Answers to Problems for Advanced Engineering Mathematics 6th Edition International Student Edition by Dennis Zill

Simplify your journey through advanced mathematical concepts with this comprehensive collection of answers to problems found in Advanced Engineering Mathematics 6th Edition by Dennis Zill. Ideal for engineering students, professionals, and educators, this resource explains challenging topics like differential equations, vector calculus, and linear algebra in a step-by-step, understandable manner.

Designed to clarify complex problems, the guide includes solutions to every exercise, offering insights into problem-solving techniques that are critical for mastering engineering mathematics. Perfect for exam preparation, project work, or simply advancing your understanding, this is your key to academic success in engineering fields.

1. Chapter 1
Introduction to Differential Equations
1.1 Definitions and Terminology
1.Second order; linear
2.Third order; nonlinear because of (dy/dx)
4
3.Fourth order; linear
4.Second order; nonlinear because of cos(r+u)
5.Second order; nonlinear because of (dy/dx)
2
or
p
1 + (dy/dx)
2
6.Second order; nonlinear because ofR
2
7.Third order; linear
8.Second order; nonlinear because of ˙x
2
9.Writing the differential equation in the formx(dy/dx) +y
2
= 1, we see that it is nonlinear
inybecause ofy
2
. However, writing it in the form (y
2
−1)(dx/dy) +x= 0, we see that it is
linear inx.
10.Writing the differential equation in the formu(dv/du)+(1+u)v=ue
u
we see that it is linear
inv. However, writing it in the form (v+uv−ue
u
)(du/dv)+u= 0, we see that it is nonlinear
inu.
11.Fromy=e
−x/2
we obtainy
′
=−
1
2
e
−x/2
. Then 2y
′
+y=−e
−x/2
+e
−x/2
= 0.
12.Fromy=
6
5
−
6
5
e
−20t
we obtaindy/dt= 24e
−20t
, so that
dy
dt
+ 20y= 24e
−20t
+ 20
θ
6
5
−
6
5
e
−20t

= 24.
13.Fromy=e
3x
cos 2xwe obtainy
′
= 3e
3x
cos 2x−2e
3x
sin 2xandy
′′
= 5e
3x
cos 2x−12e
3x
sin 2x,
so thaty
′′
−6y
′
+ 13y= 0.
1
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2. 2 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
14.Fromy=−cosxln(secx+ tanx) we obtainy
′
=−1 + sinxln(secx+ tanx) and
y
′′
= tanx+ cosxln(secx+ tanx). Theny
′′
+y= tanx.
15.The domain of the function, found by solvingx+2≥0, is [−2,∞). Fromy
′
= 1+2(x+2)
−1/2
we have
(y−x)y
′
= (y−x)[1 + (2(x+ 2)
−1/2
]
=y−x+ 2(y−x)(x+ 2)
−1/2
=y−x+ 2[x+ 4(x+ 2)
1/2
−x](x+ 2)
−1/2
=y−x+ 8(x+ 2)
1/2
(x+ 2)
−1/2
=y−x+ 8.
An interval of definition for the solution of the differential equation is (−2,∞) becausey
′
is
not defined atx=−2.
16.Since tanxis not defined forx=π/2 +nπ,nan integer, the domain ofy= 5 tan 5xis
{x


5x6=π/2 +nπ}
or{x


x6=π/10 +nπ/5}. Fromy
′
= 25 sec
2
5xwe have
y
′
= 25(1 + tan
2
5x) = 25 + 25 tan
2
5x= 25 +y
2
.
An interval of definition for the solution of the differential equation is (−π/10, π/10). Another
interval is (π/10,3π/10), and so on.
17.The domain of the function is{x


4−x
2
6= 0}or{x


x6=−2 orx6= 2}. Fromy
′
=
2x/(4−x
2
)
2
we have
y
′
= 2x
θ
1
4−x
2

2
= 2xy
2
.
An interval of definition for the solution of the differential equation is (−2,2). Other intervals
are (−∞,−2) and (2,∞).
18.The function isy= 1/
√
1−sinx, whose domain is obtained from 1−sinx6= 0 or sinx6= 1.
Thus, the domain is{x


x6=π/2 + 2nπ}. Fromy
′
=−
1
2
(1−sinx)
−3/2
(−cosx) we have
2y
′
= (1−sinx)
−3/2
cosx= [(1−sinx)
−1/2
]
3
cosx=y
3
cosx.
An interval of definition for the solution of the differential equation is (π/2,5π/2). Another
one is (5π/2,9π/2), and so on.

3. 1.1 Definitions and Terminology 3
19.Writing ln(2X−1)−ln(X−1) =tand differentiating
implicitly we obtain
2
2X−1
dX
dt
−
1
X−1
dX
dt
= 1
θ
2
2X−1
−
1
X−1

dX
dt
= 1
2X−2−2X+ 1
(2X−1) (X−1)
dX
dt
= 1
dX
dt
=−(2X−1)(X−1) = (X−1)(1−2X).
– 4–2 2 4
t
– 4
–2
2
4
x
Exponentiating both sides of the implicit solution we obtain
2X−1
X−1
=e
t
2X−1 =Xe
t
−e
t
(e
t
−1) = (e
t
−2)X
X=
e
t
−1
e
t
−2
.
Solvinge
t
−2 = 0 we gett= 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.
20.Implicitly differentiating the solution, we obtain
−2x
2
dy
dx
−4xy+ 2y
dy
dx
= 0
−x
2
dy−2xy dx+y dy= 0
2xy dx+ (x
2
−y)dy= 0.
Using the quadratic formula to solvey
2
−2x
2
y−1 = 0
fory, we gety=
−
2x
2
±
√
4x
4
+ 4
·
/2 =x
2
±
√
x
4
+ 1 .
Thus, two explicit solutions arey1=x
2
+
√
x
4
+ 1 and
y2=x
2
−
√
x
4
+ 1 . Both solutions are defined on (−∞,∞).
The graph ofy1(x) is solid and the graph ofy2is dashed.
– 4–2 2 4
x
– 4
–2
2
4
y

4. 4 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
21.DifferentiatingP=c1e
t
/
−
1 +c1e
t
·
we obtain
dP
dt
=
−
1 +c1e
t
·
c1e
t
−c1e
t
·c1e
t
(1 +c1e
t
)
2
=
c1e
t
1 +c1e
t
×−
1 +c1e
t
·
−c1e
t

1 +c1e
t
=
c1e
t
1 +c1e
t
≤
1−
c1e
t
1 +c1e
t
≥
=P(1−P).
22.Differentiatingy= 2x
2
−1 +c1e
−2x
2
we obtain
dy
dx
= 4x−4xc1e
−2x
2
, so that
dy
dx
+ 4xy= 4x−4xc1e
−2x
2
+ 8x
3
−4x+ 4c1xe
−x
2
= 8x
3
23.Fromy=c1e
2x
+c2xe
2x
we obtain
dy
dx
= (2c1+c2)e
2x
+ 2c2xe
2x
and
d
2
y
dx
2
= (4c1+ 4c2)e
2x
+
4c2xe
2x
, so that
d
2
y
dx
2
−4
dy
dx
+ 4y= (4c1+ 4c2−8c1−4c2+ 4c1)e
2x
+ (4c2−8c2+ 4c2)xe
2x
= 0.
24.Fromy=c1x
−1
+c2x+c3xlnx+ 4x
2
we obtain
dy
dx
=−c1x
−2
+c2+c3+c3lnx+ 8x,
d
2
y
dx
2
= 2c1x
−3
+c3x
−1
+ 8,
and
d
3
y
dx
3
=−6c1x
−4
−c3x
−2
,
so that
x
3
d
3
y
dx
3
+ 2x
2
d
2
y
dx
2
−x
dy
dx
+y= (−6c1+ 4c1+c1+c1)x
−1
+ (−c3+ 2c3−c2−c3+c2)x
+ (−c3+c3)xlnx+ (16−8 + 4)x
2
= 12x
2
In Problems25–28, we use the Product Rule and the derivative of an integral ((12) of this section):
d
dx
Z
x
a
g(t)dt=g(x).
25.Differentiatingy=e
3x
Z
x
1
e
−3t
t
dtwe obtain
dy
dx
=e
3x
Z
x
1
e
−3t
t
dt+
e
−3t
x
·e
3x
or
dy
dx
=e
3x
Z
x
1
e
−3t
t
dt+
1
x
, so that
x
dy
dx
−3xy=x
θ
e
3x
Z
x
1
e
−3t
t
dt+
1
x

−3x
θ
e
3x
Z
x
1
e
−3t
t
dt

=xe
3x
Z
x
1
e
−3t
t
dt+ 1−3xe
3x
Z
x
1
e
−3t
t
dt= 1

5. 1.1 Definitions and Terminology 5
26.Differentiatingy=
√
x
Z
x
4
cost
√
t
dtwe obtain
dy
dx
=
1
2
√
x
Z
x
4
cost
√
t
dt+
cosx
√
x
·
√
xor
dy
dx
=
1
2
√
x
Z
x
4
cost
√
t
dt+ cosx, so that
2x
dy
dx
−y= 2x
θ
1
2
√
x
Z
x
4
cost
√
t
dt+ cosx

−
√
x
Z
x
4
cost
√
t
dt
=
√
x
Z
x
4
cost
√
t
dt+ 2xcosx−
√
x
Z
x
4
cost
√
t
dt= 2xcosx
27.Differentiatingy=
5
x
+
10
x
Z
x
1
sint
t
dtwe obtain
dy
dx
=−
5
x
2
−
10
x
2
Z
x
1
sint
t
dt+
sinx
x
·
10
x
or
dy
dx
=−
5
x
2
−
10
x
2
Z
x
1
sint
t
dt+
10 sinx
x
2
, so that
x
2
dy
dx
+xy=x
2
θ
−
5
x
2
−
10
x
2
Z
x
1
sint
t
dt+
10 sinx
x
2

