
∆
z
= lim
∆
z
→
0
−
∆
z
(∆
z
)
z
(
z
+∆
z
)
= lim
∆
z
→
0
−
1
z
(
z
+∆
z
)
=
−
1
z
2
27.
f
i
(
z
)=12
z
2
−
(6+2
i
)
z
−
5
28.
f
i
(
z
)=20
z
3
−
3
iz
2
+ (16
−
2
i
)
z
29.
f
i
(
z
)=(2
z
+ 1)(2
z
−
4)+2(
z
2
−
4
z
+8
i
)=6
z
2
−
14
z
−
4+8
i
764

765. Exercises 17.4
30.
f
i
(
z
)=(
z
5
+3
iz
3
)(4
z
2
+3
iz
2
+4
z
−
6
i
)+(
z
4
+
iz
3
+2
z
2
−
6
iz
)(5
z
4
+9
iz
2
)
31.
f
i
(
z
)=6
z
(
z
2
−
4
i
)
2
32.
f
i
(
z
) = 6(2
z
−
1
/z
)
5
(2+1
/z
2
)
33.
f
i
(
z
)=
(2
z
+
i
)3
−
(3
z
−
4+8
i
)2
(2
z
+1)
2
=
8
−
13
i
(2
z
+
i
)
2
34.
f
i
(
z
)=
(
z
3
+ 1)(10
z
−
1)
−
(5
z
2
−
z
)3
z
2
(
z
3
+1)
2
=
−
5
z
4
+2
z
3
+10
z
−
1
(
z
3
+1)
2
35.
3
i
36.
0, 2
−
5
i
37.
−
2
i
,2
i
38.
3
−
4
i
,3+4
i
39.
We have lim
∆
z
→
0
z
+∆
z
−
z
∆
z
= lim
∆
z
→
0
∆
z
∆
z
.
If we let ∆
z
→
0 along a horizontal line then ∆
z
=∆
x
,
∆
z
=∆
x
, and
lim
∆
z
→
0
∆
z
∆
z
= lim
∆
x
→
0
∆
x
∆
x
=1
.
If we let ∆
z
→
0 along a vertical line then ∆
z
=
i
∆
y
,
∆
z
=
−
i
∆
y
, and
lim
∆
z
→
0
∆
z
∆
z
= lim
∆
y
→
0
−
i
∆
y
i
∆
y
=
−
1
.
Since these two limits are not equal,
f
(
z
)=
z
cannot be differentiable at any
z
.
40.
We have
f
i
(
z
) = lim
∆
z
→
0
(
z
+∆
z
)(
z
+
∆
z
)
−
z
z
∆
z
= lim
∆
z
→
0
z
z
+
z
∆
z
∆
z
+
∆
z
i
.
If
z
= 0, then the above limit becomes
f
i
(0) = lim
∆
z
→
0
∆
z
=0
.
If
z
t
= 0 then we first let ∆
z
→
0 along a horizontal line so that ∆
z
=∆
x
and
∆
z
=∆
x
. Thus,
f
i
(
z
) = lim
∆
z
→
0
z
z
+
z
∆
x
∆
x
+∆
x
i
=
z
+
z.
Next we let ∆
z
→
0 along a vertical line so that ∆
z
=
i
∆
y
,
∆
z
=
−
i
∆
y
.Thus
f
i
(
z
) = lim
∆
y
→
0
z
z
+
z
−
i
∆
y
i
∆
y
+
i
∆
y
i
=
z
−
z.
We must have
z
+
z
=
z
−
z
which implies
z
= 0. This is a contradiction to the assumption that
z
t
= 0. Hence
f
(
z
)=
|
z
|
2
is differentiable only at
z
=0.
41.
Each linear equation in the system
dx
dt
=2
x
,
dy
dt
=2
y
can be solved directly. We obtain
x
(
t
)=
c
1
e
2
t
and
y
(
t
)=
c
2
e
2
t
.
42.
The system
dx
dt
=
−
y
,
dy
dt
=
x
can be solved as in Section 3
.
11. We obtain
x
(
t
)=
c
1
cos
t
+
c
2
sin
t
,
y
(
t
)=
c
1
sin
t
−
c
2
cos
t
.
43.
The equations in the system
dx
dt
=
x
x
2
+
y
2
,
dy
dt
=
y
x
2
+
y
2
can be divided to give
dy
dx
=
y
x
. By separation
of variables we obtain
y
=
cx
.
44.
Each equation in the system
dx
dt
=
x
2
,
dy
dt
=
−
y
2
can be solved directly by separation of variables. We obtain
x
(
t
)=
−
1
t
+
c
1
,
y
=
1
t
+
c
2
.
765

766. Exercises 17.4
45.
If
y
=
1
2
x
2
the equations
u
=
x
2
−
y
2
,
v
=2
xy
give
u
=
x
2
−
1
4
x
4
,
v
=
x
3
. With the aid of a
computer, the graph of these parametric equations is shown.
46.
If
y
=(
x
−
1)
2
the equations
u
=
x
2
−
y
2
,
v
=2
xy
give
u
=
x
2
−
(
x
−
1)
4
,
v
=2
x
(
x
−
1)
2
. With
the aid of a computer the graph of these parametric equations is shown.
Exercises 17.5
1.
u
=
x
3
−
3
xy
2
,
v
=3
x
2
y
−
y
3
;
∂u
∂x
=3
x
2
−
3
y
2
=
∂v
∂y
,
∂u
∂y
=
−
6
xy
=
−
∂v
∂x
2.
u
=3
x
2
−
3
y
2
+5
x
,
v
=6
xy
+5
y
−
6;
∂u
∂x
=6
x
+5=
∂v
∂y
,
∂u
∂y
=
−
6
y
=
−
∂v
∂x
3.
u
=
x
,
v
=0;
∂u
∂x
=1,
∂v
∂y
= 0. Since 1
t
=0,
f
is not analytic at any point.
4.
u
=
y
,
v
=
x
;
∂u
∂x
=0=
∂v
∂y
,
∂u
∂y
=1,
−
∂v
∂x
=
−
1. Since 1
t
=
−
1,
f
is not analytic at any point.
5.
u
=
−
2
x
+3,
v
=10
y
;
∂u
∂x
=
−
2,
∂v
∂y
= 10. Since
−
2
t
= 10,
f
is not analytic at any point.
6.
u
=
x
2
−
y
2
,
v
=
−
2
xy
;
∂u
∂x
=2
x
,
∂v
∂y
=
−
2
x
;
∂u
∂y
=
−
2
y
,
−
∂v
∂x
=2
y
The Cauchy-Riemann equations hold only at (0
,
0). Since there is no neighborhood about
z
= 0 within which
f
is differentiable we conclude
f
is nowhere analytic.
7.
u
=
x
2
+
y
2
,
v
=0;
∂u
∂x
=2
x
,
∂v
∂y
=0;
∂u
∂y
=2
y
,
−
∂v
∂x
=0
The Cauchy-Riemann equations hold only at (0
,
0). Since there is no neighborhood about
z
= 0 within which
f
is differentiable we conclude
f
is nowhere analytic.
8.
u
=
x
x
2
+
y
2
,
v
=
y
x
2
+
y
2
;
∂u
∂x
=
y
2
−
x
2
(
x
2
+
y
2
)
2
,
∂v
∂y
=
x
2
−
y
2
(
x
2
+
y
2
)
2
;
∂u
∂y
=
−
2
xy
(
x
2
+
y
2
)
2
=
∂v
∂x
The Cauchy-Riemann equations hold only at (0
,
0). Since there is no neighborhood about
z
= 0 within which
f
is differentiable, we conclude
f
is nowhere analytic.
9.
u
=
e
x
cos
y
,
v
=
e
x
sin
y
;
∂u
∂x
=
e
x
cos
y
=
∂v
∂y
;
∂u
∂y
=
−
e
x
sin
y
=
−
∂v
∂x
.
f
is analytic for all
z
.
10.
u
=
x
+ sin
x
cosh
y
,
v
=
y
+ cos
x
sinh
y
;
∂u
∂x
= 1 + cos
x
cosh
y
=
∂v
∂y
;
∂u
∂y
= sin
x
sinh
y
=
−
∂v
∂x
.
f
is analytic for all
z
.
11.
u
=
e
x
2
−
y
2
cos 2
xy
,
v
=
e
x
2
−
y
2
sin 2
xy
;
∂u
∂x
=
−
2
ye
x
2
−
y
2
sin 2
xy
+2
xe
x
2
−
y
2
cos 2
xy
=
∂v
∂y
;
766

767. Exercises 17.5
∂u
∂y
=
−
2
xe
x
2
−
y
2
sin 2
xy
−
2
ye
x
2
−
y
2
cos 2
xy
=
−
∂v
∂x
.
f
is analytic for all
z
.
12.
u
=4
x
2
+5
x
−
4
y
2
+9,
v
=8
xy
+5
y
−
1;
∂u
∂x
=8
x
+5=
∂v
∂y
,
∂u
∂y
=
−
8
y
=
−
∂v
∂x
.
f
is analytic for all
z
.
13.
u
=
x
−
1
(
x
−
1)
2
+
y
2
,
v
=
−
y
(
x
−
1)
2
+
y
2
;
∂u
∂x
=
y
2
−
(
x
−
1)
2
[(
x
−
1)
2
+
y
2
]
=
∂v
∂y
,
∂u
∂y
=
−
2
y
(
x
−
1)
[(
x
−
1)
2
+
y
2
]
2
=
−
∂v
∂x
f
is analytic in any domain not containing
z
=1.
14.
u
=
x
3
+
xy
2
+
x
x
2
+
y
2
,
v
=
x
2
y
+
y
3
−
y
x
2
+
y
2
;
∂u
∂x
=
x
4
+2
x
2
y
2
−
x
2
+
y
2
+
y
4
(
x
2
+
y
2
)
2
=
∂v
∂y
,
∂u
∂y
=
−
2
xy
(
x
2
+
y
2
)
2
=
−
∂v
∂x
f
is analytic in any domain not containing
z
=0.
15.
∂u
∂x
=3=
b
=
∂v
∂y
;
∂u
∂y
=
−
1=
−
a
=
−
∂v
∂x
.
f
is analytic for all
z
when
b
=3,
a
=1.
16.
The Cauchy-Riemann equations yield the system
2
x
+
ay
=
dx
+2
y
ax
+2
by
=
−
2
cx
−
dy
or
(2
−
d
)
x
+(2
−
a
)
y
=0
(
a
+2
c
)
x
+(2
b
+
d
)
y
=0
.
The system holds for
z
=
x
+
iy
whenever 2
−
d
=0,2
−
a
=0,
a
+2
c
= 0, and 2
b
+
d
= 0. That is,
f
is analytic
for all
z
when
a
=2,
b
=
−
1,
c
=
−
1, and
d
=2.
17.
u
=
x
2
+
y
2
,
v
=2
xy
;
∂u
∂x
=2
x
,
∂v
∂y
=2
x
;
∂u
∂y
=2
y
,
−
∂v
∂x
=
−
2
y
u
and
v
are continuous and have continuous first partial derivatives. The Cauchy-Riemann equations are satisfied
for any
x
and for
y
= 0, that is, for points on the real axis. The function
f
is differentiable but not analytic
along this axis; there is no neighborhood about any point
z
=
x
within which
f
is differentiable.z
18.
u
=3
x
2
y
2
,
v
=
−
6
x
2
y
2
;
∂u
∂x
=6
xy
2
,
∂v
∂y
=
−
12
x
2
y
;
∂u
∂y
=6
x
2
y
,
−
∂v
∂x
=12
xy
2
u
and
v
are continuous and have continuous first partial derivatives. The Cauchy-Riemann equations are satisfied
whenever 6
xy
(
y
+2
x
) = 0 and 6
xy
(
x
−
2
y
) = 0. The point satisfying
y
+2
x
= 0 and
x
−
2
y
=0is
z
= 0. The
points that satisfy 6
xy
= 0 are the points along the
y
-axis (
x
= 0) or along the
x
-axis (
y
= 0). The function
f
is differentiable but not analytic on either axis; there is no neighborhood about any point
z
=
x
or
z
=
iy
within which
f
is differentiable.
19.
u
=
x
3
+3
xy
2
−
x
,
v
=
y
3
+3
x
2
y
−
y
;
∂u
∂x
=3
x
2
+3
y
2
−
1,
∂v
∂y
=3
y
2
+3
x
2
−
1;
∂u
∂y
=6
xy
,
−
∂v
∂x
=
−
6
xy
.
u
and
v
are continuous and have continuous first partial derivatives. The Cauchy-Riemann equations are
satisfied whenever 6
xy
=
−
6
xy
or 12
xy
= 0. The points satisfying 12
xy
= 0 are the points along the
y
-axis
(
x
= 0) or along the
x
-axis (
y
= 0). The function
f
is differentiable but not analytic on either axis; there is no
neighborhood about any point
z
=
x
or
z
=
iy
within which
f
is differentiable.
20.
u
=
x
2
−
x
+
y
,
v
=
y
2
−
5
y
−
x
;
∂u
∂x
=2
x
−
1,
∂v
∂y
=2
y
−
5;
∂u
∂y
=1,
−
∂v
∂x
=1
u
and
v
are continuous and have continuous first partial derivatives. The Cauchy-Riemann equations are satisfied
whenever 2
x
−
1=2
y
−
5 or for points on the line
y
=
x
+ 2. The function
f
is differentiable but not analytic
on this line; there is no neighborhood about any point
z
=
x
+(
x
+2)
i
within which
f
is differentiable.
21.
Since
f
is entire,
f
i
(
z
)=
∂u
∂x
+
i
∂v
∂x
=
e
x
cos
y
+
ie
x
sin
y
=
f
(
z
)
.
767

768. Exercises 17.5
22.
Since
f
is entire,
f
i
(
z
)=
∂u
∂x
+
i
∂v
∂x
=
−
2
ye
x
2
−
y
2
sin 2
xy
+2
xe
x
2
−
y
2
cos 2
xy
+
i
(2
ye
x
2
−
y
2
cos 2
xy
+2
xe
x
2
−
y
2
sin 2
xy
)
.
23.
∂
2
u
∂x
2
=0,
∂
2
u
∂y
2
= 0 gives
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0. Thus
u
is harmonic. Now
∂u
∂x
=1=
∂v
∂y
implies
v
=
y
+
h
(
x
),
∂u
∂y
=0=
−
∂v
∂x
implies 0 =
−
h
i
(
x
), and so
h
(
x
)=
C
(a constant.) Therefore
f
(
z
)=
x
+
i
(
y
+
C
).
24.
∂
2
u
∂x
2
=0,
∂
2
u
∂y
2
= 0 gives
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0. Thus
u
is harmonic. Now
∂u
∂x
=2
−
2
y
=
∂v
∂y
implies
v
=2
y
−
y
2
+
h
(
x
),
∂u
∂y
=
−
2
x
=
−
∂v
∂x
=
−
h
i
(
x
) implies
h
i
(
x
)=2
x
or
h
(
x
)=
x
2
+
C
. Therefore
f
(
z
)=2
x
−
2
xy
+
i
(2
y
−
y
2
+
x
2
+
C
).
25.
∂
2
u
∂x
2
=2,
∂
2
u
∂y
2
=
−
2 gives
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0. Thus
u
is harmonic. Now
∂u
∂x
=2
x
=
∂v
∂y
implies
v
=2
xy
+
h
(
x
),
∂u
∂y
=
−
2
y
=
−
∂v
∂x
=
−
2
y
−
h
i
(
x
) implies
h
i
(
x
)=0or
h
(
x
)=
C
. Therefore
f
(
z
)=
x
2
−
y
2
+
i
(2
xy
+
C
).
26.
∂
2
u
∂x
2
=
−
24
xy
,
∂
2
u
∂y
2
=24
xy
gives
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0. Thus
u
is harmonic. Now
∂u
∂x
=4
y
3
−
12
x
2
y
+1 =
∂v
∂y
implies
v
=
y
4
−
6
x
2
y
2
+
y
+
h
(
x
),
∂u
∂y
=12
xy
2
−
4
x
3
=
−
∂v
∂x
=12
xy
2
−
h
i
(
x
) implies
h
i
(
x
)=4
x
3
or
h
(
x
)=
x
4
+
C
.
Therefore
f
(
z
)=4
xy
3
−
4
x
3
y
+
x
+
i
(
y
4
−
6
x
2
y
2
+
y
+
x
4
+
C
).
27.
∂
2
u
∂x
2
=
2
y
2
−
2
x
2
(
x
2
+
y
2
)
2
,
∂
2
u
∂y
2
=
2
x
2
−
2
y
2
(
x
2
+
y
2
)
2
gives
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0. Thus
u
is harmonic. Now
∂u
∂x
=
2
x
x
2
+
y
2
=
∂v
∂y
implies
v
= 2 tan
−
1
y
x
+
h
(
x
),
∂u
∂y
=
2
y
x
2
+
y
2
=
−
∂v
∂x
=
2
y
x
2
+
y
2
−
h
i
(
x
) implies
h
i
(
x
)=0or
h
(
x
)=
C
.
Therefore
f
(
z
) = log
e
(
x
2
+
y
2
)+
i
y
tan
−
1
y
x
+
C
x
,
z
t
=0.
28.
∂
2
u
∂x
2
=2
e
x
cos
y
+
e
x
(
x
cos
y
−
y
sin
y
),
∂
2
u
∂y
2
=
e
x
(
−
x
cos
y
+
y
sin
y
−
2 cos
y
) gives
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=0. Thus
u
is
harmonic. Now
∂u
∂x
=
e
x
cos
y
+
e
x
(
x
cos
y
−
y
sin
y
)=
∂v
∂y
. Integrating by parts with respect to
y
implies
v
=
e
x
sin
y
+
e
x
(
x
sin
y
+
y
cos
y
−
sin
y
)+
h
(
x
)=
xe
x
sin
y
+
ye
x
cos
y
+
h
(
x
)
,
and
∂u
∂y
=
−
xe
x
sin
y
−
ye
x
cos
y
−
e
x
sin
y
=
−
∂v
∂x
=
−
xe
x
sin
y
−
e
x
sin
y
−
ye
x
cos
y
+
h
i
(
x
)
implies
h
i
(
x
)=0or
h
(
x
)=
C
. Therefore
f
(
z
)=
e
x
(
x
cos
y
−
y
sin
y
)+
ie
x
(
x
sin
y
+
y
cos
y
+
C
)
.
768

769. Exercises 17.6
29.
The level curves
u
(
x, y
)=
c
1
and
v
(
x, y
)=
c
2
are the families of hyperbolas
x
2
−
y
2
=
c
1
and 2
xy
=
c
2
, respectively. The graphs of these families are
displayed on the same axes in the figure.
30.
f
(
x
)=
x
x
2
+
y
2
−
i
y
x
2
+
y
2
. The level curves
u
(
x, y
)=
c
1
and
v
(
x, y
)=
c
2
are the family of circles
x
=
c
1
(
x
2
+
y
2
)
and
−
y
=
c
2
(
x
2
+
y
2
), with the exception that (0
,
0) is not on the circumference of any circle.
31.
f
(
z
)=
x
+
x
x
2
+
y
2
+
i
z
y
−
y
x
2
+
y
2
i
. The level curve
v
(
x, y
) = 0 is described by
y
−
y
x
2
+
y
2
=0or
y
(
x
2
+
y
2
−
1) = 0. We see that either
y
=0or
x
2
+
y
2
=1. Thus
v
(
x, y
) = 0 gives either the
x
-axis (without
the origin (0
,
0)) or the unit circle
x
2
+
y
2
=1.
32.
If
∇
u
=
∂u
∂x
i
+
∂u
∂y
j
and
∇
v
=
∂v
∂x
i
+
∂v
∂y
j
, then
∇
u
·∇
v
=
∂u
∂x
∂v
∂x
+
∂u
∂y
∂v
∂y
. By the Cauchy-Riemann
equations this becomes
∇
u
·∇
v
=
∂v
∂y
∂v
∂x
+
z
−
∂v
∂x
i
∂v
∂y
=0
.
Since the gradients of
u
and
v
are orthogonal vectors, the level curves
u
(
x, y
)=
c
1
and
u
(
x, y
)=
c
2
are orthogonal
families.
Exercises 17.6
1.
e
π
6
i
= cos
π
6
+
i
sin
π
6
=
√
3
2
+
1
2
i
2.
e
−
π
3
i
= cos
π
3
−
i
sin
π
3
=
1
2
−
√
3
2
i
3.
e
−
1+
π
4
i
=
e
−
1
cos
π
4
+
ie
−
1
sin
π
4
=
e
−
1

√
2
2
+
√
2
2
i

4.
e
2
−
π
2
i
=
e
2
cos
y
−
π
2
x
+
ie
2
sin
y
−
π
2
x
=
−
e
2
i
5.
e
π
+
πi
=
e
π
cos
π
+
ie
π
sin
π
=
−
e
π
6.
e
−
π
+
3
π
2
i
=
e
−
π
cos
3
π
2
+
ie
−
π
sin
3
π
2
=
−
e
−
π
i
7.
e
1
.
5+2
i
=
e
1
.
5
cos 2 +
ie
1
.
5
sin 2 =
−
1
.
8650 + 4
.
0752
i
8.
e
−
0
.
3+0
.
5
i
=
e
−
0
.
3
cos 0
.
5+
ie
−
0
.
3
sin 0
.
5=0
.
6501 + 0
.
3552
i
9.
e
5
i
= cos 5 +
i
sin 5 = 0
.
2837
−
0
.
9589
i
10.
e
−
0
.
23
−
i
=
e
−
0
.
23
cos(
−
1) +
ie
−
0
.
23
sin(
−
1) = 0
.
4293
−
0
.
6686
i
11.
e
11
π
12
i
= cos
11
π
12
+
i
sin
11
π
12
=
−
0
.
9659 + 0
.
2588
i
769

770. Exercises 17.6
12.
e
5+
5
π
2
i
=
e
5
cos
5
π
2
+
ie
5
sin
5
π
2
=
e
5
i
13.
e
−
iz
=
e
y
−
xi
=
e
y
cos
x
−
ie
y
sin
x
14.
e
2¯
z
=
e
2
x
−
2
yi
=
e
2
x
cos 2
y
−
ie
2
x
sin 2
y
15.
e
z
2
=
e
x
2
−
y
2
+2
xyi
=
e
x
2
−
y
2
cos 2
xy
+
ie
x
2
−
y
2
sin 2
xy
16.
e
1
/z
=
e
x/
(
x
2
+
y
2
)
−
iy/
(
x
2
+
y
2
)
=
e
x/
(
x
2
+
y
2
)
cos
y
x
2
+
y
2
−
ie
x/
(
x
2
+
y
2
)
sin
y
x
2
+
y
2
17.
|
e
z
|
2
=
e
2
x
cos
2
y
+
e
2
x
sin
2
y
=
e
2
x
(cos
2
y
+ sin
2
y
)=
e
2
x
implies
|
e
z
|
=
e
x
.
18.
e
z
1
e
z
2
=
e
x
1
cos
y
1
+
ie
x
1
sin
y
1
e
x
2
cos
y
2
+
ie
x
2
sin
y
2
=
(
e
x
1
cos
y
1
+
ie
x
1
sin
y
1
)(
e
x
2
cos
y
2
−
ie
x
2
sin
y
2
)
e
2
x
2
=
e
x
1
−
x
2
[(cos
y
1
cos
y
2
+ sin
y
1
sin
y
2
)+
i
(sin
y
1
cos
y
2
−
cos
y
1
sin
y
2
)]
=
e
x
1
−
x
2
[cos(
y
1
−
y
2
)+
i
sin(
y
1
−
y
2
)] =
e
x
1
−
x
2
+
i
(
y
1
−
y
2
)
=
e
(
x
1
+
iy
1
)
−
(
x
2
+
iy
2
)
=
e
z
1
−
z
2
19.
e
z
+
πi
=
e
x
+(
y
+
π
)
i
=
e
x
[cos(
y
+
π
)+
i
sin(
y
+
π
)] =
e
x
[cos(
y
−
π
)+
i
sin(
y
−
π
)] =
e
x
+(
y
−
π
)
i
=
e
z
−
πi
20.
(
e
z
)
n
=(
e
x
[cos
y
+
i
sin
y
])
n
=
e
nx
[cos
y
+
i
sin
y
]
n
=
e
nx
[cos
ny
+
i
sin
ny
]=
e
nz
,
n
an integer
21.
u
=
e
x
cos
y
,
v
=
−
e
x
sin
y
;
∂u
∂x
=
e
x
cos
y
,
∂v
∂y
=
−
e
x
cos
y
;
∂u
∂y
=
−
e
x
sin
y
,
−
∂v
∂x
=
e
x
sin
y
Since the Cauchy-Riemann equations are not satisfied at any point,
f
is nowhere analytic.
22. (a)
u
=
e
x
2
−
y
2
cos 2
xy
,
v
=
e
x
2
−
y
2
sin 2
xy
;
∂u
∂x
=
−
2
ye
x
2
−
y
2
sin 2
xy
+2
xe
x
2
−
y
2
cos 2
xy
=
∂v
∂y
;
∂u
∂y
=
−
2
xe
x
2
−
y
2
sin 2
xy
−
2
ye
x
2
−
y
2
cos 2
xy
=
−
∂v
∂x
Since
u
,
v
, and their first partial derivatives are continuous, and
u
and
v
satisfy the Cauchy-Riemann
equations everywhere, the function
f
is differentiable everywhere. Hence
f
is entire.
(b)
∂
2
u
∂x
2
=
−
4
y
2
e
x
2
−
y
2
cos 2
xy
−
4
xye
x
2
−
y
2
sin 2
xy
−
4
xye
x
2
−
y
2
sin 2
xy
+ cos 2
xy
[4
x
2
e
x
2
−
y
2
+2
e
x
2
−
y
2
];
∂
2
u
∂y
2
=
−
4
x
2
e
x
2
−
y
2
cos 2
xy
+4
xye
x
2
−
y
2
sin 2
xy
+4
xye
x
2
−
y
2
sin 2
xy
+ cos 2
xy
[4
y
2
e
x
2
−
y
2
−
2
e
x
2
−
y
2
]
Since
∂
2
u
∂x
2
+
∂
2
u
∂y
2
= 0 the function
u
is harmonic throughout the
z
-plane.
23.
ln(
−
5) = log
e
5+
i
(
π
+2
nπ
)=1
.
16094 + (
π
+2
nπ
)
i
24.
ln(
−
ei
) = log
e
e
+
i
y
−
π
2
+2
nπ
x
=1+
y
−
π
2
+2
nπ
x
i
25.
ln(
−
2+2
i
) = log
e
2
√
2+
i
z
3
π
4
+2
nπ
i
=1
.
0397 +
z
3
π
4
+2
nπ
i
i
26.
ln(1 +
i
) = log
e
√
2+
i
y
π
4
+2
nπ
x
=0
.
3466 +
y
π
4
+2
nπ
x
i
27.
ln(
√
2+
√
6
i
) = log
e
2
√
2+
i
y
π
3
+2
nπ
x
=1
.
0397 +
y
π
3
+2
nπ
x
i
28.
ln(
−
√
3+
i
) = log
e
2+
i
z
5
π
6
+2
nπ
i
=0
.
6932 +
z
5
π
6
+2
nπ
i
i
29.
Ln(6
−
6
i
) = log
e
6
√
2+
i
y
−
π
4
x
=2
.
1383
−
π
4
i
30.
Ln(
−
e
3
) = log
e
e
3
+
πi
=3+
πi
31.
Ln(
−
12 + 5
i
) = log
e
13 +
i
z
tan
−
1
z
−
5
12
i
+
π
i
=2
.
5649 + 2
.
7468
i
770

771. Exercises 17.6
32.
Ln(3
−
4
i
) = log
e
5+
i
tan
−
1
z
−
4
3
i
=1
.
6094
−
0
.
9273
i
33.
Ln(1 +
√
3
i
)
5
= Ln(16
−
16
√
3
i
) = log
e
32
−
π
3
i
=3
.
4657
−
π
3
i
34.
Ln(1 +
i
)
4
= Ln(
−
4) = log
e
4+
πi
=1
.
3863 +
πi
35.
z
= ln(4
i
) = log
e
4+
i
y
π
2
+2
nπ
x
=1
.
3863 +
y
π
2
+2
nπ
x
i
36.
1
z
= ln(
−
1) = log
e
1+
i
(
π
+2
nπ
)=(2
n
+1)
πi
and so
z
=
−
i
(2
n
+1)
π
.
37.
z
−
1 = ln(
−
ie
2
) = log
e
e
2
+
i
z
3
π
2
+2
nπ
i
=2+
z
3
π
2
+2
nπ
i
i
and so
z
=3
z
3
π
2
+2
nπ
i
i
.
38.
By the quadratic formula,
e
z
=
−
1
2
+
√
3
2
i
or
e
z
=
−
1
2
−
√
3
2
i
. Hence
z
=ln

−
1
2
+
√
3
2
i

=
z
2
π
3
+2
nπ
i
i
or
z
=ln

−
1
2
−
√
3
2
i

=
z
4
π
3
+2
nπ
i
i.
39.
(
−
i
)
4
i
=
e
4
i
ln(
−
i
)
=
e
4
i
[log
e
1+
i
(
−
π
2
+2
nπ
)]
=
e
(2
−
8
n
)
π
40.
3
1
/π
=
e
i
π
ln 3
=
e
i
π
[log
e
3+2
nπi
]
=
e
−
2
n

cos
z
1
π
log
e
3
i
+
i
sin
z
1
π
log
e
3
i.
=
e
−
2
n
[0
.
9395 + 0
.
3426
i
]
41.
(1 +
i
)
(1+
i
)
=
e
(1+
i
) ln(1+
i
)
=
e
(1+
i
)[log
e
√
2+
i
(
π
4
+2
nπ
)]
=
e
log
e
√
2
−
(
π
4
+2
nπ
)
t
cos
y
π
4
+ log
e
√
2
x
+
i
sin
y
π
4
+ log
e
√
2
xe
=
e
−
2
nπ
[0
.
2740 + 0
.
5837
i
]
42.
(1
−
i
)
2
i
=
e
2
i
ln(1
−
i
)
=
e
2
i
[log
e
√
2+
i
(
−
π
4
+2
nπ
)]
=
e
π
2
−
4
nπ
[cos(log
e
2) +
i
sin(log
e
2)] =
e
−
4
nπ
[3
.
7004 + 3
.
0737
i
]
43.
(
−
1)
−
2
i
π
=
e
−
2
i
π
Ln(
−
1)
=
e
−
2
i
π
(
πi
)
=
e
2
=7
.
3891
44.
(1
−
i
)
2
i
=
e
2
i
Ln(1
−
i
)
=
e
2
i
[log
e
√
2
−
π
4
i
]
=
e
π
2
[cos(log
e
2) +
i
sin(log
e
2)] = 3
.
7004 + 3
.
0737
i
45.
If
z
1
=
i
and
z
2
=
−
1+
i
then
Ln(
z
1
z
2
) = Ln(
−
1
−
i
) = log
e
√
2
−
3
π
4
i,
whereas
Ln
z
1
+Ln
z
2
=
π
2
i
+
z
log
e
√
2+
3
π
4
i
i
= log
e
√
2+
5
π
4
i.
46.
If
z
1
=
−
i
and
z
2
=
i
then
Ln(
z
1
/z
2
) = Ln(
−
1) =
πi,
whereas Ln
z
1
−
Ln
z
2
=
−
π
2
i
−
π
2
i
=
−
πi.
47. (a)
The statement is false.
Ln(
−
1+
i
)
2
= Ln(
−
2
i
) = log
e
2
−
π
2
i,
whereas 2Ln(
−
1+
i
)=2
z
log
e
√
2+
3
π
4
i
i
= log
e
2+
3
π
2
i.
(b)
The statement is false.
Ln
i
3
= Ln(
−
i
)=
−
π
2
i,
whereas 3Ln
i
=
3
π
2
i.
(c)
The statement is true. If we take arg(
−
i
)=
3
π
2
then ln
i
3
=ln(
−
i
)=
3
π
2
i
for
n
= 0. Also, 3 ln
i
=3
y
π
2
i
x
.
48. (a)
(
i
i
)
2
=(
e
i
ln
i
)
2
=[
e
−
(
π
2
+2
nπ
)
]
2
=
e
−
(
π
+4
nπ
)
and
i
2
i
=
e
2
i
ln
i
=
e
−
(
π
+4
nπ
)
(b)
(
i
2
)
i
=(
−
1)
i
=
e
i
ln(
−
1)
=
e
−
(
π
+2
nπ
)
, whereas
i
2
i
=
e
−
(
π
+4
nπ
)
771

772. Exercises 17.6
49.
Since
|
z
|
=
k
x
2
+
y
2
and Arg
z
= tan
−
1
y
x
for
x>
0wehave
Ln
z
= log
e
|
z
|
+
i
Arg
z
= log
e
(
x
2
+
y
2
)
1
/
2
+
i
tan
−
1
y
x
=
1
2
log
e
(
x
2
+
y
2
)+
i
tan
−
1
y
x
.
50. (a)
u
= log
e
(
x
2
+
y
2
);
∂
2
u
∂x
2
=
2(
y
2
−
x
2
)
(
x
2
+
y
2
)
2
,
∂
2
u
∂y
2
=
2(
x
2
−
y
2
)
(
x
2
+
y
2
)
2
Since
∂
2
u
∂x
2
+
∂
2
u
∂y
2
= 0 the function
u
is harmonic in any domain not containing the point (0
,
0).
(b)
v
= tan
−
1
y
x
;
∂
2
v
∂x
2
=
2
xy
(
x
2
+
y
2
)
2
,
∂
2
v
∂y
2
=
−
2
xy
(
x
2
+
y
2
)
Since
∂
2
v
∂x
2
+
∂
2
v
∂y
2
= 0 the function
v
is harmonic in any domain not containing the point (0
,
0).
Exercises 17.7
1.
cos(3
i
) = cosh 3 = 10
.
0677
2.
sin(
−
2
i
)=
i
sinh(
−
2) =
−
3
.
6269
i
3.
sin
y
π
4
+
i
x
= sin
π
4
cosh(1) +
i
cos
π
4
sinh(1) = 1
.
0911 + 0
.
8310
i
4.
cos(2
−
4
i
) = cos(2) cosh(
−
4)
−
sin(2) sinh(
−
4) =
−
11
.
3642
−
24
.
8147
i
5.
tan(
i
)=
sin(
i
)
cos(
i
)
=
i
sinh(1)
cosh(1)
=0
.
7616
i
6.
cot
y
π
2
+3
i
x
=
cos(
π
2
+3
i
)
sin(
π
2
+3
i
)
=
−
i
sinh(3)
cosh(3)
=
−
0
.
9951
i
7.
sec(
π
+
i
)=
1
cos(
π
+
i
)
=
1
−
cosh(1)
=
−
0
.
6481
8.
csc(1 +
i
)=
1
sin(1 +
i
)
=
1
sin(1) cosh(1) +
i
cos(1) sinh(1)
=0
.
6215
−
0
.
3039
i
9.
cosh(
πi
) = cos(
i
(
πi
)) = cos(
−
π
) = cos
π
=
−
1
10.
sinh
z
3
π
2
i
i
=
−
i
sin
z
i
z
3
π
2
i
ii
=
−
i
sin
z
−
3
π
2
i
=
i
sin
3
π
2
=
−
i
11.
sinh
y
1+
π
3
i
x
= sinh(1) cos
π
3
+
i
cosh(1) sin
π
3
=0
.
5876 + 1
.
3363
i
12.
cosh(2 + 3
i
) = cosh(2) cos(3) +
i
sinh(2) sin(3) =
−
3
.
7245 + 0
.
5118
i
13.
sin
y
π
2
+
i
ln 2
x
= sin
π
2
cosh(ln 2) +
i
cos
π
2
sinh(ln 2) =
e
ln 2
+
e
ln 2
−
1
2
=
2+
1
2
2
=
5
4
14.
cos
y
π
2
+
i
ln 2
x
= cos
π
2
cosh(ln 2)
−
i
sin
π
2
sinh(ln 2) =
−
i
·
e
ln 2
−
e
ln 2
−
1
2
=
−
i
·
2
−
1
2
2
=
−
3
4
i
15.
e
iz
−
e
−
iz
2
i
= 2 gives
e
2(
iz
)
−
4
ie
iz
−
1 = 0. By the quadratic formula,
e
iz
=2
i
±
√
3
i
and so
iz
= ln[(2
±
√
3)
i
]
z
=
−
i
t
log
e
(2
±
√
3)+
y
π
2
+2
nπ
x
i
e
=
π
2
+2
nπ
−
i
log
e
(2
±
√
3)
,n
=0
,
±
1
.,
±
2
,....
772

