Elementary differential equations with boundary value problems solutions

The university textbook for advance differential equations

1. STUDENT SOLUTIONS MANUAL FOR
ELEMENTARY
DIFFERENTIALEQUATIONS
AND
ELEMENTARY
DIFFERENTIALEQUATIONS
WITHBOUNDARYVALUE
PROBLEMS
William F.Trench
Andrew G. Cowles Distinguished Professor Emeritus
Department of Mathematics
Trinity University
San Antonio, Texas, USA
[email protected]
This book has been judged to meet the evaluation criteria setby the Edi-
torial Board of the American Institute of Mathematics in connection with
the Institute’sOpen Textbook Initiative. It may be copied, modified, re-
distributed, translated, and built upon subject to the Creative Commons
Attribution-NonCommercial-ShareAlike 3.0 Unported License.

2. This book was published previously by Brooks/Cole Thomson Learning
Reproduction is permitted for any valid noncommercial educational, mathematical, or scientific purpose.
However, charges for profit beyond reasonable printing costs are prohibited.

3. TOBEVERLY
Contents
Chapter 1 Introduction 1
1.2 First Order Equations 1
Chapter 2 First Order Equations 5
2.1 Linear First Order Equations 5
2.2 Separable Equations 8
2.3 Existence and Uniqueness of Solutions of Nonlinear Equations 11
2.4 Transformation of Nonlinear Equations into Separable Equations 13
2.5 Exact Equations 17
2.6 Integrating Factors 21
Chapter 3 Numerical Methods 25
3.1 Euler’s Method 25
3.2 The Improved Euler Method and Related Methods 29

4. iiContents
3.3 The Runge-Kutta Method 34
Chapter 4 Applications of First Order Equations 39
4.1 Growth and Decay 39
4.2 Cooling and Mixing 40
4.3 Elementary Mechanics 43
4.4 Autonomous Second Order Equations 45
4.5 Applications to Curves 46
Chapter 5 Linear Second Order Equations 51
5.1 Homogeneous Linear Equations 51
5.2 Constant Coefficient Homogeneous Equations 55
5.3 Nonhomgeneous Linear Equations 58
5.4 The Method of Undetermined Coefficients I 60
5.5 The Method of Undetermined Coefficients II 64
5.6 Reduction of Order 75
5.7 Variation of Parameters 79
Chapter 6 Applcations of Linear Second Order Equations 85
6.1 Spring Problems I 85
6.2 Spring Problems II 87
6.3 The RLC Circuit 89
6.4 Motion Under a Central Force 90
Chapter 7 Series Solutions of Linear Second Order Equations 108
7.1 Review of Power Series 91
7.2 Series Solutions Near an Ordinary Point I 93
7.3 Series Solutions Near an Ordinary Point II 96
7.4 Regular Singular Points; Euler Equations 102
7.5 The Method of Frobenius I 103
7.6 The Method of Frobenius II 108
7.7 The Method of Frobenius III 118
Chapter 8 Laplace Transforms 125
8.1 Introduction to the Laplace Transform 125
8.2 The Inverse Laplace Transform 127
8.3 Solution of Initial Value Problems 134
8.4 The Unit Step Function 140
8.5 Constant Coefficient Equations with Piecewise Continuous Forcing
Functions 143
8.6 Convolution 152

5. Contentsiii
8.7 Constant Cofficient Equations with Impulses 55
Chapter 9 Linear Higher Order Equations 159
9.1 Introduction to Linear Higher Order Equations 159
9.2 Higher Order Constant Coefficient Homogeneous Equations 171
9.3 Undetermined Coefficients for Higher Order Equations 17 5
9.4 Variation of Parameters for Higher Order Equations 181
Chapter 10 Linear Systems of Differential Equations 221
10.1 Introduction to Systems of Differential Equations 191
10.2 Linear Systems of Differential Equations 192
10.3 Basic Theory of Homogeneous Linear Systems 193
10.4 Constant Coefficient Homogeneous Systems I 194
10.5 Constant Coefficient Homogeneous Systems II 201
10.6 Constant Coefficient Homogeneous Systems II 245
10.7 Variation of Parameters for Nonhomogeneous Linear Systems 218
Chapter 221
11.1 Eigenvalue Problems for y
00
CyD0 221
11.2 Fourier Expansions I 223
11.3 Fourier Expansions II 229
Chapter 12 Fourier Solutions of Partial Differential Equations 239
12.1 The Heat Equation 239
12.2 The Wave Equation 247
12.3 Laplace’s Equation in Rectangular Coordinates 260
12.4 Laplace’s Equation in Polar Coordinates 270
Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations 273
13.1 Two-Point Boundary Value Problems 273
13.2 Sturm-Liouville Problems 279

6. CHAPTER1
Introduction
1.2BASIC CONCEPTS
1.2.2.(a)IfyDce
2x
, theny
0
D2ce
2x
D2y.
(b)IfyD
x
2
3
C
c
x
, theny
0
D
2x
3

c
x
2
, soxy
0
CyD
2x
2
3

c
x
C
x
2
3
C
c
x
Dx
2
.
(c)If
yD
1
2
Cce
x
2
;theny
0
D 2xce
x
2
and
y
0
C2xyD 2xce
x
2
C2x

1
2
Cce
x
2

D 2xce
x
2
CxC2cxe
x
2
Dx:
(d)If
yD
1Cce
x
2
=2
1ce
x
2
=2
then
y
0
D
.1ce
x
2
=2
/.cxe
x
2
=2
/.1Cce
x
2
=2
/cxe
x
2
=2
.1cxe
x
2
=2
/
2
D
2cxe
x
2
=2
.1ce
x
2
=2
/
2
and
y
2
1D
 
1Cce
x
2
=2
1ce
x
2
=2
!
2
1
D
.1Cce
x
2
=2
/
2
.1ce
x
2
=2
/
2
.1ce
x
2
=2
/
2
D
4ce
x
2
=2
.1ce
x
2
=2
/
2
;
1

7. 2 Chapter 1Basic Concepts
so
2y
0
Cx.y
2
1/D
4cxC4cx
.1ce
x
2
=2
/
2
D0:
(e)IfyDtan

x
3
3
Cc

, theny
0
Dx
2
sec
2

x
3
3
Cc

Dx
2

1Ctan
2

x
3
3Cc

Dx
2
.1Cy
2
/.
(f)If
yD.c1Cc2x/e
x
CsinxCx
2
;then
y
0
D.c1C2c2x/e
x
CcosxC2x;
y
0
D.c1C3c2x/e
x
sinxC2;
and y
00
2y
0
CyDc1e
x
.12C1/Cc2xe
x
.34C1/
sinx2cosxCsinxC24xCx
2
D 2cosxCx
2
4xC2:
(g)IfyDc1e
x
Cc2xC
2
x
, theny
0
Dc1e
x
Cc2
2
x
2
andy
00
Dc1e
x
C
4
x
3
, so.1x/y
00
Cxy
0
yD
c1.1xCx1/Cc2.xx/C
4.1x/
x
3

2
x

2
x
D
4.1xx
2
/
x
3
(h)IfyD
c1sinxCc2cosx
x
1=2
C4xC8theny
0
D
c1cosxc2sinx
x
1=2

c1sinxCc2cosx
2x
3=2
C4and
y
00
D
c1sinxCc2cosx
x
1=2

c1sinxc2cosx
x
3=2
C
3
4
c1sinxCc2cosx
x
5=2
, sox
2
y
00
Cxy
0
C

x
2

1
4

yD
c1

x
3=2
sinxx
1=2
cosxC
3
4
x
1=2
sinxCx
1=2
cosx
1
2
x
1=2
sinxCx
3=2
sinx
1
4
x
1=2
sinx

Cc2

x
3=2
cosxCx
1=2
sinxC
3
4
x
1=2
cosx
x
1=2
sinx
1
2
x
1=2
cosxCx
3=2
cosx
1
4
x
1=2
cosx

C4xC

x
2

1
4

.4xC8/D4x
3
C8x
2
C
3x2.
1.2.4.(a)Ify
0
D xe
x
, thenyD xe
x
C
R
e
x
dxCcD.1x/e
x
Cc, andy.0/D1)1D1Cc,
socD0andyD.1x/e
x
.
(b)Ify
0
Dxsinx
2
, thenyD
1
2
cosx
2
Cc;y
r

2

D1)1D0Cc, socD1and
yD1
1
2
cosx
2
.
(c)Writey
0
DtanxD
sinx
cosx
D
1
cosx
d
dx
.cosx/. Integrating this yieldsyD lnjcosxj Cc;
y.=4/D3)3D ln.cos.=4//Cc, or3Dln
p
2Cc, socD3ln
p
2, soyD ln.jcosxj/C
3ln
p
2D3ln.
p
2jcosxj/.
(d)Ify
00
Dx
4
, theny
0
D
x
5
5
Cc1;y
0
.2/D 1)
32
5
Cc1D 1)c1D
37
15
, soy
0
D
x
5
5

37
15
. Therefore,yD
x
6
30

37
15
.x2/Cc2;y.2/D 1)
64
30
Cc2D 1)c2D
47
15
, so
yD
47
15

37
5
.x2/C
x
6
30
.
(e)(A)
R
xe
2x
dxD
xe
2x
2

1
2
Z
e
2x
dxD
xe
2x
2

e
2x
4
. Therefore,y
0
D
xe
2x
2

e
2x
4
Cc1;
y
0
.0/D1)
1
4
Cc1D
5
4
)c1D
5
4
, soy
0
D
xe
2x
2

e
2x
4
C
5
4
; Using (A) again,yD
xe
2x
4

e
2x
8

e
2x
8
C
5
4
xCc2D
xe
2x
4

e
2x
4
C
5
4
xCc2;y.0/D7)
1
4
Cc2D7)c2D
29
4
, so
yD
xe
2x
4

e
2x
4
C
5
4
xC
29
4
.
(f)(A)
R
xsinx dxD xcosxC
R
cosx dxD xcosxCsinxand (B)
R
xcosx dxDxsinx
R
sinx dxDxsinxCcosx. Ify
00
D xsinx, then (A) implies thaty
0
DxcosxsinxCc1;y
0
.0/D
3)cD 3, soy
0
Dxcosxsinx3. Now (B) implies thatyDxsinxCcosxCcosx3xCc2D
xsinxC2cosx3xCc2;y.0/D1)2Cc2D1)c2D 1, soyDxsinxC2cosx3x1.

8. Section 1.2Basic Concepts3
(g)Ify
000
Dx
2
e
x
, theny
00
D
R
x
2
e
x
dxDx
2
e
x
2
R
xe
x
dxDx
2
e
x
2xe
x
C2e
x
Cc1;
y
00
.0/D3)2Cc1D3)c1D1, so (A)y
00
D.x
2
2xC2/e
x
C1. Since
R
.x
2
2xC2/e
x
dxD
.x
2
2xC2/e
x

R
.2x2/e
x
dxD.x
2
2xC2/e
x
.2x2/e
x
C2e
x
D.x
2
4xC6/e
x
,
(A) implies thaty
0
D.x
2
4xC6/e
x
CxCc2;y
0
.0/D 2)6Cc2D 2)c2D 8, so (B)
y
0
D.x
2
4xC6/e
x
Cx8; Since
R
.x
2
4xC6/e
x
dxD.x
2
4xC6/e
x

R
.2x4/e
x
dxD.x
2

4xC6/e
x
.2x4/e
x
C2e
x
D.x
2
6xC12/e
x
, (B) implies thatyD.x
2
6xC12/e
x
C
x
2
2
8xCc3;
y.0/D1)12Cc3D1)c3D 11, soyD.x
2
6xC12/e
x
C
x
2
2
8x11.
(h)Ify
000
D2Csin2x, theny
00
D2x
cos2x
2
Cc1;y
00
.0/D3)
1
2
Cc1D3)c1D
7
2
,
soy
00
D2x
cos2x
2
C
7
2
. Theny
0
Dx
2

sin2x
4
C
7
2
xCc2;y
0
.0/D 6)c2D 6, so
y
0
Dx
2

sin2x
4
C
7
2
x6. ThenyD
x
3
3
C
cos2x
8
C
7
4
x
2
6xCc3;y.0/D1)
1
8
Cc3D1)c3D
7
8
,
soyD
x
3
3
C
cos2x
8
C
7
4
x
2
6xC
7
8
.
(i)Ify
000
D2xC1, theny
00
Dx
2
CxCc1;y
00
.2/D7)6Cc1D7)c1D1; soy
00
Dx
2
CxC1.
Theny
0
D
x
3
3
C
x
2
2
C.x2/Cc2;y
0
.2/D 4)
14
3
Cc2D 4)c2D
26
3
, soy
0
D
x
3
3
C
x
2
2
C
.x2/
26
3
. ThenyD
x
4
12
C
x
3
6
C
1
2
.x2/
2

26
3
.x2/Cc3;y.2/D1)
8
3
Cc3D1)c3D
5
3
,
soyD
x
4
12
C
x
3
6
C
1
2
.x2/
2

26
3
.x2/
5
3
.
1.2.6.(a)IfyDx
2
.1Clnx/, theny.e/De
2
.1Clne/D2e
2
;y
0
D2x.1Clnx/CxD3xC2xlnx,
soy
0
.e/D3eC2elneD5e; (A)y
00
D3C2C2lnxD5C2lnx. Now,3xy
0
4yD3x.3xC
2xlnx/4x
2
.1Clnx/D5x
2
C2x
2
lnxDx
2
y
00
, from (A).
(b)IfyD
x
2
3
Cx1, theny.1/D
1
3
C11D
1
3
;y
0
D
2
3
xC1, soy
0
.1/D
2
3
C1D
5
3
; (A)
y
00
D
2
3
. Nowx
2
xy
0
CyC1Dx
2
x

2
3
xC1

C
x
2
3
Cx1C1D
2
3
x
2
Dx
2
y
00
, from (A).
(c)IfyD.1Cx
2
/
1=2
, theny.0/D.1C0
2
/
1=2
D1;y
0
D x.1Cx
2
/
3=2
, soy
0
.0/D0; (A)
y
00
D.2x
2
1/.1Cx
2
/
5=2
. Now,.x
2
1/yx.x
2
C1/y
0
D.x
2
1/.1Cx
2
/
1=2
x.x
2
C1/.x/.1C
x
2
/
3=2
D.2x
2
1/.1Cx
2
/
1=2
Dy
00
.1Cx
2
/
2
from (A), soy
00
D
.x
2
1/yx.x
2
C1/y
0
.x
2
C1/
2
.
(d)IfyD
x
2
1x
, theny.1=2/D
1=4
11=2
D
1
2
;y
0
D
x.x2/
.1x/
2
, soy
0
.1=2/D
.1=2/.3=2/
.11=2/
2
D3;
(A)y
00
D
2
.1x/
3
. Now, (B)xCyDxC
x
2
1x
D
x
1x
and (C)xy
0
yD
x
2
.x2/
.1x/
2

x
2
1x
D
x
2
.1x/
2
. From (B) and (C),.xCy/.xy
0
y/D
x
3
.1x/
3
D
x
3
2
y
00
, soy
00
D
2.xCy/.xy
0
y/
x
3
.
1.2.8.(a)yD.xc/
a
is defined andxcDy
1=a
on.c;1/; moreover,y
0
Da.xc/
a1
D
a

y
1=a
a1
Day
.a1/=a
.
(b)ifa > 1ora < 0, theny0is a solution of (B) on.1;1/.
1.2.10.(a)Sincey
0
Dcwe must show that the right side of (B) reduces tocfor all values ofxin some

9. 4 Chapter 1Basic Concepts
interval. IfyDc
2
CcxC2cC1,
x
2
C4xC4yDx
2
C4xC4c
2
C4cxC8cC4
Dx
2
C4.1Cc/xC4.c
2
C2cC1/
Dx
2
C4.1Cc/C2.cC1/
2
D.xC2cC2/
2
:
Therefore,
p
x
2
C4xC4yDxC2cC2and the right side of (B) reduces tocifx >2c2.
(b)Ify1D
x.xC4/
4
, theny
0
1
D
xC2
2
andx
2
C4xC4yD0for allx. Therefore,y1satisfies
(A) on.1;1/.

10. CHAPTER2
FirstOrderEquations
2.1LINEAR FIRST ORDER EQUATIONS
2.1.2.
y
0
y
D 3x
2
;jlnjyj D x
3
Ck;yDce
x
3
.yDce
.lnx/
2
=2
.
2.1.4.
y
0
y
D
3
x
; lnjyj D 3lnjxj CkD lnjxj
3
Ck;yD
c
x
3
.
2.1.6.
y
0
y
D
1Cx
x
D
1
x
1;jlnjyj D lnjxj xCk;yD
ce
x
x
;y.1/D1)cDe;
yD
e
.x1/
x
.
2.1.8.
y
0
y
D
1
x
cotx;jlnjyj D lnjxj lnjsinxj CkD lnjxsinxj Ck;yD
c
xsinx
;
y.=2/D2)cD;yD

xsinx
.
2.1.10.
y
0
y
D
k
x
;jlnjyj D klnjxj Ck1Dlnjx
k
j Ck1;yDcjxj
k
;y.1/D3)cD3;
yD3x
k
.
2.1.12.
y
0
1
y1
D 3; lnjy1j D 3x;y1De
3x
;yDue
3x
;u
0
e
3x
D1;u
0
De
3x
;uD
e
3x
3
Cc;yD
1
3
Cce
3x
.
2.1.14.
y
0
1
y1
D 2x; lnjy1j D x
2
;y1De
x
2
;yDue
x
2
;u
0
e
x
2
Dxe
x
2
;u
0
Dx;
uD
x
2
2
Cc;yDe
x
2

x
2
2
Cc

.
2.1.16.
y
0
1
y1
D
1
x
; lnjy1j D lnjxj;y1D
1
x
;yD
u
x
;
u
0
x
D
7
x
2
C3;u
0
D
7
x
C3x;
uD7lnjxj C
3x
2
2
Cc;yD
7lnjxj
x
C
3x
2
C
c
x
.
5

11. 6 Chapter 2First Order Equations
2.1.18.
y
0
1
y1
D
1
x
2x; lnjy1j D lnjxj x
2
;y1D
e
x
2
x
;yD
ue
x
2
x
;
u
0
e
x
2
x
Dx
2
e
x
2
;
u
0
Dx
3
;uD
x
4
4
Cc;yDe
x
2

x
3
4
C
c
x

.
2.1.20.
y
0
1
y1
D tanx; lnjy1j Dlnjcosxj;y1Dcosx;yDucosx;u
0
cosxDcosx;u
0
D1;
uDxCc;yD.xCc/cosx.
2.1.22.
y
0
1
y1
D
4x3
.x2/.x1/
D
5
x2

1
x1
; lnjy1j D5lnjx2j lnjx1j Dln
ˇ
ˇ
ˇ
ˇ
.x2/
5
x1
ˇ
ˇ
ˇ
ˇ
;
y1D
.x2/
5
x1
;yD
u.x2/
5
x1
;
u
0
.x2/
5
x1
D
.x2/
2
x1
;u
0
D
1
.x2/
3
;uD
1
2
1
.x2/
2
C
c;yD
1
2
.x2/
3
.x1/
Cc
.x2/
5
.x1/
.
2.1.24.
y
0
1
y1
D
3
x
; lnjy1j D 3lnjxj Dlnjxj
3
;y1D
1
x
3
;yD
u
x
3
;
u
0
x
3
D
e
x
x
2
;u
0
Dxe
x
;
uDxe
x
e
x
Cc;yD
e
x
x
2

e
x
x
3
C
c
x
3
.
2.1.26.
y
0
1
y1
D
4x
1Cx
2
; lnjy1j D 2ln.1Cx
2
/Dln.1Cx
2
/
2
;y1D
1
.1Cx
2
/
2
;yD
u
.1Cx
2
/
2
;
u
0
.1Cx
2
/
2
D
2
.1Cx
2
/
2
;u
0
D2;uD2xCc;yD
2xCc
.1Cx
2
/
2
;y.0/D1)
cD1;yD
2xC1
.1Cx
2
/
2
.
2.1.28.
y
0
1
y1
D cotx; lnjy1j D lnjsinxj;y1D
1
sinx
;yD
u
sinx
;
u
0
sinx
Dcosx;u
0
D
sinxcosx;uD
sin
2
x
2
Cc;yD
sinx
2
Cccscx;y.=2/D1)cD
1
2
;yD
1
2
.sinxCcscx/.
2.1.30.
y
0
1
y1
D
3
x1
; lnjy1j D 3lnjx1j Dlnjx1j
3
;y1D
1
.x1/
3
;yD
u
.x1/
3
;
u
0
.x1/
3
D
1
.x1/
4
C
sinx
.x1/
3
;u
0
D
1
x1
Csinx;uDlnjx1j cosxCc;yD
lnjx1j cosxCc
.x1/
3
;y.0/D1)cD0;yD
lnjx1j cosx
.x1/
3
.
2.1.32.
y
0
1
y1
D
2
x
; lnjy1j D2lnjxj Dln.x
2
/;y1Dx
2
;yDux
2
;u
0
x
2
D x;u
0
D
1
x
;
uD lnjxj Cc;yDx
2
.clnjxj/;y.1/D1)cD1;yDx
2
.1lnx/.
2.1.34.
y
0
1
y1
D
3
x1
; lnjy1j D 3lnjx1j Dlnjx1j
3
;y1D
1
.x1/
3
;yD
u
.x1/
3
;
u
0
.x1/
3
D
1C.x1/sec
2
x
.x1/
4
;u
0
D
1
x1
Csec
2
x;uDlnjx1jCtanxCc;yD
lnjx1j CtanxCc
.x1/
3
;
y.0/D 1)cD1;yD
lnjx1j CtanxC1
.x1/
3
.

12. Section 2.1Linear First Order Equations7
2.1.36.
y
0
1
y1
D
2x
x
2
1
; lnjy1j Dlnjx
2
1j;y1Dx
2
1;yDu.x
2
1/;u
0
.x
2
1/Dx;
u
0
D
x
x
2
1
;uD
1
2
lnjx
2
1j Cc;yD.x
2
1/

1
2
lnjx
2
1j Cc

;y.0/D4)cD 4;
yD.x
2
1/

1
2
lnjx
2
1j 4

.
2.1.38.
y
0
1
y1
D 2x; lnjy1j D x
2
;y1De
x
2
;yDue
x
2
;u
0
e
x
2
Dx
2
;u
0
Dx
2
e
x
2
;uD
cC
Z
x
0
t
2
e
t
2
dt;yDe
x
2

cC
Z
x
0
t
2
e
t
2
dt

;y.0/D3)cD3;yDe
x
2

3C
Z
x
0
t
2
e
t
2
dt

.
2.1.40.
y
0
1
y1
D 1; lnjy1j D x;y1De
x
;yDue
x
;u
0
e
x
D
e
x
tanx
x
;u
0
D
tanx
x
;
uDcC
Z
x
1
tant
t
dt;yDe
x

cC
Z
x
1
tant
t
dt

;y.1/D0)cD0;yDe
x
Z
x
1
tant
t
dt.
2.1.42.
y
0
1
y1
D 1
1
x
; lnjy1j D xlnjxj;y1D
e
x
x
;yD
ue
x
x
;
u
0
e
x
x
D
e
x
2
x
;
u
0
De
x
e
x
2
;uDcC
Z
x
1
e
t
e
t
2
dt;yD
e
x
x

cC
Z
x
1
e
t
e
t
2
dt

;y.1/D2)cD2e;
yD
1
x

2e
.x1/
Ce
x
Z
x
1
e
t
e
t
2
dt

.
2.1.44.(b)Eqn. (A) is equivalent to
y
0

2
x
D
1
x
.B/
on.1; 0/and.0;1/. Here
y
0
1
y1
D
2
x
; lnjy1j D2lnjxj;y1Dx
2
;yDux
2
;u
0
x
2
D
1
x
;
u
0
D
1
x
3
;uD
1
2x
2
Cc, soyD
1
2
Ccx
2
is the general solution of (A) on.1; 0/and.0;1/.
(c)From the proof of(b), any solution of (A) must be of the form
yD
8
ˆ
<
ˆ
:
1
2
Cc1x
2
; x0;
1
2
Cc2x
2
; x < 0;
.C/
forx¤0, and any function of the form (C) satisfies (A) forx¤0. To complete the proof we must show
that any function of the form (C) is differentiable and satisfies (A) atxD0. By definition,
y
0
.0/Dlim
x!0
y.x/y.0/
x0
Dlim
x!0
y.x/1=2
x
if the limit exists. But
y.x/1=2
x
D

c1x; x > 0
c2x; x < 0;
soy
0
.0/D0. Since0y
0
.0/2y.0/D002.1=2/D 1, any function of the form (C) satisfies (A) at
xD0.
(d)From(b)any solutionyof (A) on.1;1/is of the form (C), soy.0/D1=2.

13. 8 Chapter 2First Order Equations
(e)Ifx0> 0, then every function of the form (C) withc1D
y01=2
x
2
0
andc2arbitrary is a solution
of the initial value problem on.1;1/. Since these functions are all identical on.0;1/, this does not
contradict Theorem 2.1.1, which implies that (B) (so (A)) has exactly one solution on.0;1/such that
y.x0/Dy0. A similar argument applies ifx0< 0.
2.1.46.(a)LetyDc1y1Cc2y2. Then
y
0
Cp.x/yD.c1y1Cc2y2/
0
Cp.x/.c1y1Cc2y2/
Dc1y
0
1
Cc2y
0
2
Cc1p.x/y1Cc2p.x/y2
Dc1.y
0
1
Cp.x/y1/Cc2.y2Cp.x/y2/Dc1f1.x/Cc2f2.x/:
(b)Letf1Df2Dfandc1D c2D1.
(c)Letf1Df,f2D0, andc1Dc2D1.
2.1.48.(a)If´Dtany, then´
0
D.sec
2
y/y
0
, so´
0
3´D 1;´1De
3x
;´Due
3x
;u
0
e
3x
D 1;
u
0
D e
3x
;uD
e
3x
3
Cc;´D
1
3
Cce
3x
Dtany;yDtan
1

1
3
Cce
3x

.
(b)If´De
y
2
, then´
0
D2yy
0
e
y
2
, so´
0
C
2
x
´D
1
x
2
;´1D
1
x
2
;´D
u
x
2
;
u
0
x
2
D
1
x
2
;u
0
D1;
uDxCc;´D
1
x
C
c
x
2
De
y
2
;yD ˙

ln

1
x
C
c
x
2

1=2
.
(c)Rewrite the equation as
y
0
y
C
2
x
lnyD4x. If´Dlny, then´
0
D
y
0
y
, so´
0
C
2
x
´D4x;´1D
1
x
2
;
´D
u
x
2
;
u
0
x
2
D4x;u
0
D4x
3
;uDx
4
Cc;´Dx
2
C
c
x
2
Dlny;yDexp

x
2
C
c
x
2

.
(d)If´D
1
1Cy
, then´
0
D
y
0
.1Cy/
2
, so´
0
C
1
x
´D
3
x
2
;´1D
1
x
;´D
u
x
;
u
0
x
D
3
x
2
;
u
0
D
3
x
;uD 3lnjxj c;´D
3lnjxj Cc
x
D
1
1Cy
;yD 1C
x
3lnjxj Cc
.
2.2SEPARABLE EQUATIONS
2.2.2.By inspection,yk(kDinteger) is a constant solution. Separate variables to find others:

cosy
siny

y
0
D sinx; ln.jsinyj/DcosxCc.
2.2.4.y0is a constant solution. Separate variables to find others:

lny
y

y
0
D x
2
;
.lny/
2
2
D

x
3
3
Cc.
2.2.6.y1andy 1are constant solutions. For others, separate variables:.y
2
1/
3=2
yy
0
D
1
x
2
;
.y
2
1/
1=2
D
1
x
cD

1Ccx
x

;.y
2
1/
1=2
D

x
1Ccx

;.y
2
1/D

x
1Ccx

2
;
y
2
D1C

x
1Ccx

2
;yD ˙
 
1C

x
1Ccx

2
!
1=2
.

14. Section 2.2Separable Equations9
2.2.8.By inspection,y0is a constant solution. Separate variables to find others:
y
0
y
D
x
1Cx
2
;
lnjyj D
1
2
ln.1Cx
2
/Ck;yD
c
p
1Cx
2
, which includes the constant solutiony0.
2.2.10..y1/
2
y
0
D2xC3;
.y1/
3
3
Dx
2
C3xCc;.y1/
3
D3x
2
C9xCc;yD1C

3x
2
C9xCc/
1=3
.
2.2.12.
y
0
y.yC1/
D x;

1
y

1
yC1

y
0
D x; ln
ˇ
ˇ
ˇ
ˇ
y
yC1
ˇ
ˇ
ˇ
ˇ
D
x
2
2
Ck;
y
yC1
Dce
x
2
=2
;y.2/D
1)cD
e
2
2
;yD.yC1/ce
x
2
=2
;y.1ce
x
2
=2
/Dce
x
2
=2
;yD
ce
x
2
=2
1ce
x
2
=2
; settingcD
e
2
2
yieldsyD
e
.x
2
4/=2
2e
.x
2
4/=2
.
2.2.14.
y
0
.yC1/.y1/.y2/
D
1
xC1
;

1
6
1
yC1

1
2
1
y1
C
1
3
1
y2

y
0
D
1
xC1
;

1
yC1

3
y1
C
2
y2

y
0
D

6
xC1
; lnjyC1j 3lnjy1j C2lnjy2j D 6lnjxC1j Ck;
.yC1/.y2/
2
.y1/
3
D
c
.xC1/
6
;
y.1/D0)cD 256;
.yC1/.y2/
2
.y1/
3
D
256
.xC1/
6
.
2.2.16.
y
0
y.1Cy
2
/
D2x;

1
y

y
y
2
C1

y
0
D2x; ln
 
jyj
p
y
2
C1
!
Dx
2
Ck;
y
p
y
2
C1
Dce
x
2
;
y.0/D1)cD
1
p
2
;
y
p
y
2
C1
D
e
x
2
p
2
;2y
2
D.y
2
C1/e
x
2
;y
2
.2e
x
2
/De
2x
2
;yD
1
p
2e
2x
2
1
.
2.2.18.
y
0
.y1/.y2/
D 2x;

1
y2

1
y1

y
0
D 2x; ln
ˇ
ˇ
ˇ
ˇ
y2
y1
ˇ
ˇ
ˇ
ˇ
D x
2
Ck;
y2
y1
Dce
x
2
;
y.0/D3)cD
1
2
;
y2
y1
D
e
x
2
2
;y2D
e
x
2
2
.y1/;y
 
1
e
x
2
2
!
D2
e
x
2
2
;yD
4e
x
2
2e
x
2
.
The interval of validity is.1;1/.
2.2.20.
y
0
y.y2/
D 1;
1
2

1
y2

1
y

y
0
D 1;

1
y2

1
y

y
0
D 2; ln
ˇ
ˇ
ˇ
ˇ
y2
y
ˇ
ˇ
ˇ
ˇ
D 2xCk;
y2
y
Dce
2x
;y.0/D1)cD 1;
y2
y
D e
2x
;y2D ye
2x
;y.1Ce
2x
/D2;
yD
2
1Ce
2x
. The interval of validity is.1;1/.
2.2.22.y2is a constant solution of the differential equation, and it satisfies the initial condition.
Therefore,y2is a solution of the initial value problem. The interval of validity is.1;1/.
2.2.24.
y
0
1Cy
2
D
1
1Cx
2
; tan
1
yDtan
1
xCk;yDtan.tan
1
xCk/. Now use the identity
tan.ACB/D
tanACtanB
1tanAtanB
withADtan
1
xandBDtan
1
cto rewriteyasyD
xCc
1cx
, where
cDtank.

15. 10 Chapter 2First Order Equations
2.2.26..siny/y
0
Dcosx;cosyDsinxCc;y./D

2
)cD0, so (A) cosyD sinx. To obtain
yexplicity we note thatsinxDcos.xC=2/, so (A) can be rewritten as cosyDcos.xC=2/. This
equation holds if an only if one of the following conditions holds for some integerk:
(B)yDxC

2
C2kImbox.C / yD x

2
C2k:
Among these choices the only way to satisfy the initial condition is to letkD1in (C), soyD xC
3
2
:
2.2.28.Rewrite the equation asP
0
D a˛P.P1=˛/. By inspection,P0andP1=˛are
constant solutions. Separate variables to find others:
P
0
P.P1=˛/
D a˛;

1
P1=˛

1
P

P
0
D a;
ln
ˇ
ˇ
ˇ
ˇ
P1=˛
P
ˇ
ˇ
ˇ
ˇ
D atCk; (A)
P1=˛
P
Dce
˛t
;P.1ce
˛t
/D1=˛; (B)PD
1
˛.1ce
˛t
/
.
From (A),P.0/DP0)cD
P01=˛
P0
. Substituting this into (B) yieldsPD
P0
˛P0C.1˛P0/e
at
.
From this limt!1P.t/D1=˛.
2.2.30.IfqDrSthe equation forIreduces toI
0
D rI
2
, so
I
0
I
2
D r;
1
I
D rt
1
I0
; so
ID
I0
1CrI0t
and limt!1I.t/D0. Ifq¤rS, then rewrite the equation forIasI
0
D rI.I˛/
with˛DS
q
r
. Separating variables yields
I
0
I.I˛/
D r;

1
I˛

1
I

I
0
D r˛; ln
ˇ
ˇ
ˇ
ˇ
I˛
I
ˇ
ˇ
ˇ
ˇ
D
r˛tCk; (A)
I˛
I
Dce
r ˛t
;I.1ce
r ˛t
/D˛; (B)ID
˛
1ce
r ˛t
. From (A),I.0/DI0)
cD
I0˛
I0
. Substituting this into (B) yieldsID
˛I0
I0C.˛I0/e
r ˛t
. Ifq < rS, then˛ > 0and
limt!1I.t/D˛DS
q
r
. Ifq > rS, then˛ < 0and limt!1I.t/D0.
2.2.34.The given equation is separable iffDap, whereais a constant. In this case the equation is
y
0
Cp.x/yDap.x/: . A/
LetPbe an antiderivative ofp; that is,P
0
Dp.
SOLUTION BYSEPARATION OFVARIABLES.y
0
D p.x/.ya/;
y
0
ya
D p.x/; lnjyaj D
P.x/Ck;yaDce
P .x/
;yDaCce
P .x/
.
SOLUTION BYVARIATION OFPARAMETERS.y1De
P .x/
is a solution of the complementary
equation, so solutions of (A) are of the formyDue
P .x/
whereu
0
e
P .x/
Dap.x/. Hence,u
0
D
ap.x/e
P .x/
;uDae
P .x/
Cc;yDaCce
P .x/
.
2.2.36.Rewrite the given equation as (A)y
0

2
x
yD
x
5
yCx
2
.y1Dx
2
is a solution ofy
0

2
x
yD0.
Look for solutions of (A) of the formyDux
2
. Thenu
0
x
2
D
x
5
.uC1/x
2
D
x
3
uC1
;u
0
D
x
uC1
;
.uC1/u
0
Dx;
.1Cu/
2
2
D
x
2
2
C
c
2
;uD 1˙
p
x
2
Cc;yDx
2

1˙
p
x
2
Cc

.

16. Section 2.3Existence and Uniqueness of Solutions of Nonlinear Equations 11
2.2.38.y1De
2x
is a solution ofy
0
2yD0. Look for solutions of the nonlinear equation of the
formyDue
2x
. Thenu
0
e
2x
D
xe
2x
1u
;u
0
D
x
1u
;.1u/u
0
Dx;
.1u/
2
2
D
1
2
.x
2
c/;
uD1˙
p
cx
2
;yDe
2x

1˙
p
cx
2

.
2.3EXISTENCE AND UNIQUENESS OF SOLUTIONS OF NONLINEAR EQUATIO NS
2.3.2.f .x; y/D
e
x
Cy
x
2
Cy
2
andfy.x; y/D
1
x
2
Cy
2

2y.e
x
Cy/
.x
2
Cy
2
/
2
are both continuous at all.x; y/¤
.0; 0/. Hence, Theorem 2.3.1 implies that if.x0; y0/¤.0; 0/, then the initial value problem has a a
unique solution on some open interval containingx0. Theorem 2.3.1 does not apply if.x0; y0/D.0; 0/.
2.3.4.f .x; y/D
x
2
Cy
2
lnxy
andfy.x; y/D
2y
lnxy

x
2
Cy
2
x.lnxy/
2
are both continuous at all.x; y/such
thatxy > 0andxy¤1. Hence, Theorem 2.3.1 implies that ifx0y0> 0andx0y0¤1, then the initial
value problem has unique solution on an open interval containingx0. Theorem 2.3.1 does not apply if
x0y00orx0y0D1.
2.3.6.f .x; y/D2xyandfy.x; y/D2xare both continuous at all.x; y/. Hence, Theorem 2.3.1
implies that if.x0; y0/is arbitrary, then the initial value problem has a unique solution on some open
interval containingx0.
2.3.8.f .x; y/D
2xC3y
x4y
andfy.x; y/D
3
x4y
C4
2xC3y
.x4y/
2
are both continuous at all.x; y/such
thatx¤4y. Hence, Theorem 2.3.1 implies that ifx0¤4y0, then the initial value problem has a unique
solution on some open interval containingx0. Theorem 2.3.1 does not apply ifx0D4y0.
2.3.10.f .x; y/Dx.y
2
1/
2=3
is continuous at all.x; y/, butfy.x; y/D
4
3
xy.y
2
1/
1=3
is continuous
at.x; y/if and only ify¤ ˙1. Hence, Theorem 2.3.1 implies that ify0¤ ˙1, then the initial value
problem has a unique solution on some open interval containingx0, while ify0D ˙1, then the initial
value problem has at least one solution (possibly not uniqueon any open interval containingx0).
2.3.12.f .x; y/D.xCy/
1=2
andfy.x; y/D
1
2.xCy/
1=2
are both continuous at all.x; y/such that
xCy > 0Hence, Theorem 2.3.1 implies that ifx0Cy0> 0, then the initial value problem has a unique
solution on some open interval containingx0. Theorem 2.3.1 does not apply ifx0Cy00.
2.3.14.To apply Theorem 2.3.1, rewrite the given initial value problem as (A)y
0
Df .x; y/; y.x0/Dy0,
wheref .x; y/D p.x/yCq.x/andfy.x; y/D p.x/. Ifpandfare continuous on some open
interval.a; b/containingx0, thenfandfyare continuous on some open rectangle containing.x0; y0/,
so Theorem 2.3.1 implies that (A) has a unique solution onsomeopen interval containingx0. The
conclusion of Theorem 2.1.2 is more specific: the solution of(A) exists and is unique on.a; b/. For
example, in the extreme case where.a; b/D.1;1/, Theorem 2.3.1 still implies only existence and
uniqueness onsomeopen interval containingx0, while Theorem 2.1.2 implies that the solution exists and
is unique on.1;1/.
2.3.16.First find solutions of (A)y
0
Dy
2=5
. Obviouslyy0is a solution. Ify60, then we
can separate variables on any open interval whereyhas no zeros:y
2=5
y
0
D1;
5
3
y
3=5
DxCc;
yD

3
5
.xCc/
5=3

. (Note that this solution is also defined atxD c, even thoughy.c/D0.

17. 12 Chapter 2First Order Equations
To satisfy the initial condition, letcD1. Thus,yD

3
5
.xC1/
5=3

is a solution of the initial value
problem on.1;1/; moreover, sincef .x; y/Dy
2=5
andfy.x; y/D
2
5
y
3=5
are both continuous at
all.x; y/such thaty¤0, this is the only solution on.5=3;1/, by an argument similar to that given in
Example 2.3.7, the function
yD
(
0; 1< x
5
3

3
5
xC1

5=3
;
5
3
< x <1
(To see thatysatisfiesy
0
Dy
2=5
atxD
5
3
use an argument similar to that of Discussion 2.3.15-2) For
everya
5
3
, the following function is also a solution:
yD
8
ˆ
ˆ
<
ˆ
ˆ
:

3
5
.xCa/

5=3
;1< x <a;
0; ax
5
3

3
5
xC1

5=3
;
5
3
< x <1:
2.3.18.Obviously,y11is a solution. From Discussion 2.3.18 (takingcD0in the two families of
solutions) yieldsy2D1C jxj
3
andy3D1 jxj
3
. Other solutions arey4D1Cx
3
,y5D1x
3
,
y6D

1Cx
3
; x0;
1; x < 0
Iy7D

1x
3
; x0;
1; x < 0
I
y8D

1; x0;
1Cx
3
; x < 0
Iy9D

1; x0;
1x
3
; x < 0
It is straightforward to verify that all these functions satisfyy
0
D3x.y1/
1=3
for allx¤0. Moreover,
y
0
i
.0/Dlim
x!0
yi.x/1
x
D0for1i9, which implies that they also satisfy the equation atxD0.
2.3.20.Letybe any solution of (A)y
0
D3x.y1/
1=3
; y.3/D 7. By continuity, there is some open
intervalIcontainingx0D3on whichy.x/ < 1. From Discussion 2.3.18,yD1C.x
2
Cc/
3=2
onI;
y.3/D 7)cD 5; (B)yD1.x
2
5/
3=2
. It now follows that every solution of (A) satisfies
y.x/ < 1and is given by (B) on.
p
5;1/; that is, (B) is the unique solution of (A) on.
p
5;1/. This
solution can be extended uniquely to.0;1/as
yD

1; 0 < x
p
5;
1.x
2
5/
3=2
;
p
5 < x <1
It can be extended to.1;1/in infinitely many ways. Thus,
yD

1; 1< x
p
5;
1.x
2
5/
3=2
;
p
5 < x <1
is a solution of the initial value problem on.1;1/. Moroever, if˛0, then
yD
8
<
:
1C.x
2
˛
2
/
3=2
;1< x <˛;
1; ˛x
p
5;
1.x
2
5/
3=2
;
p
5 < x <1;

18. Section 2.4Transformation of Nonlinear Equations into Separable Equations 13
and
yD
8
<
:
1.x
2
˛
2
/
3=2
;1< x <˛;
1; ˛x
p
5;
1.x
2
5/
3=2
;
p
5 < x <1;
are also solutions of the initial value problem on.1;1/.
2.4TRANSFORMATION OF NONLINEAR EQUATIONS INTO SEPARABLE EQUA TIONS
2.4.2.Rewrite asy
0

2
7x
yD
x
7y
6
. Then
y
0
1
y1
D
2
7x
; lnjy1j D
2
7
lnjxj Dlnjxj
2=7
;y1Dx
2=7
;
yDux
2=7
;u
0
x
2=7
D
1
7u
6
x
5=7
;u
6
u
0
D
1
7x
;
u
7
7
D
1
7
lnjxj C
c
7
;uD.clnjxj/
1=7
;
yDx
2=7
.clnjxj/
1=7
.
2.4.4.Rewrite asy
0
C
2x
1Cx
2
yD
1
.1Cx
2
/
2
y
. Then
y
0
1
y1
D
2x
1Cx
2
; lnjy1j D ln.1Cx
2
/;
y1D
1
1Cx
2
;yD
u
1Cx
2
;
u
0
1Cx
2
D
1
u.1Cx
2
/
;u
0
uD1;
u
2
2
DxC
c
2
;uD ˙
p
2xCc;
yD ˙
p
2xCc
1Cx
2
.
2.4.6.
y
0
1
y1
D
1
3

1
x
C1

; lnjy1j D
1
3
.lnjxj Cx/;y1Dx
1=3
e
x=3
;yDux
1=3
e
x=3
;u
0
x
1=3
e
x=3
D
x
4=3
e
4x=3
u
4
;
u
0
u
4
Dxe
x
;
1
3u
3
D.x1/e
x

c
3
;uD
1
Œ3.1x/e
x
Cc
1=3
;yD

x
3.1x/Cce
x

1=3
.
2.4.8.
y
0
1
y1
Dx; lnjy1j D
x
2
2
;y1De
x
2
=2
;yDue
x
2
=2
;u
0
e
x
2
=2
Dxu
3=2
e
3x
2
=4
;
u
0
u
3=2
Dxe
x
2
=4
;
(A)
2
u
1=2
D2e
x
2
=4
C2c;u
1=2
D
1
cCe
x
2
=4
;u
D
1
.cCe
x
2
=4
/
2
;yD
1
.1Cce
x
2
=4
/
2
. Because
of (A) we must choosecso thaty.1/D4and1Cce
1=4
< 0. This implies thatcD 3e
1=4
;
yD

1
3
2
e
.x
2
1/=4

2
.
2.4.10.
y
0
1
y1
D2; lnjy1j D2x;y1De
2x
;yDue
2x
;u
0
e
2x
D2u
1=2
e
x
;u
1=2
u
0
D2e
x
;
2u
1=2
D 2e
x
C2c;u
1=2
Dce
x
> 0;y.0/D1)u.0/D1)cD2;uD.2e
x
/
2
;
yD.2e
x
1/
2
.
2.4.12.Rewrite asy
0
C
2
x
yD
y
3
x
2
. Then
y
0
1
y1
D
2
x
; lnjy1j D 2lnjxj Dlnx
2
;y1D
1
x
2
;
yD
u
x
2
;
u
0
x
2
D
u
3
x
8
;
u
0
u
3
D
1
x
6
;
1
2u
2
D
1
5x
5
Cc;y.1/D
1
p
2
)u.1/D
1
p
2
)cD
4
5
;
uD

5x
5
2.1C4x
5
/
1=2
;yD

5x
2.1C4x
5
/

1=2
.
2.4.14.PDue
at
;u
0
e
at
D a˛u
2
e
2at
;
u
0
u
2
D a˛e
at
;
1
u
D a
Z
t
0
˛./e
a
d
1
P0
;PD
P0e
at
1CaP0
R
t
0
˛./e
a
d
, which can also be written asPD
P0
e
at
CaP0e
at
R
t
0
˛./e
a
d
. Therefore,

19. 14 Chapter 2First Order Equations
limt!1P.t/D
8
<
:
1ifLD0;
0ifLD 1;
1=aLif0 < L <1:
2.4.16.yDux;u
0
xCuDu
2
C2u; (A)u
0
xDu.uC1/. Sinceu0andu 1are constant
solutions of (A),y0andyD xare solutions of the given equation. The nonconstant solutions
of (A) satisfyD
u
0
u.uC1/
D
1
x
;

1
u

1
uC1

u
0
D
1
x
; ln
ˇ
ˇ
ˇ
ˇ
u
uC1
ˇ
ˇ
ˇ
ˇ
Dlnjxj Ck;
u
uC1
Dcx;
uD.uC1/cx;u.1cx/Dcx;uD
cx
1cx
;yD
cx
2
1cx
.
2.4.18.yDux;u
0
xCuDuCsec;u
0
xDsecu;.cosu/u
0
D
1
x
; sinuDlnjxj Cc;uD
sin
1
.lnjxj Cc/;yDxsin
1
.lnjxj Cc/.
2.4.20.Rewrite the given equation asy
0
D
x
2
C2y
2
xy
;yDux;u
0
xCuD
1
u
C2u;u
0
xD
1Cu
2
u
;
uu
0
1Cu
2
D
1
x
;
1
2
ln.1Cu
2
/DlnjxjCk; ln

1C
y
2
x
2

Dlnx
2
C2k;1C
y
2
x
2
Dcx
2
;x
2
Cy
2
Dcx
4
;
yD ˙x
p
cx
2
1.
2.4.22.yDux;u
0
xCuDuCu
2
;u
0
xDu
2
;
u
0
u
2
D
1
x
;
1
u
Dlnjxj Cc;y.1/D2)u.1/D
2)cD
1
2
;uD
2
2lnjxj C1
;yD
2x
2lnjxj C1
.
2.4.24.Rewrite the given equation asy
0
D
x
2
Cy
2
xy
;yDux;u
0
xCuD
1
u
u;u
0
xD
1C2u
2
u
;

uu
0
1C2u
2
D
1
x
;
1
4
ln.1C2u
2
/Dlnjxj Ck;x
4
.1C2u
2
/Dc;y.1/D2)u.1/D2)cD9;
x
4
.1C2u
2
/D9;u
2
D
9x
4
2x
4
;uD
1
x
2

9x
4
2
1=2
;yD
1
x

9x
4
2
1=2
.
2.4.26.Rewrite the given equation asy
0
D2C
y
2
x
2
C4
y
x
;yDux;u
0
xCuD2Cu
2
C4u;
u
0
xDu
2
C3uC2D.uC1/.uC2/;
u
0
.uC1/.uC2/
D
1
x
;

1
uC1

1
uC2

u
0
D
1
x
; ln
ˇ
ˇ
ˇ
ˇ
uC1
uC2
ˇ
ˇ
ˇ
ˇ
D
lnjxj Ck;
uC1
uC2
Dcx;y.1/D1)u.1/D1)cD
2
3
;
uC1
uC2
D
2
3
x;uC1D
2
3
x.uC2/;
u

1
2
3
x

D 1C
4
3
x;uD
4x3
2x3
;yD
x.4x3/
2x3
.
2.4.28.yDux;u
0
xCuD
1Cu
1u
;u
0
xD
1Cu
2
1u
;
.1u/u
0
1Cu
2
D
1
x
; tan
1
u
1
2
ln.1Cu
2
/DlnjxjCc;
tan
1
y
x

1
2
ln

1C
y
2
x
2

Dlnjxj Cc; tan
1
y
x

1
2
ln.x
2
Cy
2
/Dc.
2.4.30.yDux;u
0
xCuD
u
3
C2u
2
CuC1
.uC1/
2
D
u.uC1/
2
C1
.uC1/
2
DuC
1
.uC1/
2
;u
0
xD
1
.uC1/
2
;
.uC1/
2
u
0
D
1
x
;
.uC1/
3
3
Dlnjxj Cc;.uC1/
3
D3.lnjxj Cc/;

y
x
C1

3
D3.lnjxj Cc/;
.yCx/
3
D3x
3
.lnjxj Cc/.

20. Section 2.4Transformation of Nonlinear Equations into Separable Equations 15
2.4.32.yDux;u
0
xCuD
u
u2
; (A)u
0
xD
u.u3/
2u
; Sinceu0andu3are constant solutions
of (A),y0andyD3xare solutions of the given equation. The nonconstant solutions of (A) satisfy
.2u/u
0
u.u3/
D
1
x
;

1
u3
C
2
u

u
0
D
3
x
; lnju3j C2lnjuj D 3lnjxj Ck;u
2
.u3/D
c
x
3
;
y
2
.y3x/Dc.
2.4.34.yDux;u
0
xCuD
1CuC3u
3
1C3u
2
DuC
1
1C3u
2
;.1C3u
2
/u
0
D
1
x
;uCu
3
Dlnjxj Cc;
y
x
C
y
3
x
3
Dlnjxj Cc.
2.4.36.Rewrite the given equation asy
0
D
x
2
xyCy
2
xy
;yDux;u
0
xCuD
1
u
1Cu;u
0
xD
1u
u
;
uu
0
u1
D
1
x
;

1C
1
u1

u
0
D
1
x
;uClnju1j D lnjxjCk;e
u
.u1/D
c
x
;e
y=x
.yx/Dc.
2.4.38.yDux;u
0
xCuD1C
1
u
Cu; (A)u
0
xD
uC1
u
. Since (A) has the constant solutionuD 1;
yD xis a solution of the given equation. The nonconstant solutions of (A) satisfy
uu
0
uC1
D
1
x
;

1
1
uC1

u
0
D
1
x
;ulnjuC1j Dlnjxj Cc;
y
x
ln
ˇ
ˇ
ˇ
y
x
1
ˇ
ˇ
ˇDlnjxj Cc;yxlnjyxj Dcy.
2.4.40.IfxDXX0andyDYY0, then
dy
dx
D
dY
dx
D
dY
dX
dX
dx
D
dY
dX
, soyDy.x/satisfies the
given equation if and only ifYDY.X/satisfies
dY
dX
D
a.XX0/Cb.YY0/C˛
c.XX0/Cd.YY0/Cˇ
;
which reduces to the nonlinear homogeneous equation
dY
dX
D
aXCbY
cXCdY
if and only if
aX0CbY0D˛
cX0CdY0Dˇ:
.B/
We will now show that ifadbc¤0, then it is possible (for any choice of˛andˇ) to solve (B).
Multiplying the first equation in (B) bydand the second bybyields
daX0CdbY0Dd˛
bcX0CbdY0Dbˇ:
Subtracting the second of these equations from the first yields.adbc/X0D˛dˇb. Sinceadbc¤
0, this implies thatX0D
˛dˇb
adbc
. Multiplying the first equation in (B) bycand the second byayields
caX0CcbY0Dc˛
acX0CadY0Daˇ:

21. 16 Chapter 2First Order Equations
Subtracting the first of these equation from the second yields.adbc/Y0D˛cˇa. Sinceadbc¤0
this implies thatY0D
˛cˇa
adbc
.
2.4.42.For the given equation, (B) of Exercise 2.4.40 is
2X0CY0D 1
X0C2Y0D 4:
Solving this pair of equations yieldsX0D2andY0D 3. The transformed differential equation is
dY
dX
D
2XCY
XC2Y
: . A/
LetYDuX;u
0
XCUD
2Cu
1C2u
; (B)u
0
XD
2.u1/.uC1/
2uC1
. Sinceu1andu 1
satisfy (B),YDXandYD Xare solutions of (A). SinceXDxC2andYDy3, it follows
thatyDxC5andyD xC1are solutions of the given equation. The nonconstant solutions
of (B) satisfy
.2uC1/u
0
.u1/.uC1/
D
2
X
;

1
uC1
C
3
u1

u
0
D
4
X
; lnjuC1j C3lnju1j D
4lnjXj Ck;.uC1/.u1/
3
D
c
X
4
;.YCX/.YX/
3
Dc; SettingXDxC2andYDy3
yields.yCx1/.yx5/
3
Dc.
2.4.44.Rewrite the given equation asy
0
D
y
3
Cx
3xy
2
;yDux
1=3
;u
0
x
1=3
C
1
3x
2=3
uD
u
3
C1
3u
2
x
2=3
;
u
0
x
1=3
D
1
3x
2=3
u
2
;u
2
u
0
D
1
3x
;
u
3
3
D
1
3
.lnjxj Cc/;uD.lnjxj Cc/
1=3
;yDx
1=3
.lnjxj Cc/
1=3
.
2.4.46.Rewrite the given equation asy
0
D
2.y
2
Cx
2
yx
4
/
x
3
;yDux
2
;u
0
x
2
C2xuD2x.u
2
C
u1/; (A)u
0
x
2
D2x.u
2
1/. Sinceu1andu 1are constant solutions of (A),yDx
2
andyD x
2
are solutions of the given equation. The nonconstant solutions of (A) satisfy
u
0
u
2
1
D
2
x
;

1
u1

1
uC1

u
0
D
4
x
; ln
ˇ
ˇ
ˇ
ˇ
u1
uC1
ˇ
ˇ
ˇ
ˇ
D4lnjxj Ck;
u1
uC1
Dcx
4
;.u1/D.uC1/cx
4
;
u.1cx
4
/D1Ccx
4
;uD
1Ccx
4
1cx
4
;yD
x
2
.1Ccx
4
/
1cx
4
.
2.4.48.yDutanx;u
0
tanxCusec
2
xD.u
2
CuC1/sec
2
x;u
0
tanxD.u
2
C1/sec
2
x;
u
0
u
2
C1
D
sec
2
xcotxDcotxCtanx; tan
1
uDlnjsinxjlnjcosxjCcDlnjtanxjCc;uDtan.lnjtanxjC
c/;yDtanxtan.lnjtanxj Cc/.
2.4.50.Rewrite the given equation asy
0
D
.yC
p
x/
2
2x.yC2
p
x/
;yDux
1=2
;u
0
x
1=2
C
1
2
p
x
uD
.uC1/
2
2
p
x .uC2/
;
u
0
x
1=2
D
1
2
p
x .uC2/
;.uC2/u
0
D
1
2x
;
.uC2/
2
2
D
1
2
.lnjxj Cc/;.uC2/
2
Dlnjxj Cc;
uD 2˙
p
lnjxj Cc;yDx
1=2
.2˙
p
lnjxj Cc/.
2.4.52.y1D
1
x
2
is a solution ofy
0
C
2
x
yD0. LetyD
u
x
2
; then
u
0
x
2
D
3x
2
.u
2
=x
4
/C6x.u=x
2
/C2
x
2
.2x.u=x
2
/C3/
D
3.u=x/
2
C6.u=x/C2
x
2
.2.u=x/C3/
;

22. Section 2.5Exact Equations17
so (A)u
0
D
3.u=x/
2
C6.u=x/C2
2.u=x/C3
. Since (A) is a homogeneous nonlinear equation, we now substitute
uDvxinto (A). This yieldsv
0
xCvD
3v
2
C6vC2
2vC3
;v
0
xD
.vC1/.vC2/
2vC3
;
.2vC3/v
0
.vC1/.vC2/
D
1
x
;

1
vC1
C
1
vC2

v
0
D
1
x
; lnj.vC1/.vC2/j Dlnjxj Ck; (B).vC1/.vC2/Dcx. Since
y.2/D2)u.2/D8)v.2/D4, (B) implies thatcD15..vC1/.vC2/D15x;v
2
C3vC
215xD0. From the quadratic formula,vD
3C
p
1C60x
2
;uDvxD
x.3C
p
1C60x/
2
;
yD
u
x
2
D
3C
p
1C60x
2x
.
2.4.54.Differentiating (A)y1.x/D
y.ax/
a
yields (B)y
0
1
.x/D
1
a
y
0
.ax/aDy
0
.ax/. Sincey
0
.x/D
q.y.x/=x/on some intervalI, (C)y
0
.ax/Dq.y.ax/=ax/on some intervalJ. Substituting (A) and (B)
into (C) yieldsy
0
1
.x/Dq.y1.x/=x/onJ.
2.4.56.IfyD´C1, then´
0
C´Dx´
2
;´Due
x
;u
0
e
x
Dxu
2
e
2x
;
u
0
u
2
Dxe
x
;
1
u
D
e
x
.xC1/c;uD
1
e
x
.xC1/Cc
;´D
1
xC1Cce
x
;yD1C
1
xC1Cce
x
.
2.4.58.IfyD´C1, then´
0
C
2
x
´D´
2
;´1D
1
x
2
;´D
u
x
2
;
u
0
x
2
D
u
2
x
4
;
u
0
u
2
D
1
x
2
;
1
u
D

1
x
CcD
1cx
x
;uD
x
1cx
;´D
1
x.1cx/
;yD1
1
x.1cx/
.
2.5EXACT EQUATIONS
2.5.2.M.x; y/D3ycosxC4xe
x
C2x
2
e
x
;N.x; y/D3sinxC3;My.x; y/D3cosxDNx.x; y/,
so the equation is exact. We must findFsuch that (A)Fx.x; y/D3ycosxC4xe
x
C2x
2
e
x
and (B)
Fy.x; y/D3sinxC3. Integrating (B) with respect toyyields (C)F.x; y/D3ysinxC3yC .x/.
Differentiating (C) with respect toxyields (D)Fx.x; y/D3ycosxC 
0
.x/. Comparing (D) with
(A) shows that (E) 
0
.x/D4xe
x
C2x
2
e
x
. Integration by parts yields
Z
xe
x
dxDxe
x
e
x
and
Z
x
2
e
x
dxDx
2
e
x
2xe
x
C2e
x
. Substituting from the last two equations into (E) yields .x/D2x
2
e
x
.
Substituting this into (C) yieldsF.x; y/D3ysinxC3yC2x
2
e
x
. Therefore,3ysinxC3yC2x
2
e
x
Dc.
2.5.4.M.x; y/D2x2y
2
;N.x; y/D12y
2
4xy;My.x; y/D 4yDNx.x; y/, so the equation
is exact. We must findFsuch that (A)Fx.x; y/D2x2y
2
and (B)Fy.x; y/D12y
2
4xy.
Integrating (A) with respect toxyields (C)F.x; y/Dx
2
2xy
2
C.y/. Differentiating (C) with
respect toyyields (D)Fy.x; y/D 4xyC
0
.y/. Comparing (D) with (B) shows that
0
.y/D12y
2
,
so we take.y/D4y
3
. Substituting this into (C) yieldsF.x; y/Dx
2
2xy
2
C4y
3
. Therefore,
x
2
2xy
2
C4y
3
Dc.
2.5.6.M.x; y/D4xC7y;N.x; y/D3xC4y;My.x; y/D7¤3DNx.x; y/, so the equation is
not exact.
2.5.8.M.x; y/D2xCy;N.x; y/D2yC2x;My.x; y/D1¤2DNx.x; y/, so the equation is not
exact.

23. 18 Chapter 2First Order Equations
2.5.10.M.x; y/D2x
2
C8xyCy
2
;N.x; y/D2x
2
C
xy
3
3
;My.x; y/D8xC2y¤4xC
y
3
3
D
Nx.x; y/, so the equation is not exact.
2.5.12.M.x; y/DysinxyCxy
2
cosxy;N.x; y/DxsinxyCxy
2
cosxy;My.x; y/D3xycosxyC
.1x
2
y
2
/sinxy¤.xyCy
2
/cosxyC.1xy
3
/sinxyDNx.x; y/, so the equation is not exact.
2.5.14.M.x; y/De
x
.x
2
y
2
C2xy
2
/C6x;N.x; y/D2x
2
ye
x
C2;My.x; y/D2xye
x
.xC2/D
Nx.x; y/, so the equation is exact. We must findFsuch that (A)Fx.x; y/De
x
.x
2
y
2
C2xy
2
/C6xand
(B)Fy.x; y/D2x
2
ye
x
C2. Integrating (B) with respect toyyields (C)F.x; y/Dx
2
y
2
e
x
C2yC .x/.
Differentiating (C) with respect toxyields (D)Fx.x; y/De
x
.x
2
y
2
C2xy
2
/C 
0
.x/. Comparing (D)
with (A) shows that 
0
.x/D6x, so we take .x/D3x
2
. Substituting this into (C) yieldsF.x; y/D
x
2
y
2
e
x
C2yC3x
2
. Therefore,x
2
y
2
e
x
C2yC3x
2
Dc.
2.5.16.M.x; y/De
xy
.x
4
yC4x
3
/C3y;N.x; y/Dx
5
e
xy
C3x;My.x; y/Dx
4
e
xy
.xyC5/C3D
Nx.x; y/, so the equation is exact. We must findFsuch that (A)Fx.x; y/De
xy
.x
4
yC4x
3
/C3yand
(B)Fy.x; y/Dx
5
e
xy
C3x. Integrating (B) with respect toyyields (C)F.x; y/Dx
4
e
xy
C3xyC .x/.
Differentiating (C) with respect toxyields (D)Fx.x; y/De
xy
.x
4
yC4x
3
/C3yC 
0
.x/. Comparing
(D) with (A) shows that 
0
.x/D0, so we take .x/D0. Substituting this into (C) yieldsF.x; y/D
x
4
e
xy
C3xy. Therefore,x
4
e
xy
C3xyDc.
2.5.18.M.x; y/D4x
3
y
2
6x
2
y2x3;N.x; y/D2x
4
y2x
3
;My.x; y/D8x
3
y6x
2
D
Nx.x; y/, so the equation is exact. We must findFsuch that (A)Fx.x; y/D4x
3
y
2
6x
2
y2x3
and (B)Fy.x; y/D2x
4
y2x
3
. Integrating (A) with respect toxyields (C)F.x; y/Dx
4
y
2
2x
3
y
x
2
3xC.y/. Differentiating (C) with respect toyyields (D)Fy.x; y/D2x
4
y2x
3
C
0
.y/.
Comparing (D) with (B) shows that
0
.y/D0, so we take.y/D0. Substituting this into (C) yields
F.x; y/Dx
4
y
2
2x
3
yx
2
3x. Therefore,x
4
y
2
2x
3
yx
2
3xDc. Sincey.1/D3)cD 1,
x
4
y
2
2x
3
yx
2
3xC1D0is an implicit solution of the initial value problem. Solving this fory
by means of the quadratic formula yieldsyD
xC
p
2x
2
C3x1
x
2
.
2.5.20.M.x; y/D.y
3
1/e
x
;N.x; y/D3y
2
.e
x
C1/;My.x; y/D3y
2
e
x
DNx.x; y/, so the
equation is exact. We must findFsuch that (A)Fx.x; y/D.y
3
1/e
x
and (B)Fy.x; y/D3y
2
.e
x
C1/.
Integrating (A) with respect toxyields (C)F.x; y/D.y
3
1/e
x
C.y/. Differentiating (C) with respect
toyyields (D)Fy.x; y/D3y
2
e
x
C
0
.y/. Comparing (D) with (B) shows that
0
.y/D3y
2
, so we take
.y/Dy
3
. Substituting this into (C) yieldsF.x; y/D.y
3
1/e
x
Cy
3
. Therefore,.y
3
1/e
x
Cy
3
Dc.
Sincey.0/D0)cD 1,.y
3
1/e
x
Cy
3
D 1is an implicit solution of the initial value problem.
Therefore,y
3
.e
x
C1/De
x
1, soyD

e
x
1
e
x
C1

1=3
.
2.5.22.M.x; y/D.2x1/.y1/;N.x; y/D.xC2/.x3/;My.x; y/D2x1DNx.x; y/, so the
equation is exact. We must findFsuch that (A)Fx.x; y/D.2x1/.y1/and (B)Fy.x; y/D.xC
2/.x3/. Integrating (A) with respect toxyields (C)F.x; y/D.x
2
x/.y1/C.y/. Differentiating
(C) with respect toyyields (D)Fy.x; y/Dx
2
xC
0
.y/. Comparing (D) with (B) shows that

0
.y/D 6, so we take.y/D 6y. Substituting this into (C) yieldsF.x; y/D.x
2
x/.y1/6y.
Therefore,.x
2
x/.y1/6yDc. Sincey.1/D 1)cD6,.x
2
x/.y1/6yD6is an implicit
solution of the initial value problem. Therefore,.x
2
x6/yDx
2
xC6, soyD
x
2
xC6
.x3/.xC2/
.
2.5.24.M.x; y/De
x
.x
4
y
2
C4x
3
y
2
C1/;N.x; y/D2x
4
ye
x
C2y;My.x; y/D2x
3
ye
x
.xC4/D
Nx.x; y/, so the equation is exact. We must findFsuch that (A)Fx.x; y/De
x
.x
4
y
2
C4x
3
y
2
C1/

24. Section 2.5Exact Equations19
and (B)Fy.x; y/D2x
4
ye
x
C2y. Integrating (B) with respect toyyields (C)F.x; y/Dx
4
y
2
e
x
C
y
2
C .x/. Differentiating (C) with respect toxyields (D)Fx.x; y/De
x
y
2
.x
4
C4x
3
/C 
0
.x/.
Comparing (D) with (A) shows that 
0
.x/De
x
, so we take .x/De
x
. Substituting this into (C) yields
F.x; y/D.x
4
y
2
C1/e
x
Cy
2
. Therefore,.x
4
y
2
C1/e
x
Cy
2
Dc.
2.5.28.M.x; y/Dx
2
Cy
2
;N.x; y/D2xy;My.x; y/D2yDNx.x; y/, so the equation is exact.
We must findFsuch that (A)Fx.x; y/Dx
2
Cy
2
and (B)Fy.x; y/D2xy. Integrating (A) with
respect toxyields (C)F.x; y/D
x
3
3
Cxy
2
C.y/. Differentiating (C) with respect toyyields (D)
Fy.x; y/D2xyC
0
.y/. Comparing (D) with (B) shows that
0
.y/D0, so we take.y/D0.
Substituting this into (C) yieldsF.x; y/D
x
3
3
Cxy
2
. Therefore,
x
3
3
Cxy
2
Dc.
2.5.30.(a)Exactness requires thatNx.x; y/DMy.x; y/D
@
@y
.x
3
y
2
C2xyC3y
2
/D2x
3
yC2xC6y.
Hence,N.x; y/D
x
4
y
4
Cx
2
C6xyCg.x/, wheregis differentiable.
(b)Exactness requires thatNx.x; y/DMy.x; y/D
@
@y
.lnxyC2ysinx/D
1
y
C2sinx. Hence,
N.x; y/D
x
y
2cosxCg.x/, wheregis differentiable.
(c)Exactness requires thatNx.x; y/DMy.x; y/D
@
@y
.xsinxCysiny/DycosyCsiny. Hence,
N.x; y/Dx.ycosyCsiny/Cg.x/, wheregis differentiable.
2.5.32.The assumptions imply that
@M1
@y
D
@N1
@x
and
@M2
@y
D
@N2
@x
. Therefore,
@
@y
.M1CM2/D
@M1
@y
C
@M2
@y
D
@N1
@x
C
@N2
@x
D
@
@x
.N1CN2/, which implies that.M1CM2/ dxC.N1CN2/ dyD0
is exact onR.
2.5.34.HereM.x; y/DAx
2
CBxyCCy
2
andN.x; y/DDx
2
CExyCFy
2
. SinceMyDBxC2Cy
andNxD2DxCEy, the equation is exact if and only ifBD2DandED2C.
2.5.36.Differentiating (A)F.x; y/D
Z
y
y0
N.x0; s/ dsC
Z
x
x0
M.t; y/ dtwith respect toxyieldsFx.x; y/D
M.x; y/, since the first integral in (A) is independent ofxandM.t; y/is a continuous function oftfor
each fixedy. Differentiating (A) with respect toyand using the assumption thatMyDNxyields
Fy.x; y/DN.x0; y/C
Z
x
x0
@M
@y
.t; y/ dtDN.x0; y/C
Z
x
x0
@N
@x
.t; y/ dtDN.x0; y/CN.x; y/
N.x0; y/DN.x; y/.
2.5.38.y1D
1
x
2
is a solution ofy
0
C
2
x
yD0. LetyD
u
x
2
; then
u
0
x
2
D
2x.u=x
2
/
.x
2
C2x
2
.u=x
2
/C1/
D
2xu
x
2
.x
2
C2uC1/
;
sou
0
D
2xu
x
2
C2uC1
, which can be rewritten as (A)2xu dxC.x
2
C2uC1/ duD0. Since
@
@u
.2xu/D
@
@x
.x
2
C2uC1/D2x, (A) is exact. To solve (A) we must findFsuch that (A)Fx.x; u/D

25. 20 Chapter 2First Order Equations
2xuand (B)Fu.x; u/Dx
2
C2uC1. Integrating (A) with respect toxyields (C)F.x; u/Dx
2
uC.u/.
Differentiating (C) with respect touyields (D)Fu.x; u/Dx
2
C
0
.u/. Comparing (D) with (B) shows
that
0
.u/D2uC1, so we take.u/Du
2
Cu. Substituting this into (C) yieldsF.x; u/Dx
2
uC
u
2
CuDu.x
2
CuC1/. Therefore,u.x
2
CuC1/Dc. Sincey.1/D 2)u.1/D 2; cD0.
Therefore,u.x
2
CuC1/D0. Sinceu0does not satisfyu.1/D 2, it follows thatuD x
2
1
andyD 1
1
x
2
.
2.5.40.y1De
x
2
is a solution ofy
0
C2xyD0. LetyDue
x
2
; thenu
0
e
x
2
D e
x
2

3xC2u
2xC3u

, so
u
0
D
3xC2u
2xC3u
, which can be rewritten as (A).3xC2u/ dxC.2xC3u/ duD0. Since
@
@u
.3xC2u/D
@
@x
.2xC3u/D2, (A) is exact. To solve (A) we must findFsuch that (A)Fx.x; u/D3xC2uand
(B)Fu.x; u/D2xC3u. Integrating (A) with respect toxyields (C)F.x; u/D
3x
2
2
C2xuC.u/.
Differentiating (C) with respect touyields (D)Fu.x; u/D2xC
0
.u/. Comparing (D) with (B) shows
that
0
.u/D3u, so we take.u/D
3u
2
2
. Substituting this into (C) yieldsF.x; u/D
3x
2
2
C2xuC
3u
2
2
. Therefore,
3x
2
2
C2xuC
3u
2
2
Dc. Sincey.0/D 1)u.0/D 1; cD
3
2
. Therefore,
3x
2
C4xuC3u
2
D3is an implicit solution of the initial value problem. Rewriting this as3u
2
C4xuC
.3x
2
3/D0and solving foruby means of the quadratic formula yieldsuD
 
2xC
p
95x
2
3
!
, so
yD e
x
2
 
2xC
p
95x
2
3
!
.
2.5.42.SinceM dxCN dyD0is exact, (A)MyDNx. SinceN dxCM dyD0is exact, (B)
MxD Ny. Differentiating (A) with respect toyand (B) with respect toxyields (C)MyyDNxyand
(D)MxxD Nyx. SinceNxyDNyx, adding (C) and (D) yieldsMxxCMyyD0. Differentiating
(A) with respect toxand (B) with respect toyyields (E)MyxDNxxand (F)MxyD Nyy. Since
MxyDMyx, subtracting (F) from (E) yieldsNxxCNyyD0.
2.5.44.(a)IfF.x; y/Dx
2
y
2
, thenFx.x; y/D2x,Fy.x; y/D 2y,Fxx.x; y/D2, and
Fyy.x; y/D 2. Therefore,FxxCFyyD0, andGmust satisfy (A)Gx.x; y/D2yand (B)
Gy.x; y/D2x. Integrating (A) with respect toxyields (C)G.x; y/D2xyC.y/. Differentiating (C)
with respect toyyields (D)Gy.x; y/D2xC
0
.y/. Comparing (D) with (B) shows that
0
.y/D0, so
we take.y/Dc. Substituting this into (C) yieldsG.x; y/D2xyCc.
(b)IfF.x; y/De
x
cosy, thenFx.x; y/De
x
cosy,Fy.x; y/D e
x
siny,Fxx.x; y/De
x
cosy,
andFyy.x; y/D e
x
cosy. Therefore,FxxCFyyD0, andGmust satisfy (A)Gx.x; y/De
x
siny
and (B)Gy.x; y/De
x
cosy. Integrating (A) with respect toxyields (C)G.x; y/De
x
sinyC.y/.
Differentiating (C) with respect toyyields (D)Gy.x; y/De
x
cosyC
0
.y/. Comparing (D) with (B)
shows that
0
.y/D0, so we take.y/Dc. Substituting this into (C) yieldsG.x; y/De
x
sinyCc.
(c)IfF.x; y/Dx
3
3xy
2
, thenFx.x; y/D3x
2
3y
2
,Fy.x; y/D 6xy,Fxx.x; y/D6x,
andFyy.x; y/D 6x. Therefore,FxxCFyyD0, andGmust satisfy (A)Gx.x; y/D6xyand
(B)Gy.x; y/D3x
2
3y
2
. Integrating (A) with respect toxyields (C)G.x; y/D3x
2
yC.y/.
Differentiating (C) with respect toyyields (D)Gy.x; y/D3x
2
C
0
.y/. Comparing (D) with (B)
shows that
0
.y/D 3y
2
, so we take.y/D y
3
Cc. Substituting this into (C) yieldsG.x; y/D
3x
2
yy
3
Cc.
(d)IfF.x; y/Dcosxcoshy, thenFx.x; y/D sinxcoshy,Fy.x; y/Dcosxsinhy,Fxx.x; y/D

26. Section 2.6Exact Equations21
cosxcoshy, andFyy.x; y/Dcosxcoshy. Therefore,FxxCFyyD0, andGmust satisfy (A)
Gx.x; y/D cosxsinhyand (B)Gy.x; y/D sinxcoshy. Integrating (A) with respect toxyields
(C)G.x; y/D sinxsinhyC.y/. Differentiating (C) with respect toyyields (D)Gy.x; y/D
sinxcoshyC
0
.y/. Comparing (D) with (B) shows that
0
.y/D0, so we take.y/Dc. Substituting
this into (C) yieldsG.x; y/D sinxsinhyCc.
(e)IfF.x; y/Dsinxcoshy, thenFx.x; y/Dcosxcoshy,Fy.x; y/Dsinxsinhy,Fxx.x; y/D
sinxcoshy, andFyy.x; y/Dsinxcoshy. Therefore,FxxCFyyD0, andGmust satisfy (A)
Gx.x; y/D sinxsinhyand (B)Gy.x; y/Dcosxcoshy. Integrating (A) with respect toxyields
(C)G.x; y/DcosxsinhyC.y/. Differentiating (C) with respect toyyields (D)Gy.x; y/D
cosxcoshyC
0
.y/. Comparing (D) with (B) shows that
0
.y/D0, so we take.y/Dc. Substi-
tuting this into (C) yieldsG.x; y/DcosxsinhyCc.
2.6INTEGRATING FACTORS
2.6.2.(a)and(b). To show that.x; y/D
1
.xy/
2
is an integrating factor for (A) and that (B) is exact,
it suffices to observe that
@
@x

xy
xy

D
y
2
.xy/
2
and
@
@y

xy
xy

D
x
2
.xy/
2
. By Theorem 2.5.1
this also shows that (C) is an implicit solution of (B). Since.x; y/is never zero, any solution of (B) is
a solution of (A).
(c)If we interpret (A) asy
2
Cx
2
y
0
D0, then substitutingyDxyieldsx
2
Cx
2
1D0.
(NOTE: In Exercises 2.6.3–2.6.23, the given equation is multiplied by an integrating factor to produce
an exact equation, and an implicit solution is found for the latter. For a complete analysis of the relation-
ship between the sets of solutions of the two equations it is necessary to check for additional solutions of
the given equation “along which" the integrating factor is undefined, or for solutions of the exact equation
“along which" the integrating factor vanishes. In the interests of brevity we omit these tedious details
except in cases where there actually is a difference betweenthe sets of solutions of the two equations.)
2.6.4.M.x; y/D3x
2
y;N.x; y/D2x
3
;My.x; y/Nx.x; y/D3x
2
6x
2
D 3x
2
;p.x/D
My.x; y/Nx.x; y/
N.x; y/
D
3x
2
2x
3
D
3
2x
;
R
p.x/ dxD
3
2
lnjxj;.x/DP.x/Dx
3=2
; there-
fore3x
1=2
y dxC2x
3=2
dyD0is exact. We must findFsuch that (A)Fx.x; y/D3x
1=2
yand (B)
Fy.x; y/D2x
3=2
. Integrating (A) with respect toxyields (C)F.x; y/D2x
3=2
yC.y/. Differenti-
ating (C) with respect toyyields (D)Fy.x; y/D2x
3=2
C
0
.y/. Comparing (D) with (B) shows that

0
.y/D0, so we take.y/D0. Substituting this into (C) yieldsF.x; y/D2x
3=2
y, sox
3=2
yDc.
2.6.6.M.x; y/D5xyC2yC5;N.x; y/D2x;My.x; y/Nx.x; y/D.5xC2/2D5x;p.x/D
My.x; y/Nx.x; y/
N.x; y/
D
5x
2x
D
5
2
;
R
p.x/ dxD
5x
2
;.x/DP.x/De
5x=2
; thereforee
5x=2
.5xyC
2yC5/ dxC2xe
5x=2
dyD0is exact. We must findFsuch that (A)Fx.x; y/De
5x=2
.5xyC2yC5/
and (B)Fy.x; y/D2xe
5x=2
. Integrating (B) with respect toyyields (C)F.x; y/D2xye
5x=2
C .x/.
Differentiating (C) with respect toxyields (D)Fx.x; y/D5xye
5x=2
C2ye
5x=2
C 
0
.x/. Comparing
(D) with (A) shows that 
0
.x/D5e
5x=2
, so we take .x/D2e
5x=2
. Substituting this into (C) yields
F.x; y/D2e
5x=2
.xyC1/, soe
5x=2
.xyC1/Dc.
2.6.8.M.x; y/D27xy
2
C8y
3
;N.x; y/D18x
2
yC12xy
2
;My.x; y/Nx.x; y/D.54xyC
24y
2
/.36xyC12y
2
/D18xyC12y
2
;p.x/D
My.x; y/Nx.x; y/
N.x; y/
D
18xyC12y
2
18x
2
yC12y
2
x
D
1
x
;
R
p.x/ dxDlnjxj;.x/DP.x/Dx; therefore.27x
2
y
2
C8xy
3
/ dxC.18x
3
yC12x
2
y
2
/ dyD0is
exact. We must findFsuch that (A)Fx.x; y/D27x
2
y
2
C8xy
3
and (B)Fy.x; y/D18x
3
yC12x
2
y
2
.
Integrating (A) with respect toxyields (C)F.x; y/D9x
3
y
2
C4x
2
y
3
C.y/. Differentiating (C)

27. 22 Chapter 2Integrating Factors
with respect toyyields (D)Fy.x; y/D18x
3
yC12x
2
y
2
C
0
.y/. Comparing (D) with (B) shows
that
0
.y/D0, so we take.y/D0. Substituting this into (C) yieldsF.x; y/D9x
3
y
2
C4x
2
y
3
, so
x
2
y
2
.9xC4y/Dc.
2.6.10.M.x; y/Dy
2
;N.x; y/D

xy
2
C3xyC
1
y

;My.x; y/Nx.x; y/D2y.y
2
C3y/D
y.yC1/;q.y/D
Nx.x; y/My.x; y/
M.x; y/
D
y.yC1/
y
2
D1C
1
y
;
R
q.y/ dyDylnjyj;.y/D
Q.y/Dye
y
; thereforey
3
e
y
dxCe
y
.xy
3
C3xy
2
C1/ dyD0is exact. We must findFsuch that
(A)Fx.x; y/Dy
3
e
y
and (B)Fy.x; y/De
y
.xy
3
C3xy
2
C1/. Integrating (A) with respect tox
yields (C)F.x; y/Dxy
3
e
y
C.y/. Differentiating (C) with respect toyyields (D)Fy.x; y/D
xy
3
e
y
C3xy
2
e
y
C
0
.y/. Comparing (D) with (B) shows that
0
.y/De
y
, so we take.y/De
y
.
Substituting this into (C) yieldsF.x; y/Dxy
3
e
y
Ce
y
, soe
y
.xy
3
C1/Dc.
2.6.12.M.x; y/Dx
2
yC4xyC2y;N.x; y/Dx
2
Cx;My.x; y/Nx.x; y/D.x
2
C4xC2/.2xC
1/Dx
2
C2xC1D.xC1/
2
;p.x/D
My.x; y/Nx.x; y/
N.x; y/
D
.xC1/
2
x.xC1/
D1C
1
x
;
R
p.x/ dxD
xClnjxj;.x/DP.x/Dxe
x
; thereforee
x
.x
3
yC4x
2
yC2xy/ dxCe
x
.x
3
Cx
2
/ dyD0is exact.
We must findFsuch that (A)Fx.x; y/De
x
.x
3
yC4x
2
yC2xy/and (B)Fy.x; y/De
x
.x
3
Cx
2
/.
Integrating (B) with respect toyyields (C)F.x; y/Dy.x
3
Cx
2
/e
x
C .x/. Differentiating (C) with
respect toxyields (D)Fx.x; y/De
x
.x
3
yC4x
2
yC2xy/C 
0
.x/. Comparing (D) with (A) shows
that 
0
.x/D0, so we take .x/D0. Substituting this into (C) yieldsF.x; y/Dy.x
3
Cx
2
/e
x
D
x
2
y.xC1/e
x
, sox
2
y.xC1/e
x
Dc.
2.6.14.M.x; y/Dcosxcosy;N.x; y/DsinxcosysinxsinyCy;My.x; y/Nx.x; y/D
cosxsiny.cosxcosycosxsiny/D cosxcosy;q.y/D
Nx.x; y/My.x; y/
M.x; y/
D
cosxcosy
cosxcosy
D
1;
R
q.y/ dyD1;.y/DQ.y/De
y
; thereforee
y
cosxcosy dxCe
y
.sinxcosysinxsinyC
y/ dyD0is exact. We must findFsuch that (A)Fx.x; y/De
y
cosxcosyand (B)Fy.x; y/D
e
y
.sinxcosysinxsinyCy/. Integrating (A) with respect toxyields (C)F.x; y/De
y
sinxcosyC
.y/. Differentiating (C) with respect toyyields (D)Fy.x; y/De
y
.sinxcosysinxsiny/C
0
.y/.
Comparing (D) with (B) shows that
0
.y/Dye
y
, so we take.y/De
y
.y1/. Substituting this into
(C) yieldsF.x; y/De
y
.sinxcosyCy1/, soe
y
.sinxcosyCy1/Dc.
2.6.16.M.x; y/Dysiny;N.x; y/Dx.sinyycosy/;My.x; y/Nx.x; y/D.ycosyCsiny/
.sinyycosy/D2ycosy;q.y/D
Nx.x; y/My.x; y/
N.x; y/
D
2cosy
siny
;
R
q.y/ dyD 2lnjsinyj;
.y/DQ.y/D
1
sin
2
y
; therefore

y
siny

dxCx

1
siny

ycosy
sin
2
y

dyD0is exact. We must
findFsuch that (A)Fx.x; y/D
y
siny
and (B)Fy.x; y/Dx

1
siny

ycosy
sin
2
y

. Integrating (A)
with respect toxyields (C)F.x; y/D
xy
siny
C.y/. Differentiating (C) with respect toyyields (D)
Fy.x; y/Dx

1
siny

ycosy
sin
2
y

C
0
.y/. Comparing (D) with (B) shows that
0
.y/D0, so we take
.y/D0. Substituting this into (C) yieldsF.x; y/D
xy
siny
, so
xy
siny
Dc. In addition, the given
equation has the constant solutionsyDk, wherekis an integer.
2.6.18.M.x; y/D˛yCxy;N.x; y/DˇxCıxy;My.x; y/Nx.x; y/D.˛Cx/.ˇCıy/;
andp.x/N.x; y/q.y/M.x; y/Dp.x/x.ˇCıy/q.y/y.˛Cx/. so exactness requires that

28. Section 2.6Exact Equations23
.˛Cx/.ˇCıy/Dp.x/x.ˇCıy/q.y/y.˛Cx/, which holds ifp.x/xD 1andq.y/yD
1. Thusp.x/D
1
x
;q.y/D
1
y
;
R
p.x/ dxD lnjxj;
R
q.y/ dyD lnjyj;P.x/D
1
x
;
Q.y/D
1
y
;.x; y/D
1
xy
. Therefore,

˛
x
C

dxC

ˇ
y
Cı

dyD0is exact. We must findF
such that (A)Fx.x; y/D
˛
x
Cand (B)Fy.x; y/D
ˇ
y
Cı. Integrating (A) with respect toxyields
(C)F.x; y/D˛lnjxj CxC.y/. Differentiating (C) with respect toyyields (D)Fy.x; y/D
0
.y/.
Comparing (D) with (B) shows that
0
.y/D
ˇ
y
Cı, so we take.y/Dˇlnjyj Cıy. Substituting this
into (C) yieldsF.x; y/D˛lnjxj CxCˇlnjyj Cıy, sojxj
˛
jyj
ˇ
e
x
e
ıy
Dc. The given equation also
has the solutionsx0andy0.
2.6.20.M.x; y/D2y;N.x; y/D3.x
2
Cx
2
y
3
/;My.x; y/Nx.x; y/D2.6xC6xy
3
/; and
p.x/N.x; y/q.y/M.x; y/D3p.x/.x
2
Cx
2
y
3
/2q.y/y. so exactness requires that (A)26x
6xy
3
D3p.x/x.xCxy
3
/2q.y/y. To obtain similar terms on the two sides of (A) we letp.x/xDa
andq.y/yDbwhereaandbare constants such that26x6xy
3
D3a.xCxy
3
/2b, which holds if
aD 2andbD 1. Thus,p.x/D
2
x
;q.y/D
1
y
;
R
p.x/ dxD 2lnjxj;
R
q.y/ dyD lnjyj;
P.x/D
1
x
2
;Q.y/D
1
y
;.x; y/D
1
x
2
y
. Therefore,
2
x
2
dxC3

1
y
Cy
2

dyD0is exact. We must
findFsuch that (B)Fx.x; y/D
2
x
2
and (C)Fy.x; y/D3

1
y
Cy
2

. Integrating (B) with respect tox
yields (D)F.x; y/D
2
x
C.y/. Differentiating (D) with respect toyyields (E)Fy.x; y/D
0
.y/.
Comparing (E) with (C) shows that
0
.y/D3

1
y
Cy
2

, so we take.y/Dy
3
C3lnjyj. Substituting
this into (D) yieldsF.x; y/D
2
x
Cy
3
C3lnjyj, so
2
x
Cy
3
C3lnjyj Dc. The given equation also
has the solutionsx0andy0.
2.6.22.M.x; y/Dx
4
y
4
;N.x; y/Dx
5
y
3
;My.x; y/Nx.x; y/D4x
4
y
3
5x
4
y
3
D x
4
y
3
;
andp.x/N.x; y/q.y/M.x; y/Dp.x/x
5
y
3
q.y/x
4
y
4
. so exactness requires thatx
4
y
3
D
p.x/x
5
y
3
q.y/x
4
y
4
, which is equivalent top.x/xq.y/yD 1. This holds ifp.x/xDaand
q.y/yDaC1whereais an arbitrary real number. Thus,p.x/D
a
x
;q.y/D
aC1
y
;
R
p.x/ dxD
alnjxj;
R
q.y/ dyD.aC1/lnjyj;P.x/D jxj
a
;Q.y/D jyj
aC1
;.x; y/D jx
a
jjyj
aC1
. Therefore,
jxj
a
jyj
aC1

x
4
y
4
dxCx
5
y
3
dy

D0is exact for any choice ofa. For simplicity we letaD 4, so (A)
is equivalent toy dxCx dyD0. We must findFsuch that (B)Fx.x; y/Dyand (C)Fy.x; y/Dx.
Integrating (B) with respect toxyields (D)F.x; y/DxyC.y/. Differentiating (D) with respect toy
yields (E)Fy.x; y/DxC
0
.y/. Comparing (E) with (C) shows that
0
.y/D0, so we take.y/D0.
Substituting this into (D) yieldsF.x; y/Dxy, soxyDc.
2.6.24.M.x; y/Dx
4
y
3
Cy;N.x; y/Dx
5
y
2
x;My.x; y/Nx.x; y/D.3x
4
y
2
C1/.5x
4
y
2

1/D 2x
4
y
2
C2;p.x/D
My.x; y/Nx.x; y/
N.x; y/
D
2x
4
y
2
2
x
5
y
2
x
D
2
x
;
R
p.x/ dxD 2lnjxj;
.x/DP.x/D
1
x
2
; therefore

x
2
y
3
C
y
x
2

dxC

x
3
y
2

1
x

dyD0is exact. We must find
Fsuch that (A)Fx.x; y/D

x
2
y
3
C
y
x
2

and (B)Fy.x; y/D

x
3
y
2

1
x

. Integrating (A) with

29. 24 Chapter 2Integrating Factors
respect toxyields (C)F.x; y/D
x
3
y
3
3

y
x
C.y/. Differentiating (C) with respect toyyields (D)
Fy.x; y/Dx
3
y
2

1
x
C
0
.y/. Comparing (D) with (B) shows that
0
.y/D0, so we take.y/D0.
Substituting this into (C) yieldsF.x; y/D
x
3
y
3
3

y
x
, so
x
3
y
3
3

y
x
Dc.
2.6.26.M.x; y/D12xyC6y
3
;N.x; y/D9x
2
C10xy
2
;My.x; y/Nx.x; y/D.12xC18y
2
/
.18xC10y
2
/D 6xC8y
2
; andp.x/N.x; y/q.y/M.x; y/Dp.x/x.9xC10y
2
/q.y/y.12xC6y
2
/,
so exactness requires that (A)6xC8y
2
Dp.x/x.9xC10y
2
/q.y/y.12xC6y
2
/. To obtain similar
terms on the two sides of (A) we letp.x/xDaandq.y/yDbwhereaandbare constants such that
6xC8y
2
Da.9xC10y
2
/b.12xC6y
2
/, which holds if9a12bD 6,10a6bD8; that is,
aDbD2. Thusp.x/D
2
x
;q.y/D
2
y
;
R
p.x/ dxD2lnjxj;
R
q.y/ dyD2lnjyj;P.x/Dx
2
;
Q.y/Dy
2
;.x; y/Dx
2
y
2
. Therefore,.12x
3
y
3
C6x
2
y
5
/ dxC.9x
4
y
2
C10x
3
y
4
/ dyD0is exact.
We must findFsuch that (B)Fx.x; y/D12x
3
y
3
C6x
2
y
5
and (C)Fy.x; y/D9x
4
y
2
C10x
3
y
4
.
Integrating (B) with respect toxyields (D)F.x; y/D3x
4
y
3
C2x
3
y
5
C.y/. Differentiating (D)
with respect toyyields (E)Fy.x; y/D9x
4
y
2
C10x
3
y
4
C
0
.y/. Comparing (E) with (C) shows that

0
.y/D0, so we take.y/D0. Substituting this into (D) yieldsF.x; y/D3x
4
y
3
C2x
3
y
5
, so
x
3
y
3
.3xC2y
2
/Dc.
2.6.28.M.x; y/Dax
m
yCby
nC1
;N.x; y/Dcx
mC1
Cdxy
n
;My.x; y/Nx.x; y/D
fi
ax
mC1
C.nC1/by
n


Œ.mC1/cx
m
Cdy
n
;p.x/N.x; y/q.y/M.x; y/Dxp.x/.cx
m
Cdy
n
/yp.y/.ax
m
Cby
n
/. Let
(A)xp.x/D˛and (B)yp.y/Dˇ, where˛andˇare to be chosen so that
fi
ax
mC1
C.nC1/by
n


Œ.mC1/cx
m
Cdy
n
D˛.cx
m
Cdy
n
/ˇ.ax
m
Cby
n
/, which will hold if
c˛aˇD a.mC1/cD
df
A
d˛bˇD dC.nC1/bD
df
B:
.C/
Sinceadbc¤0it can be verified that˛D
aBbA
adbc
andˇD
cBdA
adbc
satisfy (C). From (A) and
(B),p.x/D
˛
x
andq.y/D
ˇ
y
, so.x; y/Dx
˛
y
ˇ
is an integrating factor for the given equation.
2.6.30.(a)SinceM.x; y/Dp.x/yf .x/andN.x; y/D1,
My.x; y/Nx.x; y/
N.x; y/
Dp.x/and
Theorem 2.6.1 implies that.x/˙e
R
p.x/ dx
is an integrating factor for (C).
(b)Multiplying (A) throughD ˙e
R
p.x/ dx
yields (D).x/y
0
C
0
.x/yD.x/f .x/, which is
equivalent to..x/y/
0
D.x/f .x/. Integrating this yields.x/yDcC
Z
.x/f .x/ dx, soyD
1
.x/

cC
Z
.x/f .x/ dx

, which is equivalent to (B) sincey1D
1

is a nontrivial solution ofy
0
C
p.x/yD0.