+x
θ
5
x
+
10
x
Z
x
1
sint
t
dt

=−5−10
Z
x
1
sint
t
dt+ 10 sinx+ 5 + 10
Z
x
1
sint
t
dt= 10 sinx
28.Differentiatingy=e
−x
2
+e
−x
2
Z
x
0
e
t
2
dtwe obtain
dy
dx
=−2xe
−x
2
−2xe
−x
2
Z
x
0
e
t
2
dt+e
x
2
·e
−x
2
or
dy
dx
=−2xe
−x
2
−2xe
−x
2
Z
x
0
e
t
2
dt+ 1, so that
dy
dx
+ 2xy=
θ
−2xe
−x
2
−2xe
−x
2
Z
x
0
e
t
2
dt+ 1

+ 2x
θ
e
−x
2
+e
−x
2
Z
x
0
e
t
2
dt

=−2xe
−x
2
−2xe
−x
2
Z
x
0
e
t
2
dt+ 1 + 2xe
−x
2
+ 2xe
−x
2
Z
x
0
e
t
2
dt= 1
29.From
y=
(
−x
2
, x <0
x
2
, x≥0
we obtain
y
′
=
(
−2x, x <0
2x, x≥0
so thatxy
′
−2y= 0.
30.The functiony(x) is not continuous atx= 0 since lim
x→0
−
y(x) = 5 and lim
x→0
+
y(x) =−5. Thus,
y
′
(x) does not exist atx= 0.
31.Force the functiony=e
mx
into the equationy
′
+ 2y= 0 to get
(e
mx
)
′
+ 2(e
mx
) = 0
me
mx
+ 2e
mx
= 0
e
mx
(m+ 2) = 0
Now sincee
mx
>0 for all values ofx, we must havem=−2 and soy=e
−2x
is a solution.

6. 6 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
32.Force the functiony=e
mx
into the equation 3y
′
−4y= 0 to get
3(e
mx
)
′
−4(e
mx
) = 0
3me
mx
−4e
mx
= 0
e
mx
(3m−4) = 0
Now sincee
mx
>0 for all values ofx, we must havem= 4/3 and soy=e
4x/3
is a solution.
33.Force the functiony=e
mx
into the equationy
′′
−5y
′
+ 6y= 0 to get
(e
mx
)
′′
−5(e
mx
)
′
+ 6(e
mx
) = 0
m
2
e
mx
−5me
mx
+ 6e
mx
= 0
e
mx
(m
2
−5m+ 6) = 0
e
mx
(m−2)(m−3) = 0
Now sincee
mx
>0 for all values ofx, we must havem= 2 andm= 3 thereforey=e
2x
and
y=e
3x
are solutions.
34.Force the functiony=e
mx
into the equation 2y
′′
+ 9y
′
−5y= 0 to get
2(e
mx
)
′′
+ 9(e
mx
)
′
−5(e
mx
) = 0
2m
2
e
mx
+ 9me
mx
−5e
mx
= 0
e
mx
(2m
2
+ 9m−5) = 0
e
mx
(m+ 5)(2m−1) = 0
Now sincee
mx
>0 for all values ofx, we must havem=−5 andm= 1/2 thereforey=e
−5x
andy=e
x/2
are solutions.
35.Force the functiony=x
m
into the equationxy
′′
+ 2y
′
= 0 to get
x·(x
m
)
′′
+ 2(x
m
)
′
= 0
x·m(m−1)x
m−2
+ 2mx
m−1
= 0
(m
2
−m)x
m−1
+ 2mx
m−1
= 0
x
m−1
[m
2
+m] = 0
x
m−1
[m(m+ 1)] = 0
The last line implies thatm= 0 andm=−1 thereforey=x
0
= 1 andy=x
−1
are solutions.

7. 1.1 Definitions and Terminology 7
36.Force the functiony=x
m
into the equationx
2
y
′′
−7xy
′
+ 15y= 0 to get
x
2
·(x
m
)
′′
−7x·(x
m
)
′
+ 15(x
m
) = 0
x
2
·m(m−1)x
m−2
−7x·mx
m−1
+ 15x
m
= 0
(m
2
−m)x
m
−7mx
m
+ 15x
m
= 0
x
m
[m
2
−8m+ 15] = 0
x
m
[(m−3)(m−5)] = 0
The last line implies thatm= 3 andm= 5 thereforey=x
3
andy=x
5
are solutions.
In Problems37–40, we substitutey=cinto the differential equations and usey
′
= 0andy
′′
= 0
37.Solving 5c= 10 we see thaty= 2 is a constant solution.
38.Solvingc
2
+ 2c−3 = (c+ 3)(c−1) = 0 we see thaty=−3 andy= 1 are constant solutions.
39.Since 1/(c−1) = 0 has no solutions, the differential equation has no constant solutions.
40.Solving 6c= 10 we see thaty= 5/3 is a constant solution.
41.Fromx=e
−2t
+ 3e
6t
andy=−e
−2t
+ 5e
6t
we obtain
dx
dt
=−2e
−2t
+ 18e
6t
and
dy
dt
= 2e
−2t
+ 30e
6t
.
Then
x+ 3y= (e
−2t
+ 3e
6t
) + 3(−e
−2t
+ 5e
6t
) =−2e
−2t
+ 18e
6t
=
dx
dt
and
5x+ 3y= 5(e
−2t
+ 3e
6t
) + 3(−e
−2t
+ 5e
6t
) = 2e
−2t
+ 30e
6t
=
dy
dt
.
42.Fromx= cos 2t+ sin 2t+
1
5
e
t
andy=−cos 2t−sin 2t−
1
5
e
t
we obtain
dx
dt
=−2 sin 2t+ 2 cos 2t+
1
5
e
t
and
dy
dt
= 2 sin 2t−2 cos 2t−
1
5
e
t
and
d
2
x
dt
2
=−4 cos 2t−4 sin 2t+
1
5
e
t
and
d
2
y
dt
2
= 4 cos 2t+ 4 sin 2t−
1
5
e
t
.
Then
4y+e
t
= 4(−cos 2t−sin 2t−
1
5
e
t
) +e
t
=−4 cos 2t−4 sin 2t+
1
5
e
t
=
d
2
x
dt
2
and
4x−e
t
= 4(cos 2t+ sin 2t+
1
5
e
t
)−e
t
= 4 cos 2t+ 4 sin 2t−
1
5
e
t
=
d
2
y
dt
2
.

8. 8 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
43.(y
′
)
2
+ 1 = 0 has no real solutions because (y
′
)
2
+ 1 is positive for all differentiable functions
y=φ(x).
44.The only solution of (y
′
)
2
+y
2
= 0 isy= 0, since ify6= 0,y
2
>0 and (y
′
)
2
+y
2
≥y
2
>0.
45.The first derivative off(x) =e
x
ise
x
. The first derivative off(x) =e
kx
iske
kx
. The
differential equations arey
′
=yandy
′
=ky, respectively.
46.Any function of the formy=ce
x
ory=ce
−x
is its own second derivative. The corresponding
differential equation isy
′′
−y= 0. Functions of the formy=csinxory=ccosxhave second
derivatives that are the negatives of themselves. The differential equation isy
′′
+y= 0.
47.We first note that
p
1−y
2
=
p
1−sin
2
x=
√
cos
2
x=|cosx|. This prompts us to consider
values ofxfor which cosx <0, such asx=π. In this case
dy
dx





x=π
=
d
dx
(sinx)





x=π
= cosx


x=π
= cosπ=−1,
but
p
1−y
2
|x=π=
p
1−sin
2
π=
√
1 = 1.
Thus,y= sinxwill only be a solution ofy
′
=
p
1−y
2
when cosx >0. An interval of
definition is then (−π/2, π/2). Other intervals are (3π/2,5π/2), (7π/2,9π/2), and so on.
48.Since the first and second derivatives of sintand costinvolve sintand cost, it is plausible that
a linear combination of these functions,Asint+Bcost, could be a solution of the differential
equation. Usingy
′
=Acost−Bsintandy
′′
=−Asint−Bcostand substituting into the
differential equation we get
y
′′
+ 2y
′
+ 4y=−Asint−Bcost+ 2Acost−2Bsint+ 4Asint+ 4Bcost
= (3A−2B) sint+ (2A+ 3B) cost= 5 sint
Thus 3A−2B= 5 and 2A+ 3B= 0. Solving these simultaneous equations we findA=
15
13
andB=−
10
13
. A particular solution isy=
15
13
sint−
10
13
cost.
49.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).
50.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 firstquadrant. 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.