773. Exercises 17.7
16.
e
iz
+
e
−
iz
2
=
−
3
i
gives
e
2(
iz
)
+6
ie
iz
+ 1 = 0. By the quadratic formula,
e
iz
=
−
3
i
±
√
10
i
and so
iz
= ln[
−
3
±
√
10 )
i
]. Hence
z
=
−
i
t
log
e
(
√
10
−
3) +
y
π
2
+2
nπ
x
i
e
z
=
π
2
+2
nπ
−
i
log
e
(
√
10
−
3)
or
z
=
−
i

log
e
(
√
10+3)+
z
3
π
2
+2
nπ
i
i

or
z
=
3
π
2
+2
nπ
−
i
log
e
(
√
10+3)
n
=0,
±
1,
±
2,
...
.
17.
e
z
−
e
−
z
2
=
i
gives
e
2
z
−
2
ie
z
−
1 = 0. By the quadratic formula,
e
z
=
−
i
and so
z
= ln(
−
i
) = log
e
1+
y
−
π
2
+2
nπ
x
i
=
y
−
π
2
+2
nπ
x
i, n
=0
,
±
1
,
±
2
,....
18.
e
z
−
e
−
z
2
=
−
1 gives
e
2
z
+2
e
z
−
1 = 0. By the quadratic formula,
e
z
=
−
1
±
√
2 , and so
z
= ln(
−
1
±
√
2)
z
= log
e
(
√
2
−
1)+2
nπi
or
z
= log
e
(
√
2+1)+(
π
+2
nπ
)
i,
n
=0,
±
1,
±
2,
...
.
19.
cos
z
= sin
z
gives tan
z
= 1. One solution is
z
=
π
4
. Since tan
z
is
π
-periodic,
z
=
π
4
+
nπ
,
n
=0,
±
1,
±
2,
...
are also solutions. That these are the only solutions can be proved by solving
e
iz
+
e
−
iz
2
=
e
iz
−
e
−
iz
2
i
by the method illustrated in Problems 15-18.
20.
cos
z
=
i
sin
z
gives
e
iz
+
e
−
iz
=
e
iz
−
e
−
iz
or
e
−
iz
= 0. Since this last equation has no solutions, the original
equation has no solutions.
21.
cos
z
= cosh 2 implies cos
x
cosh
y
−
i
sin
x
sinh
y
= cosh 2 + 0
i
and so we must have cos
x
cosh
y
= cosh 2 and
sin
x
sinh
y
= 0. The last equation has solutions
x
=
nπ
,
n
=0,
±
1,
±
2,
...
,or
y
=0. For
y
= 0 the first
equation becomes cos
x
= cosh 2. Since cosh 2
>
1 this equation has no solutions. For
x
=
nπ
the first equation
becomes (
−
1)
n
cosh
y
= cosh 2. Since cosh
y>
0wesee
n
must be even, say,
n
=2
k
,
k
=0,
±
1,
±
2,
...
.Now
cosh
y
= cosh 2 implies
y
=
±
2. Solutions of the original equation are then
z
=2
kπ
±
2
i, k
=0
,
±
1
,
±
2
,....
22.
sin
z
=
i
sinh 2 implies sin
x
cosh
y
+
i
cos
x
sinh
y
=0+
i
sinh 2 and so we must have sin
x
cosh
y
= 0 and
cos
x
sinh
y
= sinh 2. Since cosh
y>
0 for all real numbers, the first equation has only the solutions
x
=
nπ
,
n
=0,
±
1,
±
2,
...
.For
x
=
nπ
the second equation becomes (
−
1)
n
sinh
y
= sinh 2. If
n
is even, sinh
y
= sinh 2
implies
y
= 2 (sinh
y
is one-to-one.) If
n
is odd,
−
sinh
y
= sinh 2 implies sinh
y
=
−
sinh(
−
2) and so
y
=
−
2.
Solutions of the original equation are then
z
=2
kπ
+2
i, z
=(2
k
+1)
π
−
2
i, k
=0
,
±
1
,
±
2
,....
23.
cos
z
=
e
i
(
x
+
iy
)
+
e
−
i
(
x
+
i
)
2
=
1
2
(
e
−
y
e
ix
+
e
y
e
−
ix
)=
1
2
[
e
−
y
(cos
x
+
i
sin
x
)+
e
y
(cos
x
−
i
sin
x
)]
= cos
x
z
e
y
+
e
−
y
2
i
−
i
sin
x
z
e
y
−
e
−
y
2
i
= cos
x
cosh
y
−
i
sin
x
sinh
y
773

774. Exercises 17.7
24.
sinh
z
=
e
x
+
iy
−
e
−
x
−
iy
2
=
1
2
(
e
x
e
iy
−
e
−
x
e
−
iy
)=
1
2
[
e
x
(cos
y
+
i
sin
y
)
−
e
−
x
(cos
y
−
i
sin
y
)]
=
z
e
x
−
e
−
x
2
i
cos
y
+
i
z
e
x
+
e
−
x
2
i
sin
y
= sinh
x
cos
y
+
i
cosh
x
sin
y
25.
cosh
z
=
e
x
+
iy
+
e
−
x
−
iy
2
=
1
2
(
e
x
e
iy
+
e
−
x
e
−
iy
)=
1
2
[
e
x
(cos
y
+
i
sin
y
)+
e
−
x
(cos
y
−
i
sin
y
)]
=
z
e
x
+
e
−
x
2
i
cos
y
+
i
z
e
x
−
e
−
x
2
i
sin
y
= cosh
x
cos
y
+
i
sinh
x
sin
y
26.
|
sinh
z
|
2
= sinh
2
x
cos
2
y
+ cosh
2
x
sin
2
y
= sinh
2
x
cos
2
y
+ (1 + sinh
2
x
) sin
2
y
= sinh
2
x
(cos
2
y
+ sin
2
y
) + sin
2
y
= sinh
2
x
+ sin
2
y
27.
|
cosh
z
|
2
= cosh
2
x
cos
2
y
+ sinh
2
x
sin
2
y
= (1 + sinh
2
x
) cos
2
y
+ sinh
2
x
sin
2
y
= cos
2
y
+ sinh
2
x
(cos
2
y
+ sin
2
y
) = cos
2
y
+ sinh
2
x
28.
cos
2
z
+ sin
2
z
=
z
e
1
z
+
e
−
iz
2
i
2
+
z
e
iz
−
e
−
iz
2
i
i
2
=
1
4
[
e
2
iz
+2+
e
−
2
iz
−
(
e
2
iz
−
2+
e
−
2
iz
)] =
4
4
=1
29.
cosh
2
z
−
sinh
2
z
=
z
e
z
+
e
−
z
2
i
−
z
e
z
−
e
−
z
2
i
2
=
1
4
[
e
2
z
+2+
e
−
2
z
−
(
e
2
z
−
2+
e
−
2
z
)] =
4
4
=1
30.
tan
z
=
sin
z
cos
z
=
sin
z
cos
z
|
cos
z
|
2
=
[sin
x
cosh
y
+
i
cos
x
sinh
y
][cos
x
cosh
y
+
i
sin
x
sinh
y
]
cos
2
x
+ sinh
2
y
=
(sin
x
cos
x
cosh
2
y
−
sin
x
cos
x
sinh
2
y
)
cos
2
x
+ sinh
2
y
+
i
cos
2
x
sinh
y
cosh
y
+ sin
2
x
sinh
y
cosh
y
cos
2
x
+ sinh
2
y
=
sin
x
cos
x
(cosh
2
y
−
sinh
2
y
)
cos
2
x
+ sinh
2
y
+
i
sin
y
cosh
y
(cos
2
x
+ sin
2
x
)
cos
2
x
+ sinh
2
y
=
sin
x
cos
x
cos
2
x
+ sinh
2
y
+
i
sinh
y
cosh
y
cos
2
x
+ sinh
2
y
=
sin 2
x
2(cos
2
x
+ sinh
2
y
)
+
i
sinh 2
y
2(cos
2
x
+ sinh
2
y
)
But
2 cos
2
x
+ 2 sinh
2
y
= (2 cos
2
x
−
1) + (2 sinh
2
y
+ 1) = cos 2
x
+ cosh 2
y.
Therefore tan
z
=
u
+
iz
where
u
=
sin 2
x
cos 2
x
+ cosh 2
y
,v
=
sinh 2
y
cos 2
x
+ cosh 2
y
.
31.
tanh(
z
+
πi
)=
sinh(
x
+(
y
+
π
)
i
)
cosh(
x
+(
y
+
π
)
i
)
=
sinh
x
cos(
y
+
π
)+
i
cosh
x
sin(
y
+
π
)
cosh
x
cos(
y
+
π
)+
i
sinh
x
sin(
y
+
π
)
=
−
[sinh
x
cos
y
+
i
cosh
x
sinh
y
]
−
[cosh
x
cos
y
+
i
sinh
x
sin
y
]
=
−
sinh
z
−
cosh
z
= tanh
z
32. (a)
sin
z
= sin
x
cosh
y
−
i
cos
x
sinh
y
= sin
x
cosh(
−
y
)+
i
cos
x
sinh(
−
y
) = sin(
x
−
iy
) = sin ¯
z
(b)
cos
z
= cos
x
cosh
y
+
i
sin
x
sinh
y
= cos
x
cosh(
−
y
)
−
i
sin
x
sinh(
−
y
) = cos(
x
−
iy
) = cos ¯
z
774

775. Exercises 17.8
Exercises 17.8
1.
sin
−
1
(
−
i
)=
−
i
ln(1
±
√
2)=
)
2
nπ
−
i
log
e
(1 +
√
2)
(2
n
+1)
π
−
i
log
e
(
√
2
−
1)
Since
√
2
−
1=1
/
(
√
2 + 1) we can have sin
−
1
(
−
i
)=
)
2
nπ
−
i
log
e
(1 +
√
2)
(2
n
+1)
π
+
i
log
e
(1 +
√
2)
.
This can be written compactly as
sin
−
1
(
−
i
)=
nπ
+(
−
1)
n
+1
i
log
e
(1 +
√
2)
,k
=0
,
±
1
,
±
2
,....
2.
sin
−
1
√
2=
−
i
ln[
√
2
±
1)
i
]=2
nπ
+
π
2
−
i
log
e
(
√
2
±
1) = 2
nπ
+
π
2
±
i
log
e
(1 +
√
2),
n
=0,
±
1,
±
2,
...
3.
sin
−
1
0=
−
i
ln(
±
1) =
)
2
nπ
+
i
log
e
1
(2
n
+1)
π
+
i
log
e
1
=
)
2
nπ
(2
n
+1)
π
=
nπ
,
n
=0,
±
1,
±
2,
...
4.
sin
−
1
13
5
=
−
i
ln
πz
13
5
±
12
5
i
i

=
)
2
nπ
+
π
2
−
i
log
e
5
2
nπ
+
π
2
−
i
log
e
1
5
=2
nπ
+
π
2
±
i
log
e
5,
n
=0,
±
1,
±
2,
...
5.
cos
−
1
2=
−
i
ln(2
±
√
3)=
)
2
nπ
−
i
log
e
(2 +
√
3)
2
nπ
−
i
log
e
(2
−
√
3)
Since 2
−
√
3=1
/
(2 +
√
3 ) this can be written compactly as
cos
−
1
2=2
nπ
±
i
log
e
(2 +
√
3)
,k
=0
,
±
1
,
±
2
,....
6.
cos
−
1
2
i
=
−
i
ln[(2
±
√
5)
i
]=
b
2
nπ
−
π
2
+
i
log
e
(2 +
√
5)
2
nπ
+
π
2
−
i
log
e
(2 +
√
5)
,
n
=0,
±
1,
±
2,
...
7.
cos
−
1
1
2
=
−
i
ln

1
2
±
√
3
2
i

=
)
2
nπ
+
π
3
−
i
log
e
1
2
nπ
+
π
3
−
i
log
e
1
=2
nπ
±
π
2
,
n
=0,
±
1,
±
2,
...
8.
cos
−
1
5
3
=
−
i
ln
z
5
3
±
4
3
i
=
)
2
nπ
−
i
log
e
3
2
nπ
−
i
log
e
1
3
=2
nπ
±
i
log
e
3,
n
=0,
±
1,
±
2,
...
9.
tan
−
1
1=
i
2
ln
i
+1
i
−
1
=
i
2
ln(
−
i
)=
−
nπ
+
π
4
+
i
2
log
e
1=
π
4
−
nπ
,
n
=0,
±
1,
±
2,
...
Note that this can also be written as tan
−
1
1=
π
4
+
nπ
,
n
=0,
±
1,
±
2,
...
.
10.
tan
−
1
3
i
=
i
2
ln
z
4
i
−
2
i
i
=
i
2
ln(
−
2) =
−
π
2
−
nπ
+
i
log
e
√
2,
n
=0,
±
1,
±
2,
...
11.
sinh
−
1
4
3
=ln
z
4
3
±
5
3
i
=
)
log
e
3+2
nπi
log
e
1
3
+(2
n
+1)
πi
=(
−
1)
n
log
e
3+
nπi
,
n
=0,
±
1,
±
2,
...
12.
cosh
−
1
i
= ln[(1 +
±
√
2)
i
]=
b
log
e
(1 +
√
2)+(
π
2
+2
nπ
)
i
log
e
(
√
2
−
1)+(
−
π
2
+2
nπ
)
i
,
n
=0,
±
1,
±
2,
...
13.
tanh
−
1
(1+2
i
)=
1
2
ln
2+2
i
−
2
i
=
1
2
ln(
−
1+
i
)=
1
2

log
e
√
2+
z
3
π
4
+2
nπ
i
i

=
1
4
log
e
2+
z
3
π
8
+
nπ
i
i
14.
tanh
−
1
(
−
√
3
i
)=
1
2
ln
1
−
√
3
i
1+
√
3
i
=
1
2
ln

−
1
2
−
√
3
2
i

=
1
2

log
e
1+
z
4
π
3
+2
nπ
i
i

=
z
2
π
3
+
nπ
i
i
,
n
=0,
±
1,
±
2,
...
775

776. Chapter 17 Review Exercises
Chapter 17 Review Exercises
1.
0; 32
2.
third
3.
−
7
/
25
4.
−
8
i
5.
4
/
5
6.
The closed annular region between the circles
|
z
+2
|
= 1 and
|
z
+2
|
= 3. These circles have center at
z
=
−
2.
7.
False. Arg[(
−
1+
i
)+(
−
1
−
i
)] = Arg(
−
2) =
π
8.
−
5
π/
6
9.
z
= ln(2
i
) = log
e
2+
i
y
π
2
+2
nπ
x
,
n
=0,
±
1,
±
2,
...
10.
True
11.
(1 +
i
)
2+
i
=
e
(2+
i
)[log
e
2+
π
4
i
]
=
e
(log
e
2
−
π
4
)+
i
(log
e
√
2+
π
2
)
=
e
log
e
2
−
π
4
t
cos
y
log
e
√
2+
π
2
x
+
i
sin
y
log
e
√
2+
π
2
xe
=
−
0
.
3097 + 0
.
8577
i
12.
f
(
−
1+
i
)=
−
33+26
i
13.
False
14.
2
πi
15.
Ln(
−
ie
3
) = log
e
e
3
+
y
−
π
2
x
i
=3
−
π
2
i
16.
True
17.
58
−
4
i
18.
−
1
13
−
17
13
i
19.
−
8+8
i
20.
4
e
πi/
12
=4
y
cos
π
12
+
i
sin
π
12
x
=3
.
8637 + 1
.
0353
i
21.
The region satisfying
xy
≤
1 is shown in the figure.
22.
The region satisfying
y
+5
>
3or
y>
−
2 is shown in the figure.
23.
The region satisfying
|
z
|≥
1 is shown in the figure.
24.
The region satisfying
y<x
is shown in the figure.
25.
Ellipse with foci (0
,
−
2) and (0
,
2)
776

777. Chapter 17 Review Exercises
26.
c
c
c
c
z
−
w
1
−
z
¯
w
c
c
c
c
2
=
z
−
w
1
−
z
¯
w
·
¯
z
−
¯
w
1
−
¯
zw
=
z
¯
z
−
z
¯
w
−
w
¯
z
+
w
¯
w
1
−
¯
zw
−
z
¯
w
+
z
¯
zw
¯
w
=
1
−
z
¯
w
−
w
¯
z
+
|
w
|
2
1
−
¯
zw
−
z
¯
w
+
|
w
|
2
=1,
since
|
z
|
2
=
z
¯
z
= 1 and
|
w
|
t
=1.
27.
The four fourth roots of 1
−
i
are given by
w
R
=2
1
/
8

cos
z
−
π
16
+
kπ
2
i
+
i
sin
z
−
π
16
+
kπ
2
i.
,n
=0
,
1
,
2
,
3
w
0
=2
1
/
8
t
cos
y
−
π
16
x
+
i
sin
y
−
π
16
xe
=1
.
0696
−
0
.
2127
i
w
1
=2
1
/
8

cos
7
π
16
+
i
sin
7
π
16

=0
.
2127 + 1
.
0696
i
w
2
=2
1
/
8

cos
15
π
16
+
i
sin
15
π
16

=
−
1
.
0696 + 0
.
2127
i
w
3
=2
1
/
8

cos
23
π
16
+
i
sin
23
π
16

=
−
0
.
2127
−
1
.
0696
i
28.
z
3
/
2
=
2
5
+
1
5
i
implies
z
3
=
3
25
+
4
25
i
. The three cube roots of
3
25
+
4
25
i
are
w
k
=
z
1
5
i
1
/
3

cos
z
1
3
tan
−
1
z
4
3
i
+
2
kπ
3
i
+
i
sin
z
1
3
tan
−
1
z
4
3
i
+
2
kπ
3
i.
,k
=0
,
1
,
2
w
0
=
z
1
5
i
1
/
3
[cos(0
.
3091) +
i
sin(0
.
3091)] = 0
.
5571 + 0
.
1779
i
w
1
=
z
1
5
i
1
/
3
[cos(2
.
4035) +
i
sin(2
.
4035)] =
−
0
.
4326 + 0
.
3935
i
w
2
=
z
1
5
i
1
/
3
[cos(4
.
4979) +
i
sin(4
.
4979)] =
−
0
.
1245
−
0
.
5714
i.
29.
Write
1+
i
√
2
=
e
πi/
4
so that
z
24
=
e
6
πi
=1
,z
20
=
e
5
πi
=
−
1
,z
12
=
e
3
πi
=
−
1
,z
6
=
e
3
πi/
2
=
−
i.
Therefore
f
z
1+
i
√
2
i
=1
−
3(
−
1)+4(
−
1)
−
5(
−
i
)=5
i.
30.
Im(
z
−
3¯
z
)=4
y
,
z
Re(
z
2
)=(
x
3
−
xy
2
)+
i
(
x
2
y
−
y
3
). Thus,
f
(
z
)=(4
y
+
x
3
−
xy
2
−
5
x
)+
i
(
x
2
y
−
y
3
−
5
y
)
.
31.
u
=
x
2
−
y
,
v
=
y
2
−
x
. When
x
= 1 we get the parametric equations
u
=1
−
y
,
v
=
y
2
−
1. Eliminating
y
then
gives
v
=(1
−
u
)
2
−
1=
u
2
−
2
u
. This is an equation of a parabola.
32.
u
=
x/
(
x
2
+
y
2
),
v
=
−
y/
(
x
2
+
y
2
). When
x
= 1 we get the parametric equations
u
=1
/
(1+
y
2
),
v
=
−
y/
(1+
y
2
).
From this we find
u
2
+
v
2
−
u
= 0. This describes a circle with the exception that (0
,
0) is not on its circumference.
33.
z
=
z
−
1
gives
z
2
= 1 or (
z
−
1)(
z
+ 1) = 0. Thus
z
=
±
1.
34.
¯
z
=1
/z
gives
z
¯
z
=1or
|
z
|
2
= 1. All points on the circle
|
z
|
= 1 satisfy the equation.
35.
¯
z
=
−
z
gives
x
=
−
x
or
x
= 0. All complex numbers of the form
z
=0+
iy
(pure imaginaries) satisfy the
equation.
777

778. Chapter 17 Review Exercises
36.
z
2
=¯
z
2
gives
xy
=
−
xy
or
xy
= 0. This implies
x
=0or
y
= 0. All real numbers (
y
= 0) and all pure imaginary
numbers (
x
= 0) satisfy the equation.
37.
u
=
−
2
xy
−
5
x
,
v
=
x
2
−
5
y
−
y
2
;
∂u
∂x
=
−
2
y
−
5=
∂v
∂y
,
∂u
∂y
=
−
2
x
=
−
∂v
∂x
;
f
i
(
z
)=
∂u
∂x
+
i
∂v
∂x
=
−
2
y
−
5+2
xi
38.
u
=
x
3
+
xy
2
−
4
x
,
v
=4
y
−
y
3
−
x
2
y
;
∂u
∂x
=3
x
2
+
y
2
−
4,
∂v
∂y
=4
−
3
y
2
−
x
2
,
∂u
∂y
=2
xy
=
−
∂v
∂x
The Cauchy-Riemann equations are satisfied at all points on the circle
x
2
+
y
2
= 2. Continuity of
u
,
v
, and the
first partial derivatives guarantee
f
is differentiable on the circle. However,
f
is nowhere analytic.
39.
Ln(1 +
i
)(1
−
i
) = Ln(2) = log
e
2; Ln(1 +
i
) = log
e
√
2+
π
4
i
; Ln(1
−
i
) = log
e
√
2
−
π
4
i
. Therefore
Ln(1 +
i
) + Ln(1
−
i
) = 2 log
e
√
2 = log
e
2=Ln(1+
i
)(1
−
i
)
.
40.
Ln
1+
i
1
−
i
=Ln
i
= log
e
1+
π
2
i
=
π
2
i
;Ln(1+
i
) = log
e
√
2+
π
4
i
; Ln(1
−
i
) = log
e
√
2
−
π
4
i
. Therefore
Ln(1 +
i
)
−
Ln(1
−
i
)=
π
4
i
−
y
−
π
4
i
x
=
π
2
i
=Ln
1+
i
1
−
i
.
778

779. 18
Integration in the Complex Plane
Exercises 18.1
1.
i
C
(
z
+3)
dz
=(2+4
i
)
t
i
3
1
(2
t
+3)
dt
+
i
i
3
1
(4
t
−
1)
dt
π
=(2+4
i
)[14 + 14
i
]=
−
28+84
i
2.
i
C
(2¯
z
−
z
)
dz
=
i
2
0
[
−
t
−
3(
t
2
+2)
i
](
−
1+2
ti
)
dt
=
i
2
0
(6
t
3
+13
t
)
dt
+
i
i
2
0
(
t
2
+2)
dt
=50+
20
3
i
3.
i
C
z
2
dz
=(3+2
i
)
3
i
2
−
2
t
2
dt
=
16
3
(3+2
i
)
3
=
−
48 +
736
3
i
4.
i
C
(3
z
2
−
2
z
)
dz
=
i
1
0
(
−
15
t
4
+4
t
3
+3
t
2
−
2
t
)
dt
+
i
i
1
0
(
−
6
t
5
+12
t
3
−
6
t
2
)
dt
=
−
2+0
i
=
−
2
5.
Using
z
=
e
it
,
−
π/
2
≤
t
≤
π/
2, and
dz
=
ie
it
dt
,
i
C
1+
z
z
dz
=
−
i
π/
2
−
π/
2
(1 +
e
it
)
dt
=(2+
π
)
i
.
6.
i
C
|
z
|
2
dz
=
i
2
1
/
2
t
5
+
2
t
z
dt
−
i
i
2
1
/
t
2
+
1
t
4
z
dt
=21+ln4
−
21
8
i
7.
Using
z
=
e
it
= cos
t
+
i
sin
t
,
dz
=(
−
sin
t
+
i
cos
t
)
dt
and
x
= cos
t
,
x
ˇ
C
Re(
z
)
dz
=
i
2
π
0
cos
t
(
−
sin
t
+
i
cos
t
)
dt
=
−
i
2
π
0
sin
t
cos
tdt
+
i
i
2
π
0
cos
2
tdt
=
−
1
2
i
2
π
0
sin 2
tdt
+
1
2
i
i
2
π
0
(1 + cos 2
t
)
dt
=
πi.
8.
Using
z
+
i
=
e
it
,0
≤
t
≤
2
π
, and
dz
=
ie
it
dt
,
x
ˇ
C
t
1
(
z
+
i
)
3
−
5
z
+
i
+8
π
dz
=
i
i
2
π
0
[
e
−
2
it
−
5+8
e
it
]
dt
=
−
10
πi.
9.
Using
y
=
−
x
+1,0
≤
x
≤
1,
z
=
x
+(
−
x
+1)
i
,
dz
=(1
−
i
)
dx
,
i
C
(
x
2
+
iy
3
)
dz
=(1
−
i
)
i
0
1
[
x
2
+(1
−
x
)
3
i
]
dx
=
−
7
12
+
1
12
i.
10.
Using
z
=
e
it
,
π
≤
t
≤
2
π
,
dz
=
ie
it
dt
,
x
= cos
t
=(
e
it
+
e
−
it
)
/
2,
y
= sin
t
=(
e
it
−
e
−
it
)
/
2
i
,
i
C
(
x
3
−
iy
3
)
dz
=
1
8
i
i
2
π
π
(
e
3
it
+3
e
it
+3
e
−
it
+
e
−
3
it
)
e
it
dt
+
1
8
i
i
2
π
π
(
e
3
it
−
3
e
it
+3
e
−
it
−
e
−
3
it
)
e
it
dt
=
1
8
i
i
2
π
π
(2
e
4
it
+6)
dt
=
3
π
4
i.
11.
i
C
e
z
dz
=
i
C
1
e
z
dz
+
i
C
2
e
z
dz
where
C
1
and
C
2
arethelinesegments
y
=0,0
≤
x
≤
2 and
y
=
−
πx
+2
π
,
1
≤
x
≤
2, respectively. Now
i
C
1
e
z
dz
=
i
2
0
e
x
dx
=
e
2
−
1
i
C
2
e
z
dz
=(1
−
πi
)
i
1
2
e
x
+(
−
πx
+2
π
)
i
dx
=(1
−
πi
)
i
1
2
e
(1
−
πi
)
x
dx
=
e
1
−
πi
−
e
2(1
−
πi
)
=
−
e
−
e
2
.
779

780. Exercises 18.1
In the second integral we have used the fact that
e
z
has period 2
πi
.Thus
i
C
e
z
dz
=(
e
2
−
1)+(
−
e
−
e
2
)=
−
1
−
e.
12.
i
C
sin
zdz
=
i
C
1
sin
zdz
+
i
C
2
sin
zdz
where
C
1
and
C
2
arethelinesegments
y
=0,0
≤
x
≤
1, and
x
=1,
0
≤
y
≤
1, respectively. Now
i
C
1
sin
zdz
=
i
1
0
sin
xdx
=1
−
cos 1
i
C
2
sin
zdz
=
i
i
1
0
sin(1 +
iy
)
dy
= cos 1
−
cos(1 +
i
)
.
Thus
i
C
sin
zdz
=(1
−
cos 1)+(cos 1
−
cos(1+
i
)) = 1
−
cos(1+
i
)=(1
−
cos 1 cosh 1)+
i
sin 1 sinh 1 = 0
.
1663+0
.
9889
i.
13.
Wehave
i
C
Im(
z
−
i
)
dz
=
i
C
1
(
y
−
1)
dz
+
i
C
2
(
y
−
1)
dz
On
C
1
,
z
=
e
it
,0
≤
t
≤
π/
2,
dz
=
ie
it
dt
,
y
= sin
t
=(
e
it
−
e
−
it
)
/
2
i
,
i
C
1
=(
y
−
1)
dz
=
1
2
i
π/
2
0
[
e
it
−
e
−
it
−
2
i
]
e
it
dt
=
1
2
i
π/
2
0
[
e
2
it
−
1+2
ie
it
]
dt
=1
−
π
4
−
1
2
i.
On
C
2
,
y
=
x
+1,
−
1
≤
x
≤
0,
z
=
x
+(
x
+1)
i
,
dz
=(1+
i
)
dx
,
i
C
2
(
y
−
1)
dz
=(1+
i
)
i
−
1
0
xdx
=
1
2
+
1
2
i.
Thus
i
C
Im(
z
−
i
)
dz
=
/
1
−
π
4
−
1
2
i
z
+
/
1
2
+
1
2
i
z
=
3
2
−
π
4
.
14.
Using
x
= 6 cos
t
,
y
= 2 sin
t
,
π/
2
≤
t
≤
3
π/
2,
z
= 6 cos
t
+2
i
sin
t
,
dz
=(
−
6 sin
t
+2
i
cos
t
)
dt
,
i
C
dz
=
−
6
i
3
π/
2
π/
2
sin
tdt
+2
i
i
3
π/
2
π/
2
cos
tdt
=2
i
(
−
2) =
−
4
i.
15.
Wehave
x
ˇ
C
ze
z
dz
=
i
C
1
ze
z
dz
+
i
C
2
ze
z
dz
+
i
C
3
ze
z
dz
+
i
C
4
ze
z
dz
On
C
1
,
y
=0,0
≤
x
≤
1,
z
=
x
,
dz
=
dx
,
i
C
1
ze
z
dz
=
i
1
0
xe
x
dx
=
xe
x
−
e
x
y
y
y
1
0
=1
.
On
C
2
,
x
=1,0
≤
y
≤
1,
z
=1+
iy
,
dz
=
idy
,
i
C
2
ze
z
dz
=
i
i
1
0
(1 +
iy
)
e
1+
iy
dy
=
ie
i
+1
.
On
C
3
,
y
=1,0
≤
x
≤
1,
z
=
x
+
i
,
dz
=
dx
,
i
C
3
ze
z
dz
=
i
0
1
(
x
+
i
)
e
x
+
i
dx
=(
i
−
1)
e
i
−
ie
1+
i
.
780

781. Exercises 18.1
On
C
4
,
x
=0,0
≤
y
≤
1,
z
=
iy
,
dz
=
idy
,
i
C
4
ze
z
dz
=
−
i
0
1
ye
iy
dy
=(1
−
i
)
e
i
−
1
.
Thus
x
ˇ
C
ze
z
dz
=1+
ie
i
+1
+(
i
−
1)
e
i
−
ie
1+
i
+(1
−
i
)
e
i
−
1=0
.
16.
Wehave
i
C
f
(
z
)
dz
=
i
C
1
f
(
z
)
dz
+
i
C
2
f
(
z
)
dz
On
C
1
,
y
=
x
2
,
−
1
≤
x
≤
0,
z
=
x
+
ix
2
,
dz
=(1+2
xi
)
dx
,
i
C
1
f
(
z
)
dz
=
i
0
−
1
2(1+2
xi
)
dx
=2
−
2
i.
On
C
2
,
y
=
x
2
,0
≤
x
≤
1,
z
=
x
+
ix
2
,
dz
=(1+2
xi
)
dx
,
i
C
2
f
(
z
)
dz
=
i
1
0
6
x
(1+2
xi
)
dx
=3+4
i.
Thus
i
C
f
(
z
)
dz
=2
−
2
i
+3+4
i
=5+2
i.
17.
Wehave
x
ˇ
C
xdz
=
i
C
1
xdz
+
i
C
2
xdz
+
i
C
3
xdz
On
C
1
,
y
=0,0
≤
x
≤
1,
z
=
x
,
dz
=
dx
,
i
C
1
xdz
=
i
1
0
xdx
=
1
2
.
On
C
2
,
x
=1,0
≤
y
≤
1,
z
=1+
iy
,
dz
=
idy
,
i
C
2
xdz
=
i
i
1
0
dy
=
i.
On
C
3
,
y
=
x
,0
≤
x
≤
1,
z
=
x
+
ix
,
dz
=(1+
i
)
dx
,
i
C
3
xdz
=(1+
i
)
i
0
1
xdx
=
−
1
2
−
1
2
i.
Thus
x
ˇ
C
xdz
=
1
2
+
i
−
1
2
−
1
2
i
=
1
2
i.
18.
Wehave
x
ˇ
C
(2
z
−
1)
dz
=
i
C
1
(2
z
−
1)
dz
+
i
C
2
(2
z
−
1)
dz
+
i
C
3
(2
z
−
1)
dz
On
C
1
,
y
=0,0
≤
x
≤
1,
z
=
x
,
dz
=
dx
,
i
C
1
(2
z
−
1)
dz
=
i
1
0
(2
x
−
1)
dx
=0
.
On
C
2
,
x
=1,0
≤
y
≤
1,
z
=1+
iy
,
dz
=
idy
,
i
C
2
(2
z
−
1)
dz
=
−
2
i
1
0
ydy
+
i
i
1
0
dy
=
−
1+
i.
On
C
3
,
y
=
x
,
z
=
x
+
ix
,
dz
=(1+
i
)
dx
,
i
C
3
(2
z
−
1)
dz
=(1+
i
)
i
0
1
(2
x
−
1+2
ix
)
dx
=1
−
i.
781

