Student Solutions Manual
Phillip E. Bedient * Richard E. Bedient
Eighth Edition
Elementary
Differential
Equations
Earl D. Rainville
Phillip E. Bedient
Richard E. Bedient
Student Solutions Manual
Phillip E. Bedient ¢ Richard E. Bedient
Eighth Edition
Elementary
Differential
Equations
Earl D. Rainville
Late Professor of Mathematics
University of Michigan
Phillip E. Bedient
rofessor Emeritus of Mathematics
Franklin and Marshall College
Richard E. Bedient
Professor of Mathematics
Hamilton College
PRENTICE HALL, UPPER SADDLE RIVER, NJ 07458
Assistant Editor: Audra Walsh
Production Editor: Carole Suraci
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Copyright © 1997 by Prentice-Hall, Inc.
A Pearson Education Company
Upper Saddle River, NJ 07458
All rights reserved. No part of this book may be
reproduced in any form or by any means,
without permission in writing from the publisher.
Printed in the United States of America
ISBN
O-13-592783-6
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Definitions; Families of Curves
12
Definitions.
2.
peak
frank
Contents
Numerical
2000.
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The General Solution of a Linear Equation
2...
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Ch bt
2.6
Miscellaneous Exercises
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Equations of Order One
2.1 Separation of Variables...
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2.2 Homogeneous Functions .. 0...
2.3 Equations with Homogeneous Coefficients...
2.4 Exact Equations 2... 0.
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Methods
3.2
Euler’s Method
3.3.
3.4
3.5
3.6
3.7
3.8
A Modification of Euler's Method. . 0... 0
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A Method of Successive Approximation .. 2.6.0.0...
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AnImprovement 2.0.0
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The Runge-Kutta Method...
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A Continuing Method 2.00.0. ee
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Elementary Applications
4.3
Simple Chemical Conversion. 2... 00
4.4
Logistic Growth and the Price of Commodities
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Additional Topics
on Equations of Order One
5.1 Integrating Factors Found by Inspection...
...-.00 00.00.0020
5.2 The Determination of Integrating Factors .......0.00 2.00000.
5.4 Bernoulli's Equation... 0.
5.5 Coefficients Linear in the Two Variables 2... 6
5.6 Solutions Involving Nonelementary Integrals...
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21
22
22
22
23
24
24
25
25
26
CONTENTS
6
Linear Differential Equations
6.2 An Existence and Uniqueness Theorem
...........0..00000000-00.
6.4 The Wronskian 2.0...
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6.8 The Fundamental Laws of Operation... 2.0000
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6.9 Some Properties of Differential Operators .......000..
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39
39
39
40
41
7
Linear Equations
with Constant Coefficients
7.2 The Auxiliary Equation: Distinct Roots ........0..00...0. 0200005 cas
7.3 The Auxiliary Equation: Repeated Roots .......0..00.000. 0.00000008
42
42
43
7.6
A Note on Hyperbolic Functions
Miscellaneous Exercises
8
9
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20.
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45
Nonhomogeneous Equations: Undetermined Coefficients
8.1 Construction of a Homogeneous Equation from a Specific Solution ..........
8.3 The Method of Undetermined Coefficieents 2.2.
0.0 0.0. ee
8.4 Solution by Inspection...
2.1.
48
48
49
54
Variation of Parameters
56
9.2
Reduction of Order...
0.
9.4 Solution of y’+ y= f(@)
0.
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10 Applications
10.3 Resonance
70
60.
70
10.4 Damped Vibrations. ©. 0. ee
10.5 The Simple Pendulum ©. 1... ee
11 Linear Systems of Equations
11.2 First-Order Systems with Constant Coefficients
11.4 Some Matrix Algebra 20
11.5 First-Order Systems Revisited. 2. 0
11.6 Complex Eigenvalues 0 00 0
11.7 Repeated Eigenvalues 2. 0.
11.8 The Phase Plane
12 Nonhomogeneous
. 0.0
13 The Existence
00 0.
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73
77
78
78
78
79
81
83
85
0
Systems of Equations
12.1 Nonhomogeneous Systems
19.2 Arms Races...
12.4 Simple Networks
©...
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6 0
87
88
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of Solutions
13.2 An Existence and Uniqueness Theorem...
87
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and Uniqueness
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90
94
94
CONTENTS
14 The
14.3
Laplace Transform
‘Transforms of Elementary Functions
14.6
14.10
FunctionsofClassA
Periodic Functions...
2...
95
.......00000.0.......0004..,
0...
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15 Inverse Transforms
15.1
15.2
101
Definition of an Inverse Transform
Partial Fractions ©...
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2...
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15.3 Initial Value Problems .. 0.0.0...
15.4 A Step Function «2.0...
15.5
A Convolution Theorem
16.2
16.56
16.7
16.9
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2.0.
15.6 Special Integral Equations...
15.8 The Deflection of Beams...
15.9 Systems of Equations...
16 Nonlinear
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108
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111
14
115
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120
2...
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Series Solutions
18 Solutions Near Regular
....
Singular Points
Regular Singular Points
2.2...
Difference of Roots Nonintegral
184
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143
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143
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Equal Roots...
18.7
Equal Roots, an Alternative...
0.