9. 1.1 Definitions and Terminology 9
51.Differentiating (x
3
+y
3
)/xy= 3cwe obtain
xy(3x
2
+ 3y
2
y
′
)−(x
3
+y
3
)(xy
′
+y)
x
2
y
2
= 0
3x
3
y+ 3xy
3
y
′
−x
4
y
′
−x
3
y−xy
3
y
′
−y
4
= 0
(3xy
3
−x
4
−xy
3
)y
′
=−3x
3
y+x
3
y+y
4
y
′
=
y
4
−2x
3
y
2xy
3
−x
4
=
y(y
3
−2x
3
)
x(2y
3
−x
3
)
.
52.A tangent line will be vertical wherey
′
is undefined, or in this case, wherex(2y
3
−x
3
) = 0.
This givesx= 0 and 2y
3
=x
3
. Substitutingy
3
=x
3
/2 intox
3
+y
3
= 3xywe get
x
3
+
1
2
x
3
= 3x
θ
1
2
1/3
x

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 atx= 0 andx= 2
2/3
, or at (0,0) and (2
2/3
,2
1/3
).
Since 2
2/3
≈1.59, the estimates of the domains in Problem 50 were close.
53.The derivatives of the functions areφ
′
1
(x) =−x/
√
25−x
2
andφ
′
2
(x) =x/
√
25−x
2
, neither
of which is defined atx=±5.
54.To determine if a solution curve passes through (0,3) we lett= 0 andP= 3 in the equation
P=c1e
t
/(1 +c1e
t
). This gives 3 =c1/(1 +c1) orc1=−
3
2
. Thus, the solution curve
P=
(−3/2)e
t
1−(3/2)e
t
=
−3e
t
2−3e
t
passes through the point (0,3). Similarly, lettingt= 0 andP= 1 in the equation for the
one-parameter family of solutions gives 1 =c1/(1 +c1) orc1= 1 +c1. Since this equation
has no solution, no solution curve passes through (0,1).
55.For the first-order differential equation integratef(x). For the second-order differential equa-
tion integrate twice. In the latter case we gety=
R
(
R
f(x)dx)dx+c1x+c2.
56.Solving fory
′
using the quadratic formula we obtain the two differential equations
y
′
=
1
x

2 + 2
p
1 + 3x
6

andy
′
=
1
x

2−2
p
1 + 3x
6

,
so the differential equation cannot be put in the formdy/dx=f(x, y).

10. 10 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
57.The differential equationyy
′
−xy= 0 has normal formdy/dx=x. These are not equivalent
becausey= 0 is a solution of the first differential equation but not a solution of the second.
58.Differentiating we gety
′
=c1+ 3c2x
2
andy
′′
= 6c2x. Thenc2=y
′′
/6xandc1=y
′
−xy
′′
/2,
so
y=
θ
y
′
−
xy
′′
2

x+
θ
y
′′
6x

x
3
=xy
′
−
1
3
x
2
y
′′
and the differential equation isx
2
y
′′
−3xy
′
+ 3y= 0.
59. (a)Sincee
−x
2
is positive for all values ofx,dy/dx >0 for allx, and a solution,y(x), of the
differential equation must be increasing on any interval.
(b)lim
x→−∞
dy
dx
= lim
x→−∞
e
−x
2
= 0 and lim
x→∞
dy
dx
= lim
x→∞
e
−x
2
= 0. Sincedy/dxapproaches 0 as
xapproaches−∞and∞, the solution curve has horizontal asymptotes to the left and
to the right.
(c)To test concavity we consider the second derivative
d
2
y
dx
2
=
d
dx
θ
dy
dx

=
d
dx

e
−x
2

=−2xe
−x
2
.
Since the second derivative is positive forx <0 and negative forx >0, the solution
curve is concave up on (−∞,0) and concave down on (0,∞).
(d)
x
y
60. (a)The derivative of a constant solutiony=cis 0, so solving 5−c= 0 we see thatc= 5
and soy= 5 is a constant solution.
(b)A solution is increasing wheredy/dx= 5−y >0 ory <5. A solution is decreasing
wheredy/dx= 5−y <0 ory >5.
61. (a)The derivative of a constant solution is 0, so solvingy(a−by) = 0 we see thaty= 0 and
y=a/bare constant solutions.
(b)A solution is increasing wheredy/dx=y(a−by) =by(a/b−y)>0 or 0< y < a/b. A
solution is decreasing wheredy/dx=by(a/b−y)<0 ory <0 ory > a/b.

11. 1.1 Definitions and Terminology 11
(c)Using implicit differentiation we compute
d
2
y
dx
2
=y(−by
′
) +y
′
(a−by) =y
′
(a−2by).
Solvingd
2
y/dx
2
= 0 we obtainy=a/2b. Sinced
2
y/dx
2
>0 for 0< y < a/2band
d
2
y/dx
2
<0 fora/2b < y < a/b, the graph ofy=φ(x) has a point of inflection at
y=a/2b.
(d)
y = a/b
y = 0
x
y
62. (a)Ify=cis a constant solution theny
′
= 0, butc
2
+ 4 is never 0 for any real value ofc.
(b)Sincey
′
=y
2
+ 4>0 for allxwhere a solutiony=φ(x) is defined, any solution must
be increasing on any interval on which it is defined. Thus it cannot have any relative
extrema.
(c)Using implicit differentiation we computed
2
y/dx
2
= 2yy
′
= 2y(y
2
+ 4). Setting
d
2
y/dx
2
= 0 we see thaty= 0 corresponds to the only possible point of inflection.
Sinced
2
y/dx
2
<0 fory <0 andd
2
y/dx
2
>0 fory >0, there is a point of inflection
wherey= 0.
(d)
x
y

12. 12 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
63.InMathematicause
Clear[y]
y[x ]:= x Exp[5x] Cos[2x]
y[x]
y''''[x]−20y'''[x] + 158y''[x]−580y'[x] +841y[x]//Simplify
The output will showy(x) =e
5x
xcos 2x, which verifies that the correct function was entered,
and 0, which verifies that this function is a solution of the differential equation.
64.InMathematicause
Clear[y]
y[x]:= 20Cos[5Log[x]]/x−3Sin[5Log[x]]/x
y[x]
xˆ3 y'''[x] + 2xˆ2 y''[x] + 20x y'[x]−78y[x]//Simplify
The output will showy(x) = 20 cos(5 lnx)/x−3 sin(5 lnx)/x, which verifies that the correct
function was entered, and 0, which verifies that this function is a solution of the differential
equation.
1.2Initial-Value Problems
1.Solving−1/3 = 1/(1 +c1) we getc1=−4. The solution isy= 1/(1−4e
−x
).
2.Solving 2 = 1/(1 +c1e) we getc1=−(1/2)e
−1
. The solution isy= 2/(2−e
−(x+1)
) .
3.Lettingx= 2 and solving 1/3 = 1/(4 +c) we getc=−1. The solution isy= 1/(x
2
−1).
This solution is defined on the interval (1,∞).
4.Lettingx=−2 and solving 1/2 = 1/(4 +c) we getc=−2. The solution isy= 1/(x
2
−2).
This solution is defined on the interval (−∞,−
√
2 ).
5.Lettingx= 0 and solving 1 = 1/cwe getc= 1. The solution isy= 1/(x
2
+ 1). This solution
is defined on the interval (−∞,∞).
6.Lettingx= 1/2 and solving−4 = 1/(1/4 +c) we getc=−1/2. The solution isy=
1/(x
2
−1/2) = 2/(2x
2
−1). This solution is defined on the interval (−1/
√
2,1/
√
2 ).
In Problems7–10, we usex=c1cost+c2sintandx
′
=−c1sint+c2costto obtain a system of
two equations in the two unknownsc1andc2.

13. 1.2 Initial-Value Problems 13
7.From the initial conditions we obtain the system
c1=−1c2= 8
The solution of the initial-value problem isx=−cost+ 8 sint.
8.From the initial conditions we obtain the system
c2= 0−c1= 1
The solution of the initial-value problem isx=−cost.
9.From the initial conditions we obtain
√
3
2
c1+
1
2
c2=
1
2
−
1
2
c2+
√
3
2
= 0
Solving, we findc1=
√
3/4 andc2= 1/4. The solution of the initial-value problem is
x= (
√
3/4) cost+ (1/4) sint.
10.From the initial conditions we obtain
√
2
2
c1+
√
2
2
c2=
√
2
[6pt]−
√
2
2
c1+
√
2
2
c2= 2
√
2.
Solving, we findc1=−1 andc2= 3. The solution of the initial-value problem isx=
−cost+ 3 sint.
In Problems11–14, we usey=c1e
x
+c2e
−x
andy
′
=c1e
x
−c2e
−x
to obtain a system of two
equations in the two unknownsc1andc2.
11.From the initial conditions we obtain
c1+c2= 1
c1−c2= 2.
Solving, we findc1=
3
2
andc2=−
1
2
. The solution of the initial-value problem isy=
3
2
e
x
−
1
2
e
−x
.
12.From the initial conditions we obtain
ec1+e
−1
c2= 0
ec1−e
−1
c2=e.
Solving, we findc1=
1
2
andc2=−
1
2
e
2
. The solution of the initial-value problem is
y=
1
2
e
x
−
1
2
e
2
e
−x
=
1
2
e
x
−
1
2
e
2−x
.

14. 14 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
13.From the initial conditions we obtain
e
−1
c1+ec2= 5
e
−1
c1−ec2=−5.
Solving, we findc1= 0 andc2= 5e
−1
. The solution of the initial-value problem isy=
5e
−1
e
−x
= 5e
−1−x
.
14.From the initial conditions we obtain
c1+c2= 0
c1−c2= 0.
Solving, we findc1=c2= 0. The solution of the initial-value problem isy= 0.
15.Two solutions arey= 0 andy=x
3
.
16.Two solutions arey= 0 andy=x
2
. (Also, any constant multiple ofx
2
is a solution.)
17.Forf(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 wherey6= 0.
18.Forf(x, y) =
√
xywe have∂f/∂y=
1
2
p
x/y. Thus, the differential equation will have a
unique solution in any region wherex >0 andy >0 or wherex <0 andy <0.
19.Forf(x, y) =
y
x
we have
∂f
∂y
=
1
x
. Thus, the differential equation will have a unique solution
in any region wherex6= 0.
20.Forf(x, y) =x+ywe have
∂f
∂y
= 1. Thus, the differential equation will have a unique
solution in the entire plane.
21.Forf(x, y) =x
2
/(4−y
2
) we have∂f/∂y= 2x
2
y/(4−y
2
)
2
. Thus the differential equation
will have a unique solution in any region wherey <−2,−2< y <2, ory >2.
22.Forf(x, y) =
x
2
1 +y
3
we have
∂f
∂y
=
−3x
2
y
2
(1 +y
3
)
2
. Thus, the differential equation will have a
unique solution in any region wherey6=−1.
23.Forf(x, y) =
y
2
x
2
+y
2
we have
∂f
∂y
=
2x
2
y
(x
2
+y
2
)
2
. Thus, the differential equation will have a
unique solution in any region not containing (0,0).