782. Exercises 18.1
Thus
x
ˇ
C
(2
z
−
1)
dz
=0
−
1+
i
+1
−
i
=0
.
19.
Wehave
x
ˇ
C
z
2
dz
=
i
C
1
z
2
dz
+
i
C
2
z
2
dz
+
i
C
3
z
2
dz
On
C
1
y
=0,0
≤
x
≤
1,
z
=
x
,
dz
=
dx
,
i
C
1
z
2
dz
=
i
1
0
x
2
dx
=
1
3
.
On
C
2
,
x
=1,0
≤
y
≤
1,
z
=1+
iy
,
dz
=
idy
,
i
C
2
z
2
dz
=
i
1
0
(1 +
iy
)
2
idy
=
−
1+
2
3
i.
On
C
3
,
y
=
x
,0
≤
x
≤
1,
z
=
x
+
ix
,
dz
=(1+
i
)
dx
,
i
C
3
z
2
dz
=(1+
i
)
3
i
0
1
x
2
dx
=
2
3
−
2
3
i.
Thus
x
ˇ
C
z
2
dz
=
1
3
−
1+
2
3
i
+
2
3
−
2
3
i
=0
.
20.
Wehave
x
ˇ
C
¯
z
2
dz
=
i
C
1
¯
z
2
dz
+
i
C
2
¯
z
2
dz
+
i
C
3
¯
z
2
dz
On
C
1
,
y
=0,0
≤
x
≤
1,
z
=
x
,
dz
=
dx
,
i
¯
z
2
dz
=
i
1
0
x
2
dx
=
1
3
.
On
C
2
,
x
=1,0
≤
y
≤
1,
z
=1+
iy
,
dz
=
idy
,
i
C
2
¯
z
2
dz
=
−
i
1
0
(1
−
iy
)
2
(
−
idy
)=1+
2
3
i.
On
C
3
,
y
=
x
,0
≤
x
≤
1,
z
=
x
+
ix
,
dz
=(1+
i
)
dx
,
i
C
3
¯
z
2
dz
=(1
−
i
)
2
(1 +
i
)
i
0
1
x
2
dx
=
−
2
3
+
2
3
i.
Thus
x
ˇ
C
¯
z
2
dz
=
1
3
+1+
2
3
i
−
2
3
+
2
3
i
=
2
3
+
4
3
i.
21.
On
C
,
y
=
−
x
+1,0
≤
x
≤
1,
z
=
x
+(
−
x
+1)
i
,
dz
=(1
−
i
)
dx
,
i
C
(
z
2
−
z
+2)
dz
=(1
−
i
)
i
1
0
[
x
2
−
(1
−
x
)
2
−
x
+2+(3
x
−
2
x
2
−
1)
i
]
dx
=
4
3
−
5
3
i.
22.
Wehave
i
C
(
z
2
−
z
+2)
dz
=
i
C
1
(
z
2
−
z
+2)
dz
+
i
C
2
(
z
2
−
z
+2)
dz
On
C
1
,
y
=1,0
≤
x
≤
1,
z
=
x
+
i
,
dz
=
dx
,
i
C
1
(
z
2
−
z
+2)
dz
=
i
1
0
[(
x
+
i
)
2
−
x
+2
−
i
]
dx
=
5
6
.
On
C
2
,
x
=1,0
≤
y
≤
1,
z
=1+
iy
,
dz
=
idy
,
i
C
2
(
z
2
−
z
+2)
dz
=
i
i
0
1
[(1 +
iy
)
2
+1
−
iy
]
dy
=
1
2
−
5
3
i.
782

783. Exercises 18.1
Thus
i
C
(
z
2
−
z
+2)
dz
=
1
2
−
5
3
i
+
5
6
=
4
3
−
5
3
i.
23.
On
C
,
y
=1
−
x
2
,0
≤
x
≤
1,
z
=
x
+
i
(1
−
x
2
),
dz
=(1
−
2
xi
)
dx
,
i
C
(
z
2
−
z
+2)
dz
=
i
1
0
(
−
5
x
4
+2
x
3
+7
x
2
−
3
x
+1)
dx
+
i
i
1
0
(2
x
5
−
8
x
3
+3
x
2
−
1)
dx
=
4
3
−
5
3
i.
24.
On
C
,
x
= sin
t
,
y
= cos
t
,0
≤
t
≤
π/
2or
z
=
ie
−
it
,
dz
=
e
−
it
dt
,
i
C
(
z
2
−
z
+2)
dz
=
i
π/
2
0
(
−
e
−
2
it
−
ie
−
it
+2)
e
−
it
dt
=
i
π/
2
0
(
−
e
−
3
it
−
ie
−
2
it
+2
e
−
it
)
dt
=
−
1
3
ie
−
3
πi/
2
+
1
2
e
−
πi
+2
ie
−
πi/
2
+
1
3
i
−
1
2
−
2
i
=
4
3
−
5
3
i.
25.
On
C
,
y
y
y
y
e
z
z
2
+1
y
y
y
y
≤
|
e
z
|
|
z
|
2
−
1
=
e
5
24
.Thus
y
y
y
y
x
ˇ
C
e
z
z
2
+1
dz
y
y
y
y
≤
e
5
24
·
10
π
=
5
π
12
e
5
.
26.
On
C
,
y
y
y
y
1
z
2
−
2
i
y
y
y
y
≤
1
|
z
|
2
−|
2
i
|
=
1
34
.Thus
y
y
y
y
i
C
1
z
2
−
2
i
dz
y
y
y
y
≤
1
34
·
1
2
(12
π
)=
3
π
17
.
27.
The length of the line segment from
z
=0to
z
=1+
i
is
√
2 . In addition, on this line segment
|
z
2
+4
|≤|
z
|
2
+4
≤|
1+
i
|
2
+4=6
.
Thus
y
y
y
y
i
C
(
z
2
+4)
dz
y
y
y
y
≤
6
√
2.
28.
On
C
,
y
y
y
y
1
z
3
y
y
y
y
=
1
|
z
|
3
=
1
64
.Thus
y
y
y
y
i
C
1
z
3
dz
y
y
y
y
≤
1
64
·
1
4
(8
π
)=
π
32
.
29. (a)
i
C
dz
= lim
i
P
it
0
n

k
=1
∆
z
k
= lim
i
P
it
0
n

k
=1
(
z
k
−
z
k
−
1
)
= lim
i
P
it
0
[(
z
1
−
z
0
)+(
z
2
−
z
1
)+(
z
3
−
z
2
)+
···
+(
z
n
−
1
−
z
n
−
2
)+(
z
n
−
z
n
−
1
)]
= lim
i
P
it
0
(
z
n
−
z
0
)=
z
n
−
z
0
(b)
With
z
n
=
−
2
i
and
z
0
=2
i
,
i
C
dz
=
−
2
i
−
(2
i
)=
−
4
i
.
30.
With
z
∗
k
=
z
k
,
i
C
zdz
= lim
i
P
it
0
n

k
=1
z
k
(
z
k
−
z
k
−
1
)
= lim
i
P
it
0
[(
z
2
1
−
z
1
z
0
)+(
z
2
2
−
z
2
z
1
)+
···
+(
z
2
n
−
z
n
z
n
−
1
)]
.
(1)
With
z
∗
k
=
z
k
−
1
,
i
C
zdz
= lim
i
P
it
0
n

k
=1
z
k
−
1
(
z
k
−
z
k
−
1
)
= lim
i
P
it
0
[(
z
0
z
1
−
z
2
0
)+(
z
1
z
2
−
z
2
1
)+
···
+(
z
n
−
1
z
n
−
z
2
n
−
1
)]
.
(2)
Adding (1) and (2) gives
2
i
C
zdz
= lim
i
P
it
0
(
z
2
n
−
z
2
0
)or
i
C
zdz
=
1
2
(
z
2
n
−
z
2
0
)
.
783

784. Exercises 18.1
31. (a)
i
C
(6
z
+4)
dz
=6
i
C
zdz
+4
i
C
dz
=
6
2
[(2+3
i
)
2
−
(1 +
i
)
2
] + 4[(2 + 3
i
)
−
(1 +
i
)] =
−
11 + 38
i
(b)
Sincethecontour is closed,
z
0
=
z
n
and so
6
i
C
zdz
+4
i
C
dz
=6[
z
2
0
−
z
2
0
]+4[
z
0
−
z
0
]=0
.
32.
For
f
(
z
)=1
/z
,
f
(
z
)=1
/
¯
z
,soon
z
=2
e
it
,¯
z
=2
e
−
it
,
dz
=2
ie
it
dt
, and
x
ˇ
C
f
(
z
)
dz
=
i
2
π
0
1
2
e
−
it
·
2
ie
it
dt
=
1
2
e
2
it
y
y
y
y
2
π
0
=
1
2
[
e
4
πi
−
1] = 0
.
Thus circulation = Re
/
x
ˇ
C
f
(
z
)
dz
z
= 0, net flux = Im
/
x
ˇ
C
f
(
z
)
dz
z
=0.
33.
For
f
(
z
)=2
z
,
f
(
z
)=2¯
z
,soon
z
=
e
it
,¯
z
=
e
−
it
,
dz
=
ie
it
dt
, and
x
ˇ
C
f
(
z
)
dz
=
i
2
π
0
(
e
−
it
)(
ie
it
dt
)=2
i
i
2
π
0
dt
=4
πi.
Thus circulation = Re
/
x
ˇ
C
f
(
z
)
dz
z
= 0, net flux = Im
/
x
ˇ
C
f
(
z
)
dz
z
=4
π
.
34.
For
f
(
z
)=1
/
(
z
−
1),
f
(
z
)=1
/
(
z
−
1), so on
z
−
1=2
e
it
,
dz
=2
ie
it
dt
, and
x
ˇ
C
f
(
z
)
dz
=
i
2
π
0
1
2
e
it
·
2
ie
it
dt
=
i
i
2
π
0
dt
=2
πi.
Thus circulation = Re
/
x
ˇ
C
f
(
z
)
dz
z
= 0, net flux = Im
/
x
ˇ
C
f
(
z
)
dz
z
=2
π
.
35.
For
f
(
z
)=¯
z
,
f
(
z
)=
z
so on thesquarewehave
x
ˇ
C
f
(
z
)
dz
=
i
C
1
zdz
+
i
C
2
zdz
+
i
C
3
zdz
+
i
C
4
zdz
wheRe
C
1
is
y
=0,0
≤
x
≤
1,
C
2
is
x
=1,0
≤
y
≤
1,
C
3
is
y
=1,0
≤
x
≤
1, and
C
4
is
x
=0,0
≤
y
≤
1. Thus
i
C
1
zdz
=
i
1
0
xdx
=
1
2
i
C
2
zdz
=
i
i
1
0
(1 +
iy
)
dy
=
−
1
2
+
i
i
C
3
zdz
=
i
0
1
(
x
+
i
)
dx
=
−
1
2
−
i
i
C
4
zdz
=
−
i
0
1
ydy
=
1
2
and so
x
ˇ
C
f
(
z
)
dz
=
1
2
+
/
−
1
2
+
i
z
+
/
−
1
2
−
i
z
+
1
2
=0
circulation = Re
/
x
ˇ
C
f
(
z
)
dz
z
= Re(0) = 0
net flux = Im
/
x
ˇ
C
f
(
z
)
dz
z
= Im(0) = 0
.
784

785. Exercises 18.2
Exercises 18.2
1.
f
(
z
)=
z
3
−
1+3
i
is a polynomial and so is an entire function.
2.
z
2
is entire and
1
z
−
4
is analytic within and on thecircle
|
z
|
=1.
3.
f
(
z
)=
z
2
z
+3
is discontinuous at
z
=
−
3
/
2 but is analytic within and on thecircle
|
z
|
=1.
4.
f
(
z
)=
z
−
3
z
2
+2
z
+2
is discontinuous at
z
=
−
1+
i
and at
z
=
−
1
−
i
but is analytic within and on thecircle
|
z
|
=1.
5.
f
(
z
)=
sin
z
(
z
2
−
25)(
z
2
+9)
is discontinuous at
z
=
±
5 and at
z
=
±
3
i
but is analytic within and on thecircle
|
z
|
=1.
6.
f
(
z
)=
e
z
2
z
2
+11
z
+15
is discontinuous at
z
=
−
5
/
2 and at
z
=
−
3 but is analytic within and on thecircle
|
z
|
=1.
7.
f
(
z
) = tan
z
is discontinuous at
z
=
±
π
2
,
±
3
π
2
,
...
but is analytic within and on thecircle
|
z
|
=1.
8.
f
(
z
)=
z
2
−
9
cosh
z
is discontinuous at
π
2
i
,
±
3
π
2
i
,
...
but is analytic within and on thecircle
|
z
|
=1.
9.
By theprincipleof deformation of contours wecan choosethemoreconvenient circular contour
C
1
defined by
|
z
|
=1. Thus
x
ˇ
C
1
z
dz
=
x
ˇ
C
1
1
z
dz
=2
πi
by (4) of Section 18
.
2.
10.
By theprincipleof deformation of contours wecan choosethemoreconvenient circular contour
C
1
defined by
|
z
−
(
−
1
−
i
)
|
=
1
16
.Thus
x
ˇ
C
5
z
+1+
i
dz
=5
x
ˇ
C
1
1
z
−
(
−
1
−
i
)
dz
= 5(2
πi
)=10
πi
by (4) of Section 18
.
2.
11.
By Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
/
z
+
1
z
z
dz
=
x
ˇ
C
zdz
+
x
ˇ
C
1
z
dz
=0+2
πi
=2
πi.
12.
By Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
/
z
+
1
z
2
z
dz
=
x
ˇ
C
1
z
dz
+
x
ˇ
C
1
z
2
dz
=0+0=0
.
13.
Since
f
(
z
)=
z
z
2
−
π
2
is analytic within and on
C
it follows from Theorem 18
.
4 that
x
ˇ
C
z
z
2
−
π
2
dz
=0.
14.
By (4) of Section 18
.
2,
x
ˇ
C
10
(
z
+
i
)
4
dz
=0.
15.
By partial fractions,
x
ˇ
C
2
z
+1
z
(
z
+1)
dz
=
x
ˇ
C
1
z
dz
+
x
ˇ
C
1
z
+1
dz
.
785

786. Exercises 18.2
(a)
By Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
1
z
dz
+
x
ˇ
C
1
z
+1
dz
=2
πi
+0=2
πi.
(b)
By writing
x
ˇ
C
=
x
ˇ
C
1
+
x
ˇ
C
2
where
C
1
and
C
2
arethecircles
|
z
|
=1
/
2 and
|
z
+1
|
=1
/
2, respectively,
wehaveby Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
1
z
dz
+
x
ˇ
C
1
z
+1
dz
=
x
ˇ
C
1
1
z
dz
+
x
ˇ
C
1
1
z
+1
dz
+
x
ˇ
C
2
1
z
dz
+
x
ˇ
C
2
1
z
+1
dz
=2
πi
+0+0+2
πi
=4
πi.
(c)
Since
f
(
z
)=
2
z
+1
z
(
z
+1)
is analytic within and on
C
it follows from Theorem 18
.
4 that
x
ˇ
C
2
z
+1
z
2
+
z
dz
=0
.
16.
By partial fractions,
x
ˇ
C
2
z
z
2
+3
dz
=
x
ˇ
C
1
z
+
√
3
i
dz
+
x
ˇ
C
1
z
−
√
3
i
dz
.
(a)
By Theorem 18
.
4,
x
ˇ
C
1
z
+
√
3
i
dz
+
x
ˇ
C
1
z
−
√
3
i
dz
=0+0=0
.
(b)
By Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
1
z
+
√
3
i
dz
+
x
ˇ
C
1
z
−
√
3
i
dz
=0+2
πi
=2
πi.
(c)
By writing
x
ˇ
C
=
x
ˇ
C
1
+
x
ˇ
C
2
where
C
1
and
C
2
arethecircles
|
z
+
√
3
i
|
=1
/
2 and
|
z
−
√
3
i
|
=1
/
2,
respectively, we have by Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
1
z
+
√
3
i
dz
+
x
ˇ
C
1
z
−
√
3
i
dz
=
x
ˇ
C
1
1
z
+
√
3
i
dz
+
x
ˇ
C
1
1
z
−
√
3
i
dz
+
x
ˇ
C
2
1
z
+
√
3
i
dz
+
x
ˇ
C
2
1
z
−
√
3
i
dz
=2
πi
+0+0+2
πi
=4
πi.
17.
By partial fractions,
x
ˇ
C
−
3
z
+2
z
2
−
8
z
+12
dz
=
x
ˇ
C
1
z
−
2
dz
−
4
x
ˇ
C
1
z
−
6
dz
.
(a)
By Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
1
z
−
2
dz
−
4
x
ˇ
C
1
z
−
6
dz
=0
−
4(2
πi
)=
−
8
πi.
(b)
By writing
x
ˇ
C
=
x
ˇ
C
1
+
x
ˇ
C
2
where
C
1
and
C
2
arethecircles
|
z
−
2
|
= 1 and
|
z
−
6
|
= 1, respectively,
wehaveby Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
1
z
−
2
dz
−
4
x
ˇ
C
1
z
−
6
dz
=
x
ˇ
C
1
1
z
−
2
dz
−
4
x
ˇ
C
1
1
z
−
6
dz
+
x
ˇ
C
2
1
z
−
2
dz
−
4
x
ˇ
C
2
1
z
−
6
dz
=2
πi
−
4(0) + 0
−
4(2
πi
)=
−
6
πi.
786

787. Exercises 18.2
18. (a)
By writing
x
ˇ
C
=
x
ˇ
C
1
+
x
ˇ
C
2
where
C
1
and
C
2
arethecircles
|
z
+2
|
= 1 and
|
z
−
2
i
|
= 1, respectively, we
have by Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
/
3
z
+2
−
1
z
−
2
i
z
dz
=
x
ˇ
C
1
3
z
+2
dz
−
x
ˇ
C
1
1
z
−
2
i
dz
+
x
ˇ
C
2
3
z
+2
dz
−
x
ˇ
C
2
1
z
−
2
i
dz
= 3(2
πi
)
−
0+0
−
2
πi
=4
πi.
19.
By partial fractions,
x
ˇ
C
z
−
1
z
(
z
−
i
)(
z
−
3
i
)
dz
=
1
3
x
ˇ
C
1
z
dz
+
/
−
1
2
+
1
2
i
z
x
ˇ
C
1
z
−
i
dz
+
/
1
6
−
1
2
i
z
x
ˇ
C
1
z
−
3
i
dz.
By Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
z
−
1
z
(
z
−
i
)(
z
−
3
i
)
dz
=0+
/
−
1
2
+
1
2
i
z
2
πi
+0=
π
(
−
1
−
i
)
.
20.
By partial fractions,
x
ˇ
C
1
z
3
+2
iz
2
dz
=
1
4
x
ˇ
C
1
z
dz
−
1
2
i
x
ˇ
C
1
z
2
dz
−
1
4
x
ˇ
C
1
z
+2
i
dz.
By Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
1
z
3
+2
iz
2
dz
=
1
4
2
πi
−
1
2
i
(0)
−
1
4
(0) =
π
2
i.
21.
Wehave
x
ˇ
C
8
z
−
3
z
2
−
z
dz
=
x
ˇ
C
1
8
z
−
3
z
2
−
z
dz
−
x
ˇ
C
2
8
z
−
3
z
2
−
z
dz
where
C
1
and
C
2
aretheclosed portions of thecurve
C
enclosing
z
= 0 and
z
= 1, respectively. By partial
fractions, Theorem 18
.
4, and (4) of Section 18
.
2,
x
ˇ
C
1
8
z
−
3
z
2
−
z
dz
=5
x
ˇ
C
1
1
z
−
1
dz
+3
x
ˇ
C
1
1
z
dz
= 5(0) + 3(2
πi
)=6
πi
x
ˇ
C
1
8
z
−
3
z
2
−
z
dz
=5
x
ˇ
C
2
1
z
−
1
dz
+3
x
ˇ
C
2
1
z
dz
= 5(2
πi
) + 3(0) = 10
πi.
Thus
x
ˇ
C
8
z
−
3
z
2
−
z
dz
=6
πi
−
10
πi
=
−
4
πi.
22.
By choosing themoreconvenient contour
C
1
defined by
|
z
−
z
0
|
=
r
where
r
is small enough so that the circle
C
1
lies entirely within
C
wecan write
x
ˇ
C
1
(
z
−
z
0
)
n
dz
=
x
ˇ
C
1
1
(
z
−
z
0
)
n
dz.
Let
z
−
z
0
=
re
it
,0
≤
t
≤
2
π
and
dz
=
ire
it
dt
. Then for
n
=1:
x
ˇ
C
1
1
z
−
z
0
dz
=
i
2
π
0
1
re
it
ire
it
dt
=
i
i
2
π
0
dt
=2
πi.
For
n
x
=1:
x
ˇ
C
1
1
(
z
−
z
0
)
n
dz
=
i
r
n
−
1
i
2
π
0
e
(1
−
n
)
it
dt
=
i
r
n
−
1
e
(1
−
n
)
it
i
(1
−
n
)
y
y
y
y
2
π
0
=
1
r
n
−
1
(1
−
n
)
[
e
2
π
(1
−
n
)
i
−
1] = 0
since
e
2
π
(1
−
n
)
i
=1.
787

788. Exercises 18.2
23.
Write
x
ˇ
C
/
e
z
z
+3
−
3¯
z
z
dz
=
x
ˇ
C
e
z
z
+3
dz
−
3
x
ˇ
C
¯
zdz
.
By Theorem 18
.
4,
x
ˇ
C
e
z
z
+3
dz
= 0. However, since ¯
z
is not analytic,
x
ˇ
C
¯
zdz
=
i
2
π
0
e
−
it
(
ie
it
dt
)=2
πi.
Thus
x
ˇ
C
/
e
z
z
+3
−
3¯
z
z
dz
=0
−
3(2
πi
)=
−
6
πi.
24.
Write
x
ˇ
C
(
z
2
+
z
+Re(
z
))
dz
=
x
ˇ
C
(
z
2
+
z
)
dz
+
x
ˇ
C
Re(
z
)
dz
.
By Theorem 18
.
4,
x
ˇ
C
(
z
2
+
z
)
dz
= 0. However, since Re(
z
)=
x
is not analytic,
x
ˇ
C
xdz
=
x
ˇ
C
1
xdz
+
x
ˇ
C
2
xdz
+
x
ˇ
C
3
xdz
where
C
1
is
y
=0,0
≤
x
≤
1,
C
2
is
x
=1,0
≤
y
≤
2, and
C
3
is
y
=2
x
,0
≤
x
≤
1. Thus,
x
ˇ
C
xdz
=
i
1
0
xdx
+
i
i
2
0
dy
+(1+2
i
)
i
0
1
xdx
=
1
2
+2
i
−
1
2
(1+2
i
)=
i.
Exercises 18.3
1. (a)
Choosing
x
=0,
−
1
≤
y
≤
1wehave
z
=
iy
,
dz
=
idy
.Thus
i
C
(4
z
−
1)
dz
=
i
i
1
−
1
(4
iy
−
1)
dy
=
−
2
i.
(b)
i
C
(4
z
−
1)
dz
=
i
i
−
i
(4
z
−
1)
dz
=2
z
2
−
z
y
y
y
i
−
i
=
−
2
i
2. (a)
Choosing theline
y
=
1
3
x
,0
≤
x
≤
3wehave
z
=
x
+
1
3
xi
,
dz
=(1+
1
3
i
)
dx
.Thus
i
C
e
z
dz
=
i
3
0
e
(1+
1
3
i
)
x
/
1+
1
3
i
z
dx
=
e
(1+
1
3
i
)
x
y
y
y
3
0
=
e
3+
i
−
e
0
=(
e
3
cos 1
−
1) +
ie
3
sin 1
.
(b)
i
C
e
z
dz
=
i
3+
i
0
e
z
dz
=
e
z
y
y
y
3+
i
0
=
e
3+
i
−
e
0
=(
e
3
cos 1
−
1) +
ie
3
sin 1
3.
The given integral is independent of the path. Thus
i
C
2
zdz
=
i
2
−
i
−
2+7
i
2
zdz
=
z
2
y
y
y
2
−
i
−
2+7
i
=48+24
i.
4.
The given integral is independent of the path. Thus
i
C
6
z
2
dz
=
i
2
−
i
2
6
z
2
dz
=
z
3
y
y
y
2
−
i
2
=
−
15
−
24
i.
5.
i
3+
i
0
z
2
dz
=
1
3
z
3
y
y
y
y
3+
i
0
=6+
26
3
i
788

789. Exercises 18.3
6.
i
1
−
2
i
(3
z
2
−
4
z
+5
i
)
dz
=
z
3
−
2
z
2
+5
iz
y
y
y
1
−
2
i
=
−
19
−
3
i
7.
i
1+
i
1
−
i
z
3
dz
=
1
4
z
4
y
y
y
y
1+
i
1
−
i
=0
8.
i
2
i
−
3
i
(
z
3
−
z
)
dz
=
1
4
z
4
−
1
2
z
2
y
y
y
y
2
i
−
3
i
=
123
4
9.
i
1
−
i
−
i/
2
(2
z
+1)
2
dz
=
1
6
(2
z
+1)
3
y
y
y
y
1
−
i
−
i/
2
=
−
7
6
−
22
3
i
10.
i
i
1
(
iz
+1)
3
dz
=
1
4
i
(
iz
+1)
4
y
y
y
y
i
1
=
−
i
11.
i
i
i/
2
e
πz
dz
=
1
π
e
πz
y
y
y
i
i/
2
=
−
1
π
−
1
π
i
12.
i
1+2
i
1
−
i
ze
z
2
dz
=
1
2
e
z
2
y
y
y
y
1+2
i
1
−
i
=
1
2
[
e
−
3+4
i
−
e
−
2
i
]=
1
2
(
e
−
3
cos 4
−
cos 2) +
1
2
(
e
−
3
sin 4 + sin 2)
i
=0
.
1918 + 0
.
4358
i
13.
i
π
+2
i
π
sin
z
2
dz
=
−
2 cos
z
2
y
y
y
y
π
+2
i
π
=
−
2

cos

π
2
+
i

−
cos
π
2

=2
i
sin
π
2
sinh 1 = 2
.
3504
i
14.
i
πi
1
−
2
i
cos
zdz
= sin
z
y
y
y
πi
1
−
2
i
= sin
πi
−
sin(1
−
2
i
)=
i
sinh
π
−
[sinh 1 cosh 2
−
i
cos 1 sinh 2]
=
−
sin 1 cosh 2 +
i
(sinh
π
+ cos 1 sinh 2) =
−
3
.
1658 + 13
.
5083
i
15.
i
2
πi
πi
cosh
zdz
= sinh
z
y
y
y
2
πi
πi
= sinh 2
πi
−
sinh
πi
=
i
sin 2
π
−
i
sin
π
=0
16.
i
1+
π
2
i
i
sinh 3
zdz
=
1
3
cosh 3
z
y
y
y
y
1+
π
2
i
i
=
1
3
t
cosh
/
3+
3
π
2
i
z
−
cosh 3
i
π
=
1
3
t
cosh 3 cos
3
π
2
+
i
sinh 3 sin
3
π
2
−
cos 3
π
=
−
1
3
cos 3
−
1
3
i
sinh 3 = 0
.
3300
−
3
.
3393
i
17.
i
4
i
−
4
i
1
z
dz
=Ln
z
y
y
y
4
i
−
4
i
=Ln4
i
−
Ln(
−
4
i
) = log
e
4+
π
2
i
−

log
e
4
−
π
2
i

=
πi
18.
i
4+4
i
1+
i
1
z
dz
=Ln
z
y
y
y
4+4
i
1+
i
= Ln(4 + 4
i
)
−
Ln(1 +
i
) = log
e
4
√
2+
π
4
i
−

log
e
√
2+
π
4
i

= log
e
4=1
.
3863
19.
i
4
i
−
4
i
1
z
2
dz
=
−
1
z
y
y
y
y
4
i
−
4
i
=
−
t
1
4
i
−
/
1
−
4
i
zπ
=
1
2
i
20.
i
1+
√
3
i
1
−
i
/
1
z
+
1
z
2
z
dz
=Ln
z
−
1
z
y
y
y
y
1+
√
3
i
1
−
i
= log
e
2+
π
3
i
−
1
1+
√
3
i
−
/
log
e
√
2
−
π
4
i
−
1
1
−
i
z
= log
e
√
2+
1
4
+
i
a
7
π
12
+
√
3
4
+
1
2
)
=0
.
5966 + 2
.
7656
i
21.
Integration by parts gives
i
e
z
cos
zdz
=
1
2
e
z
(cos
z
+ sin
z
)+
C
789

790. Exercises 18.3
and so
i
i
π
e
z
cos
zdz
=
1
2
e
z
(cos
z
+ sin
z
)
y
y
y
i
π
=
1
2
[
e
i
(cos
i
+ sin
i
)
−
e
π
(cos
π
+ sin
π
)]
=
1
2
[(cos 1 cosh 1
−
sin 1 sinh 1 +
e
π
)+
i
(cos 1 sinh 1 + sin 1 cosh 1) = 11
.
4928 + 0
.
9667
i.
22.
Integration by parts gives
i
z
sin
zdz
=
−
z
cos
z
+ sin
z
+
C
and so
i
i
0
z
sin
zdz
=
−
z
cos
z
+ sin
z
y
y
y
i
0
=
−
i
cos
i
+ sin
i
=
−
i
cosh 1 +
i
sinh 1 =
−
0
.
3679
i.
23.
Integration by parts gives
i
ze
z
dz
=
ze
z
−
e
z
+
C
and so
i
1+
i
i
ze
z
dz
=
e
z
(
z
−
1)
y
y
y
1+
i
i
=
ie
1+
i
+
e
i
(1
−
i
) = (cos 1+sin 1
−
e
sin 1)+
i
(sin 1
−
cos 1+
e
cos 1) =
−
0
.
9056+1
.
7699
i.
24.
Integration by parts gives
i
z
2
e
z
dz
=
z
2
e
z
−
2
ze
z
+2
e
z
+
C
and so
i
πi
0
z
2
e
z
dz
=
e
z
(
z
2
−
2
z
+2)
y
y
y
πi
0
=
e
πi
(
−
π
2
−
2
πi
+2)
−
2=
π
2
−
4+2
πi.
Exercises 18.4
1.
By Theorem 18
.
9, with
f
(
z
)=4,
x
ˇ
C
4
z
−
3
i
dz
=2
πi
·
4=8
πi.
2.
By Theorem 18
.
10 with
f
(
z
)=
z
2
and
f
z
(
z
)=2
z
,
x
ˇ
C
z
2
(
z
−
3
i
)
2
dz
=
2
πi
1!
2(3
i
)=
−
12
π.
3.
By Theorem 18
.
9 with
f
(
z
)=
e
z
,
x
ˇ
C
e
z
z
−
πi
dz
=2
πie
πi
=
−
2
πi.
4.
By Theorem 18
.
9 with
f
(
z
)=1+2
e
z
,
x
ˇ
C
1+2
e
z
z
dz
=2
πi
(1+2
e
0
)=6
πi.
5.
By Theorem 18
.
9 with
f
(
z
)=
z
2
−
3
z
+4
i
,
x
ˇ
C
z
2
−
3
z
+4
i
z
−
(
−
2
i
)
dz
=2
πi
(
−
4+6
i
+4
i
)=
−
π
(20+8
i
)
.
790

791. Exercises 18.4
6.
By Theorem 18
.
9 with
f
(
z
)=
1
3
cos
z
,
x
ˇ
C
1
3
cos
z
z
−
π
3
dz
=2
πi
/
1
3
cos
π
3
z
=
π
3
i.
7. (a)
By Theorem 18
.
9 with
f
(
z
)=
z
2
z
+2
i
,
x
ˇ
C
z
2
z
+2
i
z
−
2
i
dz
=2
πi
/
−
4
4
i
z
=
−
2
π.
(b)
By Theorem 18
.
9 with
f
(
z
)=
z
2
z
−
2
i
,
x
ˇ
C
z
2
z
−
2
i
z
−
(
−
2
i
)
dz
=2
πi
/
−
4
−
4
i
z
=2
π.
8. (a)
By Theorem 18
.
9 with
f
(
z
)=
z
2
+3
z
+2
i
z
+4
,
x
ˇ
C
z
2
+3
z
+2
i
z
+4
z
−
1
dz
=2
πi
/
4+2
i
5
z
=
π
/
−
4
5
+
8
5
i
z
.
(b)
By Theorem 18
.
9 with
f
(
z
)=
z
2
+3
z
+2
i
z
−
1
,
x
ˇ
C
z
2
+3
z
+2
i
z
−
1
z
−
(
−
4)
dz
=2
πi
/
4+2
i
−
5
z
=
π
/
4
5
−
8
5
i
z
.
9.
By Theorem 18
.
9 with
f
(
z
)=
z
2
+4
z
−
i
,
x
ˇ
C
z
2
+4
z
−
i
z
−
4
i
dz
=2
πi
/
−
12
3
i
z
=
−
8
π.
10.
By Theorem 18
.
9 with
f
(
z
)=
sin
z
z
+
πi
,
x
ˇ
C
sin
z
z
+
πi
z
−
πi
dz
=2
πi
/
sin
πi
2
πi
z
=
i
sinh
π.
11.
By Theorem 18
.
10 with
f
(
z
)=
e
z
2
,
f
z
(
z
)=2
ze
z
2
, and
f
zz
(
z
)=4
z
2
e
z
2
+2
e
z
2
,
x
ˇ
C
e
z
2
(
z
−
i
)
3
dz
=
2
πi
2!
[
−
4
e
−
1
+2
e
−
1
]=
−
2
πe
−
1
i.
12.
By Theorem 18
.
10 with
f
(
z
)=
z
,
f
z
(
z
)=1,
f
zz
(
z
)=0,and
f
zzz
(
z
)=0,
x
ˇ
C
z
(
z
−
(
−
i
))
4
dz
=
2
πi
3!
(0) = 0
.
791

792. Exercises 18.4
13.
By Theorem 18
.
10 with
f
(
z
) = cos 2
z
,
f
z
(
z
)=
−
2 sin 2
z
,
f
zz
(
z
)=
−
4 cos 2
z
,
f
zzz
(
z
) = 8 sin 2
z
,
f
(4)
(
z
) = 16 cos 2
z
,
x
ˇ
C
cos 2
z
z
5
dz
=
2
πi
4!
(16 cos 0) =
4
π
3
i.
14.
By Theorem 18
.
10 with
f
(
z
)=
e
−
z
sin
z
,
f
z
(
z
)=
e
−
z
cos
z
−
e
−
z
sin
z
, and
f
zz
(
z
)=
−
2
e
−
z
cos
z
,
x
ˇ
C
e
−
z
sin
z
z
3
dz
=
2
πi
2!
(
−
2
e
0
cos 0) =
−
2
πi.
15. (a)
By Theorem 18
.
9 with
f
(
z
)=
2
z
+5
z
−
2
,
x
ˇ
C
2
z
+5
z
−
2
z
dz
=2
πi
/
−
5
2
z
=
−
5
πi.
(b)
Sincethecircle
|
z
−
(
−
1)
|
= 2 encloses only
z
= 0, thevalueof theintegral is thesameas in part
(a)
.
(c)
From Theorem 18
.
9 with
f
(
z
)=
2
z
+5
z
,
x
ˇ
C
2
z
+5
z
z
−
2
dz
=2
πi
/
9
2
z
=9
πi.
(d)
Sincethecircle
|
z
−
(
−
2
i
)
|
= 1 encloses neither
z
= 0 nor
z
= 2 it follows from theCauchy-Goursat
Theorem, Theorem 18
.
4, that
x
ˇ
C
2
z
+5
z
(
z
−
2)
dz
=0
.
16.
By partial fractions,
x
ˇ
C
z
(
z
−
1)(
z
−
2)
dz
=2
x
ˇ
C
dz
z
−
2
−
x
ˇ
C
dz
z
−
1
.
(a)
By the Cauchy-Goursat Theorem, Theorem 18
.
4,
x
ˇ
C
z
(
z
−
1)(
z
−
2)
dz
=0
.
(b)
As in part
(a)
, theintegral is 0.
(c)
By Theorem 18
.
4,
x
ˇ
C
dz
z
−
2
= 0 whereas by Theorem 18
.
9,
x
ˇ
C
dz
z
−
1
=2
πi
.Thus
x
ˇ
C
z
(
z
−
1)(
z
−
2)
dz
=
−
2
πi.
(d)
By Theorem 18
.
9,
x
ˇ
C
dz
z
−
1
=2
πi
and
x
ˇ
C
dz
z
−
2
=2
πi
.Thus
x
ˇ
C
z
(
z
−
1)(
z
−
2)
dz
= 2(2
πi
)
−
2
πi
=3
πi.
17. (a)
By Theorem 18
.
10 with
f
(
z
)=
z
+2
z
−
1
−
i
and
f
z
(
z
)=
−
3
−
i
(
z
−
1
−
i
)
2
,
x
ˇ
C
z
+2
z
−
1
−
i
z
2
dz
=
2
πi
1!
/
−
3
−
i
(
−
1
−
i
)
2
z
=
−
π
(3 +
i
)
.
792