18.8
Nonlogarithmic Case...
18.9
Logarithmic Case.
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18.10
18.11
Solution for Largez oe
Many-Term Recurrence Relations 2...
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158
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181
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185
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205
2 0
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131
143
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18.6
Miscellaneous Exercises
120
123
124
126
134
17.5 Solutions Near an Ordinary Point... ... . —
18.4
101
102
Equations
Miscellaneous Exercises
18.1
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Factoring the Left Member
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17 Power
95
96
98
165
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20 Partial Differential Equations
20.3 Method of Separation of Variables
21 Orthogonal Sets of Functions
208
21.6 Other Orthogonal Sets...
22 Fourier Series
22.3 Numerical Examples of Fourier Series
22.4 Fourier Sine Series 20
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2...
0.2
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208
210
210
213
CONTENTS
23 Boundary Value Problems
23.1 The One-Dimensional Heat Equation... .........0...
23.4 Heat Conduction ina Sphere .. 0...
2. ee
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23.5 The Simple Wave Equation ........0.-.
205. ee ue
23.6 Laplace’s Equation in Two Dimensions. .............
24 Additional Properties of the Laplace Transform
24.1 Power Series and Inverse Transforms
24.2 The Error Function. 2. 0. 0
24.3 Bessel Functions
. 2... 0
.... 0.0.0.0...
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25 Partial Differential Equations:
Transform Methods
25.1 Boundary Value Problems...
25.2 The Wave Equation
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25.5 Diffusion in a Slab of Finite Width...
25.6 Diffusion in a Quarter-Infinite Solid
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2. .......0..0..0.
Chapter
1
Definitions; Families of Curves
1.2
Definitions
All answers in this section are determined by inspection.
1 . The equation is ordinary, linear in x, and of order 2.
cn
3. The equation is ordinary, nonlinear, and of order 1.
The equation is ordinary, linear in y, and of order 3.
7. The equation is partial, linear in u, and of order 2.
9. The equation is ordinary, linear in x or y, and of order 2.
il. The equation is ordinary, linear in y, and of order 1.
13. The equation is ordinary, nonlinear, and of order 3.
15. The equation is ordinary, linear in y, and of order 2.
1.3.
Families
of Solutions
1. Rewriting the equation yields y = / a? + Idx +c. Integrating, we have y = gx! + 2% +46,
Rewriting the equation yields y = 4 / cos 6a dx + ¢. Integrating, we have y = 2 sin Oz +c.
Rewriting the equation yields y = 2 /
1
ye
dz+c. Integrating, we have y = arctan (2/2)+c.
. Rewriting the equation yields y = 3 / e* dz. Integrating, we have y = 3e* +c,
the initial conditions gives 6 = 34+ corc=
Substituting
3 so y = 3e7 +3.
As in Example 1.2, y = ce**. Substituting the initial conditions gives 3 = ce® = c so y = 3e%*,
11. Rewriting the equation yields y = 4 / sin 2a dx. Integrating, we have ~2cos2¢+c.
ing the initial conditions gives 2 = -2cos7 +e = 2+corc=0s0
y = —2c0s 2u.
Substitut-
CHAPTER 2.
Chapter
EQUATIONS OF ORDER
ONE
2
Equations of Order One
2.1
Separation of Variables
1. The variables may be separated to give
[tana
few
r
Integrating both sides we have
Inr = ~-2¢ +corr=e =2t? +e orn
=
nt?
677 6,
To simplify this we replacing, the constant e° by another constant. We could use a different
name like k to get, r = ke~?*', but the convention is to reuse the name c to obtain
r= ce
oe
2t
BF
or To = €or Tr = To exp (—2t").
The variables may be separated to give
y
wee dy =
lies,
¥
dz
fm.
x
Integrating both sides we have
din(l+y*) =Ina+cor In(i+y*)/? =Ing+e.
Exponentiating both sides yields
(1 y2)¥2 = ell 49 op (1 $y?)
ax elt tee,
As in Exercise 1 replacing e° with ¢;
(1+y?)\/? = ex or y* = cz? -lory= Vex? — 1.
From the given conditions, 3=
/4¢~-1,
9=4e-1lorc=
y= 3 V 100? - 4.
42. Thus
2.1.
SEPARATION OF VARIABLES
3
. Separating the variables we have,
2{Z=3
ay
xr
y
Integrating both sides yields, 2Ina +¢ = 3iny or y? = cz.
conditions to obtain, 1 = 4c, or c = 4. Thus y = (x/2)?/8.
Finally, we substitute the initial
. Separating the variables we have,
[eva
= [oe
dz.
<a]
Integrating both sides yields, -e7~¥ = —ten
-+-c. Finally, we substitute the initial conditions
to obtain, -1 = -i =corc= —}. Thus 2e7¥ = e~®” +1 or In2—- y = In (e-™” +1) or
y = In2 —In[1 + exp (—2")).
. Separating the variables we have,
2a? [r-8dr—
fran
= f
sins,
Integrating both sides yields, ~a?r~? —Inr = —cos@+.c.
Finally, we substitute the initial
conditions to obtain, ~a?a~? — Ina = ~—cos0 +c or -Ina = c. Thus ~a2r7? ~Inr
—cos@ ~ Ina or r*(Inr — Ina) = r? cos@ — a? or r* In(r/a) = r? cos8 — a?.