15. 1.2 Initial-Value Problems 15
24.Forf(x, y) = (y+x)/(y−x) we have∂f/∂y=−2x/(y−x)
2
. Thus the differential equation
will have a unique solution in any region wherey < xor wherey > x.
In Problems25–28, we identifyf(x, y) =
p
y
2
−9and∂f/∂y=y/
p
y
2
−9. We see thatfand
∂f/∂yare both continuous in the regions of the plane determined byy <−3andy >3with no
restrictions onx.
25.Since 4>3, (1,4) is in the region defined byy >3 and the differential equation has a unique
solution through (1,4).
26.Since (5,3) is not in either of the regions defined byy <−3 ory >3, there is no guarantee
of a unique solution through (5,3).
27.Since (2,−3) is not in either of the regions defined byy <−3 ory >3, there is no guarantee
of a unique solution through (2,−3).
28.Since (−1,1) is not in either of the regions defined byy <−3 ory >3, there is no guarantee
of a unique solution through (−1,1).
29. (a)A one-parameter family of solutions isy=cx. Sincey
′
=c,xy
′
=xc=yandy(0) =
c·0 = 0.
(b)Writing the equation in the formy
′
=y/x, we see thatRcannot contain any point on the
y-axis. Thus, any rectangular region disjoint from they-axis and containing (x0, y0) will
determine an interval aroundx0and a unique solution through (x0, y0). Sincex0= 0 in
part (a), we are not guaranteed a unique solution through (0,0).
(c)The piecewise-defined function which satisfiesy(0) = 0 is not a solution since it is not
differentiable atx= 0.
30. (a)Since
d
dx
tan (x+c) = sec
2
(x+c) = 1+tan
2
(x+c), we see thaty= tan (x+c) satisfies
the differential equation.
(b)Solvingy(0) = tanc= 0 we obtainc= 0 andy= tanx. Since tanxis 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).
31. (a)Since
d
dx

−
1
x+c

=
1
(x+c)
2
=y
2
, we see thaty=−
1
x+c
is a solution of the differ-
ential equation.
(b)Solvingy(0) =−1/c= 1 we obtainc=−1 andy= 1/(1−x). Solvingy(0) =−1/c=−1
we obtainc= 1 andy=−1/(1+x). Being sure to includex= 0, we see that the interval
of existence ofy= 1/(1−x) is (−∞,1), while the interval of existence ofy=−1/(1+x)
is (−1,∞).

16. 16 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
32. (a)Solvingy(0) =−1/c=y0we obtainc=−1/y0and
y=−
1
−1/y0+x
=y01−y0x , y06= 0
Since we must have−1/y0+x6= 0, the largest interval of existence (which must contain
0) is either (−∞,1/y0) wheny0>0 or (1/y0,∞) wheny0<0.
(b)By inspection we see thaty= 0 is a solution on (−∞,∞).
33. (a)Differentiating 3x
2
−y
2
=cwe get 6x−2yy
′
= 0 oryy
′
= 3x.
(b)Solving 3x
2
−y
2
= 3 forywe get
y=φ1(x) =
p
3(x
2
−1), 1< x <∞,
y=φ2(x) =−
p
3(x
2
−1), 1< x <∞,
y=φ3(x) =
p
3(x
2
−1), −∞< x <−1,
y=φ4(x) =−
p
3(x
2
−1),−∞< x <−1.
– 4–2 2 4
x
– 4
–2
2
4
y
(c)Onlyy=φ3(x) satisfiesy(−2) = 3.
34. (a)Settingx= 2 andy=−4 in 3x
2
−y
2
=cwe get
12−16 =−4 =c, so the explicit solution is
y=−
p
3x
2
+ 4,−∞< x <∞.
– 4–2 2 4
x
– 4
–2
2
4
y
(b)Settingc= 0 we havey=
√
3xandy=−
√
3x, both defined on (−∞,∞).
In Problems35–38, we consider the points on the graphs withx-coordinatesx0=−1,x0= 0,
andx0= 1. The slopes of the tangent lines at these points are compared with the slopes given by
y
′
(x0)in(a)through(f).
35.The graph satisfies the conditions in (b) and (f).
36.The graph satisfies the conditions in (e).
37.The graph satisfies the conditions in (c) and (d).

17. 1.2 Initial-Value Problems 17
38.The graph satisfies the conditions in (a).
39.Using the functiony=c1cos 3x+c2sin 3xand the first boundary condition we get
y(0) =c1cos 0 +c2sin 0 = 0
Thereforec1= 0. Similarly for the second boundary condition we get
y(π/6) =c2sin 3 (π/6) =−1
Thereforec2=−1. The solution to the boundary value problem isy(x) =−sin 3x.
40.Using the functiony=c1cos 3x+c2sin 3xand the first boundary condition we get
y(0) =c1cos 0 +c2sin 0 = 0
Thereforec1= 0. Thusy(x) =c2sin 3x. Similarly for the second boundary condition we get
y(π) =c2sin 3(π) = 0
But the last line is satisfied for any choice ofc2since sin 3π= 0 therefore there are infinitely
many solutions in this case.
41.The derivative of the functiony=c1cos 3x+c2sin 3xisy
′
=−3c1sin 3x+ 3c2cos 3xand
using the first boundary condition we get
y
′
(0) = 0 + 3c2= 0
Thereforec2= 0 So far then we havey
′
=−3c1sin 3x. Similarly, using the second boundary
condition we get
y
′
(π/4) =−3c1
√
2/2

= 0
Thereforec1= 0. The solution is thus the trivial solutiony= 0.
42.The derivative of the functiony=c1cos 3x+c2sin 3xisy
′
=−3c1sin 3x+ 3c2cos 3xand
using the two boundary conditions we get
y(0) =c1+ 0 = 1
Thereforec1= 1. In addition
y
′
(π) = 0−3c2= 5
Thereforec2=−5/3. The solution to this boundary value problem isy(x) = cos 3x−
5
3
sin 3x
.

18. 18 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
43.Using the functiony=c1cos 3x+c2sin 3xand the first boundary condition we get
y(0) =c1cos 0 +c2sin 0 = 0
Thereforec1= 0. Thusy(x) =c2sin 3x. Similarly for the second boundary condition we get
c2sin 3(π) = 4c2·0 = 40 = 4
The last line is obviously a contradiction and so therefore the boundary value problem has
no solution in this case.
44.The derivative of the functiony=c1cos 3x+c2sin 3xisy
′
=−3c1sin 3x+ 3c2cos 3xand
using the first boundary condition we get
y
′
(π/3) = 0−3c2= 1
Thereforec2=−1/3. Similarly, using the second boundary condition we get
y
′
(π) = 0−3c2= 0
Thereforec2= 0, which is a contradiction so the problem has no solution.
45.Integratingy
′
= 8e
2x
+ 6xwe obtain
y=
Z
(8e
2x
+ 6x)dx= 4e
2x
+ 3x
2
+c.
Settingx= 0 andy= 9 we have 9 = 4 +csoc= 5 andy= 4e
2x
+ 3x
2
+ 5.
46.Integratingy
′′
= 12x−2 we obtain
y
′
=
Z
(12x−2)dx= 6x
2
−2x+c1.
Then, integratingy
′
we obtain
y=
Z
(6x
2
−2x+c1)dx= 2x
3
−x
2
+c1x+c2.
Atx= 1 they-coordinate of the point of tangency isy=−1 + 5 = 4. This gives the initial
conditiony(1) = 4. The slope of the tangent line atx= 1 isy
′
(1) =−1. From the initial
conditions we obtain
2−1 +c1+c2= 4 or c1+c2= 3
and
6−2 +c1=−1 or c1=−5.
Thus,c1=−5 andc2= 8, soy= 2x
3
−x
2
−5x+ 8.