793. Exercises 18.4
(b)
By Theorem 18
.
9 with
f
(
z
)=
z
+2
z
2
,
x
ˇ
C
z
+2
z
2
z
−
(1 +
i
)
dz
=2
πi
/
3+
i
(1 +
i
)
2
z
=
π
(3 +
i
)
.
18. (a)
By Theorem 18
.
10 with
f
(
z
)=
1
z
−
4
,
f
z
(
z
)=
−
1
(
z
−
4)
2
, and
f
zz
(
z
)=
2
(
z
−
4)
3
,
x
ˇ
C
1
z
−
4
z
3
dz
=
2
πi
2!
/
2
−
64
z
=
−
π
32
i.
(b)
By the Cauchy-Goursat Theorem, Theorem 18
.
4,
x
ˇ
C
1
z
3
(
z
−
4)
dz
=0
.
19.
By writing
x
ˇ
C
/
e
2
iz
z
4
−
z
4
(
z
−
i
)
3
z
dz
=
x
ˇ
C
e
2
iz
z
4
dz
−
x
ˇ
C
z
4
(
z
−
i
)
3
dz
we can apply Theorem 18
.
10 to each integral:
x
ˇ
C
e
2
iz
z
4
dz
=
2
πi
3!
(
−
8
i
)=
8
π
3
,
x
ˇ
C
z
4
(
z
−
i
)
3
dz
=
2
πi
2!
(
−
12) =
−
12
πi.
Thus
x
ˇ
C
/
e
2
iz
z
4
−
z
4
(
z
−
i
)
3
z
dz
=
π
/
8
3
+12
i
z
.
20.
By writing
x
ˇ
C
/
cosh
z
(
z
−
π
)
3
−
sin
2
z
(2
z
−
π
)
3
z
dz
=
x
ˇ
C
cosh
z
(
z
−
π
)
3
dz
−
x
ˇ
C
1
8
sin
2
z
(
z
−
π
2
)
3
dz
we apply Theorem 18
.
4 to the first integral and Theorem 18
.
10 to thesecond:
x
ˇ
C
cosh
z
(
z
−
π
)
3
dz
=0
,
x
ˇ
C
1
8
sin
2
z
(
z
−
π
2
)
3
dz
=
2
πi
2!
/
−
1
4
sin
2
π
2
z
=
−
π
4
i.
Thus
x
ˇ
C
/
cosh
z
(
z
−
π
)
3
−
sin
2
z
(2
z
−
π
)
3
z
dz
=
π
4
,i.
21.
Wehave
x
ˇ
C
1
z
3
(
z
−
1)
2
dz
=
x
ˇ
C
1
1
(
z
−
1)
2
z
3
dz
+
x
ˇ
C
2
1
z
3
(
z
−
1)
2
dz
where
C
1
and
C
2
arethecircles
|
z
|
=1
/
3 and
|
z
−
1
|
=1
/
3, respectively. By Theorem 18
.
10,
x
ˇ
C
1
1
(
z
−
1)
2
z
3
dz
=
2
πi
2!
(6) = 6
πi,
x
ˇ
C
2
1
z
3
(
z
−
1)
2
dz
=
2
πi
1!
(
−
3) =
−
6
πi.
Thus
x
ˇ
C
1
z
3
(
z
−
1)
2
dz
=6
πi
−
6
πi
=0
.
22.
Wehave
x
ˇ
C
1
z
2
(
z
2
+1)
dz
=
x
ˇ
C
1
1
z
2
(
z
+
i
)
z
−
i
dz
+
x
ˇ
C
2
1
z
2
+1
z
2
dz
793

794. Exercises 18.4
where
C
1
and
C
2
arethecircles
|
z
−
i
|
=1
/
3 and
|
z
|
=1
/
8, respectively. By Theorems 18
.
9 and 18
.
10,
x
ˇ
C
1
1
z
2
(
z
+
i
)
z
−
i
dz
=2
πi
/
1
−
2
i
z
=
−
π,
x
ˇ
C
2
1
z
2
+1
z
2
dz
=
2
πi
1!
(0) = 0
.
Thus
x
ˇ
C
1
z
2
(
z
2
+1)
dz
=
−
π.
23.
Wehave
x
ˇ
C
3
z
+1
z
(
z
−
2)
2
dz
=
x
ˇ
C
1
3
z
+1
z
(
z
−
2)
2
dz
−
x
ˇ
C
2
3
z
+1
(
z
−
2)
2
z
dz
where
C
1
and
C
2
aretheclosed portions of thecurve
C
enclosing
z
= 2 and
z
= 0, respectively. By
Theorems 18
.
10 and 18
.
9,
x
ˇ
C
1
3
z
+1
z
(
z
−
2)
2
dz
=
2
πi
1!
/
−
1
4
z
=
−
π
2
i,
x
ˇ
C
2
3
z
+1
(
z
−
2)
2
z
dz
=2
πi
/
1
4
z
=
π
2
i.
Thus
x
ˇ
C
3
z
+1
z
(
z
−
2)
2
dz
=
−
π
2
i
−
π
2
i
=
−
πi.
24.
Wehave
x
ˇ
C
e
iz
(
z
2
+1)
2
dz
=
x
ˇ
C
1
e
iz
(
z
+
i
)
2
(
z
−
i
)
2
dz
−
x
ˇ
C
2
e
iz
(
z
−
i
)
2
(
z
−
(
−
i
))
2
dz
where
C
1
and
C
2
aretheclosed portions of thecurve
C
enclosing
z
=
i
and
z
=
−
i
, respectively. By
Theorem 18
.
10,
x
ˇ
C
1
e
iz
(
z
+
i
)
2
(
z
−
i
)
2
dz
=
2
πi
1!
/
−
4
e
−
1
−
8
i
z
=
πe
−
1
,
x
ˇ
C
2
e
iz
(
z
−
i
)
2
(
z
−
(
−
i
))
2
dz
=
2
πi
1!
/
0
8
i
z
=0
.
Thus
x
ˇ
C
e
iz
(
z
2
+1)
2
dz
=
πe
−
1
.
Chapter 18 Review Exercises
1.
True
2.
False
3.
True
4.
True
5.
0
6.
π
(
−
16+8
i
)
7.
π
(6
π
−
i
)
8.
a constant function
9.
True(Usepartial fractions and writethegiven integral as two integrals.)
10.
True
11.
integer not equal to
−
1;
−
1
12.
12
π
794

795. Chapter 18 Review Exercises
13.
Since
f
(
z
)=
z
is entire,
i
C
(
x
+
iy
)
dz
is independent of the path
C
.Thus
x
ˇ
C
(
x
+
iy
)
dz
=
i
3
−
4
zdz
=
z
2
2
y
y
y
y
3
−
4
=
−
7
2
.
14.
Wehave
i
C
(
x
−
iy
)
dz
=
i
C
1
(
x
−
iy
)
dz
+
i
C
2
(
x
−
iy
)
dz
+
i
C
3
(
x
−
iy
)
dz
On
C
1
,
x
=4,0
≤
y
≤
2,
z
=4+
iy
,
dz
=
idy
,
i
C
1
(4
−
iy
)
idy
=
i
i
2
0
(4
−
iy
)
dy
=
i
/
4
y
−
i
2
y
2
z
y
y
y
y
2
0
=2+8
i.
On
C
2
,
y
=2,
−
4
≤
x
≤
3,
z
=
x
+2
i
,
dz
=
dx
,
i
C
2
(
x
−
2
i
)
dx
=
i
3
−
4
(
x
−
2
i
)
dx
=
1
2
x
2
−
2
ix
y
y
y
y
3
−
4
=
−
7
2
−
14
i.
On
C
3
,
x
=3,0
≤
y
≤
2,
z
=3+
iy
,
dz
=
idy
,
i
C
3
(3
−
iy
)
idy
=
i
i
0
2
(3
−
iy
)
dy
=
i
/
3
y
−
i
2
y
2
z
y
y
y
y
0
2
=
−
2
−
6
i.
Thus
i
C
(
x
−
iy
)
dz
=2+8
i
−
7
2
−
14
i
−
2
−
6
i
=
−
7
2
−
12
i.
15.
i
C
|
z
2
|
dz
=
i
2
0
(
t
4
+
t
2
)
dt
+2
i
i
2
0
(
t
5
+
t
3
)
dt
=
136
15
+
88
3
i
16.
i
C
e
πz
dz
=
1
π
i
1+
i
i
e
πz
(
πdz
)=
1
π
e
πz
y
y
y
y
1+
i
i
=
1
π
(1
−
e
π
)
17.
By the Cauchy-Goursat Theorem, Theorem 18
.
4,
x
ˇ
C
e
πz
dz
=0.
18.
i
1
−
i
3
i
(4
z
−
6)
dz
=2
z
2
−
6
z
y
y
y
1
−
i
3
i
=12+20
i
19.
i
C
sin
zdz
=
i
1+4
i
1
sin
zdz
=
−
cos
z
y
y
y
1+4
i
1
= cos 1
−
cos(1 + 4
i
)=
−
14
.
2144 + 22
.
9637
i
20.
i
C
(4
z
3
+3
z
2
+2
z
+1)
dz
=
i
2
i
0
(4
z
3
+3
z
2
+2
z
+1)
dz
=
z
4
+
z
3
+
z
2
+
z
y
y
y
2
i
0
=12
−
6
i
21.
On
|
z
|
=1,let
z
=
e
it
,
dz
=
ie
it
dt
, so that
x
ˇ
C
(
z
−
2
+
z
−
1
+
z
+
z
2
)
dz
=
i
i
2
π
0
(
e
−
2
it
+
e
−
it
+
e
it
+
e
2
it
)
e
it
dt
=
−
e
−
it
+
it
+
1
2
e
2
it
+
1
3
e
3
it
y
y
y
y
2
π
0
=2
πi.
22.
By partial-fractions and Theorem 18
.
9,
x
ˇ
C
3
z
+4
z
2
−
1
dz
=
7
2
x
ˇ
C
1
z
−
1
dz
−
1
2
x
ˇ
C
1
z
−
(
−
1)
dz
=
7
2
(2
πi
)
−
1
2
(2
πi
)=6
πi.
23.
By Theorem 18
.
10 with
f
(
z
)=
e
−
2
z
,
f
z
(
z
)=
−
2
e
−
2
z
,
f
zz
(
z
)=4
e
−
2
z
, and
f
zzz
(
z
)=
−
8
e
−
2
z
,
x
ˇ
C
e
−
2
z
z
4
dz
=
2
πi
3!
(
−
8) =
−
8
π
3
i.
795

796. Chapter 18 Review Exercises
24.
By Theorem 18
.
10 with
f
(
z
)=
cos
z
z
−
1
and
f
z
(
z
)=
sin
z
−
cos
z
−
z
sin
z
(
z
−
1)
2
,
x
ˇ
C
cos
z
z
−
1
z
2
dz
=
2
πi
1!
/
−
1
1
z
=
−
2
πi.
25.
By Theorem 18
.
9 with
f
(
z
)=
1
2(
z
+3)
,
x
ˇ
C
1
2(
z
+3)
(
z
−
(
−
1
/
2))
dz
=2
πi
/
1
5
z
=
2
π
5
i.
26.
Sincethefunction
f
(
z
)=
z/
sin
z
is analytic within and on the given simple closed contour
C
, it follows from
the Cauchy-Goursat Theorem, Theorem 18
.
4, that
x
ˇ
C
z
csc
zdz
=0
.
27.
Using theprincipleof deformation of contours wechoose
C
to be the more convenient circular contour
|
z
+
i
|
=
1
4
.
On this circle
z
=
−
i
+
1
4
e
it
and
dz
=
1
4
ie
it
dt
.Thus
x
ˇ
C
z
z
+
i
dz
=
i
i
2
π
0
/
1
4
e
it
−
i
z
dt
=2
π.
28. (a)
By Theorem 18
.
9 with
f
(
z
)=
e
iπz
2(
z
−
2)
,
x
ˇ
C
e
iπz
2(
z
−
2)
z
−
1
/
2
dz
=2
πi
/
e
iπ/
2
−
3
z
=
2
π
3
.
(b)
By Theorem 18
.
9 with
f
(
z
)=
e
iπz
2
z
−
1
,
x
ˇ
C
e
iπz
2
z
−
1
z
−
2
dz
=2
πi
/
e
2
πi
3
z
=
2
π
3
i.
(c)
By the Cauchy-Goursat Theorem, Theorem 18
.
4,
x
ˇ
C
e
iπz
2
z
2
−
5
z
+2
dz
=0
.
29.
For
f
(
z
)=
z
n
g
(
z
)wehave
f
z
(
z
)=
z
n
g
z
(
z
)+
nz
n
−
1
g
(
z
) and so
f
z
(
z
)
f
(
z
)
=
z
n
g
z
(
z
)+
nz
n
−
1
g
(
z
)
z
n
g
(
z
)
=
g
z
(
z
)
g
(
z
)
+
n
z
.
Thus by Theorem 18
.
4 and (4) of Section 18
.
2,
x
ˇ
C
f
z
(
z
)
f
(
z
)
dz
=
x
ˇ
C
g
z
(
z
)
g
(
z
)
dz
+
n
x
ˇ
C
1
z
dz
=0+
n
(2
πi
)=2
nπi.
30.
Wehave
y
y
y
y
i
C
Ln(
z
+1)
dz
y
y
y
y
≤|
max of Ln(
z
+1)on
C
|·
2,
796

797. Chapter 18 Review Exercises
where 2 is the length of the line segment. Now
|
Ln(
z
+1)
|≤|
log
e
(
z
+1)
|
+
|
Arg(
z
+1)
|
.
But max Arg(
z
+1)=
π/
4 when
z
=
i
and max
|
z
+1
|
=
√
10 when
z
=2+
i
. Thus,
y
y
y
y
i
C
Ln(
z
+1)
dz
y
y
y
y
≤
/
1
2
log
e
10 +
π
4
z
2 = log
e
10 +
π
2
.
797

798. 19
Series and Residues
Exercises 19.1
1.
5
i
,
−
5,
−
5
i
,5,5
i
2.
2
−
i
,1,2+
i
,3,2
−
i
3.
0, 2, 0, 2, 0
4.
1+
i
,2
i
,
−
2+2
i
,
−
4,
−
4
−
4
i
5.
Converges. To see this write the general term as
3
i
+2
/n
1+
i
.
6.
Converges. To see this write the general term as
k
2
5
z
n
1+
n
2
−
n
i
1+3
n
5
−
n
i
.
7.
Converges. To see this write the general term as
(
i
+2
/n
)
2
i
.
8.
Diverges. To see this consider the term
n
n
+1
i
n
and take
n
to be an odd positive integer.
9.
Diverges. To see this write the general term as
√
n
k
1+
1
√
n
i
n
z
.
10.
Converges. The real part of the general term converges to 0 and the imaginary part of the general term converges
to
π
.
11.
Re(
z
n
)=
8
n
2
+
n
4
n
2
+1
→
2as
n
→∞
, and Im(
z
n
)=
6
n
2
−
4
n
4
n
2
+1
→
3
2
as
n
→∞
.
12.
Write
z
n
=
k
1
4
+
1
4
i
z
n
in polar form as
z
n
=
π
√
2
4
i
n
cos
nθ
+
i
π
√
2
4
i
n
sin
nθ
.Now
Re(
z
n
)=
π
√
2
4
i
n
cos
nθ
→
0as
n
→∞
and Im(
z
n
)=
π
√
2
4
i
n
sin
nθ
→
0as
n
→∞
since
√
2
/
4
<
1.
13.
S
n
=
1
1+2
i
−
1
2+2
i
+
1
2+2
i
−
1
3+2
i
+
1
3+2
i
−
1
4+2
i
+
···
+
1
n
+2
i
−
1
n
+1+2
i
=
1
1+2
i
−
1
n
+1+2
i
Thus, lim
n
→∞
S
n
=
1
1+2
i
=
1
5
−
2
5
i
.
14.
By partial fractions,
i
k
(
k
+1)
=
i
k
−
i
k
+1
and so
S
n
=
i
−
i
2
+
i
2
−
i
3
+
i
3
−
i
4
+
···
+
i
n
−
i
n
+1
=
i
−
i
n
+1
.
Thus lim
n
→∞
S
n
=
i
.
15.
We identify
a
= 1 and
z
=1
−
i
. Since
|
z
|
=
√
2
>
1 the series is divergent.
16.
We identify
a
=4
i
and
z
=1
/
3. Since
|
z
|
=1
/
3
<
1 the series converges to
4
i
1
−
1
/
3
=6
i.
17.
We identify
a
=
i/
2 and
z
=
i/
2. Since
|
z
|
=1
/
2
<
1 the series converges to
i/
2
1
−
i/
2
=
−
1
5
+
2
5
i.
798

799. Exercises 19.1
18.
We identify
a
=1
/
2 and
z
=
i
. Since
|
z
|−
1 the series is divergent.
19.
We identify
a
= 3 and
z
=2
/
(1+2
i
). Since
|
z
|
=2
/
√
5
<
1 the series converges to
3
1
−
2
1+2
i
=
9
5
−
12
5
i.
20.
We identify
a
=
−
1
/
(1 +
i
) and
z
=
i/
(1 +
i
). Since
|
z
|
=1
/
√
2
<
1 the series converges to
−
1
1+
i
1
−
i
1+
i
=
−
1
.
21.
From lim
n
→∞
/
/
/
/
/
/
/
/
1
(1
−
2
i
)
n
+2
1
(1
−
2
i
)
n
+1
/
/
/
/
/
/
/
/
=
1
|
1
−
2
i
|
=
1
√
5
we see that the radius of convergence is
R
=
√
5 . The circle of convergence is
|
z
−
2
i
|
=
√
5.
22.
From lim
n
→∞
/
/
/
/
/
/
/
/
/
1
n
+1
k
i
1+
i
z
n
+1
1
n
k
i
1+
i
z
n
/
/
/
/
/
/
/
/
/
= lim
n
→∞
n
n
+1
/
/
/
/
i
1+
i
/
/
/
/
=
1
√
2
we see that the radius of convergence is
R
=
√
2 . The circle of convergence is
|
z
|
=
√
2.
23.
From lim
n
→∞
/
/
/
/
/
/
/
/
/
(
−
1)
n
+1
(
n
+ 1)2
n
+1
(
−
1)
n
n
2
n
/
/
/
/
/
/
/
/
/
= lim
n
→∞
n
2(
n
+1)
=
1
2
we see that the radius of convergence is
R
= 2. The circle of convergence is
|
z
−
1
−
i
|
=2.
24.
From lim
n
→∞
/
/
/
/
/
/
/
/
1
(
n
+1)
2
(3 + 4
i
)
n
+1
1
n
2
(3+4
i
)
n
/
/
/
/
/
/
/
/
= lim
n
→∞
k
n
n
+1
z
2
1
|
3+4
i
|
=
1
5
we see that the radius of convergence is
R
= 5. The circle of convergence is
|
z
+3
i
|
=5.
25.
From lim
n
→∞
n
e
|
1+3
i
|
n
=
|
1+3
i
|
=
√
10
we see that the radius of convergence is
R
=1
/
√
10 . The circle of convergence is
|
z
−
i
|
=1
/
√
10 .
26.
From lim
n
→∞
n
t
/
/
/
/
1
n
n
/
/
/
/
= lim
n
→∞
1
n
=0
we see that the radius of convergence is
∞
. The power series with center 0 converges absolutely for all
z
.
27.
From lim
n
→∞
n
t
/
/
/
/
1
5
2
n
/
/
/
/
= lim
n
→∞
1
25
=
1
25
we see that the radius of convergence is
R
= 25. The circle of convergence is
|
z
−
4
−
3
i
|
= 25.
799

800. Exercises 19.1
28.
From lim
n
→∞
n
t
/
/
/
/
(
−
1)
n
k
1+2
i
2
z
n
/
/
/
/
= lim
n
→∞
/
/
/
/
1+2
i
2
/
/
/
/
=
√
5
2
we see that the radius of convergence is
R
=2
/
√
5 . The circle of convergence is
|
z
+2
i
|
=2
/
√
5.
29.
The circle of convergence is
|
z
−
i
|
= 2. Since the series of absolute values
∞

k
=1
/
/
/
/
(
z
−
1)
k
k
2
k
/
/
/
/
=
∞

k
=1
|
z
−
1
|
k
k
2
k
=
∞

k
=1
2
k
k
2
k
=
∞

k
=1
1
k
is the divergent harmonic series. But
z
=
−
2+
i
is on the circle of convergence and (
z
−
i
)
k
=(
−
2)
k
. The series
∞

k
=1
(
−
2)
k
k
2
k
=
∞

k
=1
(
−
1)
k
k
is convergent.
30. (a)
The circle of convergence is
|
z
|
= 1. Since the series of absolute values
∞

k
=1
/
/
/
/
z
k
k
2
/
/
/
/
=
∞

k
=1
|
z
|
k
k
2
=
∞

k
=1
1
k
2
converges, the given series is absolutely convergent for every
z
on
|
z
|
= 1. Since absolute convergence
implies convergence, the given series converges for all
z
on
|
z
|
=1.
(b)
The circle of convergence is
|
z
|
= 1. On the circle,
n
|
z
|
n
→∞
as
n
→∞
. This implies
nz
n
e
→
0as
n
→∞
.
Thus by Theorem 19
.
3 the series is divergent for every
z
on the circle
|
z
|
=1.
Exercises 19.2
1.
z
1+
z
=
z
[1
−
z
+
z
2
−
z
3
+
···
]=
z
−
z
2
+
z
3
−
z
4
+
···
=
∞

k
=1
(
−
1)
k
+1
z
k
;
R
=1
2.
1
4
−
2
z
=
1
4
·

1+
z
2
+
z
2
2
2
+
z
3
2
3
+
···

=
1
4
∞

k
=0
z
k
2
k
;
R
=2
3.
Differentiating
1
1+2
z
=1
−
2
z
+2
2
z
2
−
2
3
z
3
+
···
gives
−
2
(1 + 2
z
)
2
=
−
2+2
·
2
2
z
−
3
·
2
3
z
2
+
···
.Thus
1
(1+2
z
)
=1
−
2
·
(2
z
)+3
·
(2
z
)
2
−···
=
∞

k
=1
(
−
1)
k
−
1
k
(2
z
)
k
−
1
.R
=1
/
2
4.
Using the binomial series gives
z
(1
−
z
)
=
z

1+3
z
+
3
·
4
2!
z
+
3
·
4
·
5
3!
z
3
+
···

=
z
+3
z
2
+
3
·
4
2!
z
3
+
3
·
4
·
5
3!
z
4
+
···
.R
=1
5.
Replacing
z
in
e
z
=
∞

k
=0
z
k
k
!
by
−
2
z
gives
e
−
2
z
=
∞

k
=0
(
−
1)
k
k
!
(2
z
)
k
.
R
=
∞
6.
Replacing
z
in
e
z
=
∞

k
=0
z
k
k
!
by
−
z
2
and multiplying the result by
z
gives
ze
−
z
2
=
∞

k
=0
(
−
1)
k
k
!
z
2
k
+1
.
R
=
∞
7.
Subtracting the series for
e
z
and
e
−
z
gives sinh
z
=
1
2
(
e
z
−
e
−
z
)=
∞

k
=0
z
2
k
+1
(2
k
+ 1)!
.
R
=
∞
800

801. Exercises 19.2
8.
Adding the series for
e
z
and
e
−
z
gives cosh
z
=
1
2
(
e
z
+
e
−
z
)=
∞

k
=0
z
2
k
(2
k
)!
.
R
=
∞
9.
Replacing
z
in cos
z
=
∞

k
=0
(
−
1)
k
z
2
k
(2
k
)!
by
z/
2 gives cos
z
2
=
∞

k
=0
(
−
1)
k
(2
k
)!

z
2

2
k
.
R
=
∞
10.
Replacing
z
in sin
z
=
∞

k
=0
(
−
1)
k
z
2
k
+1
(2
k
+ 1)!
by 3
z
gives sin 3
z
=
∞

k
=0
(
−
1)
k
(3
z
)
2
k
+1
(2
k
+ 1)!
.
R
=
∞
11.
Replacing
z
in sin
z
=
∞

k
=0
(
−
1)
k
z
2
k
+1
(2
k
+ 1)!
by
z
2
gives sin
z
2
=
∞

k
=0
(
−
1)
k
z
4
k
+2
(2
k
+ 1)!
.
R
=
∞
12.
Using the identity cos
z
=
1
2
(1 + cos 2
z
) and the series cos
z
=
∞

k
=0
(
−
1)
k
z
2
k
(2
k
)!
gives
cos
2
z
=
1
2
+
1
2
∞

k
=0
(
−
1)
k
(2
z
)
2
k
(2
k
)!
=1+
∞

k
=1
(
−
1)
k
2
2
k
−
1
(2
k
)!
z
2
k
.R
=
∞
13.
Using (6) of Section 19
.
1,
1
z
=
1
1+(
z
−
1)
=1
−
(
z
−
1)+(
z
−
1)
2
−
(
z
−
1)
3
+
···
=
∞

k
=0
(
−
1)
k
(
z
−
1)
k
.R
=1
14.
Using (6) of Section 19
.
1,
1
z
=
1
1+
i
+(
z
−
1
−
i
)
=
1
1+
i
·
1
1+
z
−
1
−
i
1+
i
=
1
1+
i

1
−
(
z
−
1
−
i
)
1+
i
+
(
z
−
1
−
i
)
2
(1 +
i
)
2
−
(
z
−
1
−
i
)
3
(1 +
i
)
3
+
···

=
1
1+
i
−
(
z
−
1
−
i
)
(1 +
i
)
2
+
(
z
−
1
−
i
)
2
(1 +
i
)
3
−
(
z
−
1
−
i
)
3
(1 +
i
)
4
+
···
=
∞

k
=0
(
−
1)
k
(
z
−
1
−
i
)
k
(1 +
i
)
k
+1
.R
=
√
2
15.
Using (5) of Section 19
.
1,
1
3
−
z
=
1
3
−
2
i
−
(
z
−
2
i
)
=
1
3
−
2
i
·
1
1
−
z
−
2
i
3
−
2
i
=
1
3
−
2
i

1+
z
−
2
i
3
−
2
i
+
(
z
−
2
i
)
2
(3
−
2
i
)
2
+
(
z
−
2
i
)
3
(3
−
2
i
)
3
+
···

=
1
3
−
2
i
+
z
−
2
i
(3
−
2
i
)
2
+
(
z
−
2
i
)
2
(3
−
2
i
)
3
+
(
z
−
2
i
)
3
(3
−
2
i
)
4
+
···
=
∞

k
=0
(
z
−
2
i
)
k
(3
−
2
i
)
k
+1
.R
=
√
13
16.
Using (6) of Section 19
.
1,
1
1+
z
=
1
1
−
i
+
z
+
i
=
1
1
−
i
·
1
1+
z
+
i
1
−
i
=
1
1
−
i

1
−
z
+
i
1
−
i
+
(
z
+
i
)
2
(1
−
i
)
2
−
(
z
+
i
)
3
(1
−
i
)
3
+
···

=
1
1
−
i
−
z
+
i
(1
−
i
)
2
+
(
z
+
i
)
2
(1
−
i
)
3
−
(
z
+
i
)
3
(1
−
i
)
4
+
···
=
∞

k
=0
(
−
1)
k
(
z
+
i
)
k
(1
−
i
)
k
+1
.R
=
√
2
801

802. Exercises 19.2
17.
Using (5) of Section 19
.
1,
z
−
1
3
−
z
=(
z
−
1)
·
1
2
−
(
z
−
1)
=
(
z
−
1)
2
·
1
1
−
z
−
1
2
=
z
−
1
2

1+
z
−
1
2
+
(
z
−
1)
2
2
2
+
(
z
−
1)
3
2
3
+
···

=
z
−
1
2
+
(
z
−
1)
2
2
2
+
(
z
−
1)
3
2
3
+
(
z
−
1)
4
2
4
+
···
=
∞

k
=1
(
z
−
1)
k
2
k
.R
=2
18.
Using (5) of Section 19
.
1,
1+
z
1
−
z
=
−
1+
2
1
−
z
=
−
1+
2
1
−
i
−
(
z
−
i
)
=
−
1+
2
1
−
i
·
1
1
−
z
−
i
1
−
i
=
−
1+
2
1
−
i

1+
z
−
i
1
−
i
+
(
z
−
i
)
2
(1
−
i
)
2
+
(
z
−
i
)
3
(1
−
i
)
3
+
···

=
−
1+
2
1
−
i
+
2(
z
−
i
)
(1
−
i
)
2
+
2(
z
−
i
)
2
(1
−
i
)
3
+
2(
z
−
i
)
3
(1
−
i
)
4
+
···
=
−
1+
∞

k
=0
2(
z
−
i
)
k
(1
−
i
)
k
+1
.R
=
√
2
19.
Using (8) of Section 19
.
2,
cos
z
=
√
2
2
−
√
2
2
·
1!

z
−
π
4

−
√
2
2
·
2!

z
−
π
4

2
+
√
2
2
·
3!

z
−
π
4

3
+
···
.R
=
∞
20.
Using the identity sin
z
= cos(
z
−
π/
2) and (14) of Section 19
.
2, sin
z
=
∞

k
=0
(
−
1)
k
(
z
−
π
2
)
2
k
(2
k
)!
.
R
=
∞
21.
Using
e
z
=
e
3
i
·
e
z
−
3
i
and (12) of Section 19
.
2,
e
z
=
e
3
i
∞

k
=0
(
z
−
3
i
)
k
k
!
.
R
=
∞
22.
Using (
z
−
1)
e
−
2
z
=
e
2
(
z
−
1)
e
−
2(
z
−
1)
and (12) of Section 19
.
2,
(
z
−
1)
e
−
2
z
=
e
2
∞

k
=0
(
−
1)
k
2
k
k
!
(
z
−
1)
k
+1
.R
=
∞
23.
Using (8) of Section 19
.
2, tan
z
=
z
+
1
3
z
3
+
2
15
z
5
+
···
.
24.
Using (8) of Section 19
.
2,
e
1
/
(1+
z
)
=
e
−
ez
+
3
e
2
z
2
−···
.
25.
Using (5) of Section 19
.
1,
f
(
z
)=
1
z
−
2
i
−
1
z
−
i
=
−
1
2
i
·
1
1
−
z/
2
i
+
1
i
1
1
−
z/i
=
−
1
2
i
k
1+
z
2
i
+
z
(2
i
)
2
+
z
3
(2
i
)
3
+
···
z
+
1
i
k
1+
z
i
+
z
2
i
2
+
z
3
i
3
+
···
z
=
−
i
2
−
3
4
z
+
7
i
8
z
2
+
15
16
z
3
−···
.
The radius of convergence is
R
=1.
26.
Using (5) and (6) of Section 19
.
1,
f
(
z
)=
2
z
+1
−
1
z
−
3
=2
·
1
1+
z
+
1
3
·
1
1
−
z/
3
= 2(1
−
z
+
z
2
−
z
3
+
···
)+
1
3
k
1+
z
3
+
z
2
3
2
+
z
3
3
3
+
···
z
=
7
3
−
17
9
z
+
55
27
z
2
−
161
81
z
3
+
···
.
27.
The distance from 2 + 5
i
to
i
is
|
2+5
i
−
i
|
=
|
2+4
i
|
=2
√
5.
802

803. Exercises 19.2
28.
The distance from
πi
to0is
|
πi
|
=
π
.
29.
The Taylor series are
f
(
z
)=
∞

k
=0
(
−
1)
k
(
z
+1)
k
,R
= 1; and
f
(
z
)=
∞

k
=0
(
−
1)
k
(
z
−
i
)
k
(2 +
i
)
k
+1
,R
=
√
5
.
30.
The series are
f
(
z
)=
∞

k
=0
(
−
1)
k
(
z
−
3)
k
3
k
+1
,R
=3
and
f
(
z
)=
∞

k
=0
(
−
1)
k
(
z
−
1
−
i
)
k
(1 +
i
)
k
+1
,R
=
√
2
.
31. (a)
The distance from
z
0
to the branch cut is one unit.
(b)
The first term of the series defined by (8) of this section is
f
(
−
1+
i
) = Ln(
−
1+
i
) = log
e
√
2+
3
π
4
i
=
1
2
log
e
2+
3
π
4
i.
The subsequent terms of the series come from
f
π
(
z
)=
1
z
,
f
ππ
(
z
)=
−
1
z
2
, and so on, evaluated at
−
1+
i
.
(c)
The series converges within the circle
|
z
+1
−
i
|
=
√
2 . Although the series converges
in the shaded region, it does not converge to (or represent) Ln
z
in this region.
32. (a)
R
= 1, which is the distance from the origin to
z
=
−
1.
(b)
Using (8) of this section (or integrating the series for 1
/
(1 +
z
) we obtain for
R
=1,
Ln(1 +
z
)=
∞

k
=1
(
−
1)
k
+1
k
z
k
.
(c)
By replacing
z
in part (b) by
−
z
we obtain for
R
=1,
Ln(1
−
z
)=
−
∞

k
=0
z
k
k
.
(d)
One way of obtaining the Maclaurin series for Ln
k
1+
z
1
−
z
z
is to use (8) of this section. Alternatively, let
us write
Ln
k
1+
z
1
−
z
z
=Ln(1+
z
)
−
L
(1
−
z
)
and subtract the series in parts
(b)
and
(c)
. This gives for the common circle of convergence
|
z
|
=1,
Ln
k
1+
z
1
−
z
z
=2
z
+
2
3
z
3
+
2
5
z
5
+
2
7
z
7
+
···
=2
∞

k
=0
1
(2
k
+1)
z
2
k
+1
.
803

804. Exercises 19.2
But recall that in general Ln(
z
1
/z
2
)
e
=Ln
z
1
−
Ln
z
2
since Ln
z
1
and Ln
z
2
could differ by a constant multiple
of
i
. That is Ln
z
1
−
Ln
z
2
=
Ci
, for some
C
.So
Ln
k
1+
z
1
−
z
z
=Ln(1+
z
)
−
Ln(1
−
z
)
−
Ci.
When
z
= 0 we obtain Ln 1 = Ln 1
−
Ln 1
−
Ci
. Since Ln 1 = 0 we get
C
=0.
33.
From
e
z
≈
1+
z
+
z
2
2
we obtain
e
(1+
i
)
/
10
≈
1+
1+
i
10
+
(1 +
i
)
2
100
=1
.
1+0
.
12
i.
34.
From sin
z
≈
z
−
z
3
6
we obtain
sin
k
1+
i
10
z
≈
1+
i
10
−
1
6
k
1+
i
10
z
3
=
1
10
+
1
10
i
−
1
6
k
−
2+2
i
1000
z
=
301
3000
+
299
3000
i.
35.
Using the series
e
z
=
∞

k
=0
z
k
k
!
we obtain
e
−
t
2
=
∞

k
=0
(
−
1)
k
t
2
k
k
!
.Thus
2
√
π
α
z
0
e
−
t
2
dt
=
2
√
π
∞

k
=0
(
−
1)
k
k
!
α
z
0
t
2
k
dt
=
2
√
π
∞

k
=0
(
−
1)
k
k
!(2
k
+1)
z
2
k
+1
.
36.
e
iz
=
∞

k
=0
(
iz
)
k
k
!
=1+
i
z
1!
−
z
2
2!
−
i
z
3
3!
+
z
4
4!
+
i
z
5
5!
−
z
6
6!
−
i
z
7
7!
+
···
=
k
1
−
z
2
2!
+
z
4
4!
−
z
6
6!
+
···
z
+
i
k
z
1!
−
z
3
3!
+
z
5
5!
−
z
7
7!
+
···
z
= cos
z
+
i
sin
z
Exercises 19.3
1.
f
(
z
)=
1
z
k
1
−
z
2
2!
+
z
4
4!
−
z
6
6!
+
···
z
=
1
z
−
z
2!
+
z
3
4!
−
z
5
6!
+
···
2.
f
(
z
)=
1
z
5

z
−
k
z
−
z
3
3!
+
z
5
5!
−
z
7
7!
+
···
zr
=
1
3!
z
2
−
1
5!
+
z
2
7!
−
z
4
9!
+
···
3.
f
(
z
)=1
−
1
1!
z
2
+
1
2!
z
4
−
1
3!
z
6
+
···
4.
f
(
z
)=
1
z
2