=
11. Separating the variables we have,
[a-/
da:
yf
i-«
Integrating both sides yields, ~y~! = ~—In(1 —«)+c. Replacing ¢ by Inc, yjln|1 — el+ine] = 1.
Thus yin |e(1 — z)j = 1.
13. Separating the variables we have,
dy _
adz
Integrating both sides yields, —}y~? = hers” +c. Replacing —2e by c leaves e~™ + y7? = c.
16. Separating the variables we have,
mde _ fndy
z
¥
Integrating both sides yields, mlng = niny +c or 2™ = cy”.
17. Separating the variables we have,
[%-Vv
aP
Po
Integrating both sides yields, nV = ~ln P+corin(PV)=cor PV =e,
CHAPTER 2.
EQUATIONS
OF ORDER
ONE
19. Separating the variables we have,
[*
bsin é dé
r
|
1—bcosé’
Integrating both sides yields, In |r| = In|1 — bcos @| + ¢ or r = c(1 ~ bcos @).
ai. Separating the variables we have,
[uev=-[
Zyae=-f
(+045)
dz.
Integrating both sides yields, dy? = —(a+1)?—In (x — 1)+c or (x+1)? +y? +2 In |e(x — 1)| = 0.
23. Separating the variables we have,
[a
Integrating by parts yields, / (i +
Thus -
— i
=~
| xe* de.
| dy = —ze* + / e” dz.
= ~—ze* +e" +c or e*(a — 1) = (2y + 1)/(2y") +e.
25. Separating the variables we have,
fue
dy = [eo
dz.
Integrating both sides yields, —ye~¥ + / e-8 dy = —a7' +c. Thus ~ye7¥ ~ e74 =~
+6
or c(y +1) = (1+ czr)e¥.
27. Separating the variables we have,
J secudy = [cost xa.
Integrating both sides yields, In|secy + tany| = da + ; sin 2n + c.
Thus 4In]secy+ tanyj] = 2x + sin 2x + ¢.
29. Separating the variables we have,
[cost xa
= fe
+t?) dt.
Integrating both sides yields, 42 + 4sin 2x = 4(1+¢)? +c. Thus 2z+sin22 =e+ (1407).
2.2.
31.
HOMOGENEOUS FUNCTIONS
5
Separating the variables we have,
[tBan- [has
or
[ (E43) aa=- f (G41) a8
Integrating both sides yields, Ina + 3a = ~Inf—§8+c
caf = exp{—3a — ~).
or lnaf = ~3a~ B8+c.
Thus
33. Separating the variables we have,
rdz
/ dy
va? — x3
Integrating both sides yields, -/a? — x? = y +c. Thus y—-c¢ = ~Va? — 2”, the lower half of
the circle x? + (y — c)? = a?.
35.
Separating the variables we have,
a? dx
aye
=f
aVvx* — a?
/
LY,
y+e
Integrating both sides yields, a aresec(t/a) = y +e. Thus x = asec y
37. Separating the variables we have,
2
_fyti,
x
_
lane: [tae
/(v
_
1+
2
ty) dy
Integrating both sides yields, § In (2* +1) = $y? —y+2Inly+1j+e.
Thus In (x? + 1) = y? — 2y + 4 In |e(y + 1).
2.2
Homogeneous
Functions
All functions are homogeneous except those of Exercises 2, 5, 6, and 19 by examination.
2.3.
Equations
1. Substituting
y
with Homogeneous
=
av,
we
have
3(327
Coefficients
+ v*z*)dr
—
(9 + v) da = Qrvdv. We first separate the variables to obtain
[e-2/
udu
x
9 + 2"
Integrating yields Inc = In(9+ v7) +c.
x = [9 + (y/x)*] or 2° = c(9x? + y*),
22?u(ude
+ xdv)
=
O
or
Replacing ¢ by Inc and substituting for v leaves
CHAPTER 2.
EQUATIONS OF ORDER
3. Substituting y
= av, we have 2(227 + v?x?)dx — x*v(ude
(4+v7)dx = xcudu. We first separate the variables to obtain
/ dx
/
gc
Integrating yields Inz = 3 In 1 (4 +v*) +c.
f
+ adv)
ONE
=
0
or
vdy
442!
Replacing ¢ by Ine and substituting for v leaves
x? = c7[4 + (y/x)*] or ct = ce? (da? + y?).
. Substituting
y
=
xv,
we have
(4a* + Tux? + Quta*)dz — a?(udz + xdv)
=
0, or
(4+ 6v + Qu") dx = xdu. We first separate the variables to obtain
[e-
loss
du
4+6u+
-4/
202
1
+
1
2Q+u
d
itv
Integrating yields Ina = —}(In(2+v) —In(1+v)] +<¢ or
a
leaves ¢[1 + (y/x)] = «7[2 + (y/z)] or x?(y + 2x) = c(y + 2).