30. CHAPTER3
NumericalMethods
3.1EULER’S METHOD
3.1.2.y1D1:200000000; y2D1:440415946; y3D1:729880994
3.1.4.y1D2:962500000; y2D2:922635828; y3D2:880205639
3.1.6.
x hD0:1 hD0:05 hD0:025 Exact
0.02.0000000002.0000000002.0000000002.000000000
0.12.1000000002.1699909652.2021145182.232642918
0.22.5142772882.6493779002.7130117202.774352565
0.33.3178727523.5276725993.6284650253.726686582
0.44.6465927724.9557982265.1063793695.254226636
0.56.7197376387.1714679777.3933229917.612186259
0.69.87615561610.53838452810.86518679911.188475269
0.714.62953239715.60568610716.08863065216.567103199
0.821.75192541823.19732855023.91332853124.623248150
0.932.39911893134.54593262735.61000537736.665439956
1.048.29814736251.49282564353.07667368554.647937102
3.1.8.
x hD0:05 hD0:025hD0:0125 Exact
1.002.0000000002.0000000002.0000000002.000000000
1.052.2500000002.2592801902.2644905702.270158103
1.102.5367346942.5597247462.5727942802.587150838
1.152.8679508542.9109364262.9357233552.963263785
1.203.2536138253.3256277153.3678431173.415384615
1.253.7067506133.8209810643.8892519003.967391304
1.304.2447006414.4207818294.5284719274.654198473
1.354.8910200015.1588835035.3273485585.528980892
1.405.6784672906.0850757906.3497859436.676923077
1.456.6538459887.2755226417.6983162218.243593315
1.507.8861704378.8524637939.54803990710.500000000
25

31. 26 Chapter 3Numerical Methods
3.1.10.
x hD0:1 hD0:05 hD0:025hD0:1hD0:05hD0:025
1.01.0000000001.0000000001.0000000000.0000 0.0000 0.0000
1.10.9200000000.9218982750.922822717-0.0384-0.0189 -0.0094
1.20.8474693260.8510184640.852746371-0.0745-0.0368 -0.0183
1.30.7817794030.7867700870.789197876-0.1092-0.0540 -0.0268
1.40.7224535560.7286822090.731709712-0.1428-0.0707 -0.0351
1.50.6690378670.6762996180.679827306-0.1752-0.0868 -0.0432
1.60.6210541760.6291485850.633080163-0.2062-0.1023 -0.0509
1.70.5780004160.5867403900.590986601-0.2356-0.1170 -0.0583
1.80.5393701870.5485889020.553070392-0.2631-0.1310 -0.0653
1.90.5046742960.5142286030.518877246-0.2889-0.1441 -0.0719
2.00.4734567370.4832274700.487986391-0.3129-0.1563 -0.0781
Approximate Solutions Residuals
3.1.12.
x hD0:1 hD0:05 hD0:025 “Exact"
1.00.0000000000.0000000000.0000000000.000000000
1.1-0.100000000-0.099875000-0.099780455-0.099664000
1.2-0.199000000-0.198243434-0.197800853-0.197315517
1.3-0.294996246-0.293129862-0.292110713-0.291036003
1.4-0.386095345-0.382748403-0.380986158-0.379168221
1.5-0.470695388-0.465664569-0.463078857-0.460450590
1.6-0.547627491-0.540901018-0.537503081-0.534085626
1.7-0.616227665-0.607969574-0.603849795-0.599737720
1.8-0.676329533-0.666833345-0.662136956-0.657473792
1.9-0.728190908-0.717819639-0.712718751-0.707670533
2.0-0.772381768-0.761510960-0.756179726-0.750912371
3.1.14.
Euler’s method
x hD0:1 hD0:05 hD0:025 “Exact"
2.02.0000000002.0000000002.0000000002.000000000
2.12.4200000002.4406107642.4519620062.464119569
2.22.9224842882.9721982242.9997530463.029403212
2.33.5241044343.6140250823.6641840993.718409925
2.44.2448235724.3893801604.4705318224.558673929
2.55.1085811855.3264263965.4495034675.583808754
2.66.1440905266.4592265916.6384094116.834855438
2.77.3857952297.8289842758.0825880768.361928926
2.88.8750170019.4855448889.83713767210.226228709
2.910.66133261811.48921198711.96902090212.502494409
3.012.80422613513.91294466214.55962305515.282004826
Euler semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
2.02.0000000002.0000000002.0000000002.000000000
2.12.4672335712.4656410812.4648714352.464119569
2.23.0360626503.0326573073.0310113163.029403212
2.33.7291697253.7236680263.7210084663.718409925
2.44.5742363564.5662794704.5624326964.558673929
2.55.6050529905.5941916435.5889402765.583808754
2.66.8628741166.8485499216.8416238146.834855438
2.78.3980731018.3795955728.3706606958.361928926
2.810.27216309610.24868142010.23732619910.226228709
2.912.56026511012.53073353112.51645210612.502494409
3.015.35412228715.31725770515.29942942115.282004826

32. Section 3.1Euler’s Method27
3.1.16.
Euler’s method
x hD0:2 hD0:1 hD0:05 “Exact"
1.02.0000000002.0000000002.0000000002.000000000
1.21.7682941971.7865144991.7944123751.801636774
1.41.6030283711.6284274871.6396788221.650102616
1.61.4745804121.5025631111.5151570631.526935885
1.81.3683495491.3968536711.4098392291.422074283
2.01.2764247611.3045048181.3174217941.329664953
2.21.1942471561.2214901111.2341224581.246155344
2.41.1190881751.1453482761.1576074181.169334346
2.61.0492844101.0745536881.0864194531.097812069
2.80.9838217451.0081629931.0196520231.030719114
3.00.9220943790.9456048000.9567528680.967523153
Euler semilinear method
x hD0:2 hD0:1 hD0:05 “Exact"
1.02.0000000002.0000000002.0000000002.000000000
1.21.8069118311.8043049581.8029785261.801636774
1.41.6597386031.6549683811.6525474361.650102616
1.61.5402578611.5336529161.5303084051.526935885
1.81.4385329321.4303618001.4262325841.422074283
2.01.3487822851.3392795771.3344862491.329664953
2.21.2674974151.2568769241.2515287661.246155344
2.41.1924974941.1809587651.1751572641.169334346
2.61.1224163791.1101477771.1039883101.097812069
2.81.0564059061.0435857431.0371582371.030719114
3.00.9939547540.9807513070.9741403200.967523153
3.1.18.
Euler’s method
x hD0:2 hD0:1 hD0:05 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.21.2000000001.1865572901.1792065741.171515153
0.41.3335434091.2984418901.2808652891.263370891
0.61.3713401421.3196983281.2950820881.271251278
0.81.3263673571.2701602371.2439589801.218901287
1.01.2330563061.1818456671.1580649021.135362070
1.21.1223591361.0804774771.0608716081.042062625
1.41.0131002620.9811249890.9659174960.951192532
1.60.9140002110.8907591070.8794604040.868381328
1.80.8278485580.8116736120.8035820000.795518627
2.00.7545725600.7438698780.7383039140.732638628
Euler semilinear method
x hD0:2 hD0:1 hD0:05 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.21.1538461541.1629065991.1672666501.171515153
0.41.2369699531.2506083571.2570979241.263370891
0.61.2441884561.2582418921.2648759871.271251278
0.81.1951554561.2075240761.2133357811.218901287
1.01.1157311891.1259664371.1307686141.135362070
1.21.0259387541.0343369181.0382833921.042062625
1.40.9376457070.9446815970.9480023460.951192532
1.60.8565818230.8626841710.8655831260.868381328
1.80.7848329100.7903311830.7929635320.795518627
2.00.7226104540.7277429660.7302202110.732638628

33. 28 Chapter 3Numerical Methods
3.1.20.
Euler’s method
x hD0:1 hD0:05 hD0:025 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.10.7000000000.7258415630.7366716900.746418339
0.20.4983300000.5329824930.5479888310.561742917
0.30.3562726890.3925925620.4087243030.423724207
0.40.2545554430.2890406390.3047089420.319467408
0.50.1814405410.2123871890.2267585940.240464879
0.60.1289530690.1556872550.1683751300.180626161
0.70.0913935430.1138515160.1247449760.135394692
0.80.0646136120.0830766410.0922309660.101293057
0.90.0455851020.0605059070.0680687760.075650324
1.00.0321051170.0439970450.0501593100.056415515
Euler semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.10.7408182210.7437843200.7451435570.746418339
0.20.5558892750.5589891060.5604107190.561742917
0.30.4189364610.4214820250.4226425410.423724207
0.40.3158904390.3178044000.3186685490.319467408
0.50.2379084210.2392870950.2399020940.240464879
0.60.1788422060.1798128110.1802398880.180626161
0.70.1341655060.1348406680.1351333670.135394692
0.80.1004509390.1009181180.1011175140.101293057
0.90.0750739680.0753969740.0755326430.075650324
1.00.0560201540.0562439800.0563364910.056415515
3.1.22.
Euler’s method
x hD0:1 hD0:05 hD0:025 “Exact"
2.01.0000000001.0000000001.0000000001.000000000
2.11.0000000001.0050625001.0071008151.008899988
2.21.0205000001.0267520911.0293673671.031723469
2.31.0534898401.0590674231.0615101371.063764243
2.41.0935216851.0977805731.0997482251.101614730
2.51.1371375541.1400596541.1414966511.142903776
2.61.1822690051.1840900311.1850562761.186038851
2.71.2277450051.2287558011.2293504411.229985178
2.81.2729403091.2733991871.2737219201.274092525
2.91.3175458331.3176515541.3177865281.317967533
3.01.3614279071.3613208241.3613325891.361383810
Euler semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
2.01.0000000001.0000000001.0000000001.000000000
2.10.9824769040.9961141421.0026084351.008899988
2.20.9881053461.0105776631.0213060441.031723469
2.31.0094958131.0373588141.0507316341.063764243
2.41.0410129551.0719948161.0869644141.101614730
2.51.0786313011.1113463651.1272622851.142903776
2.61.1196325901.1533001331.1697813761.186038851
2.71.1622702871.1964887251.2133256131.229985178
2.81.2054729271.2400604561.2571460911.274092525
2.91.2486135841.2835060011.3007917721.317967533
3.01.2913455181.3265357371.3440041021.361383810

34. Section 3.2The Improved Euler Method and Related Methods29
3.2THE IMPROVED EULER METHOD AND RELATED METHODS
3.2.2.y1D1:220207973; y2D1:489578775 y3D1:819337186
3.2.4.y1D2:961317914;y2D2:920132727;y3D2:876213748.
3.2.6.
x hD0:1 hD0:05 hD0:025 Exact
0.02.0000000002.0000000002.0000000002.000000000
0.12.2571386442.2384553422.2340551682.232642918
0.22.8260046662.7866341102.7773403602.774352565
0.33.8126719263.7471672633.7316740253.726686582
0.45.3874305805.2859968035.2619690435.254226636
0.57.8132983617.6601991977.6238930647.612186259
0.611.48933775611.26034900511.20600586911.188475269
0.717.01586121116.67435291416.59326782016.567103199
0.825.29214063024.78314986224.66226273124.623248150
0.937.66249672336.90382819136.72360892836.665439956
1.056.13448000955.00339044854.73467483654.647937102
3.2.8.
x hD0:05 hD0:025 hD0:0125 Exact
1.002.0000000002.0000000002.0000000002.000000000
1.052.2683673472.2696703362.2700308682.270158103
1.102.5826072992.5859112952.5868273412.587150838
1.152.9545100222.9608707332.9626388222.963263785
1.203.4001617883.4112121503.4142939643.415384615
1.253.9420971423.9604349003.9655707923.967391304
1.304.6128797804.6427848264.6512067694.654198473
1.355.4613486195.5101885755.5240445915.528980892
1.406.5641507536.6453347566.6686008596.676923077
1.458.0485796178.1883359988.2289722158.243593315
1.5010.14196958510.39677040910.47250211110.500000000
3.2.10.
x hD0:1 hD0:05 hD0:025hD0:1hD0:05hD0:025
1.01.0000000001.0000000001.0000000000.000000.0000000.000000
1.10.9237346630.9237307430.9237305910.000040.000001-0.000001
1.20.8544756000.8544496160.8544446970.000350.0000680.000015
1.30.7916503440.7915960160.7915846340.000780.0001670.000039
1.40.7347857790.7347038260.7346860100.001250.0002770.000065
1.50.6834240950.6833186660.6832953080.001710.0003840.000091
1.60.6370970570.6369734230.6369457100.002130.0004830.000115
1.70.5953303590.5951936340.5951627400.002500.0005720.000137
1.80.5576584220.5575130000.5574799470.002830.0006500.000156
1.90.5236389390.5234883430.5234539580.003110.0007180.000173
2.00.4928629990.4927099310.4926748550.003350.0007770.000187
Approximate Solutions Residuals

35. 30 Chapter 3Numerical Methods
3.2.12.
x hD0:1 hD0:05 hD0:025 “Exact"
1.00.0000000000.0000000000.0000000000.000000000
1.1-0.099500000-0.099623114-0.099653809-0.099664000
1.2-0.196990313-0.197235180-0.197295585-0.197315517
1.3-0.290552949-0.290917718-0.291006784-0.291036003
1.4-0.378532718-0.379013852-0.379130237-0.379168221
1.5-0.459672297-0.460262848-0.460404546-0.460450590
1.6-0.533180153-0.533868468-0.534032512-0.534085626
1.7-0.598726853-0.599496413-0.599678824-0.599737720
1.8-0.656384109-0.657214624-0.657410640-0.657473792
1.9-0.706530934-0.707400266-0.707604759-0.707670533
2.0-0.749751364-0.750637632-0.750845571-0.750912371
3.2.14.
Improved Euler method
x hD0:1 hD0:05 hD0:025 “Exact"
2.02.0000000002.0000000002.0000000002.000000000
2.12.4612421442.4633444392.4639183682.464119569
2.23.0223676333.0275072373.0289110263.029403212
2.33.7055116103.7149327093.7175071703.718409925
2.44.5376595654.5530065314.5572024144.558673929
2.55.5517169605.5751504565.5815604375.583808754
2.66.7878138536.8221586656.8315581016.834855438
2.78.2948962228.3438291808.3572279478.361928926
2.810.13266713510.20095559610.21966391710.226228709
2.912.37395473212.46775880712.49347072212.502494409
3.015.10760096815.23485600015.26975507215.282004826
Improved Euler semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
2.02.0000000002.0000000002.0000000002.000000000
2.12.4642616882.4641551392.4641284642.464119569
2.23.0297060473.0294790053.0294221653.029403212
2.33.7188976633.7185319953.7184404513.718409925
2.44.5593773974.5588499904.5587179564.558673929
2.55.5847667245.5840485105.5838687095.583808754
2.66.8361162466.8351709866.8349343476.834855438
2.78.3635524648.3623352538.3620305358.361928926
2.810.22828888010.22674431210.22635764510.226228709
2.912.50508213212.50314204212.50265636112.502494409
3.015.28523172615.28281242415.28220678015.282004826
3.2.16.
Improved Euler method
x hD0:2 hD0:1 hD0:05 “Exact"
1.02.0000000002.0000000002.0000000002.000000000
1.21.8015141851.8016061351.8016291151.801636774
1.41.6499115801.6500548701.6500906801.650102616
1.61.5267117681.5268798701.5269218821.526935885
1.81.4218415701.4220161191.4220597431.422074283
2.01.3294411721.3296090201.3296509711.329664953
2.21.2459532051.2461048191.2461427131.246155344
2.41.1691629941.1692915151.1693236391.169334346
2.61.0976778701.0977785231.0978036831.097812069
2.81.0306261791.0306958801.0307133051.030719114
3.00.9674737210.9675107900.9675200620.967523153

36. Section 3.2The Improved Euler Method and Related Methods31
Improved Euler semilinear method
x hD0:2 hD0:1 hD0:05 “Exact"
1.02.0000000002.0000000002.0000000002.000000000
1.21.8015141851.8016061351.8016291151.801636774
1.41.6499115801.6500548701.6500906801.650102616
1.61.5267117681.5268798701.5269218821.526935885
1.81.4218415701.4220161191.4220597431.422074283
2.01.3294411721.3296090201.3296509711.329664953
2.21.2459532051.2461048191.2461427131.246155344
2.41.1691629941.1692915151.1693236391.169334346
2.61.0976778701.0977785231.0978036831.097812069
2.81.0306261791.0306958801.0307133051.030719114
3.00.9674737210.9675107900.9675200620.967523153
3.2.18.
Improved Euler method
x hD0:2 hD0:1 hD0:05 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.21.1667717051.1703949021.1712440371.171515153
0.41.2558351161.2616423551.2629587881.263370891
0.61.2635171571.2695282141.2708467611.271251278
0.81.2125519971.2175316481.2185854571.218901287
1.01.1308125731.1344205891.1351502841.135362070
1.21.0391043331.0414877271.0419385361.042062625
1.40.9494400520.9508889230.9511325610.951192532
1.60.8674757870.8682639990.8683648490.868381328
1.80.7951839730.7955236960.7955303150.795518627
2.00.7326792230.7327216130.7326679050.732638628
Improved Euler semilinear method
x hD0:2 hD0:1 hD0:05 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.21.1706178591.1712924521.1714595761.171515153
0.41.2616299341.2629383471.2632629191.263370891
0.61.2691732531.2707342901.2711221861.271251278
0.81.2169260141.2184093551.2187784201.218901287
1.01.1336882351.1349449601.1352578761.135362070
1.21.0407216911.0417283861.0419791261.042062625
1.40.9501457060.9509315970.9511273450.951192532
1.60.8675734310.8681799750.8683310280.868381328
1.80.7948990340.7953642450.7954800630.795518627
2.00.7321666780.7325210780.7326092670.732638628
3.2.20.
Improved Euler method
x hD0:1 hD0:05 hD0:025 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.10.7491650000.7470227420.7465611410.746418339
0.20.5659426990.5626678850.5619612420.561742917
0.30.4286183510.4248036570.4239789640.423724207
0.40.3245564260.3205909180.3197325710.319467408
0.50.2454177350.2415586580.2407230190.240464879
0.60.1852356540.1816438130.1808663030.180626161
0.70.1395460940.1363104960.1356107490.135394692
0.80.1049385060.1020963190.1014825030.101293057
0.90.0787877310.0763406450.0758130720.075650324
1.00.0590718940.0569990280.0565530230.056415515

37. 32 Chapter 3Numerical Methods
Improved Euler semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.10.7455951270.7462151640.7463680560.746418339
0.20.5608275680.5615156470.5616864920.561742917
0.30.4229220830.4235245850.4236745860.423724207
0.40.3188203390.3193062590.3194273370.319467408
0.50.2399623170.2403397160.2404337570.240464879
0.60.1802434410.1805308660.1806024700.180626161
0.70.1351064160.1353229340.1353768550.135394692
0.80.1010773120.1012393680.1012797140.101293057
0.90.0754894920.0756103100.0756403810.075650324
1.00.0562959140.0563857650.0564081240.056415515
3.2.22.
Improved Euler method
x hD0:1 hD0:05 hD0:025 “Exact"
2.01.0000000001.0000000001.0000000001.000000000
2.11.0102500001.0091857541.0089657331.008899988
2.21.0335472731.0321053221.0318110021.031723469
2.31.0655621511.0641359191.0638490941.063764243
2.41.1031453471.1019264501.1016855531.101614730
2.51.1440856931.1431401251.1429571581.142903776
2.61.1868787961.1862028541.1860756001.186038851
2.71.2305308041.2300880351.2300079431.229985178
2.81.2744043571.2741476571.2741044301.274092525
2.91.3181041531.3179875511.3179714901.317967533
3.01.3613953091.3613792591.3613822391.361383810
Improved Euler semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
2.01.0000000001.0000000001.0000000001.000000000
2.11.0128026741.0098220811.0091241161.008899988
2.21.0384318701.0333074261.0321083591.031723469
2.31.0724848341.0658214571.0642639501.063764243
2.41.1117943291.1040135341.1021971681.101614730
2.51.1541680411.1455549681.1435471981.142903776
2.61.1981401891.1888833731.1867288491.186038851
2.71.2427624591.2329845591.2307123611.229985178
2.81.2874418451.2772219411.2748508281.274092525
2.91.3318219761.3212109921.3187530471.317967533
3.01.3756999331.3647309371.3621939971.361383810
3.2.24.
x hD0:1 hD0:05 hD0:025 Exact
1.01.0000000001.0000000001.0000000001.000000000
1.11.1510192871.1532706611.1537779571.153937085
1.21.2387986181.2418844211.2425808211.242799540
1.31.2892962581.2925731281.2933133551.293546032
1.41.3176868011.3208665991.3215852421.321811247
1.51.3330738551.3360362481.3367058201.336916440
1.61.3410271701.3437320061.3443432321.344535503
1.71.3450013451.3474463891.3479986521.348172348
1.81.3471553521.3493554731.3498520821.350008229
1.91.3488393251.3508161581.3512619951.351402121
2.01.3508907361.3526675991.3530679511.353193719

38. Section 3.2The Improved Euler Method and Related Methods33
3.2.26.
x hD0:05 hD0:025 hD0:0125 Exact
1.002.0000000002.0000000002.0000000002.000000000
1.052.2684963582.2697039432.2700436282.270158103
1.102.5828973672.5859856952.5868552752.587150838
1.152.9549950342.9609923882.9626837512.963263785
1.203.4008723423.4113842943.4143558623.415384615
1.253.9430479063.9606517943.9656449653.967391304
1.304.6140394364.6430185104.6512774244.654198473
1.355.4625680515.5103573625.5240695475.528980892
1.406.5649855806.6452242366.6684729556.676923077
1.458.0478249478.1873846798.2284130448.243593315
1.5010.13632964210.39341968110.47073141110.500000000
3.2.28.
x hD0:1 hD0:05 hD0:025 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.10.9841428400.9841333020.9841309610.984130189
0.20.9650661240.9650444550.9650391170.965037353
0.30.9426485780.9426114570.9426022790.942599241
0.40.9167055780.9166485690.9166344230.916629732
0.50.8869705250.8868874640.8868667780.886859904
0.60.8530660540.8529480110.8529184970.852908668
0.70.8144582490.8142916790.8142498480.814235883
0.80.7703805710.7701437770.7700839980.770063987
0.90.7196996430.7193553850.7192679050.719238519
1.00.6606584110.6601366300.6600028400.659957689
3.2.30.
x hD0:1 hD0:05 hD0:025 “Exact"
1.00.0000000000.0000000000.0000000000.000000000
1.1-0.099666667-0.099665005-0.099664307-0.099674132
1.2-0.197322275-0.197317894-0.197316222-0.197355914
1.3-0.291033227-0.291036361-0.291036258-0.291123993
1.4-0.379131069-0.379160444-0.379166504-0.379315647
1.5-0.460350276-0.460427667-0.460445166-0.460662347
1.6-0.533897316-0.534041581-0.534075026-0.534359685
1.7-0.599446325-0.599668984-0.599721072-0.600066382
1.8-0.657076288-0.657379719-0.657450947-0.657845646
1.9-0.707175010-0.707553135-0.707641993-0.708072516
2.0-0.750335016-0.750775571-0.750879100-0.751331499
3.2.32.(a)LetxiDaCih,iD0; 1; : : : ; n. Ifyis the solution of the initial value problem
y
0
Df .x/,y.a/D0, theny.b/D
R
b
a
f .x/ dx. The improved Euler method yieldsyiC1DyiC
:5h .f .aCih/Cf .aC.iC1/h//,iD0; 1; : : : ; n1, wherey0Daandynis an approximation to
Z
b
a
f .x/ dx. But
ynD
n1
X
iD0
.yiC1yi/D:5h .f .a/Cf .b//Ch
n1
X
iD1
f .aCih/:
(c)The local truncation error is a multiple ofy
000
.Qxi/Df
00
.Qxi/, wherexi<Qxi< xiC1. Therefore,
the quadrature formula is exact iffis a polynomial of degree< 2.
(d)LetE.f /D
Z
b
a
f .x/ dxyn. Note that E is linear. Iffis a polynomial of degree2, then

39. 34 Chapter 3Numerical Methods
f .x/Df0.x/CK.xa/
2
where deg.f0/1. SinceE.f0/D0from(c)and
E..xa/
2
/D
.ba/
3
3

.ba/
2
h
2
h
3
n1
X
iD1
i
2
Dh
3

n
3
3

n
2
2

n.n1/.2n1/
6

D
nh
3
6
D
.ba/h
2
6
;
E.f /D
K.ba/h
2
6
; therefore the error is proportional toh
2
.
3.3THE RUNGE–KUTTA METHOD
3.3.2.y1D1:221551366; y2D1:492920208
3.3.4.y1D2:961316248;y2D2:920128958.
3.3.6.
x hD0:1 hD0:05 hD0:025 Exact
0.02.0000000002.0000000002.0000000002.000000000
0.12.2327525072.2326495732.2326433272.232642918
0.22.7745827592.7743666252.7743534312.774352565
0.33.7270686863.7267100283.7266880303.726686582
0.45.2548173885.2542630055.2542288865.254226636
0.57.6130770207.6122412227.6121896627.612186259
0.611.18980677811.18855754611.18848036511.188475269
0.716.56908831016.56722597516.56711080816.567103199
0.824.62620625524.62343120124.62325949624.623248150
0.936.66984868736.66571285836.66545687436.665439956
1.054.65450969954.64834401954.64796232854.647937102
3.3.8.
x hD0:05 hD0:025 hD0:0125 Exact
1.002.0000000002.0000000002.0000000002.000000000
1.052.2701537852.2701578062.2701580832.270158103
1.102.5871398462.5871500832.5871507892.587150838
1.152.9632424152.9632623172.9632636892.963263785
1.203.4153468643.4153820203.4153844453.415384615
1.253.9673270773.9673868863.9673910153.967391304
1.304.6540899504.6541910004.6541979834.654198473
1.355.5287946155.5289680455.5289800495.528980892
1.406.6765909296.6769001166.6769215696.676923077
1.458.2429606698.2435494158.2435904288.243593315
1.5010.49865819810.49990626610.49999382010.500000000
3.3.10.
x hD0:1 hD0:05 hD0:025 hD0:1 hD0:05 hD0:025
1.01.0000000001.0000000001.0000000000.0000000000.00000000000.00000000000
1.10.9237306220.9237306770.923730681-0.000000608-0.0000000389-0.00000000245
1.20.8544432530.8544433240.854443328-0.000000819-0.0000000529-0.00000000335
1.30.7915811550.7915812180.791581222-0.000000753-0.0000000495-0.00000000316
1.40.7346804970.7346805380.734680541-0.000000523-0.0000000359-0.00000000233
1.50.6832880340.6832880510.683288052-0.000000224-0.0000000178-0.00000000122
1.60.6369370460.6369370400.6369370400.0000000790.0000000006-0.00000000009
1.70.5951530530.5951530290.5951530280.0000003510.00000001710.00000000093
1.80.5574695580.5574695220.5574695200.0000005780.00000003090.00000000179
1.90.5234431290.5234430840.5234430810.0000007600.00000004210.00000000248
2.00.4926637890.4926637380.4926637360.0000009020.00000005080.00000000302
Approximate Solutions Residuals

40. Section 3.3The Runge–Kutta Method35
3.3.12.
x hD0:1 hD0:05 hD0:025 “Exact"
1.00.0000000000.0000000000.0000000000.000000000
1.1-0.099663901-0.099663994-0.099664000-0.099664000
1.2-0.197315322-0.197315504-0.197315516-0.197315517
1.3-0.291035700-0.291035983-0.291036001-0.291036003
1.4-0.379167790-0.379168194-0.379168220-0.379168221
1.5-0.460450005-0.460450552-0.460450587-0.460450590
1.6-0.534084875-0.534085579-0.534085623-0.534085626
1.7-0.599736802-0.599737663-0.599737717-0.599737720
1.8-0.657472724-0.657473726-0.657473788-0.657473792
1.9-0.707669346-0.707670460-0.707670529-0.707670533
2.0-0.750911103-0.750912294-0.750912367-0.750912371
3.3.14.
Runge–Kutta method
x hD0:1 hD0:05 hD0:025 “Exact"
2.02.0000000002.0000000002.0000000002.000000000
2.12.4641139072.4641191852.4641195442.464119569
2.23.0293893603.0294022713.0294031503.029403212
2.33.7183845193.7184081993.7184098123.718409925
2.44.5586325164.5586711164.5586737464.558673929
2.55.5837454795.5838044565.5838084745.583808754
2.66.8347626396.8348491356.8348550286.834855438
2.78.3617966198.3619199398.3619283408.361928926
2.810.22604394210.22621615910.22622789110.226228709
2.912.50224042912.50247715812.50249328512.502494409
3.015.28166003615.28198140715.28200330015.282004826
Runge–Kutta semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
2.02.0000000002.0000000002.0000000002.000000000
2.12.4641196232.4641195732.4641195702.464119569
2.23.0294033253.0294032193.0294032123.029403212
2.33.7184101053.7184099363.7184099253.718409925
2.44.5586741884.5586739454.5586739304.558673929
2.55.5838091055.5838087765.5838087555.583808754
2.66.8348558996.8348554676.8348554406.834855438
2.78.3619295168.3619289638.3619289288.361928926
2.810.22622945610.22622875610.22622871210.226228709
2.912.50249534512.50249446812.50249441312.502494409
3.015.28200599015.28200489915.28200483115.282004826
3.3.16.
Runge–Kutta method
x hD0:2 hD0:1 hD0:05 “Exact"
1.02.0000000002.0000000002.0000000002.000000000
1.21.8016367851.8016367751.8016367741.801636774
1.41.6501026331.6501026171.6501026161.650102616
1.61.5269359041.5269358861.5269358851.526935885
1.81.4220743021.4220742841.4220742831.422074283
2.01.3296649701.3296649541.3296649531.329664953
2.21.2461553571.2461553451.2461553441.246155344
2.41.1693343551.1693343471.1693343461.169334346
2.61.0978120741.0978120701.0978120691.097812069
2.81.0307191131.0307191141.0307191141.030719114
3.00.9675231470.9675231520.9675231530.967523153

41. 36 Chapter 3Numerical Methods
Runge–Kutta semilinear method
x hD0:2 hD0:1 hD0:05 “Exact"
1.02.0000000002.0000000002.0000000002.000000000
1.21.8016367851.8016367751.8016367741.801636774
1.41.6501026331.6501026171.6501026161.650102616
1.61.5269359041.5269358861.5269358851.526935885
1.81.4220743021.4220742841.4220742831.422074283
2.01.3296649701.3296649541.3296649531.329664953
2.21.2461553571.2461553451.2461553441.246155344
2.41.1693343551.1693343471.1693343461.169334346
2.61.0978120741.0978120701.0978120691.097812069
2.81.0307191131.0307191141.0307191141.030719114
3.00.9675231470.9675231520.9675231530.967523153
3.3.18.
Runge–Kutta method
x hD0:2 hD0:1 hD0:05 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.21.1715156101.1715151561.1715151521.171515153
0.41.2633658451.2633705561.2633708691.263370891
0.61.2712389571.2712505291.2712512321.271251278
0.81.2188855281.2189003531.2189012301.218901287
1.01.1353467721.1353611741.1353620161.135362070
1.21.0420495581.0420618641.0420625791.042062625
1.40.9511819640.9511919200.9511924950.951192532
1.60.8683729230.8683808420.8683812980.868381328
1.80.7955119270.7955182410.7955186030.795518627
2.00.7326332290.7326383180.7326386090.732638628
Runge–Kutta semilinear method
x hD0:2 hD0:1 hD0:05 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.21.1715173161.1715152841.1715151611.171515153
0.41.2633744851.2633711101.2633709041.263370891
0.61.2712546361.2712514851.2712512911.271251278
0.81.2189038021.2189014421.2189012971.218901287
1.01.1353638691.1353621811.1353620771.135362070
1.21.0420639521.0420627061.0420626301.042062625
1.40.9511935600.9511925950.9511925360.951192532
1.60.8683821570.8683813780.8683813310.868381328
1.80.7955193150.7955186690.7955186290.795518627
2.00.7326392120.7326386630.7326386300.732638628
3.3.20.
Runge–Kutta method
x hD0:1 hD0:05 hD0:025 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.10.7464309620.7464189920.7464183760.746418339
0.20.5617619870.5617439210.5617429750.561742917
0.30.4237460570.4237253710.4237242740.423724207
0.40.3194898110.3194686120.3194674780.319467408
0.50.2404864600.2404660460.2404649470.240464879
0.60.1806461050.1806272440.1806262250.180626161
0.70.1354125690.1353956650.1353947490.135394692
0.80.1013087090.1012939110.1012931070.101293057
0.90.0756637690.0756510590.0756503670.075650324
1.00.0564268860.0564161370.0564155520.056415515

42. Section 3.3The Runge–Kutta Method37
Runge–Kutta semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
0.01.0000000001.0000000001.0000000001.000000000
0.10.7464163060.7464182170.7464183320.746418339
0.20.5617406470.5617427800.5617429080.561742917
0.30.4237221930.4237240840.4237241990.423724207
0.40.3194657600.3194673080.3194674020.319467408
0.50.2404635790.2404648000.2404648740.240464879
0.60.1806251560.1806261000.1806261580.180626161
0.70.1353939240.1353946450.1353946890.135394692
0.80.1012924740.1012930210.1012930550.101293057
0.90.0756498840.0756502970.0756503220.075650324
1.00.0564151850.0564154950.0564155140.056415515
3.3.22.
Runge–Kutta method
x hD0:1 hD0:05 hD0:025 “Exact"
2.01.0000000001.0000000001.0000000001.000000000
2.11.0089123981.0089006361.0089000251.008899988
2.21.0317407891.0317243681.0317235201.031723469
2.31.0637818191.0637651501.0637642951.063764243
2.41.1016300851.1016155171.1016147741.101614730
2.51.1429159171.1429043931.1429038111.142903776
2.61.1860476781.1860392951.1860388761.186038851
2.71.2299910541.2299854691.2299851941.229985178
2.81.2740959921.2740926921.2740925351.274092525
2.91.3179691531.3179676051.3179675371.317967533
3.01.3613840821.3613838121.3613838091.361383810
Runge–Kutta semilinear method
x hD0:1 hD0:05 hD0:025 “Exact"
2.01.0000000001.0000000001.0000000001.000000000
2.11.0089139341.0089008431.0089000411.008899988
2.21.0317485261.0317250011.0317235641.031723469
2.31.0637983001.0637663211.0637643711.063764243
2.41.1016562641.1016172591.1016148861.101614730
2.51.1429517211.1429066911.1429039551.142903776
2.61.1860924751.1860421051.1860390511.186038851
2.71.2300439831.2299887421.2299853971.229985178
2.81.2741561721.2740963771.2740927621.274092525
2.91.3180357871.3179716581.3179677871.317967533
3.01.3614565021.3613881961.3613840791.361383810
3.3.24.
x hD:1 hD:05 hD:025 Exact
1.000.1428548410.1428570010.1428571340.142857143
1.100.0533407450.0533419890.0533420660.053342071
1.20-0.046154629-0.046153895-0.046153849-0.046153846
1.30-0.153363206-0.153362764-0.153362736-0.153362734
1.40-0.266397049-0.266396779-0.266396762-0.266396761
1.50-0.383721107-0.383720941-0.383720931-0.383720930
1.60-0.504109696-0.504109596-0.504109589-0.504109589
1.70-0.626598326-0.626598268-0.626598264-0.626598264
1.80-0.750437351-0.750437320-0.750437318-0.750437318
1.90-0.875050587-0.875050574-0.875050573-0.875050573
2.00-1.000000000-1.000000000-1.000000000-1.000000000

43. 38 Chapter 3Numerical Methods
3.3.26.
x hD:1 hD:05 hD:025 Exact
0.50-8.954103230-8.954063245-8.954060698-8.954060528
0.60-5.059648314-5.059633293-5.059632341-5.059632277
0.70-2.516755942-2.516749850-2.516749465-2.516749439
0.80-0.752508672-0.752506238-0.752506084-0.752506074
0.900.5305284820.5305292700.5305293190.530529323
1.001.5000000001.5000000001.5000000001.500000000
1.102.2565197432.2565193522.2565193282.256519326
1.202.8635430392.8635424542.8635424172.863542415
1.303.3627313793.3627307003.3627306583.362730655
1.403.7823619483.7823612313.7823611863.782361183
1.504.1421712794.1421705534.1421705084.142170505
3.3.28.(a)LetxiDaCih,iD0; 1; : : : ; n. Ifyis the solution of the initial value problem
y
0
Df .x/,y.a/D0, theny.b/D
R
b
a
f .x/ dx. The Runge-Kutta method yieldsyiC1DyiC
h
6
.f .aCih/C4f .aC.2iC1/h=2/Cf .aC.iC1/h//,iD0; 1; : : : ; n1, wherey0Daandyn
is an approximation to
Z
b
a
f .x/ dx. But
ynD
n1
X
iD0
.yiC1yi/D
h
6
.f .a/Cf .b//C
h
3
n1
X
iD1
f .aCih/C
2h
3
n
X
iD1
f .aC.2i1/h=2/ :
(c)The local truncation error is a multiple ofy
.5/
.Qxi/Df
.4/
.Qxi/, wherexi<Qxi< xiC1. Therefore,
the quadrature formula is exact iffis a polynomial of degree< 4.
(d)LetE.f /D
Z
b
a
f .x/ dxyn. Note thatEis linear. Iffis a polynomial of degree4, then
f .x/Df0.x/CK.xa/
4
where deg.f0/3andKis constant. SinceE.f0/D0from(c)and
E..xa/
4
/D
.ba/
5
5

.ba/
4
h
6

h
5
3
n1
X
iD1
i
4

2h
5
3
n
X
iD1
.i1=2/
4
Dh
5

n
5
5

n
4
6


n
5
15

n
4
6
C
n
3
9

n
90



2n
5
15

n
3
9
C
7n
360

D
nh
5
120
D
.ba/h
4
120
;
E.f /D
.ba/h
4
120
; thus, the error is proportional toh
4
.

44. CHAPTER4
ApplicationsofFirstOrderEquations
4.1GROWTH AND DECAY
4.1.2.kDln2andD2)kD
ln2
2
;Q.t/DQ0e
tln2=2
; ifQ.T /D
Q0
10
, then
Q0
10
D
Q0e
Tln2=2
; ln10D
Tln2
2
;TD
2ln10
ln2
days.
4.1.4.Lett1be the elapsed time since the tree died. Sincep.t/De
.tln2/
, it follows thatp1D
p0e
.t1ln2/=
, so ln

p1
p0

D
t1

ln2andt1D
ln.p0=p1/
ln2
.
4.1.6.QDQ0e
k t
;Q1DQ0e
k t1;Q2DQ0e
k t2;
Q2
Q1
De
k.t2t1/
; ln

Q1
Q2

Dk.t2t1/;
kD
1
t2t1
ln

Q1
Q2

.
4.1.8.Q
0
D:06Q; Q.0/DQ0;QDQ0e
:06t
. We must findsuch thatQ./D2Q0; that is,
Q0e
:06
D2Q0, so:06Dln2andD
ln2
:06
D
50ln2
3
yr.
4.1.10.(a)IfTis the time to triple the value, thenQ.T /DQ0e
:05T
D3Q0, soe
:05T
D3. Therefore,
:05TDln3andTD20ln3.
(b)IfQ.10/D100000, thenQ0e
:5
D100000, soQ0D100000e
:5
4.1.12.Q
0
D
Q
2
2
; Q.0/D50;
Q
0
Q
2
D
1
2
;
1
Q
D
t
2
Cc;Q.0/D50)cD
1
50
;
1
Q
D
t
2
C
1
50
D
1C25t
50
;QD
50
1C25t
. NowQ.T /D25)1C25TD2)25TD1)TD
1
25
years.
4.1.14.SinceD1500,kD
ln2
1500
; henceQDQ0e
.tln2/=1500
. IfQ.t1/D
3Q0
4
, thene
.t1ln2/=1500
D
3
4
;t1
ln2
1500
Dln

3
4

D ln

4
3

;t1D1500
ln

4
3

ln2
. Finally,Q.2000/DQ0e

4
3
ln2
D2
4=3
Q0.
39

45. 40 Chapter 4Applications of First Order Equations
4.1.16.(A)S
0
D1
S
10
; S.0/D20. Rewrite the differential equation in (A) as (B)S
0
C
S
10
D1. Since
S1De
t =10
is a solution of the complementary equation, the solutions of (B) are given bySDue
t =10
,
whereu
0
e
t =10
D1. Therefore,u
0
De
t =10
;uD10e
t =10
Cc;SD10Cce
t =10
. NowS.0/D20)
cD10, soSD10C10e
t =10
and limt!1S.t/D10g.
4.1.18.(A)V
0
D 750C
V
20
; V .0/D25000. Rewrite the differential equation in (A) as (B)V
0

V
20
D 750. SinceV1De
t =20
is a solution of the complementary equation, the solutions of (B) are
given byVDue
t =20
, whereu
0
e
t =20
D 750. Therefore,u
0
D 750e
t =20
;uD15000e
t =20
Cc;
VD15000Cce
t =20
;V .0/D25000)cD10000. Therefore,VD15000C10000e
t =20
.
4.1.20.p
0
D
p
2

p
2
8
D
1
8
p.p4/;
p
0
p.p4/
D
1
8
;
1
4

1
p4

1
p

p
0
D
1
8
;

1
p4

1
p

p
0
D

1
2
;
ˇ
ˇ
ˇ
ˇ
p4
p

D
t
2
Ck;
p4
p
Dce
t =2
;p.0/D100)cD
24
25
;
p4
p
D
24
25
e
t =2
;p4D
24
25
pe
t =2
;p

1
24
25
e
t =2

D4;pD
4
1
24
25
e
t =2
D
100
2424e
t =2
.
4.1.22.(a)P
0
DrP12M.
(b)PDue
r t
;u
0
e
r t
D 12M;u
0
D 12Me
r t
;uD
12M
r
e
r t
Cc;PD
12M
r
Cce
r t
;P.0/D
P0)cDP0
12M
r
;PD
12M
r
.1e
r t
/CP0e
r t
.
(c)SinceP.N /D0, the answer to(b)implies thatMD
rP0
12.1e
rN
/
4.1.24.The researcher’s salary is the solution of the initial valueproblemS
0
DaS; S.0/DS0.
Therefore,SDS0e
at
. IfPDP.t/is the value of the trust fund, thenP
0
D S0e
at
CrP, or
P
0
rPD S0e
at
. Therefore, (A)PDue
r t
, whereu
0
e
r t
D S0e
at
, so (B)u
0
D S0e
.ar /t
. If
a¤r, then (B) implies thatuD
S0
ra
e
.ar /t
Cc, so (A) implies thatPD
S0
ra
e
at
Cce
r t
. Now
P.0/DP0)cDP0
S0
ra
; thereforePD
S0
ra
e
at
C

P0
S0
ra

e
r t
. We must choose
P0so thatP.T /D0; that is,PD
S0
ra
e
aT
C

P0
S0
ra

e
rT
D0. Solving this forP0yields
P0D
S0.1e
.ar /T
/
ra
. IfaDr, then (B) becomesu
0
D S0, souD S0tCcand (A) implies that
PD.S0tCc/e
r t
. NowP.0/DP0)cDP0; thereforePD.S0tCP0/e
r t
. To makeP.T /D0
we must takeP0DS0T.
4.1.26.Q
0
D
at
1CbtQ
2
kQ; limt!1Q.t/D.a=bk/
1=3
.
4.2COOLING AND MIXING
4.2.2.SinceT0D100andTMD 10,TD 10C110e
k t
. NowT .1/D80)80D 10C110e
k
,
soe
k
D
9
11
andkDln
11
9
. Therefore,TD 10C110e
tln
11
9.
4.2.4.LetTbe the thermometer reading. SinceT0D212andTMD70,TD70C142e
k t
. Now
T .2/D125)125D70C142e
2k
, soe
2k
D
55
142
andkD
1
2
ln
142
55
. Therefore, (A)TD

46. Section 4.2Cooling and Mixing41
70C142e

t
2
ln
142
55.
(a)T .2/D70C142e
2ln
142
55D70C142

55
142

2
91:30
ı
F.
(b)Letbe the time whenT ./D72, so72D70C142e


2
ln
142
55, ore


2
ln
142
55D
1
71
. Therefore,
D2
ln71
ln
142
55
8:99min.
(c)Since (A) implies thatT > 70for allt > 0, the thermometer will never read69
ı
F.
4.2.6.SinceTMD20,TD20C.T020/e
k t
. NowT05D20C.T020/e
4k
andT07D
20C.T020/e
8k
. Therefore,
T025
T020
De
4k
and

T027
T020

De
8k
, so
T027
T020
D

T025
T020

2
,
which implies that.T020/.T027/D.T025/
2
, orT
2
0
47T0C540DT
2
0
50T0C625; hence
3T0D85andT0D.85=3/
ı
C.
4.2.8.Q
0
D3
3
40
Q; Q.0/D0. Rewrite the differential equation as (A)Q
0
C
3
40
QD3. Since
Q1De
3t =40
is a solution of the complementary equation, the solutions of (A) are given byQD
ue
3t =40
whereu
0
e
3t =40
D3. Therefore,u
0
D3e
3t =40
,uD40e
3t =40
Cc, andQD40Cce
3t =40
.
NowQ.0/D0)cD 40, soQD40.1e
3t =40
/.
4.2.10.Q
0
D
3
2

Q
20
; Q.0/D10. Rewrite the differential equation as (A)Q
0
C
Q
20
D
3
2
. Since
Q1De
t =20
is a solution of the complementary equation, the solutions of (A) are given byQDue
t =20
whereu
0
e
t =20
D
3
2
. Therefore,u
0
D
3
2
e
t =20
,uD30e
t =20
Cc, andQD30Cce
t =20
. Now
Q.0/D10)cD 20, soQD3020e
t =20
andKD
Q
100
D:3:2e
t =20
.
4.2.12.Q
0
D10
Q
5
, or (A)Q
0
C
Q
5
D10. SinceQ1De
t =5
is a solution of the complementary
equation, the solutions of (A) are given byQDue
t =5
whereu
0
e
t =5
D10. Therefore,u
0
D10e
t =5
,
uD50e
t =10
Cc, andQD50Cce
t =5
. Since limt!1Q.t/D50, the mininum capacity is 50 gallons.
4.2.14.Since there are2tC600gallons of mixture in the tank at timetand mixture is being drained
at 4 gallons/min,Q
0
D3
2
tC300
Q; Q.0/D40. Rewrite the differential equation as (A)Q
0
C
2
tC300
QD3. SinceQ1D
1
.tC300/
2
is a solution of the complementary equation, the solutions
of (A) are given byQD
u
.tC300/
2
where
u
0
.tC300/
2
D3. Therefore,u
0
D3.tC300/
2
,uD
.tC300/
3
Cc, andQDtC300C
c
.tC300/
2
. NowQ.0/D40)cD 234fi10
5
, soQD
tC300
234fi10
5
.tC300/
2
; 0t300.
4.2.16.(a)S
0
D km.STm/; S.0/D0, so (A)SDTmC.S0Tm/e
kmt
.T
0
D k.TS/D
k

TTm.S0Tm/e
kmt

, from (A). Therefore,T
0
CkTDkTmCk.S0Tm/e
kmt
;TDue
k t
;
(B)u
0
DkTme
k t
Ck.S0Tm/e
.kkm/t
;uDTme
k t
C
k
kkm
.S0Tm/e
.kkm/t
Cc;T .0/DT0)
cDT0Tm
k
kkm
.S0Tm/;uDTme
k t
C
k
kkm
.S0Tm/e
.kkm/t
CT0Tm
k
kkm
.S0Tm/;

47. 42 Chapter 4Applications of First Order Equations
TDTmC.T0Tm/e
k t
C
k.S0Tm/
.kkm/

e
kmt
e
k t

.
(b)IfkDkm(B) becomes (B)u
0
DkTme
k t
Ck.S0Tm/;uDTme
k t
Ck.S0Tm/tCc;T .0/D
T0)cDT0Tm;uDTme
k t
Ck.S0Tm/tC.T0Tm/;TDTmCk.S0Tm/te
k t
C.T0Tm/e
k t
.
(c)limt!1T .t/Dlimt!1S.t/DTmin either case.
4.2.18.V
0
DaVbV
2
D bV .Vb=a/;
V
0
V .Va=b/
D b;

1
Va=b

1
V

V
0
D a;
ln
ˇ
ˇ
ˇ
ˇ
Va=b
V
ˇ
ˇ
ˇ
ˇ
D atCk; (A)
Va=b
V
Dce
at
; (B)VD
a
b
1
1ce
at
. SinceV .0/DV0, (A)
)cD
V0a=b
V0
. Substituting this into (B) yieldsVD
a
b
V0
V0.V0a=b/ e
at
so limt!1V .t/D
a=b
4.2.20.IfQn.t/is the number of pounds of salt inTnat timet, thenQ
0
nC1
C
r
W
QnC1Drcn.t/; nD
0; 1; : : :, wherec0.t/c. Therefore,QnC1DunC1e
r t =w
; (A)u
0
nC1
Dre
r t =W
cn.t/. In particular,
withnD0,u1DcW.e
r t =W
1/, soQ1DcW.1e
r t =W
/andc1Dc.1e
r t =W
/. We will shown
by induction thatcnDc
0
@1e
r t =W
n1
X
jD0
1
j Š

rt
W

j
1
A. This is true fornD1; if it is true for a givenn,
then, from (A),
u
0
nC1
Dcre
r t =W
0
@1e
r t =W
n1
X
jD0
1
j Š

rt
W

j
1
ADcre
r t =W
cr
n1
X
jD0
1
j Š

rt
W

j
;
so (sinceQnC1.0/D0),
unC1DcW.e
r t =W
1/c
n1
X
jD0
1
.jC1/Š
r
jC1
W
j
t
jC1
:
Therefore,
cnC1D
1
W
unC1e
r t =W
Dc
0
@1e
r t =W
n
X
jD0
1
j Š

rt
W

j
1
A;
which completes the induction. From this, limt!1cn.t/Dc.
4.2.22.Since the incoming solution contains 1/2 lb of salt per gallon and there are always 600 gal-
lons in the tank, we conclude intuitively that limt!1Q.t/D300. To verify this rigorously, note
thatQ1.t/Dexp


1
150
Z
t
0
a./ d

is a solution of the complementary equation, (A)Q1.0/D1,
and (B) limt!1Q1.t/D0(since limt!1a.t/D1). Therefore,QDQ1u;Q1u
0
D2;u
0
D
2
Q1
;uDQ0C2
Z
t
0
d
Q1./
(see (A)), andQ.t/DQ0Q1.t/C2Q1.t/
Z
t
0
d
Q1./
. From (B),
lim
t!1
Q.t/D2lim
t!1
Q1.t/
Z
t
0
d
Q1./
, a0 1indeterminate form. By L’Hospital’s rule, lim
t!1
Q.t/D
2lim
t!1
1
Q1.t/

Q
0
1
.t/
Q
2
1
.t/

D 2lim
t!1
Q1.t/
Q
0
1
.t/
D300.
vspace*10pt

48. Section 4.3Elementary Mechanics43
4.3ELEMENTARY MECHANICS
4.3.2.The firefighter’s mass ismD
192
32
D6sl, so6v
0
D 192kv, or (A)v
0
C
k
6
vD 32.
Sincev1De
k t =6
is a solution of the complementary equation, the solutions of (A) arevDue
k t =6
whereu
0
e
k t =6
D 32. Therefore,u
0
D 32e
k t =6
;uD
192
k
e
k t =6
Cc;vD
192
k
Cce
k t =6
. Now
v.0/D0)cD
192
k
. Therefore,vD
192
k
.1e
k t =6
/and limt!1v.t/D
192
k
D 16ft/s, so
kD12lb-s/ft andvD 16.1e
2t
/.
4.3.3.mD
64000
32
D2000, so2000v
0
D500002000v, or (A)v
0
CvD25. Sincev1De
t
is a solution
of the complementary equation, the solutions of (A) arevDue
t
whereu
0
e
t
D25. Therefore,u
0
D
25e
t
;uD25e
t
Cc;vD25Cce
t
. Nowv.0/D0)cD 25. Therefore,vD25.1e
t
/and
limt!1v.t/D25ft/s.
4.3.4.20v
0
D10
1
2
v, or (A)v
0
C
1
20
vD
1
2
. Sincev1De
t =40
is a solution of the complementary
equation, the solutions of (A) arevDue
t =40
whereu
0
e
t =40
D
1
2
. Therefore,u
0
D
e
t =40
2
;uD
20e
t =40
Cc;vD20Cce
t =40
. Nowv.0/D 7)cD 27. Therefore,vD2027e
t =40
.
4.3.6.mD
3200
32
D100sl. The component of the gravitational force in the direction of motion is
3200cos.=3/D 1600lb. Therefore,100v
0
D 1600Cv
2
. Separating variables yields
v
0
.v40/.vC40/
D
1
100
, or

1
v40

1
vC40

D
4
5
. Therefore, ln
ˇ
ˇ
ˇ
ˇ
v40
vC40
ˇ
ˇ
ˇ
ˇ
D
4t
5
Ckand
v40
vC40
Dce
4t =5
. Nowv.0/D
64)cD
13
3
; therefore
v40
vC40
D
13e
4t =5
3
, sovD
40.3C13e
4t =5
/
313e
4t =5
, orvD
40.13C3e
4t =5
/
133e
4t =5
.
4.3.8.From Example 4.3.1, (A)vD
mg
k
C

v0C
mg
k

e
k t =m
. Integrating this yields (B)yD

mgt
k

m
k

v0C
mg
k

e
k t =m
Cc. Nowy.0/Dy0)cDy0C
m
k

v0C
mg
k

. Substituting this
into (B) yields
yD
mgt
k

m
k

v0C
mg
k

e
k t =m
Cy0C
m
k

v0C
mg
k

Dy0C
m
k

v0gtC
mg
k


v0C
mg
k

e
k t =m

Dy0C
m
k
.v0vgt/
where the last equality follows from (A).
4.3.10.mD
256
32
D8sl. Since the resisting force is 1 lb whenjvj D4ft/s,kD
1
16
. Therefore,
8v
0
D 256C
1
16
v
2
D
1
16

v
2
.64/
2

. Separating variables yields
v
0
.v64/.vC64/
D
1
128
, or

1
v64

1
vC64

v
0
D1. Therefore,ln
ˇ
ˇ
ˇ
ˇ
v64
vC64
ˇ
ˇ
ˇ
ˇ
DtCkand
v64
vC64
Dce
t
. Nowv.0/D0)cD
1; therefore
v64
vC64
D e
t
, sovD
64.1e
t
/
1Ce
t
, orvD
64.1e
t
/
1Ce
t
. Therefore, limt!1v.t/D
64.

49. 44 Chapter 4Applications of First Order Equations
4.3.12.(a)mv
0
D mgkv
2
D mg.1C
2
v
2
/, whereD
s
k
mg
. Therefore,(A)
v
0
1C
2
v
2
D g.
With the substitutionuDv,
Z
dv
1C
2
v
2
D
1

Z
du
1Cu
2
D
1

tan
1
uD
1

tan
1
.v/. There-
fore,
1

tan
1
.v/D gtCc. Nowv.0/Dv0)cD
1

tan
1
.v0/, so
1

tan
1
.v/D gtC
1

tan
1
.v0/. Sincev.T /D0, it follows thatTD
1
g
tan
1
v0D
r
m
kg
tan
1
 
v0
s
k
mg
!
.
(b)ReplacingtbytTand settingv0D0in the answer to the previous exercise yieldsvD

r
mg
k
1e
2
q
gk
m
.tT /
1Ce
2
q
gk
m
.tT /
.
4.3.14.(a)mv
0
D mgCf .jvj/; sincesD jvj D v, (A)ms
0
Dmgf .s/.
(b)Sincefis increasing and limt!1f .s/mg,mgf .s/ > 0for alls. This and (A) imply thats
is an increasing function oft, so either (B) limt!1s.t/D 1or (C ) limt!1s.t/Ds <1. However,
(A) and (C) imply thats
0
.t/ > KDgf .s/=mfor allt > 0. Consequently,s.t/ > s0CKtfor all
t > 0, which contradicts (C) becauseK > 0.
(c)There is a unique positive numbersTsuch thatf .sT/Dmg, andssTis a constant solution of
(A). Now suppose thats.0/ < sT. Then Theorem 2.3.1 implies that (D)s.t/ < sTfor allt > 0, so (A)
implies thatsis strictly increasing. This and (D) imply that limt!1s.t/DssT. Ifs < sTthen (A)
implies thats
0
.t/ > KDgf .s/=m. Consequently,s.t/ > s.0/CKt, which contradicts (D) because
K > 0. Therefore,s.0/ < sT)limt!1s.t/DsT. A similar proof with inequlities reversed shows
thats.0/ > sT)limt!1s.t/DsT.
4.3.16.(a)(A)mv
0
D mgCk
p
jvj; since the magnitude of the resistance is 64 lb whenvD16ft/s,
4kD64, sokD16lbs
1=2
=ft
1=2
. SincemD2andgD32, (A) becomes2v
0
D 64C16
p
jvj, or
v
0
D 32C8
p
jvj.
(b)From Exercise 4.3.14(c),vTis the negative number such that32C8
p
jvTj D0; thus,vTD 16
ft/s.
4.3.18.WithhD0,veD
p
2gR, whereRis the radius of the moon andgis the acceleration due to
gravity at the moon’s surface. With length in miles,gD
5:31
5280
mi/s
2
, soveD
r
25:311080
5280
1:47
miles/s.
4.3.20.Suppose that there is a numberymsuch thaty.t/ymfor allt0and let˛D
gR
2
.ymCR/
2
.
Then
d
2
y
dt
2
˛for allt0. Integrating this inequality fromtD0totDT > 0yieldsv.T /v0
˛T, orv.T /v0˛T, sov.T / < 0forT >
v0
˛
. This implies that the vehicle must eventually fall
back to Earth, which contradicts the assumption that it continues to climb forever.
4.4AUTONOMOUS SECOND ORDER EQUATIONS
4.4.1.yD0is a stable equilibrium. The phase plane equivalent isv
dv
dy
Cy
3
D0, so the trajectories are
v
2
C
y
4
4
Dc.

50. Section 4.4Autonomous Second Order Equations45
4.4.2.yD0is an unstable equilibrium. The phase plane equivalent isv
dv
dy
Cy
2
D0, so the trajectories
arev
2
C
2y
3
3
Dc.
4.4.4.yD0is a stable equilibrium. The phase plane equivalent isv
dv
dy
Cye
y
D0, so the trajectories
arev
2
e
y
.yC1/Dc.
4.4.6.p.y/Dy
3
4yD.yC2/y.y2/, so the equilibria are2; 0; 2. Since
y.y2/.yC2/ < 0ify <2or0 < y < 2;
> 0if2 < y < 0ory > 2;
0is unstable and2; 2are stable. The phase plane equivalent isv
dv
dy
Cy
3
4yD0, so the trajectories
are2v
2
Cy
4
8y
2
Dc. Setting.y; v/D.0; 0/yieldscD0, so the equation of the separatrix is
2v
2
y
4
C8y
2
D0.
4.4.8.p.y/Dy.y2/.y1/.yC2/, so the equilibria are2; 0; 1; 2. Since
y.y2/.y1/.yC2/ > 0ify <2or0 < y < 1ory > 2;
< 0if2 < y < 0or1 < y < 2;
0; 2are stable and2; 1are unstable. The phase plane equivalent isv
dv
dy
Cy.y2/.y1/.yC2/D0, so
the trajectories are30v
2
Cy
2
.12y
3
15y
2
80yC120/Dc. Setting.y; v/D.2; 0/and.y; v/D.1; 0/
yieldscD496andcD37respectively, so the equations of the separatrices are30v
2
Cy
2
.12y
3
15y
2

80yC120/D496and30v
2
Cy
2
.12y
3
15y
2
80yC120/D37.
4.4.10.p.y/Dy
3
ay. Ifa0, thenp.0/D0,p.y/ > 0ify > 0, andp.y/ < 0ify < 0, so0is
stable. Ifa > 0, then
y
3
ayDy.y
p
a/.yC
p
a/ > 0if
p
a < y < 0ory >
p
a;
< 0ify <
p
aor0 < y <
p
a;
so
p
aand
p
aare stable and0is unstable. We say thataD0is a critical value because it separates
the two cases.
4.4.12.p.y/Dyay
3
. Ifa0, thenp.0/D0,p.y/ > 0ify > 0, andp.y/ < 0ify < 0, so0is
stable. Ifa > 0, then
yay
3
D ay.y1=
p
a/.yC1=
p
a/ > 0ify <1=
p
a < y < 0or0 < y < 1=
p
a
< 0if1=
p
a < y < 0ory > 1=
p
a;
so
p
aand
p
aare unstable and0is stable. We say thataD0is a critical value because it separates
the two cases.
4.4.24.(a)Sincev
0
D p.y/kandv.0/D0,vktand thereforeyy0Ckt
2
=2for0t < T.
(b)Let0 < < . Suppose thatyis the solution of the initial value problem (A)y
00
Cp.y/D
0; y.0/Dy0; y
0
.0/D0, wherey < y0<yC. Now letYDyyandP.Y /Dp.YC
y/. ThenP.0/D0andP.Y / < 0if0 < Y. Morover,Yis the solution ofY
00
Cp.Y /D

51. 46 Chapter 4Applications of First Order Equations
0; Y.0/DY0; Y
00
.0/D0, whereY0Dy0y, so0 < Y0< . From(a),Y.t/for somet > 0.
Therefore,y.t/ >yCfor somet > 0, soyis an unstable equilibrium ofy
00
Cp.y/D0.
4.5APPLICATIONS TO CURVES
4.5.2.Differentiating (A)e
xy
Dcyyields (B).xy
0
Cy/e
xy
Dcy
0
. From (A),cD
e
xy
y
. Substituting
this into (B) and cancellinge
xy
yieldsxy
0
CyD
y
0
y
, soy
0
D
y
2
.xy1/
.
4.5.4.SolvingyDx
1=2
CcxforcyieldscD
y
x
x
1=2
, and differentiating yields0D
y
0
x

y
x
2
C
x
3=2
2
,
orxy
0
yD
x
1=2
2
.
4.5.6.RewritingyDx
3
C
c
x
asxyDx
4
Ccand differentiating yieldsxy
0
CyD4x
3
.
4.5.8.RewritingyDe
x
Cc.1Cx
2
/as
y
1Cx
2
D
e
x
1Cx
2
Ccand differentiating yields
y
0
1Cx
2

2xy
.1Cx
2
/
2
D
e
x
1Cx
2

2xe
x
.1Cx
2
/
2
, so.1Cx
2
/y
0
2xyD.1x/
2
e
x
.
4.5.10.If (A)yDfCcg, then (B)y
0
Df
0
Ccg
0
. Mutiplying (A) byg
0
and (B) bygyields (C)
yg
0
Dfg
0
Ccgg
0
and (D)y
0
gDf
0
gCcg
0
g, and subtracting (C) from (D) yieldsy
0
gyg
0
Df
0
gfg
0
.
4.5.12.Let.x0; y0/be the center andrbe the radius of a circle in the family. Since.1; 0/and.1; 0/are
on the circle,.x0C1/
2
Cy
2
0
D.x01/
2
Cy
2
0
, which implies thatx0D0. Therefore,the equation of
the circle is (A)x
2
C.yy0/
2
Dr
2
. Since.1; 0/is on the circle,r
2
D1Cy
2
0
. Substituting this into
(A) shows that the equation of the circle isx
2
Cy
2
2yy0D1, so2y0D
x
2
Cy
2
1
y
. Differentiating
y.2xC2yy
0
/y
0
.x
2
Cy
2
1/D0, soy
0
.y
2
x
2
C1/C2xyD0.
4.5.14.From Example 4.5.6 the equation of the line tangent to the parabola at.x0; x
2
0
/is (A)yD
x
2
0
C2x0x.
(a)From (A),.x; y/D.5; 9/is on the tangent line through.x0; x
2
0
/if and only if9D x
2
0
C10x0,
orx
2
0
10x0C9D.x01/.x09/D0. Lettingx0D1in (A) yields the lineyD 1C2x, tangent
to the parabola at.x0; x
2
0
/D.1; 1/. Lettingx0D9in (A) yields the lineyD 81C18x, tangent to the
parabola at.x0; x
2
0
/D.9; 81/.
(b)From (A),.x; y/D.6; 11/is on the tangent line through.x0; x
2
0
/if and only if11D x
2
0
C12x0,
orx
2
0
12x0C11D.x01/.x011/D0. Lettingx0D1in (A) yields the lineyD 1C2x, tangent
to the parabola at.x0; x
2
0
/D.1; 1/. Lettingx0D11in (A) yields the lineyD 121C22x, tangent to
the parabola at.x0; x
2
0
/D.11; 21/.
(c)From (A),.x; y/D.6; 20/is on the tangent line through.x0; x
2
0
/if and only if20D x
2
0
12x0,
orx
2
0
C12x0C20D.x0C2/.x0C10/D0. Lettingx0D 2in (A) yields the lineyD 44x,
tangent to the parabola at.x0; x
2
0
/D.2; 4/. Lettingx0D 10in (A) yields the lineyD 10020x,
tangent to the parabola at.x0; x
2
0
/D.10; 100/.
(d)From (A),.x; y/D.3; 5/is on the tangent line through.x0; x
2
0
/if and only if5D x
2
0
6x0,
orx
2
0
C6x0C5D.x0C1/.x0C5/D0. Lettingx0D 1in (A) yields the lineyD 12x, tangent
to the parabola at.x0; x
2
0
/D.1; 1/. Lettingx0D 5in (A) yields the lineyD 2510x, tangent to
the parabola at.x0; x
2
0
/D.5; 25/.

52. Section 4.5Applications to Curves47
4.5.15.(a)If.x0; y0/is any point on the circle such thatx0¤ ˙1(and thereforey0¤0), then
differentiating (A) yields2x0C2y0y
0
0
D0, soy
0
0
D
x0
y0
. Therefore,the equation of the tangent line is
yDy0
x0
y0
.xx0/. Sincex
2
0
Cy
2
0
D1, this is equivalent to (B).
(b)Sincey
0
D
x0
y0
on the tangent line, we can rewrite (B) asyxy
0
D
1
y0
. Hence (F)
1
.yxy
0
/
2
D
y
2
0
and (G)x
2
0
D1y
2
0
D
.yxy
0
/
2
1
.yxy
0
/
2
. Since.y
0
/
2
D
x
2
0
y
2
0
, (F) and (G) imply that.y
0
/
2
D
.yxy
0
/
2
1, which implies (C).
(c)Using the quadratic formula to solve (C) fory
0
yields
y
0
D
xy˙
p
x
2
Cy
2
1
x
2
1
.H/
if.x; y/is on a tangent line with slopey
0
. IfyD
1x0x
y0
, thenx
2
Cy
2
1Dx
2
C

1x0x
y0

2
1D

xx0
y0

2
(sincex
2
0
Cy
2
0
D1). Sincey
0
D
x0
y0
, this implies that (H) is equivalent to
x0
y0
D
1
x
2
1

x.1x0x/
y0
˙
ˇ
ˇ
ˇ
ˇ
xx0
y0
ˇ
ˇ
ˇ
ˇ

, which holds if and only if we choose the “˙" so that˙
ˇ
ˇ
ˇ
ˇ
xx0
y0
ˇ
ˇ
ˇ
ˇ
D


xx0
y0

. Therefore,we must choose˙ D if
xx0
y0
> 0, so (H) reduces to (D), or˙ D Cif
xx0
y0
< 0, so (H) reduces to (E).
(d)Differentiating (A) yields2xC2yy
0
D0, soy
0
D
x
y
on either semicircle. Since (D) and (E)
both reduce toy
0
D
xy
1x
2
D
x
y
(sincex
2
Cy
2
D1) on both semicircles, the conclusion follows.
(e)From (D) and (E) the slopes of tangent lines from (5,5) tangent to the circle arey
0
D
25˙
p
49
24
D
3
4
;
4
3
. Therefore, tangent lines areyD5C
3
4
.x5/D
1C3x=5
4=5
andyD5C
4
3
.x5/D
14x=5
3=5
,
which intersect the circle at.3=5; 4=5/ .4=5;3=5/, respectively. (See (B)).
4.5.16.(a)If.x0; y0/is any point on the parabola such thatx0> 0(and thereforey0¤0), then
differentiating (A) yields1D2y0y
0
0
, soy
0
0
D
1
2y0
. Therefore,the equation of the tangent line isyD
y0C
1
2y0
.xx0/. Sincex0Dy
2
0
, this is equivalent to (B).
(b)Sincey
0
D
1
2y0
on the tangent line, we can rewrite (B) as
y0
2
Dyxy
0
. Substituting this into (B)
yieldsyD.yxy
0
/C
x
4.yxy
0
/
, which implies (C).
(c)Using the quadratic formula to solve (C) fory
0
yields
y
0
D
y˙
p
y
2
x
2x
.F/
if.x; y/is on a tangent line with slopey
0
. IfyD
y0
2
C
x
2y0
, theny
2
xD
1
4

y0
x
y0

2
so (F)

53. 48 Chapter 4Applications of First Order Equations
is equivalent to
1
2y0
D
y0C
x
y0
˙
ˇ
ˇ
ˇ
ˇ
y0
x
y0
ˇ
ˇ
ˇ
ˇ
4x
which holds if and only if we choose the “˙" so that
˙
ˇ
ˇ
ˇ
ˇ
y0
x
y0
ˇ
ˇ
ˇ
ˇ
D

y0
x
y0

. Therefore,we must choose˙ D Cifx > y
2
0
Dx0, so (F) reduces to (D),
or˙ D ifx < y
2
0
Dx0, so (F) reduces to (E).
(d)Differentiating (A) yields1D2yy
0
, soy
0
D
1
2y
on either half of the parabola. Since (D) and (E)
both reduce to this ifxDy
2
, the conclusion follows.
4.5.18.The equation of the line tangent to the curve at.x0; y.x0//isyDy.x0/Cy
0
.x0/.xx0/.
Sincey.x0=2/D0,y.x0/
y
0
.x0/x0
2
D0. Sincex0is arbitrary, it follows thaty
0
D
2y
x
, so
y
0
y
D
2
x
,
lnjyj D2lnjxj Ck, andyDcx
2
. Since.1; 2/is on the curve,cD2. Therefore,yD2x
2
.
4.5.20.The equation of the line tangent to the curve at.x0; y.x0//isyDy.x0/Cy
0
.x0/.xx0/.
Since.x1; y1/is on the line,y.x0/Cy
0
.x0/.x1x0/Dy1. Sincex0is arbitrary, it follows that
yCy
0
.x1x/Dy1, so
y
0
yy1
D
1
xx1
, lnjyy1j Dlnjxx1j Ck, andyy1Dc.xx1/.
4.5.22.The equation of the line tangent to the curve at.x0; y.x0//isyDy.x0/Cy
0
.x0/.xx0/. Since
y.0/Dx0,x0Dy.x0/y
0
.x0/x0. Sincex0is arbitrary, it follows thatxDyxy
0
, so (A)y
0

y
x
D 1.
The solutions of (A) are of the formyDux, whereu
0
xD 1, sou
0
D
1
x
. Therefore,uD lnjxj Cc
andyD xlnjxj Ccx.
4.5.24.The equation of the line normal to the curve at.x0; y0/isyDy.x0/
xx0
y
0
.x0/
. Sincey.0/D
2y.x0/,y.x0/C
x0
y
0
.x0/
D2y.x0/. Sincex0is arbitrary, it follows thaty
0
yDx, so (A)
y
2
2
D
x
2
2
C
c
2
andy
2
Dx
2
Cc. Nowy.2/D1$cD 3. Therefore,yD
p
x
2
3.
4.5.26.Differentiating the given equation yields2xC4yC4xy
0
C2yy
0
D0, soy
0
D
xC2y
2xCy
is a differential equation for the given family, and (A)y
0
D
2xCy
xC2y
is a differential equation for the
orthogonal trajectories. SubstitutingyDuxin (A) yieldsu
0
xCuD
2Cu
1C2u
, sou
0
xD
2.u
2
1/
1C2u
and
1C2u
.u1/.uC1/
u
0
D
2
x
, or

3
u1
C
1
uC1

u
0
D
4
x
. Therefore,3lnju1jClnjuC1j D 4lnjxjC
K, so.u1/
3
.uC1/D
k
x
4
. SubstitutinguD
y
x
yields the orthogonal trajectories.yx/
3
.yCx/Dk.
4.5.28.Differentiating yieldsye
x
2
.1C2x
2
/Cxe
x
2
y
0
D0, soy
0
D
y.1C2x
2
/
x
is a differential equation
for the given family. Therefore,(A)y
0
D
x
y.1C2x
2
/
is a differential equation for the orthogonal
trajectories. From (A),yy
0
D
x
1C2x
2
, so
y
2
2
D
1
4
ln.1C2x
2
/C
k
2
, and the orthogonal trajectories
are given byy
2
D
1
2
ln.1C2x
2
/Ck.

54. Section 4.5Applications to Curves49
4.5.30.Differentiating (A)yD1Ccx
2
yields (B)y
0
D2cx. From (C),cD
y1
x
2
. Substituting this
into (B) yields the differential equationy
0
D
2.y1/
x
for the given family of parabolas. Therefore,y
0
D

x
2.y1/
is a differential equation for the orthogonal trajectories. Separating variables yields2.y
1/y
0
D x, so.y1/
2
D
x
2
2
Ck. Nowy.1/D3$kD
9
2
, so.y1/
2
D
x
2
2
C
9
2
. Therefore,(D)
yD1C
r
9x
2
2
. This curve interesects the parabola (A) if and only if the equation (C)cx
2
D
r
9x
2
2
has a solutionx
2
in.0; 9/. Therefore,c > 0is a necessaary condition for intersection. We will
show that it is also sufficient. Squaring both sides of (C) andsimplifying yields2c
2
x
4
Cx
2
9D0. Using
the quadratic formula to solve this forx
2
yieldsx
2
D
1C
p
1C72c
2
4c
2
. The conditionx
2
< 9holds if
and only if1C
p
1C72c
2
< 36c
4
, which is equivalent to1C72c
2
< .1C36c
2
/
2
D1C72c
2
C1296c
4
,
which holds for allc > 0.
4.5.32.The anglesand1from thex-axis to the tangents toCandC1satisfy tanDf .x0; y0/and
tan1D
f .x0; y0/Ctan˛
1f .x0; y0/tan˛
D
tanCtan˛
1tantan˛
Dtan.C˛/. Therefore, assumingand1are both in
Œ0; 2/,1DC˛.
4.5.34.Circles centered at the origin are given byx
2
Cy
2
Dr
2
. Differentiating yields2xC2yy
0
D0,
soy
0
D
x
y
is a differential equation for the given family, andy
0
D
.x=y/Ctan˛
1C.x=y/tan˛
is a differential
equation for the desired family. SubstitutingyDuxyieldsu
0
xCuD
1=uCtan˛
1C.1=u/tan˛
D
1Cutan˛
uCtan˛
.
Therefore,u
0
xD
1Cu
2
uCtan˛
,
uCtan˛
1Cu
2
u
0
D
1
x
and
1
2
ln.1Cu
2
/Ctan˛tan
1
uD lnjxj Ck.
SubstitutinguD
y
x
yields
1
2
ln.x
2
Cy
2
/C.tan˛/tan
1
y
x
Dk.

56. CHAPTER5
LinearSecondOrderEquations
5.1HOMOGENEOUS LINEAR EQUATIONS
5.1.2.(a)Ify1De
x
cosx, theny
0
1
De
x
.cosxsinx/andy
00
1
De
x
.cosxsinxsinxcosx/D
2e
x
sinx, soy
00
1
2y
0
1
C2y1De
x
.2sinx2cosxC2sinxC2cosx/D0. Ify2De
x
sinx, then
y
0
2
De
x
.sinxCcosx/andy
00
2
De
x
.sinxCcosxCcosxsinx/D2e
x
cosx, soy
00
2
2y
0
2
C2y2D
e
x
.2cosx2sinx2cosxC2sinx/D0.
(b)If (B)yDe
x
.c1cosxCc2sinx/, then
y
0
De
x
.c1.cosxsinx/Cc2.sinxCcosx// . C/
and
y
00
Dc1e
x
.cosxsinxsinxcosx/
Cc2e
x
.sinxCcosxCcosxsinx/
D2e
x
.c1sinxCc2cosx/;
so
y
00
2y
0
C2yDc1e
x
.2sinx2cosxC2sinxC2cosx/
Cc2e
x
.2cosx2sinx2cosxC2sinx/D0:
(c)We must choosec1andc2in (B) so thaty.0/D3andy
0
.0/D 2. SettingxD0in (B) and (C)
shows thatc1D3andc1Cc2D 2, soc2D 5. Therefore,yDe
x
.3cosx5sinx/.
(d)We must choosec1andc2in (B) so thaty.0/Dk0andy
0
.0/Dk1. SettingxD0in (B) and (C)
shows thatc1Dk0andc1Cc2Dk1, soc2Dk1k0. Therefore,yDe
x
.k0cosxC.k1k0/sinx/.
5.1.4.(a)Ify1D
1
x1
, theny
0
1
D
1
.x1/
2
andy
00
1
D
2
.x1/
3
, so
.x
2
1/y
00
1
C4xy
0
1
C2y1D
2.x
2
1/
.x1/
3

4x
.x1/
2
C
2
x1
D
2.xC1/4xC2.x1/
.x1/
2
D0:
Similar manipulations show that.x
2
1/y
00
2
C4xy
0
2
C2y2D0. The general solution on each of the
intervals.1;1/,.1; 1/, and.1;1/is (B)yD
c1
x1
C
c2
xC1
.
51

57. 52 Chapter 5Linear Second Order Equations
(b)Differentiating (B) yields (C)y
0
D
c1
.x1/
2

c2
.xC1/
2
. We must choosec1andc2in (B) so
thaty.0/D 5andy
0
.0/D1. SettingxD0in (B) and (C) shows thatc1Cc2D 5;c1c2D1.
Therefore,c1D2andc2D 3, soyD
2
x1

3
xC1
on.1; 1/.
(d)The Wronskian offy1; y2gis
W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
x1
1
xC1

1
.x1/
2

1
.xC1/
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
2
.x
2
1/
2
; . D/
soW.0/D2. Sincep.x/D
4x
x
2
1
, so
Z
x
0
p.t/ dtD
Z
x
0
4t
t
2
1
dtDln.x
2
1/
2
, Abel’s formula
implies thatW.x/DW.0/e
ln.x
2
1/
2
D
2
.x
2
1/
2
, consistent with (D).
5.1.6.From Abel’s formula,W.x/DW./e
3
R
x

.t
2
C1/ dt
D0e
3
R
x

.t
2
C1/ dt
D0.
5.1.8.p.x/D
1
x
; therefore
Z
x
1
p.t/ dtD
Z
x
1
dt
t
Dlnx, so Abel’s formula yieldsW.x/DW.1/e
lnx
D
1
x
.
5.1.10.p.x/D 2;P.x/D 2x;y2Duy1Due
3x
;u
0
D
Ke
P .x/
y
2
1
.x/
D
Ke
2x
e
6x
DKe
4x
;
uD
K
4
e
4x
. ChooseKD 4; theny2De
4x
e
3x
De
x
.
5.1.12.p.x/D 2a;P.x/D 2ax;y2Duy1Due
ax
;u
0
D
Ke
P .x/
y
2
1
.x/
D
Ke
2ax
e
2ax
DK;uDKx.
ChooseKD1; theny2Dxe
ax
.
5.1.14.p.x/D
1
x
;P.x/D lnx;y2Duy1Dux;u
0
D
Ke
P .x/
y
2
1
.x/
D
Kx
x
2
D
K
x
;uDKlnx.
ChooseKD1; theny2Dxlnx.
5.1.16.p.x/D
1
x
;P.x/D lnjxj;y2Duy1Dux
1=2
e
2x
;u
0
D
Ke
P .x/
y
2
1
.x/
D
Kx
xe
4x
De
4x
;
uD
Ke
4x
4
. ChooseKD 4; theny2De
4x
.x
1=2
e
2x
/Dx
1=2
e
2x
.
5.1.18.p.x/D
2
x
;P.x/D 2lnjxj;y2Duy1Duxcosx;u
0
D
Ke
P .x/
y
2
1
.x/
D
Kx
2
x
2
cos
2
x
D
Ksec
2
x;uDKtanx. ChooseKD1; theny2Dtanx.xcosx/Dxsinx.
5.1.20.p.x/D
3xC2
3x1
D 1
3
3x1
;P.x/D xlnj3x1j;y2Duy1Due
2x
;
u
0
D
Ke
P .x/
y
2
1
.x/
D
K.3x1/e
x
e
4x
DK.3x1/e
3x
;uD Kxe
3x
. ChooseKD 1; then
y2Dxe
3x
e
2x
Dxe
x
.

58. Section 5.1Homogeneous Linear Equations53
5.1.22.p.x/D
2.2x
2
1/
x.2xC1/
D 2
2
2xC1
C
2
x
;P.x/D 2xlnj2xC1j C2lnjxj;y2D
uy1D
u
x
;u
0
D
Ke
P .x/
y
2
1
.x/
D
K.2xC1/e
2x
x
2
x
2
DK.2xC1/e
2x
;uDKxe
2x
. ChooseKD1; then
y2D
xe
2x
x
De
2x
.
5.1.24.Suppose thaty0on.a; b/. Theny
0
0andy
00
0on.a; b/, soyis a solution of (A)
y
00
Cp.x/y
0
Cq.x/yD0; y.x0/D0; y
0
.x0/D0on.a; b/. Since Theorem 5.1.1 implies that (A) has
only one solution on.a; b/, the conclusion follows.
5.1.26.Iff´1; ´2gis a fundamental set of solutions of (A) on.a; b/, then every solutionyof (A) on.a; b/
is a linear combination off´1; ´2g; that is,yDc1´1Cc2´2Dc1.˛y1Cˇy2/Cc2.y1Cıy2/D
.c1˛Cc2/y1C.c1ˇCc2ı/y2, which shows that every solution of (A) on.a; b/can be written as a
linear combination offy1; y2g. Therefore,fy1; y2gis a fundamental set of solutions of (A) on.a; b/.
5.1.28.The Wronskian offy1; y2gis
WD
ˇ
ˇ
ˇ
ˇ
ˇ
y1y2
y
0
1
y
0
2
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
y1ky1
y
0
1
ky
0
1
ˇ
ˇ
ˇ
ˇ
ˇ
Dk.y1y
0
1
y
0
1
y1/D0:
nory2can be a solution ofy
00
Cp.x/y
0
Cq.x/yD0on.a; b/.
5.1.30.W.x0/D

y1.x0/y
0
2
.x0/y
0
1
.x0/y2.x0/

D0if eithery1.x0/Dy2.x0/D0ory
0
1
.x0/D
y
0
2
.x0/D0, and Theorem 5.1.6 implies thatfy1; y2gis linearly dependent on.a; b/.
5.1.32.Letx0be an arbitrary point in.a; b/. By the motivating argument preceding Theorem 5.1.4, (B)
W.x0/Dy1.x0/y
0
2
.x0/y
0
1
.x0/y2.x0/¤0. Now letybe the solution ofy
00
Cp.x/y
0
Cq.x/yD
0; y.x0/Dy1.x0/; y
0
.x0/Dy
0
1
.x0/. By assumption,yis a linear combination offy1; y2gon.a; b/;
that is,yDc1y1Cc2y2, where
c1y1.x0/Cc2y2.x0/Dy1.x0/
c1y
0
1
.x0/Cc2y
0
2
.x0/Dy
0
1
.x0/:
Solving this system by Cramers’ rule yields
c1D
1
W.x0/
ˇ
ˇ
ˇ
ˇ
ˇ
y1.x0/ y2.x0/
y
0
1
.x0/ y
0
2
.x0/
ˇ
ˇ
ˇ
ˇ
ˇ
D1andc2D
1
W.x0/
ˇ
ˇ
ˇ
ˇ
ˇ
y1.x0/ y1.x0/
y
0
1
.x0/ y1.x0/
ˇ
ˇ
ˇ
ˇ
ˇ
D0:
Therefore,yDy1, which shows thaty1is a solution of (A). A similar argument shows thaty2is a
solution of (A).
5.1.34.Expanding the determinant by cofactors of its first column shows that the first equation in the
exercise can be written as
y
W
ˇ
ˇ
ˇ
ˇ
ˇ
y
0
1
y
0
2
y
00
1
y
00
2
ˇ
ˇ
ˇ
ˇ
ˇ

y
0
W
ˇ
ˇ
ˇ
ˇ
ˇ
y1y2
y
00
1
y
00
2
ˇ
ˇ
ˇ
ˇ
ˇ
C
y
00
W
ˇ
ˇ
ˇ
ˇ
ˇ
y1y2
y
0
1
y
0
2
ˇ
ˇ
ˇ
ˇ
ˇ
D0;
which is of the form (A) with
pD
1
W
ˇ
ˇ
ˇ
ˇ
ˇ
y1y2
y
00
1
y
00
2
ˇ
ˇ
ˇ
ˇ
ˇ
andqD
1
W
ˇ
ˇ
ˇ
ˇ
ˇ
y
0
1
y
0
2
y
00
1
y
00
2
ˇ
ˇ
ˇ
ˇ
ˇ
:

59. 54 Chapter 5Linear Second Order Equations
5.1.36.Theorem 5.1.6 implies that that there are constantsc1andc2such that (B)yDc1y1Cc2y2on
.a; b/. To see thatc1andc2are unique, assume that (B) holds, and letx0be a point in.a; b/. Then (C)
y
0
Dc1y
0
1
Cc2y
0
2
. SettingxDx0in (B) and (C) yields
c1y1.x0/Cc2y2.x0/Dy.x0/
c1y
0
1
.x0/Cc2y
0
2
.x0/Dy
0
.x0/:
Since Theorem 5.1.6 implies thaty1.x0/y
0
2
.x0/y
0
1
.x0/y2.x0/¤0, the argument preceding Theo-
rem 5.1.4 implies thatc1andc2are given uniquely by
c1D
y
0
2
.x0/y.x0/y2.x0/y
0
.x0/
y1.x0/y
0
2
.x0/y
0
1
.x0/y2.x0/
c2D
y1.x0/y
0
.x0/y
0
1
.x0/y.x0/
y1.x0/y
0
2
.x0/y
0
1
.x0/y2.x0/
:
5.1.38.The general solution ofy
00
D0isyDc1Cc2x, soy
0
Dc2. Imposing the stated initial
conditions ony1Dc1Cc2xyieldsc1Cc2x0D1andc2D0; thereforec1D1, soy1D1. Imposing
the stated initial conditions ony2Dc1Cc2xyieldsc1Cc2x0D0andc2D1; thereforec1D x0, so
y2Dxx0. The solution of the general initial value problem isyDk0Ck1.xx0/.
5.1.40.Lety1Da1cos!xCa2sin!xandy2Db1cos!xCb2sin!x. Then
a1cos!x0Ca2sin!x0D1
!.a1sin!x0Ca2cos!x0/D0
and
b1cos!x0Cb2sin!x0D0
!.b1sin!x0Cb2cos!x0/D1:
Solving these systems yieldsa1Dcos!x0,a2Dsin!x0,b1D
sin!x0
!
, andb2D
cos!x0
!
.
Therefore,y1Dcos!x0cos!xCsin!x0sin!xDcos!.xx0/andy2D
1
!
.sin!x0cos!xC
cos!x0sin!x/D
1
!
sin!.xx0/. The solution of the general initial value problem isyDk0cos!.x
x0/C
k1
!
sin!.xx0/.
5.1.42.(a)Ify1Dx
2
, theny
0
1
D2xandy
00
1
D2, sox
2
y
00
1
4xy
0
1
C6y1Dx
2
.2/4x.2x/C6x
2
D0
forxin.1;1/. Ify2Dx
3
, theny
0
2
D3x
2
andy
00
2
D6x, sox
2
y
00
2
4xy
0
2
C6y2Dx
2
.6x/
4x.3x
2
/C6x
3
D0forxin.1;1/. Ifx¤0, theny2.x/=y1.x/Dx, which is nonconstant on
.1; 0/and.0;1/, so Theorem 5.1.6 implies thatfy1; y2gis a fundamental set of solutions of (A) on
each of these intervals.
(b)Theorem 5.1.6 and(a)imply thatysatisfies (A) on.1; 0/and on.0;1/if and only ifyD

a1x
2
Ca2x
3
; x > 0;
b1x
2
Cb2x
3
; x < 0:
Sincey.0/D0we can complete the proof thatyis a solution of (A) on
.1;1/by showing thaty
0
.0/andy
00
.0/both exist if and only ifa1Db1. Since
y.x/y.0/
x0
D

a1xCa2x
2
;ifx > 0;
b1xCb2x
2
;ifx < 0;
it follows thaty
0
.0/Dlim
x!0
y.x/y.0/
x0
D0. Therefore,y
0
D

2a1xC3a2x
2
; x0;
2b1xC3b2x
2
; x < 0:
Since
y
0
.x/y
0
.0/
x0
D

2a1C3a2x;ifx > 0;
2b1C3b2x;ifx < 0;
it follows thaty
00
.0/Dlim
x!0
y
0
.x/y
0
.0/
x0
exists if and

60. Section 5.2Constant Coefficient Homogeneous Equations55
only ifa1Db1. By renaminga1Db1Dc1,a2Dc2, andb2Dc3we see thatyis a solution of (A) on
.1;1/if and only ifyD

c1x
2
Cc2x
3
; x0;
c1x
2
Cc3x
3
; x < 0:
(c)We have shown thaty.0/Dy
0
.0/D0for any choice ofc1andc2in (C). Therefore,the given
initial value problem has a solution if and only ifk0Dk1D0, in which case every function of the form
(C) is a solution.
(d)Ifx0> 0, thenc1andc2in (C) are uniquely determined byk0andk1, butc3can be chosen
arbitrarily. Therefore,(B) has a unique solution on.0;1/, but infinitely many solutions on.1;1/. If
x0< 0, thenc1andc3in (C) are uniquely determined byk0andk1, butc2can be chosen arbitrarily.
Therefore,(B) has a unique solution on.1; 0/, but infinitely many solutions on.1;1/.
5.1.44.(a)Ify1Dx
3
, theny
0
1
D3x
2
andy
00
1
D6x, sox
2
y
00
1
6xy
0
1
C12y1Dx
2
.6x/6x.3x
2
/C
12x
3
D0forxin.1;1/. Ify2Dx
4
, theny
0
2
D4x
3
andy
00
2
D12x
2
, sox
2
y
00
2
6xy
0
2
C12y2D
x
2
.12x
2
/6x.4x
3
/C12x
4
D0forxin.1;1/. Ifx¤0, theny2.x/=y1.x/Dx, which is
nonconstant on.1; 0/and.0;1/, so Theorem 5.1.6 implies thatfy1; y2gis a fundamental set of
solutions of (A) on each of these intervals.
(b)Theorem 5.1.2 and(a)imply thatysatisfies (A) on.1; 0/and on.0;1/if and only if (C)
yD

a1x
3
Ca2x
4
; x > 0;
b1x
3
Cb2x
4
; x < 0:
Sincey.0/D0we can complete the proof thatyis a solution of
(A) on.1;1/by showing thaty
0
.0/andy
00
.0/both exist for any choice ofa1,a2,b1, andb2.
Since
y.x/y.0/
x0
D

a1x
2
Ca2x
3
;ifx > 0;
b1x
2
Cb2x
3
;ifx < 0;
it follows thaty
0
.0/Dlim
x!0
y.x/y.0/
x0
D0.
Therefore,y
0
D

3a1x
2
C4a2x
3
; x0;
3b1x
2
C4b2x
3
; x < 0:
Since
y
0
.x/y
0
.0/
x0
D

3a1xC4a2x
2
;ifx > 0;
3b1xC4b2x
2
;ifx < 0;
it follows thaty
00
.0/Dlim
x!0
y
0
.x/y
0
.0/
x0
D0. Therefore,(B) is a solution of (A) on.1;1/.
(c)We have shown thaty.0/Dy
0
.0/D0for any choice ofa1,a2,b1, andb2in (B). Therefore,the
given initial value problem has a solution if and only ifk0Dk1D0, in which case every function of the
form (B) is a solution.
(d)Ifx0> 0, thena1anda2in (B) are uniquely determined byk0andk1, butb1andb2can be chosen
arbitrarily. Therefore,(C) has a unique solution on.0;1/, but infinitely many solutions on.1;1/.
Ifx0< 0, thenb1andb2in (B) are uniquely determined byk0andk1, buta1anda2can be chosen
arbitrarily. Therefore,(C) has a unique solution on.1; 0/, but infinitely many solutions on.1;1/.
5.2CONSTANT COEFFICIENT HOMOGENEOUS EQUATIONS
5.2.2.p.r/Dr
2
4rC5D.r2/
2
C1;yDe
2x
.c1cosxCc2sinx/.
5.2.4.p.r/Dr
2
4rC4D.r2/
2
;yDe
2x
.c1Cc2x/.
5.2.6.p.r/Dr
2
C6rC10D.rC3/
2
C1;yDe
3x
.c1cosxCc2sinx/.
5.2.8.p.r/Dr
2
CrDr.rC1/;yDc1Cc2e
x
.
5.2.10.p.r/Dr
2
C6rC13yD.rC3/
2
C4;yDe
3x
.c1cos2xCc2sin2x/.
5.2.12.p.r/D10r
2
3r1D.2r1/.5rC1/D10.r1=2/.rC1=5/;yDc1e
x=5
Cc2e
x=2
.
5.2.14.p.r/D6r
2
r1D.2r1/.3rC1/D6.r1=2/.rC1=3/;yDc1e
x=3
Cc2e
x=2
;
y
0
D
c1
3
e
x=3
C
c2
2
e
x=2
;y.0/D10)c1Cc2D10; y
0
.0/D0)
c1
3
C
c2
2
D0;c1D6; c2D4;
yD4e
x=2
C6e
x=3
.

61. 56 Chapter 5Linear Second Order Equations
5.2.16.p.r/D4r
2
4r3D.2r3/.2rC1/D4.r3=2/.rC1=2/;yDc1e
x=2
Cc2e
3x=2
;
y
0
D
c1
2
e
x=2
C
3c2
2
e
3x=2
;y.0/D
13
12
)c1Cc2D
13
12
; y
0
.0/D
23
24
)
c1
2
C
3c2
2
D
23
24
;
c1D
1
3
; c2D
3
4
;yD
e
x=2
3
C
3e
3x=2
4
.
5.2.18.p.r/Dr
2
C7rC12D.rC3/.rC4/;yDc1e
4x
Cc2e
3x
;y
0
D 4c1e
4x
3c2e
3x
;
y.0/D 1)c1Cc2D 1; y
0
.0/D0) 4c13c2D0;c1D3,c2D 4;yD3e
4x
4e
3x
.
5.2.20.p.r/D36r
2
12rC1D.6r1/
2
D36.r1=6/
2
;yDe
x=6
.c1Cc2x/;y
0
D
e
x=6
6
.c1C
c2x/Cc2e
x=6
;y.0/D3)c1D3; y
0
.0/D
5
2
)
c1
6
Cc2D
5
2
)c2D2;yDe
x=6
.3C2x/.
5.2.22.(a)From (A),ay
00
.x/Cby
0
.x/Ccy.x/D0for allx. Replacingxbyxx0yields (C)
ay
00
.xx0/Cby
0
.xx0/Ccy.xx0/D0. If´.x/Dy.xx0/, then the chain rule implies that
´
0
.x/Dy
0
.xx0/and´
00
.x/Dy
00
.xx0/, so (C) is equivalent toa´
00
Cb´
0
Cc´D0.
(b)Iffy1; y2gis a fundamental set of solutions of (A) then Theorem 5.1.6 implies thaty2=y1is
nonconstant. Therefore,
´2.x/
´1.x/
D
y2.xx0/
y1.xx0/
is also nonconstant, so Theorem 5.1.6 implies thatf´1; ´2g
is a fundamental set of solutions of (A).
(c)Letp.r/Dar
2
CbrCcbe the characteristic polynomial of(A). Then:
Ifp.r/D0has distinct real rootsr1andr2, then the general solution of(A)is
yDc1e
r1.xx0/
Cc2e
r2.xx0/
:
Ifp.r/D0has a repeated rootr1, then the general solution of(A)is
yDe
r1.xx0/
.c1Cc2.xx0//:
Ifp.r/D0has complex conjugate rootsr1DCi!andr2Di! .where! > 0/, then the
general solution of(A)is
yDe
.xx0/
.c1cos!.xx0/Cc2sin!.xx0//:
5.2.24.p.r/Dr
2
6r7D.r7/.rC1/;
yDc1e
.x2/
Cc2e
7.x2/
I
y
0
D c1e
.x2/
C7c2e
7.x2/
I
y.2/D
1
3
)c1Cc2D
1
3
; y
0
.2/D 5) c1C7c2D 5;c1D
1
3
; c2D
2
3
;yD
1
3
e
.x2/

2
3
e
7.x2/
.
5.2.26.p.r/D9r
2
C6rC1D.3rC1/
2
D9.rC1=3/
2
;
yDe
.x2/=3
.c1Cc2.x2//I
y
0
D
1
3
e
.x2/=3
.c1Cc2.x2//Cc2e
.x2/=3
I

62. Section 5.2Constant Coefficient Homogeneous Equations57
y.2/D2)c1D2; y
0
.2/D
14
3
)
c1
3
Cc2D
14
3
)c2D 4;yDe
.x2/=3
.24.x2//.
5.2.28.p.r/Dr
2
C3;
yDc1cos
p
3

x

3

Cc2sin
p
3

x

3

I
y
0
D
p
3c1sin
p
3

x

3

C
p
3c2cos
p
3

x

3

I
y.=3/D2)c1D2; y
0
.=3/D 1)c2D
1
p
3
;
yD2cos
p
3

x

3


1
p
3
sin
p
3

x

3

:
5.2.30.yis a solution ofay
00
Cby
0
CcyD0if and only if
yD c1e
r1.xx0/
Ce
r2.xx0/
y
0
Dr1c1e
r1.xx0/
Cr2e
r2
.xx0/:
Nowy1.x0/Dk0andy
0
1
.x0/Dk1)c1Cc2Dk0; r1c1Cr2c2Dk1. Therefore,c1D
r2k0k1
r2r1
andc2D
k1r1k0
r2r1
. Substitutingc1andc2into the above equations foryand y’ yields
yD
r2k0k1
r2r1
e
r1.xx0/
C
k1r1k0
r2r1
e
r2.xx0/
D
k0
r2r1

r2e
r1.xx0/
r1e
r2.xx0/

C
k1
r2r1

e
r2.xx0/
e
r1.xx0/

:
5.2.32.yis a solution ofay
00
Cby
0
CcyD0if and only if
yDe
.xx0/
.c1cos!.xx0/Cc2sin!.xx0// (A)
and
y
0
De
.xx0/
.c1cos!.xx0/Cc2sin!.xx0//
C!e
.xx0/
.c1sin!.xx0/Cc2cos!.xx0// :
Nowy1.x0/Dk0)c1Dk0andy
0
1
.x0/Dk1)c1C!c2Dk1, soc2D
k1k0
!
. Substituting
c1andc2into (A) yields
yDe
.xx0/

k0cos!.xx0/C

k1k0
!

sin!.xx0/

:
5.2.34.(b)
e
i1
e
i2
D.cos1Cisin1/.cos2Cisin2/
D.cos1cos2sin1sin2/Ci.sin1cos2Ccos1sin2/
Dcos.1C2/Cisin.1C2/De
i.1C2/
:

63. 58 Chapter 5Linear Second Order Equations
(c)
e
´1C´2
De
.˛1Ciˇ1/C.˛2Ciˇ2/
De
.˛1C˛2/Ci.ˇ1Cˇ2/
De
.˛1C˛2/
e
i.ˇ1Cˇ2/
(from (F) with˛D˛1C˛2andˇDˇ1Cˇ2)
De
˛1
e
˛2
e
i.ˇ1Cˇ2/
(property of the real–valued exponential function)
De
˛1
e
˛2
e
iˇ1
e
iˇ2
(from(b))
De
˛1
e
iˇ1
e
˛2
e
iˇ2
De
˛1Ciˇ1
e
˛2Ciˇ2
De
´1
e
´2
:
(d)The real and imaginary parts of´1De
.Ci !/x
areu1De
x
cos!xandv1De
x
sin!x, which
are both solutions ofay
00
Cby
0
CcyD0, by Theorem 5.2.1(c). Similarly, the real and imaginary parts
of´2De
.i !/x
areu2De
x
cos.!x/De
x
cos!xandv1De
x
sin.!x/D e
x
sin!x, which
are both solutions ofay
00
Cby
0
CcyD0, by Theorem 5.2.1,(c).
5.3NONHOMOGENEOUS LINEAR EQUATIONS
5.3.2.The characteristic polynomial of the complementary equation isp.r/Dr
2
4rC5D.r2/
2
C1,
sofe
2x
cosx; e
2x
sinxgis a fundamental set of solutions for the complementary equation. LetypD
ACBx; theny
00
p
4y
0
p
C5ypD 4BC5.ACBx/D1C5x. Therefore,5BD5;4BC5AD1, so
BD1,AD1. Therefore,ypD1Cxis a particular solution andyD1CxCe
2x
.c1cosxCc2sinx/
is the general solution.
5.3.4.The characteristic polynomial of the complementary equation isp.r/Dr
2
4rC4D.r2/
2
, so
fe
2x
; xe
2x
gis a fundamental set of solutions for the complementary equation. LetypDACBxCCx
2
;
theny
00
p
4y
0
p
C4ypD2C4.BC2Cx/C4.ACBxCCx
2
/D.2C4BC4A/C.8CC4B/xC
4Cx
2
D2C8x4x
2
. Therefore,4CD 4;8CC4BD8; 2C4BC4AD2, soCD 1,
BD0, andAD1. Therefore,ypD1x
2
is a particular solution andyD1x
2
Ce
2x
.c1Cc2x/is
the general solution.
5.3.6.The characteristic polynomial of the complementary equation isp.r/Dr
2
C6rC10D.rC
3/
2
C1, sofe
3x
cosx; e
3x
sinxgis a fundamental set of solutions for the complementary equation.
LetypDACBx; theny
00
p
C6y
0
p
C10ypD6BC10.ACBx/D22C20x. Therefore,10BD
20; 6BC10AD22, soBD2,AD1. Therefore,ypD1C2xis a particular solution and (A)
yD1C2xCe
3x
.c1cosxCc2sinx/is the general solution. Nowy.0/D2)2D1Cc1)c1D1.
Differentiating (A) yieldsy
0
D23e
3x
.c1cosxCc2sinx/Ce
3x
.c1sinxCc2cosx/, soy
0
.0/D
2) 2D23c1Cc2)c2D 1.yD1C2xCe
3x
.cosxsinx/is the solution of the initial
value problem.
5.3.8.IfypD
A
x
, thenx
2
y
00
p
C7xy
0
p
C8ypDA

x
2

2
x
3

C7x

1
x
2

C

8
x

D
3A
x
D
6
x
if
AD2. Therefore,ypD
2
x
is a particular solution.
5.3.10.IfypDAx
3
, thenx
2
y
00
p
xy
0
p
CypDA

x
2
.6x/x.3x
2
/Cx
3

D4Ax
3
D2x
3
ifAD
1
2
.
Therefore,ypD
x
3
2
is a particular solution.
5.3.12.IfypDAx
1=3
, thenx
2
y
00
p
Cxy
0
p
CypDA
 
x
2
 
2x
5=3
9
!
Cx
 
x
2=3
3
!
Cx
1=3
!
D
10A
9
x
1=3
D10x
1=3
ifAD9. Therefore,ypD9x
1=3
is a particular solution.

64. Section 5.3Nonhomogeneous Linear Equations59
5.3.14.IfypD
A
x
3
, thenx
2
y
00
p
C3xy
0
p
3ypDA

x
2

12
x
5

C3x

3
x
4

C
3
x
3

D0. Therefore,yp
is not a solution of the given equation for any choice ofA.
5.3.16.The characteristic polynomial of the complementary equation isp.r/Dr
2
C5r6D.rC6/.r
1/, sofe
6x
; e
x
gis a fundamental set of solutions for the complementary equation. LetypDAe
3x
; then
y
00
p
C5y
0
p
6ypDp.3/Ae
3x
D18Ae
3x
D6e
3x
ifAD
1
3
. Therefore,ypD
e
3x
3
is a particular solution
andyD
e
3x
3
Cc1e
6x
Cc2e
x
is the general solution.
5.3.18.The characteristic polynomial of the complementary equation isp.r/Dr
2
C8rC7D.rC1/.rC
7/, sofe
7x
; e
x
gis a fundamental set of solutions for the complementary equation. LetypDAe
2x
;
theny
00
p
C8y
0
p
C7ypDp.2/Ae
2x
D 5Ae
2x
D10e
2x
ifAD 2. Therefore,ypD 2e
2x
is
a particular solution and (A)yD 2e
2x
Cc1e
7x
Cc2e
x
is the general solution. Differentiating (A)
yieldsy
0
D4e
2x
7c1e
7x
c2e
x
. Nowy.0/D 2) 2D 2Cc1Cc2andy
0
.0/D10)
10D47c1c2. Therefore,c1D 1andc2D1, soyD 2e
2x
e
7x
Ce
x
is the solution of the
initial value problem.
5.3.20.The characteristic polynomial of the complementary equation isp.r/Dr
2
C2rC10D.rC
1/
2
C9, sofe
x
cos3x; e
x
sin3xgis a fundamental set of solutions for the complementary equation.
IfypDAe
x=2
, theny
00
p
C2y
0
p
C10ypDp.1=2/Ae
x=2
D
45
4
Ae
x=2
De
x=2
ifAD
4
45
. Therefore,
ypD
4
45
e
x=2
is a particular solution andyD
4
45
e
x=2
Ce
x
.c1cos3xCc2sin3x/is the general solution.
5.3.22.The characteristic polynomial of the complementary equation isp.r/Dr
2
7rC12D.r
4/.r3/. IfypDAe
4x
, theny
00
p
7y
0
p
C12ypDp.4/Ae
4x
D0e
4x
D0, soy
00
p
7y
0
p
C12yp¤5e
4x
for any choice ofA.
5.3.24.The characteristic polynomial of the complementary equation isp.r/Dr
2
8rC16D.r4/
2
,
sofe
4x
; xe
4x
gis a fundamental set of solutions for the complementary equation. IfypDAcosxC
Bsinx, theny
00
p
8y
0
p
C16ypD .AcosxCBsinx/8.AsinxCBcosx/C16.AcosxCBsinx/D
.15A8B/cosxC.8AC15B/sinx, so15A8BD23; 8AC15BD 7, which implies thatAD1
andBD 1. HenceypDcosxsinxandyDcosxsinxCe
4x
.c1Cc2x/is the general solution.
5.3.26.The characteristic polynomial of the complementary equation isp.r/Dr
2
2rC3D.r1/
2
C2,
sofe
x
cos
p
2x; e
x
sin
p
2xgis a fundamental set of solutions for the complementary equation. IfypD
Acos3xCBsin3x, theny
00
p
2y
0
p
C3ypD 9.Acos3xCBsin3x/6.Asin3xCBcos3x/C
3.Acos3xCBsin3x/D .6AC6B/cos3xC.6A6B/sin3x, so6A6BD 6; 6A6BD6,
which implies thatAD1andBD0. HenceypDcos3xis a particular solution andyDcos3xC
e
x
.c1cos
p
2xCc2sin
p
2x/is the general solution.
5.3.28.The characteristic polynomial of the complementary equation isp.r/Dr
2
C7rC12D.rC
3/.rC4/, sofe
4x
; e
3x
gis a fundamental set of solutions for the complementary equation. IfypD
Acos2xCBsin2x, theny
00
p
C7y
0
p
C12ypD 4.Acos2xCBsin2x/C14.Asin2xCBcos2x/C
12.AcosxCBsinx/D.8AC14B/cos2xC.8B14a/sin2x, so8AC14BD 2;14AC8BD36,
which implies thatAD 2andBD1. HenceypD 2cos2xCsin2xis a particular solution and
(A)yD 2cos2xCsin2xCc1e
4x
Cc2e
3x
is the general solution. Differentiating (A) yields
y
0
D2sin2xC2cos2x4c1e
4x
3c2e
3x
. Nowy.0/D 3) 3D 2Cc1Cc2andy
0
.0/D
3)3D24c13c2. Therefore,c1D2andc2D 3, soyD 2cos2xCsin2xC2e
4x
3e
3x
is the solution of the initial value problem.

65. 60 Chapter 5Linear Second Order Equations I
5.3.30.fcos!0x;sin!0xgis a fundamental set of solutions of the complementary equation. IfypD
Acos!xCBsin!x, theny
00
p
C!
2
0
ypD !
2
.Acos!xCBsin!x/C!
2
0
.Acos!xCBsin!x/D
.!
2
0
!
2
/.Acos!xCBsin!x/DMcos!xCNsin!xifAD
M
!
2
0
!
2
andBD
N
!
2
0
!
2
.
Therefore,
ypD
1
!
2
0
!
2
.Mcos!xCNsin!x/
is a particular solution of the given equation and
yD
1
!
2
0
!
2
.Mcos!xCNsin!x/Cc1cos!0xCc2sin!0x
is the general solution.
5.3.32.IfypDAcos!xCBsin!x, thenay
00
p
Cby
0
p
CcypD a!
2
.Acos!xCBsin!x/C
b!.Asin!xCBcos!x/Cc.Acos!xCBsin!x/D
fi
.ca!
2
/ACb!B

cos!xC
fi
b!AC.ca!
2
/B

sin!x.
Therefore,ypis a solution of (A) if and only if the set of equations (B).ca!
2
/ACb!BD
M;b!AC.ca!
2
/BDNhas a solution. If.ca!
2
/
2
C.b!/
2
¤0, then (B) has the so-
lutionAD
.ca!
2
/Mb!N
.ca!
2
/
2
C.b!/
2
,BD
.ca!
2
/NCb!M
.ca!
2
/
2
C.b!/
2
, andypDAcos!xCBsin!xis a
solution of (A). If.ca!
2
/
2
C.b!/
2
D0(which is true if and only if the left side of (A) is of the form
a.y
00
C!
2
y/, then the coefficients ofAandBin (B) are all zero, so (B) does not have a solution, so (A)
does not have a solution of the formypDAcos!xCBsin!x.
5.3.34.From Exercises 5.3.2 and 5.3.17,yp1
D1Cxandyp2
De
2x
are particular solutions of
y
00
4y
0
C5yD1C5xandy
00
4y
0
C5yDe
2x
respectively, andfe
2x
cosx; e
2x
sinxgis a fundamental
set of solutions of the complementary equation. Therefore,ypDyp1
Cyp2
D1CxCe
2x
is a particular
solution of the given equation, andyD1CxCe
2x
.1Cc1cosxCc2sinx/is the general solution.
5.3.36.From Exercises 5.3.4 and 5.3.19,yp1
D1x
2
andyp2
De
x
are particular solutions of
y
00
4y
0
C4yD2C8x4x
2
andy
00
4y
0
C4yDe
x
respectively, andfe
2x
; xe
2x
gis a fundamental
set of solutions of the complementary equation. Therefore,ypDyp1
Cyp2
D1x
2
Ce
x
is a particular
solution of the given equation, andyD1x
2
Ce
x
Ce
2x
.c1Cc2x/is the general solution.
5.3.38.From Exercises 5.3.6 and 5.3.21,yp1
D1C2xandyp2
De
3x
are particular solutions of
y
00
C6y
0
C10yD22C20xandy
00
C6y
0
C10yDe
3x
respectively, andfe
3x
cosx; e
3x
sinxgis a
fundamental set of solutions of the complementary equation. Therefore,ypDyp1
Cyp2
D1C2xCe
3x
is a particular solution of the given equation, andyD1C2xCe
3x
.1Cc1cosxCc2sinx/is the
general solution.
5.3.40.Lettingc1Dc2D0shows that (A)y
00
p
Cp.x/y
0
p
Cq.x/ypDf. Lettingc1D1andc2D0
shows that (B).y1Cyp/
00
Cp.x/.y1Cyp/
0
Cq.x/.y1Cyp/Df. Now subtract (A) from (B) to see
thaty
00
1
Cp.x/y
0
1
Cq.x/y1D0. Lettingc1D0andc2D1shows that (C).y2Cyp/
00
Cp.x/.y2C
yp/
0
Cq.x/.y2Cyp/Df. Now subtract (A) from (C) to see thaty
00
2
Cp.x/y
0
2
Cq.x/y2D0.
5.4THE METHOD OF UNDETERMINED COEFFICIENTS I
5.4.2.IfyDue
3x
, theny
00
6y
0
C5yDe
3x
Œ.u
00
6u
0
C9u/6.u
0
3u/C5uDe
3x
.358x/,
sou
00
12u
0
C32uD358xandupDACBx, where12BC32.ACBx/D358x. Therefore,32BD
8,32A12BD35, soBD
1
4
,AD1, andupD1
x
4
. Therefore,ypDe
3x

1
x
4

.