19. 1.2 Initial-Value Problems 19
47.Whenx= 0 andy=
1
2
,y
′
=−1, so the only plausible solution curve is the one with negative
slope at (0,
1
2
), or the black curve.
48.If the solution is tangent to thex-axis at (x0,0), theny
′
= 0 whenx=x0andy= 0.
Substituting these values intoy
′
+ 2y= 3x−6 we get 0 + 0 = 3x0−6 orx0= 2.
49.The theorem guarantees a unique (meaning single) solution through any point. Thus, there
cannot be two distinct solutions through any point.
50.Wheny=
1
16
x
4
,y
′
=
1
4
x
3
=x(
1
4
x
2
) =xy
1/2
, andy(2) =
1
16
(16) = 1. When
y=





0, x < 0
1
16
x
4
, x≥0
we have
y
′
=



0, x <0
1
4
x
3
, x≥0
=x



0, x <0
1
4
x
2
, x≥0
=xy
1/2
,
andy(2) =
1
16
(16) = 1. The two different solutions are the same on the interval (0,∞), which
is all that is required by Theorem 1.2.1.
51.We note that the initial conditiony(0) = 0,
0 =
Z
y
0
1
√
t
3
+ 1
dt
is satisfied only wheny= 0. For anyy >0, necessarily
Z
y
0
1
√
t
3
+ 1
dt >0
because the integrand is positive on the interval of integration. Then from (12) of Section 1.1
and the Chain Rule we have:
d
dx
x=
d
dx
Z
y
0
1
√
t
3
+ 1
dt
1 =
1
p
y
3
+ 1
dy
dx
and
dy
dx
=
p
y
3
+ 1
y
′
(0) =
dy
dx



x=0
=
q
(y(0))
3
+ 1 =
√
0 + 1 = 1.
Computing the second derivative, we see that:
d
2
y
dx
2
=
d
dx
p
y
3
+ 1 =
3y
2
2
p
y
3
+ 1
dy
dx
=
3y
2
2
p
y
3
+ 1
·
p
y
3
+ 1 =
3
2
y
2
d
2
y
dx
2
=
3
2
y
2
.
This is equivalent to 2
d
2
y
dx
2
−3y
2
= 0.

20. 20 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
1.3 Differential Equations as Mathematical Models
1.
dP
dt
=kP+r;
dP
dt
=kP−r
2.Letbbe the rate of births anddthe rate of deaths. Thenb=k1Pandd=k2P. Since
dP/dt=b−d, the differential equation isdP/dt=k1P−k2P.
3.Letbbe the rate of births anddthe rate of deaths. Thenb=k1Pandd=k2P
2
. Since
dP/dt=b−d, the differential equation isdP/dt=k1P−k2P
2
.
4.
dP
dt
=k1P−k2P
2
−h, h >0
5.From the graph in the text we estimateT0= 180
◦
andTm= 75
◦
. We observe that when
T= 85,dT/dt≈ −1. From the differential equation we then have
k=
dT/dt
T−Tm
=
−1
85−75
=−0.1.
6.By inspecting the graph in the text we takeTmto beTm(t) = 80−30 cos (πt/12). Then the
temperature of the body at timetis determined by the differential equation
dT
dt
=k
h
T−

80−30 cos

π
12
t
i
, t >0.
7.The number of students with the flu isxand the number not infected is 1000−x, sodx/dt=
kx(1000−x).
8.By analogy, with the differential equation modeling the spread of a disease, we assume that
the rate at which the technological innovation is adopted isproportional to the number of
people who have adopted the innovation and also to the numberof people,y(t), who have
not yet adopted it. If one person who has adopted the innovation is introduced into the
population, thenx+y=n+ 1 and
dx
dt
=kx(n+ 1−x), x(0) = 1.
9.The rate at which salt is leaving the tank is
Rout(3 gal/min)·
θ
A
300
lb/gal

=
A
100
lb/min.
ThusdA/dt=A/100. The initial amount isA(0) = 50.

21. 1.3 Differential Equations as Mathematical Models 21
10.The rate at which salt is entering the tank is
Rin= (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 = 1 gal/min. Aftertminutes there are 300 +tgallons of brine in the tank. The
rate at which salt is leaving is
Rout= (2 gal/min)·
θ
A
300 +t
lb/gal

=
2A
300 +t
lb/min.
The differential equation is
dA
dt
= 6−
2A
300 +t
.
11.The rate at which salt is entering the tank is
Rin= (3 gal/min)·(2 lb/gal) = 6 lb/min.
Since the tank loses liquid at the net rate of
3 gal/min−3.5 gal/min =−0.5 gal/min,
aftertminutes the number of gallons of brine in the tank is 300−
1
2
tgallons. Thus the rate
at which salt is leaving is
Rout=
θ
A
300−t/2
lb/gal

·(3.5 gal/min) =
3.5A
300−t/2
lb/min =
7A
600−t
lb/min.
The differential equation is
dA
dt
= 6−
7A
600−t
or
dA
dt
+
7
600−t
A= 6.
12.The rate at which salt is entering the tank is
Rin= (cinlb/gal)·(ringal/min) =cinrinlb/min.
Now letA(t) denote the number of pounds of salt andN(t) the number of gallons of brine
in the tank at timet. The concentration of salt in the tank as well as in the outflowis
c(t) =x(t)/N(t). But the number of gallons of brine in the tank remains steady, is increased,
or is decreased depending on whetherrin=rout,rin> rout, orrin< rout. In any case, the
number of gallons of brine in the tank at timetisN(t) =N0+ (rin−rout)t. The output rate
of salt is then
Rout=
θ
A
N0+ (rin−rout)t
lb/gal

·(routgal/min) =rout
A
N0+ (rin−rout)t
lb/min.
The differential equation for the amount of salt,dA/dt=Rin−Rout, is
dA
dt
=cinrin−rout
A
N0+ (rin−rout)t
or
dA
dt
+
rout
N0+ (rin−rout)t
A=cinrin.

22. 22 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
13.The volume of water in the tank at timetisV=Awh. The differential equation is then
dh
dt
=
1
Aw
dV
dt
=
1
Aw

−cAh
p
2gh

=−
cAh
Aw
p
2gh .
UsingAh=π
θ
2
12

2
=
π
36
,Aw= 10
2
= 100, andg= 32, this becomes
dh
dt
=−
cπ/36
100
√
64h=−
cπ
450
√
h .
14.The volume of water in the tank at timetisV=
1
3
πr
2
hwhereris the radius of the tank
at heighth. From the figure in the text we see thatr/h= 8/20 so thatr=
2
5
hand
V=
1
3
π
−
2
5
h
·
2
h=
4
75
πh
3
. Differentiating with respect totwe havedV/dt=
4
25
πh
2
dh/dtor
dh
dt
=
25
4πh
2
dV
dt
.
From Problem 13 we havedV/dt=−cAh
√
2ghwherec= 0.6,Ah=π
−
2
12
·
2
, andg= 32.
ThusdV/dt=−2π
√
h/15 and
dh
dt
=
25
4πh
2

−
2π
√
h
15
!
=−
5
6h
3/2
.
15.Sincei=dq/dtandL d
2
q/dt
2
+R dq/dt=E(t), we obtainL di/dt+Ri=E(t).
16.By Kirchhoff’s second law we obtainR
dq
dt
+
1
C
q=E(t).
17.From Newton’s second law we obtainm
dv
dt
=−kv
2
+mg.
18.Since the barrel in Figure 1.3.17(b) in the text is submergedan additionalyfeet below
its equilibrium position the number of cubic feet in the additional submerged portion is
the volume of the circular cylinder:π×(radius)
2
×height orπ(s/2)
2
y. Then 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= 15.6πs
2
y.
It then follows from Newton’s second law that
w
g
d
2
y
dt
2
=−15.6πs
2
y or
d
2
y
dt
2
+
15.6πs
2
g
w
y= 0,
whereg= 32 andwis the weight of the barrel in pounds.

23. 1.3 Differential Equations as Mathematical Models 23
19.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 ismg=ks, the differential equation is
m
d
2
x
dt
2
=−kx.
20.From Problem 19, without a damping force, the differential equation ism d
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
orm
d
2
x
dt
2
+β
dx
dt
+kx= 0.
21.As the rocket climbs (in the positive direction), it spends its amount of fuel and therefore the
mass of the fuel changes with time. The air resistance acts inthe opposite direction of the
motion and the upward thrustRworks in the same direction. Using Newton’s second law we
get
d
dt
(mv) =−mg−kv+R
Now because the mass is variable, we must use the product ruleto expand the left side of the
equation. Doing so gives us the following:
d
dt
(mv) =−mg−kv+R
v×
dm
dt
+m×
dv
dt
=−mg−kv+R
The last line is the differential equation we wanted to find.
22. (a)Since the mass of the rocket ism(t) =mp+mv+mf(t), take the time rate-of-change
and get, by straight-forward calculation,
d
dt
m(t) =
d
dt
(mp+mv+mf(t)) = 0 + 0 +m
′
f
(t) =
d
dt
mf(t)
Therefore the rate of change of the mass of the rocket is the same as the rate of change
of the mass of the fuel which is what we wanted to show.
(b)The fuel is decreasing at the constant rate ofλand so from part (a) we have
d
dt
m(t) =
d
dt
mf(t) =−λ
m(t) =−λt+c

24. 24 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
Using the given condition to solve forc,m(0) = 0 +c=m0and som(t) =−λt+m0.
The differential equation in Problem 21 now becomes
v
dm
dt
+m
dv
dt
+kv=−mg+R
−λv+ (−λt+m0)
dv
dt
+kv=−mg+R
(−λt+m0)
dv
dt
+ (k−λ)v=−mg+R
dv
dt
+
k−λ
−λt+m0
v=
−mg
−λt+m0
+
R
−λt+m0
dv
dt
+
k−λ
−λt+m0
v=−g+
R
−λt+m0
(c)From part (b) we have that
d
dt
mf(t) =−λand so by integrating this result we get
mf(t) =−λt+c. Now at timet= 0,mf(0) = 0+c=cthereforemf(t) =−λt+mf(0) .
At some later timetbwe then havemf(tb) =−λtb+mf(0) = 0 and solving this equation
for that time we gettb=mf(0)/λwhich is what we wanted to show.
23.Fromg=k/R
2
we findk=gR
2
. Usinga=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.
24.The gravitational force onmisF=−kMrm/r
2
. SinceMr= 4πδr
3
/3 andM= 4πδR
3
/3 we
haveMr=r
3
M/R
3
and
F=−k
Mrm
r
2
=−k
r
3
Mm/R
3
r
2
=−k
mM
R
3
r.
Now fromF=ma=d
2
r/dt
2
we have
m
d
2
r
dt
2
=−k
mM
R
3
ror
d
2
r
dt
2
=−
kM
R
3
r.
25.The differential equation is
dA
dt
=k(M−A).
26.The differential equation is
dA
dt
=k1(M−A)−k2A.
27.The differential equation isx
′
(t) =r−kx(t) wherek >0.