°-
= c*z*. Substituting for v
. Substituting y = cv, we have (2 ~ xv)(4z + av)da + x(5a — xv)(vdx + xdv)
(—2u? + 2u + 4) dx = (—5a + xv) du. We first separate the variables to obtain
dx
ve 5
1
[¢ =| sete’
1
0 or
2
-1f (=5+sh)
e(v
— 24/2
Integrating yields Inz = —}In(v ~ 2) —In(v+1) +c or
=
vt
dv.
= z, Substituting for v
leaves e[(y/x) — 2]!/? = a[(y/x) +1] or z(y + z)* = c*(y — 22).
. Substituting z = yu, we have (v2y? + 2y?v — 4y*)(udy + y du) — (uy? — 8y?u — 4y”) dy = 0 or
(v? + v? + 4u + 4) dy = ~y(v? + 2u — 4) dv. We first separate the variables to obtain
[*--[
yo
i
he-/
we+vrtdu+4 0 —
Integrating yields Ina = In (v + 1) ~ In(v? + 4) +c, or
1
v+t1
vt
ed
ay
vi+4
=: y. Substituting for v leaves
yl(y/a)? + 4] = c[(y/x) + 1] or 2? + 4y? = e(a + y).
i. Substituting
y
=
av,
we
have
(2?
+
v*z*)de
/
vw
1+2v?
+
z*u(udz
(1 + 2v”) dz = —zu dv. We first separate the variables to obtain
/ dx
ce
J
+
cdv)
=
0
or
Integrating yields ~Inz = 41In(1+?) +e, or (1 + 2u*)/4z = c. Substituting for v leaves
[1 + 2(y/x)?]a4 = ef or 2?(2* + 2y?) = ct.
2.3.
EQUATIONS
WITH HOMOGENEOUS
COEFFICIENTS
7
13. Substituting 2
=
vy,
we have v*(udy + ydv) +
(y? + 2y) du = —udy. We first separate the variables to obtain
- [=
at
ve
=a
Jf w+oy 2
yu(yy
il
yo
\,
y+2
+
v)dv
=
O
or
y.
1/2
}
=>
Substituting for y leaves
a
Integrating yields — Inv = g(Iny —In(y+2)}+eor € 2
c{(z/v) + 2] = (x/v)u* or xv? = efx + Qu).
16. Substituting yo = rv, we have 3 + susin 2) dx — asin Foy dx + xdv)
=
(2 + zusin®
v — usin? v) dz = x* sin? udu. We first separate the variables to obtain
dx
x
O
or
= | sin* vav.
1
Integrating yields Ing = 3Ui sin2v +c. Replacing c by Inc and substituting for v leaves
In |z/el= 4(y/x) — 4 sin (2(y/x)] or 4a in |x/e| ~ 2y + vsin (2y/x) = 0.
17, Substituting y = xv, we have (« — zvarctanu)dr + xarctanv(vdr + xdv) = 0 or
—(i — varctanv + varetanv) dz = xarctanudv. We first separate the variables to obtain
-{2
x
Integrating
yields
2
2
In eee)
~Inz
=
+c
=
= f asctanv do.
varetany
Qvarctanyv.
Then
—
4 In(l
+7).
substituting
for
Replacing
vu gives
¢ by
Inc
leaves
=
0 or
2
2
In (Eege*)
x
= 2(y/x) arctan (y/x) or 2y arctan (y/x) = x In [c?(x? + y*)/24}.
19. Substituting
t = sv, we have vs(s* + v%s?)ds — s(s? — v?s?)(uds + sdv)
(v +3 —v +") ds = s(1 — v*) dv. We first separate the variables to obtain
ds
seme
/ 8
SE
fi-v?
1
—————1) =
“3 yn
we
9 ~ 3 / (eno)
de
.
Integrating yields In |s| tes= 4(—4u7?—In[u}) or Injesv!/?|= ~jv*. Substituting for v leaves
—2t? in |es?(t/s)|= s? or s? = ~2#? In jest}.
al. Substituting
y
=
xv,
we
have
(3x?
— 227y
+ 327v")
de + 4z*u(udz
+ zdv)
(3— 2u — v*) dx = drvdv.
We first separate the variables to obtain
dz
/?
du
-/[sE=3
god
i”
—3
1
=-/[(St)
do.
Integrating yields Ing = ~3In(34+v)—-In(l—v)4+e
Substituting for v leaves z[3 + (y/x)P{1 ~ (y/x)] = c or (y — x)(y + 3x)? = ex.
=
0
or
CHAPTER 2.
EQUATIONS OF ORDER
23. Substituting
y = «xv, we have (x — zu)dz + (382 + zv)(ude + adv)
(l—v+3u +07) dz = —(3+ v)zdv. We first separate the variables to obtain
dz
-{¢
3+4
-/ ase
1
=f
for v leaves
Gs
or
2
ino™
dy
In[z(1 +v)| +c. Substituting
= Infa(1 + [y/2])] + ¢ or 22 = (2 + y) In (aw + y) + e(a + y). The initial
conditions give 6 = (1) In@ij)+e=c,
25,
0
2
(+a)
Integrating yields —Inx = In(1+v)~2(1+u)7!+cor
=
ONE
Thus 2(2r + 3y) + (@ + y) Infe+y) = 0.
Substituting y =
zu, we have (rv + Ja? +a%v?)dx — a(udr
(v+ V1 + vu? — v)dz = adv. We first separate the variables to obtain
+ xdv)
=
0
or
dy
Jitu?