66. Section 5.4The Method of Undetermined Coefficients I61
5.4.4.IfyDue
2x
, theny
00
C2y
0
CyDe
2x
Œ.u
00
C4u
0
C4u/C2.u
0
C2u/CuDe
2x
.715xC
9x
2
/sou
00
C6u
0
C9uD 715xC9x
2
andupDACBxCCx
2
, where2CC6.BC2Cx/C
9.ACBxCCx
2
/D 715xC9x
2
. Therefore,9CD9,9BC12CD 15,9AC6BC2CD 7,
soCD1,BD 3,AD1, andupD13xCx
2
. Therefore,ypDe
2x
.13xCx
2
/.
5.4.6.IfyDue
x
, theny
00
y
0
2yDe
x
Œ.u
00
C2u
0
Cu/.u
0
Cu/2uDe
x
.9C2x4x
2
/so
u
00
Cu
0
2uD9C2x4x
2
, andupDACBxCCx
2
, where2CC.BC2Cx/2.ACBxCCx
2
/D
9C2x4x
2
. Therefore,2CD 4,2BC2CD2,2ACBC2CD9, soCD2,BD1,AD 2,
andupD 2CxC2x
2
. Therefore,ypDe
x
.2CxC2x
2
/.
5.4.8.IfyDue
x
, theny
00
3y
0
C2yDe
x
Œ.u
00
C2u
0
Cu/3.u
0
Cu/C2uDe
x
.34x/, so
u
00
u
0
D34xandupDAxCBx
2
, where2B.AC2Bx/D34x. Therefore,2BD 4,
AC2BD3, soBD2,AD1, andupDx.1C2x/. Therefore,ypDxe
x
.1C2x/.
5.4.10.IfyDue
2x
, then2y
00
3y
0
2yDe
2x
Œ2.u
00
C4u
0
C4u/3.u
0
C2u/2uDe
2x
.6C
10x/, so2u
00
C5u
0
D 6C10xandupDAxCBx
2
, where2.2B/C5.AC2Bx/D 6C10x.
Therefore,10BD10,5AC4BD 6, soBD1,AD 2, andupDx.2Cx/. Therefore,ypD
xe
2x
.2Cx/.
5.4.12.IfyDue
x
, theny
00
2y
0
CyDe
x
Œ.u
00
C2u
0
Cu/2.u
0
Cu/CuDe
x
.16x/, so
u
00
D16xIntegrating twice and taking the constants of integration to be zero yieldsupDx
2

1
2
x

.
Therefore,ypDx
2
e
x

1
2
x

.
5.4.14.IfyDue
x=3
, then9y
00
C6y
0
CyDe
x=3

9

u
00

2u
0
3
C
u
9

C6

u
0

u
3

Cu

D
e
x=3
.24xC4x
2
/, so9u
00
D24xC4x
2
, oru
00
D
1
9
.24xC4x
2
/. Integrating twice
and taking the constants of integration to be zero yieldsupD
x
2
27
.32xCx
2
/. Therefore,ypD
x
2
e
x=3
27
.32xCx
2
/.
5.4.16.IfyDue
x
, theny
00
6y
0
C8yDe
x
Œ.u
00
C2u
0
Cu/6.u
0
Cu/C8uDe
x
.116x/, so
u
00
4u
0
C3uD116xandupDACBx, where4BC3.ACBx/D116x. Therefore,3BD 6,
3A4BD11, soBD 2,AD1andupD12x. Therefore,ypDe
x
.12x/. The characteristic
polynomial of the complementary equation isp.r/Dr
2
6rC8D.r2/.r4/, sofe
2x
; e
4x
gis a
fundamental set of solutions of the complementary equation. Therefore,yDe
x
.12x/Cc1e
2x
Cc2e
4x
is the general solution of the nonhomogeneous equation.
5.4.18.IfyDue
x
, theny
00
C2y
0
3yDe
x
Œ.u
00
C2u
0
Cu/C2.u
0
Cu/3uD 16xe
x
, so
u
00
C4u
0
D 16xandupDAxCBx
2
, where2BC4.AC2Bx/D 16x. Therefore,8BD 16,
4AC2BD0, soBD 2,AD1, andupDx.12x/. Therefore,ypDxe
x
.12x/. The characteristic
polynomial of the complementary equation isp.r/Dr
2
C2r3D.rC3/.r1/, sofe
x
; e
3x
gis a
fundamental set of solutions of the complementary equation. Therefore,yDxe
x
.12x/Cc1e
x
Cc2e
3x
is the general solution of the nonhomogeneous equation.
5.4.20.IfyDue
2x
, theny
00
4y
0
5yDe
2x
Œ.u
00
C4u
0
C4u/4.u
0
C2u/5uD9e
2x
.1Cx/, so
u
00
9uD9C9xandupDACBx, where9.ACBx/D9C9x. Therefore,9BD 9,9AD9,
soBD 1,AD 1, andupD 1x. Therefore,ypD e
2x
.1Cx/. The characteristic polynomial
of the complementary equation isp.r/Dr
2
4r5D.r5/.rC1/, sofe
x
; e
5x
gis a fundamental

67. 62 Chapter 5Linear Second Order Equations I
set of solutions of the complementary equation. Therefore,(A)yD e
2x
.1Cx/Cc1e
x
Cc2e
5x
is
the general solution of the nonhomogeneous equation. Differentiating (A) yieldsy
0
D 2e
2x
.1Cx/
e
2x
c1e
x
C5c2e
5x
. Nowy.0/D0; y
0
.0/D 10)0D 1Cc1Cc2;10D 3c1C5c2, so
c1D2,c2D 1. Therefore,yD e
2x
.1Cx/C2e
x
e
5x
is the solution of the initial value problem.
5.4.22.IfyDue
x
, theny
00
C4y
0
C3yDe
x
Œ.u
00
2u
0
Cu/C4.u
0
u/C3uD e
x
.2C8x/, so
u
00
C2u
0
D 28xandupDAxCBx
2
, where2BC2.AC2Bx/D 28x. Therefore,4BD 8,
2AC2BD 2, soBD 2,AD1, andupDx.12x/. Therefore,ypDxe
x
.12x/. The
characteristic polynomial of the complementary equation isp.r/Dr
2
C4rC3D.rC3/.rC1/,
sofe
x
; e
3x
gis a fundamental set of solutions of the complementary equation. Therefore,(A)yD
xe
x
.12x/Cc1e
x
Cc2e
3x
is the general solution of the nonhomogeneous equation. Differentiating
(A) yieldsy
0
D xe
x
.12x/Ce
x
.14x/c1e
x
3c2e
3x
. Nowy.0/D1; y
0
.0/D2)1D
c1Cc2; 2D1c13c2, soc1D2,c2D 1. Therefore,yDe
x
.2Cx2x
2
/e
3x
is the solution
of the initial value problem.
5.4.24.We must find particular solutionsyp1
andyp2
of (A)y
00
Cy
0
CyDxe
x
and (B)y
00
C
y
0
CyDe
x
.1C2x/, respectively. To find a particular solution of (A) we writeyDue
x
. Then
y
00
Cy
0
CyDe
x
Œ.u
00
C2u
0
Cu/C.u
0
Cu/CuDxe
x
sou
00
C3u
0
C3uDxandupDACBx,
where3BC3.ACBx/Dx. Therefore,3BD1,3AC3BD0, soBD
1
3
,AD
1
3
, andupD

1
3
.1x/, soyp1
D
e
x
3
.1x/. To find a particular solution of (B) we writeyDue
x
. Then
y
00
Cy
0
CyDe
x
Œ.u
00
2u
0
Cu/C.u
0
u/CuDe
x
.1C2x/, sou
00
u
0
CuD1C2xand
upDACBx, whereBC.ACBx/D1C2x. Therefore,BD2,ABD1, soAD3, and
upD2C3x, soyp2
De
x
.3C2x/. NowypDyp1
Cyp2
D
e
x
3
.1x/Ce
x
.3C2x/.
5.4.26.We must find particular solutionsyp1
andyp2
of (A)y
00
8y
0
C16yD6xe
4x
and (B)y
00

8y
0
C16yD2C16xC16x
2
, respectively. To find a particular solution of (A) we writeyDue
4x
. Then
y
00
8y
0
C16yDe
x
Œ.u
00
C8u
0
C16u/8.u
0
C4u/C16uD6xe
4x
, sou
00
D6x,upDx
3
. and
yp1
Dx
3
e
4x
. To find a particular solution of (B) we writeypDACBxCCx
2
. Theny
00
p
8y
0
p
C16ypD
2C8.BC2Cx/C16.ACBxCCx
2
/D.16A8BC2C /C.16B16C /xC16Cx
2
D2C16xC16x
2
if16CD16,16B16CD16,16A8BC2CD2. Therefore,CD1,BD2,AD1, and
yp2
D1C2xCx
2
. NowypDyp1
Cyp2
Dx
3
e
4x
C1C2xCx
2
.
5.4.28.We must find particular solutionsyp1
andyp2
of (A)y
00
2y
0
C2yDe
x
.1Cx/and (B)
y
00
2y
0
C2yDe
x
.28xC5x
2
/, respectively. To find a particular solution of (A) we writeyDue
x
.
Theny
00
2y
0
C2yDe
x
Œ.u
00
C2u
0
Cu/2.u
0
Cu/C2uDe
x
.1Cx/, sou
00
CuD1Cxand
upD1Cx, soyp1
De
x
.1Cx/. To find a particular solution of (B) we writeyDue
x
. Then
y
00
2y
0
C2yDe
x
Œ.u
00
2u
0
Cu/2.u
0
u/C2uDe
x
.28xC5x
2
/, sou
00
4u
0
C5uD
28xC5x
2
andupDACBxCCx
2
, where2C4.BC2Cx/C5.ACBxCCx
2
/D28xC5x
2
.
Therefore,5CD5,5B8CD 8,5A4BC2CD2, soCD1,BD0,AD0, andupDx
2
.
Therefore,yp2
Dx
2
e
x
. NowypDyp1
Cyp2
De
x
.1Cx/Cx
2
e
x
.
5.4.30.(a)IfyDue
˛x
, thenay
00
Cby
0
CcyDe
˛x
fi
a.u
00
C2˛u
0
C˛
2
u/Cb.u
0
C˛u/Ccu

D
e
˛x
fi
au
00
C.2a˛Cb//u
0
C.a˛
2
Cb˛Cc/u

De
˛x
.au
00
Cp
0
.˛/u
0
Cp.˛/u/. Therefore,ay
00
Cby
0
C
cyDe
˛x
G.x/if and only ifau
00
Cp
0
.˛/u
0
Cp.˛/uDG.x/.
(b)SubstitutingupDACBxCCx
2
CDx
3
into (B) yields
a.2CC6Dx/Cp
0
.˛/.BC2CxC3Dx
2
/Cp.˛/.ACBxCCx
2
CDx
3
/
DŒp.˛/ACp
0
.˛/BC2aC CŒp.˛/BC2p
0
.˛/CC6aDx
CŒp.˛/CC3p
0
.˛/Dx
2
Cp.˛/Dx
3
Dg0Cg1xCg2x
2
Cg3x
3

68. Section 5.4The Method of Undetermined Coefficients I63
if
p.˛/DDg3
p.˛/CC3p
0
.˛/DDg2
p.˛/BC2p
0
.˛/CC6aDDg1
p.˛/ACp
0
.˛/BC2aC Dg0:
.C/
Sincee
˛x
is not a solution of the complementary equation,p.˛/¤0. Therefore,the triangular system
(C) can be solved successively forD,C,BandA.
(c)Sincee
˛x
is a solution of the complementary equation whilexe
˛x
is not,p.˛/D0andp
0
.˛/¤0.
Therefore, (B) reduces to (D)au
00
Cp
0
.˛/uDG.x/. SubstitutingupDAxCBx
2
CCx
3
CDx
4
into
(D) yields
a.2BC6CxC12Dx
2
/Cp
0
.˛/.AC2BxC3Cx
2
C4Dx
3
/
D.p
0
.˛/AC2aB/C.2p
0
.˛/BC6aC /xC.3p
0
.˛/CC12aD/x
2
C4p
0
.˛/Dx
3
Dg0Cg1xCg2x
2
Cg3x
3
if
4p
0
.˛/DDg3
3p
0
.˛/CC12aDDg2
2p
0
.˛/BC6aC Dg1
p
0
.˛/AC2aB Dg0:
Sincep
0
.˛/¤0this triangular system can be solved successively forD,C,BandA.
(d)Sincee
˛x
andxe
˛x
are solutions of the complementary equation,p.˛/D0andp
0
.˛/D0.
Therefore, (B) reduces to (D)au
00
DG.x/, sou
00
D
G.x/
a
. Integrating this twice and taking the
constants of integration yields the particular solutionupDx
2

g0
2
C
g1
6
xC
g2
12
x
2
C
g3
20
x
3

.
5.4.32.IfypDAxe
4x
, theny
00
p
7y
0
p
C12ypDŒ.8C16x/7.1C4x/C12xAe
4x
DAe
4x
D5e
4x
ifAD1, soypD5xe
4x
.
5.4.34.IfypDe
3x
.ACBxCCx
2
/, then
y
00
p
3y
0
p
C2ypDe
3x
Œ.9AC6BC2C /C.9BC12C /xC9Cx
2

3e
3x
Œ.3ACB/C.3BC2C /xC3Cx
2

C2e
3x
.ACBxCCx
2
/
De
3x
Œ.2AC3BC2C /C.2BC6C /xC2Cx
2

De
3x
.1C2xCx
2
/
if2CD1; 2BC6CD2; 2AC3BC2CD 1. Therefore,CD
1
2
,BD
1
2
,AD
1
4
, and
ypD
e
3x
4
.1C2x2x
2
/.
5.4.36.IfypDe
x=2
.Ax
2
CBx
3
CCx
4
/, then
4y
00
p
C4y
0
p
CypDe
x=2
Œ8A.8A24B/xC.A12BC48C /x
2

Ce
x=2
Œ.B16C /x
3
CCx
4

Ce
x=2
Œ8Ax.2A12B/x
2
.2B16C /x
3
2Cx
4

Ce
x=2
.Ax
2
CBx
3
CCx
4
/
De
x=2
.8AC24BxC48Cx
2
/De
x=2
.8C48xC144x
2
/

69. 64 Chapter 5Linear Second Order Equations
if48CD144,24BD48, and8AD 8. Therefore,CD3,BD2,AD 1, andypDx
2
e
x=2
.1C
2xC3x
2
/.
5.4.38.IfyD
R
e
˛x
P.x/ dx, theny
0
De
˛x
P.x/. LetyDue
˛x
; then.u
0
C˛u/e
˛x
De
˛x
P.x/, which
implies (A). We must show that it is possible to chooseA0; : : : ; Akso that (B).A0CA1x CAkx
k
/
0
C
˛.A0CA1x CAkx
k
/Dp0Cp1xC Cpkx
k
. By equating the coefficients ofx
k
; x
k1
; : : : ; 1
(in that order) on the two sides of (B), we see that (B) holds ifand only if˛AkDpkand.kjC
1/AkjC1C˛AkDpkj; 1jk.
5.4.40.IfyD
R
x
k
e
˛x
dx, theny
0
Dx
k
e
˛x
. LetyDue
˛x
; then.u
0
C˛u/e
˛x
Dx
k
e
˛x
, so
u
0
C˛uDx
k
. This equation has a particular solutionupDA0CA1x CAkx
k
, where (A).A0C
A1x CAkx
k
/
0
C˛.A0CA1x CAkx
k
/Dx
k
. By equating the coefficients ofx
k
; x
k1
; : : : ; 1
on the two sides of (A), we see that (A) holds if and only if˛AkD1and.kjC1/AkjC1C
˛AkjD0; 1jk. Therefore,AkD
1
˛
,Ak1D
k
˛
2
,Ak2D
k.k1/
˛
3
, and, in general,
AkjD.1/
j
k.k1/ .kjC1/
˛
jC1
D
.1/
j
kŠ
˛
jC1
.kj /Š
; 1jk. By introducing the index
rDkjwe can rewrite this asArD
.1/
kr
kŠ
˛
krC1
rŠ
; 0rk. Therefore,upD
.1/
k
kŠ
˛
kC1
k
X
rD0
.˛x/
r
rŠ
andyD
.1/
k
kŠe
˛x
˛
kC1
k
X
rD0
.˛x/
r
rŠ
Cc.
5.5THE METHOD OF UNDETERMINED COEFFICIENTS II
5.5.2.Let
ypD.A0CA1x/cosxC.B0CB1x/sinxIthen
y
0
p
D.A1CB0CB1x/cosxC.B1A0A1x/sinx
y
00
p
D.2B1A0A1x/cosx.2A1CB0CB1x/sinx;so
y
00
p
C3y
0
p
CypD.3A1C3B0C2B1C3B1x/cosx
C.3B13A02A13A1x/sinx
D.26x/cosx9sinx
if3B1D 6,3A1D0,3B0C3A1C2B1D2,3A0C3B1C2A1D 9. Therefore,A1D0,
B1D 2,A0D1,B0D2, andypDcosxC.22x/sinx.
5.5.4.LetyDue
2x
. Then
y
00
C3y
0
2yDe
2x
fi
.u
00
C4u
0
C4u/C3.u
0
C2u/2u

De
2x
.u
00
C7u
0
C8u/D e
2x
.5cos2xC9sin2x/
ifu
00
C7u
0
C8uD 5cos2x9sin2x. Now letupDAcos2xCBsin2x. Then
u
00
p
C7u
0
p
C8upD 4.Acos2xCBsin2x/C14.Asin2xCBcos2x/
C8.Acos2xCBsin2x/
D.4AC14B/cos2x.14A4B/sin2x
D 5cos2x9sin2x

70. Section 5.5The Method of Undetermined Coefficients II65
if4AC14BD 5,14AC4BD 9. Therefore,AD
1
2
,BD
1
2
,
upD
1
2
.cos2xsin2x/;andypD
e
2x
2
.cos2xsin2x/:
5.5.6.LetyDue
2x
. Then
y
00
C3y
0
2yDe
2x
fi
.u
00
4u
0
C4u/C3.u
0
2u/2u

De
2x
.u
00
u
0
4u/
De
2x
Œ.4C20x/cos3xC.2632x/sin3x
ifu
00
u
0
4uD.4C20x/cos3xC.2632x/sin3x. Let
upD.A0CA1x/cos3xC.B0CB1x/sin3xIthen
u
0
p
D.A1C3B0C3B1x/cos3xC.B13A03A1x/sin3x
u
00
p
D.6B19A09A1x/cos3x.2A1C9B0C9B1x/sin3x;so
u
00
p
u
0
p
4upD Œ13A0CA1C3B06B1C.13A1C3B1/xcos3x
Œ13B0CB13A0C6A1C.13B13A1/xsin3x
D.4C20x/cos3xC.2632x/sin3xif
13A13B1D 20
3A113B1D 32
and
13A03B0A1C6B1D4
3A013B06A1B1D26:
From the first two equations,A1D 2,B1D2. Substituting these in the last two equations yields
13A03B0D 10,3A013B0D16. Solving this pair yieldsA0D1,B0D 1. Therefore,
upD.12x/.cos3xsin3x/andypDe
2x
.12x/.cos3xsin3x/:
5.5.8.Let
ypD.A0xCA1x
2
/cosxC.B0xCB1x
2
/sinxIthen
y
0
p
D
fi
A0C.2A1CB0/xCB1x
2

cosxC
fi
B0C.2B1A0/xB1x
2

sinx
y
00
p
D
fi
2A1C2B0.A04B1/xA1x
2

cosx
C
fi
2B12A0.B0C4A1/xB1x
2

sinx;so
y
00
p
CypD.2A1C2B0C4B1x/cosxC.2B12A04A1x/sinx
D.4C8x/cosxC.84x/sinx
if4B1D8,4A1D 4,2B0C2A1D 4,2A0C2B1D8. Therefore,A1D1,B1D2,A0D 2,
B0D 3, andypD x Œ.2x/cosxC.32x/sinx.
5.5.10.LetyDue
x
. Then
y
00
C2y
0
C2yDe
x
fi
.u
00
2u
0
Cu/C2.u
0
u/C2u

De
x
.u
00
Cu/De
x
.8cosx6sinx/

71. 66 Chapter 5Linear Second Order Equations
ifu
00
CuD8cosx6sinx. Now let
upDAxcosxCBxsinxIthen
u
0
p
D.ACBx/cosxC.BAx/sinx
u
00
p
D.2BAx/cosx.2ACBx/sinx;so
u
00
p
CupD2Bcosx2AsinxD8cosx6sinx
if2BD8,2AD 6. Therefore,AD3,BD4,upDx.3cosxC4sinx/, andypDxe
x
.3cosxC
4sinx/.
5.5.12.Let
ypD.A0CA1xCA2x
2
/cosxC.B0CB1xCB2x
2
/sinxIthen
y
0
p
D
fi
A1CB0C.2A2CB1/xCB2x
2

cosx
C
fi
B1A0C.2B2A1/xA2x
2

sinx;
y
00
p
D
fi
A0C2A2C2B1.A14B2/xA2x
2

cosx
C
fi
B0C2B22A1.B1C4A2/xB2x
2

sinx;so
y
00
p
C2y
0
p
CypD2
fi
A1CA2CB0CB1C.2A2CB1C2B2/xCB2x
2

cosx
C2
fi
B1CB2A0A1C.2B2A12A2/xA2x
2

sinx
D8x
2
cosx4xsinxif
(i)
2B2D8
2A2D0
;(ii)
2B1C4A2C4B2D 0
2A14A2C4B2D 4
;
(iii)
2B0C2A1C2B1C2A2D0
2A02A1C2B1C2B2D0
:
From (i),A2D0,B2D4. Substituting these into (ii) and solving forA1andB1yieldsA1D10,
B1D 8. Substituting the known coefficients into (iii) and solvingforA0andB0yieldsA0D 14,
B0D 2. Therefore,ypD .1410x/cosx.2C8x4x
2
/sinx.
5.5.14.Let
ypD.A0CA1xCA2x
2
/cos2xC.B0CB1xCB2x
2
/sin2xIthen
y
0
p
D
fi
A1C2B0C.2A2C2B1/xC2B2x
2

cos2x
C
fi
B12A0C.2B22A1/x2A2x
2

sin2x
y
00
p
D
fi
4A0C2A2C4B1.4A18B2/x4A2x
2

cos2x
C
fi
4B02B24A1.4B1C8A2/x4B2x
2

sin2x;so
y
00
p
C3y
0
p
C2ypDŒ2A0C3A1C4A2C6B0C4B1
.2A16A26B18B2/x.2A26B2/x
2

cos2x
CŒ2B0C3B1C4B26A04A1
.2B16B2C6A1C8A2/x.2B2C6A2/x
2

sin2x
D.1x4x
2
/cos2x.1C7xC2x
2
/sin2xif

72. Section 5.5The Method of Undetermined Coefficients II67
(i)
2A2C6B2D 4
6A22B2D 2
;(ii)
2A1C6B1C6A2C8B2D 1
6A12B18A2C6B2D 7
;
(iii)
2A0C6B0C3A1C4B1C2A2D 1
6A02B04A0C3B1C2B2D 1
:
From (i),A2D
1
2
,B2D
1
2
. Substituting these into (ii) and solving forA1andB1yieldsA1D0,
B1D0. Substituting the known coefficients into (iii) and solvingforA0andB0yieldsA0D0,B0D0.
Therefore,ypD
x
2
2
.cos2xsin2x/.
5.5.16.LetyDue
x
. Then
y
00
2y
0
CyDe
x
fi
.u
00
C2u
0
Cu/2.u
0
Cu/Cu

De
x
u
00
D e
x
fi
.3C4xx
2
/cosxC.34xx
2
/sinx

ifu
00
D .3C4xx
2
/cosx.34xx
2
/sinx. Now let
upD.A0CA1xCA2x
2
/cosxC.B0CB1xCB2x
2
/sinxIthen
u
0
p
D
fi
A1CB0C.2A2CB1/xCB2x
2

cosx
C
fi
B1A0C.2B2A1/xA2x
2

sinx;
u
00
p
D
fi
A0C2A2C2B1.A14B2/xA2x
2

cosx
C
fi
B0C2B22A1.B1C4A2/xB2x
2

sinx
D .3C4xx
2
/cosx.34xx
2
/sinxif
(i)
A2D1
B2D1
;(ii)
A1C4B2D 4
B14A2D 4
;
(iii)
A0C2B1C2A2D 3
B02A1C2B2D 3
:
From (i),A2D 1,B2D 1. Substituting these into (ii) and solving forA1andB1yieldsA1D0,
B1D0. Substituting the known coefficients into (iii) and solvingforA0andB0yieldsA0D1,B0D1.
Therefore,upD.1x
2
/.cosxCsinx/andypDe
x
.1x
2
/.cosxCsinx/.
5.5.18.LetyDue
x
. Then
y
00
C2y
0
CyDe
x
fi
.u
00
2u
0
Cu/C2.u
0
u/Cu

De
x
u
00
De
x
Œ.52x/cosx.3C3x/sinx
ifu
00
D.52x/cosx.3C3x/sinx. Let
upD.A0CA1x/cosxC.B0CB1x/sinxIthen
u
0
p
D.A1CB0CB1x/cosxC.B1A0A1x/sinx
u
00
p
D.2B1A0A1x/cosx.2A1CB0CB1x/sinx
D.52x/cosx.3C3x/sinx
ifA1D 2,B1D 3,A0C2B1D5,B02A1=-3. Therefore,A1D2,B1D3,A0D1,
B0D 1,upDe
x
Œ.1C2x/cosx.13x/sinx, andypDe
x
Œ.1C2x/cosx.13x/sinx.

73. 68 Chapter 5Linear Second Order Equations
5.5.20.Let
ypD.A0xCA1x
2
CA2x
3
/cosxC.B0xCB1x
2
CB2x
3
/sinxIthen
y
0
p
D
fi
A0C.2A1CB0/xC.3A2CB1/x
2
CB2x
3

cosx
C
fi
B0C.2B1A0/xC.3B2A1/x
2
A2x
3

sinx
y
00
p
D
fi
2A1C2B0.A06A24B1/x.A16B2/x
2
A2x
3

cosx
C
fi
2B12A0.B0C6B2C4A1/x.B1C6A2/x
2
B2x
3

sinx;so
y
00
p
CypD
fi
2A1C2B0C.6A2C4B1/xC6B2x
2

cosx
C
fi
2B12A0C.6B24A1/x6A2x
2

sinx
D.2C2x/cosxC.4C6x
2
/sinxif
(i)
6B2D0
6A2D6
;(ii)
4B1C6A2D2
4A1C6B2D0
;(iii)
2B0C2A1D2
2A0C2B1D4
:
From (i),A2D 1,B2D0. Substituting these into (ii) and solving forA1andB1yieldsA1D0,
B1D2. Substituting the known coefficients into (iii) and solvingforA0andB0yieldsA0D0,B0D2.
Therefore,ypD x
3
cosxC.xC2x
2
/sinx.
5.5.22.LetyDue
x
. Then
y
00
7y
0
C6yDe
x
fi
.u
00
C2u
0
Cu/7.u
0
Cu/C6u

De
x
.u
00
5u
0
/D e
x
.17cosx7sinx/
ifu
00
5u
0
D 17cosxC7sinx. Now letupDAcosxCBsinx. Then
u
00
p
5u
0
p
D .AcosxCBsinx/5.AsinxCBcosx/
D.A5B/cosx.B5A/sinxD 17cosxC7sinx
ifA5BD 17,5ABD7. Therefore,AD2,BD3,upD2cosxC3sinx, andypD
e
x
.2cosxC3sinx/. The characteristic polynomial of the complementary equation isp.r/Dr
2

7rC6D.r1/.r6/, sofe
x
; e
6x
gis a fundamental set of solutions of the complementary equation.
Therefore, (A)yDe
x
.2cosxC3sinx/Cc1e
x
Cc2e
6x
is the general solution of the nonhomogeneous
equation. Differentiating (A) yieldsy
0
De
x
.2cosxC3sinx/Ce
x
.2sinxC3cosx/Cc1e
x
C6c2e
6x
,
soy.0/D4; y
0
.0/D2)4D2Cc1Cc2; 2D2C3Cc1C6c2)c1Cc2D2; c1C6c2D 3,
soc1D3,c2D 1, andyDe
x
.2cosxC3sinx/C3e
x
e
6x
.
5.5.24.LetyDue
x
. Then
y
00
C6y
0
C10yDe
x
fi
.u
00
C2u
0
Cu/C6.u
0
Cu/C10u

De
x
.u
00
C8u
0
C17u/D 40e
x
sinx
ifu
00
C8u
0
C17uD 40sinx. LetupDAcosxCBsinx. Then
u
00
p
C6u
0
p
C17upD .AcosxCBsinx/C8.AsinxCBcosx/
C17.AcosxCBsinx/
D.16AC8B/cosx.8A16B/sinxD 40sinx

74. Section 5.5The Method of Undetermined Coefficients II69
if16AC8BD0,8AC16BD 40. Therefore,AD1,BD 2, andypDe
x
.cosx2sinx/.
The characteristic polynomial of the complementary equation isp.r/Dr
2
C6rC10D.rC3/
2
C1,
sofe
3x
cosx; e
3x
sinxgis a fundamental set of solutions of the complementary equation, and (A)
yDe
x
.cosx2sinx/Ce
3x
.c1cosxCc2sinx/is the general solution of the nonhomogeneous
equation. Therefore,y.0/D2)2D1Cc1, soc1D1. Differentiating (A) yieldsy
0
De
x
.cosx
2sinx/e
x
.sinxC2cosx/3e
3x
.c1cosxCc2sinx/Ce
3x
.c1sinxCc2cosx/. Therefore,y
0
.0/D
3) 3D123c1Cc2, soc2D1, andyDe
x
.cosx2sinx/Ce
3x
.cosxCsinx/.
5.5.26.LetyDue
3x
. Then
y
00
3y
0
C2yDe
3x
fi
.u
00
C6u
0
C9u/3.u
0
C3u/C2u

De
3x
.u
00
C3u
0
C2u/De
3x
Œ21cosx.11C10x/sinx
ifu
00
C3u
0
C2uD21cosx.11C10x/sinx. Now let
upD.A0CA1x/cosxC.B0CB1x/sinxIthen
u
0
p
D.A1CB0CB1x/cosxC.B1A0A1x/sinx
u
00
p
D.2B1A0A1x/cosx.2A1CB0CB1x/sinx;so
u
00
C3u
0
C2uDŒA0C3A1C3B0C2B1C.A1C3B1/xcosx
CŒB0C3B13A02A1C.B13A1/xsinx
D21cosx.11C10x/sinxif
A1C3B1D 0
3A1CB1D 10
and
A0C3B0C3A1C2B1D 21
3A0CB02A1C3B1D 11
:
From the first two equationsA1D3,B1D 1. Substituting these in last two equations yields and
solving forA0andB0yieldsA0D2,B0D4. Therefore,upD.2C3x/cosxC.4x/sinxand
ypDe
3x
Œ.2C3x/cosxC.4x/sinx. The characteristic polynomial of the complementary equation
isp.r/Dr
2
3rC2D.r1/.r2/, sofe
x
; e
2x
gis a fundamental set of solutions of the complementary
equation, and (A)yDe
3x
Œ.2C3x/cosxC.4x/sinxCc1e
x
Cc2e
2x
is the general solution of the
nonhomogeneous equation. Differentiating (A) yields
y
0
D3e
3x
Œ.2C3x/cosxC.4x/sinx
Ce
3x
Œ.7x/cosx.3C3x/sinxCc1e
x
C2c2e
2x
:
Therefore,y.0/D0; y
0
.0/D6)0D2Cc1Cc2; 6D6C7Cc1C2c2, soc1Cc2D 2; c1C2c2D
7. Therefore,c1D3,c2D 5, andyDe
3x
Œ.2C3x/cosxC.4x/sinxC3e
x
5e
2x
.
5.5.28.We must find particular solutionsyp1
,yp2
, andyp3
of (A)y
00
CyD4cosx2sinxand (B)
y
00
CyDxe
x
, and (C)y
00
CyDe
x
, respectively. To find a particular solution of (A) we write
yp1
DAxcosxCBxsinxIthen
y
0
p1
D.ACBx/cosxC.BAx/sinx
y
00
p1
D.2BAx/cosx.2ACBx/sinx;so
y
00
p1
Cyp1
D2Bcosx2AsinxD4cosx2sinxif2BD4,2AD 2. Therefore,AD1,BD2,
andyp1
Dx.cosxC2sinx/. To find a particular solution of (B) we writeyDue
x
. Then
y
00
CyDe
x
fi
.u
00
C2u
0
Cu/Cu

De
x
.u
00
C2u
0
C2u/Dxe
x

75. 70 Chapter 5Linear Second Order Equations
ifu
00
C2u
0
C2uDx. NowupDACBx, where2BC2.ACBx/Dx. Therefore,2BD1,
2AC2BD0, soBD
1
2
,AD
1
2
,upD
1
2
.1x/, andyp2
D
e
x
2
.1x/. To find a particular
solution of (C) we writeyp3
DAe
x
. Theny
00
p3
Cyp3
D2Ae
x
De
x
if2AD1, soAD
1
2
and
yp3
D
e
x
2
NowypDyp1
Cyp2
Cyp3
Dx.cosxC2sinx/
e
x
2
.1x/C
e
x
2
.
5.5.30.We must find particular solutionsyp1
,yp2
andyp3
of (A)y
00
2y
0
C2yD4xe
x
cosx, (B)
y
00
2y
0
C2yDxe
x
, and (C)y
00
2y
0
C2yD1Cx
2
, respectively. To find a particular solution of
(A) we writeyDue
x
. Theny
00
2y
0
C2yDe
x
Œ.u
00
C2u
0
Cu/2.u
0
Cu/C2uDe
x
.u
00
Cu/D
4xe
x
cosxifu
00
CuD4xcosx. Now let
upD.A0xCA1x
2
/cosxC.B0xCB1x
2
/sinxIthen
u
0
p
D
fi
A0C.2A1CB0/xCB1x
2

cosxC
fi
B0C.2B1A0/xB1x
2

sinx
u
00
p
D
fi
2A1C2B0.A04B1/xA1x
2

cosx
C
fi
2B12A0.B0C4A1/xB1x
2

sinx;so
u
00
p
CupD.2A1C2B0C4B1x/cosxC.2B12A04A1x/sinx
D4xcosx
if4B1D4,4A1D0,2B0C2A1D0,2A0C2B1D0. Therefore,A1D0,B1D1,A0D1,
B0D0,upDx.cosxCxsinx/, andyp1
Dxe
x
.cosxCxsinx/. To find a particular solution of (B)
we writeyDue
x
. Then
y
00
2y
0
C2yDe
x
fi
.u
00
2u
0
Cu/2.u
0
u/C2u

De
x
.u
00
4u
0
C5u/Dxe
x
ifu
00
4u
0
C5uDx. NowupDACBxwhere4BC5.ACBx/Dx. Therefore,5BD1,
5A4BD0,BD
1
5
,AD
4
25
,upD
1
25
.4C5x/, andyp2
D
e
x
25
.4C5x/. To find a particular
solution of (C) we writeyp3
DACBxCCx
2
. Then
y
00
p3
2y
0
p3
C2yp3
D2C2.BC2Cx/C2.ACBxCCx
2
/
D.2A2BC2C /C.2B4C /xC2Cx
2
D1Cx
2
if2A2BC2CD1,2B4CD0,2CD1. Therefore,CD
1
2
,BD1,AD1, andyp3
D1CxC
x
2
2
.
NowypDyp1
Cyp2
Cyp3
Dxe
x
.cosxCxsinx/C
e
x
25
.4C5x/C1CxC
x
2
2
.
5.5.32.We must find particular solutionsyp1
andyp2
of (A)y
00
4y
0
C4yD6e
2x
and (B)y
00
4y
0
C4yD
25sinx, respectively. To find a particular solution of (A), letyDue
2x
. Then
y
00
4y
0
C4yDe
2x
fi
.u
00
C4u
0
C4u/4.u
0
C2u/C4u

De
2x
u
00
D6e
2x
ifu
00
D6. Integrating twice and taking the constants of integrationto be zero yieldsupD3x
2
, so
yp1
D3x
2
e
2x
. To find a particular solution of (B), letyp2
DAcosxCBsinx. Then
y
00
p2
4y
0
p2
C4yp2
D .AcosxCBsinx/4.AsinxCBcosx/
C4.AcosxCBsinx/
D.3A4B/cosxC.4AC3B/sinxD25sinx

76. Section 5.5The Method of Undetermined Coefficients II71
if3A4BD0,4AC3BD25. Therefore,AD4,BD3, andyp2
D4cosxC3sinx. Now
ypDyp1
Cyp2
D3x
2
e
2x
C4cosxC3sinx. The characteristic polynomial of the complementary
equation isp.r/Dr
2
4rC4D.r2/
2
, sofe
2x
; xe
2x
gis a fundamental set of solutions of the
complementary equation. Therefore,(C)yD3x
2
e
2x
C4cosxC3sinxCe
2x
.c1Cc2x/is the general
solution of the nonhomogeneous equation. Nowy.0/D5)5D4Cc1, soc1D1. Differentiating (C)
yieldsy
0
D6e
2x
.xCx
2
/4sinxC3cosxC2e
2x
.c1Cc2x/Cc2e
2x
, soy
0
.0/D3)3D3C2Cc2.
Therefore,c2D 2, andyD.12xC3x
2
/e
2x
C4cosxC3sinx.
5.5.34.We must find particular solutionsyp1
andyp2
of (A)y
00
C4y
0
C4yD2cos2xC3sin2x
and (B)y
00
C4y
0
C4yDe
x
, respectively. To find a particular solution of (A) we writeyp1
D
Acos2xCBsin2x. Then
y
00
p1
C4y
0
p1
C4yp1
D 4.Acos2xCBsin2x/C8.Asin2xCBcos2x/
C4.Acos2xCBsin2x/D 8Asin2xC8Bcos2x
D2cos2xC3sin2x
if8BD2,8AD3. Therefore,AD
3
8
,BD
1
4
, andyp1
D
3
8
cos2xC
1
4
sin2x. To find a
particular solution of (B) we writeyp2
DAe
x
. Theny
00
p2
C4y
0
p2
C4yp2
DA.14C4/e
x
D
Ae
x
De
x
ifAD1. Therefore,yp2
De
x
. NowypDyp1
Cyp2
D
3
8
cos2xC
1
4
sin2xCe
x
.
The characteristic polynomial of the complementary equation isp.r/Dr
2
C4rC4D.r2/
2
,
sofe
2x
; xe
2x
gis a fundamental set of solutions of the complementary equation. Therefore,(C)yD

3
8
cos2xC
1
4
sin2xCe
x
Ce
2x
.c1Cc2x/is the general solution of the nonhomogeneous equation.
Nowy.0/D 1) 1D
3
8
C1Cc1, soc1D
13
8
. Differentiating (C) yieldsy
0
D
3
4
sin2xC
1
2
cos2xe
x
2e
2x
.c1Cc2x/Cc2e
2x
, soy
0
.0/D2)2D
1
2
12c1Cc2. Therefore,c2D
3
4
,
andyD
3
8
cos2xC
1
4
sin2xCe
x

13
8
e
2x

3
4
xe
2x
.
5.5.36.(a),(b), and(c)require only routine manipulations.(d)The coefficients of sin!xiny
0
p
,y
00
p
,
ay
00
p
Cby
0
p
Ccyp, andy
00
p
C!
2
ypcan be obtained by replacingAbyBandBbyAin the corresponding
coefficients of cos!x.
5.5.38.LetyDue
x
. Then
ay
00
Cby
0
CcyDe
x
fi
a.u
00
C2u
0
C
2
u/Cb.u
0
Cu/Ccu

De
x
fi
au
00
C.2aCb/u
0
C.a
2
CbCc/u

De
x
fi
au
00
Cp
0
./u
0
Cp./u

De
x
.P.x/cos!xCQ.x/sin!x/if
(A)au
00
Cp
0
./u
0
Cp./uP.x/cos!xCQ.x/sin!x, wherep.r/Dar
C
brCcis that characteristic
polynomial of the complementary equation (B)ay
00
Cby
0
CcyD0. Ife
x
cos!xande
x
sin!xare
not solutions of (B), then cos!xand sin!xare not solutions of the complementary equation for (A).
Then Theorem 5.5.1 implies that (A) has a particular solution
upD.A0CA1xC CAkx
k
/cos!xC.B0CB1xC CBkx
k
/sin!x;
andypDupe
x
is a particular solution of the stated form for the given equation. Ife
x
cos!xand
e
x
sin!xare solutions of (B), then cos!xand sin!xare solutions of the complementary equation for

77. 72 Chapter 5Linear Second Order Equations
(A). Then Theorem 5.5.1 implies that (A) has a particular solution
upD.A0xCA1x
2
C CAkx
kC1
/cos!xC.B0xCB1x
2
C CBkx
kC1
/sin!x;
andypDupe
x
is a particular solution of the stated form for the given equation.
5.5.40.(a)LetyD
R
x
2
cosx dx; theny
0
Dx
2
cosxNow let
ypD.A0CA1xCA2x
2
/cosxC.B0CB1xCB2x
2
/sinxIthen
y
0
p
D
fi
A1CB0C.2A2CB1/xCB2x
2

cosx
C
fi
B1A0C.2B2A1/xA2x
2

sinxDx
2
cosxif
(i)
B2D1
A2D0
;(ii)
B1C2A2D0
A1C2B2D0
;(iii)
B0CA1D0
A0CB1D0
:
Solving these equations yieldsA2D0,B2D1,A1D2,B1D0,A0D0,B0D 2. Therefore,ypD
2xcosx.2x
2
/sinxandyD2xcosx.2x
2
/sinxCc.
(b)LetyD
R
x
2
e
x
cosx dxDue
x
; theny
0
D.u
0
Cu/e
x
Dx
2
e
x
cosxifu
0
CuDx
2
cosx. Now
let
upD.A0CA1xCA2x
2
/cosxC.B0CB1xCB2x
2
/sinxIthen
u
0
p
D
fi
A1CB0C.2A2CB1/xCB2x
2

cosx
C
fi
B1A0C.2B2A1/xA2x
2

sinx;so
u
00
p
CupD
fi
A0CA1CB0C.A1C2A2CB1/xC.A2CB2/x
2

cosx
C
fi
B0CB1A0C.B1C2B2A1/xC.B2A2/x
2

sinx
Dx
2
cosxif
(i)
A2CB2D1
A2CB2D0
;(ii)
A1CB1C2A2D0
A1CB1C2B2D0
;
(iii)
A0CB0CA1D0
A0CB0CB1D0
:
From (i),A2D
1
2
,B2D
1
2
. Substituting these into (ii) and solving forA1andB1yieldsA1D0,
B1D 1. Substituting these into (iii) and solving forA0andB0yieldsA0D
1
2
,B0D
1
2
. Therefore,
upD
1
2
fi
.1x
2
/cosx.1x/
2
sinx

andyD
e
x
2
fi
.1x
2
/cosx.1x/
2
sinx

.
(c)LetyD
R
xe
x
sin2x dxDue
x
; theny
0
D.u
0
u/e
x
Dxe
x
sin2xifu
0
uDxsin2x.
Now let
upD.A0CA1x/cos2xC.B0CB1x/sin2xIthen
u
0
p
DŒ.A1C2B0/C2B1xcos2xCŒ.B12A0/2A1xsin2x;so
u
00
p
upDŒA0CA1C2B0.A12B1/xcos2x
CŒB0CB12A0.B1C2A1/xsin2xDxsin2xif

78. Section 5.5The Method of Undetermined Coefficients II73
(i)
A1C2B1D0
2A1B1D1
;(ii)
A0C2B0CA1D0
2A0B0CB1D0
:
From (i),A1D
2
5
,B1D
1
5
. Substituting these into (ii) and solving forA0andB0yieldsA0D
4
25
,
B0D
3
25
. Therefore,
upD
1
25
Œ.4C10x/cos2x.35x/sin2xCcand
ypD
e
x
25
Œ.4C10x/cos2x.35x/sin2xCc:
(d)LetyD
R
x
2
e
x
sinx dxDue
x
; theny
0
D.u
0
u/e
x
Dx
2
e
x
sinxifu
0
uDx
2
sinx.
Now let
upD.A0CA1xCA2x
2
/cosxC.B0CB1xCB2x
2
/sinxIthen
u
0
p
D
fi
A1CB0C.2A2CB1/xCB2x
2

cosx
C
fi
B1A0C.2B2A1/xA2x
2

sinx;so
u
00
p
upD
fi
A0CA1CB0.A12A2B1/x.A2B2/x
2

cosx
C
fi
B0CB1A0.B12B2CA1/x.B2CA2/x
2

sinx
Dx
2
sinxif
(i)
A2CB2D0
A2B2D1
;(ii)
A1CB1C2A2D0
A1B1C2B2D0
;
(iii)
A0CB0CA1D0
A0B0CB1D0
:
From (i),A2D
1
2
,B2D
1
2
. Substituting these into (ii) and solving forA1andB1yieldsA1D 1,
B1D0. Substituting these into (iii) and solving forA0andB0yieldsA0D
1
2
,B0D
1
2
. Therefore,
upD
e
x
2
fi
.1Cx/
2
cosx.1x
2
/sinx

and
yD
e
x
2
fi
.1Cx/
2
cosx.1x
2
/sinx

Cc:
(e)LetyD
R
x
3
e
x
sinx dxDue
x
; theny
0
D.u
0
Cu/e
x
Dx
3
e
x
sinxifu
0
CuDx
3
sinx. Now let
upD.A0CA1xCA2x
2
CA3x
3
/cosxC.B0CB1xCB2x
2
CB3x
3
/sinxIthen
u
0
p
D
fi
A1CB0C.2A2CB1/xC.3A3CB2/x
2
CB3x
3

cosx
C
fi
B1A0C.2B2A1/xC.3B3A2/x
2
A3x
3

sinx;so
u
00
p
CupDŒA0CA1CB0C.A1C2A2CB1/x
C.A2C3A3CB2/x
2
C.A3CB3/x
3

cosx
CŒB0CB1A0C.B1C2B2A1/x
C.B2C3B3A2/x
2
C.B3A3/x
3

sinxDx
3
sinxif

79. 74 Chapter 5Linear Second Order Equations
(i)
A3CB3D0
A3CB3D1
;(ii)
A2CB2C3A3D0
A2CB2C3B3D0
;
(iii)
A1CB1C2A2D0
A1CB1C2B2D0
;(iv)
A0CB0CA1D0
A0CB0CB1D0
:
From (i),A3D
1
2
,B3D
1
2
. Substituting these into (ii) and solving forA2andB2yieldsA2D
3
2
,
B2D0. Substituting these into (iii) and solving forA1andB1yieldsA1D
3
2
,B1D
3
2
. Substituting
these into (iv) and solving forA0andB0yieldsA0D0,B0D
3
2
. Therefore,
upD
1
2
fi
x.33xCx
2
/cosx.33xCx
3
/sinx

and
yD
e
x
2
fi
x.33xCx
2
/cosx.33xCx
3
/sinx

Cc:
(f)LetyD
R
e
x
Œxcosx.1C3x/sinx dxDue
x
; theny
0
D.u
0
Cu/e
x
De
x
Œxcosx.1C3x/sinx
ifu
0
CuDxcosx.1C3x/sinx. Now let
upD.A0CA1x/cosxC.B0CB1x/sinxIthen
u
0
p
DŒA1CB0CB1xcosxCŒB1A0A1xsinx;so
u
00
p
CupDŒA0CA1CB0C.A1CB1/xcosx
CŒB0CB1A0C.B1A1/xsinx
Dxcosx.1C3x/sinxif
(i)
A1CB1D 1
A1CB1D 3
;(ii)
A0CB0CA1D 0
A0CB0CB1D 1
:
From (i),A1D2,B1D 1. Substituting these into (ii) and solving forA0andB0yieldsA0D 1,B0D
1. Therefore,upD Œ.12x/cosxC.1Cx/sinxandyD e
x
Œ.12x/cosxC.1Cx/sinxC
c.
(g)LetyD
R
e
x
fi
.1Cx
2
/cosxC.1x
2
/sinx

dxDue
x
; then
y
0
D.u
0
u/e
x
De
x
fi
.1Cx
2
/cosxC.1x
2
/sinx

ifu
0
uD.1Cx
2
/cosxC.1x
2
/sinx. Now let
upD.A0CA1xCA2x
2
/cosxC.B0CB1xCB2x
2
/sinxIthen
u
0
p
D
fi
A1CB0C.2A2CB1/xCB2x
2

cosx
C
fi
B1A0C.2B2A1/xA2x
2

sinx;so
u
00
p
upD
fi
A0CA1CB0.A12A2B1/x.A2B2/x
2

cosx
C
fi
B0CB1A0.B12B2CA1/x.B2CA2/x
2

sinx
D.1Cx
2
/cosxC.1x
2
/sinxif
(i)
A2CB2D 1
A2B2D 1
;(ii)
A1CB1C2A2D0
A1B1C2B2D0
;

80. Section 5.6Reduction of Order75
(iii)
A0CB0CA1D1
A0B0CB1D1
:
From (i),A2D0,B2D1. Substituting these into (ii) and solving forA1andB1yieldsA1D1,
B1D1. Substituting these into (iii) and solving forA0andB0yieldsA0D0,B0D0. Therefore,upD
xcosxCx.1Cx/sinxandyDe
x
ŒxcosxCx.1Cx/sinxCc.
5.6REDUCTION OF ORDER
(NOTE: The termuy
00
1
is indicated by “ " in some of the following solutions, wherey
00
1
is complicated.
Since this term always drops out of the differential equation foru, it is not necessary to include it.)
5.6.2.IfyDux, theny
0
Du
0
xCuandy
00
Du
00
xC2u
0
, sox
2
y
00
Cxy
0
yDx
3
u
00
C3x
2
u
0
D
4
x
2
ifu
0
D´, where (A)´
0
C
3
x
´D
4
x
5
. Since
Z
3
x
dxD3lnjxj,´1D
1
x
3
is a solution of the
complementary equation for (A). Therefore,the solutions of (A) are of the form (B)´D
v
x
3
, where
v
0
x
3
D
4
x
5
, sov
0
D
4
x
2
. Hence,vD
4
x
CC1;u
0
D´D
4
x
4
C
C1
x
3
(see (B));uD
4
3x
3

C1
2x
2
CC2;
yDuxD
4
3x
2

C1
2x
CC2x, oryD
4
3x
2
Cc1xC
c2
x
. As a byproduct,fx; 1=xgis a fundamental set
of solutions of the complementary equation.
5.6.4.IfyDue
2x
, theny
0
D.u
0
C2u/e
2x
andy
00
D.u
00
C4u
0
C4u/e
2x
, soy
00
3y
0
C2yD
.u
00
Cu
0
/e
2x
D
1
1Ce
x
ifu
0
D´, where (A)´
0
C´D
e
2x
1Ce
x
. Since´1De
x
is a solution of
the complementary equation for (A), the solutions of (A) areof the form (B)´Dve
x
, wherev
0
e
x
D
e
2x
1Ce
x
, sov
0
D
e
x
1Ce
x
. Hence,vD ln.1Ce
x
/CC1;u
0
D´D e
x
ln.1Ce
x
/CC1e
x
(see (B));uD.1Ce
x
/ln.1Ce
x
/1e
x
C1e
x
CC2;yDue
2x
D.e
2x
Ce
x
/ln.1Ce
x
/
.C1C1/e
x
C.C21/e
2x
, oryD.e
2x
Ce
x
/ln.1Ce
x
/Cc1e
2x
Cc2e
x
. As a byproduct,fe
2x
; e
x
g
is a fundamental set of solutions of the complementary equation.
5.6.6.IfyDux
1=2
e
x
, theny
0
Du
0
x
1=2
e
x
Cu
 
x
1=2
C
x
1=2
2
!
e
x
andy
00
Du
00
x
1=2
e
x
C2u
0
 
x
1=2
C
x
1=2
2
!
e
x
C
so4x
2
y
00
C.4x8x
2
/y
0
C.4x
2
4x1/yDe
x
.4x
5=2
u
00
C8x
3=2
u
0
/D4x
1=2
e
x
.1C4x/if
u
0
D´, where (A)´
0
C
2
x
´D
1C4x
x
2
. Since
Z
2
x
dxD2lnjxj,´1D
2
x
2
is a solution of the
complementary equation for (A). Therefore,the solutions of (A) are of the form (B)´D
v
x
2
, where
v
0
x
2
D
1C4x
x
2
, sov
0
D1C4x. Hence,vDxC2x
2
CC1;u
0
D´D
1
x
C2C
C1
x
2
(see
(B));uDlnxC2x
C1
x
CC2;yDux
1=2
e
x
De
x
.2x
3=2
Cx
1=2
lnxC1x
1=2
CC2x
1=2
/, or
yDe
x
.2x
3=2
Cx
1=2
lnxCc1x
1=2
Cc2x
1=2
/. As a byproduct,fx
1=2
e
x
; x
1=2
e
x
gis a fundamental
set of solutions of the complementary equation.
5.6.8.IfyDue
x
2
, theny
0
Du
0
e
x
2
2xue
x
2
andy
00
Du
00
e
x
2
4xu
0
e
x
2
C , soy
00
C4xy
0
C
.4x
2
C2/yDu
00
e
x
2
D8e
x.xC2/
D8e
x
2
e
2x
ifu
00
D8e
2x
. Therefore,u
0
D 4e
2x
CC1;
uD2e
2x
CC1xCC2, andyDue
x
2
De
x
2
.2e
2x
CC1xCC2/, oryDe
x
2
.2e
2x
Cc1Cc2x/.
As a byproduct,fe
x
2
; xe
x
2
gis a fundamental set of solutions of the complementary equation.

81. 76 Chapter 5Linear Second Order Equations
5.6.10.IfyDuxe
x
, theny
0
Du
0
xe
x
ue
x
.x1/andy
00
Du
00
xe
x
2u
0
e
x
.x1/C ,
sox
2
y
00
C2x.x1/y
0
C.x
2
2xC2/yDx
3
u
00
Dx
3
e
2x
ifu
00
De
3x
. Therefore,u
0
D
e
3x
3
CC1;
uD
e
3x
9
CC1xCC2, andyDuxe
x
D
xe
2x
9
Cxe
x
.C1xCC2/, oryD
xe
2x
9
Cxe
x
.c1Cc2x/.
As a byproduct,fxe
x
; x
2
e
x
gis a fundamental set of solutions of the complementary equation.
5.6.12.IfyDue
x
, theny
0
D.u
0
Cu/e
x
andy
00
D.u
00
C2u
0
Cu/e
x
, so.12x/y
00
C2y
0
C
.2x3/yDe
x
Œ.12x/u
00
C.44x/u
0
D.14xC4x
2
/e
x
ifu
0
D´, where (A)´
0
C
44x
12x
´D
12x. Since
Z
44x
12x
dxD
Z
2C
2
12x

dxD2xlnj12xj,´1D.12x/e
2x
is
a solution of the complementary equation for (A). Therefore,the solutions of (A) are of the form (B)
´Dv.12x/e
2x
, wherev
0
.12x/e
2x
D.12x/, sov
0
De
2x
. Hence,vD
e
2x
2
CC1;
u
0
D´D

1
2
CC1e
2x

.12x/(see (B));uD
.2x1/
2
8
CC1xe
2x
CC2;yDue
x
D

.2x1/
2
e
x
8
CC1xe
x
CC2e
x
, oryD
.2x1/
2
e
x
8
Cc1e
x
Cc2xe
x
. As a byproduct,fe
x
; xe
x
g
is a fundamental set of solutions of the complementary equation.
5.6.14.IfyDue
x
, theny
0
D.u
0
u/e
x
andy
00
D.u
00
2u
0
Cu/e
x
, so2xy
00
C.4xC
1/y
0
C.2xC1/yDe
x
.2xu
00
Cu
0
/D3x
1=2
e
x
ifu
0
D´, where (A)´
0
C
1
2x
´D
3
2
x
1=2
. Since
Z
1
2x
dxD
1
2
lnjxj,´1Dx
1=2
is a solution of the complementary equation for (A). Therefore,the
solutions of (A) are of the form (B)´Dvx
1=2
, wherev
0
x
1=2
D
3
2
x
1=2
, sov
0
D
3
2
. Hence,
vD
3x
2
CC1;u
0
D´D
3
2
x
1=2
CC1x
1=2
(see (B));uDx
3=2
C2C1x
1=2
CC2;yDue
x
D
e
x
.x
3=2
C2C1x
1=2
CC2/, oryDe
x
.x
3=2
Cc1Cc2x
1=2
/As a byproduct, is afe
x
; x
1=2
e
x
g
fundamental set of solutions of the complementary equation.
5.6.16.IfyDux
1=2
, theny
0
Du
0
x
1=2
C
u
2x
1=2
andy
00
Du
00
x
1=2
C
u
0
x
1=2
C so4x
2
y
00

4x.xC1/y
0
C.2xC3/yD4x
5=2
.u
00
u
0
/D4x
5=2
e
2x
ifu
0
D´, where (A)´
0
´De
2x
. Since
´1De
x
is a solution of the complementary equation for (A), the solutions of (A) are of the form (B)
´Dve
x
, wherev
0
e
x
De
2x
, sov
0
De
x
. Hence,vDe
x
CC1;u
0
D´De
2x
CC1e
x
(see (B));
uD
e
2x
2
CC1e
x
CC2;yDux
1=2
Dx
1=2

e
2x
2
CC1e
x
CC2

, oryDx
1=2

e
2x
2
Cc1Cc2e
x

.
As a byproduct,fx
1=2
; x
1=2
e
x
gis a fundamental set of solutions of the complementary equation.
5.6.18.IfyDue
x
, theny
0
D.u
0
Cu/e
x
andy
00
D.u
00
C2u
0
Cu/e
x
, soxy
00
C.22x/y
0
C
.x2/yDe
x
.xu
00
C2u
0
/D0if
u
00
u
0
D
2
x
; lnju
0
j D 2lnjxj Ck;u
0
D
C1
x
2
;uD
C1
x
CC2.
Therefore,yDue
x
De
x


C1
x
CC2

is the general solution, andfe
x
; e
x
=xgis a fundamental set of
solutions.
5.6.20.IfyDulnjxj, theny
0
Du
0
lnjxj C
u
x
andy
00
Du
00
lnjxj C
2u
0
x
, sox
2
.lnjxj/
2
y
00

.2xlnjxj/y
0
C.2Clnjxj/yDx
2
.lnjxj/
3
u
00
D0ifu
00
D0;u
0
DC1;uDC1xCC2. Therefore,yD
ulnjxj D.C1xCC2/lnjxjis the general solution, andflnjxj; xlnjxjgis a fundamental set of solutions.

82. Section 5.6Reduction of Order77
5.6.22.IfyDue
x
, theny
0
Du
0
e
x
Cue
x
andy
00
Du
00
e
x
C2u
0
e
x
Cue
x
, soxy
00
.2xC2/y
0
C.xC2/yD
e
x
.xu
00
2u
0
/D0if
u
00
u
0
D
2
x
; lnju
0
j D2lnjxj Ck;u
0
DC1x
2
;uD
C1x
3
3
CC2. Therefore,
yDue
x
D

C1x
3
3
CC2

e
x
is the general solution, andfe
x
; x
3
e
x
gis a fundamental set of solutions.
5.6.24.IfyDuxsinx, theny
0
Du
0
xsinxCu.xcosxCsinx/andy
00
Du
00
xsinxC2u
0
.xcosxC
sinx/C , sox
2
y
00
2xy
0
C.x
2
C2/yD.x
3
sinx/u
00
C2.x
3
cosx/u
0
D0if
u
00
u
0
D
2cosx
sinx
; lnju
0
j D
2lnjsinxjCk;u
0
D
C1
sin
2
x
;uD C1cotxCC2. Therefore,yDuxsinxDx.C1cosxCC2sinx/
is the general solution, andfxsinx; xcosxgis a fundamental set of solutions.
5.6.26.IfyDux
1=2
, theny
0
Du
0
x
1=2
C
u
2x
1=2
andy
00
Du
00
x
1=2
C
u
0
x
1=2
C so4x
2
.sinx/y
00

4x.xcosxCsinx/y
0
C.2xcosxC3sinx/yD4x
5=2
.u
00
sinxu
0
cosx/D0if
u
00
u
0
D
cosx
sinx
; lnju
0
j D
lnjsinxj Ck;u
0
DC1sinx;uD C1cosxCC2. Therefore,yDux
1=2
D.C1cosxCC2/x
1=2
is
the general solution, andfx
1=2
; x
1=2
cosxgis a fundamental set of solutions.
5.6.28.IfyD
u
x
, theny
0
D
u
0
x

u
x
2
andy
00
D
u
00
x

2u
0
x
2
C , so.2xC1/xy
00
2.2x
2
1/y
0
4.xC
1/yD.2xC1/u
00
.4xC4/u
0
D0if
u
00
u
0
D
4xC4
2xC1
D2C
2
2xC1
; lnju
0
j D2xClnj2xC1j Ck;
u
0
DC1.2xC1/e
2x
;uDC1xe
2x
CC2. Therefore,yD
u
x
DC1e
2x
C
C2
x
is the general solution, and
f1=x; e
2x
gis a fundamental set of solutions.
5.6.30.IfyDue
2x
, theny
0
D.u
0
C2u/e
2x
andy
00
D.u
00
C4u
0
C4u/e
2x
, soxy
00
.4xC1/y
0
C
.4xC2/yDe
2x
.xu
00
u
0
/D0if
u
00
u
0
D
1
x
; lnju
0
j Dlnjxj Ck;u
0
DC1x;uD
C1x
2
2
CC2.
Therefore,yDue
2x
De
2x

C1x
2
2
CC2

is the general solution, andfe
2x
; x
2
e
2x
gis a fundamental
set of solutions.
5.6.32.IfyDue
2x
, theny
0
D.u
0
C2u/e
2x
andy
00
D.u
00
C4u
0
C4u/e
2x
, so.3x1/y
00

.3xC2/y
0
.6x8/yDe
2x
Œ.3x1/u
00
C.9x6/u
0
D0if
u
00
u
0
D
9x6
3x1
D 3C
3
3x1
.
Therefore,lnju
0
j D 3xClnj3x1jCk, sou
0
DC1.3x1/e
3x
,uD C1xe
3x
CC2. Therefore,the
general solution isyDue
2x
D C1xe
x
CC2e
2x
, or (A)yDc1e
2x
Cc2xe
x
. Nowy.0/D2)
c1D2. Differentiating (A) yieldsy
0
D2c1e
2x
Cc2.e
x
xe
x
/. Nowy
0
.0/D3)3D2c1Cc2, so
c2D 1andyD2e
2x
xe
x
.
5.6.34.IfyDux, theny
0
Du
0
xCuandy
00
Du
00
xC2u
0
, sox
2
y
00
C2xy
0
2yDx
3
u
00
C4x
2
u
0
Dx
2
if
u
0
D´, where (A)´
0
C
4
x
´D
1
x
. Since
Z
4
x
dxD4lnjxj,´1D
1
x
4
is a solution of the complementary
equation for (A). Therefore,the solutions of (A) are of the form (B)´D
v
x
4
, where
v
0
x
4
D
1
x
, sov
0
Dx
3
.
Hence,vD
x
4
4
CC1;u
0
D´D
1
4
C
C1
x
4
(see (B));uD
x
4

C1
3x
3
CC2. Therefore,the general solution is
yDuxD
x
2
4

C1
3x
2
CC2x, or (C)yD
x
2
4
Cc1xC
c2
x
2
. Differentiating (C) yieldsy
0
D
x
2
Cc12
c2
x
3
:

83. 78 Chapter 5Linear Second Order Equations
Nowy.1/D
5
4
; y
0
.1/D
3
2
)c1Cc2D1; c12c2D1, soc1D1,c2D0andyD
x
2
4
Cx.
5.6.36.IfyDuy1, theny
0
Du
0
y1Cuy
0
1
andy
00
Du
00
y1C2u
0
y
0
1
Cuy
00
1
, soy
00
Cp1.x/y
0
Cp2.x/yD
y1u
00
C.2y
0
1
Cp1y1/u
0
D0ifuis any function such that (B)
u
00
u
0
D 2
y
0
1
y1
p1. If lnju
0
.x/j D
2lnjy1.x/j
Z
x
x0
p1.t/ dt, thenusatisfies (B); therefore, if (C)u
0
.x/D
1
y
2
1
.x/
exp


Z
x
x0
p1.s/ ds

,
thenusatisfies (B). Sinceu.x/D
Z
x
x0
1
y
2
1
.t/
exp


Z
t
x0
p1.s/ds

satisfies (C),y2Duy1is a solution
of (A) on.a; b/. Since
y2
y1
Duis nonconstant, Theorem 5.1.6 implies thatfy1; y2gis a fundamental set
of solutions of (A) on.a; b/.
5.6.38.(a)The associated linear equation is (A)´
00
Ck
2
´D0, with characteristic polynomialp.r/D
r
2
Ck
2
. The general solution of (A) is´Dc1coskxCc2sinkx. Since´
0
D kc1sinkxCkc2coskx,
yD
´
0
´
D
kc1sinkxCkc2coskx
c1coskxCc2sinkx
.
(b)The associated linear equation is (A)´
00
3´
0
C2´D0, with characteristic polynomialp.r/D
r
2
3rC2D.r1/.r2/. The general solution of (A) is´Dc1e
x
Cc2e
2x
. Since´
0
Dc1e
x
Cc2e
2x
,
yD
´
0
´
D
c1C2c2e
x
c1Cc2e
x
.
(c)The associated linear equation is (A)´
00
C5´
0
6´D0, with characteristic polynomialp.r/Dr
2
C
5r6D.rC6/.r1/. The general solution of (A) is´Dc1e
6x
Cc2e
x
. Since´
0
D 6c1e
6x
Cc2e
x
,
yD
´
0
´
D
6c1Cc2e
7x
c1Cc2e
7x
.
(d)The associated linear equation is (A)´
00
C8´
0
C7´D0, with characteristic polynomialp.r/D
r
2
C8rC7D.rC7/.rC1/. The general solution of (A) is´Dc1e
7x
Cc2e
x
. Since´
0
D
7c1e
7x
2c2e
x
,yD
´
0
´
D
7c1Cc2e
6x
c1Cc2e
6x
.
(e)The associated linear equation is (A)´
00
C14´
0
C50´D0, with characteristic polynomialp.r/D
r
2
C14rC50D.rC7/
2
C1. The general solution of (A) is´De
7x
.c1cosxCc2sinx/. Since
´
0
D 7e
7x
.c1cosxCc2sinx/Ce
7x
.c1sinxCc2cosx/D .7c1c2/cosx.c1C7c2/sinx,
yD
´
0
´
D
.7c1c2/cosxC.c1C7c2/sinx
c1cosxCc2sinx
.
(f)The given equation is equivalent to (A)y
0
Cy
2

1
6
y
1
6
D0. The associated linear equation is
(B)´
00

1
6
´
0

1
6
´D0, with characteristic polynomialp.r/Dr
2

1
6
r
1
6
D

rC
1
3

r
1
2

.
The general solution of (B) is´Dc1e
x=3
Cc2e
x=2
. Since´
0
D
c1
3
e
x=3
C
c2
2
e
x=2
,yD
´
0
´
D
2c1C3c2e
5x=6
6.c1Cc2e
5x=6
/
.
(g)The given equation is equivalent to (A)y
0
Cy
2

1
3
yC
1
36
D0. The associated linear equation is
(B)´
00

1
3
´
0
C
1
36
´D0, with characteristic polynomialp.r/Dr
2

1
3
rC
1
36
D

r
1
6

2
. The general
solution of (B) is´De
x=6
.c1Cc2x/. Since´
0
D
e
x=6
6
.c1Cc2x/Cc2e
x=6
D
e
x=6
6
.c1Cc2.xC6//,

84. Section 5.7Variation of Parameters79
yD
´
0
´
D
c1Cc2.xC6/
6.c1Cc2x/
.
5.6.40.(a)Suppose that´is a solution of (B) and letyD
´
0
r´
. Then (D)
´
00
r´
C

p.x/
r
0
.x/
r.x/

yCq.x/D
0andy
0
D
´
00
r´

1
r

´
0
´

2

r
0
´
0
r
2
´
D
´
00
r´
ry
2

r
0
r
y, so
´
00
r´
Dy
0
Cry
2
C
r
0
r
y. Therefore, (D)
implies thatysatisfies (A). Now suppose thatyis a solution of (A) and let´be any function such that
´
0
Dry´. Then´
00
Dr
0
y´Cry
0
´Cry´
0
D
r
0
r
´
0
C.y
0
Cry
2
/r´D
r
0
r
´
0
.p.x/yCq.x//r´, so
´
00

r
0
r
´
0
Cp.x/ry´Cq.x/r´D0, which implies that´satisfies (B), sincery´D´
0
.
(b)Iff´1; ´2gis a fundamental set of solutions of (B) on.a; b/, then´Dc1´1Cc2´2is the general
solution of (B) on.a; b/. This and(a)imply that (C) is the general solution of (A) on.a; b/.
5.7VARIATION OF PARAMETERS
5.7.2.(A)ypDu1cos2xCu2sin2x;
u
0
1
cos2xCu
0
2
sin2xD0 (B)
2u
0
1
sin2xC2u
0
2
cos2xDsin2xsec
2
x:(C)
Multiplying (B) by2sin2xand (C) by cos2xand adding the resulting equations yields2u
0
2
Dtan2x,
sou
0
2
D
tan2x
2
. Then (B) implies thatu
0
1
D u
0
2
tan.2x/D
tan
2
2x
2
D
1sec
2
2x
2
. Therefore,u1D
x
2

tan2x
4
andu2D
lnjcos2xj
4
. Now (A) yieldsypD
sin2xlnjcos2xj
4
C
xcos2x
2

sin2x
4
.
Since sin2xsatisfies the complementary equation we redefineypD
sin2xlnjcos2xj
4
C
xcos2x
2
.
5.7.4.(A)ypDu1e
x
cosxCu2e
x
sinx;
u
0
1
e
x
cosxCu
0
2
e
x
sinxD0 (B)
u
0
1
.e
x
cosxe
x
sinx/Cu
0
2
.e
x
sinxCe
x
cosx/D3e
x
secx:(C)
Subtracting (B) from (C) and cancellinge
x
from the resulting equations yields
u
0
1
cosxCu
0
2
sinxD0 (D)
u
0
1
sinxCu
0
2
cosxD3secx:(E)
Multiplying (D) by sinxand (E) by cosxand adding the results yieldsu
0
2
D3. From (D),u
0
1
D
u
0
2
tanxD 3tanx. Thereforeu1D3lnjcosxj,u2D3x. Now (A) yieldsypD3e
x
.cosxlnjcosxjC
xsinx/.
5.7.6.(A)ypDu1e
x
Cu2e
x
;
u
0
1
e
x
Cu
0
2
e
x
D0 (B)
u
0
1
e
x
u
0
2
e
x
D
4e
x
1Ce
2x
:(C)
Adding (B) to (C) yields2u
0
1
e
x
D
4e
x
1Ce
2x
, sou
0
1
D
2e
2x
1e
2x
. From (B),u
0
2
D e
2x
u
0
1
D

2
1e
2x
D
2e
2x
1e
2x
. Using the substitutionvDe
2x
we integrateu
0
1
to obtainu1Dln.1e
2x
/.

85. 80 Chapter 5Linear Second Order Equations
Using the substitutionvDe
2x
we integrateu
0
2
to obtainu1Dln.1e
2x
/. Now (A) yieldsypD
e
x
ln.1e
2x
/e
x
ln.e
2x
1/.
5.7.8.(A)ypDu1e
x
Cu2
e
x
x
;
u
0
1
e
x
Cu
0
2
e
x
x
D0(B)
u
0
1
e
x
Cu
0
2

e
x
x

e
x
x
2

D
e
2x
x
:(C)
Subtracting (B) from (C) yields
u
0
2
e
x
x
2
D
e
2x
x
, sou
0
2
D xe
x
. From (B),u
0
1
D
u
0
2
x
De
x
. Therefore
u1De
x
,u2D xe
x
Ce
x
. Now (A) yieldsypD
e
2x
x
.
5.7.10.(A)ypDu1e
x
2
Cu2xe
x
2
;
u
0
1
e
x
2
Cu
0
2
xe
x
2
D0 (B)
2xu
0
1
e
x
2
Cu
0
2
.e
x
2
2x
2
e
x
2
/D4e
x.xC2/
:(C)
Multiplying (B) by2xand adding the result to (C) yieldsu
0
2
e
x
2
D4e
x.xC2/
, sou
0
2
D4e
2x
. From
(B),u
0
1
D u
0
2
xD 4xe
2x
. Thereforeu1D.2xC1/e
2x
,u2D 2e
2x
. Now (A) yieldsypD
e
x.xC2/
.
5.7.12.(A)ypDu1xCu2x
3
;
u
0
1
xCu
0
2
x
3
D0(B)
u
0
1
C3u
0
2
x
2
D
2x
4
sinx
x
2
D2x
2
sinx (C):
Multiplying (B) by
1
x
and subtracting the result from (C) yields2x
2
u
0
2
D2x
2
sinx, sou
0
2
Dsinx. From
(B),u
0
1
D u
0
2
x
2
D x
2
sinx. Thereforeu1D.x
2
2/cosx2xsinx,u2D cosx. Now (A) yields
ypD 2x
2
sinx2xcosx.
5.7.14.(A)ypDu1cos
p
xCu2sin
p
x;
u
0
1
cos
p
xCu
0
2
sin
p
xD0 (B)
u
0
1
sin
p
x
2
p
x
Cu
0
2
cos
p
x
2
p
x
D
sin
p
x
4x
(C):
Multiplying (B) by
sin
p
x
2
p
x
and (C) by cos
p
xand adding the resulting equations yields
u
0
2
2
p
x
D
sin
p
xcos
p
x
4x
, sou
0
2
D
sin
p
xcos
p
x
2
p
x
. From (B),u
0
1
D u
0
2
tan
p
xD
sin
2
p
x
2
p
x
. Therefore,
u1D
sin
p
xcos
p
x
2

p
x
2
,u2D
sin
2
p
x
2
. Now (A) yieldsypD
sin
p
x
2

p
xcos
p
x
2
. Since
sin
p
xsatisfies the complementary equation we redefineypD
p
xcos
p
x
2
.

86. Section 5.7Variation of Parameters81
5.7.16.(A)ypDu1x
a
Cu2x
a
lnx;
u
0
1
x
a
Cu
0
2
x
a
lnxD0 (B)
au
0
1
x
a1
Cu
0
2
.ax
a1
lnxCx
a1
/D
x
aC1
x
2
Dx
a1
(C):
Multiplying (B) by
a
x
and subtracting the result from (C) yieldsu
0
2
x
a1
Dx
a1
, sou
0
2
D1. From (B),
u
0
1
D u
0
2
lnxD lnx. Therefore,u1Dxlnx,u2Dx. Now (A) yieldsypDx
aC1
.
5.7.18.ypDu1e
x
2
Cu2e
x
2
;
u
0
1
e
x
2
Cu
0
2
e
x
2
D0(B)
2u
0
1
xe
x
2
2u
0
2
xe
x
2
D
8x
5
x
D8x
4
:(B)
Multiplying (B) by2xand adding the result to (C) yields4u
0
1
xe
x
2
D8x
4
, sou
0
1
D2x
3
e
x
2
. From (B),
u
0
2
D u
0
1
e
2x
2
D 2x
3
e
x
2
. Thereforeu1D e
x
2
.x
2
C1/,u2D e
x
2
.x
2
1/. Now (A) yields
ypD 2x
2
.
5.7.20.(A)ypDu1
p
xe
2x
Cu2
p
xe
2x
;
u
0
1
p
xe
2x
Cu
0
2
p
xe
2x
D0 (B)
u
0
1
e
2x

2
p
xC
1
2
p
x

u
0
2
e
2x

2
p
x
1
2
p
x

D
8x
5=2
4x
2
D2
p
x (C):
Multiplying (B) by
1
2x
, subtracting the result from (C), and cancelling common factors from the resulting
equations yields
u
0
1
e
2x
Cu
0
2
e
2x
D0 (D)
u
0
1
e
2x
u
0
2
e
2x
D1:(E)
Adding (D) to (E) yields2u
0
1
e
2x
D1, sou
0
1
D
e
2x
2
. From (D),u
0
2
D u
0
1
e
4x
D
e
2x
2
. Therefore,
u1D
e
2x
4
,u2D
e
2x
4
. Now (A) yieldsypD
p
x
2
.
5.7.22.(A)ypDu1xe
x
Cu2xe
x
;
u
0
1
xe
x
Cu
0
2
xe
x
D0(B)
u
0
1
.xC1/e
x
u
0
2
.x1/e
x
D
3x
4
x
2
D3x
2
(C):
Multiplying (B) by
1
x
, subtracting the resulting equation from (C), and cancelling common factors yields
u
0
1
e
x
Cu
0
2
e
x
D0 (D)
u
0
1
e
x
u
0
2
e
x
D3x:(E)
Adding (D) to (E) yields2u
0
1
e
x
D3x, sou
0
1
D
3xe
x
2
. From (D),u
0
2
D u
0
1
e
2x
D
3xe
x
2
. Therefore
u1D
3e
x
.xC1/
2
,u2D
3e
x
.x1/
2
. Now (A) yieldsypD 3x
2
.

87. 82 Chapter 5Linear Second Order Equations
5.7.24.(A)ypD
u1
x
Cu2x
3
;
u
0
1
x
Cu
0
2
x
3
D0 (B)

u
0
1
x
2
C3u
0
2
x
2
D
x
3=2
x
2
Dx
1=2
:(C)
Multiplying (B) by
1
x
and adding the result to (C) yields4u
0
2
x
2
Dx
1=2
, sou
0
2
D
x
5=2
4
. From (B),
u
0
1
D u
0
2
x
4
D
x
3=2
4
. Thereforeu1D
x
5=2
10
,u2D
x
3=2
6
. Now (A) yieldsypD
4x
3=2
15
.
5.7.26.(A)ypDu1x
2
e
x
Cu2x
3
e
x
;
u
0
1
x
2
e
x
Cu
0
2
x
3
e
x
D0(B)
u
0
1
.x
2
e
x
C2xe
x
/Cu
0
2
.x
3
e
x
C3x
2
e
x
/D
2xe
x
x
2
D
2e
x
x
:(C)
Subtracting (B) from (C) and cancelling common factors in the resulting equations yields
u
0
1
Cu
0
2
xD0 (D)
2u
0
1
xC3u
0
2
x
2
D
2
x
:(E)
Multiplying (D) by2xand subtracting the result from (E) yieldsx
2
u
0
2
D
2
x
, sou
0
2
D
2
x
3
. From (D),
u
0
1
D u
0
2
xD
2
x
2
. Thereforeu1D
2
x
,u2D
1
x
2
. Now (A) yieldsypDxe
x
.
5.7.28.(A)ypDu1xCu2e
x
;
u
0
1
xCu
0
2
e
x
D0 (B)
u
0
1
Cu
0
2
e
x
D
2.x1/
2
e
x
x1
D2.x1/e
x
:(C)
Subtracting (B) from (C) yieldsu
0
1
.1x/D2.x1/e
x
, sou
0
1
D 2e
x
. From (B),u
0
2
D u
0
1
xe
x
D2x.
Therefore,u1D 2e
x
,u2Dx
2
. Now (A) yieldsypDxe
x
.x2/.
5.7.30.(A)ypDu1e
2x
Cu2xe
x
;
u
0
1
e
2x
Cu
0
2
xe
x
D0 (B)
2u
0
1
e
2x
Cu
0
2
.e
x
xe
x
/D
.3x1/
2
e
2x
3x1
D.3x1/e
2x
:(C)
Multiplying (B) by2and subtracting the result from (C) yieldsu
0
2
.13x/e
x
D.3x1/e
2x
, so
u
0
2
D e
3x
. From (B),u
0
1
D u
0
2
xe
3x
Dx. Thereforeu1D
x
2
2
,u2D
e
3x
3
. Now (A) yields
ypD
xe
2x
.3x2/
3
. The general solution of the given equation isyD
xe
2x
.3x2/
3
Cc1e
2x
Cc2xe
x
.
Differentiating this yieldsy
0
D
e
2x
.3x
2
Cx1/
3
C2c1e
2x
Cc2.1x/e
x
. Nowy.0/D1; y
0
.0/D
2)c1D1; 2D
1
3
C2c1Cc2, soc2D
1
3
, andyD
e
2x
.3x
2
2xC6/
6
C
xe
x
3
.

88. Section 5.7Variation of Parameters83
5.7.32.(A)ypDu1.x1/e
x
Cu2.x1/;
u
0
1
.x1/e
x
Cu
0
2
.x1/D0 (B)
u
0
1
xe
x
Cu
0
2
D
.x1/
3
e
x
.x1/
2
D.x1/e
x
:(C)
From (B),u
0
1
D u
0
2
e
x
. Substituting this into (C) yieldsu
0
2
.x1/D.x1/e
x
, sou
0
2
D e
x
,u
0
1
D1.
Thereforeu1Dx,u2De
x
. Now (A) yieldsypDe
x
.x1/
2
. The general solution of the given equation
isyD.x1/
2
e
x
Cc1.x1/e
x
Cc2.x1/. Differentiating this yieldsy
0
D.x
2
1/e
x
Cc1xe
x
Cc2.
Nowy.0/D4; y
0
.0/D 6)4D1c1c2;6D 1Cc2, soc1D2; c2D 5and
yD.x
2
1/e
x
5.x1/.
5.7.34.(A)ypDu1xC
u2
x
2
;
u
0
1
xC
u
0
2
x
2
D0 (B)
u
0
1

2u
0
2
x
3
D
2x
2
x
2
D 2:(C)
Multiplying (B) by
2
x
and adding the result to (C) yields3u
0
1
D 2, sou
0
1
D
2
3
. From (B),u
0
2
D
u
0
1
x
3
D
2x
3
3
. Thereforeu1D
2x
3
,u2D
x
4
6
. Now (A) yieldsypD
x
2
2
. The general solution
of the given equation isyD
x
2
2
Cc1xC
c2
x
2
. Differentiating this yieldsy
0
D xCc1
2c2
x
3
. Now
y.1/D1; y
0
.1/D 1)1D
1
2
Cc1Cc2;1D 1Cc12c2, soc1D1; c2D
1
2
, and
yD
x
2
2
CxC
1
2x
2
.
5.7.36.SinceyDypa1y1a2y2,
P0.x/y
00
CP1.x/y
0
CP2.x/yDP0.x/.ypa1y1a2y2/
00
CP1.x/.ypa1y1a2y2/
0
CP2.x/.ypa1y1a2y2/
D.P0.x/y
00
p
CP1.x/y
0
p
CP2.x/yp/
a1
fi
P0.x/y
00
1
CP1.x/y
0
1
CP2.x/y1

a2
fi
P0.x/y
00
2
CP1.x/y
0
2
CP2.x/y2

DF.x/a10a20DF.x/I
henceyis a particular solution of (A).
5.7.38.(a)ypDu1e
x
Cu2e
x
is a solution of (A) on.a;1/ifu
0
1
e
x
Cu
0
2
e
x
D0andu
0
1
e
x

u
0
2
e
x
Df .x/. Solving these two equations yieldsu
0
1
D
e
x
f
2
,u
0
2
D
e
x
f
2
. The functionsu1.x/D
1
2
Z
x
0
e
t
f .t/ dtandu2.x/D
1
2
Z
x
0
e
t
f .t/ dtsatisfy these conditions. Therefore,
yp.x/D
e
x
2
Z
x
0
e
t
f .t/ dt
e
x
2
Z
x
0
e
t
f .t/ dt
D
1
2
Z
x
0
f .t/

e
.xt /e
.xt /

dtD
Z
x
0
f .t/sinh.xt/ dt:

89. 84 Chapter 5Linear Second Order Equations
is a particular solution ofy
00
yDf .x/. Differentiatingypyields
y
0
p
.x/D
e
x
2
Z
x
0
e
t
f .t/ dtC
e
x
2
e
x
C
e
x
2
Z
x
0
e
t
f .t/ dt
e
x
2
e
x
D
e
x
2
Z
x
0
e
t
f .t/ dtC
e
x
2
Z
x
0
e
t
f .t/ dt
D
1
2
Z
x
0
f .t/

e
.xt /
Ce
.xt /

dtD
Z
x
0
f .t/cosh.xt/ dt:
Sinceyp.x0/Dy
0
p
.x0/D0, the solution of the initial value problem is
yDypCk0coshxCk1sinhx
Dk0coshxCk1sinhxC
Z
x
0
sinh.xt/f .t/ dt:
The derivative of the solution is
y
0
Dy
0
p
Ck0sinhxCk1coshx
Dk0sinhxCk1coshxC
Z
x
0
cosh.xt/f .t/ dt:

90. CHAPTER6
ApplicationsofLinearSecondOrder
Equations
6.1SPRING PROBLEMS I
6.1.2.Since
k
m
D
g
l
D
32
:1
D320the equation of motion is (A)y
00
C320yD0. The general
solution of (A) isyDc1cos8
p
5tCc2sin8
p
5t, soy
0
D8
p
5.c1sin8
p
5tCc2cos8
p
5t/. Now
y.0/D
1
4
)c1D
1
4
andy
0
.0/D 2)c2D
1
4
p
5
, soyD
1
4
cos8
p
5t
1
4
p
5
sin8
p
5tft.
6.1.4.Since
k
m
D
g
l
D
32
:5
D64the equation of motion is (A)y
00
C64yD0. The general solution
of (A) isyDc1cos8tCc2sin8t, soy
0
D8.c1sin8tCc2cos8t/. Nowy.0/D
1
4
)c1D
1
4
and
y
0
.0/D
1
2
)c2D
1
16
, soyD
1
4
cos8t
1
16
sin8tft;RD
p
17
16
ft;!0D8rad/s;TD=4s;
:245rad 14:04
ı
.
6.1.6.SincekD
mg
l
D
.9:8/10
:7
D140, the equation of motion of the 2 kg mass is (A)y
00
C70yD
0. The general solution of (A) isyDc1cos
p
70tCc2sin
p
70t, soy
0
D
p
70.c1sin
p
70tC
c2cos
p
70t/. Nowy.0/D
1
4
)c1
1
4
Dandy
0
.0/D2)c2D
2
p
70
, soyD
1
4
cos
p
70 tC
2
p
70
sin
p
70 t
m;RD
1
4
r
67
35
m;!0D
p
70rad/s;TD2=
p
70s;2:38rad136:28
ı
.
6.1.8.Since
k
m
D
g
l
D
32
1=2
D64the equation of motion is (A)y
00
C64yD0. The general solution
of (A) isyDc1cos8tCc2sin8t, soy
0
D8.c1sin8tCc2cos8t/. Nowy.0/D
1
2
)c1D
1
2
and
y
0
.0/D 3)c2D
3
8
, soyD
1
2
cos8t
3
8
sin8tft.
6.1.10.mD
64
32
D2, so the equation of motion is2y
00
C8yD2sint, or (A)y
00
C4yDsint. Let
ypDAcostCBsint; theny
00
p
D AcostBsint, soy
00
p
C4ypD3AcostC3BsintDsintif
3AD0,3BD1. Therefore,AD0,BD
1
3
, andypD
1
3
sint. The general solution of
85

91. 86 Chapter 6Applications of Linear Second Order Equations
(A) is (B)yD
1
3
sintCc1cos2tCc2sin2t, soy.0/D
1
2
)c1D
1
2
. Differentiating (B) yields
y
0
D
1
3
cost2c1sin2tC2c2cos2t, soy
0
.0/D2)2D
1
3
C2c2, soc2D
5
6
. Therefore,
yD
1
3
sintC
1
2
cos2tC
5
6
sin2tft.
6.1.12.mD
4
32
D
1
8
andkD
mg
l
D4, so the equation of motion is
1
8
y
00
C4yD
1
4
sin8t, or (A)
y
00
C32yD2sin8t. LetypDAcos8tCBsin8t; theny
00
p
D 64Acost64Bsin8t, soy
00
p
C32ypD
32Acos8t32Bsin8tD2sin8tif32AD0,32BD2. Therefore,AD0,BD
1
16
, and
ypD
1
16
sin8t. The general solution of (A) is (B)yD
1
16
sin8tCc1cos4
p
2tCc2sin4
p
2t, so
y.0/D
1
3
)c1D
1
3
. Differentiating (B) yieldsy
0
D
1
2
cos8tC4
p
2.c1sin4
p
2tCc2cos4
p
2t/, so
y
0
.0/D 1) 1D
1
2
C4
p
2c2, soc2D
1
8
p
2
. Therefore,yD
1
16
sin8tC
1
3
cos4
p
2t
1
8
p
2
sin4
p
2tft.
6.1.14.SinceTD
2
!0
D2
r
m
k
the period is proportional to the square root of the mass. Therefore,
doubling the mass mutiplies the period by
p
2; hence the period of the system with the 20 gm mass is
TD4
p
2s.
6.1.16.mD
6
32
D
3
16
andkD
mg
l
D
6
1=3
D18so the equation of motion is
3
16
y
00
C18yD
4sin!t6cos!t, or (A)y
00
C96yD
64
3
sin!t32cos!t. The displacement will be unbounded if
!D
p
96D4
p
6, in which case (A) becomes (B)y
00
C96yD
64
3
sin4
p
6t32cos4
p
6t. Let
ypDAtcos4
p
6tCBtsin4
p
6tIthen
y
0
p
D.AC4
p
6Bt/cos4
p
6tC.B4
p
6At/sin4
p
6t
y
00
p
D.8
p
6B96At/cos4
p
6t.8
p
6AC96Bt/sin4
p
6t;so
y
00
p
C96ypD8
p
6Bcos4
p
6t8
p
6Asin4
p
6tD
64
3
sin4
p
6t32cos4
p
6t
if8
p
6BD 32,8
p
6AD
64
3
. Therefore,AD
8
3
p
6
,BD
4
p
6
, andypD
t
p
6

8
3
cos4
p
6tC4sin4
p
6t

.
The general solution of (B) is
yD
t
p
6

8
3
cos4
p
6tC4sin4
p
6t

Cc1cos4
p
6tCc2sin4
p
6t; . C/
soy.0/D0)c1D0. Differentiating (C) yields
y
0
D

8
3
p
6
cos4
p
6tC
4
p
6
sin4
p
6t

4t


8
3
sin4
p
6tC4cos4
p
6t

C4
p
6.c1sin4
p
6tCc2cos
p
6t/;
soy
0
.0/D0)0D
8
3
p
6
C4
p
6c2, andc2D
1
9
. Therefore,
yD
t
p
6

8
3
cos4
p
6tC4sin4
p
6t

C
1
9
sin4
p
6tft:

92. Section 6.2Spring Problems II87
6.1.18.The equation of motion is (A)y
00
C!
2
0
yD0. The general solution of (A) isyDc1cos!0tC
c2sin!0t. Nowy.0/Dy0)c1Dy0. Sincey
0
D!0.c1sin!0tCc2cos!0t/,y
0
.0/Dv0)c2D
v0
!0
. Therefore,yDy0cos!0tC
v0
!0
sin!0t;
RD
1
!0
p
.!0y0/
2
C.v0/
2
IcosD
y0!0
p
.!0y0/
2
C.v0/
2
IsinD
v0
p
.!0y0/
2
C.v0/
2
:
Discussion 6.1.1In Exercises 19, 20, and 21 we use the fact that in a spring–mass system with massm
and spring constantkthe period of the motion isTD2
r
m
k
. Therefore, if we have two systems with
massesm1andm2and spring constantsk1andk2, then the periods are related by
T2
T1
D
s
m2k1
m1k2
. We
will use this formula in the solutions of these exercises.
6.1.20.Letm2D2m1. Sincek1Dk2,
T2
T1
D
s
2m1
m1
D
p
2, soT2D
p
2T1.
6.1.21.Suppose thatT2D3T1. Sincem1Dm2,
s
k1
k2
D3,k1D9k2.
6.2SPRING PROBLEMS II
6.2.2.SincekD
mg
l
D
16
3:2
D5the equation of motion is
1
2
y
00
Cy
0
C5yD0, or (A)y
00
C2y
0
C10yD0.
The characteristic polynomial of (A) isp.r/Dr
2
C2rC10D.rC1/
2
C9. Therefore,the general solution
of (A) isyDe
t
.c1cos3tCc2sin3t/, soy
0
D yC3e
t
.c1sin3tCc2cos3t/. Nowy.0/D 3
andy
0
.0/D2)c1D 3and2D3C3c2, orc2D
1
3
. Therefore,yD e
t

3cos3tC
1
3
sin3t

ft. The time–varying amplitude is
p
82
3
e
t
ft.
6.2.4.SincekD
mg
l
D
96
3:2
D30the equation of motion is3y
00
C18y
0
C30yD0, or (A)y
00
C6y
0
C
10yD0. The characteristic polynomial of (A) isp.r/Dr
2
C6rC10D.rC3/
2
C1. Therefore,the
general solution of (A) isyDe
3t
.c1costCc2sint/, soy
0
D 3yCe
3t
.c1sintCc2cost/.
Nowy.0/D
5
4
andy
0
.0/D 12)c1D
5
4
and12D
15
4
Cc2, orc2D
63
4
. Therefore,
yD
e
3t
4
.5costC63sint/ft.
6.2.6.SincekD
mg
l
D
8
:32
D25the equation of motion is
1
4
y
00
C
3
2
y
0
C25yD0, or (A)y
00
C6y
0
C
100yD0. The characteristic polynomial of (A) isp.r/Dr
2
C6rC100D.rC3/
2
C91. Therefore,the
general solution of (A) isyDe
3t
.c1cos
p
91tCc2sin
p
91t/, soy
0
D 3yC
p
91e
3t
.c1sin
p
91tC
c2cos
p
91t/. Nowy.0/D
1
2
andy
0
.0/D4)c1D
1
2
and4D
3
2
C
p
91c2, orc2D
11
2
p
91
.
Therefore,yD
1
2
e
3t

cos
p
91tC
11
p
91
sin
p
91t

ft.
6.2.8.SincekD
mg
l
D
20980
5
D3920the equation of motion is20y
00
C400y
0
C3920yD0,
or (A)y
00
C20y
0
C196yD0. The characteristic polynomial of (A) isp.r/Dr
2
C20rC196D

93. 88 Chapter 6Applications of Linear Second Order Equations
.rC10/
2
C96. Therefore,the general solution of (A) isyDe
10t
.c1cos4
p
6tCc2sin4
p
6t/, so
y
0
D 10yC4
p
6e
10t
.c1sin4
p
6tCc2cos4
p
6t/. Nowy.0/D9andy
0
.0/D0)c1D9and
0D 90C4
p
6c2, orc2D
45
2
p
6
. Therefore,yDe
10t

9cos4
p
6tC
45
2
p
6
sin4
p
6t

cm.
6.2.10.SincekD
mg
l
D
32
1
D32the equation of motion is (A)y
00
C3y
0
C32yD0. The characteris-
tic polynomial of (A) isp.r/Dr
2
C3rC32D

rC
3
2

2
C
119
4
. Therefore,the general solution of (A) is
yDe
3t =2
 
c1cos
p
119
2
tCc2sin
p
119
2
t
!
, soy
0
D
3
2
yC
p
119
2
e
3t =2
 
c1sin
p
119
2
tCc2cos
p
119
2
t
!
.
Nowy.0/D
1
2
andy
0
.0/D 3)c1D
1
2
and3D
3
4
C
p
119
2
c2, orc2D
9
2
p
119
. Therefore,
yDe

3
2
t
 
1
2
cos
p
119
2
t
9
2
p
119
sin
p
119
2
t
!
ft.
6.2.12.SincekD
mg
l
D
2
:32
D
25
4
the equation of motion is
1
16
y
00
C
1
8
y
0
C
25
4
yD0, or (A)
y
00
C2y
0
C100yD0. The characteristic polynomial of (A) isp.r/Dr
2
C2rC100D.rC
1/
2
C99. Therefore,the general solution of (A) isyDe
t
.c1cos3
p
11tCc2sin3
p
11t/, soy
0
D
yC3
p
11e
t
.c1sin3
p
11tCc2cos3
p
11t/. Nowy.0/D
1
3
andy
0
.0/D5)c1D
1
3
and
5D
1
3
C3
p
11c2, orc2D
14
9
p
11
. Therefore,yDe
t


1
3
cos3
p
11tC
14
9
p
11
sin3
p
11t

ft.
6.2.14.SincekD
mg
l
D32the equation of motion is (A)y
00
C12y
0
C32yD0. The characteristic
polynomial of (A) isp.r/Dr
2
C12rC32D.rC8/.rC4/. Therefore, the general solution of (A) is
yDc1e
8t
Cc2e
4t
, soy
0
D 8c1e
8t
4c2e
4t
. Nowy.0/D
2
3
andy
0
.0/D0)c1Cc2D
2
3
;-
8c14c2D0, soc1D
2
3
,c2D
4
3
, andyD
2
3
.e
8t
2e
4t
/.
6.2.16.SincekD
mg
l
D
100980
98
D100the equation of motion is100y
00
C600y
0
C1000yD0, or
(A)y
00
C6y
0
C10yD0. The characteristic polynomial of (A) isp.r/Dr
2
C6rC10D.rC3/
2
C1.
Therefore, the general solution of (A) isyDe
3t
.c1costCc2sint/, soy
0
D 3yCe
t
.c1sintC
c2cost/. Nowy.0/D10andy
0
.0/D 100)c1D10and100D 30Cc2, orc2D 70.
Therefore,yDe
3t
.10cost70sint/cm.
6.2.18.The equation of motion is (A)2y
00
C4y
0
C20yD3cos4t5sin4t. The steady state component
of the solution of (A) is of the formypDAcos4tCBsin4t; thereforey
0
p
D 4Asin4tC4Bcos4tand
y
00
p
D 16Acos4t16Bsin4t, so2y
00
p
C4y
0
p
C20ypD.12AC16B/cos4t.16AC12B/sin4tD
3cos4t5sin4tif12AC16BD3,16A12BD 5; thereforeAD
11
100
,BD
27
100
, and
ypD
11
100
cos4tC
27
100
sin4tcm.
6.2.20.SincekD
mg
l
D
9:8
:49
D20the equation of motion is (A)y
00
C4y
0
C20yD8sin2t6cos2t.
The steady state component of the solution of (A) is of the formypDAcos2tCBsin2t; therefore

94. Section 6.3The RLC Circuit89
y
0
p
D 2Asin2tC2Bcos2tandy
00
p
D 4Acos2t4Bsin2t, soy
00
p
C4y
0
p
C20ypD.16AC
8B/cos2t.8A16B/sin2tD8sin2t6cos2tif16AC8BD 6,8AC16BD8; therefore
AD
1
2
,BD
1
4
, andyD
1
2
cos2tC
1
4
sin2tm.
6.2.22.Ife
r1t
.c1Cc2t/D0, then (A)c1Cc2tD0. Ifc2D0, thenc1¤0(by assumption), so (A) is
impossible. Ifc1¤0, then the left side of (A) is strictly monotonic and therefore cannot have the same
value for two distinct values oft.
6.2.24.IfyDe
ct =2m
.c1cos!1tCc2sin!1t/, theny
0
D
c
2m
yC!1e
ct =2m
.c1sin!1tC
c2cos!1t/, soy.0/Dy0andy
0
.0/Dv0)c1Dy0andv0D
cy0
2m
Cc2!1, soc2D
1
!1

v0C
cy0
2m
t

andyDe
ct =2m

y0cos!1tC
1
!1

v0C
cy0
2m

sin!1t

.
6.2.26.IfyDe
r1t
.c1Cc2t/, theny
0
Dr1yCc2e
r1t
, soy.0/Dy0andy
0
.0/Dv0)c1Dy0and
v0Dr1y0Cc2, soc2Dv0r1y0. Therefore,yDe
r1t
.y0C.v0r1y0/t/.
6.3THE RLC CIRCUIT
6.3.2.
1
20
Q
00
C2Q
0
C100QD0;Q
00
C40Q
0
C2000QD0;r
2
C40rC2000D.rC20/
2
C
1600D0;rD 20˙40i;QDe
20t
.2cos40tCc2sin40t/(sinceQ0D2);IDQ
0
D
e
20t
..40c240/cos40t.20c2C80/sin40t/;I0D2)40c240D2)c2D
21
20
, so
20c2C80D101;IDe
20t
.2cos40t101sin40t/.
6.3.4.
1
10
Q
00
C6Q
0
C250QD0;Q
00
C60Q
0
C2500QD0;r
2
C60rC2500D.rC30/
2
C
1600D0;rD 30˙40i;QDe
30t
.3cos40tCc2sin40t/(sinceQ0D3);IDQ
0
D
e
30t
..40c290/cos40t.30c2C120/sin40t/;I0D 10)40c290D 10)c2D2, so
30c2120D 180;ID 10e
30t
.cos40tC18sin40t/.
6.3.6.QpDAcos10tCBsin10t;Q
0
p
D10Bcos10t10Asin10t;Q
00
p
D 100Acos10t
100Bsin10t;
1
10
Q
00
p
C3Q
0
p
C100Q9D.90AC30B/cos10t.30A90B/sin10tD5cos10t
5sin10t, so90AC30BD5,30AC90BD 5. Therefore,AD1=15,BD 1=30,QpD
cos10t
15

sin10t
30
, andIpD
1
3
.cos10tC2sin10t/.
6.3.8.QpDAcos50tCBsin50t;Q
0
p
D50Bcos50t50Asin50t;Q
00
p
D 2500Acos50t
2500Bsin50t;
1
10
Q
00
p
C2Q
0
p
C100Qp.150AC100B/cos50t.100AC150B/sin50tD3cos50t
6sin50t, so150AC100BD3,100AC1500BD 6. Therefore,AD3=650,BD12=325,
QpD
3
650
.cos50tC8sin50t/, andIpD
3
13
.8cos50tsin50t/.
6.3.10.QpDAcos30tCBsin30t;Q
0
p
D30Bcos30t30Asin30t;Q
00
p
D 900Acos30t
900Bsin30t;
1
20
Q
00
p
C4Q
0
p
C125QpD.80AC120B/cos30t.120A80B/sin30tD15cos30t
30sin30t, so80AC120BD15,120AC80BD 30,AD3=13,BD 3=104,QpD
3
104
.8cos30t
sin30t/, andIpD
45
52
.cos30tC8sin30t/.

95. 90 Chapter 6Applications of Linear Second Order Equations
6.3.12.LetD.!/be the amplitude ofIp. From the solution of Exercise 6.3.11,QpDAcos!tC
Bcos!t, whereAD
.1=CL!
2
/UR!V

,BD
R!UC.1=CL!
2
/V

, andD.1=C
L!
2
/
2
CR
2
!
2
. SinceIpDQ
0
p
D!.Asin!tCBcos!t/, it follows that
2
.!/D!
2
.A
2
CB
2
/D
U
2
CV
2
.!/
, with.!/D

!
2
D.1=C!L!/
2
CR
2
, which attains it mininmum valueR
2
when
!D!0D
1
p
LC
. The maximum amplitude ofIpis.!/D
p
U
2
CV
2
R
.
6.4MOTION UNDER A CENTRAL FORCE
6.4.2.LethDr
2
0

0
0
; thenD
h
2
k
. SincerD

1Cecos./
, it follows that (A)ecos./D

r
1. Differentiating this with respect totyieldsesin./
0
D
r
0
r
2
, so (B)esin./D
r
0
h
, sincer
2

0
h. Squaring and adding (A) and (B) and settingtD0in the result yieldseD
"


r0
1

2
C

r
0
0
h

2
#
1=2
. IfeD0, then0is undefined, but also irrelevant; ife¤0, then settD0
in (A) and (B) to see thatD0˛, where˛ < , cos˛D
1
e


r0
1

and sin˛D
r
0
0
eh
6.4.4.Recall that (A)
d
2
u
d

2
D
1
mh
2
u
2
f .1=u/. LetuD
1
r
D
1
c
2
; then
d
2
u
d
2
D
6
c
4
D6cu
2
.
6cu
2
CuD
1
mh
2
u
2
f .1=u/, sof .1=u/D mh
2
.6cu
4
Cu
3
/andf .r/D mh
2

6c
r
4
C
1
r
3

.
6.4.6.(a)Withf .r/D
mk
r
3
, Eqn. 6.4.11 becomes (A)
d
2
u
d
2
C

1
k
h
2

uD0. The initial conditions
imply thatu.0/D
1
r0
and
du
d
.0/D
r
0
0
h
(see Eqn. (6.4.)
(b)LetD
ˇ
ˇ
ˇ
ˇ
1
k
h
2
ˇ
ˇ
ˇ
ˇ
1=2
. (i) Ifh
2
< k, then (A) becomes
d
2
u
d
2

2
uD0, and the solution
of the initial value problem foruisuD
1
r0
cosh.0/
r
0
0
h
sinh.0/; thereforerD
r0

cosh.0/
r0r
0
0
h
sinh.0/

1
. (ii) Ifh
2
Dk, then (A) becomes
d
2
u
d
2
D0, and the solu-
tion of the initial value problem foruisuD
1
r0

r
0
0
h
.0/; thereforerDr0

1
r0r
0
0
h
.0/

1
.
(iii) Ifh
2
> k, then (A) becomes
d
2
u
d
2
C
2
uD0, and the solution of the initial value problem foruis
uD
1
r0
cos.0/
r
0
0
h
sin.0/; thereforerDr0

cos.0/
r0r
0
0
h
sin.0/

1
.

96. CHAPTER7
SeriesSolutionsofLinearSecond
Equations
7.1REVIEW OF POWER SERIES
7.1.2.From Theorem 7.1.3,
P
1
mD0
bm´
m
converges ifj´j< 1=Land diverges ifj´j> 1=L. Therefore,
P
1
mD0
bm.xx0/
2
converges ifjxx0j< 1=
p
Land diverges ifjxx0j> 1=
p
L.
7.1.4.From Theorem 7.1.3,
P
1
mD0
bm´
m
converges ifj´j< 1=Land diverges ifj´j> 1=L. Therefore,
P
1
mD0
bm.xx0/
km
converges ifjxx0j< 1=
k
p
Land diverges ifjxx0j> 1=
k
p
L.
7.1.12..1C3x
2
/y
00
C3x
2
y
0
2yD
1
X
nD2
n.n1/anx
n2
C3
1
X
nD2
n.n1/anx
n
C3
1
X
nD1
nanx
nC1

2
1
X
nD0
anx
n
D
1
X
nD0
.nC2/.nC1/anC2x
n
C3
1
X
nD1
n.n1/nanx
n
C3
1
X
nD1
.n1/an1x
n
2
1
X
nD0
anx
n
D
2a22a0C
1
X
nD1
Œ.nC2/.nC1/anC2C.3n.n1/2/anC3.n1/an1x
n
.
7.1.13..1C2x
2
/y
00
C.23x/y
0
C4yD
1
X
nD2
n.n1/anx
n2
C2
1
X
nD2
n.n1/anx
n
C2
1
X
nD1
nanx
n1

2
1
X
nD0
anx
n
C4
1
X
nD0
anx
n
D
1
X
nD0
.nC2/.nC1/anC2x
n
C2
1
X
nD0
n.n1/anx
n
C2
1
X
nD0
.nC1/anC1x
n

3
1
X
nD0
anx
n
C4
1
X
nD0
anx
n
1
X
nD0
fi
.nC2/.nC1/anC2C2.nC1/anC1C.2n
2
5nC4/an

x
n
.
7.1.14..1Cx
2
/y
00
C.2x/y
0
C3yD
1
X
nD2
n.n1/anx
n2
C
1
X
nD2
n.n1/anx
n
C2
1
X
nD1
nanx
n1

1
X
nD1
nanx
n
C3
1
X
nD0
anx
n
D
1
X
nD0
.nC2/.nC1/anC2x
n
C
1
X
nD0
n.n1/anx
n
C2
1
X
nD0
.nC1/anC1x
n

1
X
nD0
nanx
n
C3
1
X
nD0
anx
n
D
1
X
nD0
fi
.nC2/.nC1/anC2C2.nC1/anC1C.n
2
2nC3/an

x
n
.
91

97. 92 Chapter 7Series Solutions of Linear Second Equations
7.1.16.LettDxC1; thenxy
00
C.4C2x/y
0
C.2Cx/yD.1Ct/y
00
C.2C2t/y
0
C.1Ct/yD
1
X
nD2
n.n
1/ant
n2
C
1
X
nD2
n.n1/ant
n1
C2
1
X
nD1
nant
n1
C2
1
X
nD1
nant
n
C
1
X
nD0
ant
n
C
1
X
nD0
ant
nC1
D
1
X
nD0
.nC
2/.nC1/anC2t
n
C
1
X
nD0
.nC1/nanC1t
n
C2
1
X
nD0
.nC1/anC1t
n
C2
1
X
nD0
nant
n
C
1
X
nD0
ant
n
C
1
X
nD1
an1t
n
D
.2a2C2a1Ca0/C
1
X
nD1
Œ.nC2/.nC1/anC2C.nC1/.nC2/anC1C.2nC1/anCan1 .xC2/
n
.
7.1.20.y
0
.x/Dx
r
1
X
nD0
nanx
n1
Crx
r1
1
X
nD0
anx
n
D
1
X
nD0
.nCr/x
nCr1
y
00
D
d
dx
y
0
.x/D
d
dx
"
x
r1
1
X
nD0
.nCr/anx
n
#
Dx
r1
1
X
nD0
.nCr/nanx
n1
C.r1/x
r2
1
X
nD0
.nC
r/anx
n
D
1
X
nD0
.nCr/.nCr1/anx
nCr2
.
7.1.22.x
2
.1Cx/y
00
Cx.1C2x/y
0
.4C6x/yD.x
2
y
00
Cxy
0
4y/Cx.x
2
y
00
C2xy
0
6y/D
1
X
nD0
Œ.nCr/.nCr1/C.nCr/4anx
nCr
C
1
X
nD0
Œ.nCr/.nCr1/C2.nCr/6anx
nCrC1
D
1
X
nD0
.nCr2/.nCrC2/anx
nCr
C
1
X
nD0
.nCrC3/.nCr2/anx
nCrC1
D
1
X
nD0
.nCr2/.nC
rC2/anx
nCr
C
1
X
nD1
.nCrC2/.nCr3/an1x
nCr
Dx
r
1
X
nD0
bnx
n
withb0D.r2/.rC2/a0and
bnD.nCr2/.nCrC2/anC.nCrC2/.nCr3/an1,n1.
7.1.24.x
2
.1C3x/y
00
Cx.2C12xCx
2
/y
0
C2x.3Cx/yD.x
2
y
00
C2xy
0
/Cx.3x
2
y
00
C12xy
0
C
6y/Cx
2
.xy
0
C2y/D
1
X
nD0
Œ.nCr/.nCr1/C2.nCr/anx
nCr
C
1
X
nD0
Œ3.nCr/.nCr1/C12.nC
r/C6anx
nCrC1
C
1
X
nD0
Œ.nCr/C2anx
nCrC2
D
1
X
nD0
.nCr/.nCrC1/anx
nCr
C3
1
X
nD0
.nCrC1/.nC
rC2/anx
nCrC1
C
1
X
nD0
.nCrC2/anx
nCrC2
D
1
X
nD0
.nCr/.nCrC1/anx
nCr
C3
1
X
nD1
.nCr/.nCrC
1/an1x
nCr
C
1
X
nD2
.nCr/an2x
nCr
Dx
r
1
X
nD0
bnx
n
withb0Dr.rC1/a0,b1D.rC1/.rC2/a1C
3.rC1/.rC2/a0,bnD.nCr/.nCrC1/anC3.nCr/.nCrC1/an1C.nCr/an2,n2.
7.1.26.x
2
.2Cx
2
/y
00
C2x.5Cx
2
/y
0
C2.3x
2
/yD.2x
2
y
00
C10xy
0
C6y/Cx
2
.x
2
y
00
C2xy
0
2y/D
1
X
nD0
Œ2.nCr/.nCr1/C10.nCr/C6anx
nCr
C
1
X
nD0
Œ.nCr/.nCr1/C2.nCr/2anx
nCrC2
D
2
1
X
nD0
.nCrC1/.nCrC3/anx
nCr
C
1
X
nD0
.nCr1/.nCrC2/anx
nCrC2
D2
1
X
nD0
.nCrC1/.nC

98. Section 7.2Series Solutions Near an Ordinary Point I93
rC3/anx
nCr
C
1
X
nD2
.nCr3/.nCr/an2x
nCr
Dx
r
1
X
nD0
bnx
n
withb0D2.rC1/.rC3/a0,
b1D2.rC2/.rC4/a1bnD2.nCrC1/.nCrC3/anC.nCr3/.nCr/an2,n2.
7.2SERIES SOLUTIONS NEAR AN ORDINARY POINT I
7.2.2.p.n/Dn.n1/C2n2D.nC2/.n1/;anC2D
n1
nC1
an;a2mC2D
2m1
2mC1
a2m, so
a2mD
.1/
m
2m1
a0;a2mC3D
m
mC2
a2mC1D0ifm0;yDa0
1
X
mD0
.1/
mC1
x
2m
2m1
Ca1x.
7.2.4.p.n/D n.n1/8n12D .nC3/.nC4/;anC2D
.nC3/.nC4/
.nC2/.nC1/
an;a2mC2D

.mC2/.2mC3/
.mC1/.2mC1/
a2m, soa2mD.mC1/.2mC1/a0;a2mC3D
.mC2/.2mC5/
.mC1/.2mC3/
a2mC1soa2mC1D
.mC1/.2mC3/
3
a1;yDa0
1
X
mD0
.mC1/.2mC1/x
2m
C
a1
3
1
X
mD0
.mC1/.2mC3/x
2mC1
.
7.2.6.p.n/Dn.n1/C2nC
1
4
D
.2nC1/
2
4
;anC2D
.2nC1/
2
4.nC2/.nC1/
an;a2mC2D
.4mC1/
2
8.mC1/.2mC1/
a2m,
soa2mD.1/
m
2
4
m1
Y
jD0
.4jC1/
2
2jC1
3
5
1
8
m
mŠ
a0;a2mC3D
.4mC3/
2
8.2mC3/.mC1/
a2mC1soa2mC1D
.1/
m
2
4
m1
Y
jD0
.4jC3/
2
2jC3
3
5
1
8
m
mŠ
a1;yDa0
1
X
mD0
.1/
m
2
4
m1
Y
jD0
.4jC1/
2
2jC1
3
5
x
2m
8
m
mŠ
Ca1
1
X
mD0
.1/
m
2
4
m1
Y
jD0
.4jC3/
2
2jC3
3
5
x
2mC1
8
m
mŠ
7.2.8.p.n/Dn.n1/10nC28D.n7/.n4/;anC2D
.n7/.n4/
.nC2/.nC1/
an;a2mC2D
2.2m7/.m2/
2.mC1/.2mC1/
a2m,
soa2D 14a0,a4D
5
6
a2D
35
3
a0,a2mD0ifm3;a2mC3D
.m3/.2m3/
.2mC3/.mC1/
a2mC1, so
a3D 3a1,a5D
1
5
a3D
3
5
a1,a7D
1
21
a5D
1
35
a1;
yDa0

114x
2
C
35
3
x
4

Ca1

x3x
3
C
3
5
x
5
C
1
35
x
7

.
7.2.10.p.n/D2nC3;anC2D
2nC3
.nC2/.nC1/
an;a2mC2D
4mC3
2.mC1/.2mC1/
a2m, soa2mD
2
4
m1
Y
jD0
4jC3
2jC1
3
5
.1/
m
2
m
mŠ
a0;a2mC3D
4mC5
2.2mC3/.mC1/
a2mC1soa2mC1D
2
4
m1
Y
jD0
4jC5
2jC3
3
5
.1/
m
2
m
mŠ
a1;
yDa0
1
X
mD0
.1/
m
2
4
m1
Y
jD0
4jC3
2jC1
3
5
x
2m
2
m
mŠ
Ca1
1
X
mD0
.1/
m
2
4
m1
Y
jD0
4jC5
2jC3
3
5
x
2mC1
2
m
mŠ
7.2.12.p.n/D2n.n1/9n6D.n6/.2nC1/;anC2D
.n6/.2nC1/
.nC2/.nC1/
an;a0Dy.0/D1;
a1Dy
0
.0/D 1.
7.2.13.p.n/D8n.n1/C2D2.2n1/
2
;anC2D
2.2n1/
2
.nC2/.nC1/
an;a0Dy.0/D2;a1D
y
0
.0/D 1.