1
−
k
1+
z
1!
+
z
2
2!
+
z
3
3!
+
···
zr
=
−
1
1!
z
−
1
2!
−
z
3!
−
z
2
4!
−···
5.
f
(
z
)=
e
·
e
z
−
1
z
−
1
=
e
z
−
1
k
1+
(
z
−
1)
1!
+
(
z
−
1)
2
2!
+
(
z
−
1)
3
3!
+
···
z
=
e
z
−
1
+
e
1!
+
e
(
z
−
1)
2!
+
e
(
z
−
1)
2
3!
+
···
6.
f
(
z
)=
z
k
1
−
1
2!
z
2
+
1
4!
z
4
−
1
6!
z
6
+
···
z
=
z
−
1
2!
z
+
1
4!
z
3
−
1
6!
z
5
+
···
7.
f
(
z
)=
−
1
3
z
·
1
1
−
z
3
=
−
1
3
z

1+
z
3
+
z
2
3
2
+
z
3
3
3
+
···

=
−
1
3
z
−
1
3
2
−
z
3
3
−
z
2
3
4
−···
804

805. Exercises 19.3
8.
f
(
z
)=
1
z
2
·
1
1
−
3
z
=
1
z
2

1+
3
z
+
3
2
z
2
+
3
3
z
3
+
···

=
1
z
2
+
3
z
3
+
3
2
z
4
+
3
3
z
5
+
···
9.
f
(
z
)=
1
z
−
3
·
1
3+
z
−
3
=
1
3(
z
−
3)
·
1
1+
z
−
3
3
=
1
3(
z
−
3)

1
−
z
−
3
3
+
(
z
−
3)
2
3
2
−
(
z
−
3)
3
3
3
+
···

=
1
3(
z
−
3)
−
1
3
2
+
z
−
3
3
3
−
(
z
−
3)
2
3
4
+
···
10.
f
(
z
)=
1
z
−
3
·
1
z
−
3+3
=
1
(
z
−
3)
2
·
1
1+
3
z
−
3
=
1
(
z
−
3)
2

1
−
3
z
−
3
+
3
2
(
z
−
3)
2
−
3
3
(
z
−
3)
3
+
···

=
1
(
z
−
3)
2
−
3
(
z
−
3)
3
+
3
2
(
z
−
3)
4
−
3
3
(
z
−
3)
5
+
···
11.
f
(
z
)=
1
3

1
z
−
3
−
1
z

=
1
3

1
z
−
4+1
−
1
4+
z
−
4

=
1
3



1
z
−
4
·
1
1+
1
z
−
4
−
1
4
·
1
1+
z
−
4
4



=
1
3

1
z
−
4
1
1
−
1
z
−
4
+
1
(
z
−
4)
2
−
1
(
z
−
4)
3
+
···
.
−
1
4
1
1
−
z
−
4
4
+
(
z
−
4)
2
4
2
−
(
z
−
4)
3
4
3
+
···
.9
=
···−
1
3(
z
−
4)
2
+
1
3(
z
−
1)
−
1
12
+
z
−
4
3
·
4
2
−
(
z
−
4)
2
3
·
4
3
+
···
12.
f
(
z
)=
1
3

1
z
−
3
−
1
z

=
1
3

1
−
4+
z
+1
−
1
z
+1
−
1

=
1
3



−
1
4
·
1
1
−
z
+1
4
−
1
z
+1
·
1
1
−
1
z
+1



=
1
3

−
1
4
1
1+
z
+1
4
+
(
z
+1)
2
4
2
+
(
z
+1)
3
4
3
+
···
.
−
1
z
+1
1
1+
1
z
+1
+
1
(
z
+1)
2
+
1
(
z
+1)
3
+
···
.9
=
···−
1
(
z
+1)
2
−
1
z
+1
−
1
12
−
z
+1
3
·
4
2
−
(
z
+1)
2
3
·
4
3
−···
13.
f
(
z
)=
1
z
−
2
−
1
z
−
1
=
−
1
2
·
1
1
−
z
2
−
1
z
·
1
1
−
1
z
=
−
1
2
1
1+
z
2
+
z
2
2
2
+
z
3
2
3
+
···
.
−
1
z
1
1+
1
z
+
1
z
2
+
1
z
3
+
···
.
=
···−
1
z
2
−
1
z
−
1
2
−
z
2
2
−
z
2
2
3
−···
14.
f
(
z
)=
1
z
−
2
−
1
z
−
1
=
1
z
·
1
1
−
2
z
−
1
z
·
1
1
−
1
z
=
1
z
1
1+
2
z
+
2
2
z
2
+
2
3
z
3
+
···
.
−
1
z
1
1+
1
z
+
1
z
2
+
1
z
3
+
···
.
=
1
z
2
+
2
2
−
1
z
3
+
2
3
−
1
z
4
+
2
4
−
1
z
5
+
···
15.
f
(
z
)=
1
z
−
1
·
−
1
1
−
(
z
−
1)
=
−
1
z
−
1
[1+(
z
−
1) + (
z
−
1)
2
+(
z
−
1)
3
+
···
]=
−
1
z
−
1
−
1
−
(
z
−
1)
−
(
z
−
1)
2
−···
16.
f
(
z
)=
1
z
−
2
·
1
1+(
z
−
2)
=
1
z
−
2
[1
−
(
z
−
2)+(
z
−
2)
2
−
(
z
−
2)
3
+
···
]=
1
z
−
2
−
1+(
z
−
2)
−
(
z
−
2)
2
+
···
805

806. Exercises 19.3
17.
f
(
z
)=
1
/
3
z
+1
+
2
/
3
z
−
2
=
1
3(
z
+1)
+
2
3
·
1
−
3+(
z
+1)
=
1
3(
z
+1)
−
2
9
·
1
1
−
z
+1
3
=
1
3(
z
+1)
−
2
9

1+
z
+1
3
+
(
z
+1)
2
3
2
+
(
z
+1)
3
3
3
+
···

=
1
3(
z
+1)
−
2
9
−
2(
z
+1)
3
3
−
2(
z
+1)
2
3
4
−···
18.
f
(
z
)=
1
3(
z
+1)
+
2
3
·
1
(
z
+1)
−
3
=
1
3(
z
+1)
+
2
3(
z
+1)
·
1
1
−
3
z
+1
=
1
3(
z
+1)
+
2
3(
z
+1)
1
1+
3
z
+1
+
3
2
(
z
+1)
2
+
3
3
(
z
+1)
3
+
···
.
=
1
z
+1
+
2
(
z
+1)
2
+
2
·
3
(
z
+1)
3
+
2
·
3
2
(
z
+1)
4
+
···
19.
f
(
z
)=
1
/
3
z
+1
+
2
/
3
z
−
2
=
1
3
z
·
1
1+
1
z
−
1
3
·
1
1
−
z
2
=
1
3
z
1
1
−
1
z
+
1
z
2
−
1
z
3
+
···
.
−
1
3
1
1+
z
2
+
z
2
2
2
+
z
3
2
3
+
···
.
=
···−
1
3
z
2
+
1
3
z
−
1
3
−
z
3
·
2
−
z
2
3
·
2
2
−···
20.
f
(
z
)=
2
/
3
z
−
2
+
1
3
1
3+(
z
−
2)
=
2
/
3
z
−
2
+
1
9
·
1
1+
z
−
2
3
=
2
/
3
z
−
2
+
1
9
1
1+
z
−
2
3
+
(
z
−
2)
2
3
2
+
(
z
−
2)
3
3
3
+
···
.
=
2
3(
z
−
2)
+
1
9
+
z
−
2
3
3
+
(
z
−
2)
2
3
4
+
···
21.
f
(
z
)=
1
z
(1
−
z
)
−
2
=
1
z
1
1+(
−
2)(
−
z
)+
(
−
2)(
−
3)
z
!
(
−
z
)
2
+
(
−
2)(
−
3)(
−
4)
3!
(
−
z
)
3
+
···
.
=
1
z
+2+3
z
+4
z
2
+
···
22.
f
(
z
)=
1
z
3
(1
−
1
z
)
2
=
1
z
3
1
1
−
1
z
.
−
2
=
1
z
3
2
1+(
−
2)
1
−
1
z
.
+
(
−
2)(
−
3)
2!
1
−
1
z
.
2
+
(
−
2)(
−
3)(
−
4)
3!
1
−
1
z
.
3
+
···
3
=
1
z
3
+
2
z
4
+
3
z
5
+
4
z
6
+
···
23.
f
(
z
)=
1
(
z
−
2)[1 + (
z
−
2)]
3
=
1
z
−
2
[1+(
z
−
2)]
−
3
=
1
z
−
2
1
1+(
−
3)(
z
−
2) +
(
−
3)(
−
4)
2!
(
z
−
2)
2
+
(
−
3)(
−
4)(
−
5)
3!
(
z
−
2)
3
+
···
.
=
1
z
−
2
−
3+6(
z
−
2)
−
10(
z
−
2)
2
+
···
24.
f
(
z
)=
1
(
z
−
3)
3
·
−
1
1
−
(
z
−
1)
=
−
1
(
z
−
1)
3
[1+(
z
−
1)+(
z
−
1)
2
+(
z
−
1)
3
+
···
]
=
−
1
(
z
−
1)
3
−
1
(
z
−
1)
2
−
1
z
−
1
−
1
−
(
z
−
1)
−···
25.
f
(
z
)=
3
z
+
4
z
−
1
=
3
z
−
4
·
1
1
−
z
=
3
z
−
4(1 +
z
+
z
2
+
z
3
+
···
)=
3
z
−
4
−
4
z
−
4
z
2
−···
806

807. Exercises 19.4
26.
f
(
z
)=
4
z
−
1
+3
·
1
1+(
z
−
1)
=
4
z
−
1
+3(1
−
(
z
−
1)+(
z
−
1)
2
−
(
z
−
1)
3
+
···
)=
4
z
−
1
+3
−
3(
z
−
1)+3(
z
−
1)
2
−···
27.
f
(
z
)=
z
+
2
z
−
2
=1+(
z
−
1) +
2
−
1+
z
−
1
=1+(
z
−
1) +
2
z
−
1
·
1
1
−
1
z
−
1
=1+(
z
−
1) +
2
z
−
1
k
1+
1
z
−
1
+
1
(
z
−
1)
2
+
1
(
z
−
1)
3
+
···
z
=
···
+
2
(
z
−
1)
2
+
2
z
−
1
+1+(
z
−
1)
28.
f
(
z
)=
z
+
2
z
−
2
=
2
z
−
2
+2+(
z
−
2)
Exercises 19.4
1.
Using
e
2
z
=
∞

k
=0
2
k
z
k
k
!
we obtain
e
2
z
−
1
z
=
k
1+
2
1!
z
+
2
2
2!
z
2
+
2
3
3!
z
3
+
···
z
−
1
z
=
1
z
k
2
1!
z
+
2
2
2!
z
2
+
2
3
3!
z
3
+
···
z
=
2
1!
+
2
2
2!
z
+
2
3
3!
z
2
+
···
.
From the form of the last series we see that
z
= 0 is a removable singularity. Define
f
(0) = 2.
2.
Using sin 4
z
∞

k
=0
(
−
1)
k
(4
z
)
2
k
+1
(2
k
+ 1)!
we obtain
sin 4
z
−
4
z
z
2
=
k
4
1!
z
−
4
3
3!
z
3
+
4
5
5!
z
5
−
4
7
7!
z
7
+
···
z
−
4
z
z
2
=
1
z
2
k
−
4
3
3!
z
3
+
4
5
5!
z
5
−
4
7
7!
z
7
+
···
z
=
−
4
3
3!
z
+
4
5
5!
z
3
−
4
7
7!
z
5
+
···
.
From the form of the last series we see that
z
= 0 is a removable singularity. Define
f
(0) = 0.
3.
Since
f
(
−
2+
i
)=
f
π
(
−
2+
i
) = 0 and
f
ππ
(
z
) = 2 for all
z
,
z
=
−
2+
i
is a zero of order two.
4.
Write
f
(
z
)=
z
4
−
16=(
z
2
−
4)(
z
2
+4)=(
z
−
2)(
z
+ 2)(
z
−
2
i
)(
z
+2
i
) to see that 2,
−
2, 2
i
, and
−
2
i
are
zeros of
f
.Now
f
π
(
z
)=4
z
3
and
f
π
(2)
e
=0,
f
π
(
−
2)
e
=0,
f
π
(2
i
)
e
= 0, and
f
π
(
−
2
i
)
e
= 0. This indicates that each
zero is of order one.
5.
Write
f
(
z
)=
z
2
(
z
2
+1)=
z
2
(
z
−
i
)(
z
+
i
) to see that 0,
i
, and
−
i
are zeros of
f
.Now
f
π
(
z
)=4
z
3
+2
z
and
f
π
(
i
)
e
= 0 and
f
π
(
−
i
)
e
= 0. This indicates that
z
=
i
and
z
=
−
i
are zeros of order one. However
f
π
(0) = 0, but
f
ππ
(0) = 2
e
= 0. Hence
z
= 0 is a zero of order two.
6.
Write
f
(
z
)=(
z
2
+9)
/z
=(
z
−
3
i
)(
z
+3
i
)
/z
to see that 3
i
and
−
3
i
are zeros of
f
.Now
f
π
(
z
)=1
−
9
/z
2
and
f
π
(3
i
)=
f
π
(
−
3
i
)=2
e
= 0. This indicates that each zero is of order one.
7.
Write
f
(
z
)=
e
z
(
e
z
−
1) to see that 2
nπi
,
n
=0,
±
1,
±
2,
...
are zeros of
f
.Now
f
π
(
z
)=2
e
2
z
−
e
z
and
f
π
(2
nπi
)=2
e
4
nπi
−
e
2
nπi
=1
e
= 0. This indicates that each zero is of order one.
8.
The zeros of
f
are the zeros of sin
z
, that is,
nπ
,
n
=0,
±
1,
±
2,
...
. From
f
π
(
z
) = 2 sin
z
cos
z
we see
f
π
(
nπ
)=0.
From
f
ππ
(
z
)=2(
−
sin
2
z
+ cos
2
z
)wesee
f
ππ
(
nπ
)
e
= 0. This indicates that each zero is of order two.
9.
From
f
(
z
)=
z
(1
−
cos
z
2
)=
z
k
−
z
4
2!
+
z
8
4!
−···
z
=
z
5
k
−
1
2!
+
z
4
4!
−···
z
we see that
z
= 0 is a zero of order five.
807

808. Exercises 19.4
10.
From
f
(
z
)=
z
−
sin
z
=
z
3
3!
−
z
5
5!
+
···
=
z
3
k
1
3!
−
z
2
5!
+
···
z
we see that
z
= 0 is a zero of order three.
11.
From
f
(
z
)=1
−
e
z
−
1
=
−
z
−
1
1!
−
(
z
−
1)
2
2!
−···
=(
z
−
1)
k
1
−
z
−
1
2!
−···
z
we see that
z
= 1 is a zero of order one.
12.
From the series
e
z
=
−
∞

k
=0
(
z
−
πi
)
k
k
!
centered at
πi
and
f
(
z
)=1
−
πi
+
z
+
e
z
=1
−
πi
+
z
+
k
−
1
−
z
−
πi
1!
−
(
z
−
πi
)
2
2!
−
(
z
−
πi
)
3
3!
−···
z
=
−
(
z
−
πi
)
2
2!
−
(
z
−
πi
)
3
3!
−···
=(
z
−
πi
)
2
k
−
1
2!
−
z
−
πi
3!
−···
z
we see that
z
=
πi
is a zero of order two.
13.
From
f
(
z
)=
3
z
−
1
[(
z
−
(
−
1+2
i
)][
z
−
(
−
1
−
2
i
)]
and Theorem 19
.
11 we see that
−
1+2
i
and
−
1
−
2
i
are simple poles.
14.
From
f
(
z
)=
5
z
2
−
6
z
2
and Theorem 19
.
11 we see that 0 is a pole of order two.
15.
From
f
(
z
)=
1+4
i
(
z
+ 2)(
z
+
i
)
4
and Theorem 19
.
11 we see that
−
2 is a simple pole and
−
i
is a pole of order four.
16.
From
f
(
z
)=
z
−
1
(
z
+1)
2
F
z
−
(
1
2
+
√
3
2
i
)
xF
z
−
(
1
2
−
√
3
2
i
)
x
and Theorem 19
.
11 we see that
−
1 is a pole of order two and
1
2
+
√
3
2
i
and
1
2
−
√
3
2
i
are simple poles.
17.
Since sin
z
and cos
z
are analytic at
nπ
,
n
=0,
±
1,
±
2,
...
, sin
z
has zeros of order one at
nπ
, and cos
nπ
e
=0,
it follows from Theorem 19
.
11 that the numbers
nπ
,
n
=0,
±
1,
±
2,
...
are simple poles of
f
(
z
) = tan
z
.
18.
From
z
2
sin
πz
=
z
3
k
π
−
π
3
z
2
3!
+
···
z
we see
z
= 0 is a zero of order three. From
f
(
z
)=
cos
πz
z
2
sin
πz
and
Theorem 19
.
11 we see 0 is a pole of order three. The numbers
n
,
n
±
1,
±
2,
...
are simple poles.
19.
From the Laurent series
f
(
z
)=
1
−
cosh
z
z
4
=
1
−
k
1+
z
2
2!
+
z
4
4!
+
z
6
6!
+
···
z
z
4
=
−
1
2!
z
2
−
1
4!
−
z
2
6!
−···
we see that 0 is a pole of order two.
20.
From the Laurent series
f
(
z
)=
e
z
z
2
=
k
1+
z
1!
+
z
2
2!
+
···
z
z
2
=
1
z
2
+
1
z
+
1
2!
+
···
we see that 0 is a pole of order two.
21.
From 1
−
e
z
=1
−
k
1+
z
1!
+
z
2
2!
+
···
z
=
z

−
1
−
z
2!
−···

we see that
z
= 0 is a zero of order one. By
periodicity of
e
z
it follows that
z
=2
nπi
,
n
=0,
±
1,
±
2,
...
are zeros of order one. From
f
(
z
)=
1
1
−
e
z
and
Theorem 19
.
11 we see that the numbers 2
nπi
,
n
=0,
±
1,
±
2,
...
are simple poles.
808

809. Exercises 19.5
22.
z
= 0 is a removable singularity of the function (sin
z
)
/z
. From
f
(
z
)=
sin
z
z
(
z
−
1)
we see that only 1 is a (simple)
pole.
23.
The function
f
(
z
)=
sin(1
/z
)
cos(1
/z
)
fails to be defined at
z
= 0 and at the solutions of cos
1
z
= 0, that is, at
1
z
=(2
n
+1)
π
2
,
n
=0,
±
1,
±
2,
...
. Since
z
=
2
(2
n
+1)
π
,
n
=0,
±
1,
±
2,
...
we see that in any neighborhood
of
z
= 0 there are points at which
f
is not defined and thus not analytic. Hence
z
= 0 is a non-isolated
singularity.
24.
From the Laurent series
f
(
z
)=
z
3
r
1
z
−
1
3!
k
1
z
z
3
+
1
5!
k
1
z
z
5
−
1
7!
k
1
z
7
z
+
···
b
=
z
2
−
1
3!
+
1
5!
z
2
−
1
7!
z
4
+
···
,
0
<
|
z
|
,
we see that the principal part contains an infinite number of nonzero terms. Hence
z
= 0 is an essential
singularity.
Exercises 19.5
1.
f
(
z
)=
2
5(
z
−
1)
·
1
1+
z
−
1
5
=
2
5(
z
−
1)
k
1
−
z
−
1
5
+
(
z
−
1)
2
5
2
−
(
z
−
1)
3
5
3
+
···
z
=
2
/
5
z
−
1
−
2
25
+
2(
z
−
1)
5
3
−
2(
z
−
1)
2
5
4
+
···
Res(
f
(
z
)
,
1) = 2
/
5
2.
f
(
z
)=
1
z
3
(1
−
z
)
−
3
=
1
z
3
k
1+(
−
3)(
−
z
)+
(
−
3)(
−
4)
2!
(
−
z
)
2
+
(
−
3)(
−
4)(
−
5)
3!
(
−
z
)
3
+
···
z
=
1
z
3
+
3
z
2
+
6
z
+10+
···
Res(
f
(
z
)
,
0) = 6
3.
f
(
z
)=
−
3
z
−
1
z
−
2
=
−
3
z
+
1
2
·
1
1
−
z
2
=
−
3
z
+
1
2
k
1+
z
2
+
z
2
2
2
+
z
3
2
3
+
···
z
=
−
3
z
+
1
2
+
z
2
2
+
z
2
2
3
+
···
Res(
f
(
z
)
,
0) =
−
3
4.
f
(
z
)=(
z
+3)
2
k
2
z
+3
−
2
3
3!(
z
+3)
3
+
2
5
5!(
z
+3)
5
+
···
z
=
···
+
2
5
5!(
z
+3)
3
−
2
3
3!(
z
+3)
+2(
z
+3)
Res(
f
(
z
)
,
−
3) =
−
4
3
5.
f
(
z
)=
e
−
2
/z
2
=
∞

k
=0
(
−
2
/z
2
)
k
k
!
=
···−
2
3
3!
z
6
+
2
2
2!
z
4
−
2
1!
z
2
+ 1; Res(
f
(
z
)
,
0) = 0
6.
f
(
z
)=
e
−
2
(
z
−
2)
2
e
−
(
z
−
2)
=
e
−
2
(
z
−
2)
2
k
1
−
z
−
2
1!
+
(
z
−
2)
2
2!
−
(
z
−
2)
3
3!
+
···
z
=
e
−
2
(
z
−
2)
2
−
e
−
2
z
−
2
+
e
−
2
2
−
e
−
2
(
z
−
2)
3!
+
···
Res(
f
(
z
)
,
2) =
−
e
−
2
809

810. Exercises 19.5
7.
Res(
f
(
z
)
,
4
i
) = lim
z
→
4
i
(
z
−
4
i
)
·
z
(
z
−
4
i
)(
z
+4
i
)
= lim
z
→
4
i
z
z
+4
i
=
1
2
Res(
f
(
z
)
,
−
4
i
) = lim
z
→−
4
i
(
z
+4
i
)
·
z
(
z
−
4
i
)(
z
+4
i
)
= lim
z
→−
4
i
z
z
−
4
i
=
1
2
8.
Res(
f
(
z
)
,
1
/
2) = lim
z
→
1
/
2
(
z
−
1
/
2)
4
z
+8
2(
z
−
1
/
2)
= lim
z
→
1
/
2
(2
z
+4)=5
9.
Res(
f
(
z
)
,
1) = lim
z
→
1
(
z
−
1)
1
z
2
(
z
+ 2)(
z
−
1)
= lim
z
→
1
1
z
2
(
z
+2)
=
1
3
Res(
f
(
z
)
,
−
2) = lim
z
→−
2
(
z
+2)
1
z
2
(
z
+ 2)(
z
−
1)
= lim
z
→−
2
1
z
2
(
z
−
1)
=
−
1
12
Res(
F
(
z
)
,
0) =
1
1!
lim
z
→
0
d
dz

z
2
·
1
z
2
(
z
+ 2)(
z
−
1)

= lim
z
→
0
−
2
z
−
1
(
z
+2)
2
(
z
−
1)
2
=
−
1
4
10.
Res(
f
(
z
)
,
1+
i
)=
1
1!
lim
z
→
1+
i
d
dz

(
z
−
1
−
i
)
2
·
1
(
z
−
1
−
i
)
2
(
z
−
1+1)
2

= lim
z
→
1+
i
−
2
(
z
−
1+
i
)
3
=
−
1
4
i
Res(
f
(
z
)
,
1
−
i
)=
1
1!
lim
z
→
1
−
i
d
dz

(
z
−
1+
i
)
2
·
1
(
z
−
1
−
i
)
2
(
z
−
1+
i
)
2

= lim
z
→
1
−
i
−
2
(
z
−
1
−
i
)
3
=
1
4
i
11.
Res(
f
(
z
)
,
−
1) = lim
z
→−
1
(
z
+1)
·
5
z
2
−
4
z
+3
(
z
+ 1)(
z
+ 2)(
z
+3)
= lim
z
→−
1
5
z
2
−
4
z
+3
(
z
+ 2)(
z
+3)
=6
Res(
f
(
z
)
,
−
2) = lim
z
→−
2
(
z
+2)
·
5
z
2
−
4
z
+3
(
z
+ 1)(
z
+ 2)(
z
+3)
= lim
z
→−
2
5
z
2
−
4
z
+3
(
z
+ 1)(
z
+3)
=
−
31
Res(
f
(
z
)
,
−
3) = lim
z
→−
3
(
z
+3)
·
5
z
2
−
4
z
+3
(
z
+ 1)(
z
+ 2)(
z
+3)
= lim
z
→−
3
5
z
2
−
4
z
+3
(
z
+ 1)(
z
+2)
=30
12.
Res(
f
(
z
)
,
−
3) = lim
z
→−
3
(
z
+3)
·
2
z
−
1
(
z
−
1)
4
(
z
+3)
= lim
z
→−
3
2
z
−
1
(
z
−
1)
4
=
−
7
256
Res(
f
(
z
)
,
1) =
1
3!
lim
z
→
1
d
3
dz
3

(
z
−
1)
4
·
2
z
−
1
(
z
−
1)
4
(
z
+3)

=
1
6
lim
z
→
1
−
42
(
z
+3)
4
=
−
7
256
13.
Res(
f
(
z
)
,
0) =
1
1!
lim
z
→
0
d
dz

z
2
·
cos
z
z
2
(
z
−
π
)
3

= lim
z
→
0
−
(
z
−
π
) sin
z
−
3 cos
z
(
z
−
π
)
4
=
−
3
π
4
Res(
f
(
z
)
,π
)=
1
2!
lim
z
→
π
d
2
dz
2

(
z
−
π
)
3
·
cos
z
z
2
(
z
−
π
)
3

=
1
2
lim
z
→
π
−
z
2
cos
z
+4
z
sin
z
+ 6 cos
z
z
4
=
π
2
−
6
2
π
4
14.
Using
d
dz
(
e
z
−
1) =
e
z
and the result in (4),
Res(
f
(
z
)
,
2
nπi
)=
e
z
e
z
/
/
/
/
z
=2
nπi
=1
.
15.
Using
d
dz
cos
z
=
−
sin
z
and the result in (4),
Res

f
(
z
)
,
(2
n
+1)
π
2

=
1
−
sin
z
/
/
/
/
z
=(2
n
+1)
π
2
=
1
−
sin(2
n
+1)
π
2
=(
−
1)
n
+1
.
16.
z
= 0 is a pole of order two. Thus by (2) and L’Hˆ
opital’s rule,
Res(
f
(
z
)
,
0) =
1
1!
lim
z
→
0
d
dz

z
2
·
1
z
sin
z

= lim
z
→
0
sin
z
−
z
cos
z
sin
2
z
= lim
z
→
0
cos
z
+
z
sin
z
−
cos
z
2 sin
z
cos
z
= lim
z
→
0
z
2 cos
z
=0
.
810

811. Exercises 19.5
For the simple poles at
z
=
nπ
,
n
=
±
1,
±
2,
...
we have from (4),
Res(
f
(
z
)
,nπ
)=
1
z
cos
z
+ sin
z
/
/
/
/
z
=
nπ
=
(
−
1)
n
nπ
.
17. (a)
y
ˇ
C
1
(
z
−
1)(
z
+2)
2
dz
= 0 by Theorem 18
.
4.
(b)
y
ˇ
C
1
(
z
−
1)(
z
+2)
2
dz
=2
πi
Res(
f
(
z
)
,
1) =
2
π
9
i
(c)
y
ˇ
C
1
(
z
−
1)(
z
+2)
2
dz
=2
πi
[Res(
f
(
z
)
,
1) + Res(
f
(
z
)
,
−
2)] = 2
πi

1
9
+
k
−
1
9
zr
=0
18. (a)
y
ˇ
C
z
+1
z
2
(
z
−
2
i
)
dz
=2
πi
Res(
f
(
z
)
,
0) =
π
k
−
1+
1
2
i
z
(b)
y
ˇ
C
z
+1
z
2
(
z
−
2
i
)
dz
=2
πi
Res(
f
(
z
)
,
2
i
)=
π
k
1
−
1
2
i
z
(c)
y
ˇ
C
z
+1
z
2
(
z
−
2
i
)
dz
=2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
2
i
)] = 2
πi

1
4
+
1
2
i
+
k
−
1
4
−
1
2
i
zr
=0
19. (a)
From the Laurent series
z
3
e
−
1
/z
2
=
···−
1
3!
z
3
+
1
2!
z
−
z
+
z
3
we see Res(
f
(
z
)
,
0) = 1
/
2. Hence
y
ˇ
C
z
3
e
−
1
/z
2
dz
=2
πi
Res(
f
(
z
)
,
0) =
πi.
(b)
y
ˇ
C
z
3
e
−
1
/z
2
dz
=2
πi
Res(
f
(
z
)
,
0) =
πi
(c)
y
ˇ
C
z
3
e
1
/z
2
dz
= 0 by Theorem 18
.
4.
20. (a)
y
ˇ
C
1
z
sin
z
dz
= 0 by Theorem 18
.
4.
(b)
z
= 0 is a pole of order two (see Problem 16). Thus
y
ˇ
C
1
z
sin
z
dz
=2
πi
Res(
f
(
z
)
,
0) = 2
πi
(0) = 0
.
(c)
y
ˇ
C
1
z
sin
z
dz
=2
πi
[Res(
f
(
z
)
,
−
π
) + Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,π
)] = 2
πi

1
π
+0+
k
−
1
π
zr
=0
21.
y
ˇ
C
1
z
2
+4
z
+13
dz
=2
πi
Res(
f
(
z
)
,
−
2+3
i
)=
π
3
22.
y
ˇ
C
1
z
3
(
z
−
1)
4
dz
=2
πi
Res(
f
(
z
)
,
1) =
−
20
πi
23.
y
ˇ
C
z
z
4
−
1
dz
=2
πi
[Res(
f
(
z
)
,
−
1) + Res(
f
(
z
)
,
1) + Res(
f
(
z
)
,
−
i
) + Res(
f
(
z
)
,i
)] = 2
πi

1
4
+
1
4
−
1
4
−
1
4

=0
24.
y
ˇ
C
z
(
z
+ 1)(
z
2
+1)
dz
=2
πi
[Res(
f
(
z
)
,i
) + Res(
f
(
z
)
,
−
i
)] = 2
πi

1
4
−
1
4
i
+
1
4
+
1
4
i

=
πi
25.
y
ˇ
C
ze
z
z
2
−
1
dz
=2
πi
[Res(
f
(
z
)
,
1) + Res(
f
(
z
)
,
−
1)] = 2
πi

e
2
+
e
−
1
2

=2
πi
cosh 1
811

812. Exercises 19.5
26.
y
ˇ
C
e
z
z
3
+2
z
2
dz
=2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
−
2)] = 2
πi

1
2
+
e
−
2
4

=
πi
k
1+
1
2
e
−
2
z
27.
y
ˇ
C
tan
z
z
dz
=2
πi
Res

f
(
z
)
,
π
2

)=
−
4
i
. Note:
z
= 0 is not a pole. See Example 1, Section 19
.
4.
28.
y
ˇ
C
cot
πz
z
2
dz
=2
πi
Res(
f
(
z
)
,
0) = 2
πi

−
π
3

=
−
2
π
2
3
i
Note:
z
= 0 is a pole of order three. Use L’Hˆ
opital’s rule (or
Mathematica
) to show that
Res(
f
(
z
)
,
0) =
1
2
lim
z
→
0
d
2
dz
2
z
cot
πz
=
1
2
lim
z
→
0
[
−
2
π
csc
2
πz
+2
π
2
z
cot
πz
csc
2
πz
]=
1
2
k
−
2
π
3
z
=
−
π
3
.
29.
y
ˇ
C
cot
πz dz
=2
πi
[Res(
f
(
z
)
,
1) + Res(
f
(
z
)
,
2) + Res(
f
(
z
)
,
3)] = 2
πi

1
π
+
1
π
+
1
π

=6
i
30.
y
ˇ
C
2
z
−
1
z
2
(
z
3
+1)
dz
=2
πi
r
Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
−
1) + Res
π
f
(
z
)
,
1
2
+
√
3
2
i
zb
=2
πi