25. 1.3 Differential Equations as Mathematical Models 25
28.Notice from the diagram that the segment from the waterskierto the boat is tangent to the
curve at the pointP(x, y). For the sake of simplicity, let’s label the coordinates ofthe boat
on the x-axis as (a,0). The equation of this tangent line isy=y
′
(x)(x−a) from which we
get
y=y
′
(x)(x−a)
y
y
′
=x−a
−
y
y
′
=a−x
By the Pythagorean Theorem we havey
2
+ (a−x)
2
=s
2
from which we make a substitution
and then solve for the derivative to get
y
2
+ (a−x)
2
=s
2
y
2
+ (y
◦
y
′
)
2
=s
2
(y
◦
y
′
)
2
=s
2
−y
2
y
y
′
=−
p
s
2
−y
2
y
′
=−
y
p
s
2
−y
2
Notice that the sign of the derivative is negative because asthe boat proceeds along the
positivex-axis, they-coordinate decreases.
29.We see from the figure that 2θ+α=π. Thus
y
−x
= tanα= tan(π−2θ) =−tan 2θ=−
2 tanθ
1−tan
2
θ
.
Since the slope of the tangent line isy
′
= tanθwe have
y/x= 2y
′
[1−(y
′
)
2
] ory−y(y
′
)
2
= 2xy
′
, which is the quadratic
equationy(y
′
)
2
+ 2xy
′
−y= 0 iny
′
. Using the quadratic
formula, we get
y
′
=
−2x±
p
4x
2
+ 4y
2
2y
=
−x±
p
x
2
+y
2
y
.
(x, y)
x
y
α
α
θ
θ
θ
φ
x
y
Sincedy/dx >0, the differential equation is
dy
dx
=
−x+
p
x
2
+y
2
y
ory
dy
dx
−
p
x
2
+y
2
+x= 0.

26. 26 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
30.The differential equation isdP/dt=kP, so from Problem 41 in Exercises 1.1, a one-parameter
family of solutions isP=ce
kt
.
31.The differential equation in (3) isdT/dt=k(T−Tm). When the body is cooling,T > Tm,
soT−Tm>0. SinceTis decreasing,dT/dt <0 andk <0. When the body is warming,
T < Tm, soT−Tm<0. SinceTis increasing,dT/dt >0 andk <0.
32.The differential equation in (8) isdA/dt= 6−A/100. IfA(t) attains a maximum, then
dA/dt= 0 at this time andA= 600. IfA(t) continues to increase without reaching a
maximum, thenA
′
(t)>0 fort >0 andAcannot exceed 600. In this case, ifA
′
(t) approaches
0 astincreases to infinity, we see thatA(t) approaches 600 astincreases to infinity.
33.This differential equation could describe a population that undergoes periodic fluctuations.
34. (a)As shown in Figure 1.3.23(a) in the text, the resultant of thereaction force of magnitude
Fand the weight of magnitudemgof the particle is the centripetal force of magnitude
mω
2
x. The centripetal force points to the center of the circle of radiusxon which the
particle rotates about they-axis. Comparing parts of similar triangles gives
Fcosθ=mgandFsinθ=mω
2
x.
(b)Using the equations in part (a) we find
tanθ=
Fsinθ
Fcosθ
=
mω
2
x
mg
=
ω
2
x
g
or
dy
dx
=
ω
2
x
g
.
35.From Problem 23,d
2
r/dt
2
=−gR
2
/r
2
. SinceRis a constant, ifr=R+s, thend
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
−2sR
−3
+· · ·] =−g+
2gs
R
+· · ·.
Thus, forRmuch larger thans, the differential equation is approximated byd
2
s/dt
2
=−g.
36. (a)Ifρis the mass density of the raindrop, thenm=ρVand
dm
dt
=ρ
dV
dt
=ρ
d
dt
h
4
3
πr
3
i
=ρ

4πr
2
dr
dt

=ρS
dr
dt
.
Ifdr/dtis a constant, thendm/dt=kSwhereρ dr/dt=kordr/dt=k/ρ. Since the
radius is decreasing,k <0. Solvingdr/dt=k/ρwe getr= (k/ρ)t+c0. Sincer(0) =r0,
c0=r0andr=kt/ρ+r0.
(b)From Newton’s second law,
d
dt
[mv] =mg, wherevis the velocity of the raindrop. Then
m
dv
dt
+v
dm
dt
=mg orρ

4
3
πr
3

dv
dt
+v(k4πr
2
) =ρ

4
3
πr
3

g.
Dividing by 4ρπr
3
/3 we get
dv
dt
+
3k
ρr
v=g or
dv
dt
+
3k/ρ
kt/ρ+r0
v=g, k <0.

27. 1.3 Differential Equations as Mathematical Models 27
37.We assume that the plow clears snow at a constant rate ofkcubic miles per hour. Lettbe the
time in hours after noon,x(t) the depth in miles of the snow at timet, andy(t) the distance
the plow has moved inthours. Thendy/dtis the velocity of the plow and the assumption
gives
wx
dy
dt
=k,
wherewis the width of the plow. Each side of this equation simply represents the volume
of snow plowed in one hour. Now lett0be the number of hours before noon when it started
snowing and letsbe the constant rate in miles per hour at whichxincreases. Then for
t >−t0,x=s(t+t0). The differential equation then becomes
dy
dt
=
k
ws
1
t+t0
.
Integrating, we obtain
y=
k
ws
[ ln(t+t0) +c]
wherecis a constant. Now whent= 0,y= 0 soc=−lnt0and
y=
k
ws
ln
θ
1 +
t
t0

.
Finally, from the fact that whent= 1,y= 2 and whent= 2,y= 3, we obtain
θ
1 +
2
t0

2
=
θ
1 +
1
t0

3
.
Expanding and simplifying givest
2
0
+t0−1 = 0. Sincet0>0, we findt0≈0.618 hours≈37
minutes. Thus it started snowing at about 11:23 in the morning.
38.
(1) :
dP
dt
=kPis linear (2) :
dA
dt
=kAis linear
(3) :
dT
dt
=k(T−Tm) is linear (5) :
dx
dt
=kx(n+ 1−x) is nonlinear
(6) :
dX
dt
=k(α−X) (beta−X) is nonlinear (8) :
dA
dt
= 6−
A
100
is linear
(10) :
dh
dt
=−
Ah
Aw
p
2ghis nonlinear (11) : L
d
2
q
dt
2
+R
dq
dt
+
1
C
q=E(t) is linear
(12) :
d
2
s
dt
2
=−gis linear (14) : m
dv
dt
=mg−kvis linear
(15) :m
d
2
s
dt
2
+k
ds
dt
=mgis linear (16) :
d
2
x
dt
2
−
64
L
x= 0 is linear
(17) : linearity or nonlinearity is determined by the mannerin whichWandT1involvex.

28. 28 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
39.At timet, when the population is 2 million cells, the differential equationP
′
(t) = 0.15P(t)
gives the rate of increase at timet. Thus, whenP(t) = 2 (million cells), the rate of increase
isP
′
(t) = 0.15(2) = 0.3 million cells per hour or 300,000 cells per hour.
40.SettingA
′
(t) =−0.002 and solvingA
′
(t) =−0.0004332A(t) forA(t), we obtain
A(t) =
A
′
(t)
−0.0004332
=
−0.002
−0.0004332
≈4.6 grams.
Chapter 1 in Review
1.
d
dx
c1e
kx
=k
y
z}|{
c1e
kx
;
dy
dx
=ky
2.
d
dx
(5 +c1e
−2x
) =−2c1e
−2x
=−2(
y
z}|{
5 +c1e
−2x
−5);
dy
dx
=−2(y−5) or
dy
dx
=−2y+ 10
3.
d
dx
(c1coskx+c2sinkx) =−kc1sinkx+kc2coskx;
d
2
dx
2
(c1coskx+c2sinkx) =−k
2
c1coskx−k
2
c2sinkx=−k
2
(
y
z }| {
c1coskx+c2sinkx);
d
2
y
dx
2
=−k
2
yor
d
2
y
dx
2
+k
2
y= 0
4.
d
dx
(c1coshkx+c2sinhkx) =kc1sinhkx+kc2coshkx;
d
2
dx
2
(c1coshkx+c2sinhkx) =k
2
c1coshkx+k
2
c2sinhkx=k
2
(
y
z }| {
c1coshkx+c2sinhkx);
d
2
y
dx
2
=k
2
yor
d
2
y
dx
2
−k
2
y= 0
5.y=c1e
x
+c2xe
x
;y
′
=c1e
x
+c2xe
x
+c2e
x
;y
′′
=c1e
x
+c2xe
x
+ 2c2e
x
;
y
′′
+y= 2(c1e
x
+c2xe
x
) + 2c2e
x
= 2(c1e
x
+c2xe
x
+c2e
x
) = 2y
′
;y
′′
−2y
′
+y= 0
6.y
′
=−c1e
x
sinx+c1e
x
cosx+c2e
x
cosx+c2e
x
sinx;
y
′′
=−c1e
x
cosx−c1e
x
sinx−c1e
x
sinx+c1e
x
cosx−c2e
x
sinx+c2e
x
cosx+c2e
x
cosx+
c2e
x
sinx
=−2c1e
x
sinx+ 2c2e
x
cosx;
y
′′
−2y
′
=−2c1e
x
cosx−2c2e
x
sinx=−2y;y
′′
−2y
′
+ 2y= 0