99. 94 Chapter 7Series Solutions of Linear Second Order Equations
7.2.16.p.n/D 1;anC2D
1
.nC2/.nC1/
an;a2mC2D
1
.2mC2/.2mC1/
a2m, soa2mD
1
.2m/Š
a0;
a2mC3D
1
.2mC3/.2mC1/
a2mC1, soa2mC1D
1
.2mC1/Š
a1;yDa0
1
X
mD0
.x3/
2m
.2m/Š
Ca1
1
X
mD0
.x3/
2mC1
.2mC1/Š
.
7.2.18.LettDx1; then.12t
2
/y
00
10ty
0
6yD0;p.n/D 2n.n1/10n6D
2.nC1/.nC3/;anC2D
2.nC3/
nC2
an;a2mC2D
2mC3
mC1
a2m, soa2mD
1
mŠ
2
4
m1
Y
jD0
.2jC3/
3
5a0;
a2mC3D
4.mC2/
2mC3
a2mC1, soa2mC1D
4
m
.mC1/Š
Q
m1
jD0
.2jC3/
a1;yDa0
1
X
mD0
2
4
m1
Y
jD0
.2jC3/
3
5
.x1/
2m
mŠ
Ca1
1
X
mD0
4
m
.mC1/Š
Q
m1
jD0
.2jC3/
.x1/
2mC1
.
7.2.20.LettDxC1; then

1C
3t
2
3

y
00

9t
2
y
0
C
3
2
yD0;p.n/D
3
2
n.n1/C
9
2
nC
3
2
D
3
2
.nC
1/
2
;anC2D
3.nC1/
2.nC2/
an;a2mC2D
3.2mC1/
4.mC1/
a2m, soa2mD.1/
m
2
4
m1
Y
jD0
.2jC1/
3
5
3
m
4
m
mŠ
a0;
a2mC3D
3.mC1/
2mC3
a2mC1, soa2mC1D.1/
m
3
m
mŠ
Q
m1
jD0
.2jC3/
a1;
yDa0
1
X
mD0
.1/
m
2
4
m1
Y
jD0
.2jC1/
3
5
3
m
4
m
mŠ
.xC1/
2m
Ca1
1
X
mD0
.1/
m
3
m
mŠ
Q
m1
jD0
.2jC3/
.xC1/
2mC1
.
7.2.22.p.n/DnC3;anC2D
nC3
.nC2/.nC1/
an;a0Dy.3/D 2;a1Dy
0
.3/D3.
7.2.24.LettDx3;.1C4t
2
/y
00
CyD0;p.n/D.4n.n1/C1D2n1/
2
;anC2D

.2n1/
2
.nC2/.nC1/
an;a0Dy.3/D4;a1Dy
0
.3/D 6.
7.2.26.LettDxC1;

1C
2t
2
3

y
00

20
3
ty
0
C20yD0;p.n/D
2
3
n.n1/
20
3
nC20D
2.n6/.n5/
3
;anC2D
2.n6/.n5/
3.nC2/.nC1/
an;a0Dy.1/D3;a1Dy
0
.1/D 3.
7.2.28.From Theorem 7.2.2,anC2D
p.n/
.nC2/.nC1/
an;a2mC2D
p.2m/
.2mC2/.2mC1/
a2m, so
a2mD
2
4
m1
Y
jD0
p.2j /
3
5
.1/
m
.2m/Š
a0;a2mC3D
p.2mC1/
.2mC3/.2mC2/
a2m, soa2mC1D
2
4
m1
Y
jD0
p.2jC1/
3
5
.1/
m
.2mC1/Š
a1.
7.2.30.(a)Herep.n/D Œn.n1/C2bn˛.˛C2b1/D .n˛/.nC˛C2b1/, so
Exercise 7.2.28 implies thaty1andy2have the stated forms. If˛D2k, then
y1D
1
X
mD0
2
4
m1
Y
jD0
.2j2k/.2jC2kC2b1/
3
5
x
2m
.2m/Š
.C/:

100. Section 7.2Series Solutions Near an Ordinary Point I95
If˛D2kC1, then
y2D
1
X
mD0
2
4
m1
Y
jD0
.2j2k/.2jC2kC2b/
3
5
x
2mC1
.2mC1/Š
: . D/
Since2bis not a negative integer and
Q
m1
jD0
.2j2k/D0ifm > k,y1in (C) andy2in (D) have the
stated properties. This implies the conclusions regardingPn.
(b)Multiplying (A) through by.1x
2
/
b1
yields
Œ.1x
2
/
b
P
0
n

0
D n.nC2b1/.1x
2
/
b1
Pn: . E/
(c)Therefore,
Œ.1x
2
/
b
P
0
m

0
D m.mC2b1/.1x
2
/
b1
Pm: . F/
SubtractPntimes (F) fromPmtimes (E) to obtain (B).
(d)Integrating the left side of (B) by parts overŒ1; 1yields zero, which implies the conclusion.
7.2.32.(a)LetLyD.1C˛x
3
/y
00
Cˇx
2
y
0
Cxy. IfyD
1
X
nD0
anx
n
, thenLyD
1
X
nD2
n.n1/anx
n2
C
1
X
nD0
p.n/anx
nC1
D2a2C
1
X
nD0
Œ.nC3/.nC2/anC3Cp.n/anx
nC1
D0if and only ifa2D0and
anC3D
p.n/
.nC3/.nC2/
anforn0.
7.2.34.p.r/D 2r.r1/10r8D 2.rC2/
2
; (A)
m1
Y
jD0
p.3j /
3jC2
D
m1
Y
jD0
.2/.3jC2/
2
3jC2
D
.1/
m
2
m
m1
Y
jD0
.3jC2/; (B)
m1
Y
jD0
p.3jC1/
3jC4
D
m1
Y
jD0
.2/.3jC3/
2
3jC4
D
.1/
m
2
m
.mŠ/
2
Q
m1
jD0
.3jC4/
. Substituting (A)
and (B) into the result of Exercise 7.2.32(c)yields
yDa0
1
X
mD0

2
3

m
2
4
m1
Y
jD0
.3jC2/
3
5
x
3m
mŠ
Ca1
1
X
mD0
6
m
mŠ
Q
m1
jD0
.3jC4/
x
3mC1
.
7.2.36.p.r/D 2r.r1/C6rC24D 2.r6/.rC2/; (A)
m1
Y
jD0
p.3j /
3jC2
D
m1
Y
jD0
.6/.j2/.
(B)
m1
Y
jD0
p.3jC1/
3jC4
D
m1
Y
jD0
.6/.jC1/.3j5/
3jC4
D.1/
m
6
m
m
m1
Y
jD0
3j5
3jC4
. Substituting (A) and
(B) into the result of Exercise 7.2.32(c)yields
yDa0.14x
3
C4x
6
/Ca1
1
X
mD0
2
m
2
4
m1
Y
jD0
3j5
3jC4
3
5x
3mC1
.
7.2.38.(a)LetLyD.1C˛x
kC2
/y
00
Cˇx
kC1
y
0
Cx
k
y. IfyD
1
X
nD0
anx
n
, thenLyD
1
X
nD2
n.n
1/anx
n2
C
1
X
nD0
p.n/anx
nCk
D
1
X
nDk
.nCkC2/.nCk1/anCkC2x
nCk
C
1
X
nD0
Œ.nCkC2/.nC

101. 96 Chapter 7Series Solutions of Linear Second Order Equations
kC1/anCkC2Cp.n/anx
nCk
D0if and only ifakD0for2nkC1and (A)anCkC1D

p.n/
.nCkC2/.nCkC1/
anforn0.
(b)IfanD0theanC.kC2/mD0for allm0, from (A).
7.2.40.kD2andp.r/D1; (A)
m1
Y
jD0
p.4j /
4jC3
D
1
Q
m1
jD0
.4jC3/
; (B)
m1
Y
jD0
p.4jC1/
.4jC5/
D
1
Q
m1
jD0
4jC5
.
Substituting (A) and (B) into the result of Exercise t.2.38(c)yields
yDa0
1
X
mD0
.1/
m
x
4m
4
m
mŠ
Q
m1
jD0
.4jC3/
Ca1
1
X
mD0
.1/
m
x
4mC1
4
m
mŠ
Q
m1
jD0
.4jC5/
.
7.2.42.kD6andp.r/Dr.r1/16rC72D.r9/.r8/; (A)
m1
Y
jD0
p.8j /
8jC7
D
m1
Y
jD0
8.j1/.8j9/
8jC7
;
(B)
m1
Y
jD0
p.8jC1/
.8jC9/
D
m1
Y
jD0
8.j1/.8j7/
8jC9
;
Substituting (A) and (B) into the result of Exercise 7.2.38(c)yields
yDa0

1
9
7
x
8

Ca1

x
7
9
x
9

.
7.2.44.kD4andp.r/DrC6; (A)
m1
Y
jD0
p.6j /
6jC5
D
m1
Y
jD0
6.jC1/
6jC5
D
6
m
mŠ
Q
m1
jD0
.6jC5/
;
(B)
m1
Y
jD0
p.6jC1/
.6jC7/
D1;
Substituting (A) and (B) into the result of Exercise 7.2.38(c)yields
yDa0
1
X
mD0
.1/
m
x
6m
Q
m1
jD0
.6jC5/
Ca1
1
X
mD0
.1/
m
x
6mC1
6
m
mŠ
.
7.3SERIES SOLUTIONS NEAR AN ORDINARY POINT II
7.3.2.IfyD
1
X
nD0
anx
n
, then.1CxC2x
2
/y
00
C.2C8x/y
0
C4yD
1
X
nD2
n.n1/anx
n2
C
1
X
nD2
n.n
1/anx
n1
C2
1
X
nD2
n.n1/anx
n
C2
1
X
nD1
nanx
n1
C8
1
X
nD1
nanx
n
C4
1
X
nD0
anx
n
D
1
X
nD0
.nC2/.nC
1/.anC2CanC1C2an/x
n
D0ifanC2D anC12an,an0. Starting witha0D 1anda1D2
yieldsyD 1C2x4x
3
C4x
4
C4x
5
12x
6
C4x
7
C .
7.3.4.IfyD
1
X
nD0
anx
n
, then.1CxC3x
2
/y
00
C.2C15x/y
0
C12yD
1
X
nD2
n.n1/anx
n2
C
1
X
nD2
n.n
1/anx
n1
C3
1
X
nD2
n.n1/anx
n
C2
1
X
nD1
nanx
n1
C15
1
X
nD1
nanx
n
C12
1
X
nD0
anx
n
D
1
X
nD0
Œ.nC2/.nC
1/anC2C.nC1/.nC2/anC1C3.nC2/
2
anx
n
D0ifanC2D anC1
3.nC2/
nC1
an,an0. Starting
witha0D0anda1D1yieldsyDxx
2

7
2
x
3
C
15
2
x
4
C
45
8
x
5

261
8
x
6
C
207
16
x
7
C .

102. Section 7.3Series Solutions Near an Ordinary Point II97
7.3.6.IfyD
1
X
nD0
anx
n
, then.3C3xCx
2
/y
00
C.6C4x/y
0
C2yD3
1
X
nD2
n.n1/anx
n2
C
3
1
X
nD2
n.n1/anx
n1
C
1
X
nD2
n.n1/anx
n
C6
1
X
nD1
nanx
n1
C4
1
X
nD1
nanx
n
C2
1
X
nD0
anx
n
D
1
X
nD0
.nC
2/.nC1/Œ3anC2C3anC1Canx
n
D0ifanC2D anC1an=3,an0. Starting witha0D7and
a1D3yieldsyD7C3x
16
3
x
2
C
13
3
x
3

23
9
x
4
C
10
9
x
5

7
27
x
6

1
9
x
7
C .
7.3.8.The equation is equivalent to.1CtC2t
2
/y
00
C.2C6t/y
0
C2yD0withtDx1. If
yD
1
X
nD0
ant
n
, then.1CtC2t
2
/y
00
C.2C6t/y
0
C2yD
1
X
nD2
n.n1/ant
n2
C
1
X
nD2
n.n1/ant
n1
C
2
1
X
nD2
n.n1/ant
n
C2
1
X
nD1
nant
n1
C6
1
X
nD1
nant
n
C2
1
X
nD0
ant
n
D
1
X
nD0
.nC1/Œ.nC2/anC2C.nC
2/anC1C2.nC1/ant
n
D0ifanC2D anC1
2.nC1/
nC2
an,an0. Starting witha0D1anda1D 1
yieldsyD1.x1/C
4
3
.x1/
3

4
3
.x1/
4

4
5
.x1/
5
C
136
45
.x1/
6

104
63
.x1/
7
C .
7.3.10.The equation is equivalent to.1CtCt
2
/y
00
C.3C4t/y
0
C2yD0withtDx1. If
yD
1
X
nD0
ant
n
, then.1CtCt
2
/y
00
C.3C4t/y
0
C2yD
1
X
nD2
n.n1/ant
n2
C
1
X
nD2
n.n1/ant
n1
C
1
X
nD2
n.n1/ant
n
C3
1
X
nD1
nant
n1
C4
1
X
nD1
nant
n
C2
1
X
nD0
ant
n
D
1
X
nD0
.nC1/Œ.nC2/anC2C.nC3/anC1C
.nC2/ant
n
D0ifanC2D
nC3
nC2
anC1an,an0. Starting witha0D2anda1D 1yieldsyD
2.x1/
1
2
.x1/
2
C
5
3
.x1/
3

19
12
.x1/
4
C
7
30
.x1/
5
C
59
45
.x1/
6

1091
630
.x1/
7
C
7.3.12.The equation is equivalent to.1C2tCt
2
/y
00
C.1C7t/y
0
C8yD0withtDx1. IfyD
1
X
nD0
ant
n
, then.1C2tCt
2
/y
00
C.1C7t/y
0
C8yD
1
X
nD2
n.n1/ant
n2
C2
1
X
nD2
n.n1/ant
n1
C
1
X
nD2
n.n
1/ant
n
C
1
X
nD1
nant
n1
C7
1
X
nD1
nant
n
C8
1
X
nD0
ant
n
D
1
X
nD0
Œ.nC2/.nC1/anC2C.nC1/.2nC1/anC1C
.nC2/.nC4/ant
n
D0ifanC2D
2nC1
nC2
anC1
nC4
nC1
an,an0. Starting witha0D1anda1D 2
yieldsyD12.x1/3.x1/
2
C8.x1/
3
4.x1/
4

42
5
.x1/
5
C19.x1/
6

604
35
.x1/
7
C
7.3.16.IfyD
1
X
nD0
anx
n
, then.1x/y
00
.2x/y
0
CyD
1
X
nD2
n.n1/anx
n2

1
X
nD2
n.n1/anx
n1

2
1
X
nD1
nanx
n1
C
1
X
nD1
nanx
n
C
1
X
nD0
anx
n
D
1
X
nD0
Œ.nC2/.nC1/anC2.nC2/.nC1/anC1C.nC1/anx
n
D
0ifanC2DanC1
an
nC2
,an0.

103. 98 Chapter 7Series Solutions of Linear Second Order Equations
7.3.18.IfyD
1
X
nD0
anx
n
, then.1Cx
2
/y
00
Cy
0
C2yD
1
X
nD2
n.n1/anx
n2
C
1
X
nD2
n.n1/anx
n
C
1
X
nD1
nanx
n1
C2
1
X
nD0
anx
n
D
1
X
nD0
Œ.nC2/.nC1/anC2C.nC1/anC1C.n
2
nC2/anx
n
D0if
anC2D
1
nC2
anC1
n
2
nC2
.nC2/.nC1/
an.
7.3.20.The equation is equivalent to.3C2t/y
00
C.1C2t/y
0
.12t/yD0withtDx1. If
yD
1
X
nD0
ant
n
, then.3C2t/y
00
C.1C2t/y
0
.12t/yD3
1
X
nD2
n.n1/ant
n2
C2
1
X
nD2
n.n
1/ant
n1
C
1
X
nD1
nant
n1
C2
1
X
nD1
nant
n

1
X
nD0
ant
n
C2
1
X
nD0
ant
nC1
D.6a2Ca1a0/C
1
X
nD1
Œ3.nC
2/.nC1/anC2C.nC1/.2nC1/anC1C.2n1/anC2an1t
n
D0ifa2D
a1a0
6
andanC2D

2nC1
3.nC2/
anC1
2n1
3.nC2/.nC1/
an
2
3.nC2/.nC1/
an1,n1. Starting witha0D1and
a1D 2yieldsyD12.x1/C
1
2
.x1/
2

1
6
.x1/
3
C
5
36
.x1/
4

73
1080
.x1/
5
C .
7.3.22.The equation is equivalent to.1Ct/y
00
C.22t/y
0
C.3Ct/yD0withtDxC3. IfyD
1
X
nD0
ant
n
,
then.1Ct/y
00
C.22t/y
0
C.3Ct/yD
1
X
nD2
n.n1/ant
n2
C
1
X
nD2
n.n1/ant
n1
C2
1
X
nD1
nant
n1

2
1
X
nD1
nant
n
C3
1
X
nD0
ant
n
C
1
X
nD0
ant
nC1
D.2a2C2a1C3a0/C
1
X
nD1
Œ.nC2/.nC1/anC2C.nC2/.nC
1/anC1.2n3/anCan1t
n
D0ifa2D
2a1C3a0
2
andanC2D anC1C
.2n3/anan1
.nC2/.nC1/
,
n1. Starting witha0D2anda1D 2yields
yD22.xC3/.xC3/
2
C.xC3/
3

11
12
.xC3/
4
C
67
60
.xC3/
5
C .
7.3.24.The equation is equivalent to.1C2t/y
00
C3y
0
C.1t/yD0withtDxC1. IfyD
1
X
nD0
ant
n
, then
.1C2t/y
00
C3y
0
C.1t/yD
1
X
nD2
n.n1/ant
n2
C2
1
X
nD2
n.n1/ant
n1
C3
1
X
nD1
nant
n1
C
1
X
nD0
ant
n

1
X
nD0
ant
nC1
D.2a2C3a1Ca0/C
1
X
nD1
Œ.nC2/.nC1/anC2C.2nC3/.nC1/anC1Canan1t
n
D0
ifa2D
3a1Ca0
2
andanC2D
2nC3
nC2
anC1
anan1
.nC2/.nC1/
,n1. Starting witha0D2and
a1D 3yieldsyD23.xC1/C
7
2
.xC1/
2
5.xC1/
3
C
197
24
.xC1/
4

287
20
.xC1/
5
C .
7.3.26.The equation is equivalent to.62t/y
00
C.3Ct/yD0withtDx2. IfyD
1
X
nD0
ant
n
, then

104. Section 7.3Series Solutions Near an Ordinary Point II99
.62t/y
00
C.3Ct/yD6
1
X
nD2
n.n1/ant
n2
2
1
X
nD2
n.n1/ant
n1
C3
1
X
nD0
ant
n
C
1
X
nD0
ant
nC1
D
.12a2C3a0/C
1
X
nD1
Œ6.nC2/.nC1/anC22.nC1/nanC1C3anCan1t
n
D0ifa2D
a0
4
andanC2D
n
3.nC2/
anC1
3anCan1
6.nC2/.nC1/
,n1. Starting witha0D2anda1D 4yields
yD24.x2/
1
2
.x2/
2
C
2
9
.x2/
3
C
49
432
.x2/
4
C
23
1080
.x2/
5
C .
7.3.28.The equation is equivalent to.2C4t/y
00
.12t/yD0withtDxC4. IfyD
1
X
nD0
ant
n
, then
.2C4t/y
00
.12t/yD2
1
X
nD2
n.n1/ant
n2
C4
1
X
nD2
n.n1/ant
n1

1
X
nD0
ant
n
C2
1
X
nD0
ant
nC1
D
.4a2a0/C
1
X
nD1
Œ2.nC2/.nC1/anC2C4.nC1/nanC1anC2an1t
n
D0ifa2D
a0
4
and
anC2D
2n
nC2
anC1C
an2an1
2.nC2/.nC1/
,n1. Starting witha0D 1anda1D2yieldsyD
1C2.xC1/
1
4
.xC1/
2
C
1
2
.xC1/
3

65
96
.xC1/
4
C
67
80
.xC1/
5
C .
N=5; b=zeros(N,1); b(1)=-1;b(2)=2; b(3)=b(1)/4; for n=1:N-2 b(n+3)=-2*n*b(n+2)/(n+2)+(b(n+1)-
2*b(n))/(2*(n+2)*(n+1)); end
7.3.29.LetLyD.1C˛xCˇx
2
/y
00
C.Cıx/y
0
Cy. IfyD
1
X
nD0
anx
n
, thenLyD
1
X
nD2
n.n
1/anx
n2
C˛
1
X
nD2
n.n1/anx
n1
Cˇ
1
X
nD2
n.n1/anx
n
C
1
X
nD1
nanx
n1
Cı
1
X
nD1
nanx
n
C
1
X
nD0
anx
n
D
1
X
nD0
.nC2/.nC1/anC2x
n
C˛
1
X
nD0
.nC1/nanC1x
n
Cˇ
1
X
nD0
n.n1/anx
n
C
1
X
nD0
.nC1/anC1x
n
C
ı
1
X
nD0
nanx
n
C
1
X
nD0
anx
n
D
1
X
nD0
bnx
n
, wherebnD.nC1/.nC2/anC2C.nC1/.˛nC/anC1C
Œˇn.n1/CınCan, which implies the conclusion.
7.3.30.(a)LetD2˛,ıD4ˇ, andD2ˇin Exercise 7.3.29 to obtain (B).
(b)IfanDc1r
n
1
Cc2r
n
2
, thenanC2C˛anC1CˇanDc1r
n
1
.r
2
1
C˛rCˇ/Cc2r
n
2
.r
2
2
C˛r2C
ˇ/Dc1r
n
1
P0.r1/Cc2r
n
2
P0.r2/D0, sofangsatisfies (B). Since1=r1and1=r2are the zeros ofP0,
Theorem 7.2.1 implies that
P
1
nD0
.c1r
n
1
Cc2r
n
2
/x
n
is a solution of (A) on.; /.
(c)Ifjxj< , thenjr1xj< andjr2xj< 1, so
1
X
nD0
r
n
i
x
n
D
1
1rix
Dyi,iD1; 2. Therefore,(b)
implies thatfy1; y2gis a fundamental set of solutions of (A) on.; /.
(d)(A) can written asP0y
00
C2P
0
0
y
0
CP
00
0
yD.P0y/
00
D0. Therefore,P0yDaCbxwhereaand
bare arbitrary constants, and a partial fraction expansion shows that the general solution of (A) on any
interval not containing1=r1or1=r2isyD
aCbx
P0.x/
D
c1
1r1x
C
c2
1r2x
Dc1y1Cc2y2.
(e)IfanDc1r
n
1
Cc2r
n
2
, thenanC2C˛anC1CˇanDc1r
n
1
.r
2
1
C˛rCˇ/Cc2r
n
1
Œ.nC2/r
2
1
C˛.nC
1/r2Cˇn/D.c1Cnc2/r
n
1
P0.r1/Cc2r
n
1
P
0
0
.r1/D0, sofangsatisfies (B). Since1=r1is the only zero

105. 100 Chapter 7Series Solutions of Linear Second Order Equations
ofP0, Theorem 7.2.1 implies that
P
1
nD0
.c1Cc2n/r
n
1
/x
n
is a solution of (A) on.; /.
(f)Ifjxj< , thenjr1xj< , so
1
X
nD0
r
n
1
x
n
D
1
1r1x
Dy1. Differentiating this and multiplying
the result byxshows that
1
X
nD0
nr
n
1
x
n
D
r1x
.1r1x/
2
Dr1y2. Therefore,(e)implies thatfy1; y2gis a
fundamental set of solutions of (A) on.; /.
(g)The argument is the same as in(c), but now the partial fraction expansion can be written asyD
aCbx
P0.x/
D
c1
1r1x
C
c2x
.1r2x/
2
Dc1y1Cc2y2.
7.3.32.IfyD
1
X
nD0
anx
n
, theny
00
C2xy
0
C.3C2x
2
/yD
1
X
nD2
n.n1/anx
n2
C2
1
X
nD1
nanx
n
C
3
1
X
nD0
anx
n
C2
1
X
nD0
anx
nC2
D.2a2C3a0/C.6a3C5a1/xC
1
X
nD2
Œ.nC2/.nC1/anC2C.2nC3/anC
2an2x
n
D0ifa2D 3a0=2,a3D 5a1=6, andanC2D
.2nC3/anC2an2
.nC2/.nC1/
,n2. Starting
witha0D1anda1D 2yieldsyD12x
3
2
x
2
C
5
3
x
3
C
17
24
x
4

11
20
x
5
C .
7.3.34.IfyD
1
X
nD0
anx
n
, theny
00
C5xy
0
.3x
2
/yD
1
X
nD2
n.n1/anx
n2
C5
1
X
nD1
nanx
n
3
1
X
nD0
anx
n
C
1
X
nD0
anx
nC2
D.2a23a0/C.6a3C2a1/xC
1
X
nD2
Œ.nC2/.nC1/anC2C.5n3/anCan2x
n
D0
ifa2D3a0=2,a3D a1=3, andanC2D
.5n3/anCan2
.nC2/.nC1/
,n2. Starting witha0D6and
a1D 2yieldsyD62xC9x
2
C
2
3
x
3

23
4
x
4

3
10
x
5
C .
7.3.36.IfyD
1
X
nD0
anx
n
, theny
00
3xy
0
C.2C4x
2
/yD
1
X
nD2
n.n1/anx
n2
3
1
X
nD1
nanx
n
C
2
1
X
nD0
anx
n
C4
1
X
nD0
anx
nC2
D.2a2C2a0/C.6a3a1/xC
1
X
nD2
Œ.nC2/.nC1/anC2.3n2/anC
4an2x
n
D0ifa2D a0,a3Da1=6, andanC2D
.3n2/an4an2
.nC2/.nC1/
,n2. Starting witha0D3
anda1D6yieldsyD3C6x3x
2
Cx
3
2x
4

17
20
x
5
C .
7.3.38.IfyD
1
X
nD0
anx
n
, then3y
00
C2xy
0
C.4x
2
/yD3
1
X
nD2
n.n1/anx
n2
C2
1
X
nD1
nanx
n
C
4
1
X
nD0
anx
n

1
X
nD0
anx
nC2
D.6a2C4a0/C.18a3C6a1/xC
1
X
nD2
Œ3.nC2/.nC1/anC2C.2nC4/an
an2x
n
D0ifa2D 2a0=3,a3D a1=3, andanC2D
.2nC4/anan2
3.nC2/.nC1/
,n2. Starting with
a0D 2anda1D3yieldsyD 2C3xC
4
3
x
2
x
3

19
54
x
4
C
13
60
x
5
C .

106. Section 7.3Series Solutions Near an Ordinary Point II101
7.3.40.IfyD
1
X
nD0
anx
n
, then.1Cx/y
00
Cx
2
y
0
C.1C2x/yD
1
X
nD2
n.n1/anx
n2
C
1
X
nD2
n.n
1/anx
n1
C
1
X
nD1
nanx
nC1
C
1
X
nD0
anx
n
C2
1
X
nD0
anx
nC1
D.2a2Ca0/C
1
X
nD1
Œ.nC2/.nC1/anC2C.nC
1/nanC1CanC.nC1/an1x
n
D0ifa2D a0=2andanC2D
.nC1/nanC1CanC.nC1/an1
.nC2/.nC1/
,
n1. Starting witha0D 2anda1D3yieldsyD 2C3xCx
2

1
6
x
3

3
4
x
4
C
31
120
x
5
C .
7.3.42.IfyD
1
X
nD0
anx
n
, then.1Cx
2
/y
00
C.2Cx
2
/y
0
CxyD
1
X
nD2
n.n1/anx
n2
C
1
X
nD2
n.n1/anx
n
C
2
1
X
nD1
nanx
n1
C
1
X
nD1
nanx
nC1
C
1
X
nD1
anx
nC1
D.2a2C2a1/C
1
X
nD1
Œ.nC2/.nC1/anC2C2.nC1/anC1C
n.n1/anCnan1x
n
D0ifa2D a1andanC2D
Œ2.nC1/anC1Cn.n1/anCnan1
.nC2/.nC1/
,n1.
Starting witha0D 3anda1D5yieldsyD 3C5x5x
2
C
23
6
x
3

23
12
x
4
C
11
30
x
5
C .
7.3.44.The equation is equivalent toy
00
C.1C3t
2
/y
0
C.1C2t/yD0withtDx2. IfyD
1
X
nD0
ant
n
,
theny
00
C.1C3t
2
/y
0
C.1C2t/yD
1
X
nD2
n.n1/ant
n2
C
1
X
nD1
nant
n1
C3
1
X
nD1
nant
nC1
C
1
X
nD0
ant
n
C
2
1
X
nD0
ant
nC1
D.2a2Ca1Ca0/C
1
X
nD1
Œ.nC2/.nC1/anC2C.nC1/anC1CanC.3n1/an1t
n
D0
ifa2D .a1Ca0/=2andanC2D
Œ.nC1/anC1CanC.3n1/an1
.nC2/.nC1/
,n1. Starting witha0D2
anda1D 3yields
yD23.xC2/C
1
2
.xC2/
2

1
3
.xC2/
3
C
31
24
.xC2/
4

53
120
.xC2/
5
C .
7.3.46.The equation is equivalent to.1t
2
/y
00
.78tCt
2
/y
0
CtyD0withtDxC2. IfyD
1
X
nD0
ant
n
,
then.1t
2
/y
00
.78tCt
2
/y
0
CtyD
1
X
nD2
n.n1/ant
n2

1
X
nD2
n.n1/ant
n
7
1
X
nD1
nant
n1
C
8
1
X
nD1
nant
n

1
X
nD1
nant
nC1
C
1
X
nD0
ant
nC1
D.2a27a1/C
1
X
nD1
Œ.nC2/.nC1/anC27.nC1/anC1n.n
9/an.n2/an1t
n
D0ifa2D7a1=2andanC2D
Œ7.nC1/anC1Cn.n9/anC.n2/an1
.nC2/.nC1/
,
n1. Starting witha0D2anda1D 1yields
yD2.xC2/
7
2
.xC2/
2

43
6
.xC2/
3

203
24
.xC2/
4

167
30
.xC2/
5
C .
7.3.48.The equation is equivalent to.1C3tC2t
2
/y
00
.3Ctt
2
/y
0
.3Ct/yD0withtDx1. If
yD
1
X
nD0
ant
n
, then.1C3tC2t
2
/y
00
.3Ctt
2
/y
0
.3Ct/yD
1
X
nD2
n.n1/ant
n2
C3
1
X
nD2
n.n

107. 102 Chapter 7Series Solutions of Linear Second Order Equations
1/ant
n1
C2
1
X
nD2
n.n1/ant
n
3
1
X
nD1
nant
n1

1
X
nD1
nant
n
C
1
X
nD1
nant
nC1
3
1
X
nD0
ant
n

1
X
nD0
ant
nC1
D
.2a23a13a0/C
1
X
nD1
Œ.nC2/.nC1/anC2C3.n
2
1/anC1C.2n
2
3n3/.nC1/anC.n2/an1t
n
D0
ifa2D3.a1Ca0/=2andanC2D
Œ3.n
2
1/anC1C.2n
2
3n3/.nC1/anC.n2/an1
.nC2/.nC1/
,n
1. Starting witha0D1anda1D0yieldsyD1C
3
2
.x1/
2
C
1
6
.x1/
3

1
8
.x1/
5
C .
7.4REGULAR SINGULAR POINTS; EULER EQUATIONS
7.4.2.p.r/Dr.r1/7rC7D.r7/.r1/;yDc1xCc2x
7
.
7.4.4.p.r/Dr.r1/C5rC4D.rC2/
2
;yDx
2
.c1Cc2lnx/
7.4.6.p.r/Dr.r1/3rC13D.r2/
2
C9;yDx
2
Œc1cos.3lnx/Cc2sin.3lnx/.
7.4.8.p.r/D12r.r1/5rC6D.3r2/.4r3/;yDc1x
2=3
Cc2x
3=4
.
7.4.10.p.r/D3r.r1/rC1D.r1/.3r1/;yDc1xCc2x
1=3
.
7.4.12.p.r/Dr.r1/C3rC5D.rC1/
2
C4;yD
1
x
Œc1cos.2lnx/Cc2sin.2lnx
7.4.14.p.r/Dr.r1/rC10D.r1/
2
C9;yDx Œc1cos.3lnx/Cc2sin.3lnx/.
7.4.16.p.r/D2r.r1/C3r1D.rC1/.2r1/;yD
c1
x
Cc2x
1=2
.
7.4.18.p.r/D2r.r1/C10rC9D2.rC2/
2
C1;yD
1
x
2

c1cos

1
p
2
lnx

Cc2sin

1
p
2
lnx

.
7.4.20.Ifp.r/Dar.r1/CbrCcDa.rr1/
2
, then (A)p.r1/Dp
0
.r1/D0. IfyDux
r1
, then
y
0
Du
0
x
r1Cr1ux
r11
andy
00
Du
00
x
r1C2r1u
0
x
r11
1
Cr1.r11/x
r12
1
, so
ax
2
y
00
Cbxy
0
CcyDax
r1C2
u
00
C.2ar1Cb/x
r1C1
u
0
C.ar1.r11/Cbr1Cc/ x
r1
u
Dax
r1C2
u
00
Cp
0
.r1/x
r1C1
1
u
0
Cp.r/x
r1
uDax
r1C2
u
00
;
from (A). Therefore,u
00
D0, souDc1Cc2xandyDx
r1.c1Cc2x/.
7.4.22.(a)IftDx1andY.t/Dy.tC1/Dy.x/, then.1x
2
/y
00
2xy
0
C˛.˛C1/yD
t.2Ct/
d
2
Y
dt
2
2.1Ct/
dY
dt
C˛.˛C1/YD0, soysatisfies Legendre’s equation if and only ifY
satisfies (A)t.2Ct/
d
2
Y
dt
2
C2.1Ct/
dY
dt
˛.˛C1/YD0. Since (A) can be rewritten ast
2
.2Ct/
d
2
Y
dt
2
C
2t.1Ct/
dY
dt
˛.˛C1/tYD0, (A) has a regular singular point attD00.
(b)IftDxC1andY.t/Dy.t1/Dy.x/, then.1x
2
/y
00
2xy
0
C˛.˛C1/yDt.2t/
d
2
Y
dt
2
C
2.1t/
dY
dt
C˛.˛C1/Y, soysatisfies Legendre’s equation if and only ifYsatisfies (B)t.2t/
d
2
Y
dt
2
C

108. Section 7.5The Method of Frobenius I103
2.1t/
dY
dt
C˛.˛C1/Y, Since (B) can be rewritten as (B)t
2
.2t/
d
2
Y
dt
2
C2t.1t/
dY
dt
C˛.˛C1/tY,
(B) has a regular singular point attD00.
7.5The Method of Frobenius I
7.5.2.p0.r/Dr.3r1/;p1.r/D2.rC1/;p2.r/D 4.rC2/.
a1.r/D
2
3rC2
;an.r/D
2an1.r/4an2.r/
3nC3r1
,n1.
r1D1=3;a1.1=3/D 2=3;an.1=3/D
2an1.1=3/4an2.1=3/
3n
,n1;
y1Dx
1=3

1
2
3
xC
8
9
x
2

40
81
x
3
C

.
r2D0;a1.0/D 1;an.0/D
2an1.0/4an2.0/
3n1
,n1;
y2D1xC
6
5
x
2

4
5
x
3
C .
7.5.4.p0.r/D.rC1/.4r1/;p1.r/D2.rC2/;p2.r/D4rC7.
a1.r/D
2
4rC3
;an.r/D
2
4nC4r1
an1.r/
1
nCrC1
an2.r/,n1.
r1D1=4;a1.1=4/D 1=2;an.1=4/D
1
2n
an1.1=4/
4
4nC5
an2.1=4/,n1;
y1Dx
1=4

1
1
2
x
19
104
x
2
C
1571
10608
x
3
C

.
r2D 1;a1.1/D2;an.1/D
2
4n5
an1.1/
1
n
an2.1/,n1;
y2Dx
1

1C2x
11
6
x
2

1
7
x
3
C

.
7.5.6.p0.r/Dr.5r1/;p1.r/D.rC1/
2
;p2.r/D2.rC2/.5rC9/.
a1.r/D
rC1
5rC4
;an.r/D
nCr
5nC5r1
an1.r/2an2.r/,n1.
r1D1=5;a1.1=5/D 6=25;an.1=5/D
5nC1
25n
an1.1=5/2an2.1=5/,n1;
y1Dx
1=5

1
6
25
x
1217
625
x
2
C
41972
46875
x
3
C

.
r2D0;a1.0/D 1=4;an.0/D
n
5n1
an1.0/2an2.0/,n1;
y2Dx
1
4
x
2

35
18
x
3
C
11
12
x
4
C .
7.5.8.p0.r/D.3r1/.6rC1/;p1.r/D.3rC2/.6rC1/;p2.r/D3rC5.
a1.r/D
6rC1
6rC7
;an.r/D
6nC6r5
6nC6rC1
an1.r/
1
6nC6rC1
an2.r/,n1.
r1D1=3;a1.1=3/D 1=3;an.1=3/D
2n1
2nC1
an1.1=3/
1
6nC3
an2.1=3/,n1;
y1Dx
1=3

1
1
3
xC
2
15
x
2

5
63
x
3
C

.
r2D 1=6;a1.1=6/D0;an.1=6/D
n1
n
an1.1=6/
1
6n
an2.1=6/,n1;
y2Dx
1=6

1
1
12
x
2
C
1
18
x
3
C

.

109. 104 Chapter 7Series Solutions of Linear Second Order Equations
7.5.10.p0.r/D.2rC1/.5r1/;p1.r/D.2r1/.5rC4/;p2.r/D2.2rC5/.5r1/.
a1.r/D
2r1
2rC3
;an.r/D
2nC2r3
2nC2rC1
an1.r/
10nC10r22
5nC5r1
an2.r/,n1.
r1D1=5;a1.1=5/D3=17;an.1=5/D
10n13
10nC7
an1.1=5/
2n4
n
an2.1=5/,n1;
y1Dx
1=5

1C
3
17
x
7
153
x
2

547
5661
x
3
C

.
r2D 1=2;a1.1=2/D1;an.1=2/D
n2
n
an1.1=2/
20n54
10n7
an2.1=2/,n1;
y2Dx
1=2

1CxC
14
13
x
2

556
897
x
3
C

.
7.5.14.p0.r/D.rC1/.2r1/;p1.r/D2rC1;an.r/D
1
nCrC1
an1.r/.
r1D1=2;an.1=2/D
2
2nC3
an1.1=2/;y1Dx
1=2
1
X
nD0
.2/
n
Q
n
jD1
.2jC3/
x
n
.
r2D 1;an.1/D
1
n
an1.1/;y2Dx
1
1
X
nD0
.1/
n
nŠ
x
n
.
7.5.16.p0.r/D.rC2/.2r1/;p1.r/DrC3;an.r/D
1
2nC2r1
an1.r/.
r1D1=2;an.1=2/D
1
2n
an1.1=2/;y1Dx
1=2
1
X
nD0
.1/
n
2
n
nŠ
x
n
.
r2D 2;an.2/D
1
2n5
an1.2/;y2D
1
x
2
1
X
nD0
.1/
n
Q
n
jD1
.2j5/
x
n
.
7.5.18.p0.r/D.r1/.2r1/;p1.r/D 2;an.r/D
2
.nCr1/.2nC2r1/
an1.r/.
r1D1;an.1/D
2
n.2nC1
/an1.1/;y1Dx
1
X
nD0
2
n
nŠ
Q
n
jD1
.2jC1/
x
n
.
r2D1=2;an.1=2/D
2
n.2n1
/an1.1=2/;y2Dx
1=2
1
X
nD0
2
n
nŠ
Q
n
jD1
.2j1/
x
n
.
7.5.20.p0.r/D.r1/.3rC1/;p1.r/Dr3;an.r/D
nCr4
.nCr1/.3nC3rC1/
an1.r/.
r1D;an.1/D
n3
n.3nC4/
an1.1/;y1Dx

1C
2
7
xC
1
70
x
2

.
r2D 1=3;an.1=3/D
3n13
3n.3n4/
an1.1=3/;y2Dx
1=3
1
X
nD0
.1/
n
3
n
nŠ
0
@
n
Y
jD1
3j13
3j4
1
Ax
n
.
7.5.22.p0.r/D.r1/.4r1/;p1.r/Dr.rC2/;an.r/D
nCrC1
4nC4r1
an1.r/.
r1D1;an.1/D
nC2
4nC3
an1.1/;y1Dx
1
X
nD0
.1/
n
.nC2/Š
2
Q
n
jD1
.4jC3/
x
n
.
r2D1=4;an.1=4/D
4nC5
16n
an1.1=4/;y2Dx
1=4
1
X
nD0
.1/
n
16
n
nŠ
n
Y
jD1
.4jC5/x
n

110. Section 7.5The Method of Frobenius I105
7.5.24.p0.r/D.rC1/.3r1/;p1.r/D2.rC2/.2rC3/;an.r/D 2
2nC2rC1
3nC3r1
an1.r/.
r1D1=3;an.1=3/D 2
6nC5
9n
an1.1=3/;y1Dx
1=3
1
X
nD0
.1/
n
nŠ

2
9

n
0
@
n
Y
jD1
.6jC5/
1
Ax
n
;
r2D 1;an.1/D 2
2n1
3n4
an1.1/;y2Dx
1
1
X
nD0
.1/
n
2
n
0
@
n
Y
jD1
2j1
3j4
1
Ax
n
7.5.28.p0.r/D.2r1/.4r1/;p1.r/D.rC1/
2
;an.r/D
.nCr/
2
.2nC2r1/.4nC4r1/
an1.r/.
r1D1=2;an.1=2/D
4n
2
C4nC1
8n.4nC1/
an1.1=2/;y1Dx
1=2

1
9
40
xC
5
128
x
2

245
39936
x
3
C

.
r2D1=4;an.1=4/D
16n
2
C8nC1
32n.4n1/
an1.1=4/;y2Dx
1=4

1
25
96
xC
675
14336
x
2

38025
5046272
x
3
C

.
7.5.30.p0.r/D.2r1/.2rC1/;p1.r/D.2rC1/.3rC1/;an.r/D
.3nC3r2/
.2nC2rC1/
an.r/.
r1D1=2;an.1=2/D
6n1
4.nC1/
an1.1=2/;y1Dx
1=2

1
5
8
xC
55
96
x
2

935
1536
x
3
C

.
r2D 1=2;an.1=2/D
6n7
4n
an1.1=2/;y2Dx
1=2

1C
1
4
x
5
32
x
2

55
384
x
3
C

.
7.5.32.p0.r/D.2rC1/.3rC1/;p1.r/D.rC1/.rC2/;an.r/D
.nCr/.nCrC1/
.2nC2rC1/.3nC3rC1/
an.r/.
r1D 1=3;an.1=3/D
.3n1/.3nC2/
9n.6nC1/
an1.1=3/;y1Dx
1=3

1
10
63
xC
200
7371
x
2

17600
3781323
x
3
C

.
r2D 1=2;an.1=2/D
.2n1/.2nC1/
4n.6n1/
an1.1=2/;y2Dx
1=2

1
3
20
xC
9
352
x
2

105
23936
x
3
C

.
7.5.34.p0.r/D.2r1/.4r1/;p2.r/D .2rC3/.4rC3/;a2m.r/D
8mC4r5
8mC4r1
a2m2.r/.
r1D1=2;a2m.1=2/D
8m3
8mC1
a2m2.1=2/;y1Dx
1=2
1
X
mD0
0
@
m
Y
jD1
8j3
8jC1
1
Ax
2m
.
r2D1=4;a2m.1=4/D
2m1
2m
a2m2.1=4/;y2Dx
1=4
1
X
mD0
1
2
m
mŠ
0
@
m
Y
jD1
.2j1/
1
Ax
2m
7.5.36.p0.r/Dr.3r1/;p2.r/D.r4/.rC2/;a2m.r/D
2mCr6
6mC3r1
a2m2.r/.
r1D1=3;a2m.1=3/D
6m17
18m
a2m2.1=3/;y1Dx
1=3
1
X
mD0
.1/
m
18
m
mŠ
0
@
m
Y
jD1
.6j17/
1
Ax
2m
.
r2D0;a2m.0/D
2m6
6m1
a2m2.0/;y2D1C
4
5
x
2
C
8
55
x
4
7.5.38.p0.r/D.2r1/.3r1/;p2.r/D .rC1/.3rC5/;a2m.r/D
2mCr1
4mC2r1
a2m2.r/.

111. 106 Chapter 7Series Solutions of Linear Second Order Equations
r1D1=2;a2m.1=2/D
4m1
8m
a2m2.1=2/;y1Dx
1=2
1
X
mD0
1
8
m
mŠ
0
@
m
Y
jD1
.4j1/
1
Ax
2m
.
r2D1=3;a2m.1=3/D
6m2
12m1
a2m2.1=3/;y2Dx
1=3
1
X
mD0
2
m
0
@
m
Y
jD1
3j1
12j1
1
Ax
2m
.
7.5.40.p0.r/D.2r1/.2rC1/;p1.r/D.rC1/.2rC3/;a2m.r/D
2mCr1
4mC2rC1
a2m2.r/.
r1D1=2;a2m.1=2/D
4m1
4.2mC1/
a2m2.1=2/;y1Dx
1=2
1
X
mD0
.1/
m
4
m
0
@
m
Y
jD1
4j1
2jC1
1
Ax
2m
.
r2D 1=2;a2m.1=2/D
4m3
8m
a2m2.1=2/;y2Dx
1=2
1
X
mD0
.1/
m
8
m
mŠ
0
@
m
Y
jD1
.4j3/
1
Ax
2m
7.5.42.p0.r/D.rC1/.3r1/;p1.r/D.r1/.3rC5/;a2m.r/D
2mCr3
2mCrC1
a2m2.r/.
r1D1=3;a2m.1=3/D
3m4
3mC2
a2m2.1=3/;y1Dx
1=3
1
X
mD0
.1/
m
0
@
m
Y
jD1
3j4
3jC2
1
Ax
2m
.
r2D 1;a2m.1/D
m2
m
a2m2.1/;y2Dx
1
.1Cx
2
/
7.5.44.p0.r/D.rC1/.2r1/;p1.r/Dr
2
;a2m.r/D
.2mCr2/
2
.2mCrC1/.4mC2r1/
a2m2.r/.
r1D1=2;a2m.1=2/D
.4m3/
2
8m.4mC3/
a2m2.1=2/;y1Dx
1=2
1
X
mD0
.1/
m
8
m
mŠ
0
@
m
Y
jD1
.4j3/
2
4jC3
1
Ax
2m
.
r2D 1;a2m.1/D
.2m3/
2
2m.4m3/
a2m2.1/;y2Dx
1
1
X
mD0
.1/
m
2
m
mŠ
0
@
m
Y
jD1
.2j3/
2
4j3
1
Ax
2m
.
7.5.46.p0.r/D.3r1/.3rC1/;p1.r/D3rC5;a2m.r/D
1
6mC3rC1
a2m2.r/.
r1D1=3;a2m.1=3/D
1
2.3mC1/
a2m2.1=3/;y1Dx
1=3
1
X
mD0
.1/
m
2
m
Q
m
jD1
.3jC1/
x
2m
.
r2D 1=3;a2m.1=3/D
1
6m
a2m2.1=3/;y2Dx
1=3
1
X
mD0
.1/
m
6
m
mŠ
x
2m
7.5.48.p0.r/D2.rC1/.4r1/;p2.r/D.rC3/
2
;a2m.r/D
2mCrC1
2.8mC4r1/
a2m2.r/.
r1D1=4;a2m.1=4/D
8mC5
64m
a2m2.1=4/;y1Dx
1=4

1
13
64
x
2
C
273
8192
x
4

2639
524288
x
6
C

.
r2D 1;a2m.1/D
m
8m5
a2m2.1/;y2Dx
1

1
1
3
x
2
C
2
33
x
4

2
209
x
6
C

.
7.5.50.p0.r/D.2r1/.2rC1/;p2.r/D.2rC5/
2
;a2m.r/D
4mC2rC1
4mC2r1
a2m2.r/.

112. Section 7.5The Method of Frobenius I107
r1D1=2;a2m.1=2/D
2mC1
2m
a2m2.1=2/;y1Dx
1=2

1
3
2
x
2
C
15
8
x
4

35
16
x
6
C

.
r2D 1=2;a2m.1=2/D
2m
2m1
a2m2.1=2/;y2Dx
1=2

12x
2
C
8
3
x
4

16
5
x
6
C

.
7.5.52.(a)Multiplying (A)c1y1Cc2y20byx
r2yieldsc1x
r1r2
P
1
nD0
anx
n
Cc2
P
1
nD0
bnx
n
D0,
0 < x < . Lettingx!0Cshows thatc2D0, sinceb0D1. Now (A) reduces toc1y10, soc1D0.
Therefore,y1andy2are linearly independent on.0; /.
(b)Sincey1D
P
1
nD0
an.r1/x
n
andy2D
P
1
nD0
an.r2/x
n
are linearly independent solutions of
LyD0 .0; /,fy1; y2gis a fundamental set of solutions ofLyD0on.0; /, by Theorem 5.1.6.
7.5.54.(a)Ifx > 0, thenjxj
r
x
n
Dx
nCr
, so the assertions are obvious. Ifx < 0, thenjxj
r
D.x/
r
, so
d
dx
jxj
r
D r.x/
r1
D
r.x/
r
x
D
rjxj
r
x
. Therefore,(A)
d
dx
.jxj
r
x
n
/D
rjxj
r
x
x
n
C jxj
r
.nx
n1
/D
.nCr/jxj
r
x
n1
and
d
2
dx
2
.jxj
r
x
n
/D.nCr/
d
dx
.jxj
r
x
n1
/D.nCr/.nCr1/jxj
r
x
n2
, from (A)
withnreplaced byn1.
7.5.56.(a)Herep10, so Eqn. (7.5.12) reduces toa0.r/D1,a1.r/D0,an.r/D
p2.nCr2/
p0.nCr/
an2.r/,
r0, which implies thata2mC1.r/D0formD1; 2; 3; : : :. Therefore,Eqn. (7.5.12) actually reduces
toa0.r/D1,a2m.r/D
p2.2mCr2/
p0.2mCr/
, which holds because of condition (A).
(b)Similar to the proof of Exercise 7.5.55(a).
(c)p0.2mCr1/D2m˛0.2mCr1r2/, which is nonzero ifm > 0, sincer1r20. Therefore,
the assumptions of Theorem 7.5.2 hold withrDr1, andLy1Dp0.r1/x
r1D0. Ifr1r2is not an
even integer, thenp0.2mCr2/D2m˛0.2mr1Cr2/¤0,mD1; 2; . Hence, the assumptions
of Theorem 7.5.2 hold withrDr2andLy2Dp0.r2/x
r2D0. From Exercise 7.5.52,fy1; y2gis a
fundamental set of solutions.
(d)Similar to the proof of Exercise 7.5.55(c).
7.5.58.(a)From Exercise 7.5.57,bnD0forn1.
7.5.60.(a).˛0C˛1xC˛2x
2
/
1
X
nD0
anx
n
D˛0a0C.˛0a1C˛1a0/xC
1
X
nD2
.˛0anC˛1an1C˛2an2/x
n
D
1, so
1
X
nD0
anx
n
D
˛0a0
˛0C˛1xC˛2x
2
.
(b)If
p1.r1/
p0.r/
D
˛1
˛0
and
p2.r2/
p0.r/
D
˛2
˛0
, then Eqn. (7.5.12) is equivalent toa0.r/D1,˛0a1.r/C
˛1a0.r/D0,˛0an.r/C˛1an1.r/C˛2an2.r/D0,n2. Therefore,Theorem 7.5.2 implies the
conclusion.
7.5.62.p0.r/D.2r1/.3r1/;p1.r/D0;p2.r/D2.2rC3/.3rC5/;
p1.r1/
p0.r/
D0D
˛1
˛0
;
p2.r2/
p0.r/
D2D
˛2
˛0
;y1D
x
1=3
1C2x
2
;y2D
x
1=2
1C2x
2
.
7.5.64.p0.r/D5.3r1/.3rC1/;p1.r/D.3rC2/.3rC4/;p2.r/D0;
p1.r1/
p0.r/
D
1
5
D
˛1
˛0
;
p2.r2/
p0.r/
D0D
˛2
˛0
;y1D
x
1=3
5Cx
;y2D
x
1=3
5Cx
.

113. 108 Chapter 7Series Solutions of Linear Second Order Equations
7.5.66.p0.r/D.2r3/.2r1/;p1.r/D3.2r1/.2rC1/;p2.r/D.2rC1/.2rC3/;
p1.r1/
p0.r/
D
3D
˛1
˛0
;
p2.r2/
p0.r/
D1D
˛2
˛0
;y1D
x
1=2
1C3xCx
2
;y2D
x
3=2
1C3xCx
2
.
7.5.68.p0.r/D3.r1/.4r1/;p1.r/D2r.4rC3/;p2.r/D.rC1/.4rC7/;
p1.r1/
p0.r/
D
2
3
D
˛1
˛0
;
p2.r2/
p0.r/
D
1
3
D
˛2
˛0
;y1D
x
3C2xCx
2
;y2D
x
1=4
3C2xCx
2
.
7.6THE METHOD OF FROBENIUS II
7.6.2.p0.r/D.rC1/
2
;p1.r/D.rC2/.rC3/;p2.r/D.rC3/.2r1/;
a1.r/D
rC3
rC2
;an.r/D
nCrC2
nCrC1
an1.r/
2nC2r5
nCrC1
an2.r/,n2.
a
0
1
.r/D
1
.rC2/
2
;a
0
n
.r/D
nCrC2
nCrC1
a
0
n1
.r/
2nC2r5
nCrC1
a
0
n2
.r/C
1
.nCrC1/
2
an1.r/
7
.nCrC1/
2
an2.r/,n2.
r1D 1;a1.1/D 2;an.1/D
nC1
n
an1.1/
2n7
n
an2.1/,n2;
y1Dx
1

12xC
9
2
x
2

20
3
x
3
C

;
a
0
1
.1/D1;a
0
n
.1/D
nC1
n
an1.1/
2n7
n
an2.1/C
1
n
2
an1.1/
7
n
2
an2.1/,n2;
y2Dy1lnxC1
15
4
xC
133
18
x
2
C .
7.6.4.p0.r/D.2r1/
2
;p1.r/D.2rC1/.2rC3/;p2.r/D.2rC1/.2rC3/;
a1.r/D
2rC3
2rC1
;an.r/D
.2nC2rC1/an1.r/.2nC2r3/an2.r/
2nC2r1
,n2.
a
0
1
.r/D
4
.2rC1/
2
;a
0
n
.r/D
.2nC2rC1/a
0
n1
.r/.2nC2r3/a
0
n2
.r/
2nC2r1
C
4.an1.r/an2.r//
.2nC2r1/
2
;
n2.
r1D1=2;a1.1=2/D 2;an.1=2/D
.nC1/an1.1=2/C.n1/an2.1=2/
n
;n2;
y1Dx
1=2

12xC
5
2
x
2
2x
3
C

;
a
0
1
.1=2/D1;a
0
n
.1=2/D
.nC1/a
0
n1
.1=2/C.n1/a
0
n2
.1=2/
n
C
an1.1=2/an2.1=2/
n
2
,n
2;
y2Dy1lnxCx
3=2

1
9
4
xC
17
6
x
2
C

.
7.6.6.p0.r/D.3rC1/
2
;p1.r/D3.3rC4/;p2.r/D 2.3rC7/;
a1.r/D
3
3rC4
;an.r/D
3an1.r/C2an2.r/
3nC3rC1
;n2;
a
0
1
.r/D
9
.3rC4/
2
;a
0
n
.r/D
3a
0
n1
.r/C2a
0
n2
.r/
3nC3rC1
C
9an1.r/6an2.r/
.3nC3rC1/
2
;n2.
r1D 1=3;a1.1=3/D 1;an.1=3/D
3an1.1=3/C2an2.1=3/
3n
,n2;
y1Dx
1=3

1xC
5
6
x
2

1
2
x
3
C

;

114. Section 7.6The Method of Frobenius II109
a
0
1
.1=3/D1;a
0
n
.1=3/D
3a
0
n1
.r/C2a
0
n2
.r/
3n
C
3an1.r/2an2.r/
3n
2
;n2;
y2Dy1lnxCx
2=3

1
11
12
xC
25
36
x
2
C

.
7.6.8.p0.r/D.rC2/
2
;p1.r/D2.rC3/
2
;p2.r/D3.rC4/;
a1.r/D 2;an.r/D 2an1.r/
3an2.r/
nCrC2
;n2;
a
0
1
.r/D0;a
0
n
.r/D 2a
0
n1
.r/
3a
0
n2
.r/
nCrC2
C
3an2.r/
.nCrC2/
2
,n2.
r1D 2;a1.2/D 2;an.2/D 2an1.2/
3an2.2/
n
,n2;
y1Dx
2

12xC
5
2
x
2
3x
3
C

;
a
0
1
.2/D0;a
0
n
.2/D 2an1.2/
3an2.2/
n
C
3an2.2/
n
2
;n2;
y2Dy1lnxC
3
4

13
6
xC .
7.6.10.p0.r/D.4rC1/
2
;p1.r/D4rC5;p2.r/D2.4rC9/;
a1.r/D
1
4rC5
;an.r/D
an1.r/C2an2.r/
4nC4rC1
;n2;
a
0
1
.r/D
4
.4rC5/
2
;a
0
n
.r/D
a
0
n1
.r/C2a
0
n2
.r/
4nC4rC1
C
4an1.r/C8an2.r/
.4nC4rC1/
2
;n2.
r1D 1=4;a1.1=4/D 1=4;an.1=4/D
an1.1=4/C2an2.1=4/
4n
;n2;
y1Dx
1=4

1
1
4
x
7
32
x
2
C
23
384
x
3
C

;
a
0
1
.1=4/D1=4;a
0
n
.1=4/D
a
0
n1
.1=4/C2a
0
n2
.1=4/
4n
C
an1.1=4/C2an2.1=4/
4n
2
;n
2;
y2Dy1lnxCx
3=4

1
4
C
5
64
x
157
2304
x
2
C

.
7.6.12.p0.r/D.2r1/
2
;p1.r/D4;
an.r/D
4
.2nC2r1/
2
an1.r/;
an.r/D
.4/
n
Q
n
jD1
.2jC2r1/
.
By logarithmic differentiation,a
0
n
.r/Dan.r/
n
X
jD1
2
2jC2r1
;
r1D1=2;an.1=2/D
.1/
n
.nŠ/
2
;
a
0
n
.1=2/Dan.1=2/
0
@2
n
X
jD1
1
j
1
A;
y1Dx
1=2
1
X
nD0
.1/
n
.nŠ/
2
x
n
;

115. 110 Chapter 7Series Solutions of Linear Second Order Equations
y2Dy1lnx2x
1=2
1
X
nD1
.1/
n
.nŠ/
2
0
@
n
X
jD1
1
j
1
Ax
n
;
7.6.14.p0.r/D.r2/
2
;p1.r/Dr
2
;an.r/D
.nCr1/
2
.nCr2/
2
an1.r/;an.r/D.1/
n
.nCr1/
2
.r1/
2
;
a
0
n
.r/D.1/
nC1
2n.rCn1/
.r1/
3
;r1D2;an.2/D.1/
n
.nC1/
2
;a
0
n
.2/D.1/
nC1
2n.nC1/;
y1Dx
2
1
X
nD0
.1/
n
.nC1/
2
x
n
;y2Dy1lnx2x
2
1
X
nD1
.1/
n
n.nC1/x
n
.
7.6.16.p0.r/D.5r1/
2
;p1.r/DrC1;
an.r/D
.nCr/
.5nC5r1/
2
an1.r/;
an.r/D.1/
n
n
Y
jD1
.jCr/
.5jC5r1/
2
;
By logarithmic differentiation,
a
0
n
.r/D an.r/
n
X
jD1
.5jC5rC1/
.jCr/.5jC5r1/
;
r1D1=5;an.1=5/D.1/
n
n
Y
jD1
.5jC1/
125
n
.nŠ/
2
;
a
0
n
.1=5/Dan.1=5/
n
X
jD1
5jC2
j.5jC1/
;
y1Dx
1=5
1
X
nD0
.1/
n
Q
n
jD1
.5jC1/
125
n
.nŠ/
2
x
n
;
y2Dy1lnxx
1=5
1
X
nD1
.1/
n
Q
n
jD1
.5jC1/
125
n
.nŠ/
2
0
@
n
X
jD1
5jC2
j.5jC1/
1
Ax
n
.
7.6.18.p0.r/D.3r1/
2
;p1.r/D.2r1/
2
;
an.r/D
.2nC2r3/
2
.3nC3r1/
2
an1.r/;
an.r/D.1/
n
n
Y
jD1
.2jC2r3/
2
.3jC3r1/
2
;
By logarithmic differentiation,
a
0
n
.r/D14an.r/
n
X
jD1
1
.2jC2r3/.3jC3r1/
;
r1D1=3;an.1=3/D
.1/
n
Q
n
jD1
.6j7/
2
81
n
.nŠ/
2
;
a
0
n
.1=3/D14an.1=3/
n
X
jD1
1
j.6j7/
/;
y1Dx
1=3
1
X
nD0
.1/
n
Q
n
jD1
.6j7/
2
81
n
.nŠ/
2
x
n
;

116. Section 7.6The Method of Frobenius II111
y2Dy1lnxC14x
1=3
1
X
nD1
.1/
n
Q
n
jD1
.6j7/
2
81
n
.nŠ/
2
0
@
n
X
jD1
1
j.6j7/
/
1
Ax
n
.
7.6.20.p0.r/D.rC1/
2
;p1.r/D 2.rC2/.2rC3/;
an.r/D
2.2nC2rC1/
nCrC1
an1.r/,n1;an.r/D2
n
n
Y
jD1
2jC2rC1
jCrC1
;
By logarithmic differentiation,
a
0
n
.r/Dan.r/
n
X
jD1
1
.jCrC1/.2jC2rC1/
;
r1D 1;an.1/D
2
n
Q
n
jD1
.2j1/
nŠ
;
a
0
n
.1/Dan.1/
n
X
jD1
1
j.2j1/
;
y1D
1
x
1
X
nD0
2
n
Q
n
jD1
.2j1/
nŠ
x
n
;
y2Dy1lnxC
1
x
1
X
nD1
2
n
Q
n
jD1
.2j1/
nŠ
0
@
n
X
jD1
1
j.2j1/
1
Ax
n
.
7.6.22.p0.r/D2.r2/
2
;p1.r/D.r1/.2rC1/;
an.r/D
2nC2r1
2.nCr2/
an1.r/;
an.r/D
.1/
n
2
n
n
Y
jD1
2jC2r1
jCr2
;
By logarithmic differentiation,
a
0
n
.r/D 3an.r/
n
X
jD1
1
.jCr2/.2jC2r1/
;
r1D2;an.2/D
.1/
n
Q
n
jD1
.2jC3/
2
n
nŠ
;
a
0
n
.2/D 3an.2/
n
X
jD1
1
j.2jC3/
;
y1Dx
2
1
X
nD0
.1/
n
Q
n
jD1
.2jC3/
2
n
nŠ
x
n
;
y2Dy1lnx3x
2
1
X
nD0
.1/
n
Q
n
jD1
.2jC3/
2
n
nŠ
0
@
n
X
jD1
1
j.2jC3/
1
Ax
n
.
7.6.24.p0.r/D.r3/
2
;p1.r/D 2.r1/.rC2/;
an.r/D
2.nCr2/.nCrC1/
.nCr3/
2
an1.r/;
a
0
n
.r/D
2.nCr2/.nCrC1/
.nCr3/
2
a
0
n1
.r/
2.5nC5r7/
.nCr3/
3
an1.r/;
r1D3;an.3/D
2.nC1/.nC4/
n
2
an1.3/;

117. 112 Chapter 7Series Solutions of Linear Second Order Equations
y1Dx
3
.1C20xC180x
2
C1120x
3
C ;
a
0
n
.3/D
2.nC1/.nC4/
n
2
a
0
n1
.3/
2.5nC8/
n
3
an1.3/;
y2Dy1lnxx
4

26C324xC
6968
3
x
2
C

7.6.26.p0.r/Dr
2
;p1.r/Dr
2
CrC1;
an.r/D
.n
2
Cn.2r1/Cr
2
rC1/
.nCr/
2
an1.r/;
a
0
n
.r/D
.n
2
Cn.2r1/Cr
2
rC1/
.nCr/
2
a
0
n1
.r/
.nCr2/
.nCr/
3
an1.r/;
r1D0;an.0/D
.n
2
nC1/
n
2
an1.0/;
y1D1xC
3
4
x
2

7
12
x
3
C ;
a
0
n
.0/D
.n
2
nC1/
n
2
a
0
n1
.0/
.n2/
n
3
an1.0/;
y2Dy1lnxCx

1
3
4
xC
5
9
x
2
C

.
7.6.28.p0.r/D.r1/
2
;p2.r/DrC1;
a2m.r/D
1
2mCr1
a2m2.r/,n1;a2m.r/D
.1/
m
Q
m
jD1
.2jCr1/
;
By logarithmic differentiation,
a
0
2m
.r/Da2m.r/
m
X
jD1
;
r1D1;a2m.1/D
.1/
m
2
m
mŠ
;
a
0
2m
.1/D
1
2
a2m.1/
m
X
jD1
1
j
;
y1Dx
1
X
mD0
.1/
m
2
m
mŠ
x
2m
;
y2Dy1lnx
x
2
1
X
mD1
.1/
m
2
m
mŠ
0
@
m
X
jD1
1
j
1
Ax
2m
.
7.6.30.p0.r/D.2r1/
2
;p2.r/D2rC3;
a2m.r/D
1
4mC2r1
a2m2.r/;
a2m.r/D
.1/
m
Q
m
jD1
.4jC2r1/
;
By logarithmic differentiation,
a
0
2m
.r/D 2a2m.r/
m
X
jD1
1
4jC2r1
;
r1D1=2;a2m.1=2/D
.1/
m
4
m
mŠ
;

118. Section 7.6The Method of Frobenius II113
a
0
2m
.1=2/D
1
2
a2m.1=2/
m
X
jD1
1
j
;
y1Dx
1=2
1
X
mD0
.1/
m
4
m
mŠ
x
2m
;
y2Dy1lnx
x
1=2
2
1
X
mD1
.1/
m
4
m
mŠ
0
@
m
X
jD1
1
j
1
Ax
2m
.
7.6.32.p0.r/D.2r1/
2
;p2.r/D.rC1/.2rC3/;a2m.r/D
2mCr1
4mC2r1
a2m2.r/;
a2m.r/D.1/
m
m
Y
jD1
2jCr1
4jC2r1
;
By logarithmic differentiation,
a
0
2m
.r/Da2m.r/
m
X
jD1
1
.2jCr1/.4jC2r1/
;
r1D1=2;a2m.1=2/D
.1/
m
Q
m
jD1
.4j1/
8
m
mŠ
;
a
0
2m
.1=2/Da2m.1=2/
m
X
jD1
1
2j.4j1/
;
y1Dx
1=2
1
X
mD0
.1/
m
Q
m
jD1
.4j1/
8
m
mŠ
x
2m
;
y2Dy1lnxC
x
1=2
2
1
X
mD1
.1/
m
Q
m
jD1
.4j1/
8
m
mŠ
0
@
m
X
jD1
1
j.4j1/
1
Ax
2m
.
7.6.34.p0.r/D.4rC1/
2
;p2.r/D.r1/.4rC9/;
a2m.r/D
2mCr3
8mC4rC1
a2m2.r/;
a2m.r/D.1/
m
m
Y
jD1
2jCr3
8jC4rC1
;
By logarithmic differentiation,
a
0
2m
.r/Da2m.r/
m
X
jD1
13
.2jCr3/.8jC4rC1/
;
r1D 1=4;a2m.1=4/D
.1/
m
Q
m
jD1
.8j13/
.32/
m
mŠ
;
a
0
2m
.1=4/Da2m.1=4/
m
X
jD1
13
2j.8j13/
;
y1Dx
1=4
1
X
mD0
.1/
m
Q
m
jD1
.8j13/
.32/
m
mŠ
x
2m
;
y2Dy1lnxC
13
2
x
1=4
1
X
mD1
.1/
m
Q
m
jD1
.8j13/
.32/
m
mŠ
0
@
m
X
jD1
1
j.8j13/
1
Ax
2m
.
7.6.36.p0.r/D.2r1/
2
;p2.r/D16r.rC1/;

119. 114 Chapter 7Series Solutions of Linear Second Order Equations
a2m.r/D
16.2mCr2/.2mCr1/
.4mC2r1/
2
a2m2.r/;
a2m.r/D.16/
m
m
Y
jD1
.2jCr2/.2jCr1/
.4jC2r1/
2
;
By logarithmic differentiation,
a
0
2m
.r/Da2m.r/
m
X
jD1
8jC4r5
.2jCr2/.2jCr1/.4jC2r1/
;
r1D1=2;a2m.1=2/D
.1/
m
Q
m
jD1
.4j3/.4j1/
4
m
.mŠ/
2
;
a
0
2m
.1=2/Da2m.1=2/
m
X
jD1
8j3
j.4j3/.4j1/
;
y1Dx
1=2
1
X
mD0
.1/
m
Q
m
jD1
.4j3/.4j1/
4
m
.mŠ/
2
x
2m
;
y2Dy1lnxCx
1=2
1
X
mD1
.1/
m
Q
m
jD1
.4j3/.4j1/
4
m
.mŠ/
2
0
@
m
X
jD1
8j3
j.4j3/.4j1/
1
Ax
2m
.
7.6.38.p0.r/D.rC1/
2
;p2.r/D.rC3/.2r1/;
a2m.r/D
4mC2r5
2mCrC1
a2m2.r/;
a2m.r/D.1/
m
m
Y
jD1
4jC2r5
2jCrC1
;
By logarithmic differentiation,
a
0
2m
.r/Da2m.r/
m
X
jD1
7
.2jCrC1/.4jC2r5/
;
r1D 1;a2m.1/D
.1/
m
Q
m
jD1
.4j7/
2
m
mŠ
;
a
0
2m
.1/Da2m.1/
m
X
jD1
7
2j.4j7/
;
y1D
1
x
1
X
mD0
.1/
m
Q
m
jD1
.4j7/
2
m
mŠ
x
2m
;
y2Dy1lnxC
7
2x
1
X
mD1
.1/
m
Q
m
jD1
.4j7/
2
m
mŠ
0
@
m
X
jD1
1
j.4j7/
1
Ax
2m
.
7.6.40.p0.r/D.r1/
2
;p2.r/DrC1;
a2m.r/D
1
2mCr1
a2m2.r/;
a
0
2m
.r/D
1
2mCr1
a
0
2m2
.r/C
1
.2mCr1/
2
a2m2.r/;
r1D1;a2m.1/D
1
2m
a2m2.1/;
y1Dx

1
1
2
x
2
C
1
8
x
4

1
48
x
6
C

;

120. Section 7.6The Method of Frobenius II115
a
0
2m
.1/D
1
2m
a
0
2m2
.1/C
1
4m
2
a2m2.1/,m1;
y2Dy1lnxCx
3

1
4

3
32
x
2
C
11
576
x
4
C

.
7.6.42.p0.r/D2.rC3/
2
;p2.r/Dr
2
2rC2;
a2m.r/D
4m
2
C4m.r3/Cr
2
6rC10
2.2mCrC3/
2
a2m2.r/;
a
0
2m
.r/D
4m
2
C4m.r3/Cr
2
6rC10
2.2mCrC3/
2
a
0
2m2
.r/
12mC6r19
.2mCrC3/
3
a2m2.r/;
r1D 3;a2m.3/D
4m
2
24mC37
8m
2
a2m2.3/;
y1Dx
3

1
17
8
x
2
C
85
256
x
4

85
18432
x
6
C

;
a
0
2m
.3/D
4m
2
24mC37
8m
2
a
0
2m2
.3/C
3712m
8m
3
a2m2.3/,m1;
y2Dy1lnxCx
1

25
8

471
512
x
2
C
1583
110592
x
4
C

.
7.6.44.p0.r/D.rC1/
2
;p1.r/D2.2r/.rC1/;r1D 1.
an.r/D
2.nCr/.nCr3/
.nCrC1/
2
an1.r/;an.r/D2
n
n
Y
jD1
.jCr/.jCr3/
.jCrC1/
2
,n0. Therefore,an.1/D
0ifn1andy1D1=x. Ifn4, thenan.r/D.rC1/
2
bn.r/, whereb
0
n
.1/exists; thereforea
0
n
.1/D
0ifn4. ForrD1; 2; 3,an.r/D.rC1/cn.r/, wherec1.r/D
2.r2/
.rC2/
2
,c2.r/D
4.r2/.r1/
.rC2/.rC3/
2
,
c3.r/D
8r.r2/.r1/
.rC2/.rC3/.rC4/
2
. Hence,a
0
1
.1/Dc1.1/D 6,a
0
2
.1/Dc2.1/D6,a
0
3
.1/D
c3.1/D 8=3, andy2Dy1lnx6C6x
8
3
x
2
.
7.6.46.p0.r/D.rC1/
2
;p1.r/D .r1/.rC2/;r1D 1.
an.r/D
nCr2
nCrC1
an1.r/;an.r/D
m
Y
jD1
jCr2
jCrC1
,n0. Therefore,a1.1/D 2,a2.1/D1,
andan.1/D0ifn3, soy1D
.x1/
2
x
.
a1.r/D
r1
rC2
,a
0
1
.r/D
3
.rC2/
2
,a
0
1
.1/D3;a2.r/D
r.r1/
.rC2/.rC3/
,a
0
2
.r/D
6.r
2
C2r1/
.rC2/
2
.rC3/
2
,
a
0
2
.1/D 3; ifn3 an.r/D.rC1/cn.r/wherecn.r/D
r.r1/
.nCr/.nCr1/.nCrC1/
, so
a
0
n
.1/Dcn.1/D
2
n.n2/.n1/
andy2Dy1lnxC33xC2
1
X
nD2
1
n.n
2
1/
x
n
.
7.6.48.p0.r/D.r2/
2
;p1.r/D .r5/.r1/;r1D2.
an.r/D
nCr6
nCr2
an1.r/;
an.r/D
m
Y
jD1
jCr6
jCr2
,n0. Therefore,a1.2/D 3,a2.2/D3,a3.2/D 1, andan.2/D0if
n4, soy1Dx
2
.1x/
3
.

121. 116 Chapter 7Series Solutions of Linear Second Order Equations
a1.r/D
r5
r1
,a
0
1
.r/D
4
.r1/
2
,a
0
1
.2/D4;
a2.r/D
.r5/.r4/
r.r1/
,a
0
2
.r/D
4.2r
2
10rC5/
r
2
.r1/
2
,a
0
2
.2/D 7;
a3.r/D
.r5/.r4/.r3/
r.r1/.rC1/
,a
0
3
.r/D
12.r
4
8r
3
C16r
2
5/
r
2
.r1/
2
.rC1/
2
,a
0
3
.2/D11=3; ifn4, then
an.r/D.r2/cn.r/wherecn.r/D
.r5/.r4/.r3/
.nCr5/.nCr4/.nCr3/.nCr2/
, soa
0
n
.2/D
cn.2/D
6
n.n2/.n
2
1/
and
y2Dy1lnxCx
3
 
47xC
11
3
x
2
6
1
X
nD3
1
n.n2/.n
2
1/
x
n
!
.
7.6.50.p0.r/D.3r1/
2
;p2.r/D73r;r1D1=3.
a2m.r/D
6mC3r13
.6mC3r1/
2
a2m2.r/;
a2m.r/D
m
Y
jD1
6jC3r13
.6jC3r1/
2
,m0. Therefore,a2.1=3/D1=6anda2m.1=3/D0ifm2, so
y1Dx
1=3

1
1
6
x
2

.
a2.r/D
3r7
.3rC5/
2
;a
0
2
.r/D
3.193r/
.3rC5/
3
;a
0
2
.1=3/D1=4. Ifm2, thena2m.r/D.r1=3/c2m.r/
wherec2m.r/D
3.3r7/
.6mC3r7/.6mC3r1/
Q
m
jD1
.6jC3r1/
, soa
0
2m
.1=3/Dc2m.1=3/D

1
12
1
6
m1
.m1/m mŠ
, and
y2Dy1lnxCx
7=3
 
1
4

1
12
1
X
mD1
1
6
m
m.mC1/.mC1/Š
x
2m
!
.
7.6.52.p0.r/D.2rC1/
2
;p2.r/D72r;r1D 1=2.
a2m.r/D
4mC2r11
.4mC2rC1/
2
a2m2.r/;
a2m.r/D
m
Y
jD1
4jC2r11
.4jC2rC1/
2
,m0. Therefore,a2.1=2/D 1=2,a4.1=2/D1=32, and
a2m.1=2/D0ifm3, soy1Dx
1=2

1
1
2
x
2
C
1
32
x
4

.
a2.r/D
2r7
.2rC5/
2
,a
0
2
.r/D
2.192r/
.2rC5/
3
,a
0
2
.1=2/D5=8,
a4.r/D
.2r7/.2r3/
.2rC5/
2
.2rC9/
2
,a
0
4
.r/D
4.8r
3
60r
2
146rC519/
.2rC5/
3
.2rC9/
3
,a
0
4
.1=2/D 9=128; if
m3, thena2m.r/D.rC1=2/c2m.r/where
c2m.r/D
2.2r7/.2r3/
.4mC2r7/.4mC2r3/.4mC2rC1/
Q
m
jD1
.4jC2rC1/
, soa
0
2m
.1=2/Dc2m.1=2/D
1
4
m
.m2/.m1/m mŠ
, and

122. Section 7.6The Method of Frobenius II117
y2Dy1lnxCx
3=2
 
5
8

9
128
x
2
C
1
X
mD2
1
4
mC1
.m1/m.mC1/.mC1/Š
x
2m
!
.
7.6.54.(a)Ifp0.r/D˛0.rr1/
2
, then (A)an.r/D
.1/
n
˛
n
0
n
Y
jD1
p1.jCr1/
.jCrr1/
2
. Therefore,an.r1/D
.1/
n
˛
n
0
.nŠ/
2
n
Y
jD1
p1.jCr11/. Theorem 7.6.2 impliesLy1D0.
(b)From (A), lnjan.r/j D nlnj˛0j C
n
X
jD1
.lnjp1.jCr1/j 2lnjjCrr1j/, soa
0
n
.r/D
an.r/
n
X
jD1

p
0
1
.jCr1/
p1.jCr1/

2
jCrr1

anda
0
n
.r1/Dan.r1/
n
X
jD1

p
0
1
.jCr11/
p1.jCr11/

2
j

. Theo-
rem 7.6.2 implies thatLy2D0.
(c)Sincep1.r/D1,y1andy2reduce to the stated forms. If1D0, theny1Dx
r1andy2D
x
r1lnx, which are solutions of the Euler equation˛0x
2
y
00
Cˇ0xy
0
C0y.
7.6.54.(a)Ly1Dp0.r1/x
r1D0. Now use the fact thatp0.jCr1/D˛0j
2
, so
Q
n
jD1
p0.jCr1/D
˛
n
0
.nŠ/
2
.
(b)From Theorem 7.6.2,y2Dy1lnxCx
r1
1
X
nD1
a
0
n
.r1/x
n
is a second solution ofLyD0. Since
an.r/D
.1/
n
˛
n
0
n
Y
jD1
p1.jCr1/
.jCrr1/
2
, (A) lnjan.r/j D nlnj˛0j C
n
X
jD1
lnjp1.jCr1/j 2
n
X
jD1
lnjjC
rr1j, provided thatp1.jCr1/andjCrr1are nonzero for all positive integersj. Differentiating (A)
and then settingrDr1yields
a
0
n
.r1/
an.r1/
D
n
X
jD1
p
0
1
.jCr11/
p2.jCr11/
2
n
X
jD1
1
j
, which implies the conclusion.
(c)In this casep1.r/D1andp
0
1
.r/D0, soan.r1/D
.1/
n
.nŠ/
2

1
˛0

n
andJnD 2
n
X
jD1
1
j
. If
1D0, theny1Dx
r1andy2Dx
r1lnx, while the differential equation is an Euler equation with
indicial polynomial˛0.rr
2
1
/. See Theorem 7.4.3.
7.6.56.p0.r/Dr
2
;p1.r/D1;r1D0.a2m.r/D
a2m1.r/
.2mCr/
2
,m1;a2m.r/D
.1/
m
Q
m
jD1
.2jCr/
2
,
m0. Therefore,a2m.0/D
.1/
m
4
m
.mŠ/
2
, soy1D
1
X
mD0
.1/
m
4
m
.mŠ/
2
x
2m
.
By logarithmic differentiation,a
0
2m
.r/D 2a2m.r/
m
X
jD1
1
2mCr
, soa
0
2m
.0/D a2m.0/
m
X
jD1
1
j
and
y2Dy1lnx
1
X
mD1
.1/
m
4
m
.mŠ/
2
0
@
m
X
jD1
1
j
1
Ax
2m
.
7.6.58.p0.r/D.2r1/
2
;p1.r/D.2rC1/
2
;p2.r/D0;
p1.r1/
p0.r/
D1D
˛1
˛0
;
p2.r2/
p0.r/
D0D
˛2
˛0
;
y1D
x
1=2
1Cx
;y2D
x
1=2
lnx
1Cx
.

123. 118 Chapter 7Series Solutions of Linear Second Order Equations
7.6.60.p0.r/D2.r1/
2
;p1.r/D0;p2.r/D .rC1/
2
;
p1.r1/
p0.r/
D0D
˛1
˛0
;
p2.r2/
p0.r/
D
1=2D
˛2
˛0
;y1D
x
2x
2
;y2D
xlnx
2x
2
.
7.6.62.p0.r/D4.r1/
2
;p1.r/D3r
2
;p2.r/D0;
p1.r1/
p0.r/
D3=4D
˛1
˛0
;
p2.r2/
p0.r/
D0D
˛2
˛0
;
y1D
x
4C3x
;y2D
xlnx
4C3x
.
7.6.64.p0.r/D.r1/
2
;p1.r/D 2r
2
;p2.r/D.rC1/
2
;
p1.r1/
p0.r/
D 2D
˛1
˛0
;
p2.r2/
p0.r/
D1D
˛2
˛0
;y1D
x
.1x/
2
;y2D
xlnx
.1x/
2
.
7.6.66.See the proofs of Theorems 7.6.1 and 7.6.2.
7.7THE METHOD OF FROBENIUS III
7.7.2.p0.r/Dr.r1/;p1.r/D1;r1D1;r2D0;kDr1r2D1;
an.r/D
1
.nCr/.nCr1/
an1.r/;
an.r/D
.1/
n
Q
n
jD1
.jCr/.jCr1/
;
an.1/D
.1/
n
nŠ.nC1/Š
;
y1Dx
1
X
nD0
.1/
n
nŠ.nC1/Š
x
n
;
´D1;CD p1.0/a0.0/D 1.
By logarithmic differentiation,
a
0
n
.r/D an.r/
n
X
jD1
2nC2r1
.nCr/.nCr1/
;
a
0
n
.1/D an.1/
n
X
jD1
2jC1
j.jC1/
;
y2D1y1lnxCx
1
X
nD1
.1/
n
nŠ.nC1/Š
0
@
n
X
jD1
2jC1
j.jC1/
1
Ax
n
.
7.7.4.p0.r/Dr.r1/;p1.r/DrC1;r1D1;r2D0;kDr1r2D1;an.r/D
an1.r/
nCr1
;an.r/D
.1/
n
Q
n
jD1
.jCr1/
;an.1/D
.1/
n
nŠ
;y1Dx
1
X
nD0
.1/
n
nŠ
x
n
Dxe
x
;´D1;CD p1.0/a0.0/D 1.
By logarithmic differentiation,a
0
n
.r/D an.r/
n
X
jD1
1
jCr1
;a
0
n
.1/D an.1/
n
X
jD1
1
j
;y2D1y1lnxCx
1
X
nD1
.1/
n
nŠ
0
@
n
X
jD1
1
j
1
A
7.7.6.p0.r/D.r1/.rC2/;p1.r/DrC3;r1D1;r2D 2;kDr1r2D3.an.r/D

1
nCr1
an1.r/;an.r/D
.1/
n
Q
n
jD1
.jCr1/
;an.1/D
.1/
n
nŠ
;y1Dx
1
X
nD0
.1/
n
nŠ
x
n
Dxe
x
;

124. Section 7.7The Method of Frobenius III119
´Dx
2

1C
1
2
xC
1
2
x
2

;CD
p1.0/
3
a2.2/D 1=2. By logarithmic differentiation,a
0
n
.r/D
an.r/
n
X
jD1
1
jCr1
;a
0
n
.1/Dan.1/
n
X
jD1
1
j
;y2Dx
2

1C
1
2
xC
1
2
x
2


1
2
0
@y1lnxx
1
X
nD1
.1/
n
nŠ
0
@
n
X
jD1
1
j
1
Ax
n
1
A;
7.7.8.p0.r/D.rC2/.rC7/;p1.r/D1;r1D 2;r2D 7;kDr1r2D5;an.r/D

an1.r/
.nCrC2/.nCrC7/
;an.r/D
.1/
n
Q
n
jD1
.jCrC2/.jCrC7/
;an.2/D120
n
Y
jD1
.1/
n
nŠ.nC5/Š
;
y1D
120
x
2
1
X
nD0
.1/
n
nŠ.nC5/Š
x
n
;
´Dx
7

1C
1
4
xC
1
24
x
2
C
1
144
x
3
C
1
576
x
4

;CD
p1.3/
5
a4.7/D 1=2880. By logarith-
mic differentiation,a
0
n
.r/D an.r/
n
X
jD1

2jC2rC9
.jCrC2/.jCrC7/
;a
0
n
.2/D an.2/
n
X
jD1
2jC5
j.jC5/
;
y2Dx
7

1C
1
4
xC
1
24
x
2
C
1
144
x
3
C
1
576
x
4


1
2880
0
@y1lnx
120
x
2
1
X
nD1
.1/
n
nŠ.nC5/Š
0
@
n
X
jD1
2jC5
j.jC5/
1
Ax
n
1
A.
7.7.10.p0.r/Dr.r4/;p1.r/D.r6/.r5/;r1D4;r2D0;kDr1r2D4;an.r/D

.nCr7/.nCr6/
.nCr/.nCr4/
an1.r/;an.r/D.1/
n
n
Y
jD1
.jCr7/.jCr6/
.jCr/.jCr4/
. SettingrD4yields
y1Dx
4

1
2
5
x

.´D1C10xC50x
2
C200x
3
;CD
p1.3/
4
a3.0/D 300.a1.r/D
.r6/.r5/
.r3/.rC1/
;
a
0
1
.r/D
3.3r
2
22rC31/
.r3/
2
.rC1/
2
;a
0
1
.4/D27=25.a2.r/D.r4/c2.r/, withc2.r/D
.r6/.r5/
2
.r3/.r2/.rC1/.rC2/
,
soa
0
2
.4/Dc2.4/D 1=30. Ifn3, thenan.r/D.r4/
2
bn.r/whereb
0
n
.4/exists, soa
0
n
.4/D0and
y2D1C10xC50x
2
C200x
3
300

y1lnxC
27
25
x
5

1
30
x
6

.
7.7.12.p0.r/D.r2/.rC2/;p1.r/D 2r1;r1D2;r2D 2;kDr1r2D4;an.r/D
2jC2r1
.jCr2/.jCrC2/
an1.r/;an.r/D
n
Y
jD1
2nC2r1
.nCr2/.nCrC2/
;an.2/D
1
nŠ

Q
n
jD1
2jC3
jC4

;
y1Dx
2
1
X
nD0
1
nŠ
0
@
n
Y
jD1
2jC3
jC4
1
Ax
n
;´Dx
2

1CxC
1
4
x
2

1
12
x
3

;CD
p1.1/
4
a3.2/D 1=16.
By logarithmic differentiation,
a
0
n
.r/D 2an.r/
n
X
jD1
j
2
Cj.2r1/Cr
2
rC4
.jCr2/.jCrC2/.2jC2r1/
;a
0
n
.2/D 2an.2/
n
X
jD1
.j
2
C3jC6/
j.jC4/.2jC3/
;
y2Dx
2

1CxC
1
4
x
2

1
12
x
3


1
16
y1lnxC
x
2
8
1
X
nD1
1
nŠ
0
@
n
Y
jD1
2jC3
jC4
1
A
0
@
n
X
jD1
.j
2
C3jC6/
j.jC4/.2jC3/
1
Ax
n
.
7.7.14.p0.r/D.rC1/.rC7/;p1.r/D.rC5/.2rC1/;r1D 1;r2D 7;kDr1
r2D6;an.r/D
.nCrC4/.2nC2r1/
.nCrC1/.nCrC7/
an1.r/;an.r/D.1/
n
n
Y
jD1
.jCrC4/.2jC2r1/
.jCrC1/.jCrC7/
;

125. 120 Chapter 7Series Solutions of Linear Second Order Equations
an.1/D
.1/
n
nŠ
0
@
n
Y
jD1
.jC3/.2j3/
jC6
1
A;y1D
1
x
1
X
nD0
.1/
n
nŠ
0
@
n
Y
jD1
.jC3/.2j3/
jC6
1
Ax
n
;´Dx
7

1C
26
5
xC
143
20
x
2

;
CD
p1.2/
6
a5.7/D0;y2Dx
7

1C
26
5
xC
143
20
x
2

.
7.7.16.p0.r/D.3r10/.3rC2/;p1.r/Dr.3r4/;r1D10=3;r2D 2=3;kDr1r2D4;
an.r/D
.nCr1/.3nC3r7/
.3nC3r10/.3nC3rC2/
an1.r/;an.r/D.1/
n
n
Y
jD1
.jCr1/.3jC3r7/
.3jC3r10/.3jC3rC2/
;
an.10=3/D
.1/
n
.nC1/
9
n
0
@
n
Y
jD1
3jC7
jC4
1
A;y1Dx
10=3
1
X
nD0
.1/
n
.nC1/
9
n
0
@
n
Y
jD1
3jC7
jC4
1
Ax
n
;´D
x
2=3

1C
4
27
x
1
243
x
2

;CD
p1.7=3/
36
a3.2=3/D0;y2Dx
2=3

1C
4
27
x
1
243
x
2

.
7.7.18.p0.r/D.r3/.rC2/;p1.r/D.rC1/
2
;r1D3;r2D 2;kDr1r2D5;
an.r/D
.nCr/
2
.nCr3/.nCrC2/
an1.r/;an.r/D.1/
n
n
Y
jD1
.jCr/
2
.jCr3/.jCrC2/
;an.3/D
.1/
n
nŠ

Q
n
jD1
.jC3/
2
jC5

;y1Dx
3
1
X
nD0
.1/
n
nŠ
0
@
n
Y
jD1
.jC3/
2
jC5
1
Ax
n
;´Dx
2

1C
1
4
x

;CD
p1.2/
5
a4.2/D
0;y2Dx
2

1C
1
4
x

.
7.7.20.p0.r/D.r6/.r1/;p1.r/D.r8/.r4/;r1D6;r2D1;kDr1r2D5;an.r/D

.nCr9/.nCr5/
.nCr6/.nCr1/
an1.r/;y1Dx
6

1C
2
3
xC
1
7
x
2

;´Dx

1C
21
4
xC
21
2
x
2
C
35
4
x
3

;
CD
p1.5/
6
a5.1/D0;y2Dx

1C
21
4
xC
21
2
x
2
C
35
4
x
3

.
7.7.22.p0.r/Dr.r10/;p1.r/D2.r6/.rC1/;r1D10;r2D0;kDr1r2D10;an.r/D

2.nCr7/
nCr10
an1.r/;an.r/D.2/
n
.nCr9/.nCr8/.nCr7/
.r9/.r8/.r7/
;an.10/D
.1/
n
2
n
.nC1/.nC2/.nC3/
6
;
y1D
x
10
6
1
X
nD0
.1/
n
2
n
.nC1/.nC2/.nC3/x
n
;´D

1
4
3
xC
5
3
x
2

40
21
x
3
C
40
21
x
4

32
21
x
5
C
16
21
x
6

;
CD
p1.9/
10
a9.0/D0;y2D

1
4
3
xC
5
3
x
2

40
21
x
3
C
40
21
x
4

32
21
x
5
C
16
21
x
6

.
Note: in the solutions to Exercises 7.7.23–7.7.40,´Dx
r2
P
k1
mD0
a2m.r2/x
2m
.
7.7.24.p0.r/D.r6/.r2/;p2.r/Dr;r1D6;r2D2;kD.r1r2/=2D2;a2m.r/D

a2m2.r/
2mCr6
;a2m.r/D
.1/
m
Q
m
jD1
.2jCr6/
;a2m.6/D
.1/
m
2
m
mŠ
;y1Dx
6
1
X
mD0
.1/
m
2
m
mŠ
x
2m
Dx
6
e
x
2
=2
;
´Dx
2

1C
1
2
x
2

;CD
p2.4/
4
a2.2/D 1=2. By logarithmic differentiation,a
0
2m
.r/D a2m.r/
m
X
jD1
1
2jCr6
;
a
0
2m
.6/D a2m.6/
m
X
jD1
1
2j
;y2Dx
2

1C
1
2
x
2


1
2
y1lnxC
x
6
4
1
X
mD1
.1/
m
2
m
mŠ
0
@
m
X
jD1
1
j
1
Ax
2m
.

126. Section 7.7The Method of Frobenius III121
7.7.26.p0.r/D.r1/.rC1/;p2.r/D2rC10;r1D1;r2D 1;kD.r1r2/=2D1;
a2m.r/D
2.2mCrC3/
.2mCr1/.2mCrC1/
a2m2.r/;a2m.r/D.2/
m
m
Y
jD1
2jCrC3
.2jCr1/.2jCrC1/
;
a2m.1/D
.1/
m
.mC2/
2 mŠ
;y1D
x
2
1
X
mD0
.1/
m
.mC2/
mŠ
x
2m
;´Dx
1
;CD
p2.1/
2
a0.1/D 4.
By logarithmic differentiation,
a
0
2m
.r/D a2m.r/
m
X
jD1
.4j
2
C4j.rC3/Cr
2
C6rC1/
.2jCr1/.2jCrC1/.2jCrC3/
;
a
0
2m
.1/D a2m.1/
m
X
jD1
j
2
C4jC2
2j.jC1/.jC2/
;
y2Dx
1
4y1lnxCx
1
X
mD1
.1/
m
.mC2/
mŠ
0
@
m
X
jD1
j
2
C4jC2
j.jC1/.jC2/
1
Ax
2m
.
7.7.28.p0.r/D.2rC1/.2rC5/;p2.r/D2rC3;r1D 1=2;r2D 5=2;kD.r1r2/=2D1;
a2m.r/D
.4mC2r1/
.4mC2rC1/.4mC2rC5/
a2m2.r/;a2m.r/D.1/
m
m
Y
jD1
.4jC2r1/
.4jC2rC1/.4jC2rC5/
;
a2m.1=2/D
.1/
m
Q
m
jD1
.2j1/
8
m
mŠ.mC1/Š
;y1Dx
1=2
1
X
mD0
.1/
m
Q
m
jD1
.2j1/
8
m
mŠ.mC1/Š
x
2m
;´Dx
5=2
;CD

p2.5=2/
8
a0.5=2/D1=4. By logarithmic differentiation,
a
0
2m
.r/D 2a2m.r/
m
X
jD1
.16j
2
C8j.2r1/C4r
2
4r11/
.4jC2r1/.4jC2rC1/.4jC2rC5/
;a
0
2m
.1=2/D a2m.1=2/
m
X
jD1
2j
2
2j1
2j.jC1/.2j1/
;
y2Dx
5=2
C
1
4
y1lnxx
1=2
1
X
mD1
.1/
m
Q
m
jD1
.2j1/
8
mC1
mŠ.mC1/Š
0
@
m
X
jD1
2j
2
2j1
j.jC1/.2j1/
1
Ax
2m
.
7.7.30.p0.r/D.r2/.rC2/;p2.r/D 2.rC4/;r1D2;r2D 2;kD.r1r2/=2D2;a2m.r/D
2
2mCr2
a2m2.r/;a2m.r/D
2
m
Q
m
jD1
.2jCr2/
;a2m.2/D
1
mŠ
;y1Dx
2
1
X
mD0
1
mŠ
x
2m
Dx
2
e
x
2
;
´Dx
2
.1x
2
/;CD
p2.0/
4
a2.2/D 2. By logarithmic differentiation,a
0
2m
.r/D a2m.r/
m
X
jD1
1
2jCr2
;
a
0
2m
.2/D a2m.2/
m
X
jD1
1
2j
;y2Dx
2
.1x
2
/2y1lnxCx
2
1
X
mD1
1
mŠ
0
@
m
X
jD1
1
j
1
Ax
2m
.
7.7.32.p0.r/D.3r13/.3r1/;p2.r/D2.53r/;r1D13=3;r2D1=3;kD.r1r2/=2D2;
a2m.r/D
2.6mC3r11/
.6mC3r13/.6mC3r1/
a2m2.r/;a2m.r/D2
m
m
Y
jD1
.6jC3r11/
.6jC3r13/.6jC3r1/
;
a2m.13=3/D
Q
m
jD1
.3jC1/
9
m
mŠ.mC2/Š
;y1D2x
13=3
1
X
mD0
Q
m
jD1
.3jC1/
9
m
mŠ.mC2/Š
x
2m
;´Dx
1=3

1C
2
9
x
2

;CD

p2.7=3/
36
a2.1=3/D2=81. By logarithmic differentiation,a
0
2m
.r/D 9a2m.r/
m
X
jD1
.12j
2
C4j.3r11/C3r
2
22rC47/
.6jC3r13/.6jC3r11/.6jC3r1/
;

127. 122 Chapter 7Series Solutions of Linear Second Order Equations
a
0
2m
.13=3/D a2m.13=3/
m
X
jD1
3j
2
C2jC2
2j.jC2/.3jC1/
;
y2Dx
1=3

1C
2
9
x
2

C
2
81
0
@y1lnxx
13=3
1
X
mD0
Q
m
jD1
.3jC1/
9
m
mŠ.mC2/Š
0
@
m
X
jD1
3j
2
C2jC2
j.jC2/.3jC1/
1
Ax
2m
1
A.
7.7.34.p0.r/D.r2/.rC2/;p2.r/D 3.r4/;r1D2;r2D 2;kD.r1r2/=2D2;
a2m.r/D
3.2mCr6/
.2mCr2/.2mCrC2/
a2m2.r/;y1Dx
2

1
1
2
x
2

;´Dx
2

1C
9
2
x
2

;CD

p2.0/
4
a2.2/D 27=2;a2.r/D
3.r4/
r.rC4/
,a
0
2
.r/D
3.r
2
8r16/
r
2
.rC4/
2
,a
0
2
.2/D7=12. Ifm2,
thena2m.r/D.r2/c2m.r/wherec2m.r/D
3
m
.r4/
.2mCr4/.2mCr2/
Q
m
jD1
.2jCrC2/
, so
a
0
2m
.2/Dc2m.2/D

3
2

m
m.m1/.mC2/Š
;y2Dx
2

1C
9
2
x
2


27
2
 
y1lnxC
7
12
x
4
x
2
1
X
mD2

3
2

m
m.m1/.mC2/Š
x
2m
!
.
7.7.36.p0.r/D.2r5/.2rC7/;p2.r/D.2r1/
2
;r1D5=2;r2D 7=2;kD.r1r2/=2D3;
a2m.r/D
4mC2r5
4mC2rC7
a2m2.r/;a2m.r/D
.2r1/.2rC3/.2rC7/
.4mC2r1/.4mC2rC3/.4mC2rC7/
;a2m.5=2/D
.1/
m
.mC1/.mC2/.mC3/
;y1Dx
5=2
1
X
mD0
.1/
m
.mC1/.mC2/.mC3/
x
2m
;´Dx
7=2
.1Cx
2
/
2
CD

p2.1=2/
24
a4.7=2/D0;y2Dx
7=2
.1Cx
2
/
2
.
7.7.38.p0.r/D.r3/.rC7/;p2.r/Dr.rC1/;r1D3;r2D 7;kD.r1r2/=2D5;a2m.r/D

.2mCr2/.2mCr1/
.2mCr3/.2mCrC7/
a2m2.r/;a2m.r/D.1/
m
m
Y
jD1
.2jCr2/.2jCr1/
.2jCr3/.2jCrC7/
;a2m.3/D
.1/
m
mC1
2
m
0
@
m
Y
jD1
2jC1
jC5
1
A;y1Dx
3
1
X
mD0
.1/
m
mC1
2
m
0
@
m
Y
jD1
2jC1
jC5
1
Ax
2m
;´Dx
7

1C
21
8
x
2
C
35
16
x
4
C
35
64
x
6

CD
p2.1/
10
a8.7/D0;y2Dx
7

1C
21
8
x
2
C
35
16
x
4
C
35
64
x
6

.
7.7.40.p0.r/D.2r3/.2rC5/;p2.r/D.2r1/.2rC1/;r1D3=2;r2D 5=2;kD
.r1r2/=2D2;a2m.r/D
4mC2r5
4mC2rC5
a2m2.r/;a2m.r/D.1/
m
m
Y
jD1
4jC2r5
4jC2rC5
;a2m.3=2/D
.1/
m
Q
m
jD1
.2j1/
2
m1
.mC2/Š
;y1Dx
3=2
1
X
mD0
.1/
m
Q
m
jD1
.2j1/
2
m1
.mC2/Š
x
2m
;´Dx
5=2

1C
3
2
x
2

CD
p2.1=2/
16
a2.5=2/D
0;y2Dx
5=2

1C
3
2
x
2

.
7.7.42.p0.r/Dr
2

2
;p2.r/D1;r1D;r2D ;kD.r1r2/=2D;a2m.r/D

a2m2.r/
.2mCrC/.2mCr/
;a2m.r/D
.1/
m
Q
m
jD1
.2jCrC/.2jCr/
;a2m./D
.1/
m
4
m
mŠ
Q
m
jD1
.jC/
;
a2m./D
.1/
m
4
m
mŠ
Q
m
jD1
.j/
,jD0; : : : ; 1;y1Dx

1
X
mD0
.1/
m
4
m
mŠ
Q
m
jD1
.jC/
x
2m
;´D

128. Section 7.7The Method of Frobenius III123
x

1
X
mD0
.1/
m
4
m
mŠ
Q
m
jD1
.j/
x
2m
;CD
p2.2/
2
a22./D
a22./
2
D
2
4

Š.1/Š
. By
logarithmic differentiation,a
0
2m
.r/D 2a2m.r/
m
X
jD1
2jC
.2jCrC/.2jCr/
;a
0
2m
./D a2m./
m
X
jD1
2jC
2j.jC/
;
y2Dx

1
X
mD0
.1/
m
4
m
mŠ
Q
m
jD1
.j/
x
2m

2
4

Š.1/Š
0
@y1lnx
x

2
1
X
mD1
.1/
m
4
m
mŠ
Q
m
jD1
.jC/
0
@
m
X
jD1
2jC
j.jC/
1
Ax
2m
1
A.
7.7.44.Sincean.r2/D
p1.nCr21/
p0.nCr2/
an1.r2/,1nk1,ak1.r2/D.1/
k1
k1
Y
jD1
p1.r2Cj1/
p0.r2Cj /
.
ButCD
p1.r11/
k˛0
ak1.r2/D
p1.r2Ck1/
k˛0
ak1.r2/D./
k
Q
k
jD1
p1.r2Cj1/
k˛0
Q
k1
jD1
p0.r2Cj /
D0if
and only if
k
Y
jD1
p1.r2Cj1/D0.
7.7.46.Sincep1.r/D1,an.r/D
1
˛0.nCrr1/.nCrr2/
an1.r/and (A)an.r/D.1/
n

1
˛0

n
1
Q
n
jD1
.jCrr1/.jC
Therefore,an.r1/D
.1/
n
nŠ

1
˛0

n
1
Q
n
jD1
.jCk/
forn0(soLy1D0) andan.r2/D
.1/
n
nŠ

1
˛0

n
1
Q
n
jD1
.jk/
fornD0; : : : ; k1.Ly2D0ify2Dx
r2
k1
X
nD0
an.r2/x
n
CC
 
y1lnxCx
r1
1
X
nD1
a
0
n
.r1/x
n
!
ifCD

1
k˛0
ak1.r2/D
1
k˛0
.1/
k1
.k1/Š

1
˛0

k1
.1/
k1
.k1/Š
D
1
kŠ.k1/Š

1
˛0

k
. From (A), lnjan.r/j D
nln
ˇ
ˇ
ˇ
ˇ
1
˛0
ˇ
ˇ
ˇ
ˇ

n
X
jD1
..lnjjCrr1j ClnjjCrr2j/, soa
0
n
.r/D an.r/
n
X
jD1

1
jCrr1
C
1
jCrr2

anda
0
n
.r1/D an.r1/
n
X
jD1
2jCk
j.jCk/
.
7.7.48.(a)From Exercise 7.6.66(a)of Section 7.6,L

@y
@r
.x; r/

Dp
0
0
.r/x
r
Cx
r
p0.r/lnx. SettingrD
r1yieldsL
 
y1lnxCx
r1
1
X
nD1
a
0
n
.r1/
!
Dp
0
0
.r1/x
r1
. Sincep
0
0
.r/D˛0.2rr1r2/,p
0
0
.r1/Dk˛0.
(b)From Exercise 7.5.57 of Section 7.5,L
 
x
r2
1
X
nD0
an.r2/x
n
!
Dx
r2
1
X
nD0
bnx
n
, whereb0Dp0.r2/D
0andbnD
n
X
jD0
pj.nCr2j /anj.r2/ifn1. From the definition offan.r2/g,bnD0ifn¤k,
whilebkD
k
X
jD0
pk.kCr2j /akj.r2/D
k
X
jD1
pj.r1j /akj.r2/.
(d)Letf Qan.r2/gbe the coefficients that would obtained ifQak.r2/D0. Thenan.r2/D Qan.r2/ifnD

129. 124 Chapter 7Series Solutions of Linear Second Order Equations
0; : : : ; k1, and (A)an.r2/ Qan.r2/D
1
p0.nCr2/
nk
X
jD0
pj.nCr2j /.anj.r2/ Qanjr2/ifn > k.
Now letcmDakCm.r2/ QakCm.r2/. SettingnDmCkin (A) and recalling thekCr2Dr1yields (B)
cmD
1
p0.mCr1/
m
X
jD0
pj.mCr1j /cmj. SinceckDak.r2/, (B) implies thatcmDak.r2/am.r1/
for allm0, which implies the conclusion.

130. CHAPTER8
LaplaceTransforms
8.1INTRODUCTION TO THE LAPLACE TRANSFORM
8.1.2.(a)coshtsintD
1
2

e
t
sintCe
t
sint

$
1
2

1
.s1/
2
C1
C
1
.sC1/
2
C1

D
s
2
C2
Œ.s1/
2
C1Œ.sC1/
2
C1
.
(b)sin
2
tD
1cos2t
2
$
1
2

1
s

s
.s
2
C4/

D
2
s.s
2
C4/
.
(c)cos
2
2tD
1
2

1
s
C
s
s
2
C16

D
s
2
C8
s.s
2
C16/
.
(d)cosh
2
tD
.e
t
Ce
t
/
2
/
4
D
.e
2t
C2Ce
2t
/
4
$
1
4

1
s2
C
2
s
C
1
sC2

D
s
2
2
s.s
2
4/
.
(e)tsinh2tD
te
2t
te
2t
2
$
1
2

1
.s2/
2

1
.sC2/
2

D
4s
.s
2
4/
2
.
(f)sintcostD
sin2t
2
$
1
s
2
C4
.
(g)sin.tC=4/Dsintcos.=4/Ccostcos.=4/$
1
p
2
sC1
s
2
C1
.
(h)cos2tcos3t$
s
s
2
C4

s
s
2
C9
D
5s
.s
2
C4/.s
2
C9/
.
(i)sin2tCcos4t$
2
s
2
C4
C
s
s
2
C16
D
s
3
C2s
2
C4sC32
.s
2
C4/.s
2
C16/
.
8.1.6.IfF.s/D
Z
1
0
e
st
f .t/ dt, thenF
0
.s/D
Z
1
0
.te
st
/f .t/ dtD
Z
1
0
e
st
.tf .t// dt. Apply-
ing this argument repeatedly yields the assertion.
8.1.8.Letf .t/D1andF.s/D1=s. From Exercise 8.1.6,t
n
$.1/
n
F
.n/
.s/DnŠ=s
nC1
.
8.1.10.Ifjf .t/j Me
s0t
fortt0, thenjf .t/e
st
j Me
.ss0/t
fortt0. Letg.t/De
st
f .t/,
w.t/DMe
.ss0/t
, andDt0. Since
R
1
t0
w.t/ dtconverges ifs > s0,F.s/is defined fors > s0.
8.1.12.
Z
T
0
e
st
Z
t
0
f ./ d

dtD
e
st
s
Z
t
0
f ./ d
ˇ
ˇ
ˇ
ˇ
T
0
C
1
s
Z
T
0
e
st
f .t/ dtD
e
sT
s
Z
T
0
f ./ dC
1
s
Z
T
0
e
st
f .t/ dt. Sincefis of exponential orders0, the second integral on the right converges to
125

131. 126 Chapter 8Laplace Transforms
1
s
L.f /asT! 1(Exercise 8.1.10). Now it suffices to show that (A) lim
T!1
e
sT
Z
T
0
f ./ dD0
ifs > s0. Suppose thatjf .t/j Me
s0t
iftt0andjf .t/j Kif0tt0, and letT > t0. Then
ˇ
ˇ
ˇ
ˇ
ˇ
Z
T
0
f ./ d
ˇ
ˇ
ˇ
ˇ
ˇ

ˇ
ˇ
ˇ
ˇ
Z
t0
0
f ./ d
ˇ
ˇ
ˇ
ˇ
C
ˇ
ˇ
ˇ
ˇ
ˇ
Z
T
t0
f ./ d
ˇ
ˇ
ˇ
ˇ
ˇ
< Kt0CM
Z
T
t0
e
s0
d < Kt0C
Me
s0T
s0
, which
proves (A).
8.1.14.(a)IfT > 0, then
Z
T
0
e
st
f .t/ dtD
Z
T
0
e
.ss0/t
.e
s0t
f .t// dt. Use integration by parts with
uDe
.ss0/t
,dvDe
s0t
f .t/ dt,duD .ss0/e
.ss0/t
, andvDgt obtain
Z
T
0
e
st
f .t/ dtD
e
.ss0/t
g.t/
ˇ
ˇ
ˇ
ˇ
T
0
C.ss0/
Z
T
0
e
.ss0/t
g.t/ dt. Sinceg.0/D0this reduces to
Z
T
0
e
st
f .t/ dtD
e
.ss0/T
g.T /C.ss0/
Z
T
0
e
.ss0/t
g.t/ dt. Sincejg.t/j Mfor allt0, we can lett! 1to
conclude that
Z
1
0
e
st
f .t/ dtD.ss0/
Z
1
0
e
.ss0/t
g.t/ dtifs > s0.
(b)IfF.s0/exists, theng.t/is bounded onŒ0;1/. Now apply(a).
(c)Sincef .t/D
1
2
d
dt
sin.e
t
2
/,
ˇ
ˇ
ˇ
ˇ
Z
t
0
f ./ d
ˇ
ˇ
ˇ
ˇ
D
jsin.e
t
2
/sin.1/j
2
1for allt0. Now apply(a)
withs0D0.
8.1.16.(a)€.˛/D
Z
1
0
x
˛1
e
x
dxD
x
˛
e
x
˛
ˇ
ˇ
ˇ
ˇ
1
0
C
1
˛
Z
1
0
x
˛
e
x
dxD
€.˛C1/
˛
.
(b)Use induction.€.1/D
Z
1
0
e
x
dxD1. If (A)€.nC1/DnŠ, then€.nC2/D.nC1/€.nC1/
(from(a))D.nC1/nŠ(from (A))D.nC1/Š.
(c)€.˛C1/D
Z
1
0
x
˛
e
x
dt. LetxDst. Then€.˛C1/D
Z
1
0
.st/
˛
e
st
s dt, so
Z
1
0
e
st
t
˛
dtD
€.˛C1/
˛
.
8.1.18.(a)
Z
2
0
e
st
f .t/ dtD
Z
1
0
e
st
t dtC
Z
2
1
e
st
.2t/ dtD

1
s
2

e
s
.sC1/
s
2

C

e
s
.s1/
s
2
C
e
2s
s
2

D

2e
s
s
2
C
e
2s
s
2
C
1
s
2
D
.1e
s
/
2
s
2
. Therefore,F.s/D
.1e
s
/
2
s
2
.1e
2s
/
D
1e
s
s
2
.1Ce
s
/
D
1
s
2
tanh
s
2
.
(b)
Z
1
0
e
st
f .t/ dtD
Z
1=2
0
e
st
dt
Z
1
1=2
e
st
dtD
1
s

e
s=2
s
C
e
s
s

e
s=2
s
D
2e
s=2
s
C
e
s
s
C
1
s
D
.1e
s=2
/
2
s
. Therefore,F.s/D
.1e
s=2
/
2
s.1e
s
/
D
1e
s=2
s.1Ce
s=2
/
D
1
s
tanh
s
4
.
(c)
Z

0
e
st
f .t/ dtD
Z

0
e
st
sint dtD
1Ce
s
.s
2
C1/
. Therefore,F.s/D
1Ce
s
.s
2
C1/.1e
s
/
1
s
2
C1
coth
s
2
.
(d)
Z
2
0
e
st
f .t/ dtD
Z

0
e
st
sint dtD
1Ce
s
.s
2
C1/
. Therefore,F.s/D
1Ce
s
.s
2
C1/1Ce
2s
D
1
.s
2
C1/.1e
s
/
.