2+(
−
1) +
k
−
1
2
−
1
6
√
3
i
zr
=
π
π
√
3
3
+
i
i
31.
y
ˇ
C
e
iz
+ sin
z
(
z
−
π
)
4
dz
=2
πi
Res(
f
(
z
)
,π
)=
π
k
−
1
3
+
1
3
i
z
32.
y
ˇ
C
cos
z
(
z
−
1)
2
(
z
2
+9)
dz
=2
πi
Res(
f
(
z
)
,
1) = 2
πi
(
−
0
.
02 cos 1
−
0
.
1 sin 1) =
−
0
.
5966
i
Exercises 19.6
1.
α
2
π
0
dθ
1+
1
2
sin
θ
=
y
ˇ
C
4
z
2
+4
iz
−
1
dz
= (4)2
πi
Res(
f
(
z
)
,
(
√
3
−
2)
i
)=
4
π
√
3
2.
α
2
π
0
dθ
10
−
6 cos
θ
=
1
2
·
k
−
2
i
z
y
ˇ
C
dz
(3
z
−
1)(
z
−
3)
=(
i
)2
πi
Res
k
f
(
z
)
,
1
3
z
=
π
4
3.
α
2
π
0
cos
θ
3 + sin
θ
dθ
=
y
ˇ
C
z
2
+1
z
(
z
2
+6
iz
−
1)
dz
=2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
−
3+2
√
2
i
)] = 0
4.
α
2
π
0
1
1+3cos
2
θ
dθ
=
4
i
y
ˇ
C
z
3
z
4
+10
z
2
+3
dz
=
k
4
i
z
2
πi
r
Res
π
f
(
z
)
,
(
√
3
3
i
i
+ Res
π
f
(
z
)
,
−
√
3
3
i
zb
=
π
5.
α
π
0
dθ
2
−
cos
θ
=
1
2
α
2
π
0
dθ
2
−
cos
θ
=
−
1
i
y
ˇ
C
dz
z
2
−
4
z
+1
=
k
−
1
i
z
2
πi
Res(
f
(
z
)
,
2
−
√
3)=
π
√
3
6.
α
π
0
dθ
1 + sin
2
θ
=
1
2
α
2
π
0
dθ
1 + sin
2
θ
=
−
2
i
y
ˇ
C
z
z
4
−
6
z
2
+1
dz
=
k
−
2
i
z
2
πi
[Res(
f
(
z
)
,
U
3
−
2
√
2 ) + Res(
f
(
z
)
,
−
U
3
−
2
√
2)]=
π
√
2
7.
α
2
π
0
sin
2
θ
5+4cos
θ
dθ
=
−
1
4
i
y
ˇ
C
(
z
2
−
1)
2
z
2
(2
z
2
+5
z
+2)
dz
=
k
−
1
4
i
z
2
πi

Res(
f
(
z
)
,
0) + Res
k
f
(
z
)
,
−
1
2
zr
=
π
4
8.
α
2
π
0
cos
2
θ
3
−
sin
θ
dθ
=
1
2
i
y
ˇ
C
z
4
+2
z
2
+1
z
2
(
iz
2
+6
z
−
i
)
dz
=
k
1
2
i
z
2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
3
−
2
√
2)
i
)] =
π
[6
−
4
√
2]
812

813. Exercises 19.6
9.
We use cos 2
θ
=(
z
2
+
z
−
2
)
/
2.
α
2
π
0
cos 2
θ
5
−
4 cos
θ
dθ
=
i
2
y
ˇ
C
z
4
+1
z
2
(2
z
2
−
5
z
+2)
dz
=
k
i
2
z
2
πi

Res(
f
(
z
)
,
0) + Res
k
f
(
z
)
,
1
2
zr
=
π
6
10.
α
2
π
0
1
cos
θ
+ 2 sin
θ
+3
dθ
=
2
i
y
ˇ
C
1
(1
−
2
i
)
z
2
+6
z
+1+2
i
=
k
2
i
z
2
πi
Res
k
f
(
z
)
,
−
1
5
−
2
5
i
z
=
π
11.
α
∞
−∞
1
x
2
−
2
x
+2
dx
=2
πi
Res(
f
(
z
)
,
1+
i
)=
π
12.
α
∞
−∞
1
x
2
−
2
x
+25
dx
=2
πi
Res(
f
(
z
)
,
1+2
√
6
i
)=
π
2
√
6
13.
α
∞
−∞
1
(
x
2
+4)
2
dx
=2
πi
Res(
f
(
z
)
,
2
i
)=
π
16
14.
α
∞
−∞
x
2
(
x
2
+1)
2
dx
=2
πi
Res(
f
(
z
)
,i
)=
π
2
15.
α
∞
−∞
1
(
x
2
+1)
3
dx
=2
πi
Res(
f
(
z
)
,i
)=
3
π
8
16.
α
∞
−∞
x
(
x
2
+4)
3
dx
= 0 (The integrand is an odd function)
17.
α
∞
−∞
2
x
2
−
1
x
4
+5
x
2
+4
dx
=2
πi
[Res(
f
(
z
)
,i
) + Res(
f
(
z
)
,
2
i
)] =
π
2
18.
α
∞
−∞
dx
(
x
2
+1)
2
(
x
2
+9)
=2
πi
[Res(
f
(
z
)
,i
) + Res(
f
(
z
)
,
3
i
)] =
5
π
96
19.
α
∞
0
x
2
+1
x
4
+1
dx
=
1
2
α
∞
−∞
x
2
+1
x
4
+1
dx
=
πi
Res(
f
(
z
)
,i
)=
π
√
2
20.
α
∞
0
1
x
6
+1
dx
=
1
2
α
∞
−∞
1
x
6
+1
dx
=
πi
r
Res
π
f
(
z
)
,
√
3
2
+
1
2
i
i
+ Res(
f
(
z
)
,i
) + Res
π
f
(
z
)
,
−
√
3
2
+
1
2
i
zb
=
π
3
21.
α
∞
−∞
e
ix
x
2
+1
dx
=2
πi
Res(
f
(
z
)
,i
)=
πe
−
1
. Therefore,
α
∞
−∞
cos
x
x
2
+1
dx
=Re
k
α
∞
−∞
e
ix
x
2
+1
dx
z
=
πe
−
1
.
22.
α
∞
−∞
e
2
ix
x
2
+1
dx
=2
πi
Res(
f
(
z
)
,i
)=
πe
−
2
. Therefore,
α
∞
−∞
cos 2
x
x
2
+1
dx
=Re
k
α
∞
−∞
e
2
ix
x
2
+1
dx
z
=
πe
−
2
.
23.
α
∞
−∞
xe
ix
x
2
+1
dx
=2
πi
Res(
f
(
z
)
,i
)=
πe
−
1
i
. Therefore,
α
∞
−∞
x
sin
x
x
2
+1
dx
=Im
k
α
∞
−∞
xe
ix
x
2
+1
dx
z
=
πe
−
1
.
24.
α
∞
−∞
e
ix
(
x
2
+4)
2
dx
=2
πi
Res(
f
(
z
)
,
2
i
)=
3
e
−
2
16
π
;
α
∞
−∞
cos
x
(
x
2
+4)
2
dx
=Re
k
α
∞
−∞
e
ix
(
x
2
+4)
2
dx
z
=
3
e
−
2
16
π
.
Therefore
α
∞
0
cos
x
(
x
2
+4)
2
dx
=
1
2
k
3
e
−
2
16
π
z
=
3
e
−
2
32
π.
25.
α
∞
−∞
e
3
ix
(
x
2
+1)
2
dx
=2
πi
Res(
f
(
z
)
,i
)=2
πe
−
3
;
α
∞
−∞
cos 3
x
(
x
2
+1)
2
dx
=Re
k
α
∞
−∞
e
3
ix
(
x
2
+1)
2
dx
z
=2
πe
−
3
.
Therefore
α
∞
0
cos 3
x
(
x
2
+1)
2
dx
=
1
2
(2
πe
−
3
)=
πe
−
3
.
813

814. Exercises 19.6
26.
α
∞
−∞
e
ix
x
2
+4
x
+5
dx
=2
πi
Res(
f
(
z
)
,
−
2+
i
)=
πe
−
1
−
2
i
. Therefore
α
∞
−∞
sin
x
x
2
+4
x
+5
dx
=Im
k
α
∞
−∞
e
ix
x
2
+4
x
+5
dx
z
=
−
πe
−
1
sin 2
27.
α
∞
−∞
e
2
ix
x
4
+1
dx
=2
πi

Res
k
f
(
z
)
,
1
√
2
+
1
√
2
i
z
+ Res
k
f
(
z
)
,
−
1
√
2
+
1
√
2
i
zr
=2
πi
rπ
−
√
2
8
−
√
2
8
i
i
e
(
−
√
2+
√
2
i
)
+
π
√
2
8
−
√
2
8
i
i
e
(
−
√
2
−
√
2
i
)
b
=
πe
−
√
2
r
√
2
2
cos
√
2+
√
2
2
sin
√
2
b
α
∞
−∞
cos 2
x
x
4
+1
dx
=Re
k
α
∞
−∞
e
2
ix
x
4
+1
dx
z
=
πe
−
√
2
r
√
2
2
cos
√
2+
√
2
2
sin
√
2
b
Therefore
α
∞
0
cos 2
x
x
4
+1
dx
=
πe
−
√
2
√
2
4
(cos
√
2 + sin
√
2)
.
28.
α
∞
−∞
xe
ix
x
4
+1
dx
=2
πi

Res
k
f
(
z
)
,
1
√
2
+
1
√
2
i
z
+ Res
k
f
(
z
)
,
−
1
√
2
+
1
√
2
i
zr
=2
πi

−
i
4
e
(
−
1
/
√
2+
i/
√
2)
+
i
4
e
(
−
1
/
√
2
−
i/
√
2)

=
k
πe
−
1
/
√
2
sin
1
√
2
z
i
α
∞
−∞
x
sin
x
x
4
+1
dx
=Im
k
α
∞
−∞
xe
ix
x
4
+1
dx
z
=
πe
−
1
/
√
2
sin
1
√
2
Therefore
α
∞
0
x
sin
x
x
4
+1
dx
=
π
2
e
−
1
/
√
2
sin
1
√
2
.
29.
α
∞
−∞
e
ix
(
x
2
+ 1)(
x
2
+9)
dx
=2
πi
[Res(
f
(
z
)
,i
) + Res(
f
(
z
)
,
3
i
)] = 2
πi

−
i
16
e
−
1
+
i
48
e
−
3

=
1
8
e
−
1
−
1
24
e
−
3
30.
α
∞
−∞
xe
ix
(
x
2
+ 1)(
x
2
+4)
dx
=2
πi
[Res(
f
(
z
)
,i
) + Res(
f
(
z
)
,
2
i
)] = 2
πi

1
6
e
−
1
−
1
6
e
−
2

=
π
3
(
e
−
1
−
e
−
2
)
i
;
α
∞
−∞
x
sin
x
(
x
2
+ 1)(
x
2
+4)
dx
=Im
k
α
∞
−∞
xe
ix
(
x
2
+ 1)(
x
2
+4)
dx
z
=
π
3
(
e
−
1
−
e
−
2
).
Therefore
α
∞
0
x
sin
x
(
x
2
+ 1)(
x
2
+4)
dx
=
1
2
F
π
3
(
e
−
1
−
e
−
2
)
x
=
π
6
(
e
−
1
−
e
−
2
)
.
31.
Consider the contour integral
y
ˇ
C
e
iz
z
dz
. The function
f
(
z
)=
1
z
has a simple pole at
z
= 0. If we use the
contour
C
shown in Figure 19
.
14, it follows from the Cauchy-Goursat Theorem that
y
ˇ
C
=
α
C
R
+
α
−
r
−
R
+
α
−
C
r
+
α
R
r
=0
.
Taking limits as
R
→∞
and as
r
→
0 and using Theorem 19
.
17 we then find
P
.
V
.
α
∞
−∞
e
ix
x
dx
−
πi
Res(
f
(
z
)
e
iz
,
0) = 0 or P
.
V
.
α
∞
−∞
e
ix
x
dx
=
πi.
814

815. Exercises 19.6
Equating the imaginary parts of
α
∞
−∞
cos
x
+
i
sin
x
x
dx
=0+
πi
gives
α
∞
−∞
sin
x
x
dx
=
π.
32.
Consider the contour integral
y
ˇ
C
e
iz
z
(
z
2
+1)
dz
. The function
f
(
z
)=
1
z
(
z
2
+1)
has simple poles at
z
= 0 and
at
z
=
i
. If we use the contour
C
shown in Figure 19
.
14, it follows from Theorem 19
.
14 that
y
ˇ
C
=
α
C
R
+
α
−
r
−
R
+
α
−
C
r
+
α
R
r
=2
πi
Res(
f
(
z
)
,i
)
.
Taking limits as
R
→∞
and as
r
→
0 and using Theorem 19
.
17 we then find
P
.
V
.
α
∞
−∞
e
ix
x
(
x
2
+1)
dx
−
πi
Res(
f
(
z
)
e
iz
,
0) = 2
πi
Res(
f
(
z
)
e
iz
,i
)
or
P
.
V
.
α
∞
−∞
e
ix
x
(
x
2
+1)
dx
=
πi
+2
πi
k
−
e
−
1
2
z
.
Equating the imaginary parts of
α
∞
−∞
cos
x
+
i
sin
x
x
(
x
2
+1)
dx
=0+
π
(1
−
e
−
1
)
i
gives
α
∞
−∞
sin
x
x
(
x
2
+1)
dx
=
π
(1
−
e
−
1
)
.
33.
α
π
0
dθ
(
a
+ cos
θ
)
2
=
1
2
α
2
π
0
dθ
(
a
+ cos
θ
)
2
=
2
i
y
ˇ
C
z
(
z
2
+2
az
+1)
2
dz
(
C
is
|
z
|
=1) =
2
i
y
ˇ
C
z
(
z
−
r
1
)
2
(
z
−
r
2
)
2
dz
where
r
1
=
−
a
+
√
a
2
−
1,
r
2
=
−
a
−
√
a
2
−
1 . Now
y
ˇ
C
z
(
z
−
r
1
)
2
(
z
−
r
2
)
2
dz
=2
πi
Res(
f
(
z
)
,r
1
)=2
πi
a
4(
√
a
2
−
1)
3
=
aπ
2(
√
a
2
−
1)
3
i.
Thus
α
π
0
dθ
(
a
+ cos
θ
)
2
=
2
i
·
aπ
2(
√
a
2
−
1)
3
i
=
aπ
(
√
a
2
−
1)
3
.
When
a
= 2 we obtain
α
π
0
dθ
(2 + cos
θ
)
2
=
2
π
(
√
3)
3
and so
α
2
π
0
dθ
(2 + cos
θ
)
2
=
4
π
3
√
3
.
34.
α
2
π
0
sin
2
θ
a
+
b
cos
θ
dθ
=
i
2
b
y
ˇ
C
z
2
−
1
z
2
(
z
−
r
1
)(
z
−
r
2
)
dz
(
C
is
|
z
|
= 1) where
r
1
=(
−
a
+
√
a
2
−
b
2
)
/b
,
r
2
=(
−
a
−
√
a
2
−
b
2
)
/b
.Now
y
ˇ
C
z
2
−
1
z
2
(
z
−
r
1
)(
z
−
r
2
)
dz
=2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,r
1
)] = 2
πi
r
−
2
a
b
+
2
√
a
2
−
b
2
b
b
.
Thus
α
2
π
0
sin
2
θ
a
+
b
cos
θ
dθ
=
2
π
b
2
(
a
−
e
a
2
−
b
2
)
, a>b>
0
.
815

816. Exercises 19.6
When
a
=5,
b
= 4 we obtain
α
2
π
0
sin
2
θ
5+4cos
θ
dθ
=
2
π
16
(5
−
√
9)=
π
4
.
35.
Consider the contour integral
y
ˇ
C
e
az
1+
e
z
dz
. The function
f
(
z
)=
e
az
1+
e
z
has simple poles at
z
=
πi
,3
πi
,
5
πi
,
...
in the upper plane. Using the contour in Figure 19
.
15 we have from Theorem 19
.
14
y
ˇ
C
=
α
r
−
r
+
α
C
2
+
α
C
3
+
α
C
4
=2
πi
Res(
f
(
z
)
,πi
)=
−
2
πie
aπi
.
On
C
2
,
z
=
r
+
iy
,0
≤
y
≤
π
,
dz
=
idy
,
/
/
/
/
α
C
2
e
az
1+
e
z
dz
/
/
/
/
≤
e
ar
e
r
−
1
(2
π
)
.
Because 0
<a<
1, this last expression goes to 0 as
r
→∞
.On
C
3
,
z
=
x
+2
πi
,
−
r
≤
x
≤
r
,
dz
=
dx
,
α
C
3
e
az
1+
e
z
dz
=
α
−
r
r
e
a
(
x
+2
πi
)
1+
e
x
+2
πi
dx
=
−
e
2
aπi
α
r
−
r
e
ax
1+
e
x
dx.
On
C
4
,
z
=
−
r
+
iy
,0
≤
y
≤
2
π
,
dz
=
idy
,
/
/
/
/
α
C
4
e
az
1+
e
z
dz
/
/
/
/
≤
e
−
ar
1
−
e
−
r
(2
π
)
.
Because 0
<a
, this last expression goes to 0 as
r
→∞
. Hence
α
r
−
r
e
ax
1+
e
x
dx
−
e
2
aπi
α
r
−
r
e
ax
1+
e
x
dx
=
−
2
πie
aπi
gives, as
r
→∞
,
(1
−
e
2
aπi
)
α
∞
−∞
e
ax
1+
e
x
dx
=
−
2
πie
aπi
.
That is
α
∞
−∞
e
ax
1+
e
x
dx
=
−
2
πie
aπi
1
−
e
2
aπi
=
π
e
aπi
−
e
−
aπi
2
i
=
π
sin
ax
.
36.
Using the Fourier sine transform with respect to
y
the partial differential equation becomes
d
2
U
dx
2
−
α
2
U
=0
and so
U
(
x, α
)=
c
1
cosh
αx
+
c
2
sinh
αx.
The boundary condition
u
(0
,y
) becomes
U
(0
,α
) = 0 and so
c
1
=0. Thus
U
(
x, α
)=
c
2
sinh
αx
. Now to evaluate
U
(
π, α
)=
α
∞
0
2
y
y
4
+4
sin
αy dy
=
α
∞
−∞
y
y
4
+4
sin
αy dy
we use the contour integral
α
C
ze
iαz
z
4
+4
dz
and
α
∞
−∞
xe
iαx
x
4
+4
dx
=2
πi
[Res(
f
(
z
)
,
1+
i
) + Res(
f
(
z
)
,
−
1+
i
)] = 2
πi

−
1
8
ie
(
−
1+
i
)
α
+
1
8
ie
(
−
1
−
i
)
α

=
π
2
(
e
−
α
sin
α
)
i
α
∞
−∞
x
sin
αx
x
4
+4
dx
=Im
k
α
∞
−∞
xe
iαx
x
4
+4
dx
z
=
π
2
e
−
α
sin
α.
816

817. Chapter 19 Review Exercises
Finally,
U
(
π, α
)=
π
2
e
−
α
sin
α
=
c
2
sinh
απ
gives
c
2
=
π
2
e
−
α
sin
α
sinh
απ
. Hence
U
(
x, α
)=
π
2
e
−
α
sin
α
sinh
απ
sinh
αx
and
u
(
x, y
)=
α
∞
0
e
−
α
sin
α
sinh
απ
sinh
αx
sin
αy dα.
Chapter 19 Review Exercises
1.
True
2.
False
3.
False
4.
True
5.
True
6.
True
7.
True
8.
five
9.
1
/π
10.
three;
−
1
/
6
11.
|
z
−
i
|
=
√
5
12.
False
13.
e
z
(1+
i
)
+
e
z
(1
−
i
)
2
=
1
2
k
1+
z
(1 +
i
)+
z
2
2!
(1 +
i
)
2
+
···
z
+
1
2
k
1+
z
(1
−
i
)+
z
2
2!
(1
−
i
)
2
+
···
z
=1+
z

(1 +
i
)+(1
−
i
)
2

+
z
2
2!

(1 +
i
)
2
+(1
−
i
)
2
2

+
···
=1+
∞

k
=1
(
√
2)
k
cos
kπ
4
k
!
z
k
Here we have used (1 +
i
)
n
=(
√
2)
n
e
nπi/
4
and (1
−
i
)
n
=(
√
2)
n
e
−
nπi/
4
so that
(1 +
i
)
n
+(1
−
i
)
n
2
=(
√
2)
n

e
nπi/
4
+
e
−
nπi/
4
2

=(
√
2)
n
cos
nπ
4
.
14.
sin
π
z
= 0 implies
z
=
1
n
,
n
=
±
1,
±
2,
...
. All singularities are isolated except the singularity
z
=0.
15.
f
(
z
)=
1
z
4

1
−
k
1+
iz
1!
+
i
2
z
2
2!
+
i
3
z
3
3!
+
i
4
z
4
4!
+
···
zr
=
−
i
z
3
+
1
2!
z
2
+
i
3!
z
−
1
4!
−
iz
5!
+
···
16.
e
z/
(
z
−
2)
=
e
·
e
2
/
(
z
−
2)
=
e
k
1+
2
z
−
2
+
2
2
2!(
z
−
2)
2
+
2
3
3!(
z
−
2)
3
+
···
z
=
e
∞

k
=0
2
k
k
!
(
z
−
2)
−
k
17.
(
z
−
i
)
2
sin
1
z
−
i
=(
z
−
i
)
2

1
z
−
i
−
1
3!(
z
−
i
)
3
+
1
5!(
z
−
i
)
5
−···

=
···
+
1
5!(
z
−
i
)
3
−
1
3!(
z
−
i
)
+(
z
−
i
)
18.
1
−
cos
z
2
z
5
=
1
z
5

1
−
k
1
−
z
4
2!
+
z
8
4!
−
z
12
6!
+
z
16
8!
−···
zr
=
1
2!
z
−
z
3
4!
+
z
7
6!
−
z
11
8!
+
···
19. (a)
f
(
z
)=
1
z
−
3
−
1
z
−
1
=
1
1
−
z
−
1
3
·
1
1
−
z
3
=(1+
z
+
z
2
+
z
4
+
···
)
−
1
3
k
1+
z
3
+
z
2
3
2
+
z
3
3
3
+
···
z
=
2
3
+
8
9
z
+
26
27
z
2
(b)
f
(
z
)=
−
1
z
·
1
1
−
1
z
−
1
3
·
1
1
−
z
3
=
−
1
z
k
1+
1
z
+
1
z
2
+
1
z
3
+
···
z
−
1
3
k
1+
z
3
+
z
2
3
2
+
z
3
3
3
+
···
z
=
···−
1
z
3
−
1
z
2
−
1
z
−
1
3
−
z
3
2
−
z
2
3
3
−···
(c)
f
(
z
)=
−
1
z
·
1
1
−
1
z
+
1
z
·
1
1
−
3
z
=
−
1
z
k
1+
1
z
+
1
z
2
+
1
z
3
+
···
z
+
1
z
k
1+
3
z
+
3
2
z
2
+
3
3
z
3
+
···
z
=
2
z
2
+
8
z
3
+
26
z
4
+
···
817

818. Chapter 19 Review Exercises
(d)
f
(
z
)=
−
1
z
−
1
−
1
2
·
1
1
−
z
−
1
2
=
−
1
z
−
1
−
1
2
k
1+
z
−
1
2
+
(
z
−
1)
2
2
2
+
(
z
−
1)
3
3!
+
···
z
=
−
1
z
−
1
−
1
2
−
z
−
1
2
2
−
(
z
−
1)
2
2
3
−···
20. (a)
f
(
z
)=
1
25

1
−
z
5

−
2
=
1
25

1+(
−
2)

−
z
5

+
(
−
2)(
−
3)
2!

−
z
5

2
+
(
−
2)(
−
3)(
−
4)
3!

−
z
5

3
+
···

=
1
25
+2
z
5
3
+3
z
2
5
4
+4
z
3
5
5
+
···
(b)
(
z
−
5)
−
2
=
1
z
2
k
1
−
5
z
z
−
2
=
1
z
2
r
1+(
−
2)
k
−
5
z
z
+
(
−
2)(
−
3)
2!
k
−
5
z
z
2
+
(
−
2)(
−
3)(
−
4)
3!
k
−
5
z
z
3
+
···
b
=
1
z
2
+2
5
z
3
+3
5
2
z
4
+4
5
3
z
5
+
···
(c)
1
(
z
−
5)
2
is the Laurent series.
21.
y
ˇ
C
2
z
+5
z
(
z
+ 2)(
z
−
1)
4
dz
=2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
−
2)] =
404
81
πi
22.
y
ˇ
C
z
2
(
z
−
1)
3
(
z
2
+4)
dx
=2
πi
Res(
f
(
z
)
,
1) =
8
π
125
i
23.
y
ˇ
C
1
2 sin
z
−
1
dz
=2
πi
Res

f
(
z
)
,
π
6

=
2
π
√
3
i
24.
y
ˇ
C
z
+1
sinh
z
dz
=2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,πi
)]=2
πi
[1+(
−
πi
−
1)] = 2
π
2
25.
y
ˇ
C
e
2
z
z
4
+2
z
3
+2
z
2
dz
=2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
−
1+
i
) + Res(
f
(
z
)
,
−
1
−
i
)]
=2
πi

1
2
+
e
−
2
4
(cos 2 +
i
sin 2) +
e
−
2
4
(cos 2
−
i
sin 2)

=
π
(1 +
e
−
2
cos 2)
i
26.
y
ˇ
C
1
z
4
−
2
z
2
+4
dz
=2
πi
r
Res
π
f
(
z
)
,
√
6
2
+
√
2
2
i
i
+ Res
π
f
(
z
)
,
−
√
6
2
+
√
2
2
i
zb
=
π
2
√
2
27.
y
ˇ
C
1
z
(
e
z
−
1)
dz
=2
πi
Res(
f
(
z
)
,
0) =
−
πi
. Note:
z
= 0 is a pole of order two, and so
Res(
f
(
z
)
,
0) = lim
z
→
0
d
dz
z
2
·
1
z
2
k
1+
z
2!
+
z
2
3!
+
···
z
= lim
z
→
0
−
k
1
2!
+
2
z
3!
+
···
z
k
1+
z
2!
+
z
2
3!
+
···
z
2
=
−
1
2
28.
y
ˇ
C
z
(
z
−
1)(
z
+1)
10
dz
=2
πi
[Res(
f
(
z
)
,
1) + Res(
f
(
z
)
,
−
1)] = 2
πi

1
2
10
+
k
−
1
2
10
zr
=0
29.
Using two integrals,
y
ˇ
C
ze
3
/z
dz
+
y
ˇ
C
sin
z
z
2
(
z
−
π
)
3
dz
=2
πi
Res(
f
(
z
)
,
0)+2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,π
)]
=2
πi
·
9
2
+2
πi

−
1
π
3
+
2
π
3

=
k
9
π
+
2
π
2
z
i.
818

819. Chapter 19 Review Exercises
Note: In the first integral
z
= 0 is an essential singularity and the residue is obtained from the Laurent series
ze
3
/z
=
···
+
3
3
3!
z
2
+
3
2
2!
z
+3+
z.
30.
y
ˇ
C
csc
πz dz
=2
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
1) + Res(
f
(
z
)
,
2)] = 2
πi

1
π
+
k
−
1
π
z
+
1
π

=2
i
31.
α
∞
−∞
x
2
(
x
2
+2
x
+ 2)(
x
2
+1)
2
dx
=2
πi
[Res(
f
(
z
)
,
−
1+
i
) + Res(
f
(
z
)
,i
)] = 2
πi

3
25
−
4
25
i
−
3
25
+
9
100
i

=
7
π
50
32.
α
∞
−∞
x
+
ai
x
2
+
a
2
e
ix
dx
=
α
∞
−∞
x
cos
x
−
a
sin
x
x
2
+
a
2
dx
+
i
α
∞
−∞
a
cos
x
+
x
sin
x
x
2
+
a
2
dx
=0+2
πi
Res(
f
(
z
)
,ai
)=2
πie
−
a
Thus
α
∞
−∞
a
cos
x
+
x
sin
x
x
2
+
a
2
dx
=2
πe
−
a
.
33.
α
2
π
0
cos
2
θ
2 + sin
θ
dθ
=
1
2
y
ˇ
C
z
4
+2
z
2
+1
z
2
(
z
2
+4
iz
−
1)
dz
(
C
is
|
z
|
=1) =
πi
[Res(
f
(
z
)
,
0) + Res(
f
(
z
)
,
(
−
2+
√
3)
i
)]
=
πi
[
−
4
i
+2
√
3
i
]=(4
−
2
√
3)
π
[Note: The answer in the text is correct but not simplified.]
34.
α
2
π
0
cos 3
θ
5
−
4 cos
θ
dθ
=
−
1
2
i
y
ˇ
C
z
6
+1
z
3
(
z
−
2)(2
z
−
1)
dz
(
C
is
|
z
|
=1) =
−
π

Res(
f
(
z
)
,
0) + Res
k
f
(
z
)
,
1
2
zr
=
−
π

21
8
+
k
−
65
24
zr
=
π
12
35.
The integrand of
y
ˇ
C
1
−
e
iz
z
2
dz
has a simple pole at
z
= 0. Using a contour as in Figure 19
.
14 of Section 19
.
6
we have
y
ˇ
C
=
α
C
R
+
α
−
r
−
R
+
α
−
C
R
+
α
R
r
=0
.
By taking limits as
R
→
0 and as
r
→
0, and using Theorem 19
.
17 we find
P
.
V
.
α
∞
−∞
1
−
e
ix
x
2
dx
−
πi
Res(
f
(
z
)
,
0) = 0
.
Thus
P
.
V
.
α
∞
−∞
1
−
cos
x
−
i
sin
x
x
2
dx
=
π.
Equating real parts gives
P
.
V
.
α
∞
−∞
1
−
cos
x
x
2
dx
=
π.
Finally
α
∞
0
1
−
cos
x
x
2
dx
=
1
2
α
∞
−∞
1
−
cos
x
x
2
dx
=
π
2
.
36.
We have
Ce
−
a
2
z
2
e
ibz
dz
=
α
r
−
r
+
α
C
1
+
α
C
2
+
α
C
3
=0
by the Cauchy-Goursat Theorem. Therefore
α
r
−
r
=
−
α
C
1
−
α
C
2
−
α
C
3
.
819

820. Chapter 19 Review Exercises
Let
C
1
and
C
3
denote the vertical sides of the rectangle. By the ML-inequality,
α
C
1
→
0 and
α
C
3
→
0as
r
→∞
.On
C
2
,
z
=
x
+
b
2
a
2
i
,
−
r
≤
x
≤
r
,
dz
=
dx
,
α
∞
−∞
e
−
ax
2
e
ibx
dx
=
−
α
−∞
∞
e
−
a
2
(
x
+
b
2
a
2
i
)
2
e
ib
(
x
+
b
2
a
2
i
)
dx
=
α
∞
−∞
e
−
a
2
x
2
e
−
b
2
/
4
a
2
dx
α
∞
−∞
e
−
ax
2
(cos
bx
+
i
sin
bx
)
dx
=
e
−
b
2
/
4
a
2
α
∞
−∞
e
−
a
2
x
2
dx.
Using the given value of
α
∞
−∞
e
−
a
2
x
2
dx
and equating real and imaginary parts gives
α
∞
−∞
e
−
ax
2
cos
bx dx
=
√
π
a
e
−
b
2
/
4
a
2
and so
α
∞
0
e
−
ax
2
cos
bx dx
=
√
π
2
a
e
−
b
2
/
4
a
2
.
37.
a
k
=
1
2
πi
y
ˇ
C
e
(
u/
2)(
z
−
z
−
1
)
z
k
+1
dz
=
1
2
πi
α
2
π
0
e
(
u/
2)(
e
it
−
e
−
it
)
(
e
it
)
k
+1
ie
it
dt
=
1
2
π
α
2
π
0
e
(
u/
2)(2
i
sin
t
)
e
−
kit
dt
=
1
2
π
α
2
π
0
e
−
i
(
kt
−
u
sin
t
)
dt
=
1
2
π
α
2
π
0
[cos(
kt
−
u
sin
t
)
−
i
sin(
kt
−
u
sin
t
)]
dt
=
1
2
π
α
2
π
0
cos(
kt
−
u
sin
t
)
dt
since
α
2
π
0
sin(
kt
−
u
sin
t
)
dt
= 0. (To obtain this last result, expand the integrand and let
t
=2
π
−
x
.)
820

821. 20
Conformal Mappings and Applications
Exercises 20.1
1.
For
w
=
1
z
,
u
=
x
x
2
+
y
2
and
v
=
−
y
x
2
+
y
2
.If
y
=
x
,
u
=
1
2
1
x
,
v
=
−
1
2
1
x
, and so
v
=
−
u
. The image is the
line
v
=
−
u
(with the origin (0
,
0) excluded.)
2.
If
y
=1,
u
=
x
x
2
+1
and
v
=
−
1
x
2
+1
. It follows that
u
2
+
v
2
=
1
x
2
+1
=
−
v
and so
u
2
+(
v
+
1
2
)
2
=(
1
2
)
2
. This
is a circle with radius
r
=
1
2
and center at (0
,
−
1
2
)=
−
1
2
i
. The circle can also be described by
|
w
+
1
2
i
|
=
1
2
.
3.
For
w
=
z
2
,
u
=
x
2
−
y
2
and
v
=2
xy
.If
xy
=1,
v
= 2 and so the hyperbola
xy
= 1 is mapped onto the line
v
=2.
4.
If
x
2
−
y
2
=4,
u
= 4 and so the hyperbola
x
2
−
y
2
= 4 is mapped onto the vertical line
u
=4.
5.
For
w
=Ln
z
,
u
= log
e
|
z
|
and
v
= Arg
z
. The semi-circle
|
z
|
=1,
y>
0 may also be described by
r
=1,
0
<θ<π
. Therefore
u
= 0 and 0
<v<π
. The image is therefore the open line segment from
z
=0to
z
=
πi
.
6.
If
θ
=
π/
4, then
v
=
θ
=
π/
4. In addition
u
= log
e
r
will vary from
−∞
to
∞
. The image is therefore the
horizontal line
v
=
π/
4.
7.
For
w
=
z
1
/
2
=(
re
iθ
)
1
/
2
=
r
1
/
2
e
iθ/
2
and
θ
=
θ
0
,
w
=
√
re
iθ
0
/
2
. Therefore Arg
w
=
θ
0
/
2 and so the image is
the ray
θ
=
θ
0
/
2.
8.
If
r
= 2 and 0
≤
θ
≤
π
2
,
w
=
√
2
e
iθ/
2
. Therefore
|
w
|
=
√
2 and 0
≤
Arg
w
≤
π/
4. This image is a circular arc.
9.
For
w
=
e
z
,
u
=
e
x
cos
y
and
v
=
e
x
sin
y
. Therefore if
e
x
cos
y
=1,
u
= 1. The curve
e
x
cos
y
= 1 is mapped
into the line
u
= 1. Since
v
=
sin
y
cos
y
= tan
y
,
v
varies from
−∞
to
∞
and the image is the line
u
=1.
10.
If
w
=
z
+
1
z
and
z
=
e
it
,
w
=
e
it
−
e
−
it
= 2 cos
t
. Therefore
u
= 2 cos
t
and
v
= 0 and so the image is the closed
interval [
−
2
,
2] on the
u
-axis.
11.
The first quadrant may be described by
r>
0, 0
<θ<π/
2. If
w
=1
/z
and
z
=
re
iθ
,
w
=
1
r
e
−
iθ
. Therefore
Arg
w
=
−
θ
and so
−
π/
2
<
Arg
w<
0. The image is therefore the fourth quadrant.
12.
For
w
=
1
z
,
u
=
x
x
2
+
y
2
and
v
=
−
y
x
2
+
y
2
. The line
y
= 0 is mapped to the line
v
= 0, and, from
Problem 2, the line
y
= 1 is mapped onto the circle
|
w
+
1
2
i
|
=
1
2
. Since
f
(
1
2
i
)=
−
2
i
, the region 0
≤
y
≤
1is
mapped onto the points in the half-plane
v
≤
0 which are on or outside the circle
|
w
+
1
2
i
|
=
1
2
. (The image
does not include the point
w
= 0.)
13.
Since
w
=
e
x
+
iy
=
e
x
e
iy
and
π/
4
≤
y
≤
π/
2,
π/
4
≤
Arg
w
≤
π/
2 and
|
w
|
=
e
x
. The image is therefore the
angular wedge defined by
π/
4
≤
Arg
w
≤
π/
2.
14.
Since
w
=
e
x
+
iy
=
e
x
e
iy
and 0
≤
x
≤
1, 0
≤
y
≤
π
, we have
|
w
|
=
e
x
and Arg
w
=
y
. Therefore 1
≤|
w
|≤
e
and 0
≤
Arg
w
≤
π
. These inequalities define a semi-angular region in the
w
-plane.
15.
The mapping
w
=
z
+4
i
is a translation which maps the circle
|
z
|
= 1 to a circle of radius
r
= 1 and with center
w
=4
i
. This circle may be described by
|
w
−
4
i
|
=1.
821