29. Chapter 1 in Review 29
7.a, d 8.c 9.b 10.a, c 11.b 12.a, b, d
13.A few solutions arey= 0,y=c, andy=e
x
.
14.Easy solutions to see arey= 0 andy= 3.
15.The slope of the tangent line at (x, y) isy
′
, so the differential equation isy
′
=x
2
+y
2
.
16.The rate at which the slope changes isdy
′
/dx=y
′′
, so the differential equation isy
′′
=−y
′
ory
′′
+y
′
= 0.
17. (a)The domain is all real numbers.
(b)Sincey
′
= 2/3x
1/3
, the solutiony=x
2/3
is undefined atx= 0. This function is a
solution of the differential equation on (−∞,0) and also on (0,∞).
18. (a)Differentiatingy
2
−2y=x
2
−x+cwe obtain 2yy
′
−2y
′
= 2x−1 or (2y−2)y
′
= 2x−1.
(b)Settingx= 0 andy= 1 in the solution we have 1−2 = 0−0 +corc=−1. Thus, a
solution of the initial-value problem isy
2
−2y=x
2
−x−1.
(c)Solving the equationy
2
−2y−(x
2
−x−1) = 0 by the quadratic formula we get
y= (2±
p
4 + 4(x
2
−x−1) )/2 = 1±
√
x
2
−x= 1±
p
x(x−1) . Sincex(x−1)≥0
forx≤0 orx≥1, we see that neithery= 1 +
p
x(x−1) nory= 1−
p
x(x−1) is
differentiable atx= 0. Thus, both functions are solutions of the differential equation,
but neither is a solution of the initial-value problem.
19.Settingx=x0andy= 1 iny=−2/x+x, we get
1 =−
2
x0
+x0 orx
2
0−x0−2 = (x0−2)(x0+ 1) = 0.
Thus,x0= 2 orx0=−1. Sincex= 0 iny=−2/x+x, we see thaty=−2/x+xis a
solution of the initial-value problemxy
′
+y= 2x,y(−1) = 1, on the interval (−∞,0) and
y=−2/x+xis a solution of the initial-value problemxy
′
+y= 2x,y(2) = 1, on the interval
(0,∞).
20.From the differential equation,y
′
(1) = 1
2
+[y(1)]
2
= 1+(−1)
2
= 2>0, soy(x) is increasing in
some neighborhood ofx= 1. Fromy
′′
= 2x+2yy
′
we havey
′′
(1) = 2(1)+2(−1)(2) =−2<0,
soy(x) is concave down in some neighborhood ofx= 1.
21. (a)
–3–2 123
x
–3
–2
1
2
3
y
–1
–1 –3–2 123
x
–3
–2
1
y = x
2
 + c
1 y = –x
2
 + c
2
2
3
y
–1
–1

30. 30 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
(b)Wheny=x
2
+c1,y
′
= 2xand (y
′
)
2
= 4x
2
. Wheny=−x
2
+c2,y
′
=−2xand
(y
′
)
2
= 4x
2
.
(c)Pasting togetherx
2
,x≥0, and−x
2
,x≤0, we gety=



−x
2
, x≤0
x
2
, x >0.
22.The slope of the tangent line isy
′


(−1,4)
= 6
√
4 + 5(−1)
3
= 7.
23.Differentiatingy=xsinx+xcosxwe get
y
′
=xcosx+ sinx−xsinx+ cosx
and
y
′′
=−xsinx+ cosx+ cosx−xcosx−sinx−sinx
=−xsinx−xcosx+ 2 cosx−2 sinx.
Thus
y
′′
+y=−xsinx−xcosx+ 2 cosx−2 sinx+xsinx+xcosx= 2 cosx−2 sinx.
An interval of definition for the solution is (−∞,∞).
24.Differentiatingy=xsinx+ (cosx) ln(cosx) we get
y
′
=xcosx+ sinx+ cosx
θ
−sinx
cosx

−(sinx) ln (cosx)
=xcosx+ sinx−sinx−(sinx) ln (cosx)
=xcosx−(sinx) ln (cosx)
and
y
′′
=−xsinx+ cosx−sinx
θ
−sinx
cosx

−(cosx) ln (cosx)
=−xsinx+ cosx+
sin
2
x
cosx
−(cosx) ln (cosx)
=−xsinx+ cosx+
1−cos
2
x
cosx
−(cosx) ln (cosx)
=−xsinx+ cosx+ secx−cosx−(cosx) ln (cosx)
=−xsinx+ secx−(cosx) ln (cosx).
Thus
y
′′
+y=−xsinx+ secx−(cosx) ln(cosx) +xsinx+ (cosx) ln (cosx) = secx.
To obtain an interval of definition we note that the domain of lnxis (0,∞), so we must have
cosx >0. Thus, an interval of definition is (−π/2, π/2).

31. Chapter 1 in Review 31
25.Differentiatingy= sin (lnx) we obtainy
′
= cos (lnx)/xandy
′′
=−[sin (lnx) + cos (lnx)]/x
2
.
Then
x
2
y
′′
+xy
′
+y=x
2
θ
−
sin (lnx) + cos (lnx)
x
2

+x
cos (lnx)
x
+ sin (lnx) = 0.
An interval of definition for the solution is (0,∞).
26.Differentiatingy= cos (lnx) ln (cos (lnx)) + (lnx) sin (lnx) we obtain
y
′
= cos (lnx)
1
cos (lnx)
θ
−
sin (lnx)
x

+ ln (cos (lnx))
θ
−
sin (lnx)
x

+ lnx
cos (lnx)
x
+
sin (lnx)
x
=−
ln (cos (lnx)) sin (lnx)
x
+
(lnx) cos (lnx)
x
and
y
′′
=−x
≤
ln (cos (lnx))
cos (lnx)
x
+ sin (lnx)
1
cos (lnx)
θ
−
sin (lnx)
x

1
x
2
+ ln (cos (lnx)) sin (lnx)
1
x
2
+x
≤
(lnx)
θ
−
sin (lnx)
x

+
cos (lnx)
x
≥
1
x
2
−(lnx) cos (lnx)
1
x
2
=
1
x
2
≤
−ln (cos (lnx)) cos (lnx) +
sin
2
(lnx)
cos (lnx)
+ ln (cos (lnx)) sin (lnx)
−(lnx) sin (lnx) + cos (lnx)−(lnx) cos (lnx)
≥
.
Then
x
2
y
′′
+xy
′
+y=−ln (cos (lnx)) cos (lnx) +
sin
2
(lnx)
cos (lnx)
+ ln (cos (lnx)) sin (lnx)−(lnx) sin (lnx)
+ cos (lnx)−(lnx) cos (lnx)−ln (cos (lnx)) sin (lnx)
+ (lnx) cos (lnx) + cos (lnx) ln (cos (lnx)) + (lnx) sin (lnx)
=
sin
2
(lnx)
cos (lnx)
+ cos (lnx) =
sin
2
(lnx) + cos
2
(lnx)
cos (lnx)
=
1
cos (lnx)
= sec (lnx).
To obtain an interval of definition, we note that the domain oflnxis (0,∞), so we must
have cos (lnx)>0. Since cosx >0 when−π/2< x < π/2, we require−π/2<lnx < π/2.
Sincee
x
is an increasing function, this is equivalent toe
−π/2
< x < e
π/2
. Thus, an interval
of definition is (e
−π/2
, e
π/2
). (Much of this problem is more easily done using a computer
algebra system such asMathematicaorMaple.)

32. 32 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
In Problems 27 - 30 we use (12) of Section 1.1 and the Product Rule.
27.
y=e
cosx
Z
x
0
te
−cost
dt
dy
dx
=e
cosx
−
xe
−cosx
·
−sinxe
cosx
Z
x
0
te
−cost
dt
dy
dx
+ (sinx)y=e
cosx
xe
−cosx
−sinxe
cosx
Z
x
0
te
−cost
dt+ sinx
θ
e
cosx
Z
x
0
te
−cost
dt

=x−sinxe
cosx
Z
x
0
te
−cost
dt+ sinxe
cosx
Z
x
0
te
−cost
dt=x
28.
y=e
x
2
Z
x
0
e
t−t
2
dt
dy
dx
=e
x
2
e
x−x
2
+ 2xe
x
2
Z
x
0
e
t−t
2
dt
dy
dx
−2xy=e
x
2
e
x−x
2
+ 2xe
x
2
Z
x
0
e
t−t
2
dt−2x
θ
e
x
2
Z
x
0
e
t−t
2
dt

=e
x
29.
y=x
Z
x
1
e
−t
t
dt
y
′
=x
e
−x
x
+
Z
x
1
e
−t
t
dt=e
−x
+
Z
x
1
e
−t
t
dt
y
′′
=−e
−x
+
e
−x
x
x
2
y
′′
+
−
x
2
−x
·
y
′
+ (1−x)y=
−
−x
2
e
−x
+xe
−x
·
+
θ
x
2
e
−x
+x
2
Z
x
1
e
−t
t
dt−xe
−x
−x
Z
x
1
e
−t
t
dt