Integrating
27.
yields
Inn
=
xz = ef(y/x) +
zg? =Qyt+l.
/1+ (y/z)*].
Substituting y
=
In(vu+Vl+v7)
+
«
Substituting
for
vu
leaves
The initial conditions give 3 = c(1+/3+1) orc = 1. Thus
xv, we have (x*v? + 7x*v + 16z*)dr + 2°(udz + 2dv)
=
0 or
(vu? + 7v +164 u)dx = ~—xdv. We first separate the variables to obtain
[¢Integrating yields Inz = —
ees
rs ie -/ oe
+c. Substituting for v leaves [(y/x) + 4]Ing = 1+ ¢[(y/x) + 4).
The initial conditions give 0 = 1+ 5¢ or c = ~—(1/5).
a2-y=5(yt+4z) ing.
Thus (y+ 4c) Ing = x — By + 4x) or
29. Substituting
c =
yv, we have yv(ydu + udy) + 2(y7u?
(u? + Qu? + 4) dy = —vy du. We first separate the variables to obtain
+ 2y2)dy
=
O
or
dy
v
-{% -f a pat
Integrating yields ~Iny = 3 In(3v? + 4)+c. Substituting for v leaves c = In ([3(x/y)? + 4]y®).
The initial conditions give C== 4. Thus y4(32? + 4y?) = 4,
31. Substituting
y
=
xv,
we
have
au(Qx — 2xv)dx
— 2(64 ~ cu)(vdz + adv)
(Qu ~ 2u? — 6v + vu?) dz = (6 ~v)adu. We first separate the variables to obtain
[2- [arene [qo=ae=
[Go sea) *
=
0 or
2.4.
EXACT EQUATIONS
9
Integrating yields Ing = 2lnv ~In(v —3)+c¢.
_
2*(y
—
32)
Th
c= ———~;-——. The
Seated
‘a2
.
Substituting for v leaves c = al(y/z) = 3]
(y/z)?
a
_
initial conditions give c = {1(1 ~3)|/1 = —2.
rn
2_.
Thus —2y*
or
2
= 2?(y — 32) or
¥
33 — zy — 2y? = 0.
33. Substituting y = av, we have (16x + 5xu)dx + (3x + rv)(udx + x2dv)
(16 + 5u + 3u + v*) dz = (~3 — v)rdv. We first separate the variables to obtain
-[S-
flatts
v+3
v= |
wtb pie.”
Integrating
yields
—-Ing
=
1
~In(ai(y/2) +4) = Gog
In(v+4)
ter
—In(-3+4)
= 1/(-3+4)+core
yt 3e = (y + 42) In (y + 42).
_t
+
ny
(v + 4)
1
1
v+4
(u4+4)?
+6
=
0
or
dv
‘
Substituting
for
vu
leaves
£
+4) = ae
= —1.
+c. The initial conditions give
Thus —(y + 4x) In(y+42) = xc - y — 42 or
35. Substituting
2 = yv, we have 2y*v(ydu + udy) — (8y2v?
(3u? — 2 — 2v*) dy = Quy dv. We first separate the variables to obtain
— 2y7)dy
=
0
or
dy
Integrating yields Iny = In(v? —2)+
¢.
Substituting for v leaves In (ars)
= ¢ OF
3
Cc=
2.4
gop
orc = 1/2. Thus 2? = 2y?(y +1).
Exact
Equations
OF
1. Let an 72 +y. Integrating with respect to z we have F = x7/2+2y+T(y).
with respect to y yields x
cor 2? +Iry
OF
Let a
¢
Differentiating
= 2+T"(y) = ¢-y so T(y) = —y?/2. Thus F = 2?/2+2y—y?/2 =
~yr=c.
2
= 2xy—32*.
.
.
4
2
3
Integrating with respect to x we have F = 2*y~2°+T(y).
:
sae
Differentiating
with respect to y yields 3 = x? +T7"(y) = 2? +y so that T(y) = y?/2.
Thus F = a2?y— 23 + y?/2 = cor x*y— 23+ $y? =e
. LetoF
ax
= c~2y. Integrating with respect to x we have F = x7/2—2zy+T(y).
with respect to y yields &
= —22+T"(y) = 2y — 2x so that T(y) = y?.
Thus F = 22/2 ~2ry + y? =c or 2? + 2y? = dry te.
Differentiating
CHAPTER
10
2.
EQUATIONS OF ORDER
ONE
This equation has homogeneous coefficients. Substituting y = av, we have (a — 2av) dx +
av —a)(vdz + adv) = 0 or (1 ~ 2u + 2v? — Qu) dz = (—2u + 2)xdu. We first separate the
variables to obtain
[e-
ze
(~2u+2)du
(2v — 2) dv
jf Q?—4u+1—
(Qu?
— 4u +1)
Integrating yields Ina = —1} In (2u? — du 4+ 1) +c or x7(2v* —4u +1) =c.
Substituting for v leaves z*(2(y/x)? ~ 4(y/x) + 1] = ¢ or 2? + 2y? = dey +e.
. Let i
= y* — Ixy + 6x. Integrating with respect to z we have F = zy” — xy + 32? + T(y).