132. Section 8.2The Inverse Laplace Transform127
8.2THE INVERSE LAPLACE TRANSFORM
8.2.2.(a)
2sC3
.s7/
4
D
2.s7/C17
.s7/
4
D
2
.s7/
3
C
17
.s7/
4
D
2Š
.s7/
3
C
17
6
3Š
.s7/
4
$e
7t

t
2
C
17
6
t
3

.
(b)
s
2
1
.s2/
6
D
Œ.s2/C2
2
1
.s2/
6
D
.s2/
2
C4.s2/C3
.s2/
6
D
1
.s2/
4
C
4
.s2/
5
C
3
.s2/
6
D
1
6
3Š
.s2/
4
C
1
6
4Š
.s2/
5
C
1
40
5Š
.s2/
6
$

1
6
t
3
C
1
6
t
4
C
1
40
t
5

e
2t
.
(c)
sC5
s
2
C6sC18
D
.sC3/
.sC3/
2
C9
C
2
3
3
.sC3/
2
C9
$e
3t

cos3tC
2
3
sin3t

.
(d)
2sC1
s
2
C9
D2
s
s
2
C9
C
1
3
3
s
2
C9
$2cos3tC
1
3
sin3t.
(e)
s
s
2
C2sC1
D
.sC1/1
.sC1/
2
D
1
sC1

1
.sC1/
2
$.1t/e
t
.
(f)
sC1
s
2
9
D
s
s
2
9
C
1
3
3
s
2
9
$cosh3tC
1
3
sinh3t.
(g)Expand the numerator in powers ofsC1:s
3
C2s
2
s3DŒ.sC1/1
3
C2Œ.sC1/1
2

Œ.sC1/13D.sC1/
3
.sC1/
2
2.sC1/1; therefore
s
3
C2s
2
s3
.sC1/
4
D
1
sC1

1
.sC1/
2

2
.sC1/
3

1
6
6
.sC1/
4
$

1tt
2

1
6
t
3

e
t
.
(h)
2sC3
.s1/
2
C4
D2
.s1/
.s1/
2
C4
C
5
2
2
.s1/
2
C4
$e
t

2cos2tC
5
2
sin2t

.
(i)
1
s

s
s
2
C1
$1cost.
(j)
3sC4
s
2
1
D
3s
s
2
1
C
4
s
2
1
$3coshtC4sinht. Alternatively,
3sC4
s
2
1
D
3sC4
.s1/.sC1/
D
1
2

7
s1

1
sC1

$
7e
t
e
t
2
.
(k)
3
s1
C
4sC1
s
2
C9
D3
1
s1
C4
s
s
2
C9
C
1
3
3
s
2
C9
$3e
t
C4cos3tC
1
3
sin3t.
(l)
3
.sC2/
2

2sC6
s
2
C4
D3
1
.sC2/
2
2
s
s
2
C4
3
2
s
2
C4
$3te
2t
2cos2t3sin2t.
8.2.4.(a)
2C3s
.s
2
C1/.sC2/.sC1/
D
A
sC2
C
B
sC1
C
C sCD
s
2
C1
;
where
A.s
2
C1/.sC1/CB.s
2
C1/.sC2/C.C sCD/.sC2/.sC1/D2C3s:
5AD 4 .setsD 2/I
2BD 1 .setsD 1/I
AC2BC2DD 2 .setsD0/I
ACBCCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD
4
5
,BD
1
2
,CD
3
10
,DD
11
10
. Therefore,
2C3s
.s
2
C1/.sC2/.sC1/
D
4
5
1
sC2

1
2
1
sC1

1
10
3s11
s
2
C1
$
4
5
e
2t

1
2
e
t

3
10
costC
11
10
sint:

133. 128 Chapter 8Laplace Transforms
(b)
3s
2
C2sC1
.s
2
C1/.s
2
C2sC2/
D
AsCB
s
2
C1
C
C.sC1/CD
.sC1/
2
C1
;
where
.AsCB/..sC1/
2
C1/C.C.sC1/CD/.s
2
C1/D3s
2
C2sC1:
2BCCCDD1 .setsD0/I
ACBC2DD2 .setsD 1/I
2BCCCDD1 .setsD0/I
ACCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD6=5,BD2=5,CD 6=5,DD7=5. Therefore,
3s
2
C2sC1
.s
2
C1/.s
2
C2sC2/
D
1
5

6sC2
s
2
C1

6.sC1/7
.sC1/
2
C1

$
6
5
costC
2
5
sint
6
5
e
t
costC
7
5
e
t
sint:
(c)s
2
C2sC5D.sC1/
2
C4;
3sC2
.s2/..sC1/
2
C4/
D
A
s2
C
B.sC1/CC
.sC1/
2
C4
;
where
A

.sC1/
2
/C4

C.B.sC1/CC / .s2/D3sC2:
13AD 8 .setsD2/I
4A3CD 1 .setsD 1/I
ACBD 0 .equate coefficients ofs
2
/:
Solving this system yieldsAD
8
13
,BD
8
13
,CD
15
13
. Therefore,
3sC2
.s2/..sC1/
2
C4/
D
1
13

8
s2

8.s1/15
.sC1/
2
C4

$
8
13
e
2t

8
13
e
t
cos2tC
15
26
e
t
sin2t:
(d)
3s
2
C2sC1
.s1/
2
.sC2/.sC3/
D
A
s1
C
B
.s1/
2
C
C
sC2
C
D
sC3
;
where
.A.s1/CB/.sC2/.sC3/C.C.sC3/CD.sC2//.s1/
2
D3s
2
C2sC1:
12BD6 .setsD1/I
9CD9 .setsD 2/I
16DD22 .setsD 3/I
ACCCDD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD3=8,BD1=2,CD1,DD 11=8. Therefore,
3s
2
C2sC1
.s1/
2
.sC2/.sC3/
D
3
8
1
s1
C
1
2
1
.s1/
2
C
1
sC2

11
8
1
sC3
$
3
8
e
t
C
1
2
te
t
Ce
2t

11
8
e
3t
:

134. Section 8.2The Inverse Laplace Transform129
(e)
2s
2
CsC3
.s1/
2
.sC2/
2
D
A
s1
C
B
.s1/
2
C
C
sC2
C
D
.sC2/
2
;
where
.A.s1/CB/.sC2/
2
C.C.sC2/CD/.s1/
2
D2s
2
CsC3:
9BD6 .setsD1/I
9DD9 .setsD 2/I
4AC4BC2CCDD3 .setsD0/I
ACCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD1=9,BD2=3,CD 1=9,DD1. Therefore,
2s
2
CsC3
.s1/
2
.sC2/
2
D
1
9
1
s1
C
2
3
1
.s1/
2

1
9
1
sC2
C
1
.sC2/
2
$
1
9
e
t
C
2
3
te
t

1
9
e
2t
Cte
2t
:
(f)
3sC2
.s
2
C1/.s1/
2
D
A
s1
C
B
.s1/
2
C
C sCD
s
2
C1
;
where
A.s1/.s
2
C1/CB.s
2
C1/C.C sCD/.s1/
2
D3sC2: . A/
SettingsD1yields2BD5, soBD
5
2
. Substituting this into (A) shows that
A.s1/.s
2
C1/C.C sCD/.s1/
2
D3sC2
5
2
.s
2
C1/
D
5s
2
6sC1
2
D
.s1/.5s1/
2
:
Therefore,
A.s
2
C1/C.C sCD/.s1/D
15s
2
:
2AD 2 .setsD1/I
ADD1=2 .setsD0/I
ACCD 0 .equate coefficients ofs
2
/:
Solving this system yieldsAD 1,CD1,DD
3
2
. Therefore,
3sC2
.s
2
C1/.s1/
2
D
1
s1
C
5
2
1
.s1/
2
C
s3=2
s
2
C1
$ e
t
C
5
2
te
t
Ccost
3
2
sint:
8.2.6.(a)
17s15
.s
2
2sC5/.s
2
C2sC10/
D
A.s1/CB
.s1/
2
C4
C
C.sC1/CD
.sC1/
2
C9
where
.A.s1/CB/..sC1/
2
C9/C.C.sC1/CD/..s1/
2
C4/D17s15:

135. 130 Chapter 8Laplace Transforms
13BC8CC4DD 2 .setsD1/I
18AC9BC8DD 32 .setsD 1/I
10AC10BC5CC5DD 15 .setsD0/I
ACCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD1,BD2,CD 1,DD 4. Therefore,
17s15
.s
2
2sC5/.s
2
C2sC10/
D
.s1/C2
.s1/
2
C4

.sC1/C4
.sC1/
2
C9
$e
t
.cos2tCsin2t/e
t

cos3tC
4
3
sin3t

:
(b)
8sC56
.s
2
6sC13/.s
2
C2sC5/
D
A.s3/CB
.s3/
2
C4
C
C.sC1/CD
.sC1/
2
C4
where
.A.s3/CB/..sC1/
2
C4/C.C.sC1/CD/..s3/
2
C4/D8sC56:
20BC16CC4DD80 .setsD3/I
16AC4BC20DD48 .setsD 1/I
15AC5BC13CC13DD56 .setsD0/I
ACCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD 1,BD3,CD1,DD1. Therefore,
8sC56
.s
2
6sC13/.s
2
C2sC5/
D
.s3/C3
.s3/
2
C4
C
.sC1/C1
.sC1/
2
C4
$e
3t

cos2tC
3
2
sin2t

Ce
t

cos2tC
1
2
sin2t

:
(c)
sC9
.s
2
C4sC5/.s
2
4sC13/
D
A.sC2/CB
.sC2/
2
C1
C
C.s2/CD
.s2/
2
C9
where
.A.sC2/CB/..s2/
2
C9/C.C.s2/CD/..sC2/
2
C1/DsC9:
25B4CCDD7 .setsD 2/I
36AC9BC17DD11 .setsD2/I
26AC13B10CC5DD9 .setsD0/I
ACCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD1=8,BD1=4,CD 1=8,DD1=4. Therefore,
sC9
.s
2
C4sC5/.s
2
4sC13/
D D

1
8
.sC2/C2
.sC2/
2
C1

.s2/2
.s2/
2
C3

$e
2t

1
8
costC
1
4
sint

e
2t

1
8
cos3t
1
12
sin3t

:
(d)
3s2
.s
2
4sC5/.s
2
6sC13/
D
A.s2/CB
.s2/
2
C1
C
C.s3/CD
.s3/
2
C4
where
.A.s2/CB/..s3/
2
C4/C.C.s3/CD/..s2/
2
C1/D3s2:

136. Section 8.2The Inverse Laplace Transform131
5BCCDD 4 .setsD2/I
4AC4BC2DD 7 .setsD3/I
26AC13B15CC5DD 2 .setsD0/I
ACCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD1,BD1=2,CD 1,DD1=2. Therefore,
3s2
.s
2
4sC5/.s
2
6sC13/
D D
1
2

2.s2/C1
.s2/
2
C1

2.s3/1
.s3/
2
C4

$e
2t

costC
1
2
sint

e
3t

cos2t
1
4
sin2t

:
(e)
3s1
.s
2
2sC2/.s
2
C2sC5/
D
A.s1/CB
.s1/
2
C1
C
C.sC1/CD
.sC1/
2
C4
where
.A.s1/CB/..sC1/
2
C4/C.C.sC1/CD/..s1/
2
C1/D3s1:
8BC2CCDD 2 .setsD1/I
8AC4BC5DD 4 .setsD 1/I
5AC5BC2CC2DD 1 .setsD0/I
AC5BCCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD1=5,BD2=5,CD 1=5,DD 4=5. Therefore,
3s1
.s
2
2sC2/.s
2
C2sC5/
D
1
5

.s1/C2
.s1/
2
C1

.sC1/C4
.sC1/
2
C4

:
$e
t

1
5
costC
2
5
sint

e
t

1
5
cos2tC
2
5
sin2t

:
(f)
20sC40
.4s
2
4sC5/.4s
2
C4sC5/
DŒ
A.s1=2/CB
.s1=2/
2
C1
C
C.sC1=2/CD
.sC1=2/
2
C1
where
.A.s1=2/CB/..sC1=2/
2
C1/C.C.sC1=2/CD/..s1=2/
2
C1/D
5sC10
4
:
2BCCCDD25=8 .setsD1=2/I
ACBC2DD15=8 .setsD 1=2/I
5AC10BC5CC10DD 20 .setsD0/I
ACCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD 1,BD9=8,CD1,DD 1=8. Therefore,
20sC40
.4s
2
4sC5/.4s
2
C4sC5/
D
1
8

8.s1=2/C9
.s1=2/
2
C1
C
8.sC1=2/
.sC1=2/
2
C1

$e
t =2

costC
9
8
sint

Ce
t =2

cost
1
8
sint

:
8.2.8.(a)
2sC1
.s
2
C1/.s1/.s3/
D
A
s1
C
B
s3
C
C sCD
s
2
C1

137. 132 Chapter 8Laplace Transforms
where
.A.s3/CB.s1//.s
2
C1/C.C sCD/.s1/.s3/D2sC1:
4AD3 .setsD1/I
20BD7 .setsD3/I
3ABC3DD1 .setsD0/I
ACBCCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD 3=4,BD7=20,CD2=5,DD 3=10. Therefore,
2sC1
.s
2
C1/.s1/.s3/
D
3
4
1
s1
C
7
20
1
s3
C
2
5
s
s
2
C1

3
10
1
s
2
C1
$
3
4
e
t
C
7
20
e
3t
C
2
5
cost
3
10
sint:
(b)
sC2
.s
2
C2sC2/.s
2
1/
D
A
s1
C
B
sC1
C
C.sC1/CD
.sC1/
2
C1
where
.A.sC1/CB.s1//..sC1/
2
C1/C.C.sC1/CD/.s
2
1/DsC2:
10AD3 .setsD1/I
2BD1 .setsD 1/I
2A2BCDD2 .setsD0/I
ACBCCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD3=10,BD 1=2,CD1=5,DD 3=5. Therefore,
sC2
.s
2
C2sC2/.s
2
1/
D
3
10
1
s1

1
2
1
sC1
C
1
5
sC1
.sC1/
2
C1

3
5
1
.sC1/
2
C1
$
3
10
e
t

1
2
e
t
C
1
5
e
t
coste
t
sint:
(c)
2s1
.s
2
2sC2/.sC1/.s2/
D
A
s2
C
B
sC1
C
C.s1/CD
.s1/
2
C1
where
.A.sC1/CB.s2//..s1/
2
C1/C.C.s1/CD/.s2/.sC1/D2s1:
6AD 3 .setsD2/I
15BD 3 .setsD 1/I
2A4BC2C2DD 1 .setsD0/I
ACBCCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD1=2,BD1=5,CD 7=10,DD 1=10. Therefore,
2s1
.s
2
2sC2/.sC1/.s2/
D D
1
2
1
s2
C
1
5
1
sC1

7
10
s1
.s1/
2
C1

1
10
1
.s1/
2
C1
$
1
2
e
2t
C
1
5
e
t

7
10
e
t
cost
1
10
e
t
sint:
(d)
s6
.s
2
1/.s
2
C4/
D
A
s1
C
B
sC1
C
C sCD
s
2
C4

138. Section 8.2The Inverse Laplace Transform133
where
.A.sC1/CB.s1//.s
2
C4/C.C sCD/.s
2
1/Ds6:
10AD 5 .setsD1/I
10BD 7 .setsD 1/I
4A4BDD 6 .setsD0/I
ACBCCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD 1=2,BD7=10,CD 1=5,DD6=5. Therefore,
s6
.s
2
1/.s
2
C4/
D D
1
2
1
s1
C
7
10
1
sC1

1
5
s
s
2
C4
C
3
5
C
1
s
2
C4
$
1
2
e
t
C
7
10
e
t

1
5
cos2tC
3
5
sin2t:
(e)
2s3
s.s2/.s
2
2sC5/
D
A
s
C
B
s2
C
C.s1/CD
.s1/
2
C4
where
.A.s2/CBs/..s1/
2
C4/C.C.s1/CD/s.s2/D2s3:
10AD 3 .setsD0/I
10BD 1 .setsD2/I
4AC4BDD 1 .setsD1/I
ACBCCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD3=10,BD1=10,CD 2=5,DD1=5. Therefore,
2s3
s.s2/.s
2
2sC5/
D D
3
10s
C
1
10
1
s2

2
5
s1
.s1/
2
C4
C
1
5
1
.s1/
2
C4
$
3
10
C
1
10
e
2t

2
5
e
t
cos2tC
1
10
e
t
sin2t:
(f)
5s15
.s
2
4sC13/.s2/.s1/
D
A
s1
C
B
s2
C
C.s2/CD
.s2/
2
C9
where
.A.s2/CB.s1//..s2/
2
C9/C.C.s2/CD/.s1/.s2/D5s15:
10AD 10 .setsD1/I
9BD 5 .setsD2/I
26A13B4CC2DD 15 .setsD0/I
ACBCCD 0 .equate coefficients ofs
3
/:
Solving this system yieldsAD1,BD 5=9,CD 4=9,DD1. Therefore,
5s15
.s
2
4sC13/.s2/.s1/
D D
1
s1

5
9
1
s2

4
9
s2
.s2/
2
C9
C
1
.s2/
2
C9
$e
t

5
9
e
2t

4
9
e
2t
cos3tC
1
3
e
2t
sin3t:

139. 134 Chapter 8Laplace Transforms
8.2.10.(a)LetiD1. (The proof foriD2; : : : ; n) is similar. Multiplying the given equation through by
ss1yields
P.s/
.ss2/ .ssn/
DA1C.ss1/

A2
ss2
C C
An
ssn

;
and settingsDs1yieldsA1D
P.s1/
.s1s2/ .s2sn/
.
(b)From calculus we know thatFhas a partial fraction expansion of the form
P.s/
.ss1/Q1.s/
D
A
ss1
CG.s/whereGis continuous ats1. Multiplying through byss1shows that
P.s/
Q1.s/
DAC
.ss1/G.s/. Now setsDs1to obtainAD
P.s1/
Q.s1/
.
(c)The result in(b)is generalization of the result in(a), since it shows that ifs1is a simple zero of the
denominator of the rational function, then Heaviside’s method can be used to determine the coefficient of
1=.ss1/in the partial fraction expansion even if some of the other zeros of the denominator are repeated
or complex.
8.3SOLUTION OF INITIAL VALUE PROBLEMS
8.3.2.
.s
2
s6/Y.s/D
2
s
Cs1D
2Cs.s1/
s
:
Since.s
2
s6/D.s3/.sC2/,
Y.s/D
2Cs.s1/
s.s3/.sC2/
D
1
3s
C
8
15
1
s3
C
4
5
1
sC2
andyD
1
3
C
8
15
e
3t
C
4
5
e
2t
.
8.3.4.
.s
2
4/Y.s/D
2
s3
C.1Cs/D
2C.s1/.s3/
s3
:
Sinces
2
4D.s2/.sC2/,
Y.s/D
2C.s1/.s3/
.s2/.sC2/.s3/
D
1
4
1
s2
C
17
20
1
sC2
C
2
5
1
s3
andyD
1
4
e
2t
C
17
20
e
2t
C
2
5
e
3t
.
8.3.6.
.s
2
C3sC2/Y.s/D
6
s1
C.1Cs/C3D
6C.s1/.sC2/
s1
:
Sinces
2
C3sC2D.sC2/.sC1/,
Y.s/D
6C.s1/.sC2/
.s1/.sC2/.sC1/
D
1
s1
C
2
sC2

2
sC1
andyDe
t
C2e
2t
2e
t
.

140. Section 8.3Solution of Initial Value Problems135
8.3.8.
.s
2
3sC2/Y.s/D
2
s3
C.1Cs/3D
2C.s3/.s4/
s3
:
Sinces
2
3sC2D.s1/.s2/,
Y.s/D
2C.s3/.s4/
.s1/.s2/.s3/
D
4
s1

4
s2
C
1
s3
andyD4e
t
4e
2t
Ce
3t
.
8.3.10.
.s
2
3sC2/Y.s/D
1
s3
C.4s/C3D
1.s3/.sC1/
s3
:
Sinces
2
3sC2D.s1/.s2/,
Y.s/D
1.s3/.sC1/
.s1/.s2/.s3/
D
5
2
1
s1

4
s2
C
1
2
1
s3
andyD
5
2
e
t
4e
2t
C
1
2
e
3t
.
8.3.12.
.s
2
Cs2/Y.s/D
4
s
C.3C2s/C2D
4Cs.5C2s/
s
:
Since.s
2
Cs2/D.sC2/.s1/,
Y.s/D
4Cs.5C2s/
s.sC2/.s1/
D
2
s

1
sC2
C
1
s1
;
andyD2e
2t
Ce
t
.
8.3.14.
.s
2
s6/Y.s/D
2
s
Cs1D
2Cs.s1/
s
:
Sinces
2
s6D.s3/.sC2/,
Y.s/D
2Cs.s1/
s.s3/.sC2/
D
1
3s
C
8
15
1
s3
C
4
5
1
sC2
andyD
1
3
C
8
15
e
3t
C
4
5
e
2t
.
8.3.16.
.s
2
1/Y.s/D
1
s
CsD
1Cs
2
s
:
Sinces
2
1D.s1/.sC1/,
Y.s/D
1Cs
2
s.s1/.sC1/
D
1
s
C
1
s1
C
1
sC1
andyD 1Ce
t
Ce
t
.

141. 136 Chapter 8Laplace Transforms
8.3.18.
.s
2
Cs/Y.s/D
2
s3
C.4s/1D
2.s3/
2
s3
:
Sinces
2
CsDs.sC1/,
Y.s/D
2.s3/
2
s.sC1/.s3/
D
7
3s

7
2
1
sC1
C
1
6
1
s3
andyD
7
3

7
2
e
t
C
1
6
e
3t
.
8.3.20.
.s
2
C1/Y.s/D
1
s
2
C2;soY.s/D
1
.s
2
C1/s
2
C
2
s
2
C1
:
SubstitutingxDs
2
into
1
.xC1/x
D
1
xC1

1
x
yields
1
.s
2
C1/s
2
D
1
s
2

1
s
2
C1
;
soY.s/D
1
s
2
C
1
s
2
C1
andyDtCsint.
8.3.22.
.s
2
C5sC6/Y.s/D
2
sC1
C.3Cs/C5D
2C.sC1/.sC8/
sC1
:
Sinces
2
C5sC6D.sC2/.sC3/,
Y.s/D
2C.sC1/.sC8/
.sC1/.sC2/.sC3/
D
1
sC1
C
4
sC2

4
sC3
andyDe
t
C4e
2t
4e
3t
.
8.3.24.
.s
2
2s3/Y.s/D
10s
s
2
C1
C.7C2s/4D
10s
s
2
C1
C.2sC3/:
Sinces
2
2s3D.s3/.sC1/,
Y.s/D
10s
.s3/.sC1/.s
2
C1/
C
2sC3
.s3/.sC1/
: . A/
2sC3
.s3/.sC1/
D
9
4
1
s3

1
4
1
sC1
$
9
4
e
3t

1
4
e
t
: . B/
10s
.s3/.sC1/.s
2
C1/
D
A
s3
C
B
sC1
C
C sCD
s
2
C1
where
.A.sC1/CB.s3//.s
2
C1/C.C sCD/.s3/.sC1/D10s:
40AD 30 .setsD3/I
8BD 10 .setsD 1/I
A3B3DD 0 .setsD0/I
ACBCCD 0 .equate coefficients ofs
3
/:

142. Section 8.3Solution of Initial Value Problems137
Solving this system yieldsAD3=4,BD5=4,CD 2,DD 1. Therefore,
10s
.s3/.sC1/.s
2
C1/
D
3
4
1
s3
C
5
4
1
sC1

2sC1
s
2
C1
$
3
4
e
3t
C
5
4
e
t
2costsint:
From this, (A), and (B),yD sint2costC3e
3t
Ce
t
.
8.3.26.
.s
2
C4/Y.s/D
16
s
2
C4
C
9s
s
2
C1
Cs;so
Y.s/D
16
.s
2
C4/
2
C
9s
.s
2
C4/.s
2
C1/
C
s
s
2
C4
:
From the table of Laplace transforms,
tcos2t$
s
2
4
.s
2
C4/
2
D
s
2
C4
.s
2
C4/
2

8
.s
2
C4/
2
D
1
s
2
C4

8
.s
2
C4/
2
:
Therefore,
8
.s
2
C4/
2
D
1
s
2
C4
L.tcos2t/;so
16
.s
2
C4/
2
$sin2t2tcos2t: . A/
SubstitutingxDs
2
into
9
.xC4/.xC1/
D
3
xC1

3
xC4
and multiplying bysyields
9s
.s
2
C4/.s
2
C1/
D
3s
s
2
C1

3s
s
2
C4
$3cost3cos2t: . B/
Finally,
s
s
2
C4
$cos2t: . C/
Adding (A), (B), and (C) yieldsyD .2tC2/cos2tCsin2tC3cost.
28.
.s
2
C2sC2/Y.s/D
2
s
2
C.7C2s/C4:
Since.s
2
C2sC2/D.sC1/
2
C1,
Y.s/D
2
s
2
..sC1/
2
C1/
C
2s3
.sC1/
2
C1
: . A/
2s3
.sC1/
2
C1
D
2.sC1/5
.sC1/
2
C1
$e
t
.2cost5sint/: . B/
2
s
2
..sC1/
2
C1/
D
A
s
C
B
s
2
C
C.sC1/CD
.sC1/
2
C1
;

143. 138 Chapter 8Laplace Transforms
where.AsCB/

.sC1/
2
C1

Cs
2
.C.sC1/CD/D2.
2BD2 .setsD0/I
ACBCDD2 .setsD 1/I
ACCD0 .equate coefficients ofs
3
/I
2ACBCCCDD0 .equate coefficients ofs
2
/:
Solving this system yieldsAD 1,BD1,CD1,DD0. Therefore,
2
s
2
..sC1/
2
C1/
D
1
s
C
1
s
2
C
.sC1/
.sC1/
2
C1
$ 1CtCe
t
cost:
From this, (A), and (B),yD 1CtCe
t
.cost5sint/.
8.3.30..s
2
C4sC5/Y.s/D
.sC1/C3
.sC1/
2
C1
C4. Since.s
2
C4sC5/D.sC2/
2
C1,
Y.s/D
sC4
..sC1/
2
C1/..sC2/
2
C1/
C
4
.sC2/
2
C1
: . A/
4
.sC2/
2
C1
$4e
2t
sint: . B/
sC4
..sC1/
2
C1/..sC2/
2
C1/
D
A.sC1/CB
.sC1/
2
C1
C
C.sC2/CD
.sC2/
2
C1
;
where.A.sC1/CB/

.sC2/
2
C1

C.C.sC2/CD/

.sC1/
2
C1

D4Cs.
5AC5BC4CC2DD4 .setsD0/I
2BCCCDD3 .setsD 1/I
ACBC2DD2 .setsD 2/I
ACCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD 1,BD1,CD1,DD0. Therefore,
sC4
..sC1/
2
C1/..sC2/
2
C1/
D
.sC1/C1
.sC1/
2
C1
C
sC2
.sC2/
2
C1
;$e
t
.costCsint/Ce
2t
cost:
From this, (A), and (B),yDe
t
.costCsint/Ce
2t
.costC4sint/.
8.3.32.
.2s
2
3s2/Y.s/D
4
s1
C2.2Cs/3D
4C.2s7/.s1/
s1
Since2s
2
3s2D.s2/.2sC1/,
Y.s/D
4C.2s7/.s1/
2.s2/.s1/.sC1=2/
D
1
5
1
s2

4
3
1
s1
C
32
15
1
sC1=2
andyD
1
5
e
2t

4
3
e
t
C
32
15
e
t =2
.
8.3.34.
.2s
2
C2sC1/Y.s/D
2
s
2
C2.1Cs/C2D
2
s
2
C2s:

144. Section 8.3Solution of Initial Value Problems139
Since2s
2
C2sC1D2..sC1=2/
2
C1=4/,
Y.s/D
1
s
2
..sC1=2/
2
C1=4/
C
s
..sC1=2/
2
C1=4/
: . A/
s
..sC1=2/
2
C1=4/
$e
t =2
.cos.t=2/sin.t=2//: . B/
1
s
2
..sC1=2/
2
C1=4/
D
A
s
C
B
s
2
C
C.sC1=2/CD
..sC1=2/
2
C1=4/
where
.AsCB/..sC1=2/
2
C1=4/C.C.sC1=2/CD/s
2
D1:
BD2 .setsD0/I
AC2BC2DD8 .setsD 1=2/I
5AC10BC2CC2DD8 .setsD1=2/I
ACCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD 4,BD2,CD4,DD0. Therefore,
1
s
2
..sC1=2/
2
C1=4/
D
4
s
C
2
s
2
C
4.sC1=2/
.sC1=2/
2
C1=4
$ 4C2tC4e
t =2
cos.t=2/:
This, (A), and (B) imply thatyDe
t =2
.5cos.t=2/sin.t=2//C2t4.
8.3.36.
.4s
2
C4sC1/Y.s/D
3Cs
s
2
C1
C4.1C2s/C8D
3Cs
s
2
C1
C4.1C2s/C8sC4:
Since4s
2
C4sC1D4.sC1=2/
2
,
Y.s/D
3Cs
4.sC1=2/
2
.s
2
C1/
C
2
sC1=2
: . A/
3Cs
4.sC1=2/
2
.s
2
C1/
D
A
sC1=2
C
B
.sC1=2/
2
C
C sCD
s
2
C1
where
.A.sC1=2/CB/.s
2
C1/C.C sCD/.sC1=2/
2
D
3Cs
4
:
10BD5 .setsD 1=2/I
2AC4BCDD3 .setsD0/I
12AC8BC9CC9DD4 .setsD1/I
ACCD0 .equate coefficients ofs
3
/:
Solving this system yieldsAD3=5,BD1=2,CD 3=5,DD 1=5. Therefore,
3Cs
4.sC1=2/
2
.s
2
C1/
D
3
5
1
sC1=2
C
1
2
1
.sC1=2/
2

1
5
3sC1
s
2
C1
:
$
3
5
e
t =2
C
1
2
te
t =2

1
5
.3costCsint/:

145. 140 Chapter 8Laplace Transforms
Since
2
sC1=2
$2e
t =2
, this and (A) imply thatyD
e
t =2
10
.5tC26/
1
5
.3costCsint/.
8.3.38.Transforming the initial value problem
ay
00
Cby
0
CcyD0; y.0/D1; y
0
.0/D0
yields.as
2
CbsCc/Y.s/DasCb, soY.s/D
asCb
as
2
CbsCc
. Therefore,y1DL
1

asCb
as
2
CbsCc

satisfies the initial conditionsy1.0/D1,y
0
1
.0/D0.
Transforming the initial value problem
ay
00
Cby
0
CcyD0; y.0/D0; y
0
.0/D1
yields.as
2
CbsCc/Y.s/Da, soY.s/D
a
as
2
CbsCc
. Therefore,y2DL
1

a
as
2
CbsCc

satisfies the initial conditionsy1.0/D0,y
0
1
.0/D1.
8.4THE UNIT STEP FUNCTION
8.4.2.
L.f /D
Z
1
0
e
st
f .t/ dtD
Z
1
0
e
st
t dtC
Z
1
1
e
st
dt: . A/
To relate the first term to a Laplace transform we add and subtract
R
1
1
e
st
t dtin (A) to obtain
L.f /D
Z
1
0
e
st
t dtC
Z
1
1
e
st
.1t/ dtDL.t/
Z
1
1
e
st
.t1/ dt: . B/
LettingtDxC1in the last integral yields
Z
1
1
e
st
.t1/ dtD
Z
1
0
e
s.xC1/
x dxDe
s
L.t/:
This and (B) imply thatL.f /D.1e
s
/L.t/D
1e
s
s
2
.
Alternatively,f .t/Dtu.t1/.t1/$.1e
s
/L.t/D
1e
s
s
2
.
8.4.4.
L.f /D
Z
1
0
e
st
f .t/ dtD
Z
1
0
e
st
dtC
Z
1
1
e
st
.tC2/ dt: . A/
To relate the first term to a Laplace transform we add and subtract
R
1
1
e
st
dtin (A) to obtain
L.f /D
Z
1
0
e
st
dtC
Z
1
1
e
st
.tC1/ dtDL.t/C
Z
1
1
e
st
.tC1/ dt: . B/
LettingtDxC1in the last integral yields
Z
1
1
e
st
.tC1/ dtD
Z
1
0
e
s.xC1/
.xC2/ dxDe
s
L.tC2/:
This and (B) imply thatL.f /DL.1/Ce
s
L.tC2/D
1
s
Ce
s

1
s
2
C
2
s

.

146. Section 8.4The Unit Step Function141
Alternatively,
f .t/D1Cu.t1/.tC1/$L.1/Ce
s
L.tC2/D
1
s
Ce
s

1
s
2
C
2
s

.
8.4.6.
L.f /D
Z
1
0
e
st
f .t/ dtD
Z
1
0
e
st
t
2
DL.t
2
/
Z
1
1
t
2
dt: . A/
LettingtDxC1in the last integral yields
Z
1
1
e
st
t
2
dtD
Z
1
0
e
s.xC1/
.t
2
C2tC1/ dxDe
s
L.t
2
C2tC1/:
This and (A) imply that
L.f /DL.t
2
/Ce
s
L.t
2
C2tC1/D
2
s
3
e
s

2
s
3
C
2
s
2
C
1
s

:
Alternatively,
f .t/Dt
2
.1u.t1//$L.t
2
/Ce
s
L.t
2
C2tC1/D
2
s
3
e
s

2
s
3
C
2
s
2
C
1
s

8.4.8.f .t/Dt
2
C2Cu.t1/.tt
2
2/. Sincet
2
C2$
2
s
3
C
2
s
and
L

u.t1/.tt
2
2/

De
s
L

.tC1/.tC1/
2
2

D e
s
L.t
2
CtC2/D e
s

2
s
3
C
1
s
2
C
2
s

;
it follows thatF.s/D
2
s
3
C
2
s
e
s

2
s
3
C
1
s
2
C
2
s

.
8.4.10.f .t/De
t
Cu.t1/.e
2t
e
t
/$L.e
t
/Ce
s
L.e
2.tC1/
/e
s
L.e
t1
/DL.e
t
/C
e
.sC2/
L.e
2t
/e
.sC1/
L.e
t
/D
1e
.sC1/
sC1
C
e
.sC2/
sC2
.
8.4.12.f .t/DŒu.t1/u.t2/ t$e
s
L.tC1/e
2s
L.tC2/
De
s

1
s
2
C
1
s

e
2s

1
s
2
C
2
s

.
8.4.14.
f .t/Dt2u.t1/.t1/Cu.t2/.tC4/$
1
s
2
2e
s
L.t/Ce
2s
L.tC6/
D
1
s
2

2e
s
s
2
Ce
2s

1
s
2
C
6
s

:
8.4.16.f .t/D22u.t1/tCu.t3/.5t2/$L.2/2e
s
L.tC1/Ce
3s
L.5tC13/D
2
s
e
s

2
s
2
C
2
s

Ce
3s

5
s
2
C
13
s

.

147. 142 Chapter 8Laplace Transforms
8.4.18.f .t/D.tC1/
2
Cu.t1/

.tC2/
2
.tC1/
2

Dt
2
C2tC1Cu.t1/.2tC3/$
L.t
2
C2tC1/Ce
s
L.2tC5/D
2
s
3
C
2
s
2
C
1
s
Ce
s

2
s
2
C
5
s

.
8.4.20.
1
s.sC1/
D
1
s

1
sC1
$1e
t
)e
s
1
s.sC1/
$u.t1/

1e
.t1/

D
(
0; 0 t < 1;
1e
.t1/
; t1:
8.4.22.
3
s

1
s
2
$3t)e
s

3
s

1
s
2

$u.t1/ .3.t1//Du.t1/.4t/I
1
s
C
1
s
2
$1Ct)e
3s

1
s
C
1
s
2

$u.t3/ .1C.t3//Du.t3/.t2/I
therefore
h.t/D2CtCu.t1/.4t/Cu.t3/.t2/D
8
ˆ
ˆ
<
ˆ
ˆ
:
2Ct; 0t < 1;
6; 1t < 3;
tC4; t3:
8.4.24.
12s
s
2
C4sC5
D
52.sC2/
.sC2/
2
C1
$e
2t
.5sint2cost/I
therefore,
h.t/Du.t/e
2.t/
.5sin.t/2cos.t//
Du.t/e
2.t/
.2cost5sint/
D
(
0; 0 t < ;
e
2.t/
.2cost5sint/; t:
:
8.4.26.DenoteF.s/D
3.s3/
.sC1/.s2/

sC1
.s1/.s2/
. Since
3.s3/
.sC1/.s2/
D
4
sC1

1
s2
and
sC1
.s1/.s2/
D
3
s2

2
s1
,F.s/D
4
sC1

4
s2
C
2
s1
$4e
t
4e
2t
C2e
t
. Therefore,e
2s
F.s/$
u.t2/

4e
.t2/
4e
2.t2/
C2e
.t2/

D
(
0; 0 t < 2;
4e
.t2/
4e
2.t2/
C2e
.t2/
; t2:
8.4.28.
3
s

1
s
3
$3
t
2
2
)e
2s

3
s

1
s
3

$u.t2/

3
.t2/
2
2

Du.t2/


t
2
2
C2tC1

I
1
s
2
$t)
e
4s
s
2
$u.t4/.t4/I

148. Section 8.5Constant Coeefficient Equations with Piecewise ContinuousForcing Functions 143
therefore
h.t/D1t
2
Cu.t2/


t
2
2
C2tC1

Cu.t4/.t4/
D
8
ˆ
ˆ
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
ˆ
ˆ
:
1t
2
; 0 t < 2

3t
2
2
C2tC2; 2t < 4;

3t
2
2
C3t2; t4:
8.4.30.LetTbe an arbitrary positive number. Since limm!1tmD 1, only finitely many members of
ftmgare inŒ0; T . Sincefmis continuous onŒtm;1/for eachm,fis piecewise continuous onŒ0; T .
IftMt < tMC1, thenu.ttm/D1ifmM, whileu.ttm/D0ifm > M. Therefore,
f .t/Df0.t/C
M
X
mD1
.fm.t/fm1.t//DfM.t/
8.4.32.Since
1
X
mD0
e
Km
converges if > 0,
1
X
mD0
e
tm
converges if > 0, by the comparison test.
Therefore,(C) of Exercise 8.3.31 holds ifs > s0Cifis any positive number. This implies that it holds
ifs > s0.
8.4.34.LettmDmandfm.t/D.1/
m
,mD0; 1; 2; : : :. Thenfm.t/fm1.t/D.1/
m
2, so
f .t/D1C2
1
X
mD1
.1/
m
u.tm/andF.s/D
1
s
 
1C2
1
X
mD1
.1/
m
e
ms
!
SubstitutingxDe
s
in the
identity
1
X
mD1
.1/
m
x
m
D
x
1Cx
(jxj< 1) yieldsF.s/D
1
s

1
2e
s
1Ce
s

D
1
s
1e
s
1Ce
s
.
8.4.36.LettmDmandfm.t/D.1/
m
m,mD0; 1; 2; : : :. Thenfm.t/fm1.t/D.1/
m
.2m1/,
sof .t/D
1
X
mD1
.1/
m
.2m1/u.tm/andF.s/D
1
s
1
X
mD1
.1/
m
.2m1/e
ms
. SubstitutingxDe
s
in the identities
1
X
mD1
.1/
m
x
m
D
x
1Cx
and
1
X
mD1
.1/
m
mx
m
D
x
.1Cx/
2
(jxj< 1) yieldsF.s/D
1
s

e
s
1Ce
s

2e
s
.1Ce
s
/
2

D
1
s
.1e
s
/
.1Ce
s
/
2
.
8.5CONSTANT COEEFFICIENT EQUATIONS WITH PIECEWISE CONTINUOU S FORCING
FUNCTIONS
8.5.2.y
00
CyD3Cu.t4/.2t8/; y.0/D1; y
0
.0/D0. Since
L .u.t4/.2t8//De
4s
L .2.tC4/8/De
4s
L.2t/D
2e
4s
s
2
;
.s
2
C1/Y.s/D
3
s
C
2e
4s
s
2
Cs:

149. 144 Chapter 8Laplace Transforms
Y.s/D
3
s.s
2
C1/
C
2e
4s
s
2
.s
2
C1/
C
s
s
2
C1
D3

1
s

s
s
2
C1

C2e
4s

1
s
2

1
s
2
C1

C
s
s
2
C1
D
3
s

2s
s
2
C1
C2e
4s

1
s
2

1
s
2
C1

:
Since
1
s
2

1
s
2
C1
$tsint)e
4s

1
s
2

1
s
2
C1

$u.t4/ .t4sin.t4// ;
yD32costC2u.t4/ .t4sin.t4//.
8.5.4.y
00
yDe
2t
Cu.t2/.1e
2t
/; y.0/D3; y
0
.0/D 1. Since
L.u.t2/.1e
2t
//De
2s
L.1e
2.tC2/
/De
2s

1
s

e
4
s2

;
.s
2
1/Y.s/D
1
s2
Ce
2s

1
s

e
4
s2

C.1C3s/:
Therefore,
Y.s/D
1
.s1/.sC1/.s2/
C
3s1
.s1/.sC1/
Ce
2s

1
s.s1/.sC1/

e
4
.s1/.sC1/.s2/

:
1
.s1/.sC1/.s2/
D
1
2
1
s1
C
1
6
1
sC1
C
1
3
1
s2
$
1
2
e
t
C
1
6
e
t
C
1
3
e
2t
I
e
2s
e
4
.s1/.sC1/.s2/
$u.t2/


1
2
e
tC2
C
1
6
e
.t6/
C
1
3
e
2t

I
1
s.s1/.sC1/
D
1
s
C
1
2
1
s1
C
1
2
1
sC1
$ 1C
1
2
e
t
C
1
2
e
t
I
e
2s
s.s1/.sC1/
$u.t2/

1C
1
2
e
t2
C
1
2
e
.t2/

I
3s1
.s1/.sC1/
D
1
s1
C
2
sC1
$e
t
C2e
t
:
Therefore,
yD
1
2
e
t
C
13
6
e
t
C
1
3
e
2t
Cu.t2/

1C
1
2
e
t2
C
1
2
e
.t2/
C
1
2
e
tC2

1
6
e
.t6/

1
3
e
2t

:
8.5.6.Note thatjsintj Dsintif0t < , whilejsintj D sintift < 2. Rewrite the initial
value problem as
y
00
C4yDsint2u.t/sintCu.t2/sint; y.0/D 3; y
0
.0/D1:

150. Section 8.5Constant Coeefficient Equations with Piecewise ContinuousForcing Functions 145
Since
L .u.t/sint/De
s
L.sin.tC//D e
s
L.sint/
and
L .u.t2/sint/De
2s
L.sin.tC2//De
2s
L.sint/;
.s
2
C4/Y.s/D
1C2e
s
Ce
2s
.s
2
C1/
C13s;soY.s/D
1C2e
s
Ce
2s
.s
2
C1/.s
2
C4/
C
13s
s
2
C4
:
1
.s
2
C1/.s
2
C4/
D

1
s
2
C1

1
s
2
C4

$
1
3
sint
1
6
sin2tI
therefore
e
s
.s
2
C1/.s
2
C4/
$u.t/

1
3
sin.t/
1
6
sin2.t/

D u.t/

1
3
sintC
1
6
sin2t

and
e
2s
.s
2
C1/.s
2
C4/
$u.t2/

1
3
sin.t2/
1
6
sin2.t2/

Du.t2/

1
3
sint
1
6
sin2t

I
therefore
yD
1
3
sin2t3cos2tC
1
3
sint2u.t/

1
3
sintC
1
6
sin2t

Cu.t2/

1
3
sint
1
6
sin2t

:
8.5.8.y
00
C9yDcostCu.t3=2/.sintcost/; y.0/D0; y
0
.0/D0. Since
L .u.t3=2/.sintcost//De
3s=2
L .sin.tC3=2/cos.tC3=2//
e
3s=2
L.costCsint/;
.s
2
C9/Y.s/D
1
s
2
C1
e
3s=2
sC1
s
2
C1
;soY.s/D
1
.s
2
C1/.s
2
C9/
e
3s=2
sC1
.s
2
C1/.s
2
C9/
:
1
.s
2
C1/.s
2
C9/
D
1
8

1
s
2
C1

1
s
2
C9

$
1
8

sint
1
3
sin3t

and
s
.s
2
C1/.s
2
C9/
D
1
8

s
s
2
C1

s
s
2
C9

$
1
8
.costcos3t/ :
sC1
.s
2
C1/.s
2
C9/
D
sC1
.s
2
C1/.s
2
C9/
D
sC1
8

sC1
s
2
C1

sC1
s
2
C9

$
1
8

costCsintcos3t
1
3
sin3t

;so

151. 146 Chapter 8Laplace Transforms
e
3s=2
sC1
.s
2
C1/.s
2
C9/
$
u.t3=2/
8
.cos.t3=2/Csin.t3=2/
cos3.t3=2/
1
3
sin3.t=2/

D
u.t3=2/
8

sintcostCsin3t
1
3
cos3t

:
Therefore,yD
1
8
.costcos3t/
1
8
u.t3=2/

sintcostCsin3t
1
3
cos3t

.
8.5.10.y
00
CyDt2u.t/t; y.0/D0; y
0
.0/D0. Since
L .u.t/t/De
s
L .tC/De
s

1
s
2
C

s

;
.s
2
C1/Y.s/D
1
s
2
2e
s

1
s
2
C

s

I
Y.s/D
1
s
2
.s
2
C1/
2e
s

1
s
2
.s
2
C1/
C

s.s
2
C1/

D

1
s
2

1
s
2
C1

2e
s

1
s
2

1
s
2
C1

2e
s

1
s

s
s
2
C1

:
Since
1
s
2

1
s
2
C1
$tsint)e
s

1
s
2

1
s
2
C1

$u.t/ .tsin.t//Du.t/.tCsint/
and
1
s

s
s
2
C1
$1cost)e
s

1
s

s
s
2
C1

$u.t/ .1cos.t//Du.t/.1Ccost/;
yDtsint2u.t/.tCsintCcost/.
8.5.12.y
00
CyDt3u.t2/t; y.0/D1; y
0
.0/D2I
L.u.t2/t/De
2s
L.tC2/De
2s

1
s
2
C
2
s

I
.s
2
C1/Y.s/D
13e
2s
s
2

6e
2s
s
C2CsI
Y.s/D
13e
2s
s
2
.s
2
C1/

6e
2s
s.s
2
C1/
C
2Cs
s
2
C1
I
1
s
2
.s
2
C1/
D
1
s
2

1
s
2
C1
$tsintI
e
2s
s
2
.s
2
C1/
$u.t2/..t2sin.t2//Du.t2/.t2sint/I

152. Section 8.5Constant Coeefficient Equations with Piecewise ContinuousForcing Functions 147
1
s.s
2
C1/
D
1
s

s
s
2
C1
$1costI
e
2s
s.s
2
C1/
$u.t2/.1cos.t2//Du.t2/.1cost/I
2Cs
s
2
C1
$2sintCcostI
yDtCsintCcostu.t2/.3t3sint6cost/.
8.5.14.y
00
4y
0
C3yD 1C2u.t1/; y.0/D0; y
0
.0/D0;
.s
2
4sC3/Y.s/D
1C2e
s
s
;Y.s/D
1C2e
s
s.s1/.s3/
;
1
s.s1/.s3/
D
1
3s
C
1
6
1
s3

1
2
1
s1
$
1
3
C
1
6
e
3t

1
2
e
t
;
e
s
s.s1/.s3/
$u.t1/

1
3
C
1
6
e
3.t1/

1
2
e
t1

;
yD
1
3

1
6
e
3t
C
1
2
e
t
Cu.t1/

2
3
C
1
3
e
3.t1/
e
t1

.
8.5.16.y
00
C2y
0
CyD4e
t
4u.t1/e
t
; y.0/D0; y
0
.0/D0. Since
L

4u.t1/e
t

De
s
L

4e
.tC1/

D
4e
sC1
s1
;
.s
2
C2sC1/Y.s/D
4
s1

4e
sC1
s1
;so
Y.s/D
4
.s1/.sC1/
2

4e
sC1
.s1/.sC1/
2
:
1
.s1/.sC1/
2
D
A
s1
C
B
sC1
C
C
.sC1/
2
;
where
A.sC1/
2
CB.s1/.sC1/CC.s1/D4:
AD1 .setsD1/I
CD 2 .setsD 1/I
ACBD0 .equate coefficients ofs
2
/:
Solving this system yieldsAD1,BD 1,CD 2. Therefore,
1
.s1/.sC1/
2
D
1
s1

1
sC1

2
.sC1/
2
and
yDe
t
e
t
2te
t
eu.t1/

e
t1
e
.t1/
2.t1/e
.t1/

De
t
e
t
2te
t
u.t1/

e
t
e
.t2/
2.t1/e
.t2/

:
8.5.18.y
00
4y
0
C4yDe
2t
2u.t2/e
2t
; y.0/D0; y
0
.0/D 1. Since
L

u.t2/e
2t

De
2s
L

e
2tC4

D
e
2sC4
s2
;

153. 148 Chapter 8Laplace Transforms
.s
2
4sC4/Y.s/D
1
s2

2e
2sC4
s2
1;so
Y.s/D
1
.s2/
3

2e
2sC4
.s2/
3

1
.s2/
2
:
1
.s2/
3
$
t
2
e
2t
2
)
e
2sC4
.s2/
3
$
e
4
2
u.t2/e
2.t2/
.t2/
2
Du.t2/
.t2/
2
e
2t
2
I
thereforeyD
t
2
e
2t
2
te
2t
u.t2/.t2/
2
e
2t
.
8.5.20.y
00
C2y
0
C2yD1Cu.t2/.t1/u.t3/.tC1/; y.0/D2; y
0
.0/D 1;
L.u.t2/.t1//De
2s
L..tC21//De
2s

1
s
2
C
21
s

I
L.u.t3/.tC1//De
3s
L..tC3C1//De
3s

1
s
2
C
3C1
s

I
.s
2
C2sC2/Y.s/D
1
s
Ce
2s

1
s
2
C
21
s

e
3s

1
s
2
C
3C1
s

C.1C2s/C4:
LetG.s/D
1
s.s
2
C2sC2/
,H.s/D
1
s.s
2
C2sC2/
; then
Y.s/DY1.s/Ce
2s
Y2.s/e
3s
Y3.s/; (A)
where
Y1.s/DG.s/C
2sC3
s
2
C2sC2
; (B)
Y2.s/DH.s/C.21/G.s/; (C)
Y3.s/DH.s/C.3C1/G.s/: (D)
Letyi.t/DL
1
.Yi.s//,.iD1; 2; 3/. From (A),
y.t/Dy1.t/Cu.t2/y2.t2/u.t3/y3.t3/: (E)
FindL
1
.G.s//:
G.s/D
A
s
C
B.sC1/CC
.sC1/
2
C1
whereA..sC1/
2
C1/C.B.sC1/CC /sD1. SettingsD0yieldsAD1=2; settingsD 1yields
ACD1, soCD 1=2; sinceACBD0(coefficient ofx
2
),BD 1=2. Therefore,
G.s/D
1
2

1
s

.sC1/C1
.sC1/
2
C1/

$
1
2

1
2
e
t
.costCsint/: (F)
FindL
1
.H.s//:
H.s/D
A
s
C
B
sC2
C
C.sC1/CD
.sC1/
2
C1

154. Section 8.5Constant Coeefficient Equations with Piecewise ContinuousForcing Functions 149
where.AsCB/..sC1/
2
C1/C.C.sC1/CD/s
2
D1.
2BD1 .setsD0/I
ACBCDD1 .setsD 1/I
5AC5BC2CCDD1 .setsD1/I
ACBD0D0 .equate coefficients ofs
3
/:
Solving this system yieldsAD 1=2,BD1=2,CD1=2,DD0; therefore
H.s/D
1
2

1
s

1
s
2

sC1
.sC1/
2
C1/

$
1
2
.1te
t
cost/: (G)
Since
2sC3
s
2
C2sC2
D
2.sC1/C1
.sC1/
2
C1
$e
t
.2costCsint/;
(B) and (F)) imply that
y1.t/D
1
2
e
t
.3costCsint/C
1
2
: (H)
From (C), (F)), and (G),
y2.t/D1C
t
2
C.1/e
t
cost
21
2
e
t
sint;
so
y2.t2/D

e
.t2/

.1/costC
21
2
sint

C1
t
2

: (I)
From (D), (F)), and (G),
y3.t/D
1
2

e
t
.3costC.3C1/sintCtC3/

;
so
y3.t3/D
1
2

e
.t3/
.3costC.3C1/sintCt/

: (J)
Now (E), (20), (I), and (J)
yD
1
2
e
t
.3costCsint/C
1
2
imply that
u.t2/

e
.t2/

.1/costC
21
2
sint

C1
t
2


1
2
u.t3/

e
.t3/
.3costC.3C1/sint/Ct

.
8.5.22.(a)f .t/D
1
X
nD0
u.tn/;F.s/D
1
s
1
X
nD0
e
ns
;Y.s/D
1
s.s
2
C1/
1
X
nD0
e
ns
;
1
s.s
2
C1/
D
1
s

s
s
2
C1
$1cost;
e
ns
s.s
2
C1/
$u.tn/.1cos.tn//Du.tn/.1.1/
n
cost/;
y.t/D
1
X
nD0
u.tn/.1.1/
n
cost/. Ifmt < .mC1/,y.t/D
m
X
nD0
.1.1/
n
cost/. Therefore,
y.t/D
(
2mC1cost; 2mt < .2mC1/ .mD0; 1; : : : /
2m; .2m 1/t < 2m .mD1; 2; : : : /
.

155. 150 Chapter 8Laplace Transforms
(b)f .t/D
1
X
nD0
u.t2n/t;F.s/D
1
X
nD0
e
2ns
L.tC2ns/D
1
X
nD0
e
2ns

1
s
2
C
2n
s

;Y.s/D
1
X
nD0
e
2ns
Yn.s/, whereYn.s/D
1
s
2
.s
2
C1/
C
2n
s.s
2
C1/
D
1
s
2

1
s
2
C1
C
2n
s

2n
s
2
C1
$yn.t/D
tsintC2n2ncost. Since cos.t2n/Dcostand sin.t2n/Dsint,e
2ns
Yn.s/$u.t
2n/yn.t/Du.t2n/.tsint2ncost/; thereforey.t/D
1
X
nD0
u.t2n/.tsint2ncost/.
If2mt < 2.mC1/, then
y.t/D
m
X
nD0
.tsint2ncost/D.mC1/.tsintmcost/:
(c)f .t/D1C2
1
X
nD1
.1/
n
u.tn/;F.s/D
1
s
 
1C2
1
X
nD1
.1/
n
e
ns
!
;Y.s/D
1
s.s
2
C1/
 
1C2
1
X
nD1
.1/
n
e
ns
!
;
1
s.s
2
C1/
D
1
s

s
s
2
C1
$1cost;
e
ns
s.s
2
C1/
$u.tn/.1cos.tn//Du.tn/.1
.1/
n
cost/;y.t/D1costC2
1
X
nD1
.1/
n
u.tn/.1.1/
n
cost/. Ifmt < .mC1/,
y.t/D1costC2
m
X
nD1
.1/
n
.1.1/
n
cost/D.1/
m
.2mC1/cost:
(d)f .t/D
1
X
nD0
u.tn/;F.s/D
1
s
1
X
nD0
e
ns
;Y.s/D
1
s.s
2
1/
1
X
nD0
e
ns
;
1
s.s
2
1/
D
1
2
1
s1
C
1
2
1
sC1

1
s
$
1
2
.e
t
Ce
t
2/;
e
ns
s.s
2
1/
$
u.tn/
2

e
t
Ce
t
2

;y.t/D
1
2
1
X
nD0
u.tn/

e
tn
Ce
.tn/
2

.
Ifmt < .mC1/,
y.t/D
1
2
m
X
nD0

e
tn
Ce
.tn/
2

D
1
2
.e
tm
Ce
t
/
m
X
nD0
e
n
m1
D
1e
mC1
2.1e/
.e
tm
Ce
t
/m1:
(e)f .t/D.sintC2cost/
1
X
nD0
u.t2n/;F.s/D
1C2s
s
2
C1
1
X
nD0
e
2ns
;Y.s/D
1C2s
.s
2
C1/.s
2
C2sC2/
1
X
nD0
e
2ns
;
1C2s
.s
2
C1/.s
2
C2sC2/
D
AsCB
s
2
C1
C
C.sC1/CD
.sC1/
2
C1
where
.AsCB/..sC1/
2
C1/C.C.sC1/CD/.s
2
C1/D1C2s:
2BCCCDD 1 .setsD0/I
ACBC2DD 1 .setsD 1/I
5AC5BC4CC2DD 3 .setsD1/I
ACCD 0 .equate coefficients ofs
3
/:

156. Section 8.5Constant Coeefficient Equations with Piecewise ContinuousForcing Functions 151
Solving this system yieldsAD0,BD1,CD0,DD 1. Therefore,
1C2s
.s
2
C1/.s
2
C2sC2/
D
1
s
2
C1

1
.sC1/
2
C1
$

1e
t

sint:
Since sin.t2n/Dsint,
e
2ns
1C2s
.s
2
C1/.s
2
C2sC2/
$u.t2n/

1e
.t2n/

sint;
so
y.t/Dsint
1
X
nD0
u.t2n/

1e
.t2n/

:
If2mt < 2.mC1/,
y.t/Dsint
m
X
nD0

1e
.t2n/

D
 
mC1
 
1e
2.mC1/
1e
2
!
e
t
!
sint:
(f)f .t/D
1
X
nD0
u.tn/;F.s/D
1
s
1
X
nD0
e
ns
;Y.s/D
1
s.s1/.s2/
;
1
s.s1/.s2/
D
1
2s

1
s1
C
1
2
1
s2
$
1
2

12e
t
Ce
2t

I
e
ns
s.s1/.s2/
$
1
2
u.tn/

12e
tn
Ce
2.tn/

I
y.t/D
1
2
1
X
nD0
u.tn/

12e
tn
Ce
2.tn/

:
Ifmt < mC1,
y.t/D
m
X
nD0

12e
tn
Ce
2.tn/

D
mC1
2
e
tm
m
X
nD0
e
n
C
1
2
e
2.tm/
m
X
nD0
e
2n
D
mC1
2
e
tm
1e
mC1
1e
C
1
2
e
2.tm/
1e
2mC2
1e
2
:
8.5.24.(a)The assumptions imply thaty
00
.t/D
f .t/by
0
.t/cy.t/
a
on.˛; t0/and.t0; ˇ/,y
00
.t0C/D
f .t0C/by
0
.t0/cy.t0/
a
, andy
00
.t0/D
f .t0/by
0
.t0/cy.t0/
a
. This implies the conclusion.
(b)Sincey
00
has a junp discontinouity att0, applying Exercise 8.4.23(c)toy
0
shows thaty
0
is not
differentiable att0. Therefore,ycannot satisfy (A) on.˛; ˇ/iffhas a jump discontinuity at somet0in
.˛; ˇ/.
8.5.26.If0t < t0, theny.t/D´0.t/. Therefore,y.0/D´0.0/Dk0andy
0
.0/D´
0
.0/Dk1, and
ay
00
Cby
0
CcyDa´
00
0
Cb´
0
0
Cc´0Df0.t/Df .t/; 0 < t < t0:

157. 152 Chapter 8Laplace Transforms
Now suppose that1mn. For convenience, definetnC1D 1. Iftmt < tmC1, then
y.t/D
P
m
kD0
´m.t/, so
ay
00
CbyCcyD
m
X
kD0
.a´
00
k
Cb´
0
k
Cc´k/Df0C
m
X
kD1
.fkfk1/DfmDf; tm< t < tmC1:
Thus,ysatisfiesay
00
Cby
0
CcyDfon any open interval that does not contain any of the pointst1,
t2,. . . ,tn.
Since´.tm/D´
0
.tm/formD1; 2; : : :,yandy
0
are continuous onŒ0;1/. Sincey
00
.t/D .by
0
.t/C
cy.t//=aift¤tm(mD1; 2; : : :),y
00
has limits from the left att1; : : : ; tn.
8.6CONVOLUTION
8.6.2.(a)sinat$
a
s
2
Ca
2
and cosbt$
s
s
2
Cb
2
, soH.s/D
as
.s
2
Ca
2
/.s
2
Cb
2
/
.
(b)e
t
$
1
s1
and sinat$
a
s
2
Ca
2
, soH.s/D
a
.s1/.s
2
Ca
2
/
.
(c)sinhat$
a
s
2
a
2
and coshat$
1
s
2
a
2
, soH.s/D
as
.s
2
a
2
/
2
.
(d)tsin!t$
2!s
.s
2
C!
2
/
2
andtcos!t$
s
2
!
2
.s
2
C!
2
/
2
, soH.s/D
2!s.s
2
!
2
/
.s
2
C!
2
/
4
.
(e)e
t
Z
t
0
sin!cos!.t/ dD
Z
t
0
.e

sin!/

e
.t /
cos!.t/

d;e
t
sin!t$
!
.s1/
2
C!
2
ande
t
cos!t$
s1
.s1/
2
C!
2
, soH.s/D
.s1/!
..s1/
2
C!
2
/
2
.
(f)e
t
Z
t
0

2
.t/e

dD
Z
t
0

2
e
2
.t/e
.t /
d;t
2
e
2t
$
2
.s2/
3
andte
t
$
1
.s1/
2
, so
H.s/D
2
.s2/
3
.s1/
2
.
(g)e
t
Z
t
0
e

cos!.t/ dD
Z
t
0
e
2
e
.t /
cos!.t/ d;te
2t
$
1
.sC2/
2
ande
t
cos!t$
sC1
.sC1/
2
C!
2
, soH.s/D
sC1
.sC2/
2
Œ.sC1/
2
C!
2

.
(h)e
t
Z
t
0
e
2
sinh.t/ dD
Z
t
0
e
3

e
.t /
sinh.t/

d;e
3t
$
1
s3
ande
t
sinht$
1
.s1/
2
1
,
soH.s/D
1
.s3/ ..s1/
2
1/
.
(i)te
2t
$
1
.s2/
2
and sin2t$
2
s
2
C4
, soH.s/D
2
.s2/
2
.s
2
C4/
.
(j)t
3
$
6
s
4
ande
t
$
1
s1
, soH.s/D
6
s
4
.s1/
.
(k)t
6
$
6Š
s
7
ande
t
sin3t$
3
.sC1/
2
C9
, soH.s/D
36Š
s
7
Œ.sC1/
2
C9
.
(l)t
2
$
2
s
3
andt
3
$
6
s
4
, soH.s/D
12
s
7
.
(m)t
7
$
7Š
s
8
ande
t
sin2t$
2
.sC1/
2
C4
, soH.s/D
27Š
s
8
Œ.sC1/
2
C4
.
(n)t
4
$
24
s
5
and sin2t$
2
s
2
C4
, soH.s/D
48
s
5
.s
2
C4/
.

158. Section 8.6Convolution153
8.6.4.(a)Y.s/D
1
s
2

Y.s/
s
2
;Y.s/

1C
1
s
2

D
1
s
2
;Y.s/
s
2
C1
s
2
D
1
s
2
;Y.s/D
1
s
2
C1
, soyDsint.
(b)Y.s/D
1
s
2
C1

2sY.s/
s
2
C1
;Y.s/

1C
2s
s
2
C1

D
1
s
2
C1
;Y.s/
.sC1/
2
s
2
C1
D
1
s
2
C1
;Y.s/D
1
.sC1/
2
, soyDte
t
.
(c)Y.s/D
1
s
C
2sY.s/
s
2
C1
;Y.s/

1
2s
s
2
C1

D
1
s
;Y.s/
.s1/
2
s
2
C1
D
1
s
;Y.s/D
.s
2
C1/
s.s1/
2
D
A
s
C
B
s
2
C
C
.s1/
2
, whereA.s1/
2
CBs.s1/CC sDs
2
C1. SettingsD0andsD1shows thatAD1
andCD2; equating coefficients ofs
2
yieldsACBD1, soBD0. Therefore,Y.s/D
1
s
C
1
.s1/
2
,
soyD1C2te
t
.
(d)Y.s/D
1
s
2
C
Y.s/
sC1
;Y.s/

1
1
sC1

D
1
s
2
;Y.s/

s
sC1

D
1
s
2
;Y.s/D
sC1
s
3
D
1
s
2
C
1
s
3
,
soyDtC
t
2
2
.
(e)sY.s/4D
1
s
2
C
sY.s/
s
2
C1
;Y.s/

s
s
s
2
C1

D4C
1
s
2
;Y.s/
s
3
s
2
C1
D
4s
2
C1
s
2
;Y.s/D
.4s
2
C1/.s
2
C1/
s
5
D
4s
4
C5s
2
C1
s
5
D
4
s
C
5
s
3
C
1
s
5
, soyD4C
5
2
t
2
C
1
24
t
4
.
(f)Y.s/D
s1
s
2
C1
C
Y.s/
s
2
C1
;Y.s/

1
1
s
2
C1

D
s1
s
2
C1
;Y.s/
s
2
s
2
C1
D
s1
s
2
C1
;Y.s/D
s1
s
2
D
1
s

1
s
2
, soyD1t.
8.6.6.SubstitutingxDtyields
Z
t
0
f .t/g./ dD
Z
0
t
f .x/g.tx/.dx/D
Z
t
0
f .x/g.t
x/ dxD
Z
t
0
f ./g.t/ d.
8.6.8.p.s/Y.s/DF.s/Ca.k1Ck0s/Cbk0, so (A)Y.s/D
F.s/
p.s/
C
k0.asCb/Ck1a
p.s/
. Since
p.s/Da.sr1/.sr2/and thereforebD a.r1Cr2/, (A) can be rewritten as
Y.s/D
F.s/
a.sr1/.sr2/
C
k0.sr1r2/
.sr1/.sr2/
C
k1
.sr1/.sr2/
:
1
.sr1/.sr2/
D
1
r2r1

1
sr2

1
sr1

$
e
r2t
e
r1t
r2r1
;
so the convolution theorem implies that
F.s/
a.sr1/.sr2/
$
1
a
Z
t
0
e
r2
e
r1
r2r1
f .t/ d:
sr1r2
.sr1/.sr2/
D
r2
r2r1
1
sr1

r1
r2r1
1
sr2
$
r2e
r1t
r1e
r2t
r2r1
:
Therefore,
y.t/Dk0
r2e
r1t
r1e
r2t
r2r1
Ck1
e
r2t
e
r1t
r2r1
C
1
a
a
Z
t
0
e
r2
e
r1
r2r1
f .t/ d:

159. 154 Chapter 8Laplace Transforms
8.6.10.p.s/Y.s/DF.s/Ca.k1Ck0s/Cbk0, so (A)Y.s/D
F.s/
p.s/
C
k0.asCb/Ck1a
p.s/
. Since
p.s/Da.s/
2
C!
2
and thereforebD 2a, (A) can be rewritten as
Y.s/D
F.s/
aŒ.s/
2
C!
2
C
k0.s2/
.s/
2
C!
2
C
k1
.s/
2
C!
2
:
1
.s/
2
C!
2
$
1
!
e
t
sin!t, so the convolution theorem implies that
F.s/
aŒ.s/
2
C!
2

$
1
a!
Z
t
0
e
t
f .t/sin! d:
s2
.s/
2
C!
2
D
.s/
.s/
2
C!
2
$e
t

cos!t

!
sin!t

:
Therefore,
y.t/De
t

k0

cos!t

!
sin!t

C
k1
!
sin!t

C
1
a!
Z
t
0
e
t
f .t/sin! d:
8.6.12.(a)
ay
00
Cby
0
CcyDf0.t/Cu.tt1/.f1.t/f0.t//; y.0/D0; y
0
.0/D0I
p.s/Y.s/DF0.s/CL.u.tt1/.f1.t/f0.t///DF0.s/Ce
st1
L.g/I
Y.s/D
F0.s/Ce
st1G.s/
p.s/
: . B/
(b)SinceF0.s/$f0.t/,G.s/$g.t/, and
1
p.s/
$w.t/, the convolution theorem implies that
F0.s/
p.s/
$
Z
t
0
w.t/f0./ dand
G.s/
p.s/
$
Z
t
0
w.t/g./ d:
Now Theorem 8.4.2 implies that
e
st1G.s/
p.s/
$u.tt1/
Z
t
0
w.tt1/g./ d, and (B) implies that
y.t/D
Z
t
0
w.t/f0./ dCu.tt1/
Z
tt1
0
w.tt1/g./ d:
(c)Let´0.t/D
R
t
0
w.t/f0./ dand´1.t/D
R
t
0
w.t/g./ d. Theny.t/D´0.t/Cu.t
t1/´1.tt1/. Using Leibniz’s rule as in the solution of Exercise 8.6.11(b)shows that
´
0
0
.t/D
Z
t
0
w
0
.t/f0./ d; ´
0
1
.t/D
Z
t
0
w
0
.t/g./ d; t > 0;
´
00
0
.t/D
f0.t/
a
C
Z
t
0
w
00
.t/f0./ d; ´
00
1
.t/D
g.t/
a
C
Z
t
0
w
00
.t/g./ d; t > 0;
ift > 0, and that
a´
00
0
Cb´
0
0
Cc´0Df0.t/anda´
00
1
Cb´
0
1
Cc´1Df1.tCt1/f0.tCt1/; t > 0:

160. Section 8.7Constant Coefficient Equations with Impulses155
This implies the stated conclusion fory
0
andy
00
on.0; t/and.t;1/, and thatay
00
Cby
0
CcyDf .t/
on these intervals.
(d)Since the functions´0.t/andh.t/Du.tt1/´1.tt1/are both continuous onŒ0;1/andh.t/D0
if0tt1,yis continuous onŒ0;1/. From(c),y
0
is continuous onŒ0; t1/and.t1;1/, so we need
only show thaty
0
is continuous att1. For this it suffices to show thath
0
.t1/D0. Sinceh.t1/D0if
tt1, (B) lim
t!t1
h.t/h.t1/
tt1
D0. Ift > t1, thenh.t/D
Z
tt1
0
w.tt1/g./ d. Sinceh.t1/D0,
ˇ
ˇ
ˇ
ˇ
h.t/h.t1/
tt1
ˇ
ˇ
ˇ
ˇ

Z
tt1
0
jw.tt1/g./jd: . B/
Sincegis continuous from the right at0, we can choose constantsT > 0andM > 0so thatjg./j< M
if0T. Then (B) implies that
ˇ
ˇ
ˇ
ˇ
h.t/h.t1/
tt1
ˇ
ˇ
ˇ
ˇ
M
Z
tt1
0
jw.tt1/jd; t1< t < t1CT: . C/
Now suppose > 0. Sincew.0/D0, we can chooseT1such that0 < T1< Tandjw.x/j< =Mif
0x < T1. Ift1< t < t1CT1and0tt1, then0tt1 < T1, so (C) implies that
ˇ
ˇ
ˇ
ˇ
h.t/h.t1/
tt1
ˇ
ˇ
ˇ
ˇ
< ; t1< t < t1CT:
Therefore, lim
t!t1C
h.t/h.t1/
tt1
D0. This and (B) imply thath
0
.t1/D0.
8.7CONSTANT COEFFICIENT EQUATIONS WITH IMPULSES
8.7.2..s
2
Cs2/OY .s/D
10
sC1
C.9C7s/C7;OY .s/D
10C.sC1/.7s2/
.s1/.sC2/.sC1/
D
2
sC2
C
5
sC1
;
OyD2e
2t
C5e
t
;
1
p.s/
D
1
.sC2/.s1/
D
1
3

1
s1

1
sC2

;wDL
1

1
p.s/

D
e
t
e
2t
3
;
yD2e
2t
C5e
t
C
5
3
u.t1/

e
.t1/
e
2.t1/

.
8.7.4..s
2
C1/OY .s/D
3
s
2
C9
1Cs;
OY .s/D
3
.s
2
C1/.s
2
C9/
C
s1
s
2
C1
D
3
8

1
s
2
C1

1
s
2
C9

C
s1
s
2
C1
D
1
8

8s5
s
2
C1

3
s
2
C9

;
OyD
1
8
.8cost5sintsin3t/;
1
p.s/
D
1
s
2
C1
;wDL
1

1
p.s/

Dsint;yD
1
8
.8cost5sint
sin3t/2u.t=2/cost.
8.7.6..s
2
1/OY .s/D
8
s
C1s;OY .s/D
8Cs.1s/
s.s1/.sC1/
D
4
s1
C
3
sC1

8
s
;OyD4e
t
C3e
t
8;
1
p.s/
D
1
.s1/.sC1/
D
1
2
1
s1

1
sC1
;wDL
1

1
p.s/

D
e
t
Ce
t
2
Dsinht;yD4e
t
C
3e
t
8C2u.t2/sinh.t2/;
8.7.8..s
2
C4/OY .s/D
8
s2
C8s; (A)OY .s/D
8
.s2/.s
2
C4/
C
8s
s
2
C4
;
8
.s2/.s
2
C4/
D
A
s2
C
BsCC
s
2
C4
whereA.s
2
C4/C.BsCC /.s2/D8. SettingsD2yieldsAD1; settingsD0

161. 156 Chapter 8Laplace Transforms
yields4A2CD8, soCD 2;ACBD0(coefficient ofx
2
), soBD AD 1; therefore
8
.s2/.s
2
C4/
D
1
s2

sC2
s
2
C4
, so (A) implies thatOyDe
2t
C7cos2tsin2t;
1
p.s/
D
1
s
2
C4
;wD
L
1

1
p.s/

D
1
2
sin2t. Since sin.2t/D sin2t,yDe
2t
C7cos2tsin2t
1
2
u.t=2/sin2t.
8.7.10..s
2
C2sC1/OY .s/D
1
s1
C.2s/2;OY .s/D
1s.s1/
.s1/.sC1/
2
D
A
s1
C
B
sC1
C
C
.sC1/
2
whereA.sC1/
2
C.B.sC1/CC /.s1/D1s.s1/. SettingsD1yieldsAD1=4; settingsD 1
yieldsCD1=2; sinceACBD 1(coefficient ofs
2
),BD 1AD 5=4. Therefore,OY .s/D
1
4
1
s1

5
4
1
sC1
C
1
2
1
.sC1/
2
;OyD
1
4
e
t
C
1
4
e
t
.2t5/;
1
p.s/
D
1
.sC1/
2
;wDL
1

1
p.s/

Dte
t
;
yD
1
4
e
t
C
1
4
e
t
.2t5/C2u.t2/.t2/e
.t2/
.
8.7.12..s
2
C2sC2/OY .s/D.2s/2;Y.s/D
.sC1/C1
.sC1/
2
C1
;OyDe
t
.sintcost/;
1
p.s/
D
1
.sC1/
2
C1
;wDL
1

1
p.s/

De
t
sint. Since sin.t/D sintand sin.t2/Dsint,
yDe
t
.sintcost/e
.t/
u.t/sint3u.t2/e
.t2/
sint.
8.7.14..2s
2
3s2/OY .s/D
1
s
C2.2s/C3;OY .s/D
1Cs.72s/
2s.sC1=2/.s2/
D
7
10
1
s2

6
5
1
sC1=2

1
2s
;
OyD
7
10
e
2t

6
5
e
t =2

1
2
;
1
p.s/
D
1
2.sC1=2/.s2/
D
1
5

1
s2

1
sC1=2

;wDL
1

1
p.s/

D
1
5
.e
2t
e
t =2
/;yD
7
10
e
2t

6
5
e
t =2

1
2
C
1
5
u.t2/

e
2.t2/
e
.t2/=2

;
8.7.16..s
2
C1/OY .s/D
s
s
2
C4
1;OY .s/D
s
.s
2
C1/.s
2
C4/

1
s
2
C1
D
1
3

s
s
2
C1

s
s
2
C4


1
s
2
C1
;OyD
1
3
.costcos2t3sint/;
1
p.s/
D
1
s
2
C1
;wDL
1

1
p.s/

Dsint. Since sin.t
=2/D costand sin.t/D sint,
yD
1
3
.costcos2t3sint/2u.t=2/costC3u.t/sint:
8.7.18..s
2
C2sC1/OY .s/D
1
s1
1; (A)OY .s/D
1
.s1/.sC1/
2

1
.sC1/
2
;
1
.s1/.sC1/
2
D
A
s1
C
B
sC1
C
C
.sC1/
2
whereA.sC1/
2
C.B.sC1/CC /.s1/D1. SettingsD1yieldsAD1=4; settingsD 1yields
CD 1=2; sinceACBD0(coefficient ofs
2
),BD AD 1=4. Therefore,
1
.s1/.sC1/
2
D
1
4
1
s1

1
4
1
sC1

1
2
1
.sC1/
2
:
This and (A) imply that
OY .s/D
1
4
1
s1

1
4
1
sC1

3
2
1
.sC1/
2
I

162. Section 8.7Constant Coefficient Equations with Impulses157
OyD
1
4

e
t
e
t
.1C6t/

;
1
p.s/
D
1
.sC1/
2
;wDL
1

1
p.s/

Dte
t
;
yD
1
4

e
t
e
t
.1C6t/

u.t1/.t1/e
.t1/
C2u.t2/.t2/e
.t2/
.
8.7.20.y
00
C4yD12u.t=2/Cı.t/3ı.t3=2/; y.0/D1; y
0
.0/D 1..s
2
C4/OY .s/D
12e
s=2
s
Cs1;OY .s/D
12e
s=2
s.s
2
C4/
C
s1
s
2
C4
. Since
1
s.s
2
C4/
D
1
4

1
s

s
s
2
C4

,OY .s/D
1
4s
C
3
4
s
s
2
C4

1
s
2
C4
2e
s=2

1
s

s
s
2
C4

.OyD
3
4
cos2t
1
2
sin2tC
1
4
C
1
4
u.t=2/.1Ccos2t/.
wDL
1

1
p.s/

D
1
2
sin2t. Since sin2.t/Dsin2tand sin2.t3=2/D sin2t,
yD
3
4
cos2t
1
2
sin2tC
1
4
C
1
4
u.t=2/.1Ccos2t/C
1
2
u.t/sin2tC
3
2
u.t3=2/sin2t.
8.7.26.w.t/De
t
sint;fh.t/D
u.tt0/u.tt0h/
h
;.s
2
C2sC2/Yh.s/D
1
h
e
st0e
s.t0Ch/
s
;
Yh.s/D
1
h
e
st0e
s.t0Ch/
s.s
2
C2sC2/
;
1
s.s
2
C2sC2/
D
.sC1/C1
2 ..sC1/
2
C1/
C
1
2s
$
1
2

1e
t
.costCsint/

;
yh.t/D
8
ˆ
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
ˆ
:
0; 0 t < t0;
1
2h
h
1e
.tt0/
.cos.tt0/Csin.tt0//
i
; t 0t < t0Ch;
e
.tt0/
2h
h
e
h
.cos.tt0h/Csin.tt0h//cos.tt0/sin.tt0/
i
; tt0Ch:
8.7.28.w.t/De
t
e
2t
;fh.t/D
u.tt0/u.tt0h/
h
;.s
2
C3sC2/Yh.s/D
1
h
e
st0e
s.t0Ch/
s
;
Yh.s/D
1
h
e
st0e
s.t0Ch/
s.sC1/.sC2/
;
1
s.sC1/.sC2/
D
1
2.sC2/

1
sC1
C
1
2s
$
e
2t
2
e
t
C
1
2
D
.e
t
1/
2
2
;
yh.t/D
8
ˆ
ˆ
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
ˆ
ˆ
:
0; 0 t < t0;
.e
.tt0/
1/
2
2h
; t 0t < t0Ch;
.e
.tt0/
1/
2
.e
.tt0h/
1/
2
2h
; tt0Ch:
8.7.30.(a).s
2
1/OY .s/D1, soOyDwDL
1

1
s
2
1

D
1
2

e
t
e
t

;yD OyC
1
X
kD0
u.t
k/w.tk/D
1
2
1
X
kD0
u.tk/

e
tk
e
tk

. Ifmt < mC1, thenyD
1
2
m
X
kD0

e
tk
e
tk

D
1
2
.e
tm
e
t
/
m
X
kD0
e
k
e
mC1
1
2.e1/
.e
tm
e
t
/.
(b).s
2
C1/OY .s/D1, soOyDwDL
1

1
s
2
C1

Dsint;yD OyC
1
X
kD0
u.t2k/w.t2k/D

163. 158 Chapter 8Laplace Transforms
sint
1
X
kD0
u.t2k/. If2mt < 2.mC1/, thenyD.mC1/sint.
(c).s
2
3sC2/OY .s/D1, soOyDwDL
1

1
.s1/.s2/

D.e
2t
e
t
/;yD OyC
1
X
kD0
u.t
k/w.tk/D
1
X
kD0
u.tk/

e
2.tk/
e
tk

. Ifmt < mC1, thenyD
m
X
kD0

e
2.tk/
e
tk

D
e
2.tm/
m
X
kD0
e
2k
e
tm
m
X
kD0
e
k
De
2.tm/
e
2mC2
1
e
2
1
e
.tm/
e
mC1
1
e1
.
(d)wDL
1

1
s
2
C1

Dsint;yD
1
X
kD1
u.tk/w.tk/Dsint
1
X
kD1
.1/
k
u.tk/, so
yD
(
0; 2m t < .2mC1/;
sint; .2mC1/t < .2mC2/;
.mD0; 1; : : : /.

164. CHAPTER9
LinearHigherOrderEquations
9.1INTRODUCTION TO LINEAR HIGHER ORDER EQUATIONS
9.1.2.From Example 9.1.1,yDc1x
2
Cc2x
3
C
c3
x
y
0
D2c1xC3c2x
2

c3
x
2
, andy
00
D2c1C6c2xC
2c3
x
3
,
where
c1c2c3D 4
2c1C3c2c3D 14
2c16c22c3D 20;
soc1D2,c2D 3,c3D1, andyD2x
2
3x
3
C
1
x
.
9.1.4.The general solution ofy
.n/
D0can be written asy.x/D
n1
X
mD0
cm.xx0/
m
. Sincey
.j /
.x/D
n1
X
mDj
m.m1/ .mjC1/cm.xx0/
mj
,y
.j /
.x0/Dj Šcj. Therefore,yiD
.xx0/
i1
.i1/Š
; 1in.
9.1.6.We omit the verification that the given functions are solutions of the given equations.
(a)The equation is normal on.1;1/.W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
xe
x
e
x
e
x
e
x
.1x/
e
x
e
x
e
x
.x2/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
;W.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 0
11 1
1 1 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
4. Apply Theorem 9.1.4.
(b)The equation is normal on.1;1/.
W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
cos2x e
x
sin2x
e
x
e
x
.cos2x2sin2x/ e
x
.2cos2xCsin2x/
e
x
e
x
.3cos2xC4sin2x/ e
x
.4cos2x3sin2x/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
;
W.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 0
1 1 2
13 4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
8. Apply Theorem 9.1.4.
(c)The equation is normal on.1; 0/and.0;1/.
W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x
e
x
e
x
1
e
x
e
x
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2x. Apply Theorem 9.1.4.
(d)The equation is normal on.1; 0/and.0;1/.
W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
=x e
x
=x 1
e
x
.1=x1=x
2
/ e
x
.xC1/=x
2
0
e
x
.1=x2=x
2
C2=x
3
/ e
x
.x
2
C2xC2/=x
3
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2=x
2
. Apply Theorem 9.1.4.
159

165. 160 Chapter 9Linear Higher Order Equations
(e)The equation is normal on.1;1/.W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
e
x
1 2x e
x
0 2 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
;W.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 0 1
1 0 1
0 2 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2; Apply
Theorem 9.1.4.
(f)The equation is normal on.1; 1=2/and.1=2;1/.
W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x e
x
e
x
e
2x
1 e
x
e
x
2e
2x
0 e
x
e
x
4e
.
2x/
0 e
x
e
x
8e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
.12x6/. Apply Theorem 9.1.4.
(g)The equation is normal on.1; 0/and.0;1/.
W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x
2
e
2x
e
2x
0 2x 2e
2x
2e
2x
0 2 4e
2x
4e
2x
0 0 8e
2x
8e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 128x. Apply Theorem 9.1.4.
9.1.8.From Abel’s formula, (A)W.x/DW.=2/exp


Z
x
=4
tant dt

;
Z
x
=4
tant dtD ln cost
ˇ
ˇ
ˇ
ˇ
x
=4
D
ln.
p
2cosx/; therefore (A) implies thatW.x/D
p
2Kcosx.
9.1.10.(a)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 e
x
e
x
0 e
x
e
x
0 e
x
e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D.e
x
/.e
x
/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1
0 11
0 1 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2.
(b)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
sinx e
x
cosx
e
x
e
x
.cosxCsinx/ e
x
.cosxsinx/
e
x
2e
x
cosx 2e
x
sinx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
De
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 sinx cosx
1cosxCsinxcosxsinx
1 2cosx 2sinx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 sinx cosx
0 cosx sinx
1 2cosxsinx2sinxcosx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1sinxcosx
0cosxsinx
0sinxcosx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
3x
(c)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2 xC1 x
2
C2
0 1 2x
0 0 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D4.
(d)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x xlnjxj 1=x
1lnjxj C11=x
2
0 1=x 2=x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj 1=x
2
1lnjxj C11=x
2
0 1 2=x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
1
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj 1
1lnjxj C11
0 1 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
1
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj1
0 1 2
0 1 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
1
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj1
0 1 2
0 0 4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D4=x
2
:

166. Section 9.1Introduction to Linear Higher Order Equations161
(e)W.x/D
2
6
6
6
6
6
6
6
4
1 x x
2
=2 x
3
=3 x
n
=nŠ
0 1 x x
2
=2 x
n1
=.n1/Š
0 0 1 x x
n2
=.n2/Š
0 0 0 1 x
n3
=.n3/Š
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
0 0 0 0 1
3
7
7
7
7
7
7
7
5
D1.
(f)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x
e
x
e
x
1
e
x
e
x
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x
11 1
1 1 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x
02 1x
0 0 x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2x.
(g)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
=x e
x
=x 1
e
x
=xe
x
=x
2
e
x
=xe
x
=x
2
0
e
x
=x2e
x
=x
2
C2e
x
=x
3
e
x
=xC2e
x
=x
2
C2e
x
=x
3
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1=x 1=x 1
1=x1=x
2
1=x1=x
2
0
1=x2=x
2
C2=x
3
1=xC2=x
2
C2=x
3
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
1=x1=x
2
1=x1=x
2
1=x2=x
2
C2=x
3
1=xC2=x
2
C2=x
3
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
1=x1=x
2
2=x
1=x2=x
2
C2=x
3
4=x
2
ˇ
ˇ
ˇ
ˇ
D2=x
2
:
(h)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
e
x
1 2x e
x
0 2 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1
1 2x 1
0 2 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
x

x

2x 1
2 1



x
2
1
2 1

De
x
.x
2
2xC2/.
(i)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
3
1=x 1=x
2
1 3x
2
1=x
2
2=x
3
0 6x 2=x
3
6=x
4
0 6 6=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
02x
3
2=x 3=x
2
1 3x
2
1=x
2
2=x
3
0 6x 2=x
3
6=x
4
0 6 6=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2x
3
2=x 3=x
2
6x 2=x
3
6=x
4
66=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2 2=x
4
3=x
5
6 2=x
4
6=x
5
66=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2 2=x
4
3=x
5
0 8=x
4
15=x
5
0 0 15=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 240=x
5
:

167. 162 Chapter 9Linear Higher Order Equations
(j)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x e
2x
e
x
e
x
1 2e
2x
e
x
e
x
0 4e
2x
e
x
e
x
0 8e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x 1
11 1 2
1 1 0 4
11 0 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1x 1Cx 0 12x
1 1 1 2
1 1 0 4
1 1 0 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1x 1Cx 12x
1 1 4
1 1 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2 1Cx 12x
2 1 4
01 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2 1Cx 12x
0x 3C2x
01 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D6e
2x
.2x1/:
(k)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
2x
e
2x
1 x
2
2e
2x
2e
2x
0 2x
4e
2x
4e
2x
0 2
8e
2x
8e
2x
0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 x
2
22 0 2x
4 4 0 2
88 0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
22 2x
4 4 2
88 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
02 2x
8 4 2
08 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 128x:
9.1.12.Letybe an arbitrary solution ofLyD0on.a; b/. Sincef´1; : : : ; ´ngis a fundamental set of
solutions ofLyD0on.a; b/, there are constantsc1,c2,. . . ,cnsuch thatyD
n
X
iD1
ciyi. Therefore,
yD
n
X
iD1
ci
n
X
jD1
aijyjD
n
X
jD1
Cjyj, withCjD
n
X
iD1
aijci. Hencefy1; : : : ; yngis a fundamental set of
solutions ofLyD0on.a; b/.
9.1.14.Letybe a given solution ofLyD0and´D
n
X
jD1
y
.j1/
.x0/yjThen´
.r /
.x0/Dy
.r /
.x0/,
rD0; : : : ; n1. Since the solution of every initial value problem is unique(Theorem 9.1.1),´Dy.
9.1.16.Iffy1; y2; : : : ; yngis linearly dependent on.a; b/there are constantsc1; : : : ; cn, not all zeros,
such thatc1y1Cc2y2C Ccnyn. Letkbe the smallest integer such thatck¤0. IfkD1, then
y1D
1
c1
.c2y2C Ccnyn/; if1 < k < n, thenykD0y1C C0yk1C
1
ck
.ckC1ykC1C Ccnyn/;
ifkDn, thenynD0, soynD0y1C0y2C C0yn.
9.1.18.SinceFD
P
˙f1i1
f2i2
; : : : ; fnin,
F
0
D
X
˙f
0
1i1
f2i2
; : : : ; fninC
X
˙f1i1
f
0
2i2
; : : : ; fninC C
X
˙f1i1
f2i2
; : : : ; f
0
nin
DF1CF2C CFn:

168. Section 9.1Introduction to Linear Higher Order Equations163
9.1.20.Sincey
.n/
j
D
n
X
kD1
.Pk=P0/y
.nk/
j
, Exercise 9.1.19 implies that
W
0
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
y1 y2 yn
y
0
1
y
0
2
y
0
n
:
:
:
:
:
:
:
:
:
:
:
:
y
.n2/
1
y
.n2/
2
y
.n2/
n
P
n
kD1
.Pk=P0/y
.nk/
1
P
n
kD1
.Pk=P0/y
.nk/
2

P
n
kD1
.Pk=P0/y
.nk/
n
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
;
so Exercise 9.1.17 implies that
W
0
D
n
X
kD1
Pk
P0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
y1 y2 yn
y
0
1
y
0
2
y
0
n
:
:
:
:
:
:
:
:
:
:
:
:
y
.n2/
1
y
.n2/
2
y
.n2/
n
y
.nk/
1
y
.nk/
2
y
.nk/
n
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
:
However, the determinants on the right each have two identical rows ifkD2; : : : ; n. Therefore,W
0
D

P1W
P0
. Separating variables yields
W
0
W
D
P1
P0
; hence ln
W.x/
W.x0/
D
Z
x
x0
P1.t/
P0.t/
dt, which implies
Abel’s formula.
9.1.22.See the proof of Theorem 5.3.3.
9.1.24.(a)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1 x
2
C1
1 2x 2x
0 2 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1 x
2
C1
1 0 0
0 2 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
x
2
1 x
2
C1
2 2
ˇ
ˇ
ˇ
ˇ
D 4I
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1 x
2
C1
1 2x 2x
0 0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D0;P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1 x
2
C1
0 2 2
0 0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D0;P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 2x 2x
0 2 2
0 0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
0. Therefore,4y
000
D0, which is equivalent toy
000
D0.
(b)
P0D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x
e
x
e
x
1
e
x
e
x
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x
11 1
1 1 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 0 x
02 1
1 1 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2xI
P1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x
e
x
e
x
1
e
x
e
x
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x
11 1
11 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 2 x
0 0 1
11 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2I
P2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x
e
x
e
x
0
e
x
e
x
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x
1 1 0
11 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 0 x
1 1 0
02 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2xI

169. 164 Chapter 9Linear Higher Order Equations
P3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
1
e
x
e
x
0
e
x
e
x
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
11 1
1 1 0
11 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 0 1
1 2 0
11 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2:
Therefore,2xy
000
C2y
00
C2xy
0
2yD0, which is equivalent toxy
000
y
00
xy
0
CyD0.
(c)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
xe
x
1
e
x
e
x
.1x/ 0
e
x
e
x
.x2/ 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1
1 1x 0
1 x2 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
1 1x
1 x2
ˇ
ˇ
ˇ
ˇ
D32xI
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
xe
x
1
e
x
e
x
.1x/ 0
e
x
e
x
.3x/ 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1
1 1x 0
1 3x 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
1 1x
1 3x
ˇ
ˇ
ˇ
ˇ
D2I
P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
xe
x
1
e
x
e
x
.x2/ 0
e
x
e
x
.3x/ 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1
1 x2 0
1 3x 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
1 x2
1 3x
ˇ
ˇ
ˇ
ˇ
D2x5I
P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
.1x/ 0
e
x
e
x
.x2/ 0
e
x
e
x
.3x/ 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D0:
Therefore,.32x/y
000
C2y
00
C.2x5/y
0
D0.
(d)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
e
x
1 2x e
x
0 2 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1
1 2x 1
0 2 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
2 0
1 2x2 0
0 2 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
x
ˇ
ˇ
ˇ
ˇ
x x
2
2
1 2x2
ˇ
ˇ
ˇ
ˇ
D e
x
.x
2
2xC2/I
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
e
x
1 2x e
x
0 0 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
x
ˇ
ˇ
ˇ
ˇ
x x
2
1 2x
ˇ
ˇ
ˇ
ˇ
Dx
2
e
x
I
P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
e
x
0 2 e
x
0 0 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
x
ˇ
ˇ
ˇ
ˇ
x x
2
0 2
ˇ
ˇ
ˇ
ˇ
D 2xe
x
I
P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 2x e
x
0 2 e
x
0 0 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
x
ˇ
ˇ
ˇ
ˇ
1 2x
0 2
ˇ
ˇ
ˇ
ˇ
D2e
x
:
Therefore,e
x
.x
2
2xC2/y
000
Cx
2
e
x
y
00
2xe
x
y
0
C2e
x
yD0; which is equivalent to.x
2
2xC
2/y
000
x
2
y
00
C2xy
0
2yD0.
(e)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1=x
1 2x1=x
2
0 2 2=x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1=x
2
1 2x1=x
2
0 2 2=x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1=x
2
0 x2=x
2
0 2 2=x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
x2=x
2
2 2=x
3
ˇ
ˇ
ˇ
ˇ
D
6
x
I

170. Section 9.1Introduction to Linear Higher Order Equations165
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1=x
1 2x1=x
2
0 0 6=x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6
x
4
ˇ
ˇ
ˇ
ˇ
x x
2
1 2x
ˇ
ˇ
ˇ
ˇ
D
6
x
2
I
P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1=x
0 2 2=x
3
0 0 6=x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6
x
4
ˇ
ˇ
ˇ
ˇ
x x
2
0 2
ˇ
ˇ
ˇ
ˇ
D
12
x
3
I
P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 2x1=x
2
0 2 2=x
3
0 0 6=x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6
x
4
ˇ
ˇ
ˇ
ˇ
1 2x
0 2
ˇ
ˇ
ˇ
ˇ
D
12
x
4
:
Therefore,
6
x
y
000

6
x
2
y
00
C
12
x
3
y
0

12
x
4
yD0, which is equivalent tox
3
y
000
Cx
2
y
00
2xy
0
C2yD0.
(f)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 e
x
e
3x
1 e
x
3e
3x
0 e
x
9e
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
4x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 1 1
1 1 3
0 1 9
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
4x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 08
1 0 6
0 1 9
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
4x
ˇ
ˇ
ˇ
ˇ
xC18
16
ˇ
ˇ
ˇ
ˇ
D2e
4x
.13x/I
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 e
x
e
3x
1 e
x
3e
3x
0 e
x
27e
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
4x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 1 1
1 1 3
0 1 27
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
4x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 026
1 0 24
0 1 27
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
4x
ˇ
ˇ
ˇ
ˇ
xC126
124
ˇ
ˇ
ˇ
ˇ
D2e
4x
.12x1/I
P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 e
x
e
3x
0 e
x
9e
3x
0 e
x
27e
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
4x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 1 1
0 1 9
0 1 27
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
4x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 1 1
0 1 9
0 0 18
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 18e
4x
.xC1/I
P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 e
x
3e
3x
0 e
x
9e
3x
0 e
x
27e
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
4x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 3
0 1 9
0 1 27
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
4x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 3
0 1 9
0 0 18
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D18e
4x
:
Therefore,
2e
4x
.13x/y
000
C2e
4x
.12x1/y
00
18e
4x
.xC1/y
0
C18e
4x
yD0;
which is equivalent to
.3x1/y
000
.12x1/y
00
C9.xC1/y
0
9yD0:

171. 166 Chapter 9Linear Higher Order Equations
(g)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
3
1=x 1=x
2
1 3x
2
1=x
2
2=x
3
0 6x 2=x
3
6=x
4
0 6 6=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x
2
1=x
2
1=x
3
1 3x
2
1=x
2
2=x
3
0 6x 2=x
3
6=x
4
0 6 6=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x
2
1=x
2
1=x
3
0 2x
2
2=x
2
3=x
3
0 6x 2=x
3
6=x
4
0 6 6=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2x
2
2=x
2
3=x
3
6x 2=x
3
6=x
4
66=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2x2=x
3
3=x
4
6x 2=x
3
6=x
4
66=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2x2=x
3
3=x
4
0 8=x
3
15=x
4
0 0 15=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
240
x
5
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
3
1=x 1=x
2
1 3x
2
1=x
2
2=x
3
0 6x 2=x
3
6=x
4
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x
2
1=x
2
1=x
3
1 3x
2
1=x
2
2=x
3
0 6x 2=x
3
6=x
4
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x
2
1=x
2
1=x
3
0 2x
2
2=x
2
3=x
3
0 6x 2=x
3
6=x
4
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2x2=x
3
3=x
4
6x 2=x
3
6=x
4
0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2x2=x
3
3=x
4
0 8=x
3
15=x
4
0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2x
3
ˇ
ˇ
ˇ
ˇ
8=x
3
15=x
4
24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
D
1200
x
6
I
P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
3
1=x 1=x
2
1 3x
2
1=x
2
2=x
3
0 6 6=x
4
24=x
5
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x
2
1=x
2
1=x
3
1 3x
2
1=x
2
2=x
3
0 6 6=x
4
24=x
5
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x
2
1=x
2
1=x
3
0 2x
2
2=x
2
3=x
3
0 6 6=x
4
24=x
5
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
22=x
4
3=x
5
66=x
4
24=x
5
0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
22=x
4
3=x
5
0 0 15=x
5
0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2x
3
ˇ
ˇ
ˇ
ˇ
0 15=x
5
24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
D
720
x
7
I
P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
3
1=x 1=x
2
0 6x 2=x
3
6=x
4
0 6 6=x
4
24=x
5
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
6 2=x
4
6=x
5
66=x
4
24=x
5
0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
6 2=x
4
6=x
5
08=x
4
30=x
5
0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 6x
2
ˇ
ˇ
ˇ
ˇ
8=x
4
30=x
5
24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
D
1440
x
6
I

172. Section 9.1Introduction to Linear Higher Order Equations167
P4.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 3x
2
1=x
2
2=x
3
0 6x 2=x
3
6=x
4
0 6 6=x
4
24=x
5
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
6 2=x
4
6=x
5
66=x
4
24=x
5
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
6 2=x
4
6=x
5
08=x
4
30=x
5
0 0 24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D6x
ˇ
ˇ
ˇ
ˇ
8=x
4
30=x
5
24=x
5
120=x
6
ˇ
ˇ
ˇ
ˇ
D
1440
x
9
:
Therefore,

240
x
5
y
.4/

1200
x
6
y
000
C
720
x
7
y
00
C
1440
x
8
y
0

1440
x
9
yD0;
which is equivalent tox
4
y
.4/
C5x
3
y
000
3x
2
y
00
6xy
0
C6yD0.
(h)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x xlnjxj1=x x
2
1lnjxj C11=x
2
2x
0 1=x 2=x
3
2
01=x
2
6=x
4
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj 1=x
2
x
1lnjxj C11=x
2
2x
0 1=x 2=x
3
2
01=x
2
6=x
4
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj1=x
2
x
0 1 2=x
2
x
0 1=x 2=x
3
2
01=x
2
6=x
4
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 2=x
2
x
1=x 2=x
3
2
1=x
2
6=x
4
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 2=x
2
x
1=x 2=x
3
2
1=x
2
6=x
4
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
12=x
2
x
0 4=x
3
1
08=x
4
1=x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
4=x
3
1
8=x
4
1=x
ˇ
ˇ
ˇ
ˇ
D
12
x
3
I
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x xlnjxj1=x x
2
1lnjxj C11=x
2
2x
0 1=x 2=x
3
2
0 2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj 1=x
2
x
1lnjxj C11=x
2
2x
0 1=x 2=x
3
2
0 2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj1=x
2
x
0 1 2=x
2
x
0 1=x 2=x
3
2
0 2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
12=x
2
x
1=x 2=x
3
2
2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
12=x
2
x
0 4=x
3
1
0 28=x
5
2=x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
4=x
3
1
28=x
5
2=x
2
ˇ
ˇ
ˇ
ˇ
D
36
x
4
I

173. 168 Chapter 9Linear Higher Order Equations
P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x xlnjxj1=x x
2
1lnjxj C11=x
2
2x
01=x
2
6=x
4
0
0 2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj 1=x
2
x
1lnjxj C11=x
2
2x
01=x
2
6=x
4
0
0 2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj1=x
2
x
0 1 2=x
2
x
01=x
2
6=x
4
0
0 2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 2=x
2
x
1=x
2
6=x
4
0
2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
12=x
2
x
08=x
4
1=x
0 28=x
5
2=x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
8=x
4
1=x
28=x
5
2=x
2
ˇ
ˇ
ˇ
ˇ
D
12
x
5
I
P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x xlnjxj1=x x
2
0 1=x 2=x
3
2
01=x
2
6=x
4
0
0 2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1=x 2=x
3
2
1=x
2
6=x
4
0
2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1=x 2=x
3
2
04=x
4
2=x
0 20=x
5
4=x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
4=x
4
2=x
20=x
5
4=x
2
ˇ
ˇ
ˇ
ˇ
D
24
x
6
I
P4.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1lnjxj C11=x
2
2x
0 1=x 2=x
3
2
01=x
2
6=x
4
0
0 2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1=x 2=x
3
2
1=x
2
6=x
4
0
2=x
3
24=x
5
0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2
ˇ
ˇ
ˇ
ˇ
1=x
2
6=x
4
2=x
3
24=x
5
ˇ
ˇ
ˇ
ˇ
D
24
x
7
:
Therefore,
12
x
3
y
.4/
C
36
x
4
y
000

12
x
5
y
00
C
24
x
6
y
0

24
x
7
yD0;
which is equivalent tox
4
y
.4/
C3x
2
y
000
x
2
y
00
C2xy
0
2yD0.
(i)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x e
2x
e
x
e
x
1 2e
2x
e
x
e
x
0 4e
2x
e
x
e
x
0 8e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x 1
11 1 2
1 1 0 4
11 0 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2 1 x 1
01 1 2
2 1 0 4
01 0 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2 1 x 1
01 1 2
0 0 x 3
01 0 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 2
0x 3
1 0 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 2
0x 3
01 6
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2e
2x
ˇ
ˇ
ˇ
ˇ
x 3
1 6
ˇ
ˇ
ˇ
ˇ
D6e
2x
.2x1/I

174. Section 9.1Introduction to Linear Higher Order Equations169
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x e
2x
e
x
e
x
1 2e
2x
e
x
e
x
0 4e
2x
e
x
e
x
0 16e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x 1
11 1 2
1 1 0 4
1 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 1 x 1
21 1 2
0 1 0 4
0 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1
1 0 4
1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1
0x 3
0x 15
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2e
2x
ˇ
ˇ
ˇ
ˇ
x 3
x 15
ˇ
ˇ
ˇ
ˇ
D 24xe
2x
I
P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x e
2x
e
x
e
x
1 2e
2x
e
x
e
x
0 8e
2x
e
x
e
x
0 16e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x 1
11 1 2
11 0 8
1 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 1 x 1
21 1 2
21 0 8
0 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 1 x 1
21 1 2
0 0 1 6
0 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1
01 6
1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1
01 6
0x 15
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2e
2x
ˇ
ˇ
ˇ
ˇ
1 6
x 15
ˇ
ˇ
ˇ
ˇ
D6e
2x
.52x/I
P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x e
2x
e
x
e
x
0 4e
2x
e
x
e
x
0 8e
2x
e
x
e
x
0 16e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 x 1
1 1 0 4
11 0 8
1 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
0 1 x 1
0 1 0 4
21 0 8
0 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x 1
0x 3
0x 15
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2e
2x
ˇ
ˇ
ˇ
ˇ
x 3
x 15
ˇ
ˇ
ˇ
ˇ
D24xe
2x
I
P4.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
1 2e
2x
e
x
e
x
0 4e
2x
e
x
e
x
0 8e
2x
e
x
e
x
0 16e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
11 1 2
1 1 0 4
11 0 8
1 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
01 1 2
2 1 0 4
01 0 8
2 1 0 16
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
01 1 2
2 1 0 4
01 0 8
0 0 0 12
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 2
1 0 8
0 0 12
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 2
01 6
0 0 12
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 24e
2x
:
Therefore,
6e
2x
.2x1/y
.4/
24xe
2x
y
000
C6e
2x
.52x/y
00
C24xe
2x
y
0
24e
2x
yD0;

175. 170 Chapter 9Linear Higher Order Equations
which is equivalent to.2x1/y
.4/
4xy
000
C.52x/y
00
C4xy
0
4yD0.
9.1.24.(j)
P0.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
2x
e
2x
1 x
2
2e
2x
2e
2x
0 2x
4e
2x
4e
2x
0 2
8e
2x
8e
2x
0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 x
2
22 0 2x
4 4 0 2
88 0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
22 2x
4 4 2
88 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
22 2x
0 8 24x
0 0 8x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 128x
P1.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
2x
e
2x
1 x
2
2e
2x
2e
2x
0 2x
4e
2x
4e
2x
0 2
16e
2x
16e
2x
0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 x
2
22 0 2x
4 4 0 2
16 16 0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
22 2x
4 4 2
16 16 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
22 2x
0 8 24x
0 0 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D128I
P2.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
2x
e
2x
1 x
2
2e
2x
2e
2x
0 2x
8e
2x
8e
2x
0 0
16e
2x
16e
2x
0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 x
2
22 0 2x
88 0 0
16 16 0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
22 2x
88 0
16 16 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2x
ˇ
ˇ
ˇ
ˇ
88
16 16
ˇ
ˇ
ˇ
ˇ
D512xI
P3.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
2x
e
2x
1 x
2
4e
2x
4e
2x
0 2
8e
2x
8e
2x
0 0
16e
2x
16e
2x
0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 x
2
4 4 0 2
88 0 0
16 16 0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
4 4 2
88 0
16 16 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2
ˇ
ˇ
ˇ
ˇ
88
16 16
ˇ
ˇ
ˇ
ˇ
D 512I
P4.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2e
2x
2e
2x
0 2x
4e
2x
4e
2x
0 2
8e
2x
8e
2x
0 0
16e
2x
16e
2x
0 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D0:
Therefore,128xy
.4/
C128y
000
C512xy
00
512yD0, which is equivalent toxy
.4/
y
000
4xy
00
C4y
0
D0.
9.2HIGHER ORDER CONSTANT COEFFICIENT HOMOGENEOUS EQUATIONS
9.2.2.p.r/Dr
4
C8r
2
9D.r1/.rC1/.r
2
C9/;yDc1e
x
Cc2e
x
Cc3cos3xCc4sin3x.
9.2.4.p.r/D2r
3
C3r
2
2r3D.r1/.rC1/.2rC3/;yDc1e
x
Cc2e
x
Cc3e
3x=2
.