822. Exercises 20.1
16.
If
w
=2
z
−
1 and
|
z
|
= 1, then, since
z
=
w
+1
2
,
/
/
/
/
w
+1
2
/
/
/
/
=1or
|
w
+1
|
= 2. The image is a circle with center
at
w
=
−
1 and with radius
r
=2.
17.
The mapping
w
=
iz
is a rotation through 90
◦
since
i
=
e
iπ/
2
. Therefore the strip 0
≤
y
≤
1 is rotated through
90
◦
and so the strip
−
1
≤
u
≤
0 is the image in the
w
-plane.
18.
Since
w
=(1+
i
)
z
=
√
2
e
iπ/
4
z
, the mapping is the composite of a rotation through 45
◦
and a magnification by
α
=
√
2 . The image of the first quadrant is therefore the angular wedge
π/
4
≤
Arg
w
≤
3
π/
4.
19.
The power function
w
=
z
3
changes the opening of the wedge 0
≤
Arg
z
≤
π/
4 by a factor of 3. Therefore the
image region is 0
≤
Arg
w
≤
3
π/
4.
20.
The power function
w
=
z
1
/
2
changes the opening of the wedge 0
≤
Arg
z
≤
π/
4 by a factor of 1
/
2. Therefore
the image region is 0
≤
Arg
w
≤
π/
8.
21.
We first let
z
1
=
z
−
i
to map the region 1
≤
y
≤
4 to the region 0
≤
y
1
≤
3. We then let
w
=
e
−
iπ/
2
z
1
to rotate
this strip through
−
90
◦
. Therefore
w
=
−
i
(
z
−
i
)=
−
iz
−
1 maps 1
≤
y
≤
4 to the strip 0
≤
u
≤
3.
22.
The mapping
w
=
z
−
i
lowers the strip 1
≤
y
≤
4 one unit so that the image is 0
≤
v
≤
3 in the
w
-plane.
23.
We first let
z
1
=
z
−
1 to map the disk
|
z
−
1
|≤
1 to the disk
|
z
1
|≤
1. We then use the magnification
w
=2
z
1
to obtain
|
w
|≤
2 as the image. The composite of these two mappings is
w
=2(
z
−
1).
24.
The mapping
w
=
iz
will rotate the strip
−
1
≤
x
≤
1 through 90
◦
so that the strip
−
1
≤
v
≤
1 results.
25.
We first use
z
1
=
e
−
iπ/
4
z
to rotate the wedge
π/
4
≤
Arg
z
≤
π/
2 to the wedge 0
≤
Arg
z
1
≤
π/
4. The power
function
w
=
z
4
1
then changes the opening of this wedge by a factor of 4 so that the strip 0
≤
Arg
w
≤
π
results.
The composite of these two mappings is
w
=(
e
−
π/
4
i
z
)
4
=
e
−
πi
z
4
=
−
z
4
.
26.
The magnification
z
1
=
π
4
z
maps the strip 0
≤
y
≤
4 to the strip 0
≤
y
1
≤
π
. By Example 1, Section 20
.
1,
w
=
e
z
1
maps this strip onto the upper half-plane
v
≥
0. The composite of these two mappings is
w
=
e
π
4
z
.
27.
By Example 1, Section 20
.
1,
z
1
=
e
z
maps the strip 0
≤
y
≤
π
onto the upper half-plane
y
1
≥
0, or
0
≤
Arg
z
1
≤
π
. The power function
w
=
z
3
/
2
1
changes the opening of this wedge by a factor of 3
/
2 so the wedge
0
≤
Arg
w
≤
3
π/
2 results. The composite of these two mappings is
w
=(
e
z
)
3
/
2
=
e
3
z/
2
.
28.
The power function
z
1
=
z
2
/
3
maps the wedge 0
≤
Arg
z
≤
3
π/
2 to the upper half-plane 0
≤
Arg
z
1
≤
π
.We
then let
w
=
e
−
iπ/
2
z
1
+2=
−
iz
1
+ 2 to rotate the upper half-plane through
−
90
◦
and then translate 2 units to
the right. Therefore the composite function is
w
=
−
iz
2
/
3
+ 2 and the image region is the half-plane
u
≥
2.
29.
We may obtain the image region by (1) rotating
R
through 180
◦
and (2) raising the resulting region one unit
in the vertical direction. Therefore
w
=
e
iπ
z
+
i
=
−
z
+
i
.
30.
The mapping
z
1
=
−
(
z
−
πi
) lowers
R
by
π
units in the vertical direction and then rotates the resulting region
through 180
◦
. The image region
R
1
is upper half-plane
y
1
≥
0. By Example 1, Section 20
.
1,
w
=Ln
z
1
maps
R
1
onto the strip 0
≤
v
≤
π
. The composite of these two mappings is
w
= Ln(
πi
−
z
).
822

823. Exercises 20.2
Exercises 20.2
1.
Since
f
i
(
z
)=3(
z
2
−
1),
f
is conformal at all points except
z
=
±
1.
2.
Since
f
i
(
z
)=
−
sin
z
,
f
is conformal at all points except
z
=
±
nπ
.
3.
f
i
(
z
)=1+
e
z
and 1 +
e
z
= 0 for
z
=
±
i
±
2
nπi
. Therefore
f
is conformal except for
z
=
πi
±
2
nπi
.
4.
f
(
z
)=
z
+Ln
z
+ 1 is analytic for all
z
except
z
=
y
≤
0. But
f
i
(
z
)=1+1
/z
and 1 + 1
/z
= 0 for
z
=
−
1.
Therefore
f
(
z
) is conformal at all points except those on the branch cut
y
≤
0.
5.
f
(
z
)=(
z
2
−
1)
1
/
2
=
e
1
2
Ln(
z
2
−
1)
is analytic for all
z
outside the interval [
−
1
,
1] on the real axis. This follows
from the fact that
z
2
−
1=(
x
2
−
y
2
−
1) + 2
xyi
and so we must exclude values of
z
for which
v
=2
xy
=0
and
u
=
x
2
−
y
2
−
1
≤
0. Therefore
y
= 0 and
x
2
≤
1.
f
i
(
z
)=
z/
(
z
2
−
1)
1
/
2
is non-zero outside this interval.
Therefore
f
is conformal except for
z
=
x
,
−
1
≤
x
≤
1.
6.
The function
f
(
z
)=
πi
−
1
2
[Ln(
z
+1)+Ln(
z
−
1)] is analytic except on the branch cut
x
−
1
≤
0or
x
≤
1, and
f
i
(
z
)=
−
1
2
i
1
z
+1
+
1
z
−
1
θ
=
−
z
z
2
−
1
is non-zero for
z
t
=0,
±
1. Therefore
f
is conformal except for
z
=
x
,
x
≤
1.
7.
f
(
z
) = cos
z
= sin(
π/
2
−
z
) is the composite of
z
1
=
π/
2
−
z
and
w
= sin
z
. The strip 0
≤
x
≤
π
is mapped
onto the strip
−
π/
2
≤
x
≤
π/
2by
z
1
=
π/
2
−
z
and
w
= sin
z
maps this strip onto the region shown in
Figure 20
.
11(b). The horizontal segment
z
(
t
)=
t
+
ib
,0
<t<π
is first mapped to the horizontal segment
z
i
(
t
)=(
π/
2
−
t
)
−
ib
,0
<t<π
. This latter segment is mapped onto the lower or upper portion of the ellipse
u
2
cosh
2
b
+
v
2
sinh
2
b
=1
according to whether
b>
0or
b<
0. See Figure 20
.
11.
8.
f
(
z
) = sinh
z
=
−
i
sin(
iz
) is the composite of
z
1
=
iz
,
z
2
= sin
z
1
, and
w
=
−
iz
2
. The strip
−
π/
2
≤
y
≤
π/
2,
x
≥
0 is rotated through 90
◦
so that the image is the strip
−
π/
2
≤
x
≤
π/
2,
y
≥
0. This latter strip is mapped
to the upper half-plane
y
2
≥
0by
z
2
= sin
z
1
, and
w
=
−
iz
2
rotates this upper half-plane through
−
90
◦
. The
final image region is the half-plane
u
≥
0. A vertical line segment in the original strip is mapped to the right
hand side of the ellipse
u
2
sinh
2
b
+
v
2
cosh
2
b
=1
.
See the figures below.
9.
f
(
z
) = (sin
z
)
1
/
4
is the composite of
z
1
= sin
z
and
w
=
z
1
/
4
1
. The region
−
π/
2
≤
x
≤
π/
2,
y
≥
0 is mapped to
the upper half-plane
y
2
≥
0by
z
1
= sin
z
(See Example 2) and the power function
w
=
z
1
/
4
1
maps this upper
half-plane to the angular wedge 0
≤
Arg
w
≤
π/
4. The real interval [
−
π/
2
,π/
2] is first mapped to [
−
1
,
1] and
then to the union of the line segments from
e
i
π
4
to 0 and 0 to 1. See the figures on the following page.
823

824. Exercises 20.2
10.
From Example 3,
u
=(
r
+1
/r
) cos
θ
and
v
=(
r
−
1
/r
) sin
θ
.If
r
=1,
u
= 2 cos
θ
and
v
= 0. Therefore the
image of the circle
|
z
|
= 1 is the real interval [
−
2
,
2]. If 0
<r<
1,
u
2
/a
2
+
v
2
/b
2
= 1 where
a
=
r
+1
/r >
2 and
b
=1
/r
−
r
. The resulting ellipse together with [
−
2
,
2] fill up the entire
w
-plane. Therefore the image of the
region
|
z
|≤
1 is the entire
w
-plane. If
z
=
x
and
−
1
≤
x
≤
1,
w
=
x
+1
/x
. Therefore
w
is real and, from an
analysis of the graph of
w
=
x
+1
/x
for
−
1
≤
x
≤
1,
|
w
|≥
2. Therefore the segment [
−
1
,
1] (with 0 excluded)
is mapped on those points
w
=
u
on the
u
-axis for which
|
u
|≥
2.
11.
Using H-4 with
a
=2,
w
= cos(
πz/
2) maps
R
onto the target region
R
i
. The image of
AB
is shown in the figure.
12.
Using C-3
w
=
e
z
maps
R
onto the target region
R
i
. The image of
AB
is shown in the
figure.
13.
Using H-5,
z
1
=
i
1+
z
1
−
z
θ
2
maps
R
onto the upper half-plane
y
1
≥
0, and
w
=
z
1
/
4
1
maps this half-plane onto the target region
R
i
. The image of
AB
is shown in the
figure, and
w
=
i
1+
z
1
−
z
θ
1
/
2
.
14.
Using H-1,
z
1
=
i
1
−
z
1+
z
maps
R
onto the upper half-plane
y
1
≥
0, and using M-4
with
a
=1,
w
=(
z
2
1
−
1)
1
/
2
maps this upper half-plane onto the target region
R
i
.
Therefore
w
=
π
−
i
1
−
z
1+
z
θ
2
−
1
z
1
/
2
and the image of
AB
is shown in the figure.
15.
Using H-6,
z
1
=
e
π/z
+
e
−
π/z
e
π/z
−
e
−
π/z
maps
R
onto the upper half-plane
y
1
≥
0, and
w
=
z
1
/
2
1
maps
this half-plane onto the target region
R
i
. Therefore
w
=
i
e
π/z
+
e
−
π/z
e
π/z
−
e
−
π/z
θ
1
/
2
and the image of
AB
is shown in the figure.
824

825. Exercises 20.2
16.
We can translate
R
to the origin, magnify by 2, and then use H-1 to reach the target region
R
.
. Therefore
z
1
=2(
z
−
1
2
),
w
=
i
1
−
z
1+
z
and so
w
=
i
1
−
(2
z
−
1)
1+(2
z
−
1)
=
i
1
−
z
z
.
The image of
AB
is shown in the figure.
17.
z
1
=Ln
z
maps
R
onto the horizontal strip 0
≤
y
1
≤
π
, and, to prepare this strip for mapping by the sine
function, we let
z
2
=
−
iz
1
−
π/
2 to obtain the vertical strip
−
π/
2
≤
x
≤
π/
2. Finally
w
= sin
z
2
maps this
vertical strip onto the target region
R
.
. Therefore
w
= sin
5
−
i
Ln
z
−
π
2
6
.
The image of
AB
is the real interval (
−∞
,
−
1].
18.
Using E-9,
z
1
= cosh
z
maps
R
onto the upper half-plane
y
1
≥
0. Using M-7,
w
=
z
1
+Ln
z
1
+ 1 maps this half-plane onto the target region
R
.
. Therefore
w
= cosh
z
+ Ln(cosh
z
) + 1 and the image of
AB
is shown in the figure.
In Exercises 19-22, we find a conformal mapping
w
=
f
(
z
) that maps the given
region
R
onto the upper half-plane
v
≥
0 and transfers the boundary conditions
so that the resulting Dirichlet problem is as shown in the figure.
19.
f
(
z
)=
z
4
and so
u
=
U
(
f
(
z
)) =
1
π
Arg
z
4
. The solution may also be written as
u
(
r, θ
)=4
θ/π
.
20.
f
(
z
)=
.
1+
z
1
−
z
2
2
, using H-5, and so
u
=
U
(
f
(
z
)) =
1
π
Arg
.
1+
z
1
−
z
2
2
.
21.
f
(
z
)=
i
1
−
z
1+
z
, using H-1, and so
u
=
U
(
f
(
z
)) =
1
π
Arg
.
1
−
z
1+
z
2
. The solution may also be written as
u
(
x, y
)=
1
π
tan
−
1
.
1
−
x
2
−
y
2
2
y
2
.
22.
f
(
z
)=
1
2
.
z
+
1
z
2
, using H-3 with
a
= 1, and so
u
=
U
(
f
(
z
)) =
1
π
Arg
1
2
.
z
+
1
z
2
. The solution may also be
written as
u
(
x, y
)=
1
π
tan
−
1

y
x
x
2
+
y
2
−
1
x
2
+
y
2
+1

.
In Exercises 23-26, we find a conformal mapping
w
=
f
(
z
) that maps the given
region
R
onto the upper half-plane
v
≥
0 and transfers the boundary conditions
so that the resulting Dirichlet problem is as shown in the figure.
23.
f
(
z
)=
z
2
and
c
0
= 1. Therefore
u
=
1
π
[Arg(
z
2
−
1)
−
Arg(
z
2
+ 1)].
825

826. Exercises 20.2
24.
The mapping
z
1
=
z
2
maps
R
onto the region
R
1
defined by
y
1
≥
0,
|
z
1
|≥
1 and shown in H-3, and
w
=
1
2
i
z
1
+
1
z
1
θ
maps
R
1
onto the upper half-plane
v
≥
0. Letting
c
0
=5,
u
=
5
π

Arg
i
1
2

z
2
+
1
z
2

−
1
θ
−
Arg
i
1
2

z
2
+
1
z
2

+1
θy
.
25.
f
(
z
)=
e
πz
, using H-2 with
a
= 1. Letting
c
0
= 10,
u
=
10
π
[Arg(
e
πz
−
1)
−
Arg(
e
πz
+ 1)]
.
26.
f
(
z
) = cos(
πz/
2) using H-4 with
a
= 2. Letting
c
0
=4,
u
=
4
π
[Arg(cos(
πz/
2)
−
1)
−
Arg(cos(
πz/
2) + 1)]
.
27. (a)
If
u
=
∂
2
φ
∂x
2
+
∂
2
φ
∂y
2
,
∂
2
u
∂x
2
+
∂
2
u
∂y
2
=
∂
4
φ
∂x
4
+2
∂
4
φ
∂x
2
∂y
2
+
∂
4
φ
∂y
4
=0
since
φ
is assumed to be biharmonic.
(b)
If
g
=
u
+
iv
, then
φ
= Re(¯
zg
(
z
)) =
xu
+
yv
.
∂
2
φ
∂x
2
=3
∂u
∂x
+
x
∂
2
u
∂x
2
+
y
∂
2
v
∂x
2
∂
2
φ
∂y
2
=2
∂v
∂y
+
x
∂
2
u
∂y
2
+
y
∂
2
v
∂y
2
.
Since
u
and
v
are harmonic and
∂u
∂x
=
∂v
∂y
,
∂
2
φ
∂x
2
+
∂
2
φ
∂y
2
=2
∂u
∂x
+2
∂v
∂y
=4
∂u
∂x
.
Now
u
1
=
∂u
∂x
is also harmonic and so
∂
2
u
1
∂x
2
+
∂
2
u
1
∂y
2
= 0. But
∂
2
u
1
∂x
2
+
∂
2
u
1
∂y
2
=
1
4

∂
4
φ
∂x
4
+2
∂
4
φ
∂x
2
∂y
2
+
∂
4
φ
∂y
4

and so
φ
is biharmonic.
Exercises 20.3
1. (a)
For
T
(
z
)=
i/z
,
T
(0) =
∞
,
T
(1) =
i
, and
T
(
∞
)=0.
(b)
If
|
z
|
=1,
|
w
|
=
|
i/z
|
=1
/
|
z
|
= 1. Therefore the image of the circle
|
z
|
= 1 is the circle
|
w
|
= 1 in the
w
-plane. The circle
|
z
−
1
|
= 1 passes through the pole at
z
= 0 and so the image is a line. Since
T
(2) =
1
2
i
and
T
(1 +
i
)=
1
2
+
1
2
i
, the image is the line
v
=
1
2
.
(c)
The disk
|
z
|≤
1 is mapped onto the disk
|
w
|≥
1.
2. (a)
For
T
(
z
)=
1
z
−
1
,
T
(0) =
−
1,
T
(1) =
∞
, and
T
(
∞
)=0.
826

827. Exercises 20.3
(b)
The circle
|
z
|
= 1 passes through the pole at
z
= 1 and so the image is a line. Since
T
(
−
1) =
−
1
2
and
T
(
i
)=
−
1
2
−
1
2
i
, the image is the line
u
=
−
1
2
.If
|
z
−
1
|
=1,
|
w
|
=1
/
|
z
−
1
|
= 1 and the image is the
circle
|
w
|
= 1 in the
w
-plane.
(c)
Since
T
(0) =
−
1, the image of the disk
|
z
|≤
1 is the half-plane
u
=
−
1
2
.
3. (a)
For
T
(
z
)=
z
+1
z
−
1
,
T
(0) =
−
1,
T
(1) =
∞
, and
T
(
∞
)=1.
(b)
The circle
|
z
|
= 1 passes through the pole at
z
= 1 and so the image is a line. Since
T
(
−
1) = 0 and
T
(
i
)=
−
i
, the image is the line
u
=0. If
|
z
−
1
|
=1,
|
w
−
1
|
=
1
1
1
1
z
+1
z
−
1
−
1
1
1
1
1
=
2
|
z
−
1
|
=2
and so the image is the circle
|
w
−
1
|
= 2 in the
w
-plane.
(c)
Since
T
(0) =
−
1, the image of the disk
|
z
|≤
1 is the half-plane
u
≤
0.
4. (a)
For
T
(
z
)=
z
−
i
z
,
T
(0) =
∞
,
T
(1) = 1
−
i
, and
T
(
∞
)=1.
(b)
T
(
i
)=0,
T
(1)=1
−
i
and
T
(
−
i
) = 2. The image of
|
z
|
= 1 is therefore a circle which passes through 0,
1
−
i
, and 2. This is the circle
|
w
−
1
|
= 1. The circle
|
z
−
1
|
= 1 passes through the pole at
z
= 0 and so
the image is a line. Since
T
(2) = 1
−
1
2
i
and
T
(1 +
i
)=
1
2
−
1
2
i
, the image is the line
v
=
−
1
2
.
(c)
T
(
1
2
)=1
−
2
i
which is exterior to the circle
|
w
−
1
|
= 1. Therefore the image of
|
z
|≤
1is
|
w
−
1
|≥
1, the
exterior of the circle
|
w
−
1
|
= 1 in the
w
-plane.
5.
S
−
1
(
T
(
z
)) =
az
+
b
cz
+
d
where

ab
cd

= adj
.7
i
1
1
−
1
82 7
10
i
−
1

=

−
1
−
i
1
−
2
−
i

.
Therefore
S
−
1
(
T
(
z
)) =
(
−
1
−
i
)
z
+1
−
2
z
−
i
=
(1 +
i
)
z
−
1
2
z
+
i
and
S
−
1
(
w
)=
−
w
−
1
−
w
+
i
=
w
+1
w
−
i
.
6.
S
−
1
(
T
(
z
)) =
az
+
b
cz
+
d
where

ab
cd

= adj
.7
21
11
82 7
i
0
1
−
2
i

=

−
1+
i
2
i
2
−
i
−
4
i

.
Therefore
S
−
1
(
T
(
z
)) =
(
−
1+
i
)
z
+2
i
(2
−
i
)
z
−
4
i
and
S
−
1
(
w
)=
w
−
1
−
w
+2
.
7.
S
−
1
(
T
(
z
)) =
az
+
b
cz
+
d
where

ab
cd

= adj
.7
1
−
2
1
−
1
82 7
2
−
3
1
−
3

=

0
−
3
−
10

.
Therefore
S
−
1
(
T
(
z
)) =
−
3
−
z
=
3
z
and
S
−
1
(
w
)=
−
w
+2
−
w
+1
=
w
−
2
w
−
1
.
8.
S
−
1
(
T
(
z
)) =
az
+
b
cz
+
d
where

ab
cd

= adj
.7
2
−
i
0
1
−
1
−
i
82 7
1
−
1+
i
i
−
2

=

−
1
−
i
2
2
i
−
3+
i

.
827

828. Exercises 20.3
Therefore
S
−
1
(
T
(
z
)) =
(
−
1
−
i
)
z
+2
2
iz
−
3+
i
and
S
−
1
(
w
)=
(
−
1
−
i
)
w
−
w
+2
−
i
=
(1 +
i
)
w
w
−
2+1
.
9.
T
(
z
)=
(
z
−
z
1
)(
z
2
−
z
3
)
(
z
−
z
3
)(
z
2
−
z
1
)
maps
z
1
,
z
2
,
z
3
to 0, 1,
∞
. Therefore
T
(
z
)=
(
z
+ 1)(
−
2)
(
z
−
2)(1)
=
−
2
z
+1
z
−
2
maps
−
1, 0,
2to0,1,
∞
.
10.
T
(
z
)=
(
z
−
i
)
i
(
z
+
i
)(
−
i
)
=
−
z
−
i
z
+
i
maps
i
,0,
−
i
to 0, 1,
∞
.
11.
S
(
w
)=
(
w
−
w
1
)(
w
2
−
w
3
)
(
w
−
w
3
)(
w
2
−
w
1
)
maps
w
1
,
w
2
,
w
3
to 0, 1,
∞
and so
S
.
maps 0, 1,
∞
to
w
1
,
w
2
,
w
3
. Therefore
z
=
(
w
−
0)(
i
−
2)
(
w
−
2)(
i
−
0)
and so
w
=
2
z
z
−
1
−
2
i
maps 0, 1,
∞
to 0,
i
,2.
12.
As in Exercise 11,
z
=
(
w
−
1
−
i
)(
−
1+
i
)
(
w
−
1+
i
)(
−
1
−
i
)
and, solving for
w
,
w
=
2
z
−
2
(1 +
i
)
z
−
1+
i
maps 0, 1,
∞
to 1 +
i
,0,
1
−
i
.
13.
Using the cross-ratio formula (7),
(
w
−
i
)(1)
w
(1)
=
(
z
+ 1)(
−
1)
(
z
−
1)(1)
and so
w
=
i
2
z
−
1
z
maps
−
1, 0, 1 to
i
,
∞
,0.
14.
Using the cross-ratio formula (7),
(1)(
−
i
−
1)
(
w
−
1)(1)
=
(
z
+ 1)(
−
1)
(
z
−
1)(1)
and so
w
=
(2 +
i
)
z
−
i
z
+1
maps
−
1, 0, 1 to
∞
,
−
i
,1.
15.
Using the cross-ratio formula (7),
S
(
w
)=
(
w
+ 1)(
−
3)
(
w
−
3)(1)
=
(
z
−
1)(2
i
)
(
z
+
i
)(
i
−
1)
=
T
(
z
)
.
We can solve for
w
to obtain
w
=3
(1 +
i
)
z
+(1
−
i
)
(
−
3+5
i
)
z
−
3
−
5
i
.
Alternatively we can apply the matrix method to compute
w
=
S
−
1
(
T
(
z
)).
16.
Using the cross-ratio formula (7),
S
(
w
)=
(
w
−
i
)(
−
i
−
1)
(
w
−
1)(
−
2
i
)
=
(
z
−
1)(2
i
)
(
z
+
i
)(
i
−
1)
=
T
(
z
)
.
We can solve for
w
to obtain
w
=
(1+2
i
)
z
−
i
iz
+1
−
2
i
.
Alternatively we can apply the matrix method to compute
w
=
S
−
1
(
T
(
z
)).
828

829. Exercises 20.3
17.
From Example 2,
z
=
w
+2
w
−
1
maps the annular region 1
<
|
w
|
<
2ontothe
region
R
and the circle
|
w
|
= 1 corresponds to the line
x
=
−
1
/
2. Solving for
w
,
w
=
z
+2
z
−
1
maps
R
onto the annulus and the transferred boundary conditions are
shown in the figure. The solution to this new Dirichlet problem is
U
= log
e
r/
log
e
2
and so
u
=
U
(
f
(
z
)) =
1
log
e
2
log
e
/
/
/
/
z
+2
z
−
1
/
/
/
/
is the solution to the Dirichlet problem in Figure 20
.
37. The level curves are the images of the level curves of
U
,
|
w
|
=
r
for 1
<r<
2 under the mapping
z
=
w
+2
w
+1
. Since these circles do not pass through the pole at
w
= 1, the images are circles.
18.
The mapping
T
(
z
)=
1
2
z
+1
z
maps
−
1, 1, 0 to 0, 1,
∞
and maps each of the
two circles in
R
to lines since both circles pass through the pole at
z
= 0. Since
T
(
1
2
+
1
2
i
)=1
−
i
and
T
(1) = 1, the circle
|
z
−
1
2
|
=
1
2
is mapped onto the line
u
= 1. Likewise, the circle
|
z
+
1
2
|
=
1
2
is mapped onto the line
u
= 0. The transferred
boundary conditions are shown in the figure and
U
(
u, v
)=
u
is the solution. The
solution to the Dirichlet problem in Figure 20
.
38 is
u
=
U
(
T
(
z
)) = Re
i
1
2
z
+1
z
θ
=
1
2
+
1
2
x
x
2
+
y
2
.
The level curves
u
=
c
are the circles with centers on the
x
-axis which pass through the origin. The level curve
u
=
1
2
, however, is the vertical line
x
=0.
19.
The linear fractional transformation that sends 1,
i
,
−
i
to 0, 1,
−
1 satisfies the cross-ratio equation
(
w
−
0)(2)
(
w
+ 1)(1)
=
(
z
−
1)(2
i
)
(
z
+
i
)(
i
−
1)
.
Solving for
w
,
w
=
i
1
−
z
1+
z
=
T
(
z
). Since
T
(0) =
i
, the image of the disk
|
z
|≤
1 is the upper half-plane
v
≥
0.
20.
The linear fractional transformation that sends 1,
i
,
−
1to
∞
,
i
,0is
T
(
z
)=
1+
z
1
−
z
and so
f
(
z
)=
i
1+
z
1
−
z
θ
2
maps 1,
i
,
−
1to
∞
,
−
1, 0. The upper semi-circle is mapped by
T
to the positive imaginary axis, and the real
interval [
−
1
,
1] is mapped to the positive real axis. Since
T
i
i
2
θ
=
3
5
+
4
5
i
, the image of
R
under
T
is the first
quadrant.
w
=
z
2
1
doubles the size of the opening so that the image under
f
is the upper half-plane
v
≥
0. See
the figures below.
829

830. Exercises 20.3
21.
T
2
(
T
1
(
z
)) =
a
2
T
1
(
z
)+
b
2
c
2
T
1
(
z
)+
d
2
=
a
2
a
1
z
+
b
1
c
1
z
+
d
1
+
b
2
c
2
a
1
z
+
b
1
c
1
z
+
d
1
+
d
2
=
a
1
a
2
z
+
a
2
b
1
+
b
2
c
1
z
+
b
2
d
1
a
1
c
2
z
+
b
1
c
2
+
c
1
d
2
z
+
d
1
d
2
=
(
a
1
a
2
+
c
1
b
2
)
z
+(
b
1
a
2
+
d
1
b
2
)
(
a
1
c
2
+
c
1
d
2
)
z
+(
b
1
c
2
+
d
1
d
2
)
22.
|
w
−
w
1
|
=
λ
|
w
−
w
2
|
=
⇒
(
u
−
u
1
)
2
+(
v
−
v
1
)
2
=
λ
2
[(
u
−
u
2
)
2
+(
v
−
v
2
)
2
]
The latter equation may be put in the form
Au
2
+
Bv
2
+
Cu
+
Dv
+
F
=0
where
A
=
B
=1
−
λ
2
.If
λ
= 1, the line
Cu
+
Dv
+
F
= 0 results. If
λ>
0 and
λ
t
= 1, then the equation
defines a circle.
Exercises 20.4
1.
arg
f
i
(
t
)=
−
1
2
Arg(
t
−
1) =

−
π/
2
,t<
1
0
,t>
1
. Since
f
(1) = 0, the image is the first quadrant.
2.
arg
f
i
(
t
)=
−
1
3
Arg(
t
+1)=

−
π/
3
,t<
−
1
0
,t>
−
1
. In (2),
α
1
=2
π/
3 and since
f
(
−
1) = 0, the image is the wedge
0
≤
Arg
w
≤
2
π/
3.
3.
arg
f
i
(
t
)=
−
1
2
Arg(
t
+1)+
1
2
Arg(
t
−
1) =



0
,t<
−
1
π/
2
,
−
1
<t<
1
0
,t>
1
and
α
1
=
π/
2 and
α
2
=3
π/
2. Since
f
(
−
1) = 0, the image of the upper half-plane
is the region shown in the figure.
4.
arg
f
i
(
t
)=
−
1
2
Arg(
t
+1)
−
3
4
Arg(
t
−
1) =



−
5
π/
4
,t<
−
1
−
3
π/
4
,
−
1
<t<
1
0
,t>
1
and
α
1
=
π/
2 and
α
2
=
π/
4. Since
f
(
−
1) = 0, the image of the upper half-plane is
the region shown in the figure.
5.
Since
α
1
=
α
2
=
α
3
=
π/
2,
α
i
/π
−
1=
−
1
/
2 and so
f
i
(
z
)=
A
(
z
+1)
−
1
/
2
z
−
1
/
2
(
z
−
1)
−
1
/
2
for some constant
A
.
6.
Since
α
1
=
π/
3 and
α
2
=
π/
2,
α
1
/π
−
1=
−
2
/
3 and
α
2
/π
−
1=
−
1
/
2 and so
f
i
(
z
)=
A
(
z
+1)
−
2
/
3
z
−
1
/
2
for
some constant
A
.
7.
Since
α
1
=
α
2
=2
π/
3,
α
i
/π
−
1=
−
1
/
3 and so
f
i
(
z
)=
A
(
z
+1)
−
1
/
3
z
−
1
/
3
for some constant
A
.
8.
Since
α
1
=3
π/
2 and
α
2
=
π/
4,
α
1
/π
−
1=1
/
2 and
α
2
/π
−
1=
−
3
/
4. Therefore
f
i
(
z
)=
A
(
z
+1)
1
/
2
z
−
3
/
4
for
some constant
A
.
9.
Since
α
1
=
α
2
=
π/
2,
f
i
(
z
)=
A
(
z
+1)
−
1
/
2
(
z
−
1)
−
1
/
2
=
A/
(
z
2
−
1)
1
/
2
. Therefore
f
(
z
)=
A
cosh
−
1
z
+
B
. But
f
(
−
1) =
πi
and
f
(1) = 0. Since cosh
−
1
1=0,
B
= 0. Since cosh
−
1
(
−
1) =
πi
,
πi
=
A
(
πi
) and so
A
= 1. Hence
f
(
z
) = cosh
−
1
z
.
10.
Since
α
1
=3
π/
2=
α
2
,
f
i
(
z
)=
A
(
z
+1)
1
/
2
(
z
−
1)
1
/
2
=
A
(
z
2
−
1)
1
/
2
. Therefore
f
(
z
)=
A

z
(
z
2
−
1)
1
/
2
2
−
1
2
Ln(
z
+(
z
2
−
1)
1
/
2

+
B.
but
f
(
−
1) =
−
ai
and
f
(1) =
ai
. It follows that
ai
=
f
(1) =
B,
−
ai
=
f
(
−
1) =
A
i
−
πi
2
θ
+
B
830

831. Exercises 20.5
and so
B
=
ai
and
A
=4
a/π
. Therefore
f
(
z
)=
4
a
π

z
(
z
2
−
1)
1
/
2
2
−
1
2
Ln(
z
+(
z
2
−
1)
1
/
2
)