Differentiating with respect to y yields
PP
a
‘
= Qay~ 224 7"(y) = ~2? + Izy ~ 2 so that
T(y) = —2y. Thus F = xy? ~ xy + 32" -2y =e.
il.
Let a
= cos 2y ~— 327y?. Integrating with respect to a we have F = xcos2y — zy" + T(y).
oF
Differentiating yields By
=
~Qrsin2y — 2x5y + T"(y)
:
= cos2y — 2xsin2y ~ 22%y so that
T(y) = 4sin 2y. Thus F = zcos2y — x*y? + § sin 2y = c or 4} sin 2y + x cos 2y — ay? = c.
,_
OF
13. Let an"
.
:
Integrating with respect to z we have F = x + ry? + tx7y? + T(y).
OF
Differentiating with respect to y yields By = Qry+c*y t+ T’(y) = x?y + y+ 2zy so that
1+y°+ay*.
T(y) = y?/2. Thus F = a+ zy? + $27y? + gy =corQet+y%(l+c)? =e
15. Let x
= 2? —xsec*y.
Integrating with respect to y we have F = x2*y — rtany + Q(z).
OF
Differentiating with respect to z yields —~- = 2xy —~ tany + Q(x)
Ox
= 2xy — tany so that
Q(z) = 0. Thus F = x°y — xtany =e.
17. Let a
= r+sin?—cos@. Integrating we have F = r?/2+r(sin@—cos0)+T7(@).
with respect to @ yields a
Differentiating
= r(cos@+sin 0) +7'(6) = r(cos @ +sin @) so that T(@) = 0. Thus
F = r?/24+r(sin@ — cos@) =c or r* + 2r(sin@ — cos) =e.
19. Let a
Fe
ir
= sin@ — 2rcos*@.
.
Integrating with respect to 2 we have F = rsin@ — r? cos?@ +
T(6). Differentiating yields a
= rcos 6 + 2r? cos @sin@ + 7" (8) = rcos O(2r sin @ +1) so that
T(@) =0. Thus F = rsin# — r? cos? @ =e.
ai. Let a
me
= Qry. Integrating with respect to « we have F = 2*y + T(y).
a
respect to y yields i
or y(3a? + y”) =e.
Differentiating with
= 2? +T'(y) = y? +2? so that T(y) = y°/3. Thus F = xy + y3/3 =c
2.6.
THE GENERAL SOLUTION OF A LINEAR EQUATION
OF
23. Let a = zy? + y ~ x. Integrating with respect to
OF
Differentiating with respect to y yields a
11
we have F = Za7y* + cy — $2? + Ty).
= ay +ao+T'(y)
= a*y +2
so that Ty)
= 0.
Thus F = $2*y? + ay — $27 = cor 2?y? + 2ay — 2? =.
2
25. Let -
= y?+27(1—cy)~*. Integrating we have F = y3/3+2(1—zy)7'!+Q(z). Differentiating
OF
with respect to a yields on = zy(1—zy)7?+(1—zy)7!+Q'(z) = (1~zy)~? so that Q(x) = 0.
Thus F = 3y°+a(1—ay)~* = c. Substituting the initial conditions gives }~2 = core = —8,
Thus 4y° + 2(1 — zy)! = —8 or ayt — y9 + Say — 32 = 5.
OF
27. Let iy
= x + 3y* + exp(—2?).
:
Integrating we have F = yz? + y° + yexp(—2?) + Q(x).
Differentiating with respect to x,
i = Qry — 2zyexp(—z?) + Q'(z) = 6x? + Iry — Ary exp (~x?) so that Q(z) = 223. Thus
Fe=x?y+y* + yexp(—27) +2 =e.
o
2.6
The
General Solution of a Linear Equation
d
1, The equation in standard form is = ~ (3/z)y = x‘. Thus the integrating factor is
exp (-3/<)
=exp(—3inz) = =
d
Multiplying by the integrating factor we have is (5)
= x. Integrating and simplifying yields
y/e®
= x7/2+.c or 2y=
2° + cx,
“
dz
The equation in standard form is dy + [4/(y + ijx = y/(y +1). Thus the integrating factor is
exp (1/4)
Multiplying
= exp (4In(y+ 1] = (y+1)4.
by the integrating factor we have
sl
+ 1)4a]
= y(y +1)*.
Integrating and
simplifying yields 20z(y + 1)4 = (4y — 1)(y+1)4 +c or 2027 = 4y~1+e(y+1)74.
da
The equation in standard form is ——- + [(1 — 3u)/u]z = 3 Thus the integrating factor is
du
exp (/ Co
au)
= exp (Inu — 3u) = uexp (—3u).
d
Multiplying by the integrating factor we have Gy ue")
= 3ue~*". Integrating and simpli
fying yields ue~**z = —ue~*" ~ ge" +c or cu = ce™ ~ uf.