176. Section 9.2Higher Order Constant Coefficient Homogeneous Equations171
9.2.6.p.r/D4r
3
8r
2
C5r1D.r1/.2r1/
2
;yDc1e
x
Ce
x=2
.c2Cc3x/.
9.2.8.p.r/Dr
4
Cr
2
Dr
2
.r
2
C1/;yDc1Cc2xCc3cosxCc4sinx.
9.2.10.p.r/Dr
4
C12r
2
C36D.r
2
C6/
2
;yD.c1Cc2x/cos
p
6xC.c3Cc4x/sin
p
6x.
9.2.12.p.r/D6r
4
C5r
3
C7r
2
C5rC1D.2rC1/.3rC1/.r
2
C1/;yDc1e
x=2
Cc2e
x=3
C
c3cosxCc4sinx.
9.2.14.p.r/Dr
4
4r
3
C7r
2
6rC2D.r1/
2
.r
2
2rC2/;yDe
x
.c1Cc2xCc3cosxCc4sinx/.
9.2.16.p.r/Dr
3
C3r
2
r3D.r1/.rC1/.rC3/;
yDc1e
x
Cc2e
x
Cc3e
3x
y
0
Dc1e
x
c2e
x
3c3e
3x
y
00
Dc1e
x
Cc2e
x
C9c3e
3x
I
c1Cc2Cc3D 0
c1c23c3D 14
c1Cc2C9c3D 40
I
c1D2,c2D3,c3D 5;yD2e
x
C3e
x
5e
3x
.
9.2.18.p.r/Dr
3
2r4D.r2/.r
2
C2rC2/;
yDe
x
.c1cosxCc2sinx/Cc3e
2x
y
0
D e
x
..c1c2/cosxC.c1Cc2/sinx/C2c3e
2x
y
00
De
x
.2c1sinx2c2cosx/C4c3e
2x
I
c1Cc3D6
c1Cc2C2c3D3
2c2C4c3D22
I
c1D2,c2D 3,c3D4;yD2e
x
cosx3e
x
sinxC4e
2x
.
9.2.20.p.r/Dr
3
6r
2
C12r8D.r2/
3
;
yDe
2x
.c1Cc2xCc3x
2
/
y
0
De
2x
.2c1Cc2C.2c2C2c3/xC2c3x
2
/
y
00
D2e
2x
.2c1C2c2Cc3C2.c2C2c3/xC2c3x
2
/
I
c1D 1
2c1Cc2D 1
4c1C4c2C2c3D 4
c1D1,c2D 3,c3D2;yDe
2x
.13xC2x
2
/.
9.2.22.p.r/D8r
3
4r
2
2rC1D.2rC1/.2r1/
2
;
yDe
x=2
.c1Cc2x/Cc3e
x=2
y
0
D
1
2
e
x=2
.c1C2c2Cc2x/
1
2
c3e
x=2
y
00
D
1
4
e
x=2
.c1C4c2Cc2x/C
1
4
c3e
x=2
I
c1Cc3D 4
1
2
c1Cc2
1
2
c3D 3
1
4
c1Cc2C
1
4
c3D 1
I
c1D1,c2D 2,c3D3;yDe
x=2
.12x/C3e
x=2
.
9.2.24.p.r/Dr
4
6r
3
C7r
2
C6r8D.r1/.r2/.r4/.rC1/;
yDc1e
x
Cc2e
2x
Cc3e
4x
Cc4e
x
y
0
Dc1e
x
C2c2e
2x
C4c3e
4x
c4e
x
y
00
Dc1e
x
C4c2e
2x
C16c3e
4x
Cc4e
x
y
000
Dc1e
x
C8c2e
2x
C64c3e
4x
c4e
x
I
c1Cc2Cc3Cc4D 2
c1C2c2C4c3c4D 8
c1C4c2C16c3Cc4D 14
c1C8c2C64c3c4D 62
I
c1D 4,c2D1,c3D 1,c4D2;yD 4e
x
Ce
2x
e
4x
C2e
x
.

177. 172 Chapter 9Linear Higher Order Equations
9.2.26.p.r/Dr
4
C2r
3
2r
2
8r8D.r2/.rC2/.r
2
C2rC2/;
yDc1e
2x
Cc2e
2x
Ce
x
.c3cosxCc4sinx/
y
0
D2c1e
2x
2c2e
2x
e
x
..c3c4/cosxC.c3Cc4/sinx/
y
00
D4c1e
2x
C4c2e
2x
Ce
x
.2c3sinx2c4cosx/
y
000
D8c1e
2x
8c2e
2x
Ce
x
..2c3C2c4/cosxC2.c4c3/sinx/
I
c1Cc2Cc3D 5
2c12c2c3Cc4D 2
4c1C4c22c4D 6
8c18c2C2c3C2c4D 8
I
c1D1,c2D1,c3D3,c4D1;yDe
2x
Ce
2x
Ce
x
.3cosxCsinx/.
9.2.28.(a)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
xe
x
e
2x
e
x
e
x
.xC1/ 2e
2x
e
x
e
x
.xC2/ 4e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
IW.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 1
1 1 2
1 2 4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 0
1 1 1
1 2 3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
1 1
2 3
ˇ
ˇ
ˇ
ˇ
D
1:
(b)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
cos2x sin2x e
3x
2sin2x 2cos2x 3e
3x
4cos2x4sin2x 9e
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
IW.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 1
0 2 3
4 0 9
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 0
0 2 3
4 0 13
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
2 3
0 13
ˇ
ˇ
ˇ
ˇ
D
26:
(c)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
cosx e
x
sinx e
x
e
x
.cosxCsinx/ e
x
.cosxsinx/ e
x
2e
x
sinx 2e
x
cosx e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
IW.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 1
1 1 1
02 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 1
0 1 2
02 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
1 2
2 1
ˇ
ˇ
ˇ
ˇ
D5:
(d)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 x x
2
e
x
0 1 2x e
x
0 0 2 e
x
0 0 0 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
IW.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 0 1
0 1 0 1
0 0 2 1
0 0 0 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D1:
(e)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
cosx sinx
e
x
e
x
sinxcosx
e
x
e
x
cosxsinx
e
x
e
x
sinxcosx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
I
W.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 0
11 0 1
1 1 1 0
11 0 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 0
11 0 1
1 1 1 0
0 0 0 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1
11 0
1 1 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1
11 0
0 0 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D4
ˇ
ˇ
ˇ
ˇ
1 1
11
ˇ
ˇ
ˇ
ˇ
D 8:
(f)W.x/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
cosx sinx e
x
cosx e
x
sinx
sinxcosx e
x
.cosxsinx/ e
x
.cosxCsinx/
cosxsinx 2e
x
sinx 2e
x
cosx
sinxcosxe
x
.2cosxC2sinx/ e
x
.2cosx2sinx/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
I

178. Section 9.2Higher Order Constant Coefficient Homogeneous Equations173
W.0/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 1 0
0 1 1 1
1 0 0 2
012 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 0 1 0
0 1 1 1
0 0 1 2
012 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1
0 1 2
12 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1
0 1 2
01 3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1
0 1 2
0 0 5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D5:
9.2.40.(a)SinceyDQ1.D/P1.D/yCQ2.D/P2.D/yandP1.D/yDP2.D/yD0, it follows that
yD0.
(b)Suppose that (A)a1u1C CarurCb1v1C CbsvsD0, wherea1; : : : ; arandb1; : : : ; bsare
constants. DenoteuDa1u1C CarurandvDb1v1C Cbsvs. Then (B)P1.D/uD0and (C)
P2.D/vD0. SinceuCvD0,P2.D/.uCv/D0. Therefore,0DP2.D/.uCv/DP2.D/uCP2.D/v.
Now (C) implies thatP2.D/uD0. This, (B), and(a)imply thatuDa1u1C CarurD0, so
a1D DarD0, sinceu1; : : : ; urare linearly independent. Now (A) reduces tob1v1C CbsvsD0,
sob1Dc DbsD0, sincev1; : : : ; vsare linearly independent. Therefore,u1; : : : ; ur; v1; : : : ; vrare
linearly independent.
(c)It suffices to show thatfy1; y2; : : : ; yngis linearly independent. Suppose thatc1y1C CcnynD0.
We may assume thaty1; : : : ; yrare linearly independent solutions ofp1.D/yD0andyrC1; : : : ; ynare
solutions ofP2.D/Dp2.D/ pk.D/yD0. Sincep1.r/andP2.r/have no common factors,(b)
implies that (A)c1y1C CcryrD0and (B)crC1yrC1C CcnynD0. Now (A) implies that
c1D DcrD0, sincey1; : : : ; yrare linearly independent. IfkD2, thenyrC1; : : : ; ynare linearly
independent, socrC1D DcnD0, and the proof is complete. Ifk > 2repeat this argument, starting
from (B), withp1replaced byp2, andP2replaced byP3Dp3 pn.
9.2.42.(a)
.cosACisinA/.cosBCisinB/D.cosAcosBsinAsinB/
C.cosAsinBCsinAcosB/
Dcos.ACB/Cisin.ACB/:
(b)Obvious fornD0. IfnD 1write
1
cosCisin
D
1
cosCisin
cosisin
cosisin
D
cosisin
cos
2
Csin
2

DcosisinDcos./Cisin./:
(d)Ifnis a negative integer, then (B).cosCisin/
n
D
1
.cosCisin/
jnj
. From the hint, (C)
1
.cosCisin/
jnj
D.cosisin/
jnj
D.cos./Cisin.//
jnj
. Replacingbyandnby
jnjin (A) shows that (D).cos./Cisin.//
jnj
Dcos.jnj/Cisin.jnj/. Sincejnj D n, (E)
cos.jnj/Cisin.jnj/DcosnCisinn. Now (B), (C), (D), and (E) imply (A).
(e)From (A),´
n
k
Dcos2kCisin2kD1and
n
k
Dcos.2kC1/Cisin.2kC1/Dcos.2kC
1/DcosD 1.
(f)From(e),
1=n
´0; : : : ;
1=n
´n1are all zeros of´
n
. Since they are distinct numbers,´
n
has
the stated factoriztion.

179. 174 Chapter 9Linear Higher Order Equations
From(e),
1=n
0; : : : ;
1=n
n1are all zeros of´
n
C. Since they are distinct numbers,´
n
Chas
the stated factoriztion.
9.2.43.(a)p.r/Dr
3
1D.r´0/.r´1/.r´2/where´kDcos
2k
3
Cisin
2k
3
,kD0; 1; 2.
Hence,´0D1,´1D
1
2
Ci
p
3
2
, and´2D
1
2
i
p
3
2
. Therefore,p.r/D.r1/
 

rC
1
2

2
C
3
4
!
,
so
(
e
x
; e
x=2
cos
 p
3
2
x
!
; e
x=2
sin
 p
3
2
x
!)
is a fundamental set of solutions.
(b)p.r/Dr
3
C1D.r0/.r1/.r2/wherekDcos
.2kC1/
3
Cisin
.2kC1/
3
,kD0; 1; 2.
Hence,0D
1
2
Ci
p
3
2
,1D 1 2D
1
2
i
p
3
2
. Therefore,p.r/D.rC1/
 

r
1
2

2
C
3
4
!
, so
(
e
x
; e
x=2
cos
 p
3
2
x
!
; e
x=2
sin
 p
3
2
x
!)
is a fundamental set of solutions.
(c)p.r/Dr
4
C64D.r2
p
20/.r2
p
21/.r2
p
22/.r2
p
23/, wherekDcos
.2kC1/
4
C
isin
.2kC1/
4
,kD0; 1; 2; 3. Therefore,0D
1Ci
p
2
,1D
1Ci
p
2
,2D
1i
p
2
, and3D
1i
p
2
,
sop.r/D..r2/
2
C4/..rC2/
2
C4/andfe
2x
cos2x; e
2x
sin2x; e
2x
cos2x; e
2x
sin2xgis a
fundamental set of solutions.
(d)p.r/Dr
6
1D.r´0/.r´1/.r´2/.r´3/.r´4/.r´5/where´kDcos
2k
6
Cisin
2k
6
,
kD0; 1; 2; 3; 4; 5. Therefore,´0D1,´1D
1
2
Ci
p
3
2
,´2D
1
2
Ci
p
3
2
,´3D 1,´4D
1
2
i
p
3
2
,
and´5D
1
2
i
p
3
2
, sop.r/D.r1/.rC1/
 

r
1
2

2
C
3
4
!  

rC
1
2

2
C
3
4
!
and
(
e
x
; e
x
; e
x=2
cos
 p
3
2
x
!
; e
x=2
sin
 p
3
2
x
!
; e
x=2
cos
 p
3
2
x
!
; e
x=2
sin
 p
3
2
x
!)
is a funda-
mental set of solutions.
(e)p.r/Dr
6
C64D.r20/.r21/.r22/.r23/.r24/.r25/wherekD
cos
.2kC1/
6
Cisin
.2kC1/
6
,kD0; 1; 2; 3; 4; 5. Therefore,0D
p
3
2
C
i
2
,1Di,2D
p
3
2
C
i
2
,
3D
p
3
2

i
2
,4D i, and5D
p
3
2

i
2
, sop.r/D.r
2
C4/..r
p
3/
2
C1/..rC
p
3/
2
C1/and
fcos2x;sin2x; e

p
3x
cosx; e

p
3x
sinx; e
p
3x
cosx; e
p
3x
sinxgis a fundamental set of solutions.
(f)p.r/D.r1/
6
1D.r1´0/.r1´1/.r1´2/.r1´3/.r1´4/.r1´5/where´kD
cos
2k
6
Cisin
2k
6
,kD0; 1; 2; 3; 4; 5. Therefore,´0D1,´1D
1
2
Ci
p
3
2
,´2D
1
2
Ci
p
3
2
,´3D
1,´4D
1
2
i
p
3
2
, and´5D
1
2
i
p
3
2
, sop.r/Dr.r2/
 

r
3
2

2
C
3
4
!  

r
1
2

2
C
3
4
!
and
(
1; e
2x
; e
3x=2
cos
 p
3
2
x
!
; e
3x=2
sin
 p
3
2
x
!
; e
x=2
cos
 p
3
2
x
!
; e
x=2
sin
 p
3
2
x
!)
is a funda-
mental set of solutions.
(g)p.r/Dr
5
Cr
4
Cr
3
Cr
2
CrC1D
r
6
1
r1
. Therefore, from the solution of(d)p.r/D

180. Section 9.3Undetermined Coefficients for Higher Order Equations175
.rC1/
 

r
1
2

2
C
3
4
!  

rC
1
2

2
C
3
4
!
and
(
e
x
; e
x=2
cos
 p
3
2
x
!
; e
x=2
sin
 p
3
2
x
!
; e
x=2
cos
 p
3
2
x
!
; e
x=2
sin
 p
3
2
x
!)
is a fundamen-
tal set of solutions.
9.3UNDETERMINED COEFFICIENTS FOR HIGHER ORDER EQUATIONS
9.3.2.IfyDu
3x
, theny
000
2y
00
5y
0
C6yDe
3x
Œ.u
000
11u
00
C34u
0
24u/2.u
00
6u
0
C
9u/5.u
0
3u/C6uDe
3x
.u
000
11u
00
C34u
0
24u/. LetupDACBxCCx
2
, where
.24AC34B22C /C.24BC68C /x24Cx
2
D3223xC6x
2
. ThenCD 1=4,BD1=4,
AD 3=4andypD
e
3x
4
.3xCx
2
/.
9.3.4.IfyDue
2x
, theny
000
C3y
00
y
0
3yDe
2x
Œ.u
000
6u
00
C12u
0
8u/C3.u
00
4u
0
C4u/
.u
0
2u/3uDe
2x
.u
000
3u
00
u
0
C3u/. LetupDACBxCCx
2
, where.3AB6C /C
.3B2C /xC3Cx
2
D217xC3x
2
. ThenCD1,BD 5,AD1, andypDe
2x
.15xCx
2
/.
9.3.6.IfyDue
x
, theny
000
Cy
00
2yDe
x
Œ.u
000
C3u
00
C3u
0
Cu/C.u
00
C2u
0
Cu/2uDe
x
.u
000
C4u
00
C
5u
0
/. LetupDx.ACBxCCx
2
/, where.5AC8BC6C /C.10BC24C /xC15Cx
2
D14C34xC15x
2
.
ThenCD1,BD1,AD0, andypDx
2
e
x
.1Cx/.
9.3.8.IfyDue
x
, theny
000
y
00
y
0
CyDe
x
Œ.u
000
C3u
00
C3u
0
Cu/.u
00
C2u
0
Cu/.u
0
Cu/CuD
e
x
.u
000
C2u
00
/. LetupDx
2
.ACBx/where.4AC6B/C12BxD7C6x. ThenBD1=2,AD1,
andypD
x
2
e
x
2
.2Cx/.
9.3.10.IfyDue
3x
, theny
000
5y
00
C3y
0
C9yDe
3x
Œ.u
000
C9u
00
C27u
0
C27u/5.u
00
C6u
0
C9u/C3.u
0
C
3u/C9uDe
3x
.u
000
C4u
00
/. LetupDx
2
.ACBxCCx
2
/, where.8AC6B/C.24BC24C /xC48Cx
2
D
2248x
2
. ThenCD 1,BD1,AD2, andypDx
2
e
3x
.2Cxx
2
/.
9.3.12.IfyDue
x=2
, then8y
000
12y
00
C6y
0
yDe
x=2
Œ8.u
000
C3u
00
=2C3u
0
=4Cu=8/12.u
00
Cu
0
C
u=4/C6.u
0
Cu=2/uD8e
x=2
u
000
, sou
000
D
1C4x
8
. Integrating three times and taking the constants
of integration to be zero yieldsupD
x
3
48
.1Cx/. Therefore,ypD
x
3
e
x=2
48
.1Cx/.
9.3.14.IfyDue
2x
, theny
.4/
C3y
000
Cy
00
3y
0
2yDe
2x
Œ.u
.4/
C8u
000
C24u
00
C32u
0
C16u/C3.u
000
C
6u
00
C12u
0
C8u/C.u
00
C4u
0
C4u/3.u
0
C2u/2uDe
2x
.u
.4/
C11u
000
C43u
00
C69u
0
C36u/. Let
upDACBxwhere.36AC69B/C36BxD 3336x. ThenBD 1,AD1, andypDe
2x
.1x/.
9.3.16.IfyDue
x
, then4y
.4/
11y
00
9y
0
2yDe
x
Œ4.u
.4/
C4u
000
C6u
00
C4u
0
Cu/11.u
00
C
2u
0
Cu/9.u
0
Cu/2uDe
x
.4u
.4/
C16u
000
C13u
00
15u
0
18u/. LetupDACBxwhere
.18AC15B/18BxD 1C6x. ThenBD 1=3,AD1=3, andypD
e
x
3
.1x/.
9.3.18.IfyDue
x
, theny
.4/
4y
000
C6y
00
4y
0
C2yDe
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/4.u
000
C3u
00
C
3u
0
Cu/C6.u
00
C2u
0
Cu/4.u
0
Cu/C2uDe
x
.u
.4/
Cu/. LetupDACBxCCx
2
CDx
3
CEx
4
where.AC24E/CBxCCx
2
CDx
3
CEx
4
D24CxCx
4
. ThenED1 DD0,CD0 BD1,
AD0, andypDxe
x
.1Cx
3
/.
9.3.20.IfyDue
2x
, theny
.4/
Cy
000
2y
00
6y
0
4yDe
2x
Œ.u
.4/
C8u
000
C24u
00
C32u
0
C16u/C
.u
000
C6u
00
C12u
0
C8u/2.u
00
C4u
0
C4u/6.u
0
C2u/4uDe
2x
.u
.4/
C9u
000
C28u
00
C30u
0
/. Let

181. 176 Chapter 9Linear Higher Order Equations
upDx.ACBxCCx
2
/where.30AC56BC54C /C.60BC168C /xC90Cx
2
D .4C28xC15x
2
/.
ThenCD 1=6,BD0,AD1=6, andypD
xe
2x
6
.1x
2
/.
9.3.22.IfyDue
x
, theny
.4/
5y
00
C4yDe
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/5.u
00
C2u
0
Cu/C4uD
e
x
.u
.4/
C4u
000
Cu
00
6u
0
/. LetupDx.ACBxCCx
2
/where.6AC2BC24C /C.12BC
6C /x18Cx
2
D3Cx3x
2
, soCD1=6,BD0,AD1=6. ThenypD
xe
x
6
.1Cx
2
/.
9.3.24.IfyDue
2x
, theny
.4/
3y
000
C4y
0
De
2x
Œ.u
.4/
C8u
000
C24u
00
C32u
0
C16u/3.u
000
C
6u
00
C12u
0
C8u/C4.u
0
C2u/De
2x
.u
.4/
C5u
000
C6u
00
/. LetupDx
2
.ACBxCCx
2
/where
.12AC30BC24C /C.36BC120C /xC72Cx
3
D15C26xC12x
2
. ThenCD1=6,BD1=6,
AD1=2, andypD
x
2
e
2x
6
.3CxCx
2
/.
9.3.26.IfyDue
x
, then2y
.4/
5y
000
C3y
00
Cy
0
yDe
x
Œ2.u
.4/
C4u
000
C6u
00
C4u
0
Cu/5.u
000
C
3u
00
C3u
0
Cu/C3.u
00
C2u
0
Cu/C.u
0
Cu/uDe
x
.2u
.4/
C3u
000
/. LetupDx
3
.ACBx/, where
.18AC48B/C72BxD11C12x. ThenBD1=6,AD1=6, andypD
x
3
e
x
6
.1Cx/.
9.3.28.IfyDue
2x
, theny
.4/
7y
000
C18y
00
20y
0
C8yDe
2x
Œ.u
.4/
C8u
000
C24u
00
C32u
0
C
16u/7.u
000
C6u
00
C12u
0
C8u/C18.u
00
C4u
0
C4u/20.u
0
C2u/C8uDe
2x
.u
.4/
Cu
000
/. Let
upDx
3
.ACBxCCx
2
/where.6AC24B/C.24BC120C /xC60Cx
2
D38x5x
2
. Then so
CD 1=12 BD1=12,AD1=6, andypD
x
3
e
2x
12
.2Cxx
2
/.
9.3.30.IfyDue
x
, theny
000
Cy
00
4y
0
4yDe
x
Œ.u
000
3u
00
C3u
0
u/C.u
00
2u
0
Cu/4.u
0

u/4uDe
x
.u
000
2u
00
3u
0
/. LetupD.A0CA1x/cos2xC.B0CB1x/sin2x, where
8A114B1D 22
14A1C8B1D 6
8A014B015A18B1D 1
14A0C8B0C8A115B1D 1:
ThenA1D 1,B1D1,A0D1,B0D1, andypDe
x
Œ.1x/cos2xC.1Cx/sin2x.
9.3.32.IfyDue
x
, theny
000
2y
00
Cy
0
2yDe
x
Œ.u
000
C3u
00
C3u
0
Cu/2.u
00
C2u
0
Cu/C.u
0
Cu/2uD
e
x
.u
000
Cu
00
2u/. LetupD.A0CA1xCA2x
2
/cos2xC.B0CB1xCB2x
2
/sin2xwhere
6A28B2D 4
8A26B2D 3
6A18B124A2C8B2D 5
8A16B18A224B2D 5
6A08B012A1C4B1C2A2C12B2D 9
8A06B04A112B112A2C2B2D 6:
ThenA2D0,B2D1=2;A1D1=2,B1D 1=2;A0D1=1,B0D1=2; andypD
e
x
2
Œ.1Cx/cos2xC
.1xCx
2
/sin2x.
9.3.34.IfyDue
x
, theny
000
y
00
C2yDe
x
Œ.u
000
C3u
00
C3u
0
Cu/.u
00
C2u
0
Cu/C2uD
e
x
.u
000
C2u
00
Cu
0
C2u/. Since cosxand sinxsatisfyu
000
C2u
00
Cu
0
C2uD0, letupDxŒ.A0C

182. Section 9.3Undetermined Coefficients for Higher Order Equations177
A1x/cosxC.B0CB1x/sinxwhere
4A1C8B1D4
8A14B1D 12
2A0C4B0C4A1C6B1D20
4A02B06A1C4B1D 12:
ThenA1D1,B1D1;A0D1,B0D3; andypDxe
x
Œ.1Cx/cosxC.3Cx/sinx.
9.3.36.IfyDue
3x
, thenDe
3x
Œ.u
000
C9u
00
C27u
0
C27u/6.u
00
C6u
0
C9u/C18.u
0
C3u/D
e
3x
.u
000
C3u
00
C9u
0
C27u/. Since cos3xand sin3xsatisfyu
000
C3u
00
C9u
0
C27uD0, letupD
xŒ.A0CA1x/cos3xC.B0CB1x/sin3xwhere
36A1C36B1D 3
36A136B1D 3
18A0C18B0C6A1C18B1D 2
18A018B018A1C6B1D 3:
ThenA1D 1=12,B1D0;A0D,B0D 1=12; andypD
xe
3x
12
.xcos3xCsin3x/.
9.3.38.IfyDue
x
, theny
.4/
3y
000
C2y
00
C2y
0
4yDe
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/
3.u
000
C3u
00
C3u
0
Cu/C2.u
00
C2u
0
Cu/C4.u
0
Cu/CuDe
x
.u
.4/
Cu
000
u
00
Cu
0
2u/. Let
upDAcos2xCBsin2xwhere18A6BD2and6AC18BD 1. ThenAD1=12,BD 1=12,
andypD
e
x
12
.cos2xsin2x/.
9.3.40.IfyDue
x
, theny
.4/
C6y
000
C13y
00
C12y
0
C4yDe
x
Œ.u
.4/
4u
000
C6u
00
4u
0
C
u/C6.u
000
3u
00
C3u
0
u/C13.u
00
2u
0
Cu/C12u
0
u/C4uDe
x
.u
.4/
C2u
000
Cu
00
/. Let
upD.A0CA1x/cosxC.B0CB1x/sinxwhere
2B1D 1
2A1D 1
2B06A12B1D 4
2A0C2A16B1D 5:
ThenA1D 1=2,B1D1=2,A0D 1=2,B0D 1, andypD
e
x
2
Œ.1Cx/cosxC.2x/sinx.
9.3.42.IfyDue
x
, theny
.4/
5y
000
C13y
00
19y
0
C10yDe
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/5.u
000
C
3u
00
C3u
0
Cu/C13.u
00
C2u
0
Cu/19.u
0
Cu/C10uDe
x
.u
.4/
u
000
C4u
00
4u
0
/. Since cos2x
and sin2xsatisfyu
.4/
u
000
C4u
00
4u
0
D0, letupDx.Acos2xCBsin2x/where8A16BD1
and16AC8BD1. ThenAD3=40,B0D 1=40, andypD
xe
x
40
.3cos2xsin2x/.
9.3.44.IfyDue
x
, theny
.4/
5y
000
C13y
00
19y
0
C10yDe
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/5.u
000
C
3u
00
C3u
0
Cu/C13.u
00
C2u
0
Cu/19.u
0
Cu/C10uDe
x
.u
.4/
u
000
C4u
00
4u
0
/. Since cos2x
and sin2xsatisfyu
.4/
u
000
C4u
00
4u
0
D0, letupDxŒ.A0CA1x/cos2xC.B0CB1x/sin2x/
where
16A132B1D 8
32A1C16B1D 4
8A016B040A112B1D 7
16A0C8B0C12A140B1D 8:

183. 178 Chapter 9Linear Higher Order Equations
ThenA1D0,B1D 1=4;A0D0,B0D 1=4, andypD
xe
x
4
.1Cx/sin2x.
9.3.46.IfyDue
2x
, theny
.4/
8y
000
C32y
00
64y
0
C64yC4yDe
2x
Œ.u
.4/
C8u
000
C24u
00
C32u
0
C
16u/8.u
000
C6u
00
C12u
0
C8u/C32.u
00
C4u
0
C4u/64.u
0
C2u/C64uDe
2x
.u
.4/
C8u
00
C16u/. Since
cos2x, sin2x,xcos2x, andxsin2xsatisfyu
.4/
C8u
00
C16uD0, letupDx
2
.Acos2xCBsin2x/
where32AD1and32BD 1. ThenAD 1=32,BD1=32, andypD
x
2
e
2x
32
.cos2xsin2x/.
9.3.48.Find particular solutions of(a)y
000
4y
00
C5y
0
2yD 4e
x
,(b)y
000
4y
00
C5y
0
2yDe
2x
,
and(c)y
000
4y
00
C5y
0
2yD 2cosxC4sinx.
(a)IfyDue
x
, theny
000
4y
00
C5y
0
2yDe
x
Œ.u
000
C3u
00
C3u
0
Cu/4.u
00
C2u
0
Cu/C5.u
0
Cu/2uD
e
x
.u
000
u
00
/. Letu1pDAx
2
where2AD 4. ThenAD2, andy1pD2x
2
e
x
.
(b)IfyDue
2x
, theny
000
4y
00
C5y
0
2yDe
2x
Œ.u
000
C6u
00
C12u
0
C8u/4.u
00
C4u
0
C4u/C
5.u
0
C2u/2uDe
2x
.u
000
C2u
00
Cu
0
/. Letu2pDx. Theny2pDxe
2x
.
(c)Ify3pDAcosxCBsinx, theny
000
3p
4y
00
3p
C5y
0
3p
2y3pD.2AC4B/cosxC.4AC2B/sinxD
2cosxC4sinxifAD 1andBD0, soy3pD cosx.
From the principle of superposition,ypD2x
2
e
x
Cxe
2x
cosx.
9.3.50.Find particular solutions of(a)y
000
y
0
D 2.1Cx/,(b)y
000
y
0
D4e
x
,(c)y
000
y
0
D 6e
x
,
and(d)y
000
y
0
D96e
3x
(a)Lety1pDx.ACBx/. Theny
000
1p
y
0
1p
D A2BxD 2.1Cx/ifAD2andBD1; therefore
y1pD2xC2x
2
.
(b)IfyDue
x
, theny
000
y
0
De
x
Œ.u
000
C3u
00
C3u
0
Cu/.u
0
Cu/De
x
.u
000
C3u
00
C2u
0
/. Let
u2pD4x. Theny2pD4xe
x
.
(c)IfyDue
x
, theny
000
y
0
De
x
Œ.u
000
3u
00
C3u
0
u/.u
0
u/De
x
.u
000
3u
00
C2u
0
/. Let
u2pD 3x. Theny2pD 6xe
x
.
(d)Sincee
3x
does not satisfy the complementary equation, lety4pDAe
3x
. Theny
000
4p
y
0
4p
D
24Ae
3x
. LetAD4; theny4pD4e
4x
.
From the principle of superposition,ypD2xCx
2
C2xe
x
3xe
x
C4e
3x
9.3.52.Find particular solutions of(a)y
000
C3y
00
C3y
0
CyD12e
x
and(b)y
000
C3y
00
C3y
0
CyD
9cos2x13sin2x.
(a)IfyDue
2x
, theny
000
C3y
00
C3y
0
CyDe
2x
Œ.u
000
3u
00
C3u
0
u/C3.u
00
2u
0
Cu/C
3.u
0
u/CuDe
x
u
000
. Letu
000
1p
D12. Integrating three times and taking the constants of integration
to be zero yieldsu1pD2x
3
. Therefore,y1pD2x
3
.
(b)Lety2pDAcos2xCBsin2xwhere11A2BD9and2A11BD 13. ThenAD 1,
BD1, andy2pD cos2xCsin2x.
From the principle of superposition,ypD2x
3
e
2x
cos2xCsin2x.
9.3.54.Find particular solutions of(a)y
.4/
5y
00
C4yD 12e
x
,(b)y
.4/
5y
00
C4yD6e
x
, and(c)
y
.4/
5y
00
C4yD10cosx.
(a)IfyDue
x
, theny
.4/
5y
00
C4yDe
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/5.u
00
C2u
0
Cu/C4uD
e
x
.u
.4/
C4u
000
Cu
00
6u
0
/. Letu1pD2x. Theny1pD2xe
x
.
(b)IfyDue
x
, theny
.4/
5y
00
C4yDe
x
Œ.u
.4/
4u
000
C6u
00
4u
0
Cu/5.u
00
2u
0
Cu/C4uD
e
x
.u
.4/
4u
000
Cu
00
C6u
0
/. Letu2pDx. Theny2pDxe
x
.
(c)Lety3pDAcosxCBsinxwhere10AD10and10BD0. ThenAD1,BD0, andy3pDcosx.
From the principle of superposition,ypD2xe
x
Cxe
x
Ccosx.
9.3.56.Find particular solutions of(a)y
.4/
C2y
000
3y
00
4y
0
C4yD2e
x
.1Cx/and(b)y
.4/
C2y
000

3y
00
4y
0
C4yDe
2x
.

184. Section 9.3Undetermined Coefficients for Higher Order Equations179
(a)IfyDue
x
, theny
.4/
C2y
000
3y
00
4y
0
C4yDe
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/C2.u
000
C
3u
00
C3u
0
Cu/3.u
00
C2u
0
Cu/4.u
0
Cu/C4uDe
x
.u
.4/
C6u
000
C9u
00
/. Letu1pDx
2
.ACBx/
where.18AC36B/C54BxD2C2x. ThenBD1=27,AD1=27, andy1pD
x
2
27
.1Cx/e
x
.
(b)IfyDue
2x
, theny
.4/
C2y
000
3y
00
4y
0
C4yDe
2x
Œ.u
.4/
8u
000
C24u
00
32u
0
C16u/C
2.u
000
6u
00
C12u
0
8u/3.u
00
4u
0
C4u/4.u
0
2u/C4uDe
2x
.u
.4/
6u
000
C9u
00
/. Let
u2pDAx
2
where18AD1. ThenAD1=18andypD
x
2
18
e
2x
.
From the principle of superposition,ypD
x
2
54
Œ.2C2x/e
x
C3e
2x
.
9.3.58.Find particular solutions of(a)y
.4/
C5y
000
C9y
00
C7y
0
C2yDe
x
.30C24x/and(b)
y
.4/
C5y
000
C9y
00
C7y
0
C2yD e
2x
.
(a)IfyDue
x
, theny
.4/
C5y
000
C9y
00
C7y
0
C2yDe
x
Œ.u
.4/
4u
000
C6u
00
4u
0
Cu/C5.u
000

3u
00
C3u
0
u/C9.u
00
2u
0
Cu/C7.u
0
u/C2uDe
x
.u
.4/
Cu
000
/. Letu1pDx
3
.ACBx/where
.6AC24B/C24BxD30C24x. TheBD1,AD1, andy1pDx
3
.1Cx/e
x
.
(b)IfyDue
2x
, theny
.4/
C5y
000
C9y
00
C7y
0
C2yDe
2x
Œ.u
.4/
8u
000
C24u
00
32u
0
C16u/C
5.u
000
6u
00
C12u
0
8u/C9.u
00
4u
0
C4u/C7.u
0
2u/C2uDe
2x
.u
.4/
3u
000
C3u
00
u
0
/. Let
u2pDx. Theny2pDxe
2x
.
From the principle of superposition,ypDx
3
.1Cx/e
x
Cxe
2x
.
9.3.60.IfyDue
2x
, theny
000
y
00
y
0
CyDe
2x
Œ.u
000
C6u
00
C12u
0
C8u/.u
00
C4u
0
C4u/.u
0
C2u/C
uDe
2x
.u
000
C5u
00
C7u
0
C3u/. LetupDACBx, where.3AC7B/C3xD10C3x. ThenBD1,
AD1andypDe
2x
.1Cx/. Sincep.r/D.rC1/.r1/
2
,yDe
2x
.1Cx/Cc1e
x
Ce
x
.c2Cc3x/
9.3.62.IfyDue
2x
, theny
000
6y
00
C11y
0
6yDe
2x
Œ.u
000
C6u
00
C12u
0
C8u/6.u
00
C4u
0
C4u/C11.u
0
C
2u/6uDe
2x
.u
000
u
0
/. LetupDx.ACBxCCx
2
/where.AC6C /2Bx3Cx
2
D54x3x
2
.
ThenCD1,BD2,AD1, andypDxe
2x
.1Cx/
2
. Sincep.r/D.r1/.r2/.r3/,
yDxe
2x
.1Cx/
2
Cc1e
x
Cc2e
2x
Cc3e
3x
.
9.3.64.IfyDue
x
, theny
000
3y
00
C3y
0
yDe
x
Œ.u
000
C3u
00
C3u
0
Cu/3.u
00
C2u
0
Cu/C
3.u
0
Cu/uDe
x
u
000
. Letu
000
D1Cx. Integrating three times and taking the constants of integration
to be zero yieldsuD
x
3
24
.4Cx/. Therefore,ypD
x
3
e
x
24
.4Cx/. Sincep.r/D.r1/
3
,yD
x
3
e
x
24
.4Cx/Ce
x
.c1Cc2xCc3x
2
/.
9.3.66.IfyDue
2x
, theny
000
C2y
00
y
0
2yDe
2x
Œ.u
000
6u
00
C12u
0
8u/C2.u
00
4u
0
C4u/
.u
0
2u/2uDe
2x
.u
000
4u
00
C3u
0
/. LetupD.A0CA1x/cosxC.B0CB1x/sinxwhere
4A1C2B1D 2
2A1C4B1D 9
4A0C2B08B1D23
2A0C4B0C8A1D 8:
ThenA1D1=2,B1D 2;A0D1,B0D3=2, andypDe
2x


1C
x
2

cosxC

3
2
2x

sinx

.
Sincep.r/D.r1/.rC1/.rC2/,yDe
2x


1C
x
2

cosxC

3
2
2x

sinx

Cc1e
x
Cc2e
x
C
c3e
2x
.

185. 180 Chapter 9Linear Higher Order Equations
9.3.68.IfyDue
x
, theny
.4/
4y
000
C14y
00
20y
0
C25yDe
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/4.u
000
C
3u
00
C3u
0
Cu/C14.u
00
C2u
0
Cu/20.u
0
Cu/C25uDe
x
.u
.4/
C8u
00
C16u/. Since cos2x, sin2x,
xcos2x, andxsin2xsatisfyu
.4/
C8u
00
C16uD0, letupDx
2
Œ.A0CA1x/cos2xC.B0CB1x/sin2x
where
96A1D6
96B1D0
32A0C48B1D2
48A132B0D3:
ThenA1D 1=16,B1D0;A0D 1=16,B0D0, andyŒD
x
2
e
x
16
.1Cx/cos2x. Sincep.r/D
Œ.r1/
2
C1
2
,yD
x
2
e
x
16
.1Cx/cos2xCe
x
Œ.c1Cc2x/cos2xC.c3Cc4x/sin2x.
9.3.70.IfyDue
x
, theny
000
y
00
y
0
CyDe
x
Œ.u
000
3u
00
C3u
0
u/.u
00
2u
0
Cu/.u
0

u/CuDe
x
.u
000
4u
00
C4u
0
/. LetupDx.ACBx/, where.4A8B/C8BxD 4C8x.
ThenBD1,AD1, andypDx.1Cx/e
x
. Sincep.r/D.rC1/.r1/
2
the general solution is
yDx.1Cx/e
x
Cc1e
x
Cc2e
x
Cc3xe
x
. Therefore,
2
4
y
y
0
y
00
3
5D
2
4
x.1Cx/e
x
e
x
.x
2
x1/
e
x
.x
2
3x/
3
5C
2
4
e
x
e
x
xe
x
e
x
e
x
e
x
.xC1/
e
x
e
x
e
x
.xC2/
3
5
2
4
c1
c2
c3
3
5:
SettingxD0and imposing the initial conditions yields
2
4
2
0
0
3
5D
2
4
0
1
0
3
5C
2
4
1 1 0
1 1 1
1 1 2
3
5
2
4
c1
c2
c3
3
5;
soc1D1,c2D1,c3D 1, andyDe
x
.1CxCx
2
/C.1x/e
x
.
9.3.72.IfyDue
x
, theny
000
2y
00
5y
0
C6yDe
x
Œ.u
.4/
4u
000
C6u
00
4u
0
Cu/C2.u
000

3u
00
C3u
0
u/C2.u
00
2u
0
Cu/C2.u
0
u/CuDe
x
.u
.4/
2u
000
C2u
00
/. LetupDx
2
.ACBx/,
where.4A12B/C12BxD2012x. ThenBD 1,AD2, andypDx
2
.2x/e
x
. Since
p.r/D.rC1/
2
.r
2
C1/, the general solution isyDx
2
.2x/e
x
Ce
x
.c1Cc2x/Cc3cosxCc4sinx.
Therefore,
2
6
6
4
y
y
0
y
00
y
000
3
7
7
5
D
2
6
6
4
x
2
.2x/e
x
x.x
2
5xC4/e
x
.x
3
8x
2
C14x4/e
x
.x
3
11x
2
C30x18/e
x
3
7
7
5
C
2
6
6
4
e
x
xe
x
cosx sinx
e
x
.1x/e
x
sinx cosx
e
x
.x2/e
x
cosxsinx
e
x
.3x/e
x
sinxcosx
3
7
7
5
2
6
6
4
c1
c2
c3
c4
3
7
7
5
SettingxD0and imposing the initial conditions yields
2
6
6
4
3
4
7
22
3
7
7
5
D
2
6
6
4
0
0
4
18
3
7
7
5
C
2
6
6
4
1 0 1 0
1 1 0 1
121 0
1 3 0 1
3
7
7
5
2
6
6
4
c1
c2
c3
c4
3
7
7
5
;
soc1D2,c2D 1,c3D1,c4D 1, andyD.2x/.x
2
C1/e
x
Ccosxsinx.
9.3.74.IfyDue
x
, theny
.4/
3y
000
C5y
00
2y
0
De
x
Œ.u
.4/
C4u
000
C6u
00
C4u
0
Cu/3.u
000
C3u
00
C
3u
0
Cu/C4.u
00
C2u
0
Cu/2.u
0
Cu/De
x
.u
.4/
Cu
000
Cu
00
Cu
0
/. Since cosxand sinxsatisfy

186. Section 9.4Variation of Parameters for Higher Order Equations181
u
.4/
Cu
000
Cu
00
Cu
0
D0, letupDx.AcosxCBsinx/where2A2BD 2and2A2BD2.
ThenAD1,BD0, andypDe
x
cosx. Sincep.r/Dr.r1/Œ.r1/
2
C1the general solution is
yDe
x
cosxCc1Ce
x
.c2Cc3cosxCc4sinx/. Therefore,
2
6
6
4
y
y
0
y
00
y
000
3
7
7
5
D
2
6
6
4
xe
x
cosx
e
x
..xC1/cosxxsinx/
e
x
.2cosx2.xC1/sinx/
e
x
.2xcosxC2.xC3/sinx/
3
7
7
5
C
2
6
6
4
1 e
x
e
x
cosx e
x
sinx
0 e
x
e
x
.cosxsinx/ e
x
.cosxCsinx/
0 e
x
2e
x
sinx 2e
x
cosx
0 e
x
e
x
.2cosxC2sinx/ e
x
.2cosx2sinx/
3
7
7
5
2
6
6
4
c1
c2
c3
c4
3
7
7
5
SettingxD0and imposing the initial conditions yields
2
6
6
4
2
0
1
5
3
7
7
5
D
2
6
6
4
0
1
2
0
3
7
7
5
C
2
6
6
4
1 1 1 0
0 1 1 1
0 1 0 2
0 12 2
3
7
7
5
2
6
6
4
c1
c2
c3
c4
3
7
7
5
;
soc1D2,c2D 1,c3D1,c4D 1, and2Ce
x
Œ.1Cx/cosxsinx1.
9.4VARIATION OF PARAMETERS FOR HIGHER ORDER EQUATIONS
9.4.2.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
2
xe
x
2
x
2
e
x
2
2xe
x
2
e
x
2
.12x
2
/ 2xe
x
2
.1x
2
/
e
x
2
.4x
2
2/ 2xe
x
2
.2x
2
3/ 2e
x
2
.2x
4
5x
2
C1/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2e
3x
2
;
W1D
ˇ
ˇ
ˇ
ˇ
ˇ
xe
x
2
x
2
e
x
2
e
x
2
.12x
2
/ 2xe
x
2
.1x
2
/
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
2
e
2x
2
;W2D
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
2
x
2
e
x
2
2xe
x
2
2xe
x
2
.1x
2
/
ˇ
ˇ
ˇ
ˇ
ˇ
D
2xe
2x
2
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
2
xe
x
2
2xe
x
2
e
x
2
.12x
2
/
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
2
;u
0
1
D
F W1
P0W
D
1
2
x
5=2
;u
0
2
D
F W2
P0W
D
x
3=2
;u
0
3
D
F W2
P0W
D
p
x=2;u1Dx
7=2
=7;u2D
2
5
x
5=2
;u3Dx
3=2
=3;ypDu1y1Cu2y2Cu3y3D
8
105
e
x
2
x
7=2
.
9.4.4.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
e
x
x
e
x
x
0
e
x
.x1/
x
2

e
x
.xC1/
x
2
0
e
x
.x
2
2xC2/
x
3
e
x
.x
2
C2xC2/
x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2=x
2
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
x
e
x
x
e
x
.x1/
x
2

e
x
.xC1/
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D

2
x
2
;W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
e
x
x
0
e
x
.xC1/
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
x
.xC1/
x
2
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
e
x
x
0
e
x
.x1/
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
x
.x1/
x
2
;u
0
1
D
F W1
P0W
D 2;u
0
2
D
F W2
P0W
De
x
.xC1/;u
0
3
D
F W2
P0W
De
x
.x1/;u1D 2x;u2D e
x
.xC2/;
u3De
x
.x2/;ypDu1y1Cu2y2Cu3y3D 2.x
2
C2/=x.
9.4.6.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
1
x
e
x
e
x
1=x
2
e
x
e
x
2
x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
2.x
2
2/
x
3
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
1
x
e
x

1
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
x
.x1/
x
2
;W2D

187. 182 Chapter 9Linear Higher Order Equations
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
1
x
e
x

1
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
x
.xC1/
x
2
;W3D
ˇ
ˇ
ˇ
ˇ
e
x
e
x
e
x
e
x
ˇ
ˇ
ˇ
ˇ
D 2;u
0
1
D
F W1
P0W
De
x
.x1/;u
0
2
D

F W2
P0W
De
x
.xC1/;u
0
3
D
F W2
P0W
D 2x
2
;ypDu1y1Cu2y2Cu3y3D 2
x
2
3
.
9.4.8.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
p
x
1
p
x
x
2
1
2
p
x

1
2x
3=2
2x

1
4x
3=2
3
4x
5=2
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
15
4x
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
p
x
x
2

1
2x
3=2
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
5
p
x
2
;W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
p
x x
2
1
2
p
x
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
3x
3=2
2
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
p
x
1
p
x
1
2
p
x

1
2x
3=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
1
x
;u
0
1
D
F W1
P0W
D 5
p
x;u
0
2
D
F W2
P0W
D3x
3=2
;u
0
3
D
F W2
P0W
D
2
x
;u1D
10
3
x
3=2
;u2D
6
5
x
5=2
;u3D2lnjxj;ypDu1y1Cu2y2Cu3y3D2x
2
lnjxj
32
15
x
2
. Since
32
15
x
2
satisfies the complementary equation we takeypDlnjxj.
9.4.10.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x 1=x
e
x
x
1
1
x
2
e
x
.x1/
x
2
0
2
x
3
e
x
.x
2
2xC2/
x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
2e
x
.1x/
x
3
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
x
e
x
x

1
x
2
e
x
.x1/
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
x
x
2
;
W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
e
x
x
1
e
x
.x1/
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
x
.x2/
x
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
1
x
1
1
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
2
x
;u
0
1
D
F W1
P0W
D1;u
0
2
D
F W2
P0W
D
x.2x/;u
0
3
D
F W2
P0W
D 2xe
x
;u1Dx;u2D
x
2
.3x/
3
;u3D2e
x
.xC1/;ypDu1y1C
u2y2Cu3y3D
2x
3
C3x
2
C6xC6
3x
. SincexC
2
x
satisfies the complementary equation we take
ypD
2x
2
C6
3
.
9.4.12.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x e
x
e
x
1 e
x
e
x
0 e
x
e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2x;W1D
ˇ
ˇ
ˇ
ˇ
e
x
e
x
e
x
e
x
ˇ
ˇ
ˇ
ˇ
D 2;W2D
ˇ
ˇ
ˇ
ˇ
x e
x
1e
x
ˇ
ˇ
ˇ
ˇ
D e
x
.xC
1/;W3D
ˇ
ˇ
ˇ
ˇ
x e
x
1 e
x
ˇ
ˇ
ˇ
ˇ
De
x
.x1/;u
0
1
D
F W1
P0W
D 1;u
0
2
D
F W2
P0W
De
x
.xC1/=2;u
0
3
D
F W2
P0W
D
e
x
.x1/=2;u1D x;u2D e
x
.xC2/=2;u3De
x
.x2/=2;ypDu1y1Cu2y2Cu3y3D x
2
2.

188. Section 9.4Variation of Parameters for Higher Order Equations183
9.4.14.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
p
x 1=
p
x x
3=2
1
x
3=2
1
2
p
x

1
2x
3=2
3
p
x
2

3
2x
5=2

1
4x
3=2
3
4x
5=2
3
4
p
x
15
4x
7=2
3
8x
5=2

15
8x
7=2

3
8x
3=2

105
8x
9=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
12
x
6
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
p
x
x
3=2
1
x
3=2

1
2x
3=2
3
p
x
2

3
2x
5=2
3
4x
5=2
3
4
p
x
15
4x
7=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6
x
7=2
;W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
p
x x
3=2
1
x
3=2
1
2
p
x
3
p
x
2

3
2x
5=2

1
4x
3=2
3
4
p
x
15
4x
7=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6
x
5=2
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
p
x
1
p
x
1
x
3=2
1
2
p
x

1
2x
3=2

3
2x
5=2

1
4x
3=2
3
4x
5=2
15
4x
7=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D

2
x
9=2
;W4D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
p
x
1
p
x
x
3=2
1
2
p
x

1
2x
3=2
3
p
x
2

1
4x
3=2
3
4x
5=2
3
4
p
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
2
x
3=2
;u
0
1
D
F W1
P0W
D 3x;u
0
2
D
F W2
P0W
D
3x
2
;u
0
3
D
F W2
P0W
D1;u
0
4
D
F W4
P0W
D x
3
;u1D
3x
2
2
;u2Dx
3
;u3Dx;u4D
x
4
4
;
ypDu1y1Cu2y2Cu3y3D
x
5=2
4
.
9.4.16.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
x
3
x
4
1 2x 3x
2
4x
3
0 2 6x 12x
2
0 0 6 24x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D12x
4
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
2
x
3
x
4
2x 3x
2
4x
3
2 6x 12x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2x
6
;W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
3
x
4
1 3x
2
4x
3
0 6x 12x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6x
5
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
x
4
1 2x 4x
3
0 2 12x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D6x
4
;W4D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
x
3
1 2x 3x
2
0 2 6x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2x
3
;u
0
1
D
F W1
P0W
D
x
2
6
;
u
0
2
D
F W2
P0W
D
x
2
;u
0
3
D
F W2
P0W
D
1
2
;u
0
4
D
F W4
P0W
D
1
6x
;u1D
x
3
18
;u2D
x
2
4
;u3D
x
2
;
u4D
lnjxj
6
;ypDu1y1Cu2y2Cu3y3D
x
4
lnjxj
6

11x
4
36
. Since
11x
4
36
satisfies the complementary
equation we takeypD
x
4
lnjxj
6
.
9.4.18.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1=x 1=x
2
1 2x1=x
2
2=x
3
0 2 2=x
3
6=x
4
0 0 6=x
4
24=x
5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 72=x
6
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
2
1=x 1=x
2
2x1=x
2
2=x
3
2 2=x
3
6=x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 12=x
4
;
W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x 1=x 1=x
2
11=x
2
2=x
3
0 2=x
3
6=x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 6=x
5
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1=x
2
1 2x2=x
3
0 2 6=x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D12=x
2
;W4D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1=x
1 2x1=x
2
0 2 2=x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6=x;u
0
1
D
F W1
P0W
D 2;u
0
2
D
F W2
P0W
D1=x;u
0
3
D
F W2
P0W
D2x
2
;u
0
4
D
F W4
P0W
D x
3
;u1D 2x;
u2Dlnjxj;u3D2x
3
=3;u4D x
4
=4;ypDu1y1Cu2y2Cu3y3Dx
2
lnjxj 19x
2
=12. Since
19x
2
=12satisfies the complementary equation we takeypDx
2
lnjxj.

189. 184 Chapter 9Linear Higher Order Equations
9.4.20.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
2x
e
x
=x
e
2x
x
e
x
2e
2x
e
x
.x1/
x
2
e
2x
.2x1/
x
2
e
x
4e
2x
e
x
.x
2
2xC2/
x
3
2e
2x
.2x
2
2xC1/
x
3
e
x
8e
2x
e
x
.x
3
3x
2
C6x6/
x
4
2e
2x
.4x
3
6x
2
C6x3/
x
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
6x
x
4
;
W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
2x
e
x
x
e
2x
x
2e
2x
e
x
.x1/
x
2
e
2x
.2x1/
x
2
4e
2x
e
x
.x
2
2xC2/
x
3
2e
2x
.2x
2
2xC1/
x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
5x
x
3
;
W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
x
x
e
2x
x
e
x
e
x
.x1/
x
2
e
2x
.2x1/
x
2
e
x
e
x
.x
2
2xC2/
x
3
2e
2x
.2x
2
2xC1/
x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
4x
x
3
;
W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
2x
e
2x
x
e
x
2e
2x
e
2x
.2x1/
x
2
e
x
4e
2x
2e
2x
.2x
2
2xC1/
x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
5x
.2x/
x
3
;
W4D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
2x
e
x
x
e
x
2e
2x
e
x
.x1/
x
2
e
x
4e
2x
e
x
.x
2
2xC2/
x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
e
4x
.xC2/
x
3
;
u
0
1
D
F W1
P0W
D3;u
0
2
D
F W2
P0W
D3e
x
;u
0
3
D
F W2
P0W
D3.2x/;u
0
4
D
F W4
P0W
D 3e
x
.xC2/;
u1D3x;u2D 3e
x
;u3D3x.4x/=2;u4D3e
x
.xC3/;ypDu1y1Cu2y2Cu3y3D
3e
x
.x
2
C4xC6/
2x
. Since
3e
x
.2xC3/
x
is a solution of the complementary equation we takeypD
3xe
x
2
.
9.4.22.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
3
xlnx
1 3x
2
1Clnx
0 6x 1=x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 4x
2
;W1D
ˇ
ˇ
ˇ
ˇ
x
3
xlnx
3x
2
1Clnx
ˇ
ˇ
ˇ
ˇ
Dx
3
2x
3
lnx;W2D
ˇ
ˇ
ˇ
ˇ
x xlnx
1 1Clnx
ˇ
ˇ
ˇ
ˇ
Dx;W3D
ˇ
ˇ
ˇ
ˇ
x x
3
1 3x
2
ˇ
ˇ
ˇ
ˇ
D2x
3
;u
0
1
D
F W1
P0W
D2lnx=x
1
x
;u
0
2
D
F W2
P0W
D
1
x
3
;
u
0
3
D
F W2
P0W
D
2
x
;u1D.lnx/
2
lnx;u2D
1
2x
2
;u3D 2lnx;ypDu1y1Cu2y2Cu3y3D
x.lnx/
2
xlnx
x
2
. Sincexlnx
x
2
satisfies the complementary equation we takeypD x.lnx/
2

190. Section 9.4Variation of Parameters for Higher Order Equations185
The general solution isyD x.lnx/
2
Cc1xCc2x
3
Cc3xlnx, so
2
4
y
y
0
y
00
3
5D
2
6
4
x.lnx/
2
.lnx/
2
2lnx

2lnx
x

2
x
3
7
5C
2
6
4
x x
3
xlnx
1 3x
2
1Clnx
0 6x
1
x
3
7
5
2
4
c1
c2
c3
3
5
SettingxD1and imposing the initial conditions yields
2
4
4
4
2
3
5D
2
4
0
0
2
3
5C
2
4
1 1 0
1 3 1
0 6 1
3
5
2
4
c1
c2
c3
3
5:
Solving this system yieldsc1D3,c2D1,c3D 2. Therefore,yD x.lnx/
2
C3xCx
3
2xlnx.
9.4.24.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
2x
xe
x
e
x
2e
2x
e
x
.1x/
e
x
4e
2x
e
x
.x2/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
2x
.6x5/;W1D
ˇ
ˇ
ˇ
ˇ
e
2x
xe
x
2e
2x
e
x
.1x/
ˇ
ˇ
ˇ
ˇ
De
x
.13x/;
W2D
ˇ
ˇ
ˇ
ˇ
e
x
xe
x
e
x
e
x
.1x/
ˇ
ˇ
ˇ
ˇ
D12x;W3D
ˇ
ˇ
ˇ
ˇ
e
x
e
2x
e
x
2e
2x
ˇ
ˇ
ˇ
ˇ
De
3x
;u
0
1
D
F W1
P0W
D13x;u
0
2
D

F W2
P0W
De
x
.2x1/;u
0
3
D
F W2
P0W
De
2x
;u1D
x.23x/
2
;u2D e
x
.2xC1/;u3D
e
2x
2
;
ypDu1y1Cu2y2Cu3y3D e
x
.3x
2
CxC2/=2. Since
e
x
2
is a solution of the complementary
equation we takeypD
e
x
.3xC1/x
2
.
The general solution isyD
e
x
.3xC1/x
2
Cc1e
x
Cc2e
2x
Cc3xe
x
, so
2
4
y
y
0
y
00
3
5D
2
6
6
6
6
6
4

e
x
.3xC1/x
2

e
x
.3x
2
C7xC1/
2

e
x
.3x
2
C13xC8/
2
3
7
7
7
7
7
5
C
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
x
e
2x
xe
x
e
x
2e
2x
e
x
.1x/
e
x
4e
2x
e
x
.x2/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2
4
c1
c2
c3
3
5:
SettingxD0and imposing the initial conditions yields
2
4
4

3
2
19
3
5D
2
4
0

1
2
4
3
5C
2
4
1 1 0
1 2 1
1 42
3
5
2
4
c1
c2
c3
3
5:
Solving this system yieldsc1D 3,c2D 1,c3D4. Therefore,yD
e
x
.3xC1/x
2
3e
x
e
2x
C
4xe
x
.
9.4.26.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
e
x
1 2x e
x
0 2 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
De
x
.x
2
2xC2/;W1D
ˇ
ˇ
ˇ
ˇ
x
2
e
x
2x e
x
ˇ
ˇ
ˇ
ˇ
De
x
.x
2
2x/;W2D
ˇ
ˇ
ˇ
ˇ
x e
x
1 e
x
ˇ
ˇ
ˇ
ˇ
D
e
x
.x1/;W3D
ˇ
ˇ
ˇ
ˇ
x x
2
1 2x
ˇ
ˇ
ˇ
ˇ
Dx
2
;u
0
1
D
F W1
P0W
Dx.x2/;u
0
2
D
F W2
P0W
D1x;u
0
3
D
F W2
P0W
D
x
2
e
x
;ypDu1y1Cu2y2Cu3y3D
x
4
C6x
2
C12xC12
6
. Since
6x
2
C12x
6
is a solution of the
complementary equations we takeypD
x
4
C12
6
.

191. 186 Chapter 9Linear Higher Order Equations
The general solution isyD
x
4
C12
6
Cc1xCc2x
2
Cc3e
x
, so
2
4
y
y
0
y
00
3
5D
2
4
.x
4
C12/=6
2x
3
=3
2x
2
3
5C
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
e
x
1 2x e
x
0 2 e
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2
4
c1
c2
c3
3
5:
SettingxD0and imposing the initial conditions yields
2
4
0
5
0
3
5D
2
4
2
0
0
3
5C
2
4
0 0 1
1 0 1
0 2 1
3
5
2
4
c1
c2
c3
3
5:
Solving this system yieldsc1D3,c2D 1,c3D2. Therefore,yD
x
4
C12
6
C3xx
2
C2e
x
.
9.4.28.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 e
x
e
3x
1 e
x
3e
3x
0 e
x
9e
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D2e
4x
.3x1/;W1D
ˇ
ˇ
ˇ
ˇ
e
x
e
3x
e
x
3e
3x
ˇ
ˇ
ˇ
ˇ
D2e
4x
;W2D
ˇ
ˇ
ˇ
ˇ
xC1 e
3x
1 3e
3x
ˇ
ˇ
ˇ
ˇ
D
e
3x
.3xC2/;W3D
ˇ
ˇ
ˇ
ˇ
xC1 e
x
1 e
x
ˇ
ˇ
ˇ
ˇ
Dxe
x
;u
0
1
D
F W1
P0W
D2e
x
;u
0
2
D
F W2
P0W
D 3x2;
u
0
3
D
F W2
P0W
Dxe
2x
;u1D2e
x
;u2D
x.3xC4/
2
;u3D
e
2x
.2xC1/
4
;ypDu1y1C
u2y2Cu3y3D
e
x
.6x
2
C2x7/
4
. Since
7e
x
4
is a solution of the complementary equation we take
ypD
xe
x
.3xC1/
2
The general solution isyD
xe
x
.3xC1/
2
Cc1.xC1/Cc2e
x
Cc3e
2x
, so
2
4
y
y
0
y
00
3
5D
2
4
xe
x
.3xC1/=2
e
x
.3x
2
C7xC1/=2
e
x
.3x
2
C13xC8/=2
3
5C
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
xC1 e
x
e
3x
1 e
x
3e
3x
0 e
x
9e
3x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
:
SettingxD0and imposing the initial conditions yields
2
4
3
4
5
4
1
4
3
5D
2
4
0

1
2
4
3
5C
2
4
1 1 1
1 1 3
0 1 9
3
5
2
4
c1
c2
c3
3
5:
Solving this system yieldsc1D
1
2
,c2D
1
4
,c3D
1
2
. Therefore,yD
xe
x
.3xC1/
2
C
xC1
2

e
x
4
C
e
2x
2
.
9.4.30.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1
x
xlnx
1 2x
1
x
2
lnxC1
0 2
2
x
3
1
x
0 0
6
x
2

1
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
12
x
3
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
2
1
x
xlnx
2x
1
x
2
lnxC1
2
2
x
3
1
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D6
lnx
x

3
x
;

192. Section 9.4Variation of Parameters for Higher Order Equations187
W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
1
x
xlnx
1
1
x
2
lnxC1
0
2
x
3
1
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
4
x
2
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
xlnx
1 2xlnxC1
0 2
1
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D x;W4D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1
x
1 2x
1
x
2
0 2
2
x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6
x
;u
0
1
D
F W1
P0W
D9lnx=2
9
4
;u
0
2
D
F W2
P0W
D
3
x
;u
0
3
D
F W2
P0W
D
3x
2
4
;u
0
4
D
F W4
P0W
D
9
2
;
u1D
9xlnx
2

27x
4
;u2D3lnx;u3D
x
3
4
;u4D
9x
2
;ypDu1y1Cu2y2Cu3y3D3x
2
lnx7x
2
.
Since7x
2
satisfies the complementary equation we takeypD3x
2
lnx.
The general solution isyD3x
2
lnxCc1xCc2x
2
C
c3
x
Cc4xlnx, so
2
6
6
4
y
y
0
y
00
y
000
3
7
7
5
D
2
6
6
6
4
3x
2
lnx
6xlnxC3x
6lnxC9
6
x
3
7
7
7
5
C
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x x
2
1
x
xlnx
1 2x
1
x
2
lnxC1
0 2
2
x
3
1
x
0 0
6
x
2

1
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2
6
6
4
c1
c2
c3
c4
3
7
7
5
:
SettingxD1and imposing the initial conditions yields
2
6
6
4
7
11
5
6
3
7
7
5
D
2
6
6
4
0
3
9
6
3
7
7
5
C
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 0
1 21 1
0 2 2 1
0 061
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2
6
6
4
c1
c2
c3
c4
3
7
7
5
:
Solving this system yieldsc1D0,c2D 7,c3D0,c4D0. Therefore,yD3x
2
lnx7x
2
.
9.4.32.WD
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
p
x 1=x 1=
p
x
1 1=2
p
x1=x
2
1=2x
3=2
01=4x
3=2
2=x
3
3=4x
5=2
0 3=8x
5=2
6=x
4
15=8x
7=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 9=8x
6
;W1D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
p
x 1=x 1=
p
x
1=2
p
x1=x
2
1=2x
3=2
1=4x
3=2
2=x
3
3=4x
5=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
3=4x
4
;W2D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x 1=x 1=
p
x
11=x
2
1=2x
3=2
0 2=x
3
3=4x
5=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D3=2x
7=2
;W3D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
p
x 1=
p
x
1 1=2
p
x1=2x
3=2
01=4x
3=2
3=4x
5=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 3=4x
2
;
W4D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
p
x 1=x
1 1=2
p
x1=x
2
01=4x
3=2
2=x
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 3=2x
5=2
;u
0
1
D
F W1
P0W
D1=x;u
0
2
D
F W2
P0W
D 2=
p
x;
u
0
3
D
F W2
P0W
D x;u
0
4
D
F W4
P0W
D2
p
x;u1Dlnx;u2D 4
p
x;u3D x
2
=2;u4D4x
3=2
=3;
ypDu1y1Cu2y2Cu3y3Dxlnx19x=6. since19x=6satisfies the complementary equation we
takeypDxlnx.
The general solution isyDxlnxCc1xCc2
p
xCc3=xCc4=
p
x, so
2
6
6
4
y
y
0
y
00
y
000
3
7
7
5
D
2
6
6
4
xlnx
lnxC1
1=x
1=x
2
3
7
7
5
C
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
p
x 1=x 1=
p
x
1 1=2
p
x1=x
2
1=2x
3=2
01=4x
3=2
2=x
3
3=4x
5=2
0 3=8x
5=2
6=x
4
15=8x
7=2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2
6
6
4
c1
c2
c3
c4
3
7
7
5
:

193. 188 Chapter 9Linear Higher Order Equations
SettingxD1and imposing the initial conditions yields
2
6
6
4
2
0
4
37
4
3
7
7
5
D
2
6
6
4
0
1
1
1
3
7
7
5
C
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 1 1 1
1 1=2 11=2
01=4 2 3=4
0 3=8 615=8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
2
6
6
4
c1
c2
c3
c4
3
7
7
5
:
Solving this system yieldsc1D1,c2D 1,c3D1,c4D1. Therefore,yDxlnxCx
p
xC
1
x
C
1
p
x
.
9.4.34.(a)Sinceu
0
j
D.1/
nj
F Wj
P0W
(1jn), the argument used in the derivation of the method of
variation of parameters implies thatypis a solution of (A).
(b)Follows immediately from(a), sinceuj.x0/D0,jD1; 2; : : : ; n.
(c)Expand the determinant in cofactors of itsnth row.
(d)Just differentiate the determinantn1times.
(e)If0jn2, then
@
j
G.x; t/
@x
j
ˇ
ˇ
ˇ
ˇ
xDt
has two identical rows, and is therefore zero, while
@
n1
G.x; t/
@x
j
ˇ
ˇ
ˇ
ˇ
xDt
DW.t/
(f)Sinceyp.x/D
Z
x
x0
G.x; t/F.t/ dt,y
0
p
.x/DG.x; x/F.x/C
Z
x
x0
@G.x; t/
@x
F.t/ dt. ButG.x; x/D
0from(e), soy
0
p
.x/D
Z
x
x0
@G.x; t/
@x
F.t/ dt. Repeating this argument forjD1; : : : ; nand invoking
(e)each time yields the conclusion.
9.4.36.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
y1.t/ y1.t/ y2.t/
y
0
1
.t/ y
0
1
.t/ y
0
2
.t/
y1.x/ y1.x/ y2.x/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
t t
2
1=t
1 2t1=t
2
x x
2
1=x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
t
2
1=t
2t1=t
2
ˇ
ˇ
ˇ
ˇ
x
2
ˇ
ˇ
ˇ
ˇ
t 1=t
11=t
2
ˇ
ˇ
ˇ
ˇ
C
1
x
ˇ
ˇ
ˇ
ˇ
t t
2
1 2t
ˇ
ˇ
ˇ
ˇ
D 3xC2
x
2
t
C
t
2
x
D
.xt/
2
.2xCt/
xt
:
SinceP0.t/Dt
3
andW.t/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
t t
2
1=t
1 2t1=t
2
x x
2
1=x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6
t
,G.x; t/D
.xt/
2
.2xCt/
6xt
3
, soypD
Z
x
x0
.xt/
2
.2xCt/
6xt
3
F.t/ dt.
9.4.38.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
y1.t/ y1.t/ y2.t/
y
0
1
.t/ y
0
1
.t/ y
0
2
.t/
y1.x/ y1.x/ y2.x/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
t 1=t e
t
=t
11=t
2
e
t
.1=t1=t
2
/
x 1=x e
x
=x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Dx
ˇ
ˇ
ˇ
ˇ
1=t e
t
=t
1=t
2
e
t
.1=t1=t
2
/
ˇ
ˇ
ˇ
ˇ

1
x
ˇ
ˇ
ˇ
ˇ
t e
t
=t
1 e
t
.1=t1=t
2
/
ˇ
ˇ
ˇ
ˇ
C
e
x
x
ˇ
ˇ
ˇ
ˇ
t 1=t
11=t
2
ˇ
ˇ
ˇ
ˇ
D
xe
t
t
2

e
t
.t2/
xt

2e
x
xt
D
x
2
e
t
e
t
t.t2/2te
x
xt
2

194. Section 9.4Variation of Parameters for Higher Order Equations189
SinceP0.t/Dt.1t/andW.t/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
t 1=t e
t
=t
11=t
2
e
t
.1=t1=t
2
/
0 2=t
3
e
t
.1=t2=t
2
C2=t
3
/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
2e
t
.1t/
t
3
,G.x; t/D
x
2
t.t2/2te
.xt /
2x.t1/
2
, soypD
Z
x
x0
x
2
t.t2/2te
.xt /
2x.t1/
2
F.t/ dt.
9.4.40.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
y1.t/ y1.t/ y2.t/ y3.t/
y
0
1
.t/ y
0
1
.t/ y
0
2
.t/ y
0
3
.t/
y
00
1
.t/ y
00
1
.t/ y
00
2
.t/ y
00
3
.t/
y1.x/ y1.x/ y2.x/ y3.x/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t t
2
1=t
0 1 2t 1=t
2
0 0 2 2=t
3
1 x x
2
1=x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
t t
2
1=t
1 2t1=t
2
0 2 2=t
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Cx
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t
2
1=t
0 2t1=t
2
0 2 2=t
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
x
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t 1=t
0 11=t
2
0 0 2=t
3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
C
1
x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t t
2
0 1 2t
0 0 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
6
t
C
6x
t
2

2x
2
t
3
C
2
x
D
2.tx/
3
xt
3
:
SinceP0.t/DtandW.t/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t t
2
1=t
0 1 2t1=t
2
0 0 2 2=t
3
0 0 0 6=t
4
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
12
t
4
,G.x; t/D
.xt/
3
6x
, soypD
Z
x
x0
.xt/
3
6x
F.t/ dt.
9.4.42.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
y1.t/ y1.t/ y2.t/ y3.t/
y
0
1
.t/ y
0
1
.t/ y
0
2
.t/ y
0
3
.t/
y
00
1
.t/ y
00
1
.t/ y
00
2
.t/ y
00
3
.t/
y1.x/ y1.x/ y2.x/ y3.x/
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t
2
e
2t
e
2t
0 2t 2e
2t
2e
2t
0 2 4e
2t
4e
2t
1 x
2
e
2x
e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
t
2
e
2t
e
2t
2t 2e
2t
2e
2t
2 4e
2t
4e
2t
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Cx
2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 e
2t
e
2t
0 2e
2t
2e
2t
0 4e
2t
4e
2t
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
e
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t
2
e
2t
0 2t2e
2t
0 2 4e
2t
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Ce
2x
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t
2
e
2t
0 2t 2e
2t
0 2 4e
2t
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .16t
2
8/C16x
2
e
2.xt /t
.8tC4/Ce
2.xt /
.8t4/:
SinceP0.t/DtandW.t/D
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 t
2
e
2t
e
2t
0 2t 2e
2t
2e
2t
0 2 4e
2t
4e
2t
0 0 8e
2t
8e
2t
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D 128t,
G.x; t/D
e
2.xt /
.1C2t/Ce
2.xt /
.12t/4x
2
C4t
2
2
32t
2
, so
ypD
Z
x
x0
e
2.xt /
.1C2t/Ce
2.xt /
.12t/4x
2
C4t
2
2
32t
2
F.t/ dt.

196. CHAPTER10
LinearSystemsofDifferential
Equations
10.1INTRODUCTION TO SYSTEMS OF DIFFERENTIAL EQUATIONS
10.1.2.Q
0
1
D.rate in/1.rate out/1andQ
0
2
D.rate in/2.rate out/2.
The volumes of the solutions inT1andT2areV1.t/D100C2tandV2.t/D100C3t, respectively.
T1receives salt from the external source at the rate of (2 lb/gal)fi(6 gal/min)D12 lb/min, and from
T2at the rate of (lb/gal inT2/fi(1 gal/min)D
1
100C3t
Q2lb/min. Therefore, (A) (rate in)
1D12C
1
100C3t
Q2. Solution leavesT1at 5 gal/min, since 3 gal/min are drained and 2 gal/min are pumped to
T2; hence (B).rate out/1D.lb/gal in T
1/fi(5 gal/min)D
1
100C2t
Q1fi5D
5
100C2t
Q1. Now (A)
and (B) imply that (C)Q
0
1
D12
5
100C2t
Q1C
1
100C3t
Q2.
T2receives salt from the external source at the rate of (1 lb/gal)fi(5 gal/min)D5 lb/min, and from
T1at the rate of (lb/gal inT1/fi(2 gal/min)D
1
100C2t
Q1fi2D
1
50Ct
Q1lb/min. Therefore, (D)
(rate in)
2D5C
1
50Ct
Q1. Solution leavesT2at 4 gal/min, since 3 gal/min are drained and 1 gal/min is
pumped toT1; hence (E).rate out/2D.lb/gal in T
2/fi(4 gal/min)D
1
100C3t
Q2fi4D
4
100C3t
Q2.
Now (D) and (E) imply that (F)Q
0
2
D5C
1
50Ct
Q1
4
100C3t
Q2. Now (C) and (F) form the desired
system.
10.1.4.mX
00
D ˛X
0
mgR
2
X
kXk
3
; see Example 10.1.3.10.1.8.
I1iDg1.ti; y1i; y2i/;
J1iDg2.ti; y1i; y2i/;
I2iDg1.tiCh; y1iChI1i; y2iChJ1i/ ;
J2iDg2.tiCh; y1iChI1i; y2iChJ1i/ ;
y1;iC1Dy1iC
h
2
.I1iCI2i/;
y2;iC1Dy2iC
h
2
.J1iCJ2i/:
191

197. 192 Chapter 10Linear Systems of Differential Equations
10.2LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS
10.2.6.LetyiDy
.i1/
,iD1; 2; : : : ; n; theny
0
i
DyiC1,iD1; 2; : : :; n1andP0.t/y
0
n
CP1.t/ynC
CPn.t/y1DF.t/, so
AD
1
P0
2
6
6
6
6
6
4
0 1 0 0
0 0 1 0
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
0 0 0 1
PnPn1Pn2 P1
3
7
7
7
7
7
5
andfD
1
P0
2
6
6
6
6
6
4
0
0
:
:
:
0
F
3
7
7
7
7
7
5
:
IfP0; P1; : : : ; PnandFare continuous andP0has no zeros on.a; b/, thenP1=P0; : : : ; Pn=P0andF=P0
are contnuous on.a; b/.
10.2.7.(a).c1PCc2Q/
0
ij
D.c1pijCc2qij/
0
Dc1p
0
ij
Cc2q
0
ij
D.c1P
0
Cc2Q
0
/ij; hence.c1PCc2Q/
0
D
c1P
0
Cc2Q
0
.
(b)LetPbekfirandQberfis; thenPQiskfisand.PQ/ijD
r
X
łD1
piłqłj. Therefore,.PQ/
0
ij
D
r
X
łD1
p
0
ił
qłjC
r
X
łD1
piłq
0
łj
D.P
0
Q/ijC.PQ
0
/ij. Therefore,.PQ/
0
DP
0
QCPQ
0
.
10.2.10.(a)From Exercise 10.2.7(b)withPDQDX,.X
2
/
0
D.XX/
0
DX
0
XCXX
0
.
(b)By starting from Exercise 10.2.7(b)and using induction it can be shown ifP1; P2; : : : ; Pnare
square matrices of the same order, then.P1P2 Pn/
0
DP
0
1
P2 PnCP1P
0
2
PnC CP1P2 P
0
n
.
TakingP1DP2D DPnDXyields (A).Y
n
/
0
DY
0
Y
n1
CY Y
0
Y
n2
CY
2
Y
0
Y
n3
C C
Y
n1
Y
0
D
n1
X
rD0
Y
r
Y
0
Y
nr1
.
(c)IfYis a scalar function, then (A) reduces to the familiar result.Y
n
/
0
DnY
n1
Y
0
.
10.2.12.From Exercise 10.2.6, the initial value problem (A)P0.x/y
.n/
CP1.x/y
.n1/
C CPn.x/yD
F.x/,y.x0/Dk0; y
0
.x0/Dk1; : : : ; y
.n1/
.x0/Dkn1is equivalent to the initial value problem (B)
y
0
DA.t/yCf.t/, with
AD
1
P0
2
6
6
6
6
6
4
0 1 0 0
0 0 1 0
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
0 0 0 1
PnPn1Pn2 P1
3
7
7
7
7
7
5
;fD
1
P0
2
6
6
6
6
6
4
0
0
:
:
:
0
F
3
7
7
7
7
7
5
;andkD
2
6
6
6
4
k0
k1
:
:
:
kn1
3
7
7
7
5
:
Since Theorem 10.2.1 implies that (B) has a unique solution on.a; b/, it follows that (A) does also.
10.3BASIC THEORY OF HOMOGENEOUS LINEAR SYSTEM
10.3.2.(a)The system equivalent of (A) is (B)y
0
D
1
P0.x/

0 1
P2.x/ P1.x/

y, whereyD

y
y
0

.
Lety1D

y1
y
0
1

andy1D

y2
y
0
2

. Then the Wronskian offy1;y2gas defined in this section is

y1y2
y
0
1
y
0
2

DW.

198. Section 10.3Basic Theory of Homogeneous Linear System193
(b)The trace of the matrix in (B) isP1.x/=P0.x/, so Eqn. 10.3.6 implies thatW.x/DW.x0/exp


Z
x
x0
P1.s/
P0.s/
ds

.
10.3.4.(a)See the solution of Exercise 9.1.18.
(c)
ˇ
ˇ
ˇ
ˇ
y
0
11
y
0
12
y21y22
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
a11y11Ca12y21a11y12Ca12y22
y21 y22
ˇ
ˇ
ˇ
ˇ
Da11
ˇ
ˇ
ˇ
ˇ
y11y12
y21y22
ˇ
ˇ
ˇ
ˇ
Ca12
ˇ
ˇ
ˇ
ˇ
y21y22
y21y22
ˇ
ˇ
ˇ
ˇ
D
a11WCa120Da11W. Similarly,

y11y12
y
0
21
y
0
22

Da22W.
10.3.6.(a)From the equivalence of Theorem 10.3.3(b)and(e),Y.t0/is invertible.
(b)From the equivalence of Theorem 10.3.3(a)and(b), the solution of the initial value problem is
yDY.t/c, wherecis a constant vector. To satisfyy.t0/Dk, we must haveY.t0/cDk, socDY
1
.t0/k
andyDY
1
.t0/Y.t/k.
10.3.8.(b) yD

e
4t
e
4t

Cc2

2e
3t
5e
3t

where
c12c2D10
c1C5c2D 4
;soc1D6,c2D 2, and
yD

6e
4t
C4e
3t
6e
4t
10e
3t

.
(c)Y.t/D

e
4t
2e
3t
e
4t
5e
3t

;Y.0/D

12
1 5

;Y
1
.0/D
1
7

5 2
1 1

;yDY.t/Y
1
.0/kD
1
7

5e
4t
C2e
3t
2e
4t
2e
3t
5e
4t
5e
3t
2e
4t
C5e
3t

k.
10.3.10.(b) y1Dc1

e
3t
e
3t

Cc2

e
t
e
t

, where
c1Cc2D2
c1c2D8
;soc1D5,c2D 3, and
yD

5e
3t
3e
t
5e
3t
C3e
t

.
(c)Y.t/D

e
3t
e
t
3e
3t
e
t

;Y.0/D

1 1
11

;
Y
1
.0/D
1
2

1 1
11

;yDY.t/Y
1
.0/kD
1
2

e
3t
Ce
t
e
3t
e
t
e
3t
e
t
e
3t
Ce
t

k.
10.3.12.(b) yDc1
2
4
e
2t
0
e
2t
3
5Cc2
2
4
e
2t
e
2t
0
3
5Cc3
2
4
e
4t
e
4t
e
4t
3
5, where
c1c2Cc3D 0
c2Cc3D 9
c1Cc3D12
;so
c1D11,c2D 10,c3D1, andyD
1
3
2
4
e
2t
Ce
4t
10e
2t
Ce
4t
11e
2t
Ce
4t
3
5.
(c)Y.t/D
2
4
e
2t
e
2t
e
4t
0 e
2t
e
4t
e
2t
0 e
4t
3
5;Y.0/D
2
4
11 1
0 1 1
1 0 1
3
5;Y
1
.0/D
1
3
2
4
11 2
1 2 1
1 1 1
3
5;
yDY.t/Y
1
.0/kD
1
3
2
4
2e
2t
Ce
4t
e
2t
Ce
4t
e
2t
Ce
4t
e
2t
Ce
4t
2e
2t
Ce
4t
e
2t
Ce
4t
e
2t
Ce
4t
e
2t
Ce
4t
2e
2t
Ce
4t
3
5k.
10.3.14.IfYandZare both fundamental matrices fory
0
DA.t/y, thenZDC Y, whereCis a constant
invertible matrix. Therefore,ZY
1
DCandY Z
1
DC
1
.
10.3.16.(a)The Wronskian offy1;y2; : : : ;yngequals one whentDt0. Apply Theorem 10.3.3.
(b)LetYbe the matrix with columnsfy1;y2; : : : ;yng. From(a),Yis a fundamental matrix for
y
0
DA.t/yon.a; b/. From Exercise 10.3.15(b), so isZDY CifCis any invertible constant matrix.

199. 194 Chapter 10Linear Systems of Differential Equations
10.3.18.(a)€
0
1
.t/DZ
0
.t/Z.s/DAZ.t/Z.s/DA€.t/and€1.0/DZ.s/, sinceZ.0/DI.€
0
2
.t/D
Z
0
.tCs/DAZ.tCs/DA€2.t/(sinceAis constant) and€2.0/DZ.s/. Applying Theorem 10.2.1 to
the columns of€1and€2shows that€1D€2.
(b)WithsD t,(a)implies thatZ.t/Z.t/DZ.0/DI; therefore.Z.t//
1
DZ.t/.
(c)e
0A
DIis analogous toe
oa
De
0
D1whenais a scalar, whilee
.tCs/A
De
tA
e
sA
is analagous
toe
.tCs/a
De
t a
e
sa
whenais a scalar.
10.4CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS I
10.4.2.
1
4
ˇ
ˇ
ˇ
ˇ
54 3
3 54
ˇ
ˇ
ˇ
ˇ
D.C1=2/.C2/. Eigenvectors associated with1D 1=2satisfy

3 3
4 4

x1
x2

D

0
0

, sox1Dx2. Takingx2D1yieldsy1D

1
1

e
t =2
. Eigenvectors
associated with2D 2satisfy

3
4
3
4
1 1

x1
x2

D

0
0

, sox1D x2. Takingx2D1yields
y2D

1
1

e
2t
. HenceyDc1

1
1

e
t =2
Cc2

1
1

e
2t
.
10.4.4.
ˇ
ˇ
ˇ
ˇ
14
11
ˇ
ˇ
ˇ
ˇ
D.1/.C3/. Eigenvectors associated with1D 3satisfy

24
12

x1
x2

D

0
0

, sox1D2x2. Takingx2D1yieldsy1D

2
1

e
3t
. Eigenvectors
associated with2D1satisfy

24
12

x1
x2

D

0
0

, sox1D 2x2. Takingx2D1yields
y2D

2
1

e
t
. HenceyDc1

2
1

e
3t
Cc2

2
1

e
t
.
10.4.6.
ˇ
ˇ
ˇ
ˇ
43
21
ˇ
ˇ
ˇ
ˇ
D.2/.1/. Eigenvectors associated with1D2satisfy

23
23

x1
x2

D

0
0

,
sox1D
3
2
x2. Takingx2D2yieldsy1D

3
2

e
2t
. Eigenvectors associated with2D1sat-
isfy

33
22

x1
x2

D

0
0

, sox1Dx2. Takingx2D1yieldsy2D

1
1

e
t
. HenceyD
c1

3
2

e
2t
Cc2

1
1

e
t
.
10.4.8.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
11 2
123
4 1 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .C3/.C1/.2/. The eigenvectors associated with
with1D 3satisfy the system with augmented matrix
2
6
6
6
4
412
:
:
: 0
1 1 3
:
:
: 0
4 1 2
:
:
: 0
3
7
7
7
5
, which is row
equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 12
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3andx2D2x3. Takingx3D1yieldsy1D

200. Section 10.4Constant Coefficient Homogeneous Systems I195
2
4
1
2
1
3
5e
3t
. The eigenvectors associated with with2D 1satisfy the system with augmented ma-
trix
2
6
6
6
4
212
:
:
: 0
113
:
:
: 0
4 1 0
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 1
:
:
: 0
0 1 4 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D x3
andx2D 4x3. Takingx3D1yieldsy2D
2
4
1
4
1
3
5e
t
. The eigenvectors associated with with
3D2satisfy the system with augmented matrix
2
6
6
6
4
112
:
:
: 0
143
:
:
: 0
4 1 3
:
:
: 0
3
7
7
7
5
, which is row equivalent
to
2
6
6
6
4
1 0 1
:
:
: 0
0 1 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D x3andx2D x3. Takingx3D1yieldsy3D
2
4
1
1
1
3
5e
2t
.
HenceyDc1
2
4
1
2
1
3
5e
3t
Cc2
2
4
1
4
1
3
5e
t
Cc3
2
4
1
1
1
3
5e
2t
.
10.4.10.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
3 5 8
112
1 11
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .1/.C2/.2/. The eigenvectors associated with
with1D1satisfy the system with augmented matrix
2
6
6
6
4
2 5 8
:
:
: 0
122
:
:
: 0
112
:
:
: 0
3
7
7
7
5
, which is row equiv-
alent to
2
6
6
6
4
1 0
2
3
:
:
: 0
0 1
4
3
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D
2
3
x3andx2D
4
3
x3. Takingx3D3yieldsy1D
2
4
2
4
3
3
5e
t
. The eigenvectors associated with with2D 2satisfy the system with augmented ma-
trix
2
6
6
6
4
5 5 8
:
:
: 0
1 1 2
:
:
: 0
11 1
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 1 0
:
:
: 0
0 0 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D x2
andx3D0. Takingx2D1yieldsy2D
2
4
1
1
0
3
5e
2t
. The eigenvectors associated with with

201. 196 Chapter 10Linear Systems of Differential Equations
3D2satisfy the system with augmented matrix
2
6
6
6
4
1 5 8
:
:
: 0
132
:
:
: 0
113
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0
7
4
:
:
: 0
0 1
5
4
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D
7
4
x3andx2D
5
4
x3. Takingx3D4yieldsy3D
2
4
7
5
4
3
5e
2t
.
HenceyDc1
2
4
2
4
3
3
5e
t
Cc2
2
4
1
1
0
3
5e
2t
Cc3
2
4
7
5
4
3
5e
2t
.
10.4.12.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
41 4
432
1 11
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .3/.C2/.C1/. The eigenvectors associated with
1D3satisfy the system with augmented matrix
2
6
6
6
4
114
:
:
: 0
462
:
:
: 0
114
:
:
: 0
3
7
7
7
5
, which is row equiva-
lent to
2
6
6
6
4
1 011
:
:
: 0
0 17
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D11x3andx2D7x3. Takingx3D1yieldsy1D
2
4
11
7
1
3
5e
3t
. The eigenvectors associated with2D 2satisfy the system with augmented matrix
2
6
6
6
4
614
:
:
: 0
412
:
:
: 0
11 1
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 12
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3andx2D
2x3. Takingx3D1yieldsy2D
2
4
1
2
1
3
5e
2t
. The eigenvectors associated with3D 1satisfy the sys-
tem with augmented matrix
2
6
6
6
4
514
:
:
: 0
422
:
:
: 0
11 0
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
.
Hencex1Dx3andx2Dx3. Takingx3D1yieldsy3D
2
4
1
1
1
3
5e
t
. HenceyDc1
2
4
11
7
1
3
5e
3t
Cc2
2
4
1
2
1
3
5e
2t
Cc3
2
4
1
1
1
3
5e
t
.
10.4.14.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
3 2 2
2 72
10 10 5
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .C5/.5/
2
. The eigenvectors associated with1D

202. Section 10.4Constant Coefficient Homogeneous Systems I197
5satisfy the system with augmented matrix
2
6
6
6
4
8 22
:
:
: 0
2 122
:
:
: 0
10 10 0
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0
1
5
:
:
: 0
0 1
1
5
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D
1
5
x3andx2D
1
5
x3. Takingx3D5yieldsy1D
2
4
1
1
5
3
5e
5t
. The
eigenvectors associated with2D5satisfy the system with augmented matrix
2
6
6
6
4
2 2 2
:
:
: 0
2 2 2
:
:
: 0
10 1010
:
:
: 0
3
7
7
7
5
,
which is row equivalent to
2
6
6
6
4
11 1
:
:
: 0
0 0 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx2x3. Takingx2D0andx3D
1yieldsy2D
2
4
1
0
1
3
5e
5t
. Takingx2D1andx3D0yieldsy3D
2
4
1
1
0
3
5e
5t
. HenceyD
c1
2
4
1
1
5
3
5e
5t
Cc2
2
4
1
0
1
3
5e
5t
Cc3
2
4
1
1
0
3
5e
5t
.
10.4.16.
ˇ
ˇ
ˇ
ˇ
7 4
6 7
ˇ
ˇ
ˇ
ˇ
D.5/.C5/. Eigenvectors associated with1D5satisfy

12 4
6 2

x1
x2

D

0
0

,
sox1D
x2
3
. Takingx2D3yieldsy1D

1
3

e
5t
. Eigenvectors associated with2D5satisfy

2 4
6 12

x1
x2

D

0
0

, sox1D2x2. Takingx2D1yieldsy2D

2
1

e
5t
. The general
solution isyDc1

1
3

e
5t
Cc2

2
1

e
5t
. Nowy.0/D

2
4

)c1

1
3

Cc2

2
1

D

2
4

,
soc1D 2andc2D2. Therefore,yD

2
6

e
5t
C

4
2

e
5t
.
10.4.18.
ˇ
ˇ
ˇ
ˇ
2112
2415
ˇ
ˇ
ˇ
ˇ
D.9/.C3/. Eigenvectors associated with1D9satisfy

1212
2424

x1
x2

D

0
0

, sox1Dx2. Takingx2D1yieldsy1D

1
1

e
9t
. Eigenvec-
tors associated with2D 3

2412
2412

x1
x2

D

0
0

, sox1D
1
2
x2. Takingx2D2yields
y2D

1
2

e
3t
. The general solution isyDc1

1
1

e
9t
Cc2

1
2

e
3t
. Nowy.0/D

5
3

)
c1

1
1

Cc2

1
2

D

5
3

, soc1D7andc2D 2. Therefore,yD

7
7

e
9t


2
4

e
3t
.

203. 198 Chapter 10Linear Systems of Differential Equations
10.4.20.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1
6

1
3
0
2
3

1
6
0
0 0
1
2

ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .C1=2/.1=2/
2
. The eigenvectors associated with with
1D 1=2satisfy the system with augmented matrix
2
6
6
6
4
2
3
1
3
0
:
:
: 0
2
3
1
3
0
:
:
: 0
0 0 1
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1
1
2
0
:
:
: 0
0 0 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D
x2
2
andx3D0. Takingx2D2yieldsy1D
2
4
1
2
0
3
5e
t =2
.
The eigenvectors associated with with2D3D1=2satisfy the system with augmented matrix
2
6
6
6
4

1
3
1
3
0
:
:
: 0
2
3

2
3
0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
11 0
:
:
: 0
0 0 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx2and
x3is arbitrary. Takingx2D1andx3D0yieldsy2D
2
4
1
1
0
3
5e
t =2
. Takingx2D0andx3D1yields
y3D
2
4
0
0
1
3
5e
t =2
. The general solution isyDc1
2
4
1
2
0
3
5e
t =2
Cc2
2
4
1
1
0
3
5e
t =2
Cc3
2
4
0
0
1
3
5e
t =2
.
Nowy.0/D
2
4
4
7
1
3
5)c1
2
4
1
2
0
3
5Cc2
2
4
1
1
0
3
5Cc3
2
4
0
0
1
3
5e
t =2
D
2
4
4
7
1
3
5, soc1D1,c2D5, and
c3D1. HenceyD
2
4
1
2
0
3
5e
t =2
C
2
4
5
5
0
3
5e
t =2
C
2
4
0
0
1
3
5e
t =2
.
10.4.22.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
63 8
2 12
3 35
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .1/.C2/.3/. The eigenvectors associated with
1D1satisfy the system with augmented matrix
2
6
6
6
4
538
:
:
: 0
2 0 2
:
:
: 0
336
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 1 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3andx2D x3. Takingx3Dyieldsy1D
2
4
1
1
1
3
5e
t
. The
eigenvectors associated with2D 2satisfy the system with augmented matrix
2
6
6
6
4
838
:
:
: 0
2 3 2
:
:
: 0
333
:
:
: 0
3
7
7
7
5
,

204. Section 10.4Constant Coefficient Homogeneous Systems I199
which is row equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 1 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3andx2D0. Takingx3D1yields
y2D
2
4
1
0
1
3
5e
2t
. The eigenvectors associated with3D3satisfy the system with augmented matrix
2
6
6
6
4
338
:
:
: 0
222
:
:
: 0
338
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
11 0
:
:
: 0
0 0 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx2andx3D
0. Takingx2D1yieldsy3D
2
4
1
1
0
3
5e
3t
. The general solution isyDc1
2
4
1
1
1
3
5e
t
Cc2
2
4
1
0
1
3
5e
2t
C
c3
2
4
1
1
0
3
5e
3t
. Nowy.0/D
2
4
0
1
1
3
5)c1
2
4
1
1
1
3
5Cc2
2
4
1
0
1
3
5Cc3
2
4
1
1
0
3
5D
2
4
0
1
1
3
5, soc1D2,
c2D 3, andc3D1. Therefore,yD
2
4
2
2
2
3
5e
t

2
4
3
0
3
3
5e
2t
C
2
4
1
1
0
3
5e
3t
.
10.4.24.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
3 0 1
112 7
1 0 3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .2/.C2/.4/. The eigenvectors associated with
with1D2satisfy the system with augmented matrix
2
6
6
6
4
1 0 1
:
:
: 0
114 7
:
:
: 0
1 0 1
:
:
: 0
3
7
7
7
5
, which is row equiv-
alent to
2
6
6
6
4
1 0 1
:
:
: 0
0 1 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D x3andx2D x3. Takingx3D1yieldsy1D
2
4
1
1
1
3
5e
2t
. The eigenvectors associated with with2D 2satisfy the system with augmented ma-
trix
2
6
6
6
4
5 0 1
:
:
: 0
11 0 7
:
:
: 0
1 0 5
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 0
:
:
: 0
0 0 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3D0
andx2is arbitrary. Takingx3D1yieldsy2D
2
4
0
1
0
3
5e
2t
. The eigenvectors associated with with

205. 200 Chapter 10Linear Systems of Differential Equations
3D4satisfy the system with augmented matrix
2
6
6
6
4
1 0 1
:
:
: 0
116 7
:
:
: 0
1 0 1
:
:
: 0
3
7
7
7
5
, which is row equivalent
to
2
6
6
6
4
1 01
:
:
: 0
0 13
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3andx2D3x3. Takingx3D1yieldsy3D
2
4
1
3
1
3
5e
4t
.
The general solution isyDc1
2
4
1
1
1
3
5e
2t
Cc2
2
4
0
1
0
3
5e
2t
Cc3
2
4
1
3
1
3
5e
4t
. Nowy.0/D
2
4
2
7
6
3
5)
c1
2
4
1
1
1
3
5Cc2
2
4
0
1
0
3
5Cc3
2
4
1
3
1
3
5D
2
4
2
7
6
3
5, soc1D2,c2D 3, andc3D4. HenceyD
2
4
2
2
2
3
5e
2t

2
4
0
3
0
3
5e
2t
C
2
4
4
12
4
3
5e
4t
10.4.26.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
31 0
42 0
4 4 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .C1/.2/
2
. The eigenvectors associated with1D 1sat-
isfy the system with augmented matrix
2
6
6
6
4
41 0
:
:
: 0
41 0
:
:
: 0
44 3
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0
1
4
:
:
: 0
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
Hencex1Dx2=4andx2Dx3. Takingx3D4yieldsy1D
2
4
1
4
4
3
5e
t
. The eigenvectors associated
with with2D3D2satisfy the system with augmented matrix
2
6
6
6
4
11 0
:
:
: 0
44 0
:
:
: 0
44 0
:
:
: 0
3
7
7
7
5
, which is row
equivalent to
2
6
6
6
4
11 0
:
:
: 0
0 0 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx2andx3is arbitrary. Takingx2D1andx3D0
yieldsy2D
2
4
1
1
0
3
5e
2t
. Takingx2D0andx3D1yieldsy3D
2
4
0
0
1
3
5e
2t
. The general solution is
yDc1
2
4
1
4
4
3
5e
t
Cc2
2
4
1
1
0
3
5e
2t
Cc3
2
4
0
0
1
3
5e
2t
. Nowy.0/D
2
4
7
10
2
3
5)c1
2
4
1
4
4
3
5Cc2
2
4
1
1
0
3
5C
c3
2
4
0
0
1
3
5D
2
4
7
10
2
3
5, soc1D1,c2D6, andc3D 2. HenceyD
2
4
1
4
4
3
5e
t
C
2
4
6
6
2
3
5e
2t
.