+
ai
=
4
a
π

z
(
z
2
−
1)
1
/
2
2
−
1
2
cosh
−
1
z

+
ai.
11.
f
i
(
z
)=
A
(
z
+1)
α
1
/π
−
1
z
α
2
/π
−
1
(
z
−
1)
α
3
/π
−
1
from (3). Since
f
(
−
1) =
πi
,
α
1
→
π
as
w
1
→∞
in the horizontal
direction. Likewise
α
2
→
0 and
α
3
→
π
. This suggests we examine
f
i
(
z
)=
Az
−
1
=
A/z
. Therefore
f
(
z
)=
A
Ln
z
+
B
. But
f
(
−
1) =
πi
and
f
(1) = 0. It follows that
A
= 1 and
B
= 0 so that
f
(
z
)=Ln
z
.We
verified in Example 1, Section 20
.
1 that
f
(
z
)=Ln
z
maps the upper half-plane
y
≥
0 to the horizontal strip
0
≤
v
≤
π
.
12.
From (3),
f
i
(
z
)=
Az
−
3
/
4
(
z
−
1)
α
2
/π
−
1
. But
α
2
→
π
as
θ
→
0. This suggests that we examine
f
i
(
z
)=
Az
−
3
/
4
.
Therefore
f
(
z
)=
A
1
z
1
/
4
+
B
1
. But
f
(0) = 0 and
f
(1) = 1 so that
B
1
= 0 and
A
1
= 1. Hence
f
(
z
)=
z
1
/
4
and
we recognize that this power function maps the upper half-plane onto the wedge 0
≤
Arg
w
≤
π/
4.
13.
From (3),
f
i
(
z
)=
A
(
z
+1)
α
1
/π
−
1
z
α
2
/π
−
1
(
z
−
1)
α
3
/π
−
1
. But as
u
1
→
0,
α
1
→
π/
2,
α
2
→
2
π
, and
α
3
→
π/
2.
This suggests that we examine
f
i
(
z
)=
A
(
z
+1)
−
1
/
2
z
(
z
−
1)
−
1
/
2
=
A
z
(
z
2
−
1)
1
/
2
.
Therefore
f
(
z
)=
A
(
z
2
−
1)
1
/
2
+
B
. But
f
(
−
1) =
f
(1) = 0 and
f
(0) =
ai
. This implies that
B
= 0 and
ai
=
Ai
or
A
=
a
. Therefore
f
(
z
)=
a
(
z
2
−
1)
1
/
2
. By expressing
f
(
z
) as the composite of
z
1
=
z
2
,
z
2
=
z
1
−
1,
z
3
=
z
1
/
2
2
and
w
=
az
3
we can show that the image of the upper half-plane is
R
i
.
14.
If
w
(
t
)=
u
(
t
)+
iv
(
t
), then
w
i
(
t
)=
u
i
(
t
)+
iv
i
(
t
) and so
tan(arg
w
i
(
t
)) =
v
i
(
t
)
u
i
(
t
)
=
dv
du
.
If arg
w
i
(
t
) is constant, then
dv/du
=
m
or
v
=
mu
+
b
for some constants
m
and
b
.
Exercises 20.5
1.
Using (3) with
x
0
=
−
1,
x
1
=0,
x
2
= 1 and
u
1
=
−
1 and
u
1
=1,
u
=
−
1
π
Arg
i
z
z
+1
θ
+
1
π
Arg
i
z
−
1
z
θ
.
2.
Using (3) with
x
0
=
−
2,
x
1
=0,
x
2
= 1 and
u
1
= 5 and
u
2
=1,
u
=
5
π
Arg
i
z
z
+2
θ
+
1
π
Arg
i
z
−
1
z
θ
.
3.
The harmonic function
u
1
=
5
π
[
π
−
Arg(
z
−
1)] =

5
,x>
1
0
,x<
1
,
and
u
2
=
−
1
π
Arg
i
z
+1
z
+2
θ
+
1
π
Arg
i
z
z
+1
θ
from (3) satisfies all boundary conditions except that
u
2
= 0 for
x>
1. Therefore
u
=
u
1
+
u
2
is the solution
to the given Dirichlet problem.
4.
The harmonic function
u
1
=
1
π
Arg(
z
+2)=

1
,x<
−
2
0
,x>
−
2
,
and
u
2
=
−
1
π
Arg
i
z
+1
z
+2
θ
+
1
π
Arg
i
z
+1
z
−
1
θ
831

832. Exercises 20.5
satisfies all boundary conditions except that
u
2
= 0 for
x<
−
2. Therefore
u
=
u
1
+
u
2
is the solution to the
given Dirichlet problem.
5.
By Theorem 20
.
5,
u
(
x, y
)=
y
π
w
1
0
t
2
(
x
−
t
)
2
+
y
2
dt.
If we let
s
=
x
−
t
, this integral can be expressed in terms of the natural log and inverse tangent. Using
Maple
we obtain
u
=
y
π
i
y
2
−
x
2
y

tan
−
1
i
x
−
1
y
θ
−
tan
−
1
i
x
y
θy
+
x
ln
(
x
−
1)
2
+
y
2
x
2
+
y
2
+1
θ
.
6.
From Theorem 20
.
5,
u
(
x, y
)=
y
π
w
∞
−∞
cos
t
(
x
−
t
)
2
+
y
2
dt
=
y
π
w
∞
−∞
cos(
x
−
s
)
s
2
+
y
2
ds
letting
s
=
x
−
t
. But cos(
x
−
s
) = cos
x
cos
x
+ sin
x
sin
s
. It follows that
u
(
x, y
)=
y
cos
x
π
w
∞
−∞
cos
s
s
2
+
y
2
ds
+
y
sin
x
π
w
∞
−∞
sin
s
s
2
+
y
2
ds
=
y
cos
x
π
i
πe
−
y
y
θ
=
e
−
y
cos
x, y >
0
.
7.
The mapping
f
(
z
)=
z
2
maps
R
onto the upper half-plane
R
i
. The corresponding
boundary value problem in
R
i
is shown in the figure. From (3),
U
=
5
π
Arg(
w
+1)+
1
π
Arg
i
w
−
1
w
θ
is the solution in
R
i
. Therefore
u
=
U
(
f
(
z
)) =
5
π
Arg(
z
2
+1)+
1
π
Arg
i
z
2
−
1
z
2
θ
is the solution to the original Dirichlet problem in
R
.
8.
Using H-4 with
a
=3,
f
(
z
) = cos(
πz/
3) maps
R
onto the upper half-plane
R
i
. The
corresponding Dirichlet problem in
R
i
is shown in the figure. From (3),
U
=
1
π
[
π
−
Arg(
w
−
1)] +
1
π
Arg(
w
+1)
is the solution in
R
i
. Therefore
u
=
U
(
f
(
z
)) = 1 +
1
π
[Arg(cos(
πz/
3)+1)
−
Arg(cos(
πz/
3)
−
1)]
is the solution to the original Dirichlet problem in
R
.
9.
Using H-1,
f
(
z
)=
i
1
−
z
1+
z
maps
R
onto the upper half-plane
R
i
. The corresponding
Dirichlet problem in
R
i
is shown in the figure. From (3),
U
=
−
1
π
Arg
i
w
w
+1
θ
+
1
π
Arg
i
w
−
1
w
θ
=
1
π

Arg
i
w
−
1
w
θ
−
Arg
i
w
w
+1
θy
.
The harmonic function
u
=
U
(
f
(
z
)) may be simplified to
u
=
1
π

Arg
i
(1
−
i
)
z
−
(1 +
i
)
1
−
z
θ
−
Arg
i
1
−
z
−
(1 +
i
)
z
+1
−
i
θy
and is the solution to the original Dirichlet problem in
R
.
832

833. Exercises 20.5
10.
Using H-5,
f
(
z
)=
i
1+
z
1
−
z
θ
2
maps
R
onto the upper half-plane
R
i
. The corre-
sponding Dirichlet problem in
R
i
is shown in the figure. From (3),
U
=
1
π
Arg
i
w
w
+1
θ
+
1
π
[
π
−
Arg(
w
−
1)]=1
−
1
π
Arg(
w
−
1) +
1
w
Arg
i
w
w
+1
θ
.
The harmonic function
u
=
U
(
f
(
z
)) may be simplified to
u
=1
−
1
π
Arg
i
4
z
(1
−
z
)
2
θ
+
1
π
Arg
i
(1 +
z
)
2
2(1 +
z
2
)
θ
.
11.
From Theorem 20
.
6,
u
(
x, y
)=
1
2
π
w
π
−
π
t
2
π
2
1
−|
z
|
2
|
e
it
−
z
|
2
dt
. Therefore,
u
(0
,
0) =
1
2
π
w
π
−
π
t
2
π
2
dt
=
1
3
.
To estimate
u
(0
.
5
,
0) and
u
(
−
0
.
5
,
0) we must use a numerical integration method. With the aid of Simpson’s
Rule,
u
(0
.
5
,
0) = 0
.
1516 and
u
(
−
0
.
5
,
0) = 0
.
5693.
12.
From Theorem 20
.
6,
u
(
x, y
)=
1
2
π
w
π
−
π
e
−|
t
|
1
−|
z
|
2
|
e
it
−
z
|
2
dt
. Therefore,
u
(0
,
0) =
1
2
π
w
π
−
π
e
−|
t
|
dt
=
1
π
w
π
0
e
−
t
dt
=
1
π
(1
−
e
−
π
)
.
With the aid of Simpson’s Rule,
u
(0
.
5
,
0) = 0
.
5128 and
u
(
−
0
.
5
,
0) = 0
.
1623.
13.
u
(0
.
0) =
1
2
π
w
π
−
π
u
(
e
it
)
dt
. The latter integral is just the average value of
u
(
e
it
) for
−
π
≤
t
≤
π
.
14.
For
u
(
e
iθ
) = cos 2
θ
, the Fourier series solution (6) reduces to
u
(
r, θ
)=
r
2
cos 2
θ
= Re(
z
2
)or
u
(
x, y
)=
x
2
−
y
2
.
The corresponding system of level curves is shown in the figure.
15.
For
u
(
e
iθ
) = sin
θ
+ cos
θ
, the Fourier series solution (6) reduces to
u
(
r, θ
)=
r
sin
θ
+
r
cos
θ
or
u
(
x, y
)=
y
+
x
.
The corresponding system of level curves is shown in the figure.
833

834. Exercises 20.6
Exercises 20.6
1.
g
(
z
) = cos
θ
0
−
i
sin
θ
0
=
e
−
iθ
0
is analytic everywhere and so div
F
= 0 and curl
F
=
0
by Theorem 20
.
7. A complex potential is
G
(
z
)=
e
−
iθ
0
z
and
φ
(
x, y
) = Re(
G
(
z
)) =
x
cos
θ
0
+
y
sin
θ
0
.
The equipotential lines (corresponding to
θ
0
=
π/
6) are shown in the figure.
2.
g
(
z
)=
−
y
+
xi
=
i
(
x
+
iy
)=
iz
is analytic everywhere and so div
F
= 0 and
curl
F
=
0
by Theorem 20
.
7. A complex potential is
G
(
z
)=
i
2
z
2
and
φ
(
x, y
)=Re
i
i
2
z
2
θ
=
−
xy.
The equipotential lines
−
xy
=
c
are shown in the figure and are hyperbolas.
3.
g
(
z
)=
x
x
2
+
y
2
−
y
x
2
+
y
2
i
=
1
z
is analytic for
z
t
= 0 and so div
F
= 0 and
curl
F
=
0
by Theorem 20
.
7. A complex potential is
G
(
z
)=Ln
z
and
φ
(
x, y
) = Re(
G
(
z
)) =
1
2
log
e
(
x
2
+
y
2
)
.
The equipotential lines
φ
(
x, y
)=
c
are circles
x
2
+
y
2
=
e
2
c
and are shown in the
figure.
4.
g
(
z
)=
x
2
−
y
2
−
2
xyi
(
x
2
+
y
2
)
2
=
1
z
2
is analytic for
z
t
= 0 and so div
F
= 0 and
curl
F
=
0
by Theorem 20
.
7. A complex potential is
G
(
z
)=
−
1
z
and
φ
(
x, y
)=Re
i
−
1
z
θ
=
−
x
x
2
+
y
2
.
The equipotential lines
−
x
x
2
+
y
2
=
c
can be writen as
i
x
+
1
2
c
θ
2
+
y
2
=
i
1
2
c
θ
2
for
c
t
= 0. See the figure.
5.
The mapping
f
(
z
)=
z
4
maps the wedge 0
≤
Arg
z
≤
π/
4 to the upper half-plane
R
i
and
U
=
1
π
Arg
w
is the
solution to the corresponding Dirichlet problem in
R
i
. Therefore
φ
(
x, y
)=
U
(
z
4
)=
1
π
Arg
z
4
=
4
π
Arg
z
is the potential in the wedge. A complex potential is
G
(
z
)=
4
π
Ln
z
and, since
φ
(
r, θ
)=
4
π
θ
, the equipotential
lines are the rays
θ
=
π
4
c
. Finally
F
=
G
i
(
z
)=
4
π
1
¯
z
=
4
π
i
x
x
2
+
y
2
,
y
x
2
+
y
2
θ
.
834

835. Exercises 20.6
6.
The function
f
(
z
)=
1
z
maps the original region
R
to the strip
−
1
2
≤
v
≤
0 (see
Example 2, Section 20
.
1). The boundary conditions transfer as shown in the figure.
U
=
−
2
v
is the solution in the horizontal strip and so
φ
(
x, y
)=
−
2Im
i
1
z
θ
=
2
y
x
2
+
y
2
is the potential in the original region
R
. The equipotential lines
2
y
x
2
+
y
2
=
c
may be written as
x
2
+
i
y
+
1
c
θ
2
=
i
1
c
θ
2
for
c
t
= 0 and are circles. If
c
= 0, we obtain the line
y
= 0. Note that
φ
(
x, y
)=Re
i
2
i
z
θ
and so
G
(
z
)=
2
i
z
is a complex potential. The corresponding vector field is
F
=
G
i
(
z
)=
2
i
¯
z
2
=
i
−
4
xy
(
x
2
+
y
2
)
2
,
2(
x
2
−
y
2
)
(
x
2
+
y
2
)
2
θ
.
7.
Using H-5,
w
=
i
1
−
z
1+
z
θ
2
=
i
z
−
1
z
+1
θ
2
maps
R
onto the upper half-plane
R
i
and
U
=
1
π
Arg
w
is the solution
to the corresponding Dirichlet problem in
R
i
. Therefore
µ
=
1
π
Arg
i
z
−
1
z
+1
θ
2
.
In
R
i
the equipotential lines are rays
θ
=
θ
0
. The inverse transformation is
z
=
S
(
T
(
w
)) where
T
(
w
)=
w
1
/
2
and
S
(
w
)=
1+
w
1
−
w
.
T
maps the ray
θ
=
θ
0
to the ray
θ
=
θ
0
/
2 and
S
maps
θ
=
θ
0
/
2 to an arc of a circle since
S
(0) = 1 and
S
(
∞
)=
−
1 and
S
is a linear fractional transformation.
8.
Using C-1 with
b
= 2 and
c
= 4, we have
a
=
3+
√
5
2
and
r
0
=
7
−
3
√
5
2
,so
T
(
z
)=
z
−
a
az
−
1
maps
R
onto
the annular region
R
i
defined by
r
0
≤|
w
|≤
1. The solution to the corresponding Dirichlet problem in
R
i
is
U
=
log
e
|
w
|
log
e
r
0
and so
φ
=
1
log
e
r
0
log
e
/
/
/
/
z
−
a
az
−
1
/
/
/
/
.
The level curves of
U
are circles
|
w
|
=
r
where
r
0
<r<
1, and the equipotential lines
φ
(
x, y
)=
c
are the images
of these circles under the inverse transformation
T
−
1
(
w
)=
−
w
+
a
−
aw
+1
.
T
−
1
has a pole at
w
=
1
a
=
2
3+
√
5
(
≈
0
.
38) and
r
0
<
1
a
<
1. All circles
|
w
|
=
r
are mapped onto circles in the
z
-plane with the exception of
|
w
|
=
1
a
which is mapped onto a line.
9. (a)
ψ
(
x, y
) = Im(
z
4
)=4
xy
(
x
2
−
y
2
) and so
ψ
(
x, y
) = 0 when
y
=
x
and
y
=0.
(b) V
=
G
i
(
z
)=
4
z
3
=4(
x
3
−
3
xy
2
,y
3
−
3
x
2
y
)
(c)
In polar coordinates
r
4
sin 4
θ
=
c
or
r
=(
c
csc 4
θ
)
1
/
4
, for 0
<θ<π/
4, are the
streamlines. See the figure.
835

836. Exercises 20.6
10. (a)
Since
G
(
re
iθ
)=
r
2
/
3
e
i
2
θ/
3
,
ψ
(
r, θ
) = Im(
G
(
re
iθ
)) =
r
2
/
3
sin
2
θ
3
. Note that
ψ
= 0 on the boundary where
θ
= 0 and
θ
=3
π/
2.
(b) V
=
G
i
(
z
)=
2
3
z
−
1
/
3
. Therefore, letting
z
=
re
iθ
,
V
=
2
3
r
−
1
/
3
(cos(
θ/
3)
,
sin(
θ/
3)) for 0
<θ<
3
π/
2.
(c)
r
2
/
3
sin(2
θ/
3) =
c
implies that
r
=[
c
csc(2
θ/
3)]
2
/
3
for 0
<θ<
3
π/
2. The
streamlines are shown in the figure.
11. (a)
ψ
(
x, y
) = Im(sin
z
) = cos
x
sinh
y
and
ψ
(
x, y
) = 0 when
x
=
±
π/
2 or when
y
=0.
(b) V
=
G
i
(
z
)=
cos
z
= (cos
x
cosh
y,
sin
x
sinh
y
).
(c)
cos
x
sinh
y
=
c
=
⇒
y
= sinh
−
1
(
c
sec
x
) and the streamlines are shown in the
figure.
12. (a)
The image of
R
under
w
=
i
sin
−
1
z
is the horizontal strip (see E-6)
−
π/
2
≤
v
≤
π/
2 and
ψ
(
x, y
) = Im(
i
sin
−
1
z
)=

π/
2
,x
≥
1
−
π/
2
,x
≤−
1
.
Each piece of boundary is therefore a streamline.
(b) V
=
G
i
(
z
)=
i
(1
−
z
2
)
1
/
2
=
−
i
(1
−
¯
z
2
)
1
/
2
(c)
The streamlines are the images of the lines
v
=
b
,
−
π/
2
<b<π/
2 under
z
=
−
i
sin
w
and are therefore hyperbolas. See Example 2, Section 20
.
2, and the
figure. Note that at
z
=0,
v
=
−
i
and the flow is downward.
13. (a)
If
z
=
re
iθ
,
G
(
re
iθ
)=
r
2
e
2
iθ
+
1
r
2
e
−
2
iθ
and so
ψ
(
r, θ
) = Im(
G
(
re
iθ
)) =
i
r
2
−
1
r
2
θ
sin 2
θ.
Note that
ψ
= 0 when
r
= 1 or when either
θ
=0or
θ
=
π/
2. Therefore
ψ
= 0 or the boundary of
R
.
(b) V
=
G
i
(
z
)=(
2
z
−
2
z
−
3
) and so in polar coordinates
V
=2
re
−
iθ
−
2
r
3
e
3
iθ
=2(
r
cos
θ
−
r
−
3
cos 3
θ,
−
r
sin
θ
−
r
−
3
sin 3
θ
)
.
(c)
In rectangular coordinates, the streamlines are
ψ
(
x, y
)=2
xy

1
−
1
(
x
2
+
y
2
)
2

=
c.
14. (a)
ψ
(
x, y
) = Im(
e
z
)=
e
x
sin
y
and so
ψ
= 0 when
y
=0or
π
. Therefore
ψ
= 0 on the boundary of
R
.
(b) V
=
G
i
(
z
)=
e
z
=(
e
x
cos
y
−
e
x
sin
y
)
(c)
The streamlines are
e
x
sin
y
=
c
and so
x
= log
e
(
c
csc
y
) for 0
<y<π
. See
the figure.
836

837. Exercises 20.6
15. (a)
For
f
(
z
)=
πi
−
1
2
[Ln(
z
+1)+Ln(
z
−
1)]
f
(
t
)=
πi
−
1
2
[log
e
|
t
+1
|
+ log
e
|
t
−
1
|
+
i
Arg(
t
+1)+
i
Arg(
t
−
1)]
and so Im(
f
(
t
)) =



0
,t<
−
1
π/
2
,
−
1
<t<
1
π, t >
1
. Hence Im(
G
(
z
)) =
ψ
(
x, y
) = 0 on the boundary of
R
.
(b)
x
=
−
1
2
[log
e
|
t
+1+
ic
|
+ log
e
|
t
−
1+
ic
|
],
y
=
π
−
1
2
[Arg(
t
+1+
ic
) + Arg(
t
−
1+
ic
)] for
c>
0
(c)
16. (a)
For
f
(
z
)=(
z
2
−
1)
1
/
2
,
f
(
t
)=
|
t
2
−
1
|
1
/
2
cos
i
1
2
Arg(
t
2
−
1)
θ
+
i
|
t
2
−
1
|
1
/
2
sin
i
1
2
Arg(
t
2
−
1)
θ
and so
f
(
t
)=

|
t
2
−
1
|
1
/
2
,
|
t
|
>
1
i
|
t
2
−
1
|
1
/
2
,
|
t
|
<
1
. Hence Im(
G
(
z
)) = 0 on the boundary of
R
.
(b)
x
=
|
(
t
+
ic
)
2
−
1
|
1
/
2
cos
i
1
2
Arg((
t
+
ic
)
2
−
1)
θ
,
y
=
|
(
t
+
ic
)
2
−
1
|
1
/
2
sin
i
1
2
Arg((
t
+
ic
)
2
−
1)
θ
for
c>
0
(c)
17. (a)
For
f
(
z
)=
1
π
[(
z
2
−
1)
1
/
2
+ cosh
−
1
z
],
f
(
t
)=
1
π
b
(
t
2
−
1)
1
/
2
+ cosh
−
1
t
t
=
1
π
x
(
t
2
−
1)
1
/
2
+ Ln(
t
+(
t
2
−
1)
1
/
2
)
c
and so Im(
f
(
t
)) =

1
,t<
−
1
0
,t>
1
and Re(
f
(
t
)) = 0, for
−
1
<t<
1. Hence Im(
G
(
z
)) =
ψ
(
x, y
) = 0 on the
boundary of
R
.
(b)
x
=Re
i
1
π
b
F
(
t
+
ic
)
2
−
1
V
1
/
2
+ cosh
−
1
(
t
+
ic
)
c
θ
,
y
=Im
i
1
π
b
F
(
t
+
ic
)
2
−
1
V
1
/
2
+ cosh
−
1
(
t
+
ic
)
c
θ
for
c>
0
(c)
837

838. Exercises 20.6
18. (a)
For
f
(
z
)=2(
z
+1)
1
/
2
+Ln
i
(
z
+1)
1
/
2
−
1
(
z
+1)
1
/
2
+1
θ
,
f
(
t
)=2(
t
+1)
1
/
2
+Ln
(
t
+1)
1
/
2
−
1
(
t
+1)
1
/
2
+1
.
If we write (
t
+1)
1
/
2
=
|
t
+1
|
1
/
2
e
(
i/
2)Arg(
t
+1)
, we may conclude that
Im(
f
(
t
)) =

0
,t>
0
π,
−
1
<t<
0
and Re(
f
(
t
)) = 0 for
t<
−
1
.
Therefore Im(
G
(
z
)) =
ψ
(
x, y
) = 0 on the boundary of
R
.
(b)
x
=Re

2(
t
+
ic
+1)
1
/
2
+Ln
(
t
+
ic
+1)
1
/
2
−
1
(
t
+
ic
+1)
1
/
2
+1

y
=Im

2(
t
+
ic
+1)
1
/
2
+Ln
(
t
+
ic
+1)
1
/
2
−
1
(
t
+
ic
+1)
1
/
2
+1

for
c>
0
(c)
19.
In Example 5,
V
=(2
x,
−
2
y
) and so the only stagnation point occurs at
z
= 0. In Example 6,
V
=1
−
1
/
¯
z
2
and so if
V
=
0
,¯
z
2
= 1. Therefore
z
=1,
−
1 are the only stagnation points.
20. (a)
ψ
(
x, y
) = Im(
G
(
z
)) =
k
Arg(
z
−
x
1
) and so if
ψ
(
x, y
)=
c
, Arg(
z
−
x
1
) is constant. This implies that the
streamlines are rays with vertex at
z
=
x
1
.
(b) V
=
G
i
(
z
)=
k
z
−
x
1
=
k
z
−
x
1
=
k
|
z
−
x
1
|
2
(
z
−
x
1
).
The direction of the flow is determined by the sign of
k
, and if
k<
0 the flow is directed towards
z
=
x
1
.
21.
f
(
z
)=
z
2
maps the first quadrant onto the upper half-plane and
f
(
ξ
0
)=
f
(1) = 1. Therefore
G
(
z
) = Ln(
z
2
−
1)
is the complex potential, and so
ψ
(
x, y
) = Arg(
z
2
−
1) = tan
−
1
i
2
xy
x
2
−
y
2
−
1
θ
is the streamline function where tan
−
1
is chosen to be between 0 and
π
.If
ψ
(
x, y
)=
c
, then
x
2
+
Bxy
−
y
2
−
1=0
where
B
=
−
2 cot
c
. Each hyperbola in the family passes through (1
,
0) and the boundary of the first quadrant
satisfies
ψ
(
x, y
)=0.
22. (a)
From E-5,
f
(
z
)=
e
z
maps the horizontal strip 0
<y<π
onto the upper half-plane and
f
(
ξ
0
)=
f
(0)=1.
Therefore
G
(
z
)=
k
Ln(
e
z
−
1) is a complex potential. To construct a sink at
ξ
0
= 0, we must have
k<
0.
838

839. Chapter 20 Review Exercises
(b)
ψ
(
x, y
) = Im(
k
Ln(
e
z
−
1)) =
k
Arg(
e
z
−
1) =
k
tan
−
1
i
e
x
sin
y
e
x
cos
y
−
1
θ
where tan
−
1
is chosen to be between 0 and
π
. If we set
k
=
−
1, then the
streamlines
ψ
(
x, y
)=
c
,
−
π<c<
0, satisfy
e
x
[
B
cos
y
−
sin
y
]=
B
where
B
=
−
tan
c
, and so
x
= log
e

B
B
cos
y
−
sin
y

.
Note that
B
will vary over all real values. The streamlines are also the images of rays through
w
= 1 under
the inverse transformation
z
=Ln
w
. See the figure.
23.
ψ
= Im(
G
(
z
)) =
k
Arg(
z
−
1)
−
k
Arg(
z
+1)=
k
Arg
i
z
−
1
z
+1
θ
.
(See Figure 20
.
55 in the text). In rectangular coordinates
ψ
(
x, y
)=
k
tan
−
1
i
2
y
x
2
+
y
2
−
1
θ
where tan
−
1
is chosen to be between 0 and
π
. Level curves
ψ
(
x, y
)=
c
can be put in the form
x
2
+
y
2
−
2
By
=1 or
x
2
+(
y
−
B
)
2
=1+
B
2
.
Each member of the family passes through (1
,
0) and (
−
1
,
0).
24. (a) V
=
a
+
ib
¯
z
=
i
ax
−
by
x
2
+
y
2
,
bx
+
ay
x
2
+
y
2
θ
and since (
x
i
(
t
)
,y
i
(
t
)) =
V
, the path of the particle satisfies
dx
dt
=
ax
−
by
x
2
+
y
2
,
dy
dt
=
bx
+
ay
x
2
+
y
2
.
(b)
Switching to polar coordinates,
dr
dt
=
1
r
i
x
dx
dt
+
y
dy
dt
θ
=
1
r
i
ax
2
−
bxy
r
2
+
bxy
+
ay
2
r
2
θ
=
a
r
dθ
dt
=
1
r
2
i
−
y
dx
dt
+
x
dy
dt
θ
=
1
r
2
i
−
axy
+
by
2
r
2
+
bx
2
+
axy
r
2
θ
=
b
r
2
.
Therefore
dr
dθ
=
a
b
r
and so
r
=
ce
aθ/b
.
(c)
dr
dt
=
a
r
<
0 if and only if
a<
0, and in this case
r
is decreasing and the curve spirals inward.
dθ
dt
=
b
r
2
<
0
if and only if
b<
0, and in this case
θ
is decreasing and the curve is traversed clockwise.
Chapter 20 Review Exercises
1.
f
(
z
)=
x
2
−
y
2
+2
xyi
and so the hyperbola
xy
= 2 is mapped onto the line
v
=4.
2.
−
i
=
e
−
iπ/
2
and so
f
(
z
)=
−
iz
is a rotation through
−
90
◦
.
3.
The wedge 0
≤
Arg
w
≤
2
π/
3. See figure 20
.
6 in the text.
4.
f
i
(
z
) = sinh
z
= 0 for
z
=
±
πi
, and so
f
(
z
) = cosh
z
is conformal except at
z
=
±
nπi
.
839

840. Chapter 20 Review Exercises
5.
True, by Theorem 20
.
2. Arg
w
is harmonic and is the upper half-plane
v
≥
0.
6.
A line, since
|
z
−
1
|
= 1 passes through the pole at
z
=2.
7.
0, 1,
∞
.
8.
True.
α
1
=
α
2
=
α
3
=
π/
2
9.
False.
g
(
z
)=
P
−
iQ
is analytic. See Theorem 20
.
7.
10.
iG
(
z
)=
−
ψ
(
x, y
)+
iφ
(
x, y
) is analytic in
R
and Im(
G
(
z
)) =
φ
(
x, y
). True
11.
If
z
=
re
iθ
and 0
<θ<π/
2,
w
=Ln
z
= log
e
r
+
iθ
. Therefore
v
=
θ
and so 0
<v<π/
2. The image of the
first quadrant is the strip 0
<v<π/
2 in the
w
-plane. Rays
θ
=
θ
0
are mapped onto horizontal lines
v
=
θ
0
.
12.
First use
z
1
=
z
2
to map the first quadrant onto the upper half-plane
y
1
≥
0, and segment
AB
to the negative
real axis. We then use
w
=
1
π
(
z
1
+Ln
z
1
+ 1) to map this half-plane onto the target region
R
i
. The composite
transformation is
w
=
1
π
[
z
2
+ Ln(
z
2
)+1]
and the image of
AB
is the ray extending to the left from
w
=
i
along the line
v
=1.
13.
From H-4,
z
1
= cos
πz
maps
R
onto the upper half-plane
y
1
≥
0, and
AB
onto the real segment (
−∞
,
−
1].
Using C-4,
w
=
i
−
z
1
1+
z
1
maps this upper half-plane onto the target region
R
i
. The composite transformation is
w
=
i
−
cos
πz
i
+ cos
πz
and the image of
AB
is the circular arc lying in quadrants II and III.
14.
The power function
w
=
z
4
maps
R
onto the upper half-plane
v
≥
0; the trans-
ferred boundary conditions are shown in the figure. From (2) of Section 20
.
5,
U
=
1
π
Arg
i
w
−
1
w
+1
θ
and so
u
=
1
π
Arg
i
z
4
−
1
z
4
+1
θ
is the solution to the original Dirichlet problem in
R
.
15.
The inversion
w
=1
/z
maps
R
onto the horizontal strip
−
1
≤
v
≤−
1
/
2 and the
transferred boundary conditions are shown in the figure. The solution in the strip
is
U
=2
v
+ 2 and so
u
=
U
i
1
z
θ
= 2Im
i
1
z
θ
+2=
−
2
y
x
2
+
y
2
+2
is the solution to the original Dirichlet problem in
R
.
16.
Using the cross ratio formula,
(
w
−
i
)(
−
i
+1)
(
w
+ 1)(
−
2
i
)
=
(
z
−
1)
·
1
1
·
(
−
2)
we see that
w
=
−
z
−
i
z
+
i
maps 1,
−
1,
∞
to
i
,
−
i
,
−
1.
17. (a)
We note that as
u
1
→
+
∞
,
α
1
→
0,
α
2
→
2
π
, and
α
3
→
0.
(b)
Since
f
i
(
z
)=
A
(
z
+1)
1
/π
−
1
z
α
2
/π
−
1
(
z
−
1)
α
3
/π
−
1
,
840

841. Chapter 20 Review Exercises
this suggests that we examine
f
i
(
z
)=
A
(
z
+1)
−
1
z
(
z
−
1)
−
1
=
A
z
z
2
−
1
.
We may write
f
i
(
z
)=
A
2
i
1
z
+1
+
1
z
−
1
θ
and so
f
(
z
)=
1
2
A
[Ln(
z
+1)+Ln(
z
−
1)] +
B.
(c)
For
t
real, we can write
f
(
t
)=
1
2
A
[log
e
|
t
+1
|
+ log
e
|
t
−
1
|
+
i
Arg(
t
+1)+
i
Arg(
t
−
1)] +
B.
Since
f
(0) =
πi
2
,
πi
2
=
πi
2
A
+
B
.For
t
real,
f
(
t
)=
R
1
2
A
[ln
|
t
2
−
1
|
+2
πi
]+
B, t <
−
1
1
2
A
ln
|
t
2
−
1
|
+
B, t >
1
.
If
A
is real and we require that Im(
f
(
t
)) = 0 for
t<
−
1, then 0 =
Aπi
+ Im(
B
). If Im(
f
(
t
)) =
π
for
t>
1,
then
π
= Im(
B
). All three equations are satisfied by letting
B
=
πi
and
A
=
−
1. Therefore
f
(
z
)=
πi
−
1
2
[Ln(
z
+1)+Ln(
z
−
1)]
.
18. (a)
From Theorem 20
.
5,
u
(
x, y
)=
y
π
w
∞
−∞
sin
t
(
x
−
t
)
2
+
y
2
dt
=
y
π
w
∞
−∞
sin(
x
−
s
)
s
2
+
y
2
]
ds
(letting
s
=
x
−
t
)
.
But sin(
x
−
s
) = sin
x
cos
s
−
cos
x
sin
s
. We now proceed as in the solution to Problem 6, Section 20
.
5to
show that
u
(
x, y
)=
e
−
y
sin
x
.
(b)
For
u
(
e
iθ
) = sin
θ
, the Fourier Series solution (6) in Section 20
.
5 reduces to
u
(
r, θ
)=
r
sin
θ
.
19.
If
f
(
w
)=
w
+
e
w
+1,
G
(
z
)=
f
−
1
(
z
) maps
R
onto the strip 0
≤
v
≤
π
and the
transferred boundary conditions are shown in the figure. The solution for the strip
is
U
=
v/π
and so the solution in
R
is
φ
(
x, y
)=
U
(
G
(
z
)) =
1
π
Im(
G
(
z
)) =
1
π
ψ
(
x, y
)
.
Therefore the equipotential lines
φ
(
x, y
)=
c
are the streamlines
ψ
(
x, y
)=
cπ
of the flow in Figure 20
.
72.
20.
G
(
re
iθ
)=
−
r
1
/
2
sin(
θ/
2) +
i
[
r
1
/
2
cos(
θ/
2)
−
1] and so
ψ
(
r, θ
)=
√
r
cos(
θ/
2) = 1. If
ψ
(
r, θ
)=0,
r
cos
2
(
θ/
2) = 1 or
r
cos
θ
+
r
= 2 (using cos
2
(
θ/
2) = (1 + cos
θ
)
/
2.) In
rectangular coordinates,
x
+
n
x
2
+
y
2
=2or
y
2
=4
−
4
x
. Therefore the boundary
of
R
is a stramline. To sketch the streamlines, note that
ψ
(
r, θ
)=
c
implies that
r
=(
c
+1)
2
sec
2
(
θ/
2). See the figure.
841