CHAPTER 2,
12
_ @
The equation in standard form is oy
ultiplying
Multiplyi
P
ONE
(cot z)y = cse x. Thus the integrating factor is
dx
ex
EQUATIONS OF ORDER
/ cota dz) = exp |[—In (sinx)| =
i
=
exp
™ sina’
2= esc”cae
Y)
df
by the integrating factor we have Ge
Gna)
oe:
-
x. Integrating and simplifying
yields y/sinz = ~—cotr-+cor y = csinz — cose.
dy
The equation in standard form is dn + (1/cosx)y = cosx. Thus the integrating factor is
exp (/ secede
= exp [In (secx + tan z)] = sec + tan x.
d
Multiplying by the integrating factor we have 7, lulsec x +tanz)| = 1+sinz. Integrating and
simplifying yields y(secz + tanz) = c+
— cosa.
dy
li, The equation in standard form is ix + (cotz +1/a})y = 1. Thus the integrating factor is
1
exp ( / (cots + *)
az)
.
d
Multiplying by the integrating factor we have az
fying yields zysina = ¢+ sing ~ xcosa.
13. The equation in standard form is
dy
a _
.
= exp [In (sin) + Ing] = vsina.
sing) = zsinz.
Integrating and simpli-
[2x /(1+27)ly = (x? +24)/(1+27). Thus the integrating
factor is
exp (-/ ar) = exp[-In(1+2”)] =(1+2%)"1
;
_,
d
Multiplying by the integrating factor we have qv
simplifying yields y(1 + 2?)~! = x — aretanz +
Bet
+2°)~"]
xe
= ite’
Integrating and
or y = (1 + 2”)(¢ + x — arctanz).
d
.
.
:
15. The equation in standard form is = ~ (ma)y = cie™*. Thus the integrating factor is
exp (/ —ma dz)
= exp (—Mmgr).
d
_m
Multiplying by the integrating factor we have ——(e~™*"y) = cye™~™2)*_
simplifying yields y = cye™” + cge™?*, where cz = e,/(m, — m2).
:
Integrating and
2.6.
THE GENERAL SOLUTION OF A LINEAR EQUATION
13
17. The equation in standard form is a + [2/(x(a? + 1)ly = (2? + 1)*/x. Thus the integrating
factor is
oo ([ garry) ores
2dx
_
wa:
:
.
Multiplying by the integrating factor we have
2
2
Ul ay
d
2?
is (ay )
=
= a? +2.
Integrating and
simplifying yields x7y/(x? +1) = 3 (0? +1)? +c or 2?y = (x? +1)9 +e(2? +1).
d
+ (2/(x? ~ 1)]y = 1. Thus the integrating factor is
19, The equation in standard form is =
ex
P
fea
gz? —]
dx}
= exp
=
exp
{in
vl
d
Multiplying by the integrating factor we have — (v3
dx
fying yields y ral
a+]
=
_ 27}
e+ij//
-I1
\"ri+l
«+1
bp 1
) =,
g+1
Integrating and simpli-
—-2In|[x+1|+ecor (x — ijy = (a@+1)(e+
¢~ 2injx +1).
d
21. The equation in standard form is = ~ {3tanz)y = 1. Thus the integrating factor is
exp(-3 / tanz ar)
= exp [~3
In (sec x)} = cos? x.
d
Multiplying by the integrating factor we have ay lu(eos” x)| = cos? a. Integrating and simpli-
fying yields ycos* « = 4(2 + cos* x) sinx + ¢ or 3ycos*x = c+ 3sinz ~ sin? z.
d
23. The equation in standard form is a
:
dx
[6a/(x? + a)]y = 20(x? + a*). Thus the integrating
factor is
exp(-6 / wept)
= exp{—3ln (2? + a?)] = (x? + a?)7?.
Multiplying by the integrating factor we have < [y(a? + a?)3}
Integrating and
* (a? $a)?
simplifying yields y(x? + a7)~* = —(a? + a?)~1 + ¢ or y = (x? + a*)"[e(a? + a?) — 1),
25. If n = 0, the equation is separable so fa
= of
x
zr+a
dx or y = ba +e~-abln |x + al.
If nm = —1, the equation in standard form is a —y/(x@+a) = be/(x+a). Thus the integrating
factor is
exp (-/
<a)
= exp[—In(z +4)} = (2 +a)7}.
CHAPTER 2.
14
d
Multiplying by the integrating factor we have que
EQUATIONS OF ORDER ONE
+a)7}}
_
bx
Integrating and
~ (x-+a)?"
simplifying yields y(x+a)~! = bIn|x + al+ab/(x+a)+c or y = ab+e(x+a)+b(2+a) In |x + al.
dy
27. The equation in standard form is in
[1/(Qx + 3)]ly = (22 +3)-/?.
Thus the integrating
factor is
a
on (-/
+ 3)] = (22 +3)7¥/?,
3) = exp[—4 In (2x
d
Multiplying by the integrating factor we have 5, lv(ant3)-
2) = (20+3)~!. Integrating yields
y(2z +3)~1/? = 4 In (2x + 3) +c. Substituting initial conditions gives 0 = § In(~2+3)+c or
c= 0 so 2y = (2c+ 3)'/?
In (2x + 3).
29, The equation in standard form is “ + (R/L)i = E/L. Thus the integrating factor is
exp (cR/2) fz)
= exp (Rt/Z).
Multiplying by the integrating factor we have < [iexp (Rt/L)] = (£/L) exp (Rt/L). Integrating yields iexp(Rt/L)
= (E/L)(L/R)exp({Rt/L) +c.