206. Section 10.5Constant Coefficient Homogeneous Systems II201
10.4.28.(a)Ify.t0/D0, thenyis the solution of the initial value problemy
0
DAy;y.t0/D0. Since
y0is a solution of this problem, Theorem 10.2.1 implies the conclusion.
(b)It is given thaty
0
1
.t/DAy1.t/for allt. Replacingtbytshows thaty
0
1
.t/DAy1.t/D
Ay2.t/for allt. Sincey
0
2
.t/Dy
0
1
.t/by the chain rule, this implies thaty
0
2
.t/DAy2.t/for allt.
(c)Ifz.t/Dy1.t/, thenz.t2/Dy1.t1/Dy2.t2/; thereforezandy2are both solutions of the initial
value problemy
0
DAy;y.t2/Dk, wherekDy2.t2/.
10.4.42.The characteristic polynomial ofAisp./D
2
.aCb/Cab˛ˇ, so the eigenvalues of
Aare1D
aCb
2
and1D
aCbC
2
, whereD
p
.ab/
2
C4˛ˇ;x1D

baC
2ˇ

and
x2D

ba
2ˇ

are associated eigenvectors. Since >jbaj, ifL1andL2are lines through
the origin parallel tox1andx2, thenL1is in the first and third quadrants andL2is in the second and
fourth quadrants. The slope ofL1isD
2ˇ
baC
> 0. IfQ0DP0there are three possibilities:
(i) if˛ˇDab, then1D0andP.t/DP0,Q.t/DQ0for allt > 0; (ii) if˛ˇ < ab, then
1> 0and limt!1P.t/Dlimt!1Q.t/D 1(monotonically); (iii) if˛ˇ > ab, then1< 0and
limt!1P.t/Dlimt!1Q.t/D0(monotonically). Now supposeQ0¤P0, so that the trajectory
cannot intersectL1, and assume for the moment that (A) makes sense for allt > 0; that is, even if
one or the other ofPandQis negative. Since2> 0it follows that either limt!1P.t/D 1or
limt!1Q.t/D 1(or both), and the trajectory is asymptotically parallel toL2. Therefore,the trajectory
must cross into the third quadrant (soP.T /D0andQ.T / > 0for some finiteT) ifQ0> P0, or into
the fourth quadrant (soQ.T /D0andP.T / > 0for some finiteT) ifQ0< P0.
10.5CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS II
10.5.2.
ˇ
ˇ
ˇ
ˇ
1
12
ˇ
ˇ
ˇ
ˇ
D.C1/
2
. Hence1D 1. Eigenvectors satisfy

11
11

x1
x2

D

0
0

,
sox1Dx2. Takingx2D1yieldsy1D

1
1

e
t
. For a second solution we need a vectorusuch that

11
11

u1
u2

D

1
1

. Letu1D1andu2D0. Theny2D

1
0

e
t
C

1
1

te
t
. The general
solution isyDc1

1
1

e
t
Cc2

1
0

e
t
C

1
1

te
t

.
10.5.4.y
0
D
ˇ
ˇ
ˇ
ˇ
3 1
1 1
ˇ
ˇ
ˇ
ˇ
D.2/
2
. Hence1D2. Eigenvectors satisfy

1 1
11

x1
x2

D

0
0

,
sox1D x2. Takingx2D1yieldsy1D

1
1

e
2t
. For a second solution we need a vectorusuch that

1 1
11

u1
u2

D

1
1

. Letu1D 1andu2D0. Theny2D

1
0

e
2t
C

1
1

te
2t
.
The general solution isyDc1

1
1

e
2t
Cc2

1
0

e
2t
C

1
1

te
2t

.
10.5.6.
ˇ
ˇ
ˇ
ˇ
10 9
4 2
ˇ
ˇ
ˇ
ˇ
D.C4/
2
. Hence1D 4. Eigenvectors satisfy

6 9
4 6

x1
x2

D

0
0

,
sox1D
3
2
x2. Takingx2D2yieldsy1D

3
2

e
4t
. For a second solution we need a vectorusuch that

6 9
4 6

u1
u2

D

3
2

. Letu1D
1
2
andu2D0. Theny2D

1
0

e
4t
2
C

3
2

te
4t
. The

207. 202 Chapter 10Linear Systems of Differential Equations
general solution isyDc1

3
2

e
4t
Cc2

1
0

e
4t
2
C

3
2

te
4t

.
10.5.8.
2
4
2 1
4 6 1
0 4 2
3
5D .4/
2
. Hence1D0and2D3D4. The eigenvectors
associated with1D0satisfy the system with augmented matrix
2
6
6
6
4
0 2 1
:
:
: 0
4 6 1
:
:
: 0
0 4 2
:
:
: 0
3
7
7
7
5
, which is
row equivalent to
2
6
6
6
4
1 0
1
2
:
:
: 0
0 1
1
2
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D
1
2
x3andx2D
1
2
x3. Takingx3D2yields
y1D
2
4
1
1
2
3
5. The eigenvectors associated with2D4satisfy the system with augmented matrix
2
6
6
6
4
4 2 1
:
:
: 0
4 2 1
:
:
: 0
0 42
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0
1
2
:
:
: 0
0 1
1
2
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D
1
2
x3and
x2D
1
2
x3. Takingx3D2yieldsy2D
2
4
1
1
2
3
5e
4t
. For a third solution we need a vectorusuch
that
2
4
4 2 1
4 2 1
0 42
3
5
2
4
u1
u2
u3
3
5D
2
4
1
1
2
3
5. The augmented matrix of this system is row equivalent to
2
6
6
6
4
1 0
1
2
:
:
: 0
0 1
1
2
:
:
:
1
2
0 0 0
:
:
: 0
3
7
7
7
5
. Letu3D0,u1D0, andu2D
1
2
. Theny3D
2
4
0
1
0
3
5
e
4t
2
C
2
4
1
1
2
3
5te
4t
. The
general solution is
yDc1
2
4
1
1
2
3
5Cc2
2
4
1
1
2
3
5e
4t
Cc3
0
@
2
4
0
1
0
3
5
e
4t
2
C
2
4
1
1
2
3
5te
4t
1
A:
10.5.10.
2
4
1 1 1
2 2
1 3 1
3
5D .2/.C2/
2
. Hence1D2and2D3D 2. The
eigenvectors associated with1D2satisfy the system with augmented matrix
2
6
6
6
4
3 1 1
:
:
: 0
22 2
:
:
: 0
1 3 3
:
:
: 0
3
7
7
7
5
,

208. Section 10.5Constant Coefficient Homogeneous Systems II203
which is row equivalent to
2
6
6
6
4
1 0 0
:
:
: 0
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D0andx2Dx3. Takingx3D1yields
y1D
2
4
0
1
1
3
5e
2t
. The eigenvectors associated with2D 2satisfy the system with augmented ma-
trix
2
6
6
6
4
1 11
:
:
: 0
2 2 2
:
:
: 0
1 3 1
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 1 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3and
x2D0. Takingx3D1yieldsy2D
2
4
1
0
1
3
5e
2t
. For a third solution we need a vectorusuch
that
2
4
1 11
2 2 2
1 3 1
3
5
2
4
u1
u2
u3
3
5D
2
4
1
0
1
3
5. The augmented matrix of this system is row equivalent to
2
6
6
6
4
1 01
:
:
:
1
2
0 1 0
:
:
:
1
2
0 0 0
:
:
: 0
3
7
7
7
5
. Letu3D0,u1D
1
2
, andu2D
1
2
. Theny3D
2
4
1
1
0
3
5
e
2t
2
C
2
4
1
0
1
3
5te
2t
.
The general solution is
yDc1
2
4
0
1
1
3
5e
2t
Cc2
2
4
1
0
1
3
5e
2t
Cc3
0
@
2
4
1
1
0
3
5
e
2t
2
C
2
4
1
0
1
3
5te
2t
1
A:
10.5.12.
2
4
65 3
21 3
2 1 1
3
5D .C2/.4/
2
. Hence1D 2and2D3D4. The
eigenvectors associated with1D 2satisfy the system with augmented matrix
2
6
6
6
4
85 3
:
:
: 0
2 1 3
:
:
: 0
2 1 3
:
:
: 0
3
7
7
7
5
,
which is row equivalent to
2
6
6
6
4
1 0 1
:
:
: 0
0 1 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D x3andx2D x3. Takingx3D1
yieldsy1D
2
4
1
1
1
3
5e
2t
. The eigenvectors associated with2D4satisfy the system with augmented
matrix
2
6
6
6
4
25 3
:
:
: 0
25 3
:
:
: 0
2 1 3
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3

209. 204 Chapter 10Linear Systems of Differential Equations
andx2Dx3. Takingx3D1yieldsy2D
2
4
1
1
1
3
5e
4t
. For a third solution we need a vectorusuch
that
2
4
25 3
25 3
2 1 3
3
5
2
4
u1
u2
u3
3
5D
2
4
1
1
1
3
5. The augmented matrix of this system is row equivalent to
2
6
6
6
4
1 01
:
:
:
1
2
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Letu3D0,u1D
1
2
, andu2D0. Theny3D
2
4
1
0
0
3
5
e
4t
2
C
2
4
1
1
1
3
5te
4t
. The
general solution isyDc1
2
4
1
1
1
3
5e
2t
Cc2
2
4
1
1
1
3
5e
4t
Cc3
0
@
2
4
1
0
0
3
5
e
4t
2
C
2
4
1
1
1
3
5te
4t
1
A.
10.5.14.
ˇ
ˇ
ˇ
ˇ
159
169
ˇ
ˇ
ˇ
ˇ
D.3/
2
. Hence1D3. Eigenvectors satisfy

129
1612

x1
x2

D

0
0

,
sox1D
3
4
x2. Takingx2D4yieldsy1D

3
4

e
3t
. For a second solution we need a vectorusuch
that

129
1612

u1
u2

D

3
4

. Letu1D
1
4
andu2D0. Theny2D

1
0

e
3t
4
C

3
4

te
3t
.
The general solution isyDc1

3
4

e
3t
Cc2

1
0

e
3t
4
C

3
4

te
3t

. Nowy.0/D

5
8

)
c1

3
4

Cc2

1
4
0

D

5
8

, soc1D2andc2D 4. Therefore,yD

5
8

e
3t


12
16

te
3t
.
10.5.16.
ˇ
ˇ
ˇ
ˇ
7 24
6 17
ˇ
ˇ
ˇ
ˇ
D.5/
2
. Hence1D5. Eigenvectors satisfy

12 24
6 12

x1
x2

D

0
0

,
sox1D2x2. Takingx2D1yieldsy1D

2
1

e
5t
. For a second solution we need a vectorusuch
that

12 24
6 12

u1
u2

D

2
1

. Letu1D
1
6
andu2D0. Theny2D

1
0

e
5t
6
C

2
1

te
5t
.
The general solution isyDc1

2
1

e
5t
Cc2

1
0

e
5t
6
C

2
1

te
5t

. Nowy.0/D

3
1

)
c1

2
1

Cc2

1
6
0

D

3
1

, soc1D1andc2D6. Therefore,yD

3
1

e
5t


12
6

te
5t
.
10.5.18.
2
4
1 1 0
1 12
1 11
3
5D .1/.C2/
2
. Hence1D1and2D3D 2. The
eigenvectors associated with1D1satisfy the system with augmented matrix
2
6
6
6
4
2 1 0
:
:
: 0
122
:
:
: 0
112
:
:
: 0
3
7
7
7
5
,
which is row equivalent to
2
6
6
6
4
1 0
2
3
:
:
: 0
0 1
4
3
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D
2
3
x3andx2D
4
3
x3. Taking

210. Section 10.5Constant Coefficient Homogeneous Systems II205
x3D3yields The eigenvectors associated with2D 2satisfy the system with augmented matrix
2
6
6
6
4
1 1 0
:
:
: 0
1 1 2
:
:
: 0
11 1
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 1 0
:
:
: 0
0 0 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D x2and
x3D0. Takingx2D1yieldsy2D
2
4
1
1
0
3
5e
2t
. For a third solution we need a vectorusuch
that
2
4
1 1 0
1 1 2
11 1
3
5
2
4
u1
u2
u3
3
5D
2
4
1
1
0
3
5. The augmented matrix of this system is row equiva-
lent to
2
6
6
6
4
1 1 0
:
:
:1
0 0 1
:
:
:1
0 0 0
:
:
: 0
3
7
7
7
5
. Letu2D0,u1D 1, andu3D 1. Theny3D
2
4
1
0
1
3
5e
2t
C
2
4
1
1
0
3
5te
2t
. The general solution isyDc1
2
4
2
4
3
3
5e
t
Cc2
2
4
1
1
0
3
5e
2t
Cc3
0
@
2
4
1
0
1
3
5e
2t
C
2
4
1
1
0
3
5te
2t
1
A.
Nowy.0/D
2
4
6
5
7
3
5)c1
2
4
2
4
3
3
5Cc2
2
4
1
1
0
3
5Cc3
2
4
1
0
1
3
5D
2
4
6
5
7
3
5, soc1D 2,c2D 3,
andc3D1. Therefore,yD
2
4
4
8
6
3
5e
t
C
2
4
2
3
1
3
5e
2t
C
2
4
1
1
0
3
5te
2t
.
10.5.20.
2
4
74 4
1 0 1
9 5 6
3
5D .C3/.1/
2
. Hence1D 3and2D3D1. The
eigenvectors associated with1D 3satisfy the system with augmented matrix
2
6
6
6
4
44 4
:
:
: 0
1 3 1
:
:
: 0
95 9
:
:
: 0
3
7
7
7
5
,
which is row equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 1 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx3andx2D0. Takingx3D1yields
y1D
2
4
1
0
1
3
5e
3t
. The eigenvectors associated with2D1satisfy the system with augmented ma-
trix
2
6
6
6
4
84 4
:
:
: 0
11 1
:
:
: 0
955
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 0
:
:
: 0
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D0

211. 206 Chapter 10Linear Systems of Differential Equations
andx2Dx3. Takingx3D1yieldsy2D
2
4
0
1
1
3
5e
t
. For a third solution we need a vectorusuch
that
2
4
84 4
11 1
95 5
3
5
2
4
u1
u2
u3
3
5D
2
4
0
1
1
3
5. The augmented matrix of this system is row equivalent to
2
6
6
6
4
1 0 0
:
:
: 1
0 11
:
:
:2
0 0 0
:
:
: 0
3
7
7
7
5
. Letu3D0,u1D1, andu2D 2. Theny3D
2
4
1
2
0
3
5e
t
C
2
4
0
1
1
3
5te
t
.
The general solution isyDc1
2
4
1
0
1
3
5e
3t
Cc2
2
4
0
1
1
3
5e
t
Cc3
0
@
2
4
1
2
0
3
5e
t
C
2
4
0
1
1
3
5te
t
1
A. Now
y.0/D
2
4
6
9
1
3
5)c1
2
4
1
0
1
3
5Cc2
2
4
0
1
1
3
5Cc3
2
4
1
2
0
3
5D
2
4
6
9
1
3
5, soc1D 2,c2D1, and
c3D 4. Therefore,yD
2
4
2
0
2
3
5e
3t
C
2
4
4
9
1
3
5e
t

2
4
0
4
4
3
5te
t
.
10.5.22.
2
4
48 4
313
1 1 9
3
5D .C4/.8/
2
. Hence1D 4and2D3D8. The
eigenvectors associated with1D 4satisfy the system with augmented matrix
2
6
6
6
4
884
:
:
: 0
3 3 3
:
:
: 0
11 13
:
:
: 0
3
7
7
7
5
,
which is row equivalent to
2
6
6
6
4
11 0
:
:
: 0
0 0 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1Dx2andx3D0. Takingx2D1yields
y1D
2
4
1
1
0
3
5e
t
. The eigenvectors associated with2D8satisfy the system with augmented matrix
2
6
6
6
4
484
:
:
: 0
393
:
:
: 0
11 1
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 1
:
:
: 0
0 1 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Hencex1D x3and
x2D0. Takingx3D1yieldsy2D
2
4
1
0
1
3
5e
8t
. For a third solution we need a vectorusuch that
2
4
484
393
11 1
3
5
2
4
u1
u2
u3
3
5D
2
4
1
0
1
3
5. The augmented matrix of this system is row equivalent to

212. Section 10.5Constant Coefficient Homogeneous Systems II207
2
6
6
6
4
1 0 1
:
:
:
3
4
0 1 0
:
:
:
1
4
0 0 0
:
:
: 0
3
7
7
7
5
. Letu3D0,u1D
3
4
, andu2D
1
4
. Theny3D
2
4
3
1
0
3
5
e
8t
4
C
2
4
1
0
1
3
5te
8t
.
The general solution isc1
2
4
1
1
0
3
5e
t
Cc2
2
4
1
0
1
3
5e
8t
Cc3
0
@
2
4
3
1
0
3
5
e
8t
4
C
2
4
1
0
1
3
5te
8t
1
A. Nowy.0/D
2
4
4
1
3
3
5)c1
2
4
1
1
0
3
5Cc2
2
4
1
0
1
3
5Cc3
2
4
3
4

1
4
0
3
5D
2
4
4
1
3
3
5, soc1D 1,c2D 3, andc3D 8.
Therefore,yD
2
4
1
1
0
3
5e
4t
C
2
4
3
2
3
3
5e
8t
C
2
4
8
0
8
3
5te
8t
.
10.5.24.
2
4
51 1
1 93
2 2 4
3
5D .6/
3
. Hence1D6. The eigenvectors satisfy the system
with augmented matrix
2
6
6
6
4
11 1
:
:
: 0
1 3 3
:
:
: 0
2 2 2
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 0
:
:
: 0
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
.
Hencex1D0andx2Dx3. Takingx3D1yieldsy1D
2
4
0
1
1
3
5e
6t
. For a second solution we need a
vectorusuch that
2
4
11 1
1 3 3
2 2 2
3
5
2
4
u1
u2
u3
3
5D
2
4
0
1
1
3
5. The augmented matrix of this system is row
equivalent to
2
6
6
6
4
1 0 0
:
:
:
1
4
0 11
:
:
:
1
4
0 0 0
:
:
: 0
3
7
7
7
5
. Letu3D0,u1D
1
4
, andu2D
1
4
. Theny2D
2
4
1
1
0
3
5
e
6t
4
C
2
4
0
1
1
3
5te
6t
. For a third solution we need a vectorvsuch that
2
4
11 1
1 3 3
2 2 2
3
5
2
4
v1
v2
v3
3
5D
2
6
4

1
4
1
4
0
3
7
5.
The augmented matrix of this system is row equivalent to
2
6
6
6
4
1 0 0
:
:
:
1
8
0 11
:
:
:
1
8
0 0 0
:
:
: 0
3
7
7
7
5
. Letv3D0,v1D
1
8
,
andv2D
1
8
. Theny3D
2
4
1
1
0
3
5
e
6t
8
C
2
4
1
1
0
3
5
te
6t
4
C
2
4
0
1
1
3
5
t
2
e
6t
2
. The general solution isyD

213. 208 Chapter 10Linear Systems of Differential Equations
c1
2
4
0
1
1
3
5e
6t
Cc2
0
@
2
4
1
1
0
3
5
e
6t
4
C
2
4
0
1
1
3
5te
6t
1
A
Cc3
0
@
2
4
1
1
0
3
5
e
6t
8
C
2
4
1
1
0
3
5
te
6t
4
C
2
4
0
1
1
3
5
t
2
e
6t
2
1
A:
10.5.26.
2
4
64 4
2 1 1
2 3 1
3
5D .C2/
3
. Hence1D 2. The eigenvectors satisfy the sys-
tem with augmented matrix
2
6
6
6
4
444
:
:
: 0
2 1 1
:
:
: 0
2 3 3
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 0
:
:
: 0
0 1 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
.
Hencex1D0andx2D x3. Takingx3D1yieldsy1D
2
4
0
1
1
3
5e
2t
. For a second solution we need a
vectorusuch that
2
4
444
2 1 1
2 3 3
3
5
2
4
u1
u2
u3
3
5D
2
4
0
1
1
3
5. The augmented matrix of this system is row
equivalent to
2
6
6
6
4
1 0 0
:
:
:1
0 1 1
:
:
: 1
0 0 0
:
:
: 0
3
7
7
7
5
. Letu3D0,u1D 1, andu2D1. Theny2D
2
4
1
1
0
3
5e
2t
C
2
4
0
1
1
3
5te
2t
. For a third solution we need a vectorvsuch that
2
4
444
2 1 1
2 3 3
3
5
2
4
v1
v2
v3
3
5D
2
4
1
1
0
3
5. The augmented matrix of this system is row equivalent to
2
6
6
6
4
1 0 0
:
:
:
3
4
0 1 1
:
:
:
1
2
0 0 0
:
:
: 0
3
7
7
7
5
. Letv3D0,
v1D
3
4
, andv2D
1
2
. Theny3D
2
4
3
2
0
3
5
e
2t
4
C
2
4
1
1
0
3
5te
2t
C
2
4
0
1
1
3
5
t
2
e
2t
2
. The general
solution isyDc1
2
4
0
1
1
3
5e
2t
Cc2
0
@
2
4
1
1
0
3
5e
2t
C
2
4
0
1
1
3
5te
2t
1
A
Cc3
0
@
2
4
3
2
0
3
5
e
2t
4
C
2
4
1
1
0
3
5te
2t
C
2
4
0
1
1
3
5
t
2
e
2t
2
1
A
10.5.28.
2
4
212 10
2 24 11
2 24 8
3
5D .C6/
3
. Hence1D 6. The eigenvectors satisfy the

214. Section 10.5Constant Coefficient Homogeneous Systems II209
system with augmented matrix
2
6
6
6
4
412 10
:
:
: 0
218 11
:
:
: 0
224 14
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 1
:
:
: 0
0 1
1
2
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
.
Hencex1D x3andx2D
x3
2
. Takingx3D2yieldsy1D
2
4
2
1
2
3
5e
6t
. For a second solu-
tion we need a vectorusuch that
2
4
412 10
218 11
224 14
3
5
2
4
u1
u2
u3
3
5D
2
4
2
1
2
3
5. The augmented matrix
of this system is row equivalent to
2
6
6
6
4
1 0 1
:
:
:1
0 1
1
2
:
:
:
1
6
0 0 0
:
:
: 0
3
7
7
7
5
. Letu3D0,u1D 1, andu2D

1
6
. Theny2D
2
4
6
1
0
3
5
e
6t
6
C
2
4
2
1
2
3
5te
6t
. For a third solution we need a vectorvsuch that
2
4
412 10
218 11
224 14
3
5
2
4
v1
v2
v3
3
5D
2
4
1

1
6
0
3
5. The augmented matrix of this system is row equivalent to
2
6
6
6
4
1 0 1
:
:
:
1
3
0 1
1
2
:
:
:
1
36
0 0 0
:
:
: 0
3
7
7
7
5
. Letv3D0,v1D
1
3
, andv2D
1
36
. Theny3D
2
4
12
1
0
3
5
e
6t
36

2
4
6
1
0
3
5
te
6t
6
C
2
4
2
1
2
3
5
t
2
e
6t
2
. The general solution isyDc1
2
4
2
1
2
3
5e
6t
Cc2
0
@
2
4
6
1
0
3
5
e
6t
6
C
2
4
2
1
2
3
5te
6t
1
A
Cc3
0
@
2
4
12
1
0
3
5
e
6t
36

2
4
6
1
0
3
5
te
6t
6
C
2
4
2
1
2
3
5
t
2
e
6t
2
1
A:
10.5.30.
2
4
4 0 1
131
1 0 2
3
5D .C3/
3
. Hence1D3. The eigenvectors satisfy the sys-
tem with augmented matrix
2
6
6
6
4
1 01
:
:
: 0
1 01
:
:
: 0
1 0 1
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 1
:
:
: 0
0 0 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
.
Hencex1D x3andx2is arbitrary. Takingx2D0andx3D1yieldsy1D
2
4
1
0
1
3
5e
3t
. Takingx2D
1andx3D0yieldsy2D
2
4
0
1
0
3
5e
3t
. For a third solution we need constants˛andˇand a vectorusuch

215. 210 Chapter 10Linear Systems of Differential Equations
that
2
4
1 01
1 01
1 0 1
3
5
2
4
u1
u2
u3
3
5D˛
2
4
1
0
1
3
5Cˇ
2
4
0
1
0
3
5. The augmented matrix of this system is row
equivalent to
2
6
6
6
4
1 0 1
:
:
: ˛
0 0 0
:
:
: ˛Cˇ
0 0 0
:
:
: 0
3
7
7
7
5
; hence the system has a solution if˛D ˇD1, which yields
the eigenvectorx3D
2
4
1
1
1
3
5. Takingu1D1andu2Du3D0yields the solutiony3D
2
4
1
0
0
3
5e
3t
C
2
4
1
1
1
3
5te
3t
. The general solution isyDc1
2
4
1
0
1
3
5e
3t
Cc2
2
4
0
1
0
3
5e
3t
Cc3
0
@
2
4
1
0
0
3
5e
3t
C
2
4
1
1
1
3
5te
3t
1
A
10.5.32.
2
4
31 0
1 1 0
1 12
3
5D .C2/
3
. Hence1D 2. The eigenvectors satisfy the
system with augmented matrix
2
6
6
6
4
11 0
:
:
: 0
1 1 0
:
:
: 0
11 0
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 1 0
:
:
: 0
0 0 0
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
.
Hencex1D x2andx3is arbitrary. Takingx2D1andx3D0yieldsy1D
2
4
1
1
0
3
5e
2t
. Takingx2D
0andx3D1yieldsy2D
2
4
0
0
1
3
5e
2t
. For a third solution we need constants˛andˇand a vectorusuch
that
2
4
11 0
1 1 0
11 0
3
5
2
4
u1
u2
u3
3
5D˛
2
4
1
1
0
3
5Cˇ
2
4
0
0
1
3
5. The augmented matrix of this system is row
equivalent to
2
6
6
6
4
1 1 0
:
:
: ˛
0 0 0
:
:
: ˛Cˇ
0 0 0
:
:
: 0
3
7
7
7
5
; hence the system has a solution if˛D ˇD1, which yields
the eigenvectorx3D
2
4
1
1
1
3
5. Takingu1D1andu2Du3D0yields the solutiony3D
2
4
1
0
0
3
5e
2t
C
2
4
1
1
1
3
5te
2t
. The general solution isyDc1
2
4
1
1
0
3
5e
2t
Cc2
2
4
0
0
1
3
5e
2t
Cc3
0
@
2
4
1
0
0
3
5e
2t
C
2
4
1
1
1
3
5te
2t
1
A.
10.5.34.
y
0
3
Ay3D.1IA/ve
1t
C.1IA/ute
1t
Cue
1t
C.1IA/x
t
2
e
1t
2
Cxte
1t
D ue
1t
xte
1t
Cue
1t
C0Cxte
1t
D0:

216. Section 10.6Constant Coefficient Homogeneous Systems III211
Now suppose thatc1y1Cc2y2Cc3y3D0. Then
c1xCc2.uCtx/Cc3

vCtuC
t
2
2
x

D0: . A/
Differentiating this twice yieldsc3xD0, soc3D0sincex¤0. Therefore,(A) reduces to (B)c1xC
c2.uCtx/D0. Differentiating this yieldsc2xD0, soc2D0sincex¤0. Therefore,(B) reduces to
c3xD0, soc1D0sincex¤0. Therefore,y1,y2, andy3are linearly independendent.
10.6CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS III
10.6.2.
ˇ
ˇ
ˇ
ˇ
11 4
26 9
ˇ
ˇ
ˇ
ˇ
D.C1/
2
C4. The augmented matrix of.A.1C2i/ I /xD0
is
2
4
102i 4
:
:
: 0
26 102i
:
:
: 0
3
5, which is row equivalent to
2
4
1
5Ci
13
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
.5i/x2=13. Takingx2D13yields the eigenvectorxD

5i
13

. Taking real and imaginary parts of
e
t
.cos2tCisin2t//

5i
13

yields
yDc1e
t

5cos2tCsin2t
13cos2t

Cc2e
t

5sin2tcos2t
13sin2t

:
10.6.4.
ˇ
ˇ
ˇ
ˇ
56
31
ˇ
ˇ
ˇ
ˇ
D.2/
2
C9. Hence,D2C3iis an eigenvalue ofA. The associated
eigenvectors satisfy.A.2C3i/ I /xD0. The augmented matrix of this system is
2
4
33i6
:
:
: 0
3 33i
:
:
: 0
3
5,
which is row equivalent to
2
4
11i
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D.1Ci/x2. Takingx2D1yields
x1D1Ci, soxD

1Ci
1

is an eigenvector. Taking real and imaginary parts ofe
2t
.cos3tC
isin3t/

1Ci
1

yieldsyDc1e
2t

cos3tsin3t
cos3t

Cc2e
2t

sin3tCcos3t
sin3t

.
10.6.6.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
3 3 1
1 53
3 7 3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .C1/

.C2/
2
C4

. The augmented matrix of.AC
I /xD0is
2
6
6
6
4
2 3 1
:
:
: 0
143
:
:
: 0
3 7 4
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 1
:
:
: 0
0 1 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D
x2D x3. Takingx3D1yieldsy1D
2
4
1
1
1
3
5e
t
. The augmented matrix of.A.2C2i/I /xD0

217. 212 Chapter 10Linear Systems of Differential Equations
is
2
6
6
6
4
12i 3 1
:
:
: 0
1 32i3
:
:
: 0
3 7 5 2i
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0
1Ci
2
:
:
: 0
0 1
1i
2
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D
.1Ci/
2
x3andx2D
.1i/
2
x3. Takingx3D2yields the eigenvectorx2D
2
4
1Ci
1Ci
2
3
5. The real
and imaginary parts ofe
2t
.cos2tCisin2t/
2
4
1Ci
1Ci
2
3
5arey2De
2t
2
4
cos2tsin2t
cos2tsin2t
2cos2t
3
5and
y3De
2t
2
4
sin2tCcos2t
sin2tCcos2t
2sin2t
3
5. Therefore,
yDc1
2
4
1
1
1
3
5e
t
Cc2e
2t
2
4
cos2tsin2t
cos2tsin2t
2cos2t
3
5Cc3e
2t
2
4
sin2tCcos2t
sin2tCcos2t
2sin2t
3
5:
10.6.8.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
3 1 3
4 1 2
4 2 3
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .1/

.C1/
2
C4

. The augmented matrix of.AI /xD0
is
2
6
6
6
4
4 1 3
:
:
: 0
42 2
:
:
: 0
42 2
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 1
:
:
: 0
0 1 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1Dx2D
x3. Takingx3D1yieldsy1D
2
4
1
1
1
3
5e
t
. The augmented matrix of.A.1C2i/I /xD0is
2
6
6
6
4
22i 1 3
:
:
: 0
4 2i 2
:
:
: 0
4 2 42i
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0
1i
2
:
:
: 0
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D

1i
2
x3andx2Dx3. Takingx3D2yields the eigenvectorx2D
2
4
1Ci
2
2
3
5. The real and
imaginary parts ofe
t
.cos2tCisin2t/
2
4
1Ci
2
2
3
5arey2De
t
2
4
sin2tcos2t
2cos2t
2cos2t
3
5andy3D
e
t
2
4
cos2tsin2t
2sin2t
2sin2t
3
5. Therefore,
yDc1
2
4
1
1
1
3
5e
t
Cc2e
t
2
4
sin2tcos2t
2cos2t
2cos2t
3
5Cc3e
t
2
4
cos2tsin2t
2sin2t
2sin2t
3
5:

218. Section 10.6Constant Coefficient Homogeneous Systems III213
10.6.10.
1
3
ˇ
ˇ
ˇ
ˇ
735
2 53
ˇ
ˇ
ˇ
ˇ
D.2/
2
C1. The augmented matrix of.A.2Ci/I /xD0is
1
3
2
4
13i5
:
:
: 0
2 13i
:
:
: 0
3
5, which is row equivalent to
2
4
1
1C3i
2
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
1C3i
2
x2. Takingx2D2yields the eigenvectorxD

1C3i
2

. Taking real and imaginary parts of
e
2t
.costCisint/

1C3i
2

yields
yDc1e
2t

cost3sint
2cost

Cc2e
2t

sintC3cost
2sint

:
10.6.12.
ˇ
ˇ
ˇ
ˇ
34 52
2030
ˇ
ˇ
ˇ
ˇ
D.2/
2
C16. The augmented matrix of.A.2C4i/ I /xD0
is
2
4
324i 52
:
:
: 0
20324i
:
:
: 0
3
5, which is row equivalent to
2
4
1
8Ci
5
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D

.8Ci/
5
x2. Takingx2D5yields the eigenvectorxD

8i
5

. Taking real and imaginary parts of
e
2t
.cos4tCisin4t/

8i
5

yieldsyDc1e
2t

sin4t8cos4t
5cos4t

Cc2e
2t

cos4t8sin4t
5sin4t

.
10.6.14.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
34 2
5 78
10 13 8
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .C2/

.2/
2
C9

. The augmented matrix of.AC
2I /xD0is
2
6
6
6
4
542
:
:
: 0
5 9 8
:
:
: 0
10 136
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 02
:
:
: 0
0 12
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D
x2D2x3. Takingx3D1yieldsy1D
2
4
2
2
1
3
5e
2t
. The augmented matrix of.A.2C3i/I /xD0is
2
6
6
6
4
13i4 2
:
:
: 0
5 53i 8
:
:
: 0
10 13 103i
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 0 1i
:
:
: 0
0 1i
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D
.1i/x3andx2Dix3. Takingx3D1yields the eigenvectorx2D
2
4
1Ci
i
1
3
5. The real
and imaginary parts ofe
2t
.cos3tCisin3t/
2
4
1Ci
i
1
3
5arey2De
2t
2
4
cos3tsin3t
sin3t
cos3t
3
5and

219. 214 Chapter 10Linear Systems of Differential Equations
y3Dc3e
2t
2
4
sin3tCcos3t
cos3t
sin3t
3
5. Therefore,
yDc1
2
4
2
2
1
3
5e
2t
Cc2e
2t
2
4
cos3tsin3t
sin3t
cos3t
3
5Cc3e
2t
2
4
sin3tCcos3t
cos3t
sin3t
3
5:
10.6.16.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
1 2 2
0 21
1 0
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .2/

.1/
2
C1

. The augmented matrix of.AI /xD0
is
2
6
6
6
4
0 22
:
:
: 0
0 11
:
:
: 0
1 01
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 01
:
:
: 0
0 11
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1Dx2D1.
Takingx3D1yieldsy1D
2
4
1
1
1
3
5e
t
. The augmented matrix of.A.1Ci/I /xD0is
2
6
6
6
4
i 2 2
:
:
: 0
0 1i1
:
:
: 0
1 0 1i
:
:
: 0
3
7
7
7
5
,
which is row equivalent to
2
6
6
6
4
1 01i
:
:
: 0
0 1
1Ci
2
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D.1Ci/x3andx2D
.1Ci/
2
x3.
Takingx3D2yields the eigenvectorx2D
2
4
2C2i
1Ci
2
3
5. The real and imaginary parts ofe
4t
.costC
isint/
2
4
2C2i
1Ci
2
3
5arey2De
t
2
4
2cost2sint
costsint
2cost
3
5andy3Dc3e
t
2
4
2sintC2cost
costCsint
2sint
3
5. Therefore,
yDc1
2
4
1
1
1
3
5e
t
Cc2e
t
2
4
2cost2sint
costsint
2cost
3
5Cc3e
t
2
4
2sintC2cost
costCsint
2sint
3
5:
10.6.18.
ˇ
ˇ
ˇ
ˇ
7 15
3 1
ˇ
ˇ
ˇ
ˇ
D.4/
2
C36. The augmented matrix of.A.4C6i/ I /xD0
is
2
4
36i 15
:
:
: 0
336i
:
:
: 0
3
5, which is row equivalent to
2
4
1 1C2i
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
.1C2i/x2. Takingx2D1yields the eigenvectorxD

12i
1

. Taking real and imaginary parts of
e
4t
.cos6tCisin6t/

12i
1

yieldsyDc1e
4t

2sin6tcos6t
cos6t

Cc2e
4t

2cos6tsin6t
sin6t

.
Nowy.0/D

5
1

)

12
1 0

c1
c2

D

5
1

, soc1D1,c2D 3, andyDe
4t

5cos6tC5sin6t
cos6t3sin6t

.

220. Section 10.6Constant Coefficient Homogeneous Systems III215
10.6.20.
1
6
ˇ
ˇ
ˇ
ˇ
462
5 26
ˇ
ˇ
ˇ
ˇ
D


1
2

2
C
1
4
. The augmented matrix of

A
1Ci
2
I

xD0
is
1
6
2
4
13i2
:
:
: 0
5 13i
:
:
: 0
3
5, which is row equivalent to
2
4
1
1C3i
5
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
1C3i
5
x2. Takingx2D5yields the eigenvectorxD

1C3i
5

. Taking real and imaginary parts of
e
t =2
.cost=2Cisint=2/

1C3i
5

yieldsyDc1e
t =2

cost=23sint=2
5cost=2

Cc2e
t =2

sint=2C3cost=2
5sint=2

.
Nowy.0/D

1
1

)

1 3
5 0

c1
c2

D

1
1

, soc1D
1
5
,c2D
2
5
, andyDe
t =2

cos.t=2/Csin.t=2/
cos.t=2/C2sin.t=2/

.
10.6.22.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
4 4 0
8 1020
2 3 2
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .8/

.2/
2
C4

. The augmented matrix of.A
8I /xD0is
2
6
6
6
4
0 4 0
:
:
: 0
8 620
:
:
: 0
2 3 6
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 02
:
:
: 0
0 12
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D
x2D2x3. Takingx3D2yieldsy1D
2
4
2
2
1
3
5e
8t
. The augmented matrix of.A.2C2i/I /xD0is
2
6
6
6
4
22i 4 0
:
:
: 0
8 82i20
:
:
: 0
2 3 42i
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 02C2i
:
:
: 0
0 1 2i
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D
.22i/x3andx2D2ix3. Takingx3D1yields the eigenvectorx2D
2
4
22i
2i
1
3
5. The real and
imaginary parts ofe
2t
.cos2tCisin2t/
2
4
22i
2i
1
3
5arey2De
2t
2
4
2cos2tC2sin2t
2sin2t
2cos2t
3
5andy3D
e
2t
2
4
2sin2tC2cos2t
2cos2t
sin2t
3
5, so the general solution isyDc1
2
4
2
2
1
3
5e
8t
Cc2e
2t
2
4
2cos2tC2sin2t
2sin2t
2cos2t
3
5C
c3e
2t
2
4
2sin2tC2cos2t
2cos2t
sin2t
3
5. Nowy.0/D
2
4
8
6
5
3
5)
2
4
2 22
2 0 2
1 1 0
3
5
2
4
c1
c2
c3
3
5D
2
4
8
6
5
3
5, soc1D2,
c2D3,c3D1, andyD
2
4
4
4
2
3
5e
8t
Ce
2t
2
4
4cos2tC8sin2t
6sin2tC2cos2t
3cos2tCsin2t
3
5.
10.6.24.
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
44 4
10 3 15
2 3 1
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
D .8/.
2
C16/. The augmented matrix of.A8I /xD0is

221. 216 Chapter 10Linear Systems of Differential Equations
2
6
6
6
4
44 4
:
:
: 0
105 15
:
:
: 0
237
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 02
:
:
: 0
0 1 1
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D2x3
andx2D x3. Takingx3D1yieldsy1D
2
4
2
1
1
3
5e
8t
. The augmented matrix of.A4iI /xD0is
2
6
6
6
4
44i4 4
:
:
: 0
10 34i 15
:
:
: 0
2 3 14i
:
:
: 0
3
7
7
7
5
, which is row equivalent to
2
6
6
6
4
1 01Ci
:
:
: 0
0 11C2i
:
:
: 0
0 0 0
:
:
: 0
3
7
7
7
5
. Therefore,x1D
.1i/x3andx2D.12i/x3. Takingx3D1yields the eigenvectorx2D
2
4
1i
12i
1
3
5. The
real and imaginary parts of.cos4tCisin4t/
2
4
1i
12i
1
3
5arey2D
2
4
cos4tCsin4t
cos4tC2sin4t
cos4t
3
5andy3D
2
4
sin4tcos4t
sin4t2cos4t
sin4t
3
5, so the general solution isyDc1
2
4
2
1
1
3
5e
8t
Cc2
2
4
cos4tCsin4t
cos4tC2sin4t
cos4t
3
5C
c3
2
4
sin4tcos4t
sin4t2cos4t
sin4t
3
5. Nowy.0/D
2
4
16
14
6
3
5)
2
4
2 11
1 12
1 1 0
3
5
2
4
c1
c2
c3
3
5D
2
4
16
14
6
3
5, soc1D3,
c2D3,c3D 7, andyD
2
4
6
3
3
3
5e
8t
C
2
4
10cos4t4sin4t
17cos4tsin4t
3cos4t7sin4t
3
5.
10.6.28.(a)From the quadratic formula the roots are
k1D
kuk
2
kv
2
k C
p
.kuk
2
kv
2
k/
2
C4.u;v/
2
2.u;v/
k2D
kuk
2
kv
2
k
p
.kuk
2
kv
2
k/
2
C4.u;v/
2
2.u;v/
:
Clearlyk1> 0andk2< 0. Moreover,
k1k2D
.kuk
2
kv
2
k/
2

fi
.kuk
2
kv
2
k/
2
C4.u;v/
2

4.u;v/
2
D 1:
(b)Sincek2D 1=k1,
u
.2/
1
Duk2vDuC
1
k1
vD
1
k1
.vCk1u/D
1
k1
v
.1/
1
v
.2/
1
DvCk2uDv
1
k1
uD
1
k1
.uk1v/D
1
k1
u
.1/
1
:

222. Section 10.6Constant Coefficient Homogeneous Systems III217
10.6.30.
ˇ
ˇ
ˇ
ˇ
15 10
25 15
ˇ
ˇ
ˇ
ˇ
D
2
C25. The augmented matrix of.A5iI /xD0is
2
4
155i 10
:
:
: 0
25 155i
:
:
: 0
3
5,
which is row equivalent to
2
4
1
3Ci
5
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
.3i/
5
x2. Takingx2D5yields the
eigenvectorxD

3i
5

, souD

3
5

andvD

1
0

. The quadratic equation is3k
2
33kC3D
0, with positive rootk:0902. Routine calculations yieldU

:5257
:8507

,V

:8507
:5257

.
10.6.32.
ˇ
ˇ
ˇ
ˇ
315
3 3
ˇ
ˇ
ˇ
ˇ
D
2
C36. The augmented matrix of.A6iI /xD0is
2
4
36i15
:
:
: 0
3 3 6i
:
:
: 0
3
5,
which is row equivalent to
2
4
1 12i
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D .12i/x2. Takingx2D1
yields the eigenvectorxD

1C2i
1

, souD

1
1

andvD

2
0

. The quadratic equation is
2k
2
C2kC2D0, with positive rootk1:6180. Routine calculations yieldU

:9732
:2298

,
V

:2298
:9732

.
10.6.34.
ˇ
ˇ
ˇ
ˇ
512
67
ˇ
ˇ
ˇ
ˇ
D.C1/
2
C36. The augmented matrix of.A.1C6i/I /xD0
is
2
4
66i12
:
:
: 0
6 66i
:
:
: 0
3
5, which is row equivalent to
2
4
1.1Ci/
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
.1Ci/x2. Takingx2D1yields the eigenvectorxD

1Ci
1

, souD

1
1

andvD

1
0

.
The quadratic equation isk
2
k1D0, with positive rootk1:6180. Routine calculations yield
U

:5257
:8507

,V

:8507
:5257

.
10.6.36.
ˇ
ˇ
ˇ
ˇ
4 9
5 2
ˇ
ˇ
ˇ
ˇ
D.C1/
2
C36. The augmented matrix of.A.1C6i/I /xD0
is
2
4
36i 9
:
:
: 0
5 36i
:
:
: 0
3
5, which is row equivalent to
2
4
1
36i
5
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
36i
5
x2. Takingx2D5yields the eigenvectorxD

36i
5

, souD

3
5

andvD

6
0

. The
quadratic equation is18k
2
C2kC18D0, with positive rootk1:0571. Routine calculations yield
U

:8817
:4719

,V

:4719
:8817

.
10.6.38.
ˇ
ˇ
ˇ
ˇ
15
201
ˇ
ˇ
ˇ
ˇ
D.C1/
2
C100. The augmented matrix of.A.1C10i/I /xD0

223. 218 Chapter 10Linear Systems of Differential Equations
is
2
4
10i5
:
:
: 0
2010i
:
:
: 0
3
5, which is row equivalent to
2
4
1
i
2
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
i
2
x2.
Takingx2D2yields the eigenvectorxD

i
2

, souD

0
2

andvD

1
0

. Since.u;v/D0we
just normalizeuandvto obtainUD

0
1

,VD

1
0

.
10.6.40.
ˇ
ˇ
ˇ
ˇ
7 6
12 5
ˇ
ˇ
ˇ
ˇ
D.C1/
2
C36. The augmented matrix of.A.1C6i/I /xD0
is
2
4
66i 6
:
:
: 0
12 66i
:
:
: 0
3
5, which is row equivalent to
2
4
1
1i
2
:
:
: 0
0 0
:
:
: 0
3
5. Therefore,x1D
1i
2
x2. Takingx2D2yields the eigenvectorxD

1i
2

, souD

1
2

andvD

1
0

. The
quadratic equation isk
2
4kC1D0, with positive rootk:2361. Routine calculations yield
U

:5257
:8507

,V

:8507
:5257

.
10.7VARIATION OF PARAMETERS FOR NONHOMOGENEOUS LINEAR SYSTEMS
10.7.2.YD

3e
t
e
2t
e
t
2e
2t

;u
0
DY
1
fD
1
5

2e
t
e
t
e
2t
3e
2t

50e
37
10e
3t

D

20e
4t
2e
2t
10e
5t
C6e
t

;
uD

e
2t
5e
4t
2e
5t
6e
t

;ypDYuD

13e
3t
C3e
3t
e
3t
11e
3t

.
10.7.4.YD

e
2t
e
t
2e
2t
e
t

;u
0
DY
1
fD

e
2t
e
2t
2e
t
e
t

2
2e
t

D

2e
2t
2e
t
2e
2t
4e
t

;
uD

2e
t
e
2t
e
2t
4e
t

;ypDYuD

53e
t
5e
t
6

.
10.7.6.YD

sintcost
costsint

;u
0
DY
1
fD

sintcost
costsint

1
t

D

tcostCsint
tsintcost

;
uD

tsint
tcost

;ypDYuD

t
0

.
10.7.8.YD
2
4
e
3t
e
2t
e
t
e
3t
03e
t
e
3t
e
2t
7e
t
3
5;u
0
DY
1
fD
1
6
2
4
3e
3t
6e
3t
3e
3t
4e
2t
6e
2t
2e
2t
e
t
0 e
t
3
5
2
4
1
e
t
e
t
3
5D
1
6
2
4
3e
3t
9e
2t
8e
t
C4e
2t
e
2t
e
t
3
5;uD
1
12
2
4
9e
2t
2e
3t
16e
t
4e
2t
e
2t
2e
t
3
5;ypDYuD
1
6
2
4
3e
t
C4
6e
t
4
10
3
5.
10.7.10.YD
2
4
e
t
e
t
te
t
e
t
e
t
3e
t
te
t
e
t
e
t
te
t
3
5;u
0
DY
1
fD
1
18
2
4
9e
t
0 9e
t
3e
t
.32t/6te
t
9e
t
6e
t
6e
t
0
3
5
2
4
e
t
e
t
e
t
3
5D
1
3
2
4
0
e
2t
.32t/
2e
2t
3
5;uD
1
3
2
4
0
e
2t
.2t/
e
2t
3
5;ypDYuD
1
3
2
4
2e
t
e
t
2e
t
3
5.

224. Section 10.7Variation of Parameters for Nonhomogeneous Linear Systems219
10.7.12.u
0
DY
1
fD
1
2t

e
t
e
t
e
t
e
t

t
t
2

D
1
2

e
t
.tC1/
e
t
.1t/

;uD
1
2

e
t
.tC2/
e
t
.2t/

;
ypDYuD
t
2

e
t
e
t
e
t
e
t

e
t
.tC2/
e
t
.2t/

D

t
2
2t

.
10.7.14.u
0
DY
1
fD
1
3

2e
t
e
t
2

e
2t
e
2t

D
1
3

2e
2t
e
3t
2e
2t
e
3t

;uD
1
9

3e
2t
Ce
3t
e
3t
3e
2t

;
ypDYuD
1
9

2 e
t
e
t
2

3e
2t
Ce
3t
e
3t
3e
2t

D
1
9

5e
2t
e
3t
e
3t
5e
2t

.
10.7.16.u
0
DY
1
fD
1
t
2
1

te
t
e
t
t

t
2
1
t
2
1

D

te
t
te
t

;uD
1
2

2e
t
Ct
2
t
2
2e
t

;
ypDYuD
1
2

t e
t
e
t
t

2e
t
Ct
2
t
2
2e
t

D
1
2

te
t
.tC2/Ct
3
2
te
t
.t2/Ct
3
C2

.
10.7.18.u
0
DY
1
fD
1
3
2
4
e
5t
e
4t
e
3t
2e
2t
e
t
1
e
2t
2e
t
1
3
5
2
4
e
t
0
0
3
5D
1
3
2
4
e
4t
2e
t
e
t
3
5;uD
1
12
2
4
e
4t
8e
t
4e
t
3
5;
ypDYuD
1
12
2
4
e
5t
e
2t
0
e
4t
0 e
t
e
3t
11
3
5
2
4
e
4t
8e
t
4e
t
3
5D
1
4
2
4
3e
t
1
e
t
3
5.
10.7.20.u
0
DY
1
fD
1
2t
2
4
e
2t
te
t
te
t
Ce
2t
1 te
t
te
t
1
2 0 2
3
5
2
4
e
t
0
e
t
3
5D
2
4
1
e
2t
0
3
5;uD
1
4
2
4
2t
e
2t
0
3
5;
ypDYuDYD
1
4t
2
4
e
t
e
t
t
e
t
e
t
e
t
e
t
e
t
0
3
5
2
4
2t
e
2t
0
3
5D
e
t
4t
2
4
2tC1
2t1
2tC1
3
5.
10.7.22.(c)IfypDYu, theny
0
p
DY
0
uCYu
0
DAYuCYu
0
, so (E)y
0
p
DAypCYu
0
. However, from
the derivation of the method of variation of parameters in Section 9.4,Yu
0
Dfas defined in the solution
of(a). This and (E) imply the conclusion.
(d)SinceYu
0
Dfwithfas defined in the solution of(a),u1; u2; : : : ; unsatisfy the conditions required
in the derivation of the method of variation of parameters inSection 9.4; hence,ypDc1y1Cc2y2C
Ccnynis a particular solution of (A).

226. CHAPTER11
BoundaryValueProblemsandFourier
Expansions
11.1EIGENVALUE PROBLEMS FOR y
00
CyD0
11.1.2.From Theorem 11.1.2 withLD,nDn
2
,ynDsinnx,nD1; 2; 3; : : :
11.1.4.From Theorem 11.1.4 withLD,nD
.2n1/
2
4
,ynDsin
.2n1/x
2
,nD1; 2; 3; : : :;
11.1.6.From Theorem 11.1.6 withLD,0D0,y0D1,nDn
2
,y1nDcosnx,y2nDsinnx,
nD1; 2; 3; : : :
11.1.8.From Theorem 11.1.5 withLD1,nD
.2n1/
2

2
4
,ynDcos
.2n1/x
2
,nD1; 2; 3; : : :
11.1.10.From Theorem 11.1.6 withLD1,0D0,y0D1,nDn
2

2
,y1nDcosnx,y2nD
sinnx,nD1; 2; 3; : : :
11.1.12.From Theorem 11.1.6 withLD2,0D0,y0D1,nD
n
2

2
4
,y1nDcos
nx
2
,y2nD
sin
nx
2
,nD1; 2; 3; : : :
11.1.14.From Theorem 11.1.5 withLD3,nD
.2n1/
2

2
36
,ynDcos
.2n1/x
6
,nD1; 2; 3; : : :
11.1.16.From Theorem 11.1.3 withLD5,nD
n
2

2
25
,ynDcos
nx
5
,nD1; 2; 3; : : :
11.1.18.From Theorem 11.1.1, any eigenvalues of Problem 11.1.4 mustbe positive. If > 0, then
every solution ofy
00
CyD0is of the formyDc1cos
p
xCc2sin
p
xwherec1andc2are
constants. Therefore,y
0
D
p
.c1sin
p
xCc2cos
p
x/. Sincey
0
.0/D0,c2D0. Therefore,
yDc1cos
p
x. Sincey.L/D0,c1cos
p
LD0. To makec1cos
p
LD0withc1¤0we must
choose
p
D
.2n1/
2L
, wherenis a positive integer. Therefore,nD
.2n1/
2

2
4L
2
is an eigenvalue
andynDcos
.2n1/x
2L
is an associated eigenfunction.
221

227. 222 Chapter 11Boundary Value Problems and Fourier Expansions
11.1.20.Ifris a positive integer, then
Z
L
L
cos
rx
L
dxD
L
r
sin
rx
L
ˇ
ˇ
ˇ
ˇ
L
L
D0, soy0D1is orthogonal
to all the other eigenfunctions. Ifmandnare distinct positive integers, then
Z
L
0
cos
mx
L
cos
nx
L
dxD
1
2
Z
L
L
cos
mx
L
cos
nx
L
dxD0, from Example 11.1.4.
11.1.22.Letmandnbe distinct positive integers. From the identity cosAcosBD
1
2
Œcos.AB/C
cos.ACB/withAD.2m1/x=2LandBD.2n1/x=2L,
Z
L
0
cos
.2m1/x
2L
cos
.2n1/x
2L
dxD
1
2
Z
L
0

cos
.mn/x
L
Ccos
.mCn1/x
L

dxD0:
11.1.24.IfyDc1Cc2x, theny
0
.0/D0implies thatc2D0, soyDc1. Now
Z
L
0
y.x/ dxD
c1
Z
L
0
dxDc1LD0only ifc1D0. Therefore,zero is not an eigenvalue.
IfyDc1coshkxCc2sinhkx, theny
0
.0/D0implies thatc2D0, soyDc1coshkx. Now
Z
L
0
y.x/ dxDc1
Z
L
0
coshkx dxDc1
sinhkL
k
D0withk > 0only ifc1D0. Therefore, there are no
negative eigenvalues.
IfyDc1coskxCc2sinkx, theny
0
.0/D0implies thatc2D0, soyDc1coskx. Now
Z
L
0
y.x/ dxDc1
Z
L
0
coskx dxDc2
sinkL
k
D0ifkD
n
L
, wherenis a positive integer. Therefore,
nD
n
2

2
L
2
andynDcos
nx
L
,nD1; 2; 3; : : :.
11.1.26.IfyDc1Cc2.xL/, theny
0
.L/D0implies thatc2D0, soyDc1. Now
Z
L
0
y.x/ dxD
c1
Z
L
0
dxDc1LD0only ifc1D0. Therefore,zero is not an eigenvalue.
IfyDc1coshk.xL/Cc2sinhk.xL/, theny
0
.L/D0implies thatc2D0, soyDc1coshk.x
L/. Now
Z
L
0
y.x/ dxDc1
Z
L
0
coshk.xL/ dxDc1
sinhkL
k
D0withk > 0only ifc1D0.
Therefore,there are no negative eigenvalues.
IfyDc1cosk.xL/Cc2sink.xL/, theny
0
.L/D0implies thatc2D0, soyDc1cosk.xL/.
Now
Z
L
0
y.x/ dxDc1
Z
L
0
cosk.xL/ dxDc2
sinkL
k
D0ifkD
n
L
, wherenis a positive integer.
Therefore,nD
n
2

2
L
2
andynDcos
n.xL/
L
, or, equivalently,ynDcos
nx
L
,nD1; 2; 3; : : :.

228. Section 11.2Fourier Expansions I223
11.2FOURIER EXPANSIONS I
11.2.2.
a0D
1
2
Z
1
1
.2x/ dxD
Z
1
0
2 dxD2I
anD
Z
1
1
.2x/cosnx dxD4
Z
1
0
cosnx dxD
4
n
sinnx
ˇ
ˇ
ˇ
ˇ
1
0
D0I
bnD
Z
1
1
.2x/sinnx dxD 2
Z
1
0
xsinnx dx
D
2
n
"
xcosnx
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
cosnx dx
#
D
2
n
"
cosn
1
n
sinnx
ˇ
ˇ
ˇ
ˇ
1
0
#
D.1/
n
2
n
I
F.x/D2C
2

1
X
nD1
.1/
n
n
sinnx. From Theorem 11.2.4,
F.x/D
8
<
:
2; x D 1;
2x;1 < x < 1;
2; x D1:
11.2.4.Sincefis even,bnD0forn1;a0D
Z
1
0
.13x
2
/ dxD.xx
3
/
ˇ
ˇ
ˇ
ˇ
1
0
D0; ifn1, then
anD2
Z
1
0
.13x
2
/cosnx dxD
2
n
"
.13x
2
/sinnx
ˇ
ˇ
ˇ
ˇ
1
0
C6
Z
1
0
xsinnx dx
#
D
12
n
2

2
"
xcosnx
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
cosnx dx
#
D
12
n
2

2
"
cosn
1
n
sinnx
ˇ
ˇ
ˇ
ˇ
1
0
#
D.1/
nC1
12
n
2

2
I
F.x/D
12

2
1
X
nD1
.1/
n
cosnx
n
2
. From Theorem 11.2.4,F.x/D13x
2
,1x1.
11.2.6.Sincefis odd,anD0ifn0;
b1D
2

Z

0
xcosxsinx dxD
1

Z

0
xsin2x dx
D
1
2

xcos2x
ˇ
ˇ
ˇ
ˇ

0

Z

0
cos2x dx

D
1
2


sin2x
2
ˇ
ˇ
ˇ
ˇ

0

D
1
2
:

229. 224 Chapter 11Boundary Value Problems and Fourier Expansions
ifn2, then
bnD
2

Z

0
xcosxsinnx dxD
1

Z

0
xŒsin.nC1/xCsin.n1/x dx
D
1


x

cos.nC1/x
nC1
C
cos.n1/x
n1
ˇ
ˇ
ˇ
ˇ

0

Z

0

cos.nC1/x
nC1
C
cos.n1/x
n1

dx

D.1/
n

1
nC1
C
1
n1

C
1


sin.nC1/x
.nC1/
2
C
sin.n1/x
.n1/
2
ˇ
ˇ
ˇ
ˇ

0
D.1/
n
2n
n
2
1
I
F.x/D
1
2
sinxC2
1
X
nD2
.1/
n
n
n
2
1
sinnx. From Theorem 11.2.4,F.x/Dxcosx,x.
11.2.8.Sincefis even,bnD0ifn1;a0D
1

Z

0
xsinx dxD
1


xcosx
ˇ
ˇ
ˇ
ˇ

0

Z

0
cosx dx

D
1C
sinx

ˇ
ˇ
ˇ
ˇ

0
D1;a1D
2

Z

0
xsinxcosx dxD
1

Z

0
xsin2x dxD
1
2

xcos2x
ˇ
ˇ
ˇ
ˇ

0

Z

0
cos2x dx

D

1
2
C
sin2x
4
ˇ
ˇ
ˇ
ˇ

0
D
1
2
; ifn2, then
anD
2

Z

0
xsinxcosnx dxD
1

Z

0
xŒsin.nC1/xsin.n1/x dx
D
1


x

cos.n1/x
n1

cos.nC1/x
nC1
ˇ
ˇ
ˇ
ˇ

0

Z

0

cos.n1/x
n1

cos.nC1/x
nC1

dx

D.1/
nC1

1
n1

1
nC1


1


sin.n1/x
.n1/
2

sin.nC1/x
.nC1/
2
ˇ
ˇ
ˇ
ˇ

0
D.1/
nC1
2
n
2
1
I
F.x/D1
1
2
cosx2
1
X
nD2
.1/
n
n
2
1
cosnx. From Theorem 11.2.4,F.x/Dxsinx,x.
11.2.10.Sincefis even,bnD0ifn0;a0D
Z
1=2
0
cosx dxD
sinx

ˇ
ˇ
ˇ
ˇ
1=2
0
D
1

;a1D
2
Z
1=2
0
cos
2
x dxD
Z
1=2
0
.1cos2x/ dxD
1
2

sin2x
2
ˇ
ˇ
ˇ
ˇ
1=2
0
D
1
2
; ifn2, then
anD2
Z
1=2
0
cosxcosnx dxD
Z
1=2
0
Œcos.nC1/xCcos.n1/x dx
D
1


sin.nC1/x
nC1
C
sin.n1/x
n1
ˇ
ˇ
ˇ
ˇ
1=2
0
D
1


1
nC1

1
n1

cos
n
2
D
2
.n
2
1/
cos
n
2
D
8
<
:
.1/
mC1
2
.4m
2
1/
ifnD2m;
0 ifnD2mC1I
F.x/D
1

C
1
2
cosx
2

1
X
nD1
.1/
n
4n
2
1
cos2nx. From Theorem 11.2.4,F.x/Df .x/,1x1.

230. Section 11.2Fourier Expansions I225
11.2.12.Sincefis odd,anD0ifn0;b1D2
Z
1=2
0
sin
2
2x dxD
Z
1=2
0
.1cos4x/ dxD
1
2

sin4x
4
ˇ
ˇ
ˇ
ˇ
1=2
0
D
1
2
; ifn2, then
bnD2
Z
1=2
0
sinxsinnx dxD
Z
1=2
0
Œcos.n1/xcos.nC1/x dx
D
1


sin.n1/x
n1

sin.nC1/x
nC1
ˇ
ˇ
ˇ
ˇ
1=2
0
D
2n
.n
2
1/
cos
n
2
D
(
.1/
mC1
4m
4m
2
1
ifnD2m;
0 ifnD2mC1I
F.x/D
1
2
sinx
4

1
X
nD1
.1/
n
n
4n
2
1
sin2nx. From Theorem 11.2.4,
F.x/D
8
ˆ
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
ˆ
:
0;1x <
1
2
;

1
2
; x D
1
2
;
sinx;
1
2
< x <
1
2
;
1
2
; x D
1
2
;
0;
1
2
< x1:
11.2.14.Sincefis even,bnD0ifn1;
a0D
Z
1=2
0
xsinx dxD
1

"
xcosx
ˇ
ˇ
ˇ
ˇ
1=2
0

Z
1=2
0
cosx dx
#
D
sinx

2
ˇ
ˇ
ˇ
ˇ
1=2
0
D
1

2
I
a1D2
Z
1=2
0
xsinxcosx dxD
Z
1=2
0
xsin2x dx
D
1
2
"
xcos2x
ˇ
ˇ
ˇ
ˇ
1=2
0

Z
1=2
0
cos2x dx
#
D
1
4
C
sin2x
4
2
ˇ
ˇ
ˇ
ˇ
1=2
0
D
1
4
I
ifn2, then
anD2
Z
1=2
0
xsinxcosnx dxD
Z
1=2
0
xŒsin.nC1/xsin.n1/x dx
D
1

"
x

cos.n1/x
n1

cos.nC1/x
nC1
ˇ
ˇ
ˇ
ˇ
1=2
0

Z
1=2
0

cos.n1/x
n1

cos.nC1/x
nC1

dx
#
D
1
2

cos.n1/=2
n1

cos.nC1/=2
nC1


1

2

sin.n1/=2
.n1/
2

sin.nC1/=2
.nC1/
2

D
1

n
n
2
1
sin
n
2
C
2

2
n
2
C1
.n
2
1/
2
cos
n
2
D
8
ˆ
ˆ
<
ˆ
ˆ
:
.1/
m
2

2
4m
2
C1
.4m
2
1/
2
ifnD2m;
.1/
m
1
4
2mC1
m.mC1/
ifnD2mC1I

231. 226 Chapter 11Boundary Value Problems and Fourier Expansions
F.x/D
1

2
C
1
4
cosxC
2

2
1
X
nD1
.1/
n
4n
2
C1
.4n
2
1/
2
cos2nxC
1
4
1
X
nD1
.1/
n
2nC1
n.nC1/
cos.2nC1/x:
From Theorem 11.2.4,
F.x/D
8
ˆ
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
ˆ
:
0; 1x <
1
2
;
1
4
; x D
1
2
;
xsinx;
1
2
< x <
1
2
;
1
4
; x D
1
2
;
0;
1
2
< x1:
11.2.16.Note that
Z
0
1
x
2
g.x/ dxD
Z
1
0
x
2
g.x/ dx; therefore,a0D
1
2
Z
0
1
x
2
dxC
Z
1
0
.1x
2
/ dx

D
1
2
Z
1
0
dxD
1
2
, and ifn1, then
anD
Z
0
1
x
2
cosnx dxC
Z
1
0
.1x
2
/cosnx dxD
Z
1
0
cosnx dxD
sinnx
n
ˇ
ˇ
ˇ
ˇ
1
0
D0
and
bnD
Z
0
1
x
2
sinnx dxC
Z
1
0
.1x
2
/sinnx dxD
Z
1
0
.12x
2
/sinnx dx
D
1
n
"
.12x
2
/cosnx
ˇ
ˇ
ˇ
ˇ
1
0
C4
Z
1
0
xcosnx dx
#
D
1Ccosn
n

4
n
2

2
"
xsinnx
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
sinnx dx
#
D
1Ccosn
n

4cosnx
n
3

3
ˇ
ˇ
ˇ
ˇ
1
0
D
1Ccosn
n
C
4.1cosn/
n
3

3
D
8
ˆ
<
ˆ
:
1
m
ifnD2m;
8
.2mC1/
3

3
ifnD2mC1I
F.x/D
1
2
C
1

1
X
nD1
1
n
sin2nxC
8

3
1
X
nD0
1
.2nC1/
3
sin.2nC1/x:
From Theorem 11.2.4,
F.x/D
8
ˆ
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
ˆ
:
1
2
; x D 1;
x
2
;1 < x < 0;
1
2
; x D0;
1x
2
; 0 < x < 1;
1
2
; x D1:

232. Section 11.2Fourier Expansions I227
11.2.18.a0D
1
6
Z
2
3
2 dxC
Z
2
2
3 dxC
Z
3
2
1 dx

D
5
2
. Ifn1, then
anD
1
3
Z
3
3
f .x/cos
nx
3
dx
D
1
3
Z
2
3
2cos
nx
3
dxC
Z
2
2
3cos
nx
3
dxC
Z
3
2
cos
nx
3
dx

D
3
n
sin
2n
3
;
bnD
1
3
Z
2
3
2sin
nx
3
dxC
Z
2
2
3sin
nx
3
dxC
Z
3
2
sin
nx
3
dx

D
1
n

cosncos
2n
3

I
F.x/D
5
2
C
3

1
X
nD1
1
n
sin
2n
3
cos
nx
3
C
1

1
X
nD1
1
n

cosncos
2n
3

sin
nx
3
:
11.2.20.(a)a0D
1
2
Z


e
x
dxD
e
x
2
ˇ
ˇ
ˇ
ˇ


D
sinh

. Ifn1then (A)anD
1

Z


e
x
cosnx dxand
(B)bnD
1

Z


e
x
sinnx dx. Integrating (B) by parts yields
bnD
1


e
x
sinnx
ˇ
ˇ
ˇ
ˇ


n
Z


e
x
cosnx dx

D nan: . C/
Integrating (A) by parts yields
anD
1


e
x
cosnx
ˇ
ˇ
ˇ
ˇ


Cn
Z


e
x
sinnx dx

D.1/
n
2sinh

CnbnD.1/
n
2sinh

n
2
an;
from (C). Therefore,anD
2sinh

.1/
n
n
2
C1
. Now (C) implies thatbnD
2sinh

.1/
nC1
n
n
2
C1
. Therefore,
F.x/D
sinh

 
1C2
1
X
nD1
.1/
n
n
2
C1
cosnx2
1
X
nD1
.1/
n
n
n
2
C1
sinnx
!
:
(b)From Theorem 11.2.4,F./Dcosh, so
sinh

 
1C2
1
X
nD1
1
n
2
C1
!
Dcosh, which implies
the stated result.
11.2.24.Sincefis even,bnD0,n1,a0D
1

Z

0
coskx dxD
sinkx
k
ˇ
ˇ
ˇ
ˇ

0
D
sink
k
; ifn1then
anD
2

Z

0
coskxcosnx dxD
1

Z

0
Œcos.nk/xCcos.nCk/x dx
D
1


sin.nk/x
nk
C
sin.nCk/x
nCk
ˇ
ˇ
ˇ
ˇ

0
D
cosnsinkx


1
nCk

1
nk

D.1/
nC1
2ksink
.n
2
k
2
/
I
F.x/D
sink

"
1
k
2k
1
X
nD1
.1/
n
n
2
k
2
cosnx
#
.

233. 228 Chapter 11Boundary Value Problems and Fourier Expansions
11.2.26.Sincefis continuous onŒL; Landf .L/Df .L/, Theorem 11.2.4 implies that
f .x/Da0C
1
X
nD1

ancos
nx
L
Cbnsin
nx
L

;LxL;
ifa0D
1
2L
Z
L
L
f .x/ dx, and, forn1,
anD
1
L
Z
L
L
f .x/cos
nx
L
dxD
1
n
"
f .x/sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
L

Z
L
L
f
0
.x/sin
nx
L
dx
#
D
L
n
2

2
"
f
0
.x/cos
nx
L
ˇ
ˇ
ˇ
ˇ
L
L

Z
L
L
f
00
.x/cos
nx
L
dx
#
D
L
n
2

2
Z
L
L
f
00
.x/cos
nx
L
dx
(sincef
0
.L/Df
0
.L/), and
bnD
1
L
Z
L
L
f .x/sin
nx
L
dxD
1
n
"
f .x/cos
nx
L
ˇ
ˇ
ˇ
ˇ
L
L

Z
L
L
f
0
.x/cos
nx
L
dx
#
D
1
n
Z
L
L
f
0
.x/cos
nx
L
dx(sincef .L/Df .L/)
D
L
n
2

2
"
f
0
.x/sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
L

Z
L
L
f
00
.x/sin
nx
L
dx
#
D
L
n
2

2
Z
L
L
f
00
.x/sin
nx
L
dx:
Iff
000
is integrable onŒL; L, then
anD
L
n
2

2
Z
L
L
f
00
.x/cos
nx
L
dxD
L
2
n
3

3
"
f
000
.x/sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
L

Z
L
L
f
000
.x/sin
nx
L
dx
#
D
L
2
n
3

3
Z
L
L
f
000
.x/sin
nx
L
dx:
11.2.28.The Fourier series isa0C
1
X
nD1

ancos
nx
L
Cbnsin
nx
L

where
a0D
1
2L
Z
L
L
f .x/ dxD
1
2L
"
Z
0
L
f .x/ dxC
Z
L
0
f .x/ dx
#
: . A/
Since
Z
0
L
f .x/ dxD
Z
0
L
f .xCL/ dxD
Z
L
0
f .x/ dx, (A) implies thata0D0. Ifn1, then
anD
1
L
Z
L
L
f .x/cos
nx
L
dxD
1
L
"
Z
0
L
f .x/cos
nx
L
dxC
Z
L
0
f .x/cos
nx
L
dx
#
: .B/
Since
Z
0
L
f .x/cos
nx
L
dxD
Z
0
L
f .xCL/cos
nx
L
dxD
Z
L
0
f .x/cos
n.xCL/
L
dx
D.1/
nC1
Z
L
0
f .x/cos
nx
L
dx;

234. Section 11.3Fourier Expansions II229
(B) implies thata2n1DAnanda2nD0,n1. A similar argument showsthatb2n1DBnand
b2nD0,n1.
11.2.30.(b)Let0D1and2mDcos
mx
L
,2m1Dsin
mx
L
,m1. Thenc0Da0andc2mDam,
c2m1Dbm,m1. Since
Z
L
L

2
0
.x/ dxD2Land
Z
L
L

2
2m1
.x/ dxD
1
2
Z
L
L

1cos
2mx
L

dxD
L,
Z
L
L

2
2m
.x/ dxD
1
2
Z
L
L

1Ccos
2mx
L

dxDL,m1, Exercise 12.2.29(d)implies the conclu-
sion.
11.3FOURIER EXPANSIONS II
11.3.2.a0D
Z
1
0
.1x/ dxD
.1x/
2
2
ˇ
ˇ
ˇ
ˇ
1
0
D
1
2
; ifn1,
anD2
Z
1
0
.1x/cosnx dxD
2
n
"
.1x/sinnx
ˇ
ˇ
ˇ
ˇ
1
0
C
Z
1
0
sinnx dx
#
D
2
n
2

2
cosnx
ˇ
ˇ
ˇ
ˇ
1
0
D
2
n
2

2
Œ1.1/
n
D
8
<
:
4
.2m1/
2

2
ifnD2m1;
0 ifnD2mI
C.x/D
1
2
C
4

2
1
X
nD1
1
.2n1/
2
cos.2n1/x:
11.3.4.a0D
1

Z

0
sinkx dxD
coskx
k
ˇ
ˇ
ˇ
ˇ

0
D
1cosk
k
; ifn1, then
anD
2

Z

0
sinkxcosnx dxD
1

Z

0
Œsin.nCk/xsin.nk/x dx
D
1


cos.nk/x
nk

cos.nCk/x
nCk
ˇ
ˇ
ˇ
ˇ

0
D
1


cos.nk/1
nk

cos.nCk/1
nCk

D
2kŒ1.1/
n
cosk
.n
2
k
2
/
I
C.x/D
1cosk
k

2k

1
X
nD1
Œ1.1/
n
cosk
n
2
k
2
cosnx:
11.3.6.a0D
1
L
Z
L
0
.x
2
L
2
/ dxD
1
L

x
3
3
L
2
x
ˇ
ˇ
ˇ
ˇ
L
0
D
2L
2
3
; ifn1,
anD
2
L
Z
L
0
.x
2
L
2
/cos
nx
L
dxD
2
n
"
.x
2
L
2
/sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
2
Z
L
0
xsin
nx
L
dx
#
D
4L
n
2

2
"
xcos
nx
L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
cos
nx
L
dx
#
D.1/
n
4L
2
n
2

2

4L
2
n
3

3
sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
D.1/
n
4L
2
n
2

2
I
C.x/D
2L
2
3
C
4L
2

2
1
X
nD1
.1/
n
n
2
cos
nx
L
:

235. 230 Chapter 11Boundary Value Problems and Fourier Expansions
11.3.8.a0D
Z

0
e
x
dxDe
x
ˇ
ˇ
ˇ
ˇ

0
D
e

1

; ifn1, then
anD
2

Z

0
e
x
cosnx dxD
2


e
x
cosnx
ˇ
ˇ
ˇ
ˇ

0
n
Z

0
e
x
sinnx dx

D
2


.1/
n
e

1ne
x
sinnx
ˇ
ˇ
ˇ
ˇ

0
n
2
Z

0
e
x
cosnx dx

D
2

Œ.1/
n
e

1n
2
anI
.1Cn
2
/anD
2

Œ.1/
n
e

1;anD
2
.n
2
C1/
Œ.1/
n
e

1;
C.x/D
e

1

C
2

1
X
nD1
Œ.1/
n
e

1
.n
2
C1/
cosnx:
11.3.10.a0D
1
L
Z
L
0
.x
2
2Lx/ dxD
1
L

x
3
3
Lx
2
ˇ
ˇ
ˇ
ˇ
L
0
D
2L
2
3
; ifn1,
anD
2
L
Z
L
0
.x
2
2Lx/cos
nx
L
dxD
2
n
"
.x
2
2Lx/sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
2
Z
L
0
.xL/sin
nx
L
dx
#
D
4L
n
2

2
"
.xL/cos
nx
L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
cos
nx
L
dx
#
D
4L
2
n
2

2

4L
3
n
3

3
sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
D
4L
2
n
2

2
I
C.x/D
2L
2
3
C
4L
2

2
1
X
nD1
1
n
2
cos
nx
L
:
11.3.12.bnD2
Z
1
0
.1x/sinnx dxD
2
n
"
.1x/cosnx
ˇ
ˇ
ˇ
ˇ
1
0
C
Z
1
0
cosnx dx
#
D
2
n
C
2
n
2

2
sinnx
ˇ
ˇ
ˇ
ˇ
1
0
D
2
n
;S.x/D
2

1
X
nD1
1
n
sinnx.
11.3.14.bnD
2
L
Z
L=2
0
sin
nx
L
dxD
2
n
cos
nx
L
ˇ
ˇ
ˇ
ˇ
L=2
0
D
2
n
h
1cos
n
2
i
;
S.x/D
2

1
X
nD1
1
n
h
1cos
n
2
i
sin
nx
L
:
11.3.16.
b1D
2

Z

0
xsin
2
x dxD
1

Z

0
x.1cos2x/ dxD
x
2
2
ˇ
ˇ
ˇ
ˇ

0

1

Z

0
xcos2x dx
D

2

1
2

xsin2x
ˇ
ˇ
ˇ
ˇ

0

Z

0
cos2x dx

D

2
C
sin2x
4
ˇ
ˇ
ˇ
ˇ

0
D

2
I

236. Section 11.3Fourier Expansions II231
ifn2, then
bnD
2

Z

0
xsinxsinnx dxD
1

Z

0
xŒcos.n1/xcos.nC1/x dx
D
1


x

sin.n1/x
n1

sin.nC1/x
nC1
ˇ
ˇ
ˇ
ˇ

0

Z

0

sin.n1/x
n1

sin.nC1/x
nC1

dx

D
1


cos.n1/x
.n1/
2

cos.nC1/x
.nC1/
2
ˇ
ˇ
ˇ
ˇ

0
D
1


1
.n1/
2

1
.nC1/
2

fi
.1/
nC1
1

D
4n
.n
2
1/
2

fi
.1/
nC1
1

D
8
<
:
0 ifnD2m1;

16m
.4m
2
1/
ifnD2mI
S.x/D

2
sinx
16

1
X
nD1
n
.4n
2
1/
2
sin2nx:
11.3.18.cnD
2
L
Z
L
0
cos
.2n1/x
2L
dxD
4
.2n1/
sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
D.1/
nC1
4
.2n1/
;
CM.x/D
4

1
X
nD1
.1/
n
2n1
cos
.2n1/x
2L
:
11.3.20.
dnD2
Z
1
0
xcos
.2n1/x
2
dx
D
4
.2n1/
"
xsin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
sin
.2n1/x
2
dx
#
D
4
.2n1/
"
.1/
nC1
C
2
.2n1/
cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
4
.2n1/

.1/
n
C
2
.2n1/

I
CM.x/D
4

1
X
nD1

.1/
n
C
2
.2n1/

cos
.2n1/x
2
:
11.3.22.
cnD
2

Z

0
cosxcos
.2n1/x
2
dxD
1

Z

0

cos.2nC1/x
2
C
cos.2n3/x
2

dx
D
2


sin.2nC1/x=2
2nC1
C
sin.2n3/x=2
2n3
ˇ
ˇ
ˇ
ˇ

0
D.1/
n
2


1
2nC1
C
1
2n3

D.1/
n
4.2n1/
.2n3/.2nC1/
I

237. 232 Chapter 11Boundary Value Problems and Fourier Expansions
CM.x/D
4

1
X
nD1
.1/
n
2n1
.2n3/.2nC1/
cos
.2n1/x
2
:
11.3.24.
cnD
2
L
Z
L
0
.Lxx
2
/cos
.2n1/x
2L
dx
D
4
.2n1/
"
.Lxx
2
/sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
.L2x/sin
.2n1/x
2L
dx
#
D
8L
.2n1/
2

2
"
.L2x/cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
C2
Z
L
0
cos
.2n1/x
2L
dx
#
D
8L
2
.2n1/
2

2
C
32L
2
.2n1/
3

3
sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
D
32L
2
.2n1/
3

3
sin
.2n1/x
2
D
8L
2
.2n1/
2

2
C.1/
n1
32L
2
.2n1/
3

3
I
CM.x/D
8L
2

2
1
X
nD1
1
.2n1/
2

1C
4.1/
n
.2n1/

cos
.2n1/x
2L
:
11.3.26.
dnD
2
L
Z
L
0
x
2
sin
.2n1/x
2L
dx
D
4
.2n1/
"
x
2
cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
2
Z
L
0
xcos
.2n1/x
2L
dx
#
D
16L
.2n1/
2

2
"
xsin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
sin
.2n1/x
2L
dx
#
D.1/
nC1
16L
2
.2n1/
2

2
C
32L
2
.2n1/
3

3
cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
D.1/
nC1
16L
2
.2n1/
2

2

32L
2
.2n1/
3

3
I
SM.x/D
16L
2

2
1
X
nD1
1
.2n1/
2

.1/
n
C
2
.2n1/

sin
.2n1/x
2L
:
11.3.28.
dnD
2

Z

0
cosxsin
.2n1/x
2
dxD
1

Z

0

sin.2nC1/x
2
C
sin.2n3/x
2

dx
D
2


cos.2nC1/x=2
2n1
C
cos.2n3/x=2
2n3
ˇ
ˇ
ˇ
ˇ

0
D
2


1
2nC1
C
1
2n3

D
4.2n1/
.2n3/.2nC1/
I

238. Section 11.3Fourier Expansions II233
SM.x/D
4

1
X
nD1
2n1
.2n3/.2nC1/
sin
.2n1/x
2
:
11.3.30.
dnD
2
L
Z
L
0
.Lxx
2
/sin
.2n1/x
2L
dx
D
4
.2n1/
"
.Lxx
2
/cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
.L2x/cos
.2n1/x
2L
dx
#
D
8L
.2n1/
2

2
"
.L2x/sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
C2
Z
L
0
sin
.2n1/x
2L
dx
#
D.1/
n
8L
2
.2n1/
2

2

32L
2
.2n1/
3

3
cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
D.1/
n
8L
2
.2n1/
2

2
C
32L
2
.2n1/
3

3
I
SM.x/D
8L
2

2
1
X
nD1
1
.2n1/
2

.1/
n
C
4
.2n1/

sin
.2n1/x
2L
:
11.3.32.a0D
1
L
Z
L
0
.3x
4
4Lx
3
/ dxD
1
L

3x
5
5
Lx
4
ˇ
ˇ
ˇ
ˇ
L
0
D
2L
4
5
. Sincef
0
.0/Df
0
.L/D0
andf
000
.x/D24.3xL/,
anD
48L
2
n
3

3
Z
L
0
.3xL/sin
nx
L
dxD
48L
3
n
4

4
"
.3xL/cos
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
3
Z
L
0
cos
nx
L
dx
#
D
48L
3
n
4

4
Œ.1/
n
2LCLC
144L
4
n
5

5
sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
D
48L
4
n
4

4
Œ1C.1/
n
2 ; n1I
C.x/D
2L
4
5

48L
4

4
1
X
nD1
1C.1/
n
2
n
4
cos
nx
L
:
11.3.34.a0D
1
L
Z
L
0
.x
4
2Lx
3
CL
2
x
2
/ dxD
1
L

x
5
5

Lx
4
2
C
L
2
x
3
3
ˇ
ˇ
ˇ
ˇ
L
0
D
L
4
30
. Sincef
0
.0/D
f
0
.L/D0andf
000
.x/D12.2xL/,
anD
24L
2
n
3

3
Z
L
0
.2xL/sin
nx
L
dxD
24L
3
n
4

4
"
.2xL/cos
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
2
Z
L
0
cos
nx
L
dx
#
D
24L
3
n
4

4
Œ.1/
n
LCLC
48L
4
n
5

5
sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
D
24L
4
n
4

4
Œ1C.1/
n

D
8
<
:
0 ifnD2m1;

3L
4
m
4

4
ifnD2m;
n1:

239. 234 Chapter 11Boundary Value Problems and Fourier Expansions
C.x/D
L
4
30

3L
4

4
1
X
nD1
1
n
4
cos
2nx
L
:
11.3.36.Sincef .0/Df .L/D0andf
00
.x/D 2,
bnD
4L
n
2

2
Z
L
0
sin
nx
L
dxD
4L
2
n
3

3
cos
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
D
4L
2
n
3

2
.cosn1/
D
8
<
:
8L
2
.2m1/
3

3
;ifnD2m1;
0; ifnD2mI
S.x/D
8L
2

3
1
X
nD1
1
.2n1/
3
sin
.2n1/x
L
:
11.3.38.Sincef .0/Df .L/D0andf
00
.x/D 6x,
bnD
12L
n
2

2
Z
L
0
xsin
nx
L
dxD
12L
2
n
3

3
"
xcos
nx
L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
cos
nx
L
dx
#
D.1/
nC1
12L
3
n
3

3
I
S.x/D
12L
3

3
1
X
nD1
.1/
n
n
3
sin
nx
L
:
11.3.40.Sincef .0/Df .L/Df
00
.0/Df
00
.L/D0andf
.4/
D360x,
bnD
720L
3
n
4

4
Z
L
0
xsin
nx
L
dxD
720L
4
n
5

5
"
xcos
nx
L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
cos
nx
L
dx
#
D.1/
nC1
720L
5
n
5

5
C
720L
5
n
6

6
sin
nx
L
ˇ
ˇ
ˇ
ˇ
L
0
D.1/
nC1
720L
5
n
5

5
I
S.x/D
720L
5

5
1
X
nD1
.1/
n
n
5
sin
nx
L
:
11.3.42.(a)Sincefis continuous onŒ0; Landf .L/D0, Theorem 11.3.3 implies that
f .x/D
1
X
nD1
cncos
.2n1/x
2L
,LxL, with
cnD
2
L
Z
L
0
f .x/cos
.2n1/x
2L
dx
D
4
.2n1/
"
f .x/sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
f
0
.x/sin
.2n1/x
2L
dx
#
D
4
.2n1/
Z
L
0
f
0
.x/sin
.2n1/x
2L
dx(sincef .L/D0)
D
8L
2
.2n1/
2

2
"
f
0
.x/cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
f
00
.x/cos
.2n1/x
2L
dx
#
D
8L
.2n1/
2

2
Z
L
0
f
00
.x/cos
.2n1/x
2L
dx(sincef
0
.0/D0):

240. Section 11.3Fourier Expansions II235
(b)Continuing the integration by parts yields
cnD
16L
2
.2n1/
3

3
"
f
00
.x/sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
f
000
.x/sin
.2n1/x
2L
dx
#
D
16L
2
.2n1/
3

3
Z
L
0
f
000
.x/sin
.2n1/x
2L
dx:
11.3.44.Sincef
0
.0/Df .L/D0andf
00
.x/D 2,
cnD
16L
.2n1/
2

2
Z
L
0
cos
.2n1/x
2L
dx
D
32L
2
.2n1/
3

3
sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
D.1/
nC1
32L
2
.2n1/
3

3
I
CM.x/D
32L
2

3
1
X
nD1
.1/
n
.2n1/
3
cos
.2n1/x
2L
:
11.3.46.Sincef
0
.0/Df .L/D0andf
00
.x/D6.2xCL/,
cnD
48L
.2n1/
2

2
Z
L
0
.2xCL/cos
.2n1/x
2L
dx
D
96L
2
.2n1/
3

3
"
.2xCL/sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
2
Z
L
0
sin
.2n1/x
2L
#
dx
D
96L
2
.2n1/
3

3
"
.1/
nC1
3L
4L
.2n1/
cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
#
D
96L
3
.2n1/
3

3

.1/
n
3C
4
.2n1/

I
CM.x/D
96L
3

3
1
X
nD1
1
.2n1/
3

.1/
n
3C
4
.2n1/

cos
.2n1/x
2L
:
11.3.48.Sincef
0
.0/Df .L/Df
00
.L/D0andf
000
.x/D12.2xL/,
cnD
192L
2
.2n1/
3

3
Z
L
0
.2xL/sin
.2n1/x
2L
dx
D
384L
3
.2n1/
4

4
"
.2xL/cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
2
Z
L
0
cos
.2n1/x
2L
#
dx
D
384L
3
.2n1/
4

4
"
L
4L
.2n1/
sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
#
D
384L
4
.2n1/
4

4

1C
.1/
n
4
.2n1/

I
CM.x/D
384L
4

4
1
X
nD1
1
.2n1/
4

1C
.1/
n
4
.2n1/

cos
.2n1/x
2L
:

241. 236 Chapter 11Boundary Value Problems and Fourier Expansions
11.3.50.(a)Sincefis continuous onŒ0; Landf .0/D0, Theorem 11.3.4 implies that
f .x/D
1
X
nD1
dnsin
.2n1/x
2L
,LxL, with
dnD
2
L
Z
L
0
f .x/sin
.2n1/x
2L
dx
D
4
.2n1/
"
f .x/cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
f
0
.x/cos
.2n1/x
2L
dx
#
D
4
.2n1/
Z
L
0
f
0
.x/cos
.2n1/x
2L
dx(sincef .0/D0)
D
8L
.2n1/
2

2
"
f
0
.x/sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
f
00
.x/sin
.2n1/x
2L
dx
#
D
8L
.2n1/
2

2
Z
L
0
f
00
.x/sin
.2n1/x
2L
dxsincef
0
.L/D0:
(b)Continuing the integration by parts yields
dnD
16L
2
.2n1/
3

3
"
f
00
.x/cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0

Z
L
0
f
000
.x/cos
.2n1/x
2L
dx
#
D
16L
2
.2n1/
3

3
Z
L
0
f
000
.x/cos
.2n1/x
2L
dx:
11.3.52.Sincef .0/Df
0
.L/D0, andf
00
.x/D6.L2x/
dnD
48L
.2n1/
2

2
Z
L
0
.L2x/sin
.2n1/x
2L
dx
D
96L
2
.2n1/
3

3
"
.L2x/cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
C2
Z
L
0
cos
.2n1/x
2L
dx
#
D
96L
2
.2n1/
3

3
"
LC
4L
.2n1/
sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
#
D
96L
3
.2n1/
3

3

1C.1/
n
4
.2n1/

I
SM.x/D
96L
3

3
1
X
nD1
1
.2n1/
3

1C.1/
n
4
.2n1/

sin
.2n1/x
2L
:
11.3.54.Sincef .0/Df
0
.L/Df
00
.0/D0andf
000
.x/D6,
dnD
96L
2
.2n1/
3

3
Z
L
0
cos
.2n1/x
2L
dx
D
192L
3
.2n1/
4

4
sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
D.1/
n
192L
3
.2n1/
4

4
I

242. Section 11.3Fourier Expansions II237
SM.x/D
192L
3

4
1
X
nD1
.1/
n
.2n1/
4
sin
.2n1/x
2L
:
11.3.56.Sincef .0/Df
0
.L/Df
00
.0/D0andf
000
.x/D12.2xL/,
dnD
192L
2
.2n1/
3

3
Z
L
0
.2xL/cos
.2n1/x
2L
dx
D
384L
3
.2n1/
4

4
"
.2xL/sin
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
2
Z
L
0
sin
.2n1/x
2L
dx
#
D
384L
3
.2n1/
4

4
"
.1/
nC1
LC
4L
.2n1/
cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L
0
#
D
384L
3
.2n1/
4

4

.1/
nC1
L
4L
.2n1/

D
384L
4
.2n1/
4

4

.1/
n
C
4
.2n1/

I
SM.x/D
384L
4

4
1
X
nD1
1
.2n1/
4

.1/
n
C
4
.2n1/

sin
.2n1/x
2L
:
11.3.58.The Fourier sine series off4onŒ0; 2Lis
1
X
nD1
Bnsin
nx
2L
, where
BnD
1
L
Z
2L
0
f4.x/sin
nx
2L
dxD
1
L
"
Z
L
0
f .x/sin
nx
2L
dxC
Z
2L
L
f .2Lx/sin
nx
2L
dx
#
:
Replacingxby2Lxyields
Z
2L
L
f .2Lx/sin
nx
2L
dxD
Z
L
0
f .x/sin
n.2Lx/
2L
dx. Since
sin
n.2Lx/
2L
D.1/
nC1
sin
nx
2L
,
Z
2L
L
f .2Lx/sin
nx
2L
dxD.1/
nC1
Z
L
0
f .x/sin
nx
2L
dx;
so
BnD
1C.1/
nC1
L
Z
L
0
f .x/sin
nx
2L
dxD
8
<
:
2
L
Z
L
0
f .x/sin
.2m1/x
2L
dxifnD2m1;
0 ifnD2m:
Therefore,the Fourier sine series off4onŒ0; 2Lis
1
X
nD1
dnsin
.2n1/x
2L
with
dnD
2
L
Z
L
0
f .x/sin
.2n1/x
2L
dx:

243. 238 Chapter 11Boundary Value Problems and Fourier Expansions
11.3.60.The Fourier cosine series off4onŒ0; 2LisA0C
1
X
nD1
Ancos
nx
2L
, where
A0D
1
2L
Z
2L
0
f4.x/ dxD
1
2L
"
Z
L
0
f .x/ dxC
Z
2L
L
f .2Lx/ dx
#
D
1
L
Z
L
0
f .x/ dx
and
AnD
1
L
Z
2L
0
f4.x/cos
nx
2L
dxD
1
L
"
Z
L
0
f .x/cos
nx
2L
dxC
Z
2L
L
f .2Lx/cos
nx
2L
dx
#
:
Replacingxby2Lxyields
Z
2L
L
f .2Lx/cos
nx
2L
dxD
Z
0
L
f .x/cos
n.2Lx/
2L
dxD
Z
L
0
f .x/cos
n.2Lx/
2L
dx:
Since cos
n.2Lx/
2L
Dcosncos
nx
2L
D.1/
n
cos
nx
2L
,
AnD
1C.1/
n
L
Z
L
0
f .x/cos
nx
2L
dxD
8
<
:
0 ifnD2m1
2
L
Z
L
0
f .x/cos
mx
L
dxifnD2m:
Therefore,the Fourier cosine series off4onŒ0; 2LisA0C
1
X
nD0
A2ncos
nx
L
=a0C
1
X
nD0
ancos
nx
L
.

244. CHAPTER12
FourierSolutionsofPartialDifferential
12.1THE HEAT EQUATION
12.1.2.X.x/T .t/satisfiesutDa
2
uxxifX
00
CXD0and (A)T
0
D a
2
Tfor the same value
of. The product also satisfies the boundary conditionsu.0; t/Dux.L; t/D0; t > 0, if and only
ifX.0/DX
0
.L/D0. Since we are interested in nontrivial solutions,Xmust be a nontrivial solution
of (B)X
00
CXD0; X.0/D0; X
0
.L/D0. From Theorem 11.1.4,nD.2n1/
2

2
=4L
2
is an
eigenvalue of (B) with associated eigenfunctionXnDsin
.2n1/x
2L
,nD1; 2; 3; : : :. Substituting
D.2n1/
2

2
=4L
2
into (A) yieldsT
0
D ..2n1/
2

2
a
2
=4L
2
/T, which has the solutionTnD
e
.2n1/
2

2
a
2
t =4L
2
.
We have now shown that the functionsun.x; t/De
.2n1/
2

2
a
2
t =4L
2
sin
.2n1/x
2L
,nD1; 2; 3; : : :
satisfyutDa
2
uxxand the boundary conditionsu.0; t/Dux.L; t/D0; t > 0. Any finite sum
m
X
nD1
dne
.2n1/
2

2
a
2
t =4L
2
sin
.2n1/x
2L
also has these properties. Therefore,it is plausible to expect
that that this is also true of the infinite series (C)u.x; t/D
1
X
nD1
dne
.2n1/
2

2
a
2
t =4L
2
sin
.2n1/x
2L
under suitable conditions on the coefficientsfdng. Sinceu.x; 0/D
1
X
nD1
dnsin
.2n1/x
2L
, iffdngare
the mixed Fourier sine coefficients offonŒ0; L, thenu.x; 0/Df .x/at all pointsxinŒ0; Lwhere the
mixed Fourier sine series converges tof .x/. In this case (C) is a formal solution of the initial-boundary
value problem of Definition 12.1.3.
12.1.8.Sincef .0/Df .1/D0andf
00
.x/D 2, Theorem 11.3.5(b)implies that
˛nD
4
n
2

2
Z
1
0
sinnx dxD
4
n
3

3
cosnx
ˇ
ˇ
ˇ
ˇ
1
0
D
4
n
3

2
.cosn1/
D
8
<
:
8
.2m1/
3

3
;ifnD2m1;
0; ifnD2mI
S.x/D
8

3
1
X
nD1
1
.2n1/
3
sin
.2n1/x
L
. From Definition 12.1.1,
u.x; t/D
8

3
1
X
nD1
1
.2n1/
3
e
.2n1/
2

2
t
sin.2n1/x:
239

245. 240 Chapter 12Fourier Solutions of Partial Differential
12.1.10.
˛1D
2

Z

0
xsin
2
x dxD
1

Z

0
x.1cos2x/ dxD
x
2
2
ˇ
ˇ
ˇ
ˇ

0

1

Z

0
xcos2x dx
D

2

1
2

xsin2x
ˇ
ˇ
ˇ
ˇ

0

Z

0
cos2x dx

D

2
C
sin2x
4
ˇ
ˇ
ˇ
ˇ

0
D

2
I
ifn2, then
˛nD
2

Z

0
xsinxsinnx dxD
1

Z

0
xŒcos.n1/xcos.nC1/x dx
D
1


x

sin.n1/x
n1

sin.nC1/x
nC1
ˇ
ˇ
ˇ
ˇ

0

Z

0

sin.n1/x
n1

sin.nC1/x
nC1

dx

D
1


cos.n1/x
.n1/
2

cos.nC1/x
.nC1/
2
ˇ
ˇ
ˇ
ˇ

0
D
1


1
.n1/
2

1
.nC1/
2

fi
.1/
nC1
1

D
4n
.n
2
1/
2

fi
.1/
nC1
1

D
8
<
:
0 ifnD2m1;

16m
.4m
2
1/
ifnD2mI
S.x/D

2
sinx
16

1
X
nD1
n
.4n
2
1/
2
sin2nx. From Definition 12.1.1,
u.x; t/D

2
e
3t
sinx
16

1
X
nD1
n
.4n
2
1/
2
e
12n
2
t
sin2nx:
12.1.12.Sincef .0/Df .L/D0andf
00
.x/D 6x, Theorem 11.3.5(b)implies that
˛nD
36
n
2

2
Z
3
0
xsin
nx
3
dxD
108
n
3

3
"
xcos
nx
3
ˇ
ˇ
ˇ
ˇ
3
0

Z
3
0
cos
nx
3
dx
#
D.1/
nC1
108
n
3

3
C
108
n
4

4
sin
nx
3
ˇ
ˇ
ˇ
ˇ
3
0
D.1/
nC1
324
n
3

3
I
S.x/D
324

3
1
X
nD1
.1/
n
n
3
sin
nx
3
. From Definition 12.1.1,u.x; t/D
324

3
1
X
nD1
.1/
n
n
3
e
4n
2

2
t =9
sin
nx
3
.
12.1.14.Sincef .0/Df .1/Df
00
.0/Df
00
.L/D0andf
.4/
D360x, Theorem 11.3.5(b)and
Exercise 35(b)of Section 11.3 imply that
˛nD
720
n
4

4
Z
1
0
xsinnx dxD
720
n
5

5
"
xcosnx
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
cosnx dx
#
D.1/
nC1
720
n
5

5
C
720
n
6

6
sin
nx
L
ˇ
ˇ
ˇ
ˇ
1
0
D.1/
nC1
720
n
5

5
I
S.x/D
720

5
1
X
nD1
.1/
n
n
5
sinnx. From Definition 12.1.1,u.x; t/D
720

5
1
X
nD1
.1/
n
n
5
e
7n
2

2
t
sinnx.

246. Section 12.1The Heat Equation241
12.1.16.Sincef .0/Df .1/Df
00
.0/Df
00
.L/D0andf
.4/
D120.3x1/, Theorem 11.3.5(b)and
Exercise 35(b)of Section 11.3 imply that
˛nD
240
n
4

4
Z
1
0
.3x1/sinnx dxD
240
n
5

5
"
.3x1/cosnx
ˇ
ˇ
ˇ
ˇ
1
0
3
Z
1
0
cosnx dx
#
D
240
n
5

5
Œ.1/
n
2C1C
720
n
6

6
sinnx
ˇ
ˇ
ˇ
ˇ
1
0
D
240
n
5

5
Œ1C.1/
n
2I
S.x/D
240

5
1
X
nD1
1C.1/
n
2
n
5
sinnx. From Definition 12.1.1,
u.x; t/D
240

5
1
X
nD1
1C.1/
n
2
n
5
e
2n
2

2
t
sinnx:
12.1.18.˛0D
1
2
Z
2
0
.x
2
4x/ dxD
1
L

x
3
3
2x
2
ˇ
ˇ
ˇ
ˇ
2
0
D
8
3
; ifn1,
˛nD
Z
2
0
.x
2
4x/cos
nx
2
dxD
2
n
"
.x
2
4x/sin
nx
2
ˇ
ˇ
ˇ
ˇ
2
0
2
Z
2
0
.x2/sin
nx
2
dx
#
D
8
n
2

2
"
.x2/cos
nx
2
ˇ
ˇ
ˇ
ˇ
2
0

Z
2
0
cos
nx
2
dx
#
D
16
n
2

2

32
n
3

3
sin
nx
2
ˇ
ˇ
ˇ
ˇ
2
0
D
16
n
2

2
I
C.x/D
8
3
C
16

2
1
X
nD1
1
n
2
cos
nx
2
. From Definition 12.1.3,u.x; t/D
8
3
C
16

2
1
X
nD1
1
n
2
e
n
2

2
t
cos
nx
2
.
12.1.20.From Example 11.3.5,C.x/D4
384

4
1
X
nD1
1
.2n1/
4
cos
.2n1/x
2
. From Definition 12.1.3,
u.x; t/D4
384

4
1
X
nD1
1
.2n1/
4
e
3.2n1/
2

2
t =4
cos
.2n1/x
2
.
12.1.22.˛0D
1
L
Z
1
0
.3x
4
4Lx
3
/ dxD
1
L

3x
5
5
x
4
ˇ
ˇ
ˇ
ˇ
1
0
D
2
5
. Sincef
0
.0/Df
0
.1/D0and
f
000
.x/D24.3x1/, Theorem 11.3.5(a)implies that
˛nD
48
n
3

3
Z
1
0
.3x1/sinnx dxD
48
n
4

4
"
.3x1/cosnx
ˇ
ˇ
ˇ
ˇ
1
0
3
Z
1
0
cosnx dx
#
D
48
n
4

4
Œ.1/
n
2C1C
144
n
5

5
sinnx
ˇ
ˇ
ˇ
ˇ
1
0
D
48
n
4

4
Œ1C.1/
n
2 ; n1I
C.x/D
2
5

48

4
1
X
nD1
1C.1/
n
2
n
4
cosnx. From Definition 12.1.3,
u.x; t/D
2
5

48

4
1
X
nD1
1C.1/
n
2
n
4
e
3n
2

2
t
cosnx:

247. 242 Chapter 12Fourier Solutions of Partial Differential
12.1.24.˛0D
1

Z

0
.x
4
2x
3
C
2
x
2
/ dxD
1


x
5
5

x
4
2
C

2
x
3
3
ˇ
ˇ
ˇ
ˇ

0
D

4
30
. Sincef
0
.0/D
f
0
./D0andf
000
.x/D12.2x/, Theorem 11.3.5(a)implies that
˛nD
24
n
3

Z

0
.2x/sinnx dxD
24
n
4


.2x/cosnx
ˇ
ˇ
ˇ
ˇ

0
2
Z

0
cosnx dx

D
24
n
4

Œ.1/
n
CC
48
n
5

sinnx
ˇ
ˇ
ˇ
ˇ

0
D
24
n
4
Œ1C.1/
n

D
(
0 ifnD2m1;

3
m
4
ifnD2m;
n1I
C.x/D

4
30
3
1
X
nD1
1
n
4
cos2nx. From Definition 12.1.3,u.x; t/D

4
30
3
1
X
nD1
1
n
4
e
4n
2
t
cos2nx.
12.1.26.
˛nD
2

Z

0
.xx
2
/sin
.2n1/x
2
dx
D
4
.2n1/

.xx
2
/cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ

0

Z

0
.2x/cos
.2n1/x
2
dx

D
8
.2n1/
2


.2x/sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ

0
C2
Z

0
sin
.2n1/x
2
dx

D.1/
n
8
.2n1/
2

32
.2n1/
3

cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ

0
D.1/
n
8
.2n1/
2
C
32
.2n1/
3

I
SM.x/D8
1
X
nD1
1
.2n1/
2

.1/
n
C
4
.2n1/

sin
.2n1/x
2
. From Definition 12.1.4,
u.x; t/D8
1
X
nD1
1
.2n1/
2

.1/
n
C
4
.2n1/

e
3.2n1/
2
t =4
sin
.2n1/x
2
:
12.1.28.Sincef .0/Df
0
.1/D0, andf
00
.x/D6.12x/, Theorem 11.3.5(d)implies that
˛nD
48
.2n1/
2

2
Z
1
0
.12x/sin
.2n1/x
2
dx
D
96
.2n1/
3

3
"
.12x/cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
C2
Z
1
0
cos
.2n1/x
2
dx
#
D
96
.2n1/
3

3
"
1C
4
.2n1/
sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
96
.2n1/
3

3

1C.1/
n
4
.2n1/

I

248. Section 12.1The Heat Equation243
SM.x/D
96

3
1
X
nD1
1
.2n1/
3

1C.1/
n
4
.2n1/

sin
.2n1/x
2
. From Definition 12.1.4,
u.x; t/D
96

3
1
X
nD1
1
.2n1/
3

1C.1/
n
4
.2n1/

e
.2n1/
2

2
t
sin
.2n1/x
2
:
12.1.30.Sincef .0/Df
0
.L/Df
00
.0/D0andf
000
.x/D6, Theorem 11.3.5(d)and Exercise 11.3.50(b)
imply that
˛nD
96
.2n1/
3

3
Z
1
0
cos
.2n1/x
2
dx
D
192
.2n1/
4

4
sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
D.1/
n
192
.2n1/
4

4
I
SM.x/D
192

4
1
X
nD1
.1/
n
.2n1/
4
sin
.2n1/x
2
. From Definition 12.1.4,
u.x; t/D
192

4
1
X
nD1
.1/
n
.2n1/
4
e
.2n1/
2

2
t
sin
.2n1/x
2
:
12.1.32.Sincef .0/Df
0
.1/Df
00
.0/D0andf
000
.x/D12.2x1/, Theorem 11.3.5(d)and Exer-
cise 11.3.50(b)imply that
˛nD
192
.2n1/
3

3
Z
1
0
.2x1/cos
.2n1/x
2
dx
D
384
.2n1/
4

4
"
.2x1/sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
2
Z
1
0
sin
.2n1/x
2
dx
#
D
384
.2n1/
4

4
"
.1/
nC1
1C
4
.2n1/
cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
384
.2n1/
4

4

.1/
nC1
1
4
.2n1/

D
384
.2n1/
4

4

.1/
n
C
4
.2n1/

I
SM.x/D
384

4
1
X
nD1
1
.2n1/
4

.1/
n
C
4
.2n1/

sin
.2n1/x
2
. From Definition 12.1.4,
u.x; t/D
384

4
1
X
nD1
1
.2n1/
4

.1/
n
C
4
.2n1/

e
.2n1/
2

2
t
sin
.2n1/x
2
:
12.1.36.From Example 11.3.3,CM.x/D
8

2
1
X
nD1
1
.2n1/
2
cos
.2n1/x
2
. From Definition 12.1.5,
u.x; t/D
8

2
1
X
nD1
1
.2n1/
2
e
3.2n1/
2

2
t =4
cos
.2n1/x
2
:

249. 244 Chapter 12Fourier Solutions of Partial Differential
12.1.38.Sincef
0
.0/Df ./D0andf
00
.x/D 2, Theorem 11.3.5(c)implies that
˛nD
16
.2n1/
2

Z

0
cos
.2n1/x
2
dxD
32
.2n1/
3

sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ

0
D.1/
nC1
32
.2n1/
3

I
CM.x/D
32

1
X
nD1
.1/
n
.2n1/
3
cos
.2n1/x
2
. From Definition 12.1.5,
u.x; t/D
32

1
X
nD1
.1/
n
.2n1/
3
e
7.2n1/
2
t =4
cos
.2n1/x
2
:
12.1.40.Sincef
0
.0/Df .1/D0andf
00
.x/D6.2xC1/, Theorem 11.3.5(c)implies that
˛nD
48L
.2n1/
2

2
Z
1
0
.2xC1/cos
.2n1/x
2
dx
D
96
.2n1/
3

3
"
.2xC1/sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
2
Z
1
0
sin
.2n1/x
2
#
dx
D
96
.2n1/
3

3
"
.1/
nC1
3
4
.2n1/
cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
96
.2n1/
3

3

.1/
n
3C
4
.2n1/

I
CM.x/D
96

3
1
X
nD1
1
.2n1/
3

.1/
n
3C
4
.2n1/

cos
.2n1/x
2
. From Definition 12.1.5,
u.x; t/D
96

3
1
X
nD1
1
.2n1/
3

.1/
n
3C
4
.2n1/

e
.2n1/
2

2
t =4
cos
.2n1/x
2
:
12.1.42.Theorem 11.3.5(c)and Exercise 11.3.42(b)imply that Sincef
0
.0/Df .1/Df
00
.1/D0and
f
000
.x/D12.2x1/,
˛nD
192
.2n1/
3

3
Z
1
0
.2x1/sin
.2n1/x
2
dx
D
384
.2n1/
4

4
"
.2x1/cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
2
Z
1
0
cos
.2n1/x
2
#
dx
D
384
.2n1/
4

4
"
1
4
.2n1/
sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
384
.2n1/
4

4

1C
.1/
n
4
.2n1/

I
CM.x/D
384

4
1
X
nD1
1
.2n1/
4

1C
.1/
n
4
.2n1/

cos
.2n1/x
2
. From Definition 12.1.5,
u.x; t/D
384

4
1
X
nD1
1
.2n1/
4

1C
.1/
n
4
.2n1/

e
.2n1/
2

2
t =4
cos
.2n1/x
2
:

250. Section 12.1The Heat Equation245
12.1.44.˛nD
2
L
Z
L=2
0
sin
nx
L
dxD
2
n
cos
nx
L
ˇ
ˇ
ˇ
ˇ
L=2
0
D
2
n
h
1cos
n
2
i
;
S.x/D
2

1
X
nD1
1
n
h
1cos
n
2
i
sin
nx
L
. From Definition 12.1.1,
u.x; t/D
2

1
X
nD1
1
n
h
1cos
n
2
i
e
n
2

2
t
2
=L
2
sin
nx
L
:
12.1.46.
˛nD
2
L
Z
L=2
0
sin
.2n1/x
2L
dxD
4
.2n1/
cos
.2n1/x
2L
ˇ
ˇ
ˇ
ˇ
L=2
0
D
4
.2n1/

1cos
.2n1//
4

I
SM.x/D
4

1
X
nD1
1
2n1

1cos
.2n1//
4

sin
.2n1/x
2L
. From Definition 12.1.4,
u.x; t/D
4

1
X
nD1
1
2n1

1cos
.2n1//
4

e
.2n1/
2

2
a
2
t =4L
2
sin
.2n1/x
2L
:
12.1.48.Letu.x; t/Dv.x; t/Cq.x/; thenutDvtanduxxDvxxCq
00
, so
vtD9vxxC9q
00
54x; 0 < x < 4; t > 0;
v.0; t/D1q.0/; v.4; t/D61q.4/; t > 0;
v.x; 0/D2xCx
3
q.x/; 0x4:
.A/
We wantq
00
.x/D6x; q.0/D1; q.4/D61;q.x/Dx
3
Ca1Ca2x;q.0/D1)a1D1;
q.x/Dx
3
C1Ca2x;q.4/D61)a2D 1;q.x/Dx
3
C1x. Now (A) reduces to
vtD9vxx; 0 < x < 4; t > 0;
v.0; t/D0; v.4; t/D0; t > 0;
v.x; 0/D1; 0x4;
which we solve by separation of variables.
˛nD
1
2
Z
L
0
sin
nx
4
dxD
2
n
cos
nx
4
ˇ
ˇ
ˇ
ˇ
4
0
D
2
n
Œ1.1/
n
D
8
<
:
4
.2m1/
ifnD2m1;
0 ifnD2mI
S.x/D
4

1
X
nD1
1
.2n1/
sin
.2n1/x
4
. From Definition 12.1.1,
v.x; t/D
4

1
X
nD1
1
.2n1/
e
9
2
n
2
t =16
sin
.2n1/x
4
:

251. 246 Chapter 12Fourier Solutions of Partial Differential
Therefore,
u.x; t/D1xCx
3
C
4

1
X
nD1
e
9
2
n
2
t =16
.2n1/
sin
.2n1/x
4
:
12.1.50.Letu.x; t/Dv.x; t/Cq.x/; thenutDvtanduxxDvxxCq
00
, so
vtD3uxxC3q
00
18x; 0 < x < 1; t > 0;
vx.0; t/D 1q
0
.0/; v.1; t/D 1q.1/; t > 0;
v.x; 0/Dx
3
2xq.x/; 0x1:
.A/
We wantq
00
.x/D6x; q
0
.0/D 1; q.1/D 1;q
0
.x/D3x
2
Ca2;q
0
.0/D 1)a2D 1;
q
0
.x/D3x
2
1;q.x/Dx
3
xCa1x;q.1/D 1)a1D 1;q.x/Dx
3
x1. Now (A) reduces
to
vtD3uxx; 0 < x < 1; t > 0;
vx.0; t/D0; v.1; t/D0; t > 0;
v.x; 0/D1x; 0x1:
From Example 11.3.3,CM.x/D
8

2
1
X
nD1
1
.2n1/
2
cos
.2n1/x
2
. From Definition 12.1.5,v.x; t/D
8

2
1
X
nD1
1
.2n1/
2
e
3.2n1/
2

2
t =4
cos
.2n1/x
2
. Therefore,
u.x; t/D 1xCx
3
C
8

2
1
X
nD1
1
.2n1/
2
e
3.2n1/
2

2
t =4
cos
.2n1/x
2
:
12.1.52.Letu.x; t/Dv.x; t/Cq.x/; thenutDvtanduxxDvxxCq
00
, so
vtDvxxCq
00
C
2
sinx; 0 < x < 1; t > 0;
v.0; t/D q.0/; vx.1; t/D q
0
.1/; t > 0;
v.x; 0/D2sinxq.x/; 0x1:
.A/
We wantq
00
.x/D
2
sinx; q.0/D0; q
0
.1/D ;q
0
.x/DcosxCa2;q
0
.1/D )a2D0;
q
0
.x/Dcosx;q.x/DsinxCa1;q.0/D0)a1D0;q.x/Dsinx. Now (A) reduces to
vtDvxx; 0 < x < 1; t > 0;
v.0; t/D0; vx.1; t/D0; t > 0;
v.x; 0/Dsinx; 0x1:
˛nD2
Z
1
0
sinxsin
.2n1/x
2
dxD
Z
1
0

cos.2n3/x
2

cos.2nC1/x
2

dx
D
2


sin.2n3/x=2
.2n3/

sin.2nC1/x=2
.2nC1/
ˇ
ˇ
ˇ
ˇ
1
0
D.1/
n
2


1
2n3

1
2nC1

D.1/
n
8

1
.2nC1/.2n3/
I
SM.x/D
8

1
X
nD1
.1/
n
.2nC1/.2n3/
sin
.2n1/x
2
. From Definition 12.1.4,
v.x; t/D
8

1
X
nD1
.1/
n
.2nC1/.2n3/
e
.2n1/
2

2
t =4
sin
.2n1/x
2
:

252. Section 12.2The Wave Equation247
Therefore,u.x; t/DsinxC
8

1
X
nD1
.1/
n
.2nC1/.2n3/
e
.2n1/
2

2
t =4
sin
.2n1/x
2
.
12.1.54.(a)Sincefis piecewise smooth ofŒ0; L, there is a constantKsuch thatjf .x/j K,0x
L. Therefore,j˛nj D
2
L
ˇ
ˇ
ˇ
ˇ
ˇ
Z
L
0
f .x/sin
nx
L
dx
ˇ
ˇ
ˇ
ˇ
ˇ