+
θ
x
Z
x
1
e
−t
t
dt−x
2
Z
x
1
e
−t
t
dt

= 0

33. Chapter 1 in Review 33
30.
y= sinx
Z
x
0
e
t
2
cost dt−cosx
Z
x
0
e
t
2
sint dt
y
′
= sinx

e
x
2
cosx

+ cosx
Z
x
0
e
t
2
cost dt−cosx

e
x
2
sinx

+ sinx
Z
x
0
e
t
2
sint dt
= cosx
Z
x
0
e
t
2
cost dt+ sinx
Z
x
0
e
t
2
sint dt
y
′′
= cosx

e
x
2
cosx

−sinx
Z
x
0
e
t
2
cost dt+ sinx

e
x
2
sinx

+ cosx
Z
x
0
e
t
2
sint dt
=e
x
2−
cos
2
x+ sin
2
x
·
−




y
z}| {
sinx
Z
x
0
e
t
2
cost dt−cosx
Z
x
0
e
t
2
sint dt




=e
x
2
−y
y
′′
+y=e
x
2
−y+y=e
x
2
31.Using implicit differentiation we get
x
3
y
3
=x
3
+ 5
3x
2
·y
3
+x
3
·3y
2
dy
dx
= 3x
2
3x
2
y
3
3x
2
y
2
+
x
3
3y
2
3x
2
y
2
dy
dx
=
3x
2
3x
2
y
2
y+x
dy
dx
=
1
y
2

34. 34 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
32.Using implicit differentiation we get
(x−7)
2
+y
2
= 1
2(x−7) + 2y
dy
dx
= 0
2y
dy
dx
=−2(x−7)
y
dy
dx
=−(x−7)
θ
y
dy
dx

2
= (x−7)
2
θ
dy
dx

2
=
(x−7)
2
y
2
Now from the original equation, isolating the first term leads to (x−7)
2
= 1−y
2
. Continuing
from the last line of our proof we now have
θ
dy
dx

2
=
(x−7)
2
y
2
=
1−y
2
y
2
=
1
y
2
−1
Adding 1 to both sides leads to the desired result.
33.Using implicit differentiation we get
y
3
+ 3y= 2−3x
3y
2
y
′
+ 3y
′
=−3
y
2
y
′
+y
′
=−1
(y
2
+ 1)y
′
=−1
y
′
=
−1
y
2
+ 1
Differentiating the last line and remembering to use the quotient rule on the right side leads
to
y
′′
=
2yy
′
(y
2
+ 1)
2
Now sincey
′
=−1
◦−
y
2
+ 1
·
we can write the last equation as
y
′′
=
2y
(y
2
+ 1)
2
y
′
=
2y
(y
2
+ 1)
2
−1
(y
2
+ 1)
= 2y
θ
−1
y
2
+ 1

3
= 2y(y
′
)
3
which is what we wanted to show.

35. Chapter 1 in Review 35
34.Using implicit differentiation we get
y=e
−xy
y
′
=e
−xy
(−y−xy
′
)
ye
−xy
+xe
−xy
y
′
+y
′
= 0
ye
−xy
+ (xe
−xy
+ 1)y
′
= 0
Now sincey=e
−xy
, substitute this into the last line to get
yy+ (xy+ 1)y
′
= 0
or (1 +xy)y
′
+y
2
= 0 which is what we wanted to show.
In Problem 35 - 38,y=c1e
−3x
+c2e
x
+ 4xis given as a two-parameter family of solutions of the
second-order differential equationy
′′
+ 2y
′
−3y=−12x+ 8.
35.Ify(0) = 0 andy
′
(0) = 0, then
c1+c2= 0
−3c1+c2=−4
subtracting the second equation from the first gives us 4c1= 4 orc1= 1, and thusc2=−1.
Thereforey=e
−3x
−e
x
+ 4x.
36.Ify(0) = 5 andy
′
(0) =−11, then
c1+c2= 5
−3c1+c2=−15
subtracting the second equation from the first gives us 4c1= 20 orc1= 5, and thusc2= 0.
Thereforey= 5e
−3x
+ 4x.
37.Ify(1) =−2 andy
′
(1) = 4, then
c1e
−3
+c2e=−6
−3c1e
−3
+c2e= 0
subtracting the second equation from the first gives us 4c1=−6 orc1=−
3
2
e
3
, and thus
c2=−
9
2
e
−1
. Thereforey=−
3
2
e
−3x+3
−
9
2
e
x−1
+ 4x.

36. 36 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
38.Ify(−1) = 1 andy
′
(−1) = 1, then
c1e
3
+c2e
−1
= 5
−3c1e
3
+c2e
−1
=−3
subtracting the second equation from the first gives us 4c1= 8 orc1= 2e
−3
, and thusc2= 3e.
Thereforey= 2e
−3x−3
+ 3e
x+1
+ 4x.
39.We are to use the Leibniz’s rule
d
dx
Z
v(x)
u(x)
F(x, t)dt=F(x, v(x))
dv
dx
−F(x, u(x))
du
dx
+
Z
v(x)
u(x)
∂
∂x
F(x, t)dt
Sincey(x) =
1
3
Z
x
0
f(t) sin (3x−3t)dt, takeF(x, t) =f(t) sin (3x−3t),u(x) = 0, and
v(x) =xto get
d
dx
y(x) =
d
dx
1
3
Z
x
0
f(t) sin (3x−3t)dt
=
1
3
≤
F(x, x)·1−F(x,0)·0 +
Z
x
0
∂
∂x
f(t) sin (3x−3t)dt
≥
=
1
3
≤
f(x) sin (3x−3x) +
Z
x
0
3f(t) cos (3x−3t)dt
≥
=
Z
x
0
f(t) cos (3x−3t)dt
Apply the Leibniz’s rule a second time toy
′
=
Z
x
0
f(t) cos (3x−3t)dtby takingF(x, t) =
f(t) cos (3x−3t),u(x) = 0, andv(x) =xto get
d
dx
y
′
(x) =
d
dx
Z
x
0
f(t) cos (3x−3t)dt
=F(x, x)·1−F(x,0)·0 +
Z
x
0
∂
∂x
f(t) cos (3x−3t)dt
=f(x) cos (3x−3x) +
Z
x
0
−3f(t) sin (3x−3t)dt
=f(x)−3
Z
x
0
f(t) sin (3x−3t)dt
Thereforey
′′
=f(x)−3
Z
x
0
f(t) sin (3x−3t)dt. Now substitutingyandy
′′
into the differential
equation we get
y
′′
+ 9y=f(x)−3
Z
x
0
f(t) sin (3x−3t)dt+ 9·
1
3
Z
x
0
f(t) sin (3x−3t)dt=f(x)

37. Chapter 1 in Review 37
40.We are to use the Leibniz’s rule
d
dx
Z
v(x)
u(x)
F(x, t)dt=F(x, v(x))
dv
dx
−F(x, u(x))
du
dx
+
Z
v(x)
u(x)
∂
∂x
F(x, t)dt
Sincey(x) =
Z
π
0
e
xcost
dt, takeF(x, t) =e
xcost
,u(x) = 0, andv(x) =πto get
d
dx
y(x) =
d
dx
Z
π
0
e
xcost
dt
=F(x, π)·0−F(x,0)·0 +
Z
π
0
∂
∂x
e
xcost
dt
=
Z
π
0
coste
xcost
dt
Applying the Leibniz’s rule a second time toy
′
(x) =
Z
π
0
coste
xcost
dtby takingF(x, t) =
coste
xcost
,u(x) = 0, andv(x) =πto get
d
dx
y
′
(x) =
Z
π
0
coste
xcost
dt
=F(x, π)·0−F(x,0)·0 +
Z
π
0
∂
∂x
coste
xcost
dt
=
Z
π
0
cos
2
te
xcost
dt
An alternative form ofy
′
can be obtained by integrating by parts with respect tot:
y
′
= sint·e
xcost



π
0
+
Z
π
0
xe
cost
sin
2
t dt=
Z
π
0
xe
xcost
sin
2
t dt=x
Z
π
0
e
xcost
sin
2
t dt.
We use this last form ofy
′
instead of the first in the differential equation:
xy
′′
+y
′
=x
Z
π
0
e
xcost
cos
2
t dt+x
Z
π
0
e
xcost
sin
2
t dt
=x
Z
π
0



1
z}|{
cos
2
t+ sin
2
t


e
xcost
dt=
y
z }| {
x
Z
π
0
e
xcost
dt=xy
xy
′′
+y
′
=xy
Therefore,xy
′′
+y
′′
−xy= 0.
41.From the graph we see that estimates fory0andy1arey0=−3 andy1= 0.

38. 38 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
42.The differential equation is
dh
dt
=−
cA0
Aw
p
2gh .
UsingA0=π(1/24)
2
=π/576,Aw=π(2)
2
= 4π, andg= 32, this becomes
dh
dt
=−
cπ/576
4π
√
64h=−
c
288
√
h .
43.Letx(t) represent the height of the top of the rope at any timetwith the positive direction
upward as indicated. The weight of the portion of the rope off the ground is given by
W= (xft)·(1 lb/ft) =x. The mass of the rope ism=W/g=x/32. The net force is
F= 5−W= 5−x. Now by Newton’s second law we get
F=
d
dt
(mv) =
d
dt

x
32
·v

= 5−x
Now expand the derivative and remember thatv=dx/dtto get
d
dt

x
32
·v

= 5−x
1
32
θ
x
dv
dt
+v
dx
dt

= 5−x
x
d
dt
θ
dx
dt

+
θ
dx
dt

dx
dt
= 160−32x
x
d
2
x
dt
2
+
θ
dx
dt

2
+ 32x= 160