2
L
Z
L
0
jf .x/jdxD2K. Hence,j˛ne
n
2

2
a
2
t =L
j
2Ke
n
2

2
a
2
t =L
, sou.x; t/converges for allxift > 0, by the comparison test.
(b)Lettbe a fixed positive number. Apply Theorem 12.1.2 with´Dxandwn.x/D˛ne
n
2

2
t =L
2
sin
nx
L
.
Thenw
0
n
.x/D

L
n˛ne
n
2

2
t =L
2
cos
nx
L
, sojw
0
n
.x/j
2K
L
ne
n
2

2
t =L
2
,1< x <1. Since
1
X
nD1
ne
n
2

2
a
2
t =L
2
converges ift > 0, Theorem 12.1.1 (with´1Dx1and´2Dx2arbitrary) implies
the concclusion.
(c)Since
1
X
nD1
n
2
e
n
2

2
a
2
t =L
2
also converges ift > 0, an argument like that in(b)withwn.x/D
n˛ne
n
2

2
t =L
2
cos
nx
L
yields the conclusion.
(d)Letxbe arbitrary, but fixed. Apply Theorem 12.1.2 with´Dtand
wn.t/D˛ne
n
2

2
a
2
t =L
2
sin
nx
L
. Thenw
0
n
.t/D

2
a
2
L
2
n
2
˛ne
n
2

2
a
2
t =L
2
sin
nx
L
, sojw
0
n
.t/j
2K
2
a
2
L
n
2
e
n
2

2
a
2
t0=L
2
ift > t0. Since
1
X
nD1
n
2
e
n
2

2
a
2
t0=L
2
converges, Theorem 12.1.1 (with´1D
t0> 0and´2Dt1arbitrary implies the conclusion fortt0. However, sincet0is an arbitrary positive
number, this holds fort > 0.
12.2THE WAVE EQUATION
12.2.1.ˇnD2
h
R
1=2
0
xsinnxC
R
1
1=2
.1x/sinnx dx
i
;
Z
1=2
0
xsinnx dxD
1
n
"
xcosnx
ˇ
ˇ
ˇ
ˇ
1=2
0

Z
1=2
0
cosnx dx
#
D
1
2n
cos
n
2
C
1
n
2

2
sinnx
ˇ
ˇ
ˇ
ˇ
1=2
0
D
1
2n
cos
n
2
C
1
n
2

2
sin
n
2
I
Z
1=2
0
.1x/sinnx dxD
1
n
"
.1x/cosnx
ˇ
ˇ
ˇ
ˇ
1
1=2
C
Z
1=2
0
cosnx dx
#
D
1
2n
cos
n
2

1
n
2

2
sinnx
ˇ
ˇ
ˇ
ˇ
1
1=2
D
1
2n
cos
n
2
C
1
n
2

2
sin
n
2
I
bnD
4
n
2

2
sin
n
2
D
8
<
:
.1/
mC1
4
.2m1/
2

2
ifnD2m1
0 ifnD2mI

253. 248 Chapter 12Fourier Solutions of Partial Differential Equations
Sg.x/D
4

2
1
X
nD1
.1/
nC1
.2n1/
2
sin.2n1/x. From Definition 12.1.1,
u.x; t/D
4
3
3
1
X
nD1
.1/
nC1
.2n1/
3
sin3.2n1/tsin.2n1/x:
12.2.2.Sincef .0/Df .1/D0andf
00
.x/D 2, Theorem 11.3.5(b)implies that
˛nD
4
n
2

2
Z
1
0
sinnx dxD
4
n
3

3
cosnx
ˇ
ˇ
ˇ
ˇ
1
0
D
4
n
3

2
.cosn1/
D
8
<
:
8
.2m1/
3

3
;ifnD2m1;
0; ifnD2mI
Sf.x/D
8

3
1
X
nD1
1
.2n1/
3
sin.2n1/x. From Definition 12.1.1,
u.x; t/D
8

3
1
X
nD1
1
.2n1/
3
cos3.2n1/tsin.2n1/x:
12.2.4.Sinceg.0/Dg.1/D0andg
00
.x/D 2, Theorem 11.3.5(b)implies that
ˇnD
4
n
2

2
Z
1
0
sinnx dxD
4
n
3

3
cosnx
ˇ
ˇ
ˇ
ˇ
1
0
D
4
n
3

2
.cosn1/
D
8
<
:
8
.2m1/
3

3
;ifnD2m1;
0; ifnD2mI
Sg.x/D
8

3
1
X
nD1
1
.2n1/
3
sin.2n1/x. From Definition 12.1.1,
u.x; t/D
8
3
4
1
X
nD1
1
.2n1/
4
sin3.2n1/tsin.2n1/x:
12.2.6.From Example 11.2.6,Sf.x/D
324

3
1
X
nD1
.1/
n
n
3
sin
nx
3
. From Definition 12.1.1,
u.x; t/D
324

3
1
X
nD1
.1/
n
n
3
cos
8 nt
3
sin
nx
3
:
12.2.8.From Example 11.2.6Sg.x/D
324

3
1
X
nD1
.1/
n
n
3
sin
nx
3
. From Definition 12.1.1,
u.x; t/D
81
2
4
1
X
nD1
.1/
n
n
4
sin
8 nt
3
sin
nx
3
:

254. Section 12.2The Wave Equation249
12.2.10.
˛1D
2

Z

0
xsin
2
x dxD
1

Z

0
x.1cos2x/ dxD
x
2
2
ˇ
ˇ
ˇ
ˇ

0

1

Z

0
xcos2x dx
D

2

1
2

xsin2x
ˇ
ˇ
ˇ
ˇ

0

Z

0
cos2x dx

D

2
C
sin2x
4
ˇ
ˇ
ˇ
ˇ

0
D

2
I
ifn2, then
˛nD
2

Z

0
xsinxsinnx dxD
1

Z

0
xŒcos.n1/xcos.nC1/x dx
D
1


x

sin.n1/x
n1

sin.nC1/x
nC1
ˇ
ˇ
ˇ
ˇ

0

Z

0

sin.n1/x
n1

sin.nC1/x
nC1

dx

D
1


cos.n1/x
.n1/
2

cos.nC1/x
.nC1/
2
ˇ
ˇ
ˇ
ˇ

0
D
1


1
.n1/
2

1
.nC1/
2

fi
.1/
nC1
1

D
4n
.n
2
1/
2

fi
.1/
nC1
1

D
8
<
:
0 ifnD2m1;

16m
.4m
2
1/
ifnD2mI
Sf.x/D

2
sinx
16

1
X
nD1
n
.4n
2
1/
2
sin2nx. From Definition 12.1.1,
u.x; t/D

2
cos
p
5 tsinx
16

1
X
nD1
n
.4n
2
1/
2
cos2n
p
5 tsin2nx:
12.2.12.
ˇ1D
2

Z

0
xsin
2
x dxD
1

Z

0
x.1cos2x/ dxD
x
2
2
ˇ
ˇ
ˇ
ˇ

0

1

Z

0
xcos2x dx
D

2

1
2

xsin2x
ˇ
ˇ
ˇ
ˇ

0

Z

0
cos2x dx

D

2
C
sin2x
4
ˇ
ˇ
ˇ
ˇ

0
D

2
I
ifn2then
ˇnD
2

Z

0
xsinxsinnx dxD
1

Z

0
xŒcos.n1/xcos.nC1/x dx
D
1


x

sin.n1/x
n1

sin.nC1/x
nC1
ˇ
ˇ
ˇ
ˇ

0

Z

0

sin.n1/x
n1

sin.nC1/x
nC1

dx

D
1


cos.n1/x
.n1/
2

cos.nC1/x
.nC1/
2
ˇ
ˇ
ˇ
ˇ

0
D
1


1
.n1/
2

1
.nC1/
2

fi
.1/
nC1
1

D
4n
.n
2
1/
2

fi
.1/
nC1
1

D
8
<
:
0 ifnD2m1;

16m
.4m
2
1/
ifnD2mI
Sg.x/D

2
sinx
16

1
X
nD1
n
.4n
2
1/
2
sin2nx. From Definition 12.1.1,
u.x; t/D

2
p
5
sin
p
5 tsinx
8

p
5
1
X
nD1
1
.4n
2
1/
2
sin2n
p
5 tsin2nx:

255. 250 Chapter 12Fourier Solutions of Partial Differential Equations
12.2.14.Sincef .0/Df .1/Df
00
.0/Df
00
.L/D0andf
.4/
D360x, Theorem 11.3.5(b)and
Exercise 35(b)of Section 11.3 imply that
˛nD
720
n
4

4
Z
1
0
xsinnx dxD
720
n
5

5
"
xcosnx
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
cosnx dx
#
D.1/
nC1
720
n
5

5
C
720
n
6

6
sin
nx
L
ˇ
ˇ
ˇ
ˇ
1
0
D.1/
nC1
720
n
5

5
I
Sf.x/D
720

5
1
X
nD1
.1/
n
n
5
sinnx. From Definition 12.1.1,u.x; t/D
720

5
1
X
nD1
.1/
n
n
5
cos3ntsinnx.
12.2.16.(a)tmust be in some interval of the formŒmL=a; .mC1/L=a. If
mL
a
t

mC
1
2

L
a
,
then(i)holds with0L=2a. If

mC
1
2

L
a
t
.mC1/L
a
, then(ii)holds with0L=2a.
(b)Suppose that(i)holds. Since
cos
.2n1/a
L

C
mL
a

Dcos
.2n1/a
L
cos.2n1/mD.1/
m
cos
.2n1/a
L
;
(A) implies thatu.x; t/D.1/
m
u.x; /.
Suppose that(ii)holds. Since
cos
.2n1/a
L

C
.mC1/L
a

Dcos
.2n1/a
L
cos.2n1/.mC1/
D.1/
mC1
cos
.2n1/a
L
;
(B) implies that thatu.x; t/D.1/
mC1
u.x; /.
12.2.18.Sincef
0
.0/Df .2/D0andf
00
.x/D 2, Theorem 11.3.5(c) implies that
˛nD
32
.2n1/
2

2
Z
2
0
cos
.2n1/x
4
dx
D
128
.2n1/
3

3
sin
.2n1/x
4
ˇ
ˇ
ˇ
ˇ
4
0
D.1/
nC1
128
.2n1/
3

3
I
CMf.x/D
128

3
1
X
nD1
.1/
n
.2n1/
3
cos
.2n1/x
4
. From Exercise 12.2.17,
u.x; t/D
128

3
1
X
nD1
.1/
n
.2n1/
3
cos
3.2n1/t
4
cos
.2n1/x
4
:
12.2.20.Sinceg
0
.0/Dg.2/D0andg
00
.x/D 2, Theorem 11.3.5(c) implies that
ˇnD
32
.2n1/
2

2
Z
2
0
cos
.2n1/x
4
dx
D
128
.2n1/
3

3
sin
.2n1/x
4
ˇ
ˇ
ˇ
ˇ
4
0
D.1/
nC1
128
.2n1/
3

3
I

256. Section 12.2The Wave Equation251
CMf.x/D
128

3
1
X
nD1
.1/
n
.2n1/
3
cos
.2n1/x
4
. From Exercise 12.2.17,
u.x; t/D
512
3
4
1
X
nD1
.1/
n
.2n1/
4
sin
3.2n1/t
4
cos
.2n1/x
4
:
12.2.22.Sincef
0
.0/Df .1/D0andf
00
.x/D6.2xC1/, Theorem 11.3.5(c)implies that
˛nD
48L
.2n1/
2

2
Z
1
0
.2xC1/cos
.2n1/x
2
dx
D
96
.2n1/
3

3
"
.2xC1/sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
2
Z
1
0
sin
.2n1/x
2
#
dx
D
96
.2n1/
3

3
"
.1/
nC1
3
4
.2n1/
cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
96
.2n1/
3

3

.1/
n
3C
4
.2n1/

I
CMf.x/D
96

3
1
X
nD1
1
.2n1/
3

.1/
n
3C
4
.2n1/

cos
.2n1/x
2
:
From Exercise 12.2.17,
u.x; t/D
96

3
1
X
nD1
1
.2n1/
3

.1/
n
3C
4
.2n1/

cos
.2n1/
p
5 t
2
cos
.2n1/x
2
:
12.2.24.Sinceg
0
.0/Dg.1/D0andg
00
.x/D6.2xC1/, Theorem 11.3.5(c)implies that
ˇnD
48L
.2n1/
2

2
Z
1
0
.2xC1/cos
.2n1/x
2
dx
D
96
.2n1/
3

3
"
.2xC1/sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
2
Z
1
0
sin
.2n1/x
2
#
dx
D
96
.2n1/
3

3
"
.1/
nC1
3
4
.2n1/
cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
96
.2n1/
3

3

.1/
n
3C
4
.2n1/

I
CMg.x/D
96

3
1
X
nD1
1
.2n1/
3

.1/
n
3C
4
.2n1/

cos
.2n1/x
2
. From Exercise 12.2.17,
u.x; t/D
192

4
p
5
1
X
nD1
1
.2n1/
4

.1/
n
3C
4
.2n1/

sin
.2n1/
p
5 t
2
cos
.2n1/x
2
:

257. 252 Chapter 12Fourier Solutions of Partial Differential Equations
12.2.26.Sincef
0
.0/Df .1/Df
00
.1/D0andf
000
.x/D24.x1/, Theorem 11.3.5(c)and Exer-
cise 42(b)of Section 11.3 imply that
˛nD
384
.2n1/
3

3
Z
1
0
.x1/sin
.2n1/x
2
dx
D
768
.2n1/
4

4
"
.x1/cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
cos
.2n1/x
2
#
dx
D
768
.2n1/
4

4
"
1
2
.2n1/
sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
768
.2n1/
4

4

1C
.1/
n
2
.2n1/

I
CMf.x/D
384

4
1
X
nD1
1
.2n1/
4

1C
.1/
n
4
.2n1/

cos
.2n1/x
2
. From Exercise 12.2.17,
u.x; t/D
384

4
1
X
nD1
1
.2n1/
4

1C
.1/
n
4
.2n1/

cos
3.2n1/t
2
cos
.2n1/x
2
:
12.2.28.Sinceg
0
.0/Dg.1/Dg
00
.1/D0andg
000
.x/D24.x1/, Theorem 11.3.5(c)and Exer-
cise 11.2.42(b)imply that
ˇnD
384
.2n1/
3

3
Z
1
0
.x1/sin
.2n1/x
2
dx
D
768
.2n1/
4

4
"
.x1/cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
cos
.2n1/x
2
#
dx
D
768
.2n1/
4

4
"
1
2
.2n1/
sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
768
.2n1/
4

4

1C
.1/
n
2
.2n1/

I
CMg.x/D
384

4
1
X
nD1
1
.2n1/
4

1C
.1/
n
4
.2n1/

cos
.2n1/x
2
:
From Exercise 12.2.17,
u.x; t/D
768
3
5
1
X
nD1
1
.2n1/
5

1C
.1/
n
4
.2n1/

sin
3.2n1/t
2
cos
.2n1/x
2
:
12.2.30.Sincef
0
.0/Df .1/Df
00
.1/D0andf
000
.x/D24.x1/, Theorem 11.3.5(c)and Exer-

258. Section 12.2The Wave Equation253
cise 42(b)of Section 11.3 imply that
˛nD
384
.2n1/
3

3
Z
1
0
.x1/sin
.2n1/x
2
dx
D
768
.2n1/
4

4
"
.x1/cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
cos
.2n1/x
2
#
dx
D
768
.2n1/
4

4
"
1
2
.2n1/
sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
768
.2n1/
4

4

1C
.1/
n
2
.2n1/

I
CMf.x/D
768

4
1
X
nD1
1
.2n1/
4

1C
.1/
n
2
.2n1/

cos
.2n1/x
2
. From Exercise 12.2.17,
u.x; t/D
768

4
1
X
nD1
1
.2n1/
4

1C
.1/
n
2
.2n1/

cos
.2n1/t
2
cos
.2n1/x
2
:
12.2.32.SettingAD.2n1/x=2LandBD.2n1/at=2Lin the identities
cosAcosBD
1
2
Œcos.ACB/Ccos.AB/and cosAsinBD
1
2
Œsin.ACB/sin.AB/yields
cos
.2n1/at
2L
cos
.2n1/x
2L
D
1
2

cos
.2n1/.xCat/
2L
Ccos
.2n1/.xat/
2L

.A/
and
sin
.2n1/at
2L
cos
.2n1/x
2L
D
1
2

sin
.2n1/.xCat/
2L
sin
.2n1/.xat/
2L

D
.2n1/
4L
Z
xCat
xat
cos
.2n1/
2L
d:
.B/
SinceCMf.x/D
1
X
nD1
˛ncos
.2n1/x
2L
, (A) implies that
1
X
nD1
˛ncos
.2n1/at
2L
cos
.2n1/x
2L
D
1
2
ŒCMf.xCat/CCMf.xat/: . C/
Since it can be shown that a mixed Fourier cosine series can beintegrated term by term between any two
limits, (B) implies that
1
X
nD1
2Lˇn
.2n1/a
sin
.2n1/at
2L
cos
.2n1/x
2L
D
1
2a
1
X
nD1
ˇn
Z
xCat
xat
cos
.2n1/
2L
d
D
1
2a
Z
xCat
xat
 
1
X
nD1
ˇncos
.2n1/
2L
!
d
D
1
2a
Z
xCat
xat
CMg./ d:

259. 254 Chapter 12Fourier Solutions of Partial Differential Equations
This and (C) imply that
u.x; t/D
1
2
ŒCMf.xCat/CCMf.xat/C
1
2a
Z
xCat
xat
CMg./ d:
12.2.34.We begin by looking for functions of the formv.x; t/DX.x/T .t/that are not identically zero
and satisfyvt tDa
2
vxx,v.0; t/D0,vx.L; t/D0for all.x; t/. As shown in the text,XandTmust
satisfyX
00
CXD0and (B)T
00
Ca
2
TD0for the same value of. Sincev.0; t/DX.0/T .t/
andvx.L; t/DX
0
.L/T .t/and we don’t wantTto be identically zero,X.0/D0andX
0
.L/D0.
Therefore,must be an eigenvalue of (C)X
00
CXD0,X.0/D0,X
0
.L/D0, andXmust be
a-eigenfunction. From Theorem 11.1.4, the eigenvalues of (C) arenD
.2n1/
2

2
=4L
2
, integer), with
associated eigenfunctionsXnDsin
.2n1/x
2L
,nD1, 2, 3,. . . . SubstitutingD
.2n1/
2

2
4L
2
into
(B) yieldsT
00
C..2n1/
2

2
a
2
=4L
2
/TD0, which has the general solution
TnD˛ncos
.2n1/at
2L
C
2ˇnL
.2n1/a
sin
.2n1/at
2L
;
where˛nandˇnare constants. Now let
vn.x; t/DXn.x/Tn.t/D

˛ncos
.2n1/at
2L
C
2ˇnL
.2n1/a
sin
.2n1/at
2L

sin
.2n1/x
2L
:
Then
@vn
@t
.x; t/D


.2n1/a
2L
˛nsin
.2n1/at
2L
Cˇncos
.2n1/at
2L

sin
.2n1/x
2L
;
so
vn.x; 0/D˛nsin
.2n1/x
2L
and
@vn
@t
.x; 0/Dˇnsin
.2n1/x
2L
:
Therefore,vnsatisfies (A) withf .x/D˛nsin
.2n1/x
2L
andg.x/Dˇnsin
.2n1/x
2L
. More gen-
erally, if˛1; ˛2; : : : ; ˛mandˇ1; ˇ2; : : : ; ˇmare constants and
um.x; t/D
m
X
nD1

˛ncos
.2n1/at
2L
C
2ˇnL
.2n1/a
sin
.2n1/at
2L

sin
.2n1/x
2L
;
thenumsatisfies (A) with
f .x/D
m
X
nD1
˛nsin
.2n1/x
2L
andg.x/D
m
X
nD1
ˇnsin
.2n1/x
2L
:
This motivates the definition.

260. Section 12.2The Wave Equation255
12.2.36.Sincef .0/Df
0
.1/D0, andf
00
.x/D6.12x/, Theorem 11.3.5(d)implies that
˛nD
48
.2n1/
2

2
Z
1
0
.12x/sin
.2n1/x
2
dx
D
96
.2n1/
3

3
"
.12x/cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
C2
Z
1
0
cos
.2n1/x
2
dx
#
D
96
.2n1/
3

3
"
1C
4
.2n1/
sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
96
.2n1/
3

3

1C.1/
n
4
.2n1/

I
SMf.x/D
96

3
1
X
nD1
1
.2n1/
3

1C.1/
n
4
.2n1/

sin
.2n1/x
2
. From Exercise 12.2.34,
u.x; t/D
96

3
1
X
nD1
1
.2n1/
3

1C.1/
n
4
.2n1/

cos
3.2n1/t
2
sin
.2n1/x
2
:
12.2.38.Sinceg.0/Dg
0
.1/D0, andg
00
.x/D6.12x/, Theorem 11.3.5(d)implies that
ˇnD
48
.2n1/
2

2
Z
1
0
.12x/sin
.2n1/x
2
dx
D
96
.2n1/
3

3
"
.12x/cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
C2
Z
1
0
cos
.2n1/x
2
dx
#
D
96
.2n1/
3

3
"
1C
4
.2n1/
sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
96
.2n1/
3

3

1C.1/
n
4
.2n1/

I
SMg.x/D
96

3
1
X
nD1
1
.2n1/
3

1C.1/
n
4
.2n1/

sin
.2n1/x
2
. From Exercise 12.2.34,
u.x; t/D
64

4
1
X
nD1
1
.2n1/
4

1C.1/
n
4
.2n1/

sin
3.2n1/t
2
sin
.2n1/x
2
:
12.2.40.Sincef .0/Df
0
./Df
00
.0/D0andf
000
.x/D6, Theorem 11.3.5(d)and Exercise 11.3.50(b)
imply that
˛nD
96
.2n1/
3

Z

0
cos
.2n1/x
2
dxD
192
.2n1/
4

sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ

0
D.1/
n
192
.2n1/
4

I
SMf.x/D
192

1
X
nD1
.1/
n
.2n1/
4
sin
.2n1/x
2
. From Exercise 12.2.34,
u.x; t/D
192

1
X
nD1
.1/
n
.2n1/
4
cos
.2n1/
p
3 t
2
sin
.2n1/x
2
:

261. 256 Chapter 12Fourier Solutions of Partial Differential Equations
12.2.42.Sinceg.0/Dg
0
./Dg
00
.0/D0andg
000
.x/D6, Theorem 11.3.5(d)and Exercise 50(b)imply
that
ˇnD
96
.2n1/
3

Z

0
cos
.2n1/x
2
dxD
192
.2n1/
4

sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ

0
D.1/
n
192
.2n1/
4

I
SMg.x/D
192

1
X
nD1
.1/
n
.2n1/
4
sin
.2n1/x
2
. From Exercise 12.2.34,
u.x; t/D
384
p
3
1
X
nD1
.1/
n
.2n1/
5
sin
.2n1/
p
3 t
2
sin
.2n1/x
2
:
12.2.44.Sincef .0/Df
0
.1/Df
00
.0/D0andf
000
.x/D12.2x1/, Theorem 11.3.5(d)and Exer-
cise 11.3.50(b)imply that
˛nD
192
.2n1/
3

3
Z
1
0
.2x1/cos
.2n1/x
2
dx
D
384
.2n1/
4

4
"
.2x1/sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
2
Z
1
0
sin
.2n1/x
2
dx
#
D
384
.2n1/
4

4
"
.1/
nC1
C
4
.2n1/
cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
384
.2n1/
4

4

.1/
nC1

4
.2n1/

D
384
.2n1/
4

4

.1/
n
C
4
.2n1/

I
SMf.x/D
384

4
1
X
nD1
1
.2n1/
4

.1/
n
C
4
.2n1/

sin
.2n1/x
2
. From Exercise 12.2.34,
u.x; t/D
384

4
1
X
nD1
1
.2n1/
4

.1/
n
C
4
.2n1/

cos.2n1/tsin
.2n1/x
2
:
12.2.46.Sinceg.0/Dg
0
.1/Dg
00
.0/D0andg
000
.x/D12.2x1/, Theorem 11.3.5(d)and Exer-
cise 11.3.50(b)imply that
ˇnD
192
.2n1/
3

3
Z
1
0
.2x1/cos
.2n1/x
2
dx
D
384
.2n1/
4

4
"
.2x1/sin
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
2
Z
1
0
sin
.2n1/x
2
dx
#
D
384
.2n1/
4

4
"
.1/
nC1
C
4
.2n1/
cos
.2n1/x
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
384
.2n1/
4

4

.1/
nC1

4
.2n1/

D
384
.2n1/
4

4

.1/
n
C
4
.2n1/

I
SMg.x/D
384

4
1
X
nD1
1
.2n1/
4

.1/
n
C
4
.2n1/

sin
.2n1/x
2
. From Exercise 12.2.34,
u.x; t/D
384

5
1
X
nD1
1
.2n1/
5

.1/
n
C
4
.2n1/

sin.2n1/tsin
.2n1/x
2
:

262. Section 12.2The Wave Equation257
12.2.48.Sincefis continuous onŒ0; Landf
0
.L/D0, Theorem 11.3.4 implies thatSMf.x/Df .x/,
0xL. From Exercise 11.3.58,SMfis the odd periodic extension (with period2L) of the function
r.x/D

f .x/; 0xL;
f .2Lx/; L < x2L;
which is continuous onŒ0; 2L. Sincer.0/Dr.2L/Df .0/D
0,SMfis continuous on.1;1/. Moreover,r
0
.x/D

f
0
.x/; 0 < x < L;
f
0
.2Lx/; L < x < 2L;
r
0
C
.0/D
f
0
C
.0/,r
0

.2L/D f
0
C
.0/, and, sincef
0

.L/D0,r
0
.L/D0. Hence,ris differentiable onŒ0; 2L.
Sincer.0/Dr.2L/Df .0/D0, Theorem 12.2.3(a)withhDr,pDSMf, andLreplaced by2L
implies thatSMfis differentiable on.1;1/. Similarly,SMgis differentiable on.1;1/.
Now we note thatr
00
.x/D

f
00
.x/; 0 < x < L;
f
00
.2Lx/; L < x < 2L;
r
00
.L/Df
00

.L/, andr
00
C
.0/Dr
00

.2L/D
f
00
C
.0/D0. SinceS
0
Mf
is the even periodic extension ofr
0
, Theorem 12.2.3(b)withhDr
0
,qDS
0
Mf
,
andLreplaced by2Limplies thatS
0
Mf
is differentiable on.1;1/. Now follow the argument used to
complete the proof of Theorem 12.2.4.
12.2.50.From Example 11.3.5,Cf.x/D4
768

4
1
X
nD1
1
.2n1/
4
cos
.2n1/x
2
. From Exercise 12.2.49,
u.x; t/D4
768

4
1
X
nD1
1
.2n1/
4
cos
p
5.2n1/t
2
cos
.2n1/x
2
:
12.2.52.˛0D
Z

0
.3x
4
4Lx
3
/ dxD

3x
5
5
x
4
ˇ
ˇ
ˇ
ˇ

0
D
2
4
5
. Sincef
0
.0/Df
0
./D0and
f
000
.x/D24.3x/, Theorem 11.3.5(a)implies that
˛nD
48
n
3

Z

0
.3x/sinnx dxD
48
n
4


.3x/cosnx
ˇ
ˇ
ˇ
ˇ

0
3
Z

0
cosnx dx

D
48
n
4

Œ.1/
n
2CC
144
n
5

sinnx
ˇ
ˇ
ˇ
ˇ

0
D
48
n
4
Œ1C.1/
n
2 ; n1I
Cf.x/D
2
4
5
48
1
X
nD1
1C.1/
n
2
n
4
cosnx. From Exercise 12.2.49,
u.x; t/D
2
4
5
48
1
X
nD1
1C.1/
n
2
n
4
cos2ntcosnx:
12.2.54.ˇ0D
Z

0
.3x
4
4Lx
3
/ dxD
1


3x
5
5
x
4
ˇ
ˇ
ˇ
ˇ

0
D
2
4
5
. Sinceg
0
.0/Dg
0
./D0and
g
000
.x/D24.3x/, Theorem 11.3.5(a)implies that
ˇnD
48
n
3

Z

0
.3x/sinnx dxD
48
n
4


.3x/cosnx
ˇ
ˇ
ˇ
ˇ

0
3
Z

0
cosnx dx

D
48
n
4

Œ.1/
n
2CC
144
n
5

sinnx
ˇ
ˇ
ˇ
ˇ

0
D
48
n
4
Œ1C.1/
n
2 ; n1I

263. 258 Chapter 12Fourier Solutions of Partial Differential Equations
Cg.x/D
2
4
5
48
1
X
nD1
1C.1/
n
2
n
4
cosnx. From Exercise 12.2.49,
u.x; t/D
2
4
t
5
24
1
X
nD1
1C.1/
n
2
n
5
sin2ntcosnx:
12.2.56.˛0D
1

Z

0
.x
4
2x
3
C
2
x
2
/ dxD
1


x
5
5

x
4
2
C

2
x
3
3
ˇ
ˇ
ˇ
ˇ

0
D

4
30
. Sincef
0
.0/D
f
0
./D0andf
000
.x/D12.2x/, Theorem 11.3.5(a)implies that
˛nD
24
n
3

Z

0
.2x/sinnx dxD
24
n
4


.2x/cosnx
ˇ
ˇ
ˇ
ˇ

0
2
Z

0
cosnx dx

D
24
n
4

Œ.1/
n
CC
48
n
5

sinnx
ˇ
ˇ
ˇ
ˇ

0
D
24
n
4
Œ1C.1/
n

D
(
0 ifnD2m1;

3
m
4
ifnD2m;
n1I
Cf.x/D

4
30
3
1
X
nD1
1
n
4
cos2nx. From Exercise 12.2.49,u.x; t/D

4
30
3
1
X
nD1
1
n
4
cos8ntcos2nx.
12.2.58.ˇ0D
1

Z

0
.x
4
2x
3
C
2
x
2
/ dxD
1


x
5
5

x
4
2
C

2
x
3
3
ˇ
ˇ
ˇ
ˇ

0
D

4
30
. Sinceg
0
.0/D
g
0
./D0andg
000
.x/D12.2x/, Theorem 11.3.5(a)implies that
ˇnD
24
n
3

Z

0
.2x/sinnx dxD
24
n
4


.2x/cosnx
ˇ
ˇ
ˇ
ˇ

0
2
Z

0
cosnx dx

D
24
n
4

Œ.1/
n
CC
48
n
5

sinnx
ˇ
ˇ
ˇ
ˇ

0
D
24
n
4
Œ1C.1/
n

D
(
0 ifnD2m1;

3
m
4
ifnD2m;
1I
Cg.x/D

4
30
3
1
X
nD1
1
n
4
cos2nx. From Exercise 12.2.49,u.x; t/D

4
t
30

3
8
1
X
nD1
1
n
5
sin8ntcos2nx.
12.2.60.SettingADnx=LandBDnat=Lin the identities cosAcosBD
1
2
Œcos.ACB/Ccos.A
B/and cosAsinBD
1
2
Œsin.ACB/sin.AB/yields
cos
nat
L
cos
nx
L
D
1
2

cos
n.xCat/
L
Ccos
n.xat/
L

.A/
and
sin
nat
L
cos
nx
L
D
1
2

sin
n.xCat/
L
sin
n.xat/
L

D
n
2L
Z
xCat
xat
cos
n
L
d:
.B/

264. Section 12.2The Wave Equation259
SinceCf.x/D˛0C
1
X
nD1
˛nsin
nx
L
, (A) implies that
˛0C
1
X
nD1
˛ncos
nat
L
cos
nx
L
D
1
2
ŒCf.xCat/CCf.xat/: . C/
Since it can be shown that a Fourier sine series can be integrated term by term between any two limits,
(B) implies that
ˇ0tC
1
X
nD1
ˇnL
na
sin
nat
L
cos
nx
L
Dˇ0tC
1
2a
1
X
nD1
ˇn
Z
xCat
xat
cos
n
L
d
D
1
2a
Z
xCat
xat
 
ˇ0C
1
X
nD1
ˇncos
n
L
!
d
D
1
2a
Z
xCat
xat
Cg./ d:
This and (C) imply that
u.x; t/D
1
2
ŒCf.xCat/CCf.xat/C
1
2a
Z
xCat
xat
Cg./ d:
12.2.62.(a). Sincejpn.x/j 1andjqn.t/j 1for allt,jknpn.x/qn.t/j jknjfor all.x; t/, and the
comparison test implies the conclusion.
(b)Iftis fixed but arbitrary, thenjknp
0
n
.x/qn.t/j jjnjknj, so Theorem 12.1.2 with´Dxand
wn.x/Dknpn.x/qn.t/justifies term by term differentiation with respect toxon.1;1/. Ifxis fixed
but arbitrary, thenjknpn.x/q
0
n
.t/j jjnjknj, so Theorem 12.1.2 with´Dtandwn.t/Dknpn.x/qn.t/
justifies term by term differentiation with respect toton.1;1/.
(c)The argument is similar to argument use in(b).
(d)Apply(b)and(c)to the series
1
X
nD1
˛ncos
nat
L
sin
nx
L
and
1
X
nD1
ˇnL
na
sin
nat
L
sin
nx
L
, recall-
ing that the individual terms in the series satisfyut tDa
2
uxxfor all.x; t/.
(d)u.x; t/D
f .xCct/Cf .xct/
2
C
1
2a
Z
xCat
xat
g.u/ du.
12.2.64.
u.x; t/D
.xCat/C.xat/
2
C
1
2a
Z
xCat
xat
4au duDxC2
Z
xCat
xat
u duDxCu
2
ˇ
ˇ
ˇ
ˇ
xCat
xat
DxC.xCat/
2
.xat/
2
Dx.1C4at/:
12.2.66.
u.x; t/D
sin.xCat/Csin.xat/
2
C
1
2a
Z
xCat
xat
acosu du
D
sin.xCat/Csin.xat/
2
C
sin.xCat/sin.xat/
2
Dsin.xCat/:

265. 260 Chapter 12Fourier Solutions of Partial Differential Equations
12.2.68.
u.x; t/D
.xCat/sin.xCat/C.xat/sin.xat/
2
C
1
2a
Z
xCat
xat
sinu du
D
xŒsin.xCat/Csin.xat/
2
C
atŒsin.xCat/sin.xat/
2
C
cos.xat/cos.xCat/
2a
DxsinxcosatCatcosxsinatC
sinxsinat
a
:
12.3LAPLACE’S EQUATION IN RECTANGULAR COORDINATES
12.3.2.Sincef .0/Df .1/D0andf
00
.x/D26x, Theorem 11.3.5(b)implies that
˛nD
4
n
2

2
Z
1
0
.46x/sin
nx
2
dxD
8
n
3

3
"
.46x/cos
nx
2
ˇ
ˇ
ˇ
ˇ
2
0
C6
Z
2
0
cos
nx
2
dx
#
D
32
n
3

3
.1C.1/
n
2/C
96
n
4

4
sin
nx
2
ˇ
ˇ
ˇ
ˇ
2
0
D
32
n
3

3
Œ1C.1/
n
2I
S.x/D
32

3
1
X
nD1
Œ1C.1/
n
2
n
3
sin
nx
2
. From Example 12.3.1,
u.x; y/D
32

3
1
X
nD1
Œ1C.1/
n
2sinhn.3y/=2
n
3
sinh3n=2
sin
nx
2
:
12.3.4.
˛1D
2

Z

0
xsin
2
x dxD
1

Z

0
x.1cos2x/ dxD
x
2
2
ˇ
ˇ
ˇ
ˇ

0

1

Z

0
xcos2x dx
D

2

1
2

xsin2x
ˇ
ˇ
ˇ
ˇ

0

Z

0
cos2x dx

D

2
C
sin2x
4
ˇ
ˇ
ˇ
ˇ

0
D

2
I
ifn2, then
˛nD
2

Z

0
xsinxsinnx dxD
1

Z

0
xŒcos.n1/xcos.nC1/x dx
D
1


x

sin.n1/x
n1

sin.nC1/x
nC1
ˇ
ˇ
ˇ
ˇ

0

Z

0

sin.n1/x
n1

sin.nC1/x
nC1

dx

D
1


cos.n1/x
.n1/
2

cos.nC1/x
.nC1/
2
ˇ
ˇ
ˇ
ˇ

0
D
1


1
.n1/
2

1
.nC1/
2

fi
.1/
nC1
1

D
4n
.n
2
1/
2

fi
.1/
nC1
1

D
8
<
:
0 ifnD2m1;

16m
.4m
2
1/
ifnD2mI
S.x/D

2
sinx
16

1
X
nD1
n
.4n
2
1/
2
sin2nx. From Example 12.3.1,
u.x; y/D

2
sinh.1y/
sinh1
sinx
16

1
X
nD1
nsinh2n.1y/
.4n
2
1/
2
sinh2n
sin2nx:

266. Section 12.3Laplace’s Equation in Rectangular Coordinates261
12.3.6.˛0D
Z
1
0
.1x/ dxD
.1x/
2
2
ˇ
ˇ
ˇ
ˇ
1
0
D
1
2
; ifn1,
˛nD2
Z
1
0
.1x/cosnx dxD
2
n
"
.1x/sinnx
ˇ
ˇ
ˇ
ˇ
1
0
C
Z
1
0
sinnx dx
#
D
2
n
2

2
cosnx
ˇ
ˇ
ˇ
ˇ
1
0
D
2
n
2

2
Œ1.1/
n
D
8
<
:
4
.2m1/
2

2
ifnD2m1;
0 ifnD2mI
C.x/D
1
2
C
4

2
1
X
nD1
1
.2n1/
2
cos.2n1/x. From Example 12.3.3,
u.x; y/D
y
2
C
4

3
1
X
nD1
sinh.2n1/y
.2n1/
3
cosh2.2n1/
cos.2n1/x:
12.3.8.˛0D
Z
1
0
.x1/
2
dxD
.x1/
3
3
ˇ
ˇ
ˇ
ˇ
1
0
D
1
3
; ifn1, then
˛nD2
Z
1
0
.x1/
2
cosnx dxD
2
n
"
.x1/
2
sinnx
ˇ
ˇ
ˇ
ˇ
1
0
2
Z
1
0
.x1/sinnx dx
#
D
4
n
2

2
"
.x1/cosnx
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
cosnx dx
#
D
4
n
2

2

4
n
3

3
sinnx
ˇ
ˇ
ˇ
ˇ
1
0
D
4
n
2

2
I
C.x/D
1
3
C
4

2
1
X
nD1
1
n
2
cosnx. From Example 12.3.3,u.x; y/D
y
3
C
4

3
1
X
nD1
sinhny
n
3
coshn
cosnx.
12.3.10.Sinceg.0/Dg
0
.1/D0, andg
00
.y/D6.12y/, Theorem 11.3.5(d)implies that
˛nD
48
.2n1/
2

2
Z
1
0
.12y/sin
.2n1/y
2
dy
D
96
.2n1/
3

3
"
.12y/cos
.2n1/y
2
ˇ
ˇ
ˇ
ˇ
1
0
C2
Z
1
0
cos
.2n1/y
2
dy
#
D
96
.2n1/
3

3
"
1C
4
.2n1/
sin
.2n1/y
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
96
.2n1/
3

3

1C.1/
n
4
.2n1/

I
SM.y/D
96

3
1
X
nD1
1
.2n1/
3

1C.1/
n
4
.2n1/

sin
.2n1/y
2
. From Example 12.3.5,
u.x; y/D
96

3
1
X
nD1

1C.1/
n
4
.2n1/

cosh.2n1/.x2/=2
.2n1/
3
cosh2.2n1/=2
sin
.2n1/y
2
:

267. 262 Chapter 12Fourier Solutions of Partial Differential Equations
12.3.12.From Example 11.3.8.3,
SM.y/D
96

3
1
X
nD1
1
.2n1/
3

3C.1/
n
4
.2n1/

sin
.2n1/y
2
:
From Example 12.3.5,
u.x; y/D
96

3
1
X
nD1

3C.1/
n
4
.2n1/

cosh.2n1/.x3/=2
.2n1/
3
cosh3.2n1/=2
sin
.2n1/y
2
:
12.3.14.
cnD
2
3
Z
3
0
.3yy
2
/cos
.2n1/y
6
dy
D
4
.2n1/
"
.3yy
2
/sin
.2n1/y
6
ˇ
ˇ
ˇ
ˇ
3
0

Z
3
0
.32y/sin
.2n1/y
6
dy
#
D
24
.2n1/
2

2
"
.32y/cos
.2n1/y
6
ˇ
ˇ
ˇ
ˇ
3
0
C2
Z
3
0
cos
.2n1/y
6
dy
#
D
72
.2n1/
2

2
C
288
.2n1/
3

3
sin
.2n1/y
6
ˇ
ˇ
ˇ
ˇ
3
0
D
288
.2n1/
3

3
sin
.2n1/y
2
D
72
.2n1/
2

2
C.1/
n1
288
.2n1/
3

3
I
CM.y/D
72

2
1
X
nD1
1
.2n1/
2

1C
4.1/
n
.2n1/

cos
.2n1/y
6
. From Example 12.3.7,
u.x; y/D
432

3
1
X
nD1

1C
4.1/
n
.2n1/

cosh.2n1/x=6
.2n1/
3
sinh.2n1/=3
cos
.2n1/y
6
:
12.3.16.Sinceg
0
.0/Dg.1/D0andg
00
.y/D 6y, Theorem 11.3.5(c)implies that
˛nD
48
.2n1/
2

2
Z
1
0
ycos
.2n1/y
2
dy
D
96
.2n1/
3

3
"
ysin
.2n1/y
2
ˇ
ˇ
ˇ
ˇ
1
0

Z
1
0
sin
.2n1/y
2
dy
#
D
96
.2n1/
3

3
"
.1/
nC1
C
2
.2n1/
cos
.2n1/y
2
ˇ
ˇ
ˇ
ˇ
1
0
#
D
96
.2n1/
3

3

.1/
n
C
2
.2n1/

dyI
CM.y/D
96

3
1
X
nD1
1
.2n1/
3

.1/
n
C
2
.2n1/

cos
.2n1/y
2
. From Example 12.3.7,
u.x; y/D
192

4
1
X
nD1
cosh.2n1/x=2
.2n1/
4
sinh.2n1/=2

.1/
n
C
2
.2n1/

cos
.2n1/y
2
:

268. Section 12.3Laplace’s Equation in Rectangular Coordinates263
12.3.18.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
CXD0,
X
0
.0/D0,X
0
.a/D0, and (B)Y
00
YD0,Y.0/D1,Y.b/D0. From Theorem 11.1.3, the eigenval-
ues of (A) areD0, with associated eigenfunctionX0D1, andnD
n
2

2
a
2
, with associated eigenfunc-
tionsYnDcos
nx
a
,nD1,2,3,. . . . SubstitutingD0into (B) yieldsY
00
0
D0,Y0.0/D1,Y0.b/D0,
soY0.y/D1
y
b
. SubstitutingD
n
2

2
a
2
into (B) yieldsY
00
n
.n
2

2
=a
2
/YnD0,Yn.0/D1,Yn.b/D
0, soYnD
sinhn.by/=a
sinhnb=a
. Thenvn.x; y/DXn.x/Yn.y/D
sinhn.by/=a
sinhnb=a
cos
nx
a
, so
vn.x; 0/Dcos
nx
a
. Therefore,vnis solution of the given problem withf .x/Dcos
nx
a
. More gener-
ally, if˛0; : : : ; ˛mare arbitrary constants, thenum.x; y/D˛0

1
y
b

C
m
X
nD1
˛n
sinhn.by/=a
sinhnb=a
cos
nx
a
is a solution of the given problem withf .x/D˛0C
m
X
nD1
˛ncos
nx
a
. Therefore, iffis an arbi-
trary piecewise smooth function onŒ0; awe define the formal solution of the given problem to be
u.x; y/D˛0

1
y
b

C
1
X
nD1
˛n
sinhn.by/=a
sinhnb=a
cos
nx
a
, whereC.x/D˛0C
1
X
nD1
˛ncos
nx
a
is
the Fourier cosine series offonŒ0; a; that is,˛0D
1
a
Z
a
0
f .x/ dxand˛nD
2
a
Z
a
0
f .x/cos
nx
a
dx,
n1.
Now consider the special case.˛0D
1
2
Z
2
0
.x
4
4x
3
C4x
2
/ dxD
1
2

x
5
5
x
4
C
4x
3
3
ˇ
ˇ
ˇ
ˇ
2
0
D
6
5
.
Sincef
0
.0/Df
0
.2/D0andf
000
.x/D12.2x2/, Theorem 11.3.5(a)implies that
˛nD
96
n
3

3
Z
2
0
.2x2/sin
nx
2
dxD
192
n
4

4
"
.2x2/cos
nx
2
ˇ
ˇ
ˇ
ˇ
2
0
2
Z
2
0
cos
nx
2
dx
#
D
192
n
4

4
Œ.1/
n
2C2C
768
n
5

5
sin
nx
2
ˇ
ˇ
ˇ
ˇ
2
0
D
384
n
4

4
Œ1C.1/
n

D
(
0 ifnD2m1;

48
m
4

4
ifnD2m;
n1:
C.x/D
8
15

48

4
1
X
nD1
1
n
4
cosnx;u.x; y/D
8.1y/
15

48

4
1
X
nD1
1
n
4
sinhn.1y/
sinhn
cosnx.
12.3.20.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
CXD0,
X.0/D0,X
0
.a/D0, and (B)Y
00
YD0,Y.0/D1,Y.b/D0. From Theorem 11.1.4, the
eigenvalues of (A) arenD
.2n1/
2

2
4a
2
, with associated eigenfunctionsYnDsin
.2n1/x
2a
,
nD1,2,3,. . . . SubstitutingD
.2n1/
2

2
4a
2
into (B) yieldsY
00
n
..2n1/
2

2
=4a
2
/YnD0,
Yn.0/D1,Yn.b/D0, soYnD
sinh.2n1/.by/=2a
sinh.2n1/b=2a
. Thenvn.x; y/DXn.x/Yn.y/D
sinh.2n1/.by/=2a
sinh.2n1/b=2a
sin
.2n1/x
2a
, sovn.x; 0/Dsin
.2n1/x
2a
. Therefore,vnis solution of

269. 264 Chapter 12Fourier Solutions of Partial Differential Equations
the given problem withf .x/Dsin
.2n1/x
2a
. More generally, if˛1; : : : ; ˛mare arbitrary constants,
thenum.x; y/D
m
X
nD1
˛n
sinh.2n1/.by/=2a
sinh.2n1/b=2a
cos
.2n1/x
2a
is a solution of the given problem
withf .x/D
m
X
nD1
˛nsin
.2n1/x
2a
. Therefore, iffis an arbitrary piecewise smooth function onŒ0; a
we define the formal solution of the given problem to beu.x; y/D
1
X
nD1
˛n
sinh.2n1/.by/=2a
sinh.2n1/b=2a
sin
.2n1/x
2a
,
whereSm.x/D
1
X
nD1
˛nsin
.2n1/x
2a
is the mixed Fourier sine series offonŒ0; a; that is,˛nD
2
a
Z
a
0
f .x/sin
.2n1/x
2a
.
Now consider the special case. Sincef .0/Df
0
.L/D0andf
00
.x/D 2, Theorem 11.3.5(d)implies
that
˛nD
48
.2n1/
2

2
Z
3
0
sin
.2n1/x
6
dxD
288
.2n1/
3

3
cos
.2n1/x
6
ˇ
ˇ
ˇ
ˇ
3
0
D
288
.2n1/
3

3
I
SM.x/D
288

3
1
X
nD1
1
.2n1/
3
sin
.2n1/x
6
I
u.x; y/D
288

3
1
X
nD1
sinh.2n1/.2y/=6
.2n1/
3
sinh.2n1/=3
sin
.2n1/x
6
:
12.3.22.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
CXD0,
X
0
.0/D0,X
0
.a/D0, and (B)Y
00
YD0,Y
0
.0/D0,Y.b/D1. From Theorem 11.1.3, the
eigenvalues of (A) areD0, with associated eigenfunctionX0D1, andnD
n
2

2
a
2
, with associated
eigenfunctionsYnDcos
nx
a
,nD1,2,3,. . . . SubstitutingD0into (B) yieldsY
00
0
D0,Y
0
0
.0/D0,
Y0.b/D1, soY0D1. SubstitutingD
n
2

2
a
2
into (B) yieldsY
00
n
.n
2

2
=a
2
/YnD0,Y
0
n
.0/D
0,Yn.b/D1, soYnD
coshny=a
coshnb=a
. Thenvn.x; y/DXn.x/Yn.y/D
sinhny=a
coshnb=a
cos
nx
a
, so
vn.x; b/Dcos
nx
a
. Therefore,vnis solution of the given problem withf .x/Dcos
nx
a
. More
generally, if˛0; : : : ; ˛mare arbitrary constants, thenum.x; y/D˛0C
m
X
nD1
˛n
coshny=a
coshnb=a
cos
nx
a
is
a solution of the given problem withf .x/D˛0C
m
X
nD1
˛ncos
nx
a
. Therefore, iffis an arbitrary
piecewise smooth function onŒ0; awe define the formal solution of the given problem to beu.x; y/D
˛0C
1
X
nD1
˛n
coshny=a
coshnb=a
cos
nx
a
, whereC.x/D˛0C
1
X
nD1
˛ncos
nx
a
is the Fourier cosine series of
fonŒ0; a; that is,˛0D
1
a
Z
a
0
f .x/ dxand˛nD
2
a
Z
a
0
f .x/cos
nx
a
dx,n1.

270. Section 12.3Laplace’s Equation in Rectangular Coordinates265
Now consider the special case.
˛0D
1

Z

0
.x
4
2x
3
C
2
x
2
/ dxD
1


x
5
5

x
4
2
C

2
x
3
3
ˇ
ˇ
ˇ
ˇ

0
D

4
30
:
Sincef
0
.0/Df
0
./D0andf
000
.x/D12.2x/, Theorem 11.3.5(a)implies that
˛nD
24
n
3

Z

0
.2x/sinnx dxD
24
n
4


.2x/cosnx
ˇ
ˇ
ˇ
ˇ

0
2
Z

0
cosnx dx

D
24
n
4

Œ.1/
n
CC
48
n
5

sinnx
ˇ
ˇ
ˇ
ˇ

0
D
24
n
4
Œ1C.1/
n

D
(
0 ifnD2m1;

3
m
4
ifnD2m;
n1I
C.x/D

4
30
3
1
X
nD1
1
n
4
cos2nx;u.x; y/D

4
30
3
1
X
nD1
1
n
4
cosh2ny
cos2n
cos2nx
12.3.24.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
XD
0,X
0
.0/D0,X.a/D1, and (B)Y
00
CYD0,Y.0/D0,Y.b/D0. From Theorem 11.1.2,
the eigenvalues of (B) arenD
n
2

2
b
2
, with associated eigenfunctionsYnDsin
ny
b
,nD1,2,
3,. . . . SubstitutingD
n
2

2
b
2
into (A) yieldsX
00
n
.n
2

2
=b
2
/XnD0,X
0
n
.0/D0,Xn.a/D1, so
XnD
coshnx=b
coshna=b
. Thenvn.x; y/DXn.x/Yn.y/D
coshnx=b
coshna=b
sin
ny
b
, sovn.a; y/Dsin
ny
b
.
Therefore,vnis solution of the given problem withg.y/Dsin
ny
b
. More generally, if˛1; : : : ; ˛m
are arbitrary constants, thenum.x; y/D
m
X
nD1
˛n
coshnx=b
coshna=b
sin
ny
b
is a solution of the given problem
withg.y/D
m
X
nD1
˛nsin
ny
b
. Therefore, ifgis an arbitrary piecewise smooth function onŒ0; bwe
define the formal solution of the given problem to beu.x; y/D
1
X
nD1
˛n
coshnx=b
coshna=b
sin
ny
b
, where
S.y/D
1
X
nD1
˛nsin
ny
b
is the Fourier sine series ofgonŒ0; b; that is,˛nD
2
b
Z
b
0
g.y/sin
ny
b
dy.
Now consider the special case. Sinceg.0/Dg.1/Dg
00
.0/Dg
00
.L/D0andf
.4/
.y/D24,
Theorem 11.3.5(b)and Exercise 35(b)of Section 11.3 imply that
˛nD
48
n
4

4
Z
1
0
sinny dyD
48
n
5

5
cosny
ˇ
ˇ
ˇ
ˇ
1
0
D
48
n
5

5
Œ.1/
n
1D
8
<
:
96
.2m1/
5

5
ifnD2m1
0 ifnD2mI
S.y/D
96

5
1
X
nD1
1
.2n1/
5
sin.2n1/y;u.x; y/D
96

5
1
X
nD1
cosh.2n1/x
.2n1/
5
cosh.2n1/
sin.2n1/y.

271. 266 Chapter 12Fourier Solutions of Partial Differential Equations
12.3.26.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
XD0,
X
0
.0/D0,X
0
.a/D1, and (B)Y
00
CYD0,Y.0/D0,Y.b/D0. From Theorem 11.1.2, the
eigenvalues of (B) arenD
n
2

2
b
2
, with associated eigenfunctionsYnDsin
ny
b
,nD1,2,3,. . . .
SubstitutingD
n
2

2
b
2
into (A) yieldsX
00
n
.n
2

2
=b
2
/XnD0,X
0
n
.0/D0,X
0
n
.a/D1, soXnD
b
n
coshnx=b
sinhna=b
. Thenvn.x; y/DXn.x/Yn.y/D
b
n
coshnx=b
sinhna=b
sin
ny
b
, so
@vn
@x
.a; y/Dsin
ny
b
.
Therefore,vnis solution of the given problem withg.y/Dsin
ny
b
. More generally, if˛1; : : : ; ˛mare
arbitrary constants, thenum.x; y/D
b

m
X
nD1
˛n
coshnx=b
nsinhna=b
sin
ny
b
is a solution of the given problem
withg.y/D
m
X
nD1
˛nsin
ny
b
. Therefore, ifgis an arbitrary piecewise smooth function onŒ0; bwe
define the formal solution of the given problem to beu.x; y/D
b

1
X
nD1
˛n
coshnx=b
nsinhna=b
sin
ny
b
, where
S.y/D
1
X
nD1
˛nsin
ny
b
is the Fourier sine series ofgonŒ0; b; that is,˛nD
2
b
Z
b
0
g.y/sin
ny
b
dy.
Now consider the special case.˛nD
1
2
Z
2
0
ysin
ny
4
C
Z
4
2
.4y/sin
ny
4
dy

;
Z
2
0
ysin
ny
4
dyD
4
n
"
ycos
ny
4
ˇ
ˇ
ˇ
ˇ
2
0

Z
2
0
cos
ny
4
dy
#
D
2
n
cos
n
2
C
4
n
2

2
sin
ny
4
ˇ
ˇ
ˇ
ˇ
2
0
D
2
n
cos
n
2
C
4
n
2

2
sin
n
2
I
Z
2
0
.4y/sin
ny
4
dyD
2
n
"
.4y/cos
ny
4
ˇ
ˇ
ˇ
ˇ
4
2
C
Z
2
0
cos
ny
4
dy
#
D
2
n
cos
n
2

4
n
2

2
sin
ny
4
ˇ
ˇ
ˇ
ˇ
4
2
D
2
n
cos
n
2
C
4
n
2

2
sin
n
2
I
˛nD
16
n
2

2
sin
n
2
D
8
<
:
.1/
mC1
16
.2m1/
2

2
ifnD2m1
0 ifnD2mI
S.y/D
16

2
1
X
nD1
.1/
nC1
.2n1/
2
sin
.2n1/y
4
I
u.x; y/D
64

3
1
X
nD1
.1/
nC1
cosh.2n1/x=4
.2n1/
3
sinh.2n1/=4
sin
.2n1/y
4
:
12.3.28.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
XD0,
X
0
.0/D1,X.a/D0, and (B)Y
00
CYD0,Y
0
.0/D0,Y
0
.b/D0. From Theorem 11.1.3, the

272. Section 12.3Laplace’s Equation in Rectangular Coordinates267
eigenvalues of (B) are0D0, with associated eigenfunctionY0D1, andnD
n
2

2
b
2
, with associ-
ated eigenfunctionsYnDcos
ny
b
,nD1,2,3,. . . . Substituting0D0into (A) yieldsX
00
0
D0,
X
0
0
.0/D1 X.a/D0, soX0Dxa. SubstitutingD
n
2

2
b
2
into (A) yieldsX
00
n
.n
2

2
=b
2
/XnD
0,X
0
n
.0/D1,Xn.a/D0, soXnD
b
n
sinhn.xa/=b
coshna=b
. Thenvn.x; y/DXn.x/Yn.y/D
b
n
sinhn.xa/=b
coshna=b
cos
ny
b
, so
@vn
@x
.0; y/Dcos
ny
b
. Therefore,vnis solution of the given prob-
lem withg.y/Dcos
ny
b
. More generally, if˛0; : : : ; ˛mare arbitrary constants, thenum.x; y/D
˛0.xa/C
b

m
X
nD1
˛n
sinhn.xa/=b
ncoshna=b
cos
ny
b
is a solution of the given problem withg.y/D
m
X
nD1
˛ncos
ny
b
. Therefore, ifgis an arbitrary piecewise smooth function onŒ0; bwe define the formal
solution of the given problem to beu.x; y/D˛0.xa/C
b

1
X
nD1
˛n
sinhn.xa/=b
ncoshna=b
cos
ny
b
where
C.y/D˛0C
1
X
nD1
˛ncos
ny
b
is the Fourier cosine series ofgonŒ0; b; that is,˛0D
1
b
Z
b
0
g.y/cos
ny
b
dy,
˛nD
2
b
Z
b
0
g.y/cos
ny
b
dy,n1.
Now consider the special case. From Example 11.3.1,
C.y/D

2

4

1
X
nD1
1
.2n1/
2
cos.2n1/yI
u.x; y/D
.x2/
2

4

1
X
nD1
sinh.2n1/.x2/
.2n1/
3
cosh2.2n1/
cos.2n1/y:
12.3.30.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
CXD0,
X
0
.0/D0,X.a/D0, and (B)Y
00
YD0,Y.0/D1, andYis bounded. From Theorem 11.1.5, the
eigenvalues of (A) arenD
.2n1/
2

2
4a
2
, with associated eigenfunctionsYnDcos
.2n1/x
2a
,nD1,
2,3,. . . . SubstitutingD
.2n1/
2

2
4a
2
into (B) yieldsY
00
n
..2n1/
2

2
=4a
2
/YnD0,Yn.0/D1, so
YnDe
.2n1/y=2a
. Thenvn.x; y/DXn.x/Yn.y/De
.2n1/y=2a
cos
.2n1/x
2a
, sovn.x; 0/D
cos
.2n1/x
2a
. Therefore,vnis solution of the given problem withf .x/Dcos
.2n1/x
2a
. More
generally, if˛1; : : : ; ˛mare arbitrary constants, thenum.x; y/D
m
X
nD1
˛ne
.2n1/y=2a
cos
.2n1/x
2a
is a solution of the given problem withf .x/D
m
X
nD1
˛ncos
.2n1/x
2a
. Therefore, iffis an ar-
bitrary piecewise smooth function onŒ0; awe define the formal solution of the given problem to be

273. 268 Chapter 12Fourier Solutions of Partial Differential Equations
u.x; y/D
1
X
nD1
˛ne
.2n1/y=2a
cos
.2n1/x
2a
, whereCm.x/D
1
X
nD1
˛ncos
.2n1/x
2a
is the mixed
Fourier cosine series offonŒ0; a; that is,˛nD
2
a
Z
a
0
f .x/cos
.2n1/x
2a
.
Now consider the special case. Sincef
0
.0/Df .L/D0andf
00
.x/D 2, Theorem 11.3.5(d)implies
that
˛nD
16L
.2n1/
2

2
Z
3
0
cos
.2n1/x
6
dx
D
288
.2n1/
3

3
sin
.2n1/x
6
ˇ
ˇ
ˇ
ˇ
3
0
D.1/
nC1
288
.2n1/
3

3
I
CM.x/D
288

3
1
X
nD1
.1/
n
.2n1/
3
cos
.2n1/x
6
I
u.x; y/D
288

3
1
X
nD1
.1/
n
.2n1/
3
e
.2n1/y=6
cos
.2n1/x
6
:
12.3.32.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
CXD
0,X.0/D0,X.a/D0, and (B)Y
00
YD0,Y
0
.0/D1, andYis bounded. From Theo-
rem 11.1.2, the eigenvalues of (A) arenD
n
2

2
a
2
, with associated eigenfunctionsYnDsin
nx
a
,
nD1,2,3,. . . . SubstitutingD
n
2

2
a
2
into (B) yieldsY
00
n
.n
2

2
=a
2
/YnD0,Y
0
n
.0/D1, so
YnD
a
n
e
ny=a
. Thenvn.x; y/DXn.x/Yn.y/D
a
n
e
ny=a
sin
nx
a
, so
@vn
@y
.x; 0/Dsin
nx
a
.
Therefore,vnis solution of the given problem withf .x/Dsin
nx
a
. More generally, if˛1; : : : ; ˛mare
arbitrary constants, thenum.x; y/D
a

m
X
nD1
˛n
n
e
ny=a
sin
nx
a
is a solution of the given problem
withf .x/D
m
X
nD1
˛nsin
nx
a
. Therefore, iffis an arbitrary piecewise smooth function onŒ0; awe
define the formal solution of the given problem to beu.x; y/D
a

1
X
nD1
˛n
n
e
ny=a
sin
nx
a
, where
C.x/D
1
X
nD1
˛nsin
nx
a
is the Fourier sine series offonŒ0; a; that is,˛nD
2
a
Z
a
0
f .x/sin
nx
a
dx.
Now consider the special case. Sincef .0/Df ./D0andf
00
.x/D26x, Theorem 11.3.5(b)
implies that
˛nD
2
n
2

Z

0
.26x/sinnx dxD
2
n
3


.26x/cosnx
ˇ
ˇ
ˇ
ˇ

0
C6
Z

0
cosnx dx

D
4
n
3
Œ1C.1/
n
2C
12
n
4

sinnx
ˇ
ˇ
ˇ
ˇ

0
D
4
n
3
Œ1C.1/
n
2I
S.x/D 4
1
X
nD1
Œ1C.1/
n
2
n
3
sinnx;u.x/D4
1
X
nD1
Œ1C.1/
n
2
n
4
e
ny
sinnx;

274. Section 12.3Laplace’s Equation in Rectangular Coordinates269
12.3.34.The boundary conditions require productsv.x; y/DX.x/Y.y/such that (A)X
00
CXD0,
X.0/D0,X
0
.a/D0, and (B)Y
00
YD0,Y
0
.0/D1, andYis bounded. From Theorem 11.1.4, the
eigenvalues of (A) arenD
.2n1/
2

2
4a
2
, with associated eigenfunctionsYnDsin
.2n1/x
2a
,nD1,
2,3,. . . . SubstitutingD
.2n1/
2

2
4a
2
into (B) yieldsY
00
n
..2n1/
2

2
=4a
2
/YnD0,Y
0
n
.0/D1, so
YnD
2a
.2n1/
e
.2n1/y=2a
. Thenvn.x; y/DXn.x/Yn.y/D
2a
.2n1/
e
.2n1/y=2a
sin
.2n1/x
2a
,
so
@vn
@y
.x; 0/Dsin
.2n1/x
2a
. Therefore,vnis solution of the given problem withf .x/Dsin
.2n1/x
2a
.
More generally, if˛1; : : : ; ˛mare arbitrary constants, thenum.x; y/D
2a

m
X
nD1
˛n
2n1
e
.2n1/y=2a
sin
.2n1/x
2a
is a solution of the given problem withf .x/D
m
X
nD1
˛nsin
.2n1/x
2a
. Therefore, iffis an ar-
bitrary piecewise smooth function onŒ0; awe define the formal solution of the given problem to be
u.x; y/D
2a

1
X
nD1
˛n
2n1
e
.2n1/y=2a
sin
.2n1/x
2a
,whereSm.x/D
1
X
nD1
˛nsin
.2n1/x
2a
is
the mixed Fourier sine series offonŒ0; a; that is,˛nD
2
a
Z
a
0
f .x/sin
.2n1/x
2a
.
Now consider the special case.
˛nD
2
5
Z
5
0
.5xx
2
/sin
.2n1/x
10
dx
D
4
.2n1/
"
.5xx
2
/cos
.2n1/x
10
ˇ
ˇ
ˇ
ˇ
5
0

Z
5
0
.52x/cos
.2n1/x
10
dx
#
D
40
.2n1/
2

2
"
.52x/sin
.2n1/x
10
ˇ
ˇ
ˇ
ˇ
5
0
C2
Z
5
0
sin
.2n1/x
10
dx
#
D.1/
n
200
.2n1/
2

2

800
.2n1/
3

3
cos
.2n1/x
10
ˇ
ˇ
ˇ
ˇ
5
0
D.1/
n
200
.2n1/
2

2
C
800
.2n1/
3

3
I
SM.x/D
200

2
1
X
nD1
1
.2n1/
2

.1/
n
C
4
.2n1/

sin
.2n1/x
10
I
u.x; y/D
2000

3
1
X
nD1
1
.2n1/
3

.1/
n
C
4
.2n1/

e
.2n1/y=10
sin
.2n1/x
10
:
12.3.36.Solving BVP.1; 1; 1; 1/.f0; 0; 0; 0/requires productsX.x/Y.y/such that
X
00
CXD0; X
0
.0/D0; X
0
.a/D0IY
00
YD0; Y
0
.b/D0; Y
0
.0/D1:
Hence,XnDcos
nx
a
,YnD
acoshn.yb/=a
nsinhnb=a
, andc1
a

1
X
nD1
An
coshn.yb/=a
nsinhnb=a
cos
nx
a
is

275. 270 Chapter 12Fourier Solutions of Partial Differential Equations
a formal solution of BVP.1; 1; 1; 1/.f0; 0; 0; 0/ifc1is any constant and
1
X
nD1
Ancos
nx
a
is the Fourier
cosine expansion off0onŒ0; a, which is possible if and only if
Z
a
0
f0.x/ dxD0.
Similarly,c2C
a

1
X
nD1
Bn
coshny=a
nsinhnb=a
cos
nx
a
is a formal solution of BVP.1; 1; 1; 1/.0; f1; 0; 0/if
c2is any constant and
1
X
nD1
Bncos
nx
a
is the Fourier cosine expansion off1onŒ0; a, which is possible
if and only if
Z
a
0
f1.x/ dxD0.
Interchangingxandyandaandbshows thatc3
b

1
X
nD1
Cn
coshn.xa/=b
nsinhna=b
cos
ny
b
is a formal so-
lution of BVP.1; 1; 1; 1/.0; 0; g0; 0/ifc3is any constant and
1
X
nD1
Cncos
ny
b
is the Fourier cosine expan-
sion ofg0onŒ0; b, which is possible if and only if
Z
b
0
g0.x/ dxD0, andc4C
b

1
X
nD1
Dn
coshnx=b
nsinhna=b
cos
ny
b
is a formal solution of BVP.1; 1; 1; 1/.0; 0; 0; g1/ifc4is any constant and
1
X
nD1
Dncos
ny
b
is the Fourier
cosine expansion ofg1onŒ0; b, which is possible if and only if
Z
b
0
g1.x/ dxD0.
Adding the four solutions yields
u.x; y/DCC
a

1
X
nD1
Bncoshny=aAncoshn.yb/=a
nsinhnb=a
cos
nx
a
C
b

1
X
nD1
Dncoshnx=bCncoshn.xa/=b
nsinhna=b
cos
ny
b
;
whereCis an arbitrary constant.
12.4LAPLACE’S EQUATION IN POLAR COORDINATES
12.4.2.v.r; /DR.r/‚./where (A)r
2
R
00
CrR
0
RD0and‚
00
C‚D0,‚.0/D0,
‚./D0. From Theorem 11.1.2,nD
n
2

2

2
,‚nDsin
n

,nD1,2,3,. . . . SubstitutingD
n
2

2

2
into (A) yields the Euler equationr
2
R
00
n
CrR
0
n

n
2

2

2
RnD0forRn. The indicial polynomial is

s
n


sC
n


, soRnDc1r
n=
Cc2r
n=
, by Theorem 7.4.3. We wantRn./D1and
Rn.0/D0, soRn.r/D

n=
0
r
n=

n=
0
r
n=

n=
0

n=

n=
0

n=
;
vn.r; /D

n=
0
r
n=

n=
0
r
n=

n=
0

n=

n=
0

n=
sin
n

I

276. Section 12.4Laplace’s Equation in Polar Coordinates271
u.r; /D
1
X
nD1
˛n

n=
0
r
n=

n=
0
r
n=

n=
0

n=

n=
0

n=
sin
n

, whereS./D
1
X
nD1
nsin
n

is the Fourier
sine series iffonŒ0; ; that is,˛nD
1

Z

0
f ./sin
n

d,nD1,2,3,. . . .
12.4.4.v.r; /DR.r/‚./where (A)r
2
R
00
CrR
0
RD0and‚
00
C‚D0,‚
0
.0/D0,‚./D0.
From Theorem 11.1.5,nD
.2n1/
2

2
4
2
,‚nDcos
.2n1/
2
,nD1,2,3,. . . . Substituting
D
.2n1/
2

2
4
2
into (A) yields the Euler equationr
2
R
00
n
CrR
0
n

.2n1/
2

2
4
2
RnD0forRn. The
indicial polynomial is

s
.2n1/
2

sC
.2n1/
2

, soRnDc1r
.2n1/=2
Cc2r
.2n1/=2
,
by Theorem 7.4.3. We wantRnto be bounded asr!0CandRn./D1, so we takeRn.r/D
r
.2n1/=2

.2n1/=2
;vn.r; /D
r
.2n1/=2

.2n1/=2
cos
.2n1/
2
;u.r; /D
1
X
nD1
˛n
r
.2n1/=2

.2n1/=2
cos
.2n1/
2
,
whereCM./D
1
X
nD1
˛ncos
.2n1/
2
is the mixed Fourier cosine series offonŒ0; ; that is,˛nD
2

Z

0
f ./cos
.2n1/
2
d,nD1,2,3,. . . .
12.4.6.v.r; /DR.r/‚./where (A)r
2
R
00
CrR
0
RD0and‚
00
C‚D0,‚
0
.0/D0,‚
0
./D0.
From Theorem 11.1.3,0D0,‚0D1;nD
n
2

2

2
,‚nDcos
n

,nD1,2,3,. . . . Substituting
D0into (A) yields the equationr
2
R
00
0
CrR
0
0
D0forR0;R0Dc1Cc2lnr. Since we wantR0to be
bounded asr!0CandR0./D1,R0.r/D1; thereforev0.r; /D1.
SubstitutingD
n
2

2

2
into (A) yields the Euler equationr
2
R
00
n
CrR
0
n

n
2

2

2
RnD0forRn. The
indicial polynomial is

s
n


sC
n


, soRnDc1r
n=
Cc2r
n=
, by Theorem 7.4.3. Since
we wantRnto be bounded asr!0CandRn./D1,Rn.r/D
r
n=

n=
;vn.r; /D
r
n=

n=
cos
n

,
nD1,2,3,. . . ;u.r; /D˛0C
1
X
nD1
˛n
r
n=

n=
cos
n

, whereF./D˛0C
1
X
nD1
˛ncos
n

is the
Fourier cosine series offonŒ0; ; that is,˛0D
1

Z

0
f ./ dand˛nD
2

Z

0
f ./cos
n

d,
nD1,2,3,. . . .

278. CHAPTER13
BoundaryValueProblemsforSecond
OrderOrdinaryDifferentialEquations
13.1BOUNDARY VALUE PROBLEMS
13.1.2.By inspection,ypD x;yD xCc1e
x
Cc2e
x
;y.0/D 2H)c1Cc2D 2;
y.1/D1H) 1Cc1eCc2=eD1;

1 1
e 1=e

c1
c2

D

2
2

;c1D
2
e1
;c2D
2e
1e
;
yD xC
2

e
x
e
.x1/

e1
13.1.4.By inspection,ypD x;yD xCc1e
x
Cc2e
x
;y
0
D 1Cc1e
x
c2e
x
;y.0/Cy
0
.0/D
3H) 1C2c1D3;c1D2;y.1/y
0
.1/D2H)
2c2
e
D2;c2De;yD xC2e
x
Ce
.x1/
13.1.6.ypDAx
2
e
x
;y
0
p
DA.x
2
C2xe
x
/;y
00
p
DA.x
2
e
x
C4xe
x
C2e
x
/;y
00
p
2y
0
p
CypD2Ae
x
De
x
ifAD1;ypDx
2
e
x
;yD.x
2
Cc1Cc2x/e
x
;y
0
D.x
2
C2xCc1Cc2Cc2x/e
x
;B1.y/D3and
B2.y/D6eH) c12c2D3,2c1C3c2D24;c1D13;c2D 8;yD.x
2
8xC13/e
x
.
13.1.8.B1.y/Dy.0/;B2.y/Dy.1/y
0
.1/. Lety1Dx,y2D1;B1.y1/DB2.y1/D0. By
variation of parameters, ifypDu1xCu2whereu
0
1
xCu
0
2
D0andu
0
1
DF, theny
00
p
DF.x/. Let
u
0
1
DF,u
0
2
D xF;u1D
Z
1
x
F.t/ dt,u2D
Z
x
0
tF.t/ dt;ypD x
Z
1
x
F.t/ dt
Z
x
0
tF.t/ dt;
y
0
p
D
Z
1
x
F.t/ dt;yDypCc1xCc2. SinceB1.yp/D0,B1.x/D0andB1.1/D1,B1.y/D
0H)c2D0; hence,yDypCc1x. SinceB2.yp/D
Z
1
0
tF.t/ dtandB2.x/D0,B2.y/D0H)
Z
1
0
tF.t/ dtD0:There is no solution if this conditions does not hold. If it does hold, then the solutions
areyDypCc1x, withc1arbitrary.
13.1.10.(a)The condition isba¤.kC1=2/(kDinteger). Lety1Dsin.xa/andy2Dcos.xb/.
Theny1.a/Dy
0
2
.b/D0andfy1; y2gis linearly independent ifba¤.kC1=2/(kDinteger),
since ˇ
ˇ
ˇ
ˇ
sin.xa/cos.xb/
cos.xa/sin.xb/
ˇ
ˇ
ˇ
ˇ
D cos.ba/¤0:
273

279. 274 Chapter 13Boundary Value Problems for Second Order Ordinary Differential Equations
Now Theorem 13.1.2 implies that (A) has a unique solution forany continuousFand constantsk1and
k2. IfyDu1sin.xa/Cu2cos.xb/where
u
0
1
sin.xa/Cu
0
2
cos.xb/D0
u
0
1
cos.xa/u
0
2
sin.xb/DF;
theny
00
CyDF.
u
0
1
DF.x/
cos.xb/
cos.ba/
; u
0
2
D F.x/
sin.xa/
cos.ba/
;
u1D
1
cos.ba/
Z
b
x
F.t/cos.tb/ dt; u2D
1
cos.ba/
Z
x
a
F.t/sin.tb/ dtI
yD
sin.xa/
cos.ba/
Z
b
x
F.t/cos.tb/ dt
cos.xb/
cos.ba/
Z
x
a
F.t/sin.ta/ dt:
(b)IfbaD.kC1=2/(kDinteger), theny1Dsin.xa/satisfies both boundary conditions
y.a/D0andy
0
.b/D0. Lety2Dcos.xa/. IfypDu1sin.xa/Cu2cos.x/where
u
0
1
sin.xa/Cu
0
2
cos.xa/D0;
u
0
1
cos.xa/u
0
2
sin.xa/DF;
theny
00
p
CypDF;u
0
1
DFcos.xa/;u
0
2
D Fsin.xa/;
u1D
Z
b
x
F.t/cos.ta/ dt; u2D
Z
x
a
F.t/sin.ta/ dtI
ypD sin.xa/
Z
b
x
F.t/cos.ta/ dtcos.xa/
Z
x
a
F.t/sin.ta/ dtI
y
0
p
D cos.xa/
Z
b
x
F.t/cos.ta/ dtCsin.xa/
Z
x
a
F.t/sin.ta/ dt:
The general solution ofy
00
CyDFisyDypCc1sin.xa/Cc2cos.xa/. SincebaD.kC1=2/,
y
0
.b/D0H)y
0
p
.b/D.1/
k
Z
b
a
F.t/sin.ta/ dtD0; therefore,Fmust satisfy
Z
b
a
F.t/sin.t
a/ dtD0. In this case, the solutions of the boundary value problem areyDypCc1sin.xa/, withc1
arbitrary.
13.1.12.Lety1Dsinh.xa/andy2Dsinh.xb/. Theny1.a/D0,y2.b/D0, and
W.x/D
ˇ
ˇ
ˇ
ˇ
sinh.xa/sinh.xb/
cosh.xa/cosh.xb/
ˇ
ˇ
ˇ
ˇ
Dsinh.ba/¤0:
(SinceWis constant (Theorem 5.1.4), evaluate it by settingxDb.) From Theorem 13.1.2, (A) has a
unique solution for any continuousFand constantsk1andk2. IfyDu1sinh.xa/Cu2sinh.xb/
where
u
0
1
sinh.xa/Cu
0
2
sinh.xb/D0
u
0
1
cosh.xa/Cu
0
2
cosh.xb/DF;
theny
00
yDF.
u
0
1
D F.x/
sinh.xb/
sinh.ba/
; u
0
2
DF.x/
sinh.xa/
sinh.ba/
;

280. Section 13.1Boundary Value Problems275
u1D
1
sinh.ba/
Z
b
x
F.t/sinh.tb/ dt; u2D
1
sinh.ba/
Z
x
a
F.t/sinh.ta/ dtI
yD
sinh.xa/
sinh.ba/
Z
b
x
F.t/sinh.tb/ dtC
sinh.xb/
sinh.ba/
Z
x
a
F.t/sinh.ta/ dt:
13.1.14.Lety1Dcosh.xa/andy2Dcosh.xb/. Theny
0
1
.a/Dy
0
2
.b/D0and
W.x/D
ˇ
ˇ
ˇ
ˇ
cosh.xa/cosh.xb/
sinh.xa/sinh.xb/
ˇ
ˇ
ˇ
ˇ
D sinh.ba/¤0:
(SinceWis constant (Theorem 5.1.40, evaluate it by settingxDb.) If
yDu1cosh.xa/Cu2cosh.xb/where
u
0
1
cosh.xa/Cu
0
2
cosh.xb/D0
u
0
1
sinh.xa/Cu
0
2
sinh.xb/DF;
theny
00
yDF.
u
0
1
DF.x/
cosh.xb/
sinh.ba/
; u
0
2
D F.x/
cosh.xa/
sinh.ba/
;
u1D
1
sinh.ba/
Z
b
x
F.t/cosh.tb/ dt; u2D
1
sinh.ba/
Z
x
a
F.t/cosh.ta/ dtI
yD
cosh.xa/
sinh.ba/
Z
b
x
F.t/cosh.tb/ dt
cosh.xb/
sinh.ba/
Z
x
a
F.t/cosh.ta/ dt:
13.1.16.Lety1Dsin!x,y2sin!.x/; theny1.0/D0,y2./D0,
W.x/D
ˇ
ˇ
ˇ
ˇ
sin!x sin!.x/
!cos!x !cos!.x/
ˇ
ˇ
ˇ
ˇ
D!sin!¤0
if and only if!is not an integer. If this is so, thenyDu1sin!xCu2sin!.x/if
u
0
1
sin!xCu
0
2
sin!.x/D0
!.u
0
1
cos!xCu
0
2
cos!.x//DFI
u
0
1
D F
sin!.x/
!sin!
; u
0
2
DF
sin!x
!sin!
I
u1D
1
!sin!
Z

x
F.t/sin!.t/ dtIu2D
1
!sin!
Z
x
0
F.t/sin!t dtI
yD
1
!sin!

sin!x
Z

x
F.t/sin!.t/ dtCsin!.x/
Z

0
F.t/sin!t dt

:
If!Dn(positive integer), theny1Dsinnxis a nontrivial solution ofy
00
CyD0,y.0/D0,y./D
0. Lety2Dcosnx; thenW.x/D
ˇ
ˇ
ˇ
ˇ
sinnx cosnx
ncosnxnsinnx
ˇ
ˇ
ˇ
ˇ
D n, andypDu1sinnxCu2cosnx
satisfiesy
00
p
Cn
2
ypD0if
u
0
1
sinnxCu
0
2
cosnxD0
nu
0
1
cosnxnu
0
2
sinnxDFI

281. 276 Chapter 13Boundary Value Problems for Second Order Ordinary Differential Equations
u
0
1
D
1
n
Fcosnx,u
0
2
D
1
n
Fsinnx;u1D
1
n
Z

x
F.t/cosnt dt;u2D
1
n
Z
x
0
F.t/sinnt dt;
ypD
1
n

sinnx
Z

x
F.t/cosnt dtCcosnx
Z
x
0
F.t/sinnt dt

I
yDypCc1sinnxCc2cosnx. SinceypD0,y.0/D0, soc2D0;yDypCc1sinnx. Since
y./D0,
Z

0
F.t/sinnt dtD0is necessary for existence of a solution. If this hold, then the solutions
areyDypCc1sinnx, withc1arbitrary.
13.1.18.Lety1Dcos!x;y2Dsin!.x/; theny
0
1
.0/Dy2./D0, and
W.x/D
ˇ
ˇ
ˇ
ˇ
cos!x sin!.x/
!sin!x !cos!.x/
ˇ
ˇ
ˇ
ˇ
D!cos!¤0
if and only if!¤nC1=2(nDinteger). If this is so, thenyDu1cos!xCu2sin!.x/satisfies
y
00
C!
2
yDF.x/if
u
0
1
cos!xCu
0
2
sin!.x/D0
!.u
0
1
sin!xCu
0
2
cos!.x//!DFI
then
u
0
1
D
Fsin!.x/
!cos!
; u
0
2
D
Fcos!x
!cos!
;
u1D
1
!cos!
Z

x
F.t/sin!.t/; dt; u2D
1
!cos!
Z
x
0
F.t/cos!t dt;
yD
1
!cos!

sin!x
Z

x
F.t/sin!.t/; dtCsin!.x/
Z
x
0
F.t/cos!t dt

:
If!DnC1=2(nDinteger), theny1Dcos.nC1=2/xis a nontrivial solutiony
00
CyD0,
y
0
.0/Dy./D0. Lety2Dsin.nC1=2/x; then
W.x/D
ˇ
ˇ
ˇ
ˇ
cos.nC1=2/x sin.nC1=2/x
.nC1=2/sin.nC1=2/x .nC1=2/cos.nC1=2/x
ˇ
ˇ
ˇ
ˇ
DnC1=2;
soypDu1cos.nC1=2/xCu2sin.nC1=2/xsatisfiesy
00
p
C.nC1=2/
2
ypDFif
u
0
1
cos.nC1=2/xCu
0
2
sin.nC1=2/xD0
.nC1=2/u
0
1
sin.nC1=2/xC.nC1=2/u
0
2
cos.nC1=2/xDF:
u
0
1
D
Fsin.nC1=2/x
nC1=2
Iu
0
2
D
Fcos.nC1=2/x
nC1=2
I
u1D
Z

x
F.t/sin.nC1=2/t
nC1=2
dtIu2D
Z
x
0
F.t/cos.nC1=2/t
nC1=2
dtI
ypD
1
nC1=2

cos.nC1=2/x
Z

x
F.t/sin.nC1=2/t dtCsin.nC1=2/x
Z
x
0
F.t/cos.nC1=2/t dt

I
y
0
p
D sin.nC1=2/x
Z

x
F.t/sin.nC1=2/t dtCcos.nC1=2/x
Z
x
0
F.t/cos.nC1=2/t dtI

282. Section 13.1Boundary Value Problems277
yDypCc1cos.nC1=2/xCc2sin.nC1=2/xI
y
0
Dy
0
p
C.nC1=2/.c1sin.nC1=2/xCc2cos.nC1=2//x:
Sincey
0
p
.0/D0,y
0
.0/D0H)c2D0;yDypCc1cos.nC1=2/x;
y
0
Dy
0
p
C.nC1=2/c1.nC1=2/x;y./D0H)yp./D0. Hence,
Z

0
F.y/cos.nC1=2/t dtD0
is a necessary condition for existence of a solution. If thisholds, then the solutions areyDypC
c1cos.nC1=2/xwithc1arbitrary.
13.1.20.SupposeyDc1´1Cc2´2is a nontrivial solution of the homogeneous boundary value prob-
lem. ThenB1.y/Dc1B1.´1/Cc2B1.´2/D0. From Theorem 13.1.1 we may assume without
loss of generality thatB1.´2/¤0. Thenc2D
B1.´1/
B1.´2/
c1. Therefore,yis constant multiple of
y0DB1.´2/´1B1.´1/´2¤0. To that checkysatisfies the boundary conditions, note thatB1.y0/D
B1.´2/B1.´1/B1.´1/B1.´2/D0 B2.y0/DB1.´2/B2.´1/B1.´1/B2.´2/D0, by Theorem 13.1.2.
13.1.22.y1Da1Ca2x;y1.0/2y
0
1
.0/Da12a2D0ifa1D2,a2D1;y1D2Cx.y2Db1Cb2x;
y2.1/D2y
0
1
.1/Db1C3b2D0ifb1D3,b2D 1;y2D3x.
W.x/D
ˇ
ˇ
ˇ
ˇ
2Cx 3x
1 1
ˇ
ˇ
ˇ
ˇ
D 5IG.x; t/D
8
ˆ
ˆ
<
ˆ
ˆ
:

.2Ct/.3x/
5
; 0tx;

.2Cx/.3t/
5
; xt1:
yD
1
5

.2Cx/
Z
1
x
.3t/F.t/ dtC.3x/
Z
x
0
.2Ct/F.t/ dt

: (B)
(a)WithF.x/D1, (B) becomes
yD
1
5

.2Cx/
Z
1
x
.3t/ dtC.3x/
Z
x
0
.2Ct/ dt

D
1
5

.2Cx/

x
2
6xC5
2

C.3x/

x
2
C4x
2

D
x
2
x2
2
:
(b)WithF.x/Dx, (B) becomes
yD
1
5

.2Cx/
Z
1
x
.3tt
2
/ dtC.3x/
Z
x
0
.2tCt
2
/ dt

D
1
5

.2Cx/

2x
3
9x
2
C7
6

C.3x/

x
3
C3x
2
3

D
5x
3
7x14
30
:
(b)WithF.x/Dx
2
, (B) becomes
yD
1
5

.2Cx/
Z
1
x
.3t
2
t
3
/ dtC.3x/
Z
x
0
.2t
2
Ct
3
/ dt

D
1
5

.2Cx/

x
4
4x
2
C3
4

C.3x/

3x
4
C8x
3
12

D
5x
4
9x18
60
:

283. 278 Chapter 13Boundary Value Problems for Second Order Ordinary Differential Equations
13.1.24.y1Dx
2
x,y2Dx
2
2x; theny1.1/D0,y2.2/D0;W.x/D
ˇ
ˇ
ˇ
ˇ
x
2
x x
2
2x
2x1 2x2
ˇ
ˇ
ˇ
ˇ
Dx
2
.
SinceP0.x/Dx
2
,G.x; t/D
8
ˆ
ˆ
<
ˆ
ˆ
:
.t1/x.x2/
t
3
; 1tx;
x.x1/.t2/
t
3
; xt2:
:
yDx.x1/
Z
2
x
t2
t
3
F.t/ dtCx.x2/
Z
x
1
F.t/ dt: (B)
(a)WithF.x/D2x
3
, (B) becomes
yD2x.x1/
Z
2
x
.t2/ dtC2x.x2/
Z
x
1
.t1/ dt
D x.x1/.x2/
2
Cx.x2/.x1/
2
Dx.x1/.x2/:
(b)WithF.x/D6x
4
, (B) becomes
yD6x.x1/
Z
2
x
.t2/t dtC6x.x2/
Z
x
1
.t1/t dt
D 2x.x1/.xC1/.x2/
2
Cx.x2/.x1/
2
.2xC1/Dx.x1/.x2/.xC3/:
13.1.26.y1Da1Ca2x;y
0
1
Da2;B1.y1/D˛a1Cˇa2D0ifa1Dˇ,a2D ˛;y1Dˇ˛x.
y2Db1Cb2x;y
0
2
Db2;B2.y2/Db1C.Cı/b2D0ifb1DCı,b2D ;y2DCıx;
W.x/D

ˇ˛x Cıx
˛

D˛.Cı/ˇ. From Theorem 13.1.2, (A) has a unique solution
if and only if˛.Cı/ˇ¤0. Then
G.x; t/D
8
ˆ
ˆ
<
ˆ
ˆ
:
.ˇ˛t/.Cıx/
˛.Cı/ˇ
; 0tx;
.ˇ˛x/.Cıt/
˛.Cı/ˇ
; xt1:
13.1.28.y1Da1cosxCa2sinx;y
0
1
D a1sinxCa2cosx;B1.y1/D˛a1Cˇa2D0ifa1Dˇ,a2D
˛.y1Dˇcosx˛sinx.y2Db1cosxCb2sinx;y
0
2
D b1sinxCb2cosx;B2.y2/Db2ıb1D
0ifb1D,b2Dı;y2DcosxCısinx;W.x/D

ˇcosx˛sinx cosxCısinx
ˇsinx˛cosxsinxCıcosx

Since
Wis constant, we can evaluate it withxD0:WD

ˇ
˛ ı

D˛Cˇı. From Theorem 13.1.2, (A)
has a unique solution if and only if˛Cˇı¤0. Then
G.x; t/D
8
ˆ
ˆ
<
ˆ
ˆ
:
.ˇcost˛sint/.cosxCısinx/
˛Cˇı
; 0tx;
.ˇcosx˛sinx/.costCısint//
˛Cˇı
; xt:
13.1.30.y1De
x
.a1cosxCa2sinx/;y
0
1
De
x
Œa1.cosxsinx/Ca2.sinxCcosx/;
B1.y1/D.˛Cˇ/a1Cˇa2D0ifa1Dˇ,a2D .˛Cˇ/.

284. Section 13.2Sturm-Liouville Problems279
y1De
x
.ˇcosx.˛Cˇ/sinx.y2De
x
.b1cosxCb2sinx/;
y
0
2
De
x
Œ.b1.cosxsinx//Cb2.sinxCcosx/;
B2.y2/D e
=2
Œ.Cı/b2ıb1D0ifb1Dı,b2D.Cı/;
y2De
x
Œ.Cı/cosxCısinx/;
To evaluateW.x/, we writey1De
x
v1andy2De
x
v2, where
v1Dˇcosx.˛Cˇ/sinxandv2D.Cı/cosxCısinx.
Theny
0
1
Dy1Ce
x
v
0
1
andy2Dy2Ce
x
y
0
2
,
W.x/D
ˇ
ˇ
ˇ
ˇ
y1 y2
y1Ce
x
v
0
1
y2Ce
x
v
0
2
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
y1y2
e
x
v
0
1
xv
0
2
ˇ
ˇ
ˇ
ˇ
De
2x
ˇ
ˇ
ˇ
ˇ
v1v2
v
0
1
xv
0
2
ˇ
ˇ
ˇ
ˇ
D
ˇ
ˇ
ˇ
ˇ
.ˇcosx.˛Cˇ/sinx .Cı/cosxCısinx
.ˇsinx.˛Cˇ/cosx ..Cı/sinxCıcosx
ˇ
ˇ
ˇ
ˇ
:
Sincev
00
i
CviD0,iD1,2, Theorem 5.1.4 implies thatW.x/DKe
2x
, where is a constant that can be
determined by settingxD0in the determinant:
W.x/e
2x
ˇ
ˇ
ˇ
ˇ
ˇ Cı
˛ˇ ı
ˇ
ˇ
ˇ
ˇ
DŒˇıC.˛Cˇ/.Cı/:
From Theorem 13.1.2, the boundary value problem has a uniquesolution if and only ifˇıC.˛Cˇ/.C
ı/¤0. In this case the Green’s function is
G.x; t/D
8
ˆ
ˆ
<
ˆ
ˆ
:
e
xt
Œˇcost.˛Cˇ/sintŒCı/cosxCısinx
ˇıC.˛Cˇ/.Cı/
; atx
e
xt
Œˇcosx.˛Cˇ/sinxŒCı/costCısint
ˇıC.˛Cˇ/.Cı/
; xt=2:
13.1.32.LetypD
Z
b
a
G.x; t/F.t/ dt. From Theorem 13.1.3,LypDF,B1.yp/D0, andB2.yp/D0.
The solution ofLyDF,B1.y/Dk1, andB2.y/Dk2is of the formyDypCc1y1Cc2y2. Since
B1.yp/D0andB1.y1/D0,B1.y/Dk1H)k1Dc2B1.y2/H)c2D
k1
B1.y2/
. Since
B2.yp/D0andB2.y2/D0,B2.y/Dk2H)k2Dc1B2.y1/H)c1D
k2
B2.y1/
.
13.2STURM-LIOUVILLE PROBLEMS
13.2.2.y
00
C
1
x
y
0
C

1

2
x
2

yD0;
p
0
p
D
1
x
; lnjpj Dlnjxj;pDx;xy
00
Cy
0
D

x

2
x

yD0;
.xy
0
/
0
C

x

2
x

yD0.
13.2.4.y
00
C
b
x
y
0
C
c
x
2
yD0;
p
0
p
D
b
x
; lnjpj Dblnjxj;pDx
b
;
x
b
y
00
Cbx
b1
y
0
Ccx
b2
yD0;.x
b
y
0
/
0
Ccx
b2
yD0.
13.2.6.xy
00
C.1x/y
0
C˛yD0;y
00
C

1
x
1

y
0
C
˛
x
yD0;
p
0
p
D
1
x
1; lnjpj Dlnjxj x;
pDxe
x
;xe
x
y
00
C.1x/y
0
C˛e
x
yD0;.xe
x
y
0
/
0
C˛e
x
yD0.

285. 280 Chapter 13Boundary Value Problems for Second Order Ordinary Differential Equations
13.2.8.Ifis an eigenvalues of (A) andyis a-eigenfunction, multiplying the differential equation in
(B) byyyields.xy
0
/yC

x
y
2
D0;

Z
2
1
y
2
.x/
x
2
dxD
Z
2
1
.xy
0
.x//
0
y.x/ dxD xy
0
.x/y.x/
ˇ
ˇ
ˇ
ˇ
2
1
C
Z
2
1
x.y
0
.x//
2
dxI
y.1/Dy.2/D0H)xy
0
.x/y.x/
ˇ
ˇ
ˇ
ˇ
2
1
D0I
Z
2
1
y
2
.x/
x
dxD
Z
2
1
x.y
0
.x//
2
dx:
Therefore0:We must still show thatD0is not an eigenvalue. To this end, suppose that.xy
0
/
0
D0;
thenxy
0
Dc1;y
0
D
c1
x
;yDc1lnjxj Cc2;y.1/D0H)c2D0;yDc1lnjxj;y.2/D0H)
c1D0;y0; thereforeD0is not an eigenvalue.
13.2.10.Characteristic equation:r
2
C2rC1CD0;rD 1˙
p
.
D0:yDe
x
.c1Cc2x/;y
0
D e
x
.c1c2Cc2x/;y
0
.0/D0H)c1Dc2;y
0
D c2xe
x
;
y
0
.1/D0H) c2=eD0H)c2D0;D0is not an eigenvalue.
D k
2
,k > 0:rD 1˙k;yDe
x
.c1coshkxCc2sinhkx/;
y
0
D c1e
x
.coshkxCksinhkx/Cc2e
x
.sinhkxCkcoshkx/.
The boundary conditions require that
c1Cc2kD0and.coshkCksinhk/c1C.sinhkCkcoshk/c2D0.
This system has a nontrivial solution if and only if.1k
2
/sinhkD0. LetkD1andc1Dc2D1; then
D 1is the only negative eigenvalue, with associated eigenfunctionyD1.
Dk
2
,k > 0:rD 1˙ik;yDe
x
.c1coskxCc2sinkx/;
y
0
Dc1e
x
.coskxksinkx/Cc2e
x
.sinxCkcoskx/. The boundary conditions require that
c1Cc2kD0and.coskksink/c1C.sinkCkcosk/c2D0.
This system has a nontrivial solution if and only if.1Ck
2
/sinkD0. LetkDn(ka positive
integer)andc1Dn,c2D1; thennDn
2

2
is an eigenvalue, with associated eigenfunctionynD
e
x
.ncosnxCsinnx/.
13.2.12.Characteristic equation:r
2
CD0.
D0WyDc1Cc2x.y.0/D0H)c1D0, soyDc2x. Nowy.1/2y
0
.1/D0H)c2D0.
ThereforeD0is not an eigenvalue.
D k
2
,k > 0:yDc1coshkxCc2sinhkx;y
0
Dk.c1sinhkxCc2coshkx/.y
0
.0/H)c2D0,
soyDc1coshkx. Nowy.1/2y
0
.1/D0H)c1.coshk2ksinhk/D0, which is possible with
c1¤0if and only if tanhkD
1
2k
. Graphing both sides of this equation on the same axes show that it
has one positive solutionk0;y0Dcoshk0xis ak
2
0
-eigefunction.
Dk
2
,k > 0:yDc1coskxCc2sinkx;y
0
Dk.c1sinkxCc2coskx/.y
0
.0/H)c2D0, so
yDc1coskx. Nowy.1/2y
0
.1/D0H)c1.coskC2ksink/D0, which is possible withc1¤0
if and only if tankD
1
2k
. Graphing both sides of this equation on the same axes shows that it has a
solutionknin..2n1/=2; n/,nD1,2,3, . . . ;ynDcosknxis ak
2
n
-eigenfunction.
13.2.14.Characteristic equation:r
2
CD0.
D0WyDc1Cc2x. The boundary conditions require thatc1C2c2D0andc1Cc2D0, which
imply thatc1Dc2D0, soD0is not an eigenvalue.
D k
2
,k > 0:yDc1coshkxCc2sinhkx;y
0
Dk.c1sinhkxCc2coshkx/. The boundary
conditions require that
c1C2kc2D0andc1coshkCc2sinh2kD0.
This system has a nontrivial solution if and only if tanhkD2k. Graphing both sides of this equation

286. Section 13.2Sturm-Liouville Problems281
on the same axes shows that it has a solutionk0in.0; /;y0D2k0coshk0xsinhk0xis ak
2
0
-
eigenfunction.
Dk
2
,k > 0:yDc1coskxCc2sinkx;y
0
Dk.c1sinkxCc2coskx/. The boundary conditions
require that
c1C2kc2D0andc1coskCc2sinkD0.
This system has a nontrivial solution if and only if tankD2k. Graphing both sides of this equation on
the same axes shows that it has a solutionknin.n; nC1=2/,nD1,2,3, . . . ;ynD2kncosknxsinknx
is ak
2
n
-eigenfunction.
13.2.16.Characteristic equation:r
2
CD0.
D0WyDc1Cc2x. The boundary conditions require thatc1Cc2D0andc1C4c2D0, so
c1Dc2D0. ThereforeD0is not an eigenvalue.
D k
2
,k > 0:yDc1coshkxCc2sinhkx;y
0
Dk.c1sinhkxCc2coshkx/. The boundary
conditions require that
c1Ckc2D0and.cosh2kC2ksinh2k/c1C.sinh2kC2kcosh2k/c2D0.
This system has a nontrivial solution if and only if tanh2kD
k
12k
2
. Graphing both sides of this
equation on the same axes shows that it has a solutionk0in.1=
p
2/;y0Dk0coshk0xsinhk0xis a
k
2
0
- eigenfunction.
Dk
2
,k > 0:yDc1coskxCc2sinkx;y
0
Dk.c1sinkxCc2coskx/. The boundary conditions
require that
c1Ckc2D0and.cos2k2ksin2k/c1C.sin2kC2kcos2k/c2D0.
This system has a nontrivial solution if and only if tan2kD
k
1C2k
2
. Graphing both sides of this
equation on the same axes shows that it has a solutionknin..2n1/=4; n=2/,nD1,2,3, . . . ;
ynDkncosknxsinknxis ak
2
n
-eigenfunction.
13.2.18.Characteristic equation:r
2
CD0.
D0WyDc1Cc2x. The boundary conditions require that3c1C2c2D0and3c1C4c2D0, so
c1Dc2D0. ThereforeD0is not an eigenvalue.
D k
2
,k > 0:yDc1coshkxCc2sinhkx;y
0
Dk.c1sinhkxCc2coshkx/. The boundary
conditions require that
3c1Ckc2D0and.3cosh2k2ksinh2k/c1C.3sinh3k2kcosh2k/c2D0.
This system has a nontrivial solution if and only if tanh2kD
9k
9C2k
2
. Graphing both sides of this equa-
tion on the same axes shows that it has solutionsy1in.1; 2/andy2in.5=2; 7=2/;ynDkncoshknx
3sinhknxis ak
2
n
-eigenfunction,kD1,2.
Dk
2
,k > 0:yDc1coskxCc2sinkx;y
0
Dk.c1sinkxCc2coskx/. The boundary conditions
require that
3c1Ckc2D0and.3cos2kC2ksin2k/c1C.3sin2k2kcos2k/c2D0.
This system has a nontrivial solution if and only if tan2kD
9k
92k
2
. Graphing both sides of this
equation on the same axes shows that it has solutionsk0in.3=
p
2; /andknin..2nC3/=4; .nC
2/=3/,nD1,2,3, . . . ;ynDkncosknx3sinknxis ak
2
n
-eigenfunction.
13.2.20.Characteristic equation:r
2
CD0.
D0WyDc1Cc2x. The boundary conditions require that5c1C2c2D0and5c1C3c2, so
c1Dc2D0. ThereforeD0is not an eigenvalue.
D k
2
,k > 0:yDc1coshkxCc2sinhkx;y
0
Dk.c1sinhkxCc2coshkx/. The boundary
conditions require that
5c1C2kc2D0and.5coshk2ksinhk/c1C.5sinhk2kcoshk/c2D0.

287. 282 Chapter 13Boundary Value Problems for Second Order Ordinary Differential Equations
This system has a nontrivial solution if and only if tanhkD
20k
25C4k
2
. Graphing both sides of this equa-
tion on the same axes shows that it has solutionsk1in.1; 2/andk2in.5=2; 7=2/;ynD2kncoshknx
sinhknxiskn-eigenfunction,nD1,2.
Dk
2
,k > 0:yDc1coskxCc2sinkx;y
0
Dk.c1sinkxCc2coskx/. The boundary conditions
require that
5c1C2kc2D0and.5coskC2ksink/c1C.5sink2kcosk/c2D0.
This system has a nontrivial solution if and only if tankD
20k
254k
2
. Graphing both sides of this
equation on the same axes shows that it has a solutionknin
..2nC1/=2; .nC1//,nD1,2,3, . . . ;ynD2kncosknx3sinknxis ak
2
n
-eigenfunction.
13.2.22.D0:x
2
y
00
2xy
0
C2yD0is an Euler equation with indicial equationr.r1/2rC2D
.r1/.r2/D0.yDx.c1Cc2x/;y.1/Dy.2/D0H)c1Cc2Dc1C2c2D0H)c1Dc2D0,
soD0is not an eigenvalue.
D k
2
;k > 0:yDx.c1coshk.x1/Cc2sinhk.x1//;y.1/D0H)c1D0;yD
c2xsinhk.x1/;y.2/D0H)2c2sinhkD0H)c2D0;is not an eigenvalue.
Dk
2
;k > 0:yDx.c1cosk.x1/Cc2sink.x1//;y.1/D0H)c1D0;yDc2xsink.x1/;
y.2/D0withc2¤0ifkDn(na positive integer);nDn
2

2
;ynDxsinn.x1/is ak
2
n
-
eigenfunction.
13.2.24.D0:x
2
y
00
2xy
0
C2yD0is an Euler equation with indicial equationr.r1/2rC2D
.r1/.r2/D0.yDx.c1Cc2x/;y
0
Dc1C2c2x;y.1/Dy
0
.2/D0H)c1Cc2Dc1C4c2D
0H)c1Dc2D0, soD0is not an eigenvalue.
D k
2
;k > 0:yDx.c1coshk.x1/Cc2sinhk.x1//;y.1/D0H)c1D0;yD
c2xsinhk.x1/;y
0
Dc2.sinhk.x1/Ckxcoshk.x1//;y
0
.2/D0H)c2.sinhkCkcoshk/H)
c2D0;is not an eigenvalue.
Dk
2
;k > 0:yDx.c1cosk.x1/Cc2sink.x1//;y.1/D0H)c1D0;yDc2xsink.x1/;
y
0
Dc2.sink.x1/Ckxcosk.x1//;y
0
.2/D0withc2¤0if and only if sinkC2kcoskD0
or, equivalently, tankD 2k. Graphing both sides of this equation on the same axes shows that it has a
solutionknin..2n1/=2; n/,nD1,2,3, . . . ;ynDxsinkn.x1/is ak
2
n
-eigenfunction.
13.2.26.D0:yDc1Cc2x. The boundary conditions require thatc1C˛c2D0andc1C.C˛/c2D
0, soc1Dc2D0. ThereforeD0is not an eigenvalue of (A).
D k
2
,k > 0:yDc1coshkxCc2sinhkx;y
0
Dk.c1sinhkxCc2coshkx/. The boundary
conditions require that
c1C˛kc2D0
.coshkC˛ksinhk/c1C.sinhkC˛kcoshk/c2D0:
(D)
This system has a nontrivial solution if and only if.1k
2
˛
2
/sinhkD0, which holds withk > 0if and
only ifk
2
D ˙1=˛. ThereforeD 1=˛
2
is the only negative eigenvalue. We can choosekD ˙1=˛.
Either way, the first equation in (D) implies thate
x=˛
is an associated eigenfunction.
Dk
2
,k > 0:yDc1coskxCc2sinkx;y
0
Dk.c1sinkxCc2coskx/. The boundary conditions
require that
c1C˛kc2D0
.cosk˛ksink/c1C.sinkC˛kcosk/c2D0:
(E)
This system has a nontrivial solution if and only if.1Ck
2
˛
2
/sinkD0. ChoosingkDnproduces
eigenvaluesnDn
2

2
. SettingkDnin the first equation in (E) yieldsc1C˛nc2D0, soynD
n˛cosnxsinnx.

288. Section 13.2Sturm-Liouville Problems283
13.2.28.yDc1Cc2x. The boundary conditions require that
˛c1Cˇc2D0andc1C.LCı/c2D0.
This system has a nontrivial solution if and only if˛.LCı/ˇD0.
13.2.30.(a)yDc1coskxCc2sinkx;y
0
Dk.c1sinkxCc2coskx/. The boundary conditions
require that
˛c1Cˇkc2D0and.coskLıksinkL/c1C.sinkLCıkcoskL/c2D0.
This system has a nontrivial solution if and only if its determinant is zero. This implies the conclusion.
(b)If˛ıˇD0, (A) reduces to
.˛Ck
2
ˇı/sinkLD0: (B)
From the solution of Exercise 13.2.29(b),˛Ck
2
ˇı > 0for allk > 0. Therefore the positive zeros of
(B) areknDn=L,nD1,2,3, . . . , so the positive eigenvalues (SL) arenDn
2

2
=L
2
,nD1,2,3,
. . . .
13.2.32.Supposeis an eigenvalue andyis an associated eigenfunction. From the solution of Exer-
cise 13.2.31,

Z
b
a
r.x/y
2
.x/ dxDp.a/y.a/y
0
.a/p.b/y.b/y
0
.b/C
Z
b
a
p.x/.y
0
.x//
2
dx: (A)
If˛ˇD0then eithery.a/D0ory
0
.a/D0, soy.a/y
0
.a/D0. If˛ˇ < 0theny.a/D
ˇ
˛
y
0
.a/, so
y.a/y
0
.a/D
ˇ
˛
.y
0
.a//
2
: (B)
Moreover,y
0
.a/¤0because ify
0
.a/D0theny.a/D0, from (B), andy0, a contradiction. Since

ˇ
˛
> 0if˛ˇ < 0, we conclude that if˛ˇ0, then
p.a/y.a/y
0
.a/0; (C)
with equality if and only ifıD0. A similar argument shows that ifı0, then
p.b/y.b/y
0
.b/0; (D)
with equality if and only if˛ˇD0. Since.˛ˇ/
2
C.ı/
2
> 0, the inequality must hold in at least one of
(C) and (D). Now (A) implies that > 0.