R
O= E/R+core=—(E/R) soi =5
Substituting initial conditions gives
E - exp (-F
|
d
31. The equation in standard form is = + 2y = 4x. Thus the integrating factor is
exp (22)
= exp 22.
d
Multiplying by the integrating factor we have qa (ve)
= 4xe**.
.
Integrating yields ye** =
Qne?* — e** + ¢. Substituting initial conditions gives -1 = —-1+corc=0soy=2z—-1.
d.
:
33. The equation in standard form is = + [(2t3)/(1 + t*)|s = 6t(1 + ¢?). Thus the integrating
factor is
exp { 2
.
‘
°
1+
:
dt ) = exp {t? — In(t? + 1)] =
d
Multiplying by the integrating factor we have i
ee
815 si}
= 3e°
2
+c.
et
say
e .
i +1
?
= 6te’.
Substituting the initial conditions gives 2 = 3+c¢
s = (1+ t*)[3 — exp (—¢?)).
.
:
Integrating yields
orc = ~1 so
2.6.
THE GENERAL SOLUTION OF A LINEAR EQUATION
Miscellaneous
15
Exercises
1. The equation is separable.
Separating the variables we have,
few
ss j[ era.
Integrating both sides yields, e¥ = $e°* + ¢. Thus 2e¥ = e?* +c.
3. The equation is linear in y.
The equation in standard form is a + [3/(a + 1)]y = 4/(c + 1)”. Thus the integrating factor
1s
exp (3 / =)
= exp (31n (2 + 1)} = (2 +1)°.
d
Multiplying by the integrating factor we have que
+1)5]
= 4(2+ 1).
simplifying yields y(x + 1)? = 2(2 +1)? + cor y = (x41)!
Integrating and
+e(24+1)79.
5. The equation is homogeneous.
Substituting
y
=
xv,
we
have
(v*z*)dz
+ (~22?
— 3x?u)(vde
+ zdv)
=
0
or
(uv? ~ 2u ~ 3u”) dz = (2+ 3v)xdu. We first separate the variables to obtain
de
,
f3v+2,
,
2
1
patfapeentl Gta) &
Integrating yields ~ Ina = Inv + $In(v+1)+c or z*v?(v +1) = c. Substituting for v leaves
a*(y/x)*[(y/z) + 1] = ¢ or yx + y) = on.
7. The equation is linear in y.
a
The equation in standard form is = + (2x)y = x3. Thus the integrating factor is
exp ( / 2x ax) = exp (x),
Multiplying by the integrating factor we have lle)
yields y(e™’) Fe 1 (p22 - ev’) +c,
=e" 23. Integrating and simplifying
= $(x?~1)+ ce~™",
Substituting initial conditions gives 2 = ce~! or c = 2e so 2y = x? ~ 14 4exp(1 — 2”).
9. The equation is homogeneous.
Substituting y
=
xv, we have «v(t + 3ev)dr + a2(vde
(vu + 3u? + v)de = —adv. We first separate the variables to obtain
- [2 -fate-i/
ze
fj 3v2?+22u
?
1
»
3
+
xdv)
=
0
or
NG
Bupa)"
Integrating yields —Inz = 3{Inv—1n (3u + 2)|+¢ or x?v = c(3v +2). Substituting for v leaves
x*(y/x) = c[3(y/x) + 2] or x*y = c(2x + 3y).
CHAPTER 2.
16
EQUATIONS
OF ORDER
ONE
=
0 or
il. The equation is separable.
Separating the variables we have,
dy
x4
[f-/aee
Integrating both sides yields, ~gy3 =m za% —a-+arctangz +c.
Thus 23y3 + 1 = y3(c + 32 — 3arctan 2).
13. The equation is separable.
Separating the variables we have,
dz
me SE / cos’ tdz.
cosx
Integrating both sides yields, In|seca + tanz| = $¢+ ; sin 2t +c.
Thus 41n| seca + tanz| = 2¢+ sin 2i+c.
15. The equation is homogeneous.
Substituting y = av, we have
(2°
— 2*v*)dx
— (2?v + x*)(ude + adv)
(vy ~v? — v? — v) de = x(v + 1) dv. We first separate the variables to obtain
ax
-{G-
vt
ye
1
aw =3
[
1
1
(3 +53)
Integrating yields —Inc = }(Inv — v7!) + ¢ or 1/v = Incx*v.
t/y = In|cx*(y/z)| or x = yln|ecy).
du
Substituting for v leaves
17. The equation is exact.
Let a
= 7+2y. Integrating with respect to 7 we have F = x?/2+2cy+T(y).
with respect to y yields 3
Differentiating
= 9n+T"(y) = 22 +y so Ty) = 7/2. Thus 2? + 4cy + yi=e.
19. The equation is exact.
Ee
Let a = £°+y3, Integrating with respect to z we have F = x‘/ 4+y%2+T(y).
with respect to y yields %
Differentiating
= 3y2x+T"(y) = 3y22 + ky? so T(y) = ky*/4.
Thus ky! + 4zy? + 24 =.
21. The equation is exact.
Let a
= 3y + 2zy'. Integrating with respect to x we have F = 3xry + xy? + T(y).
0
entiating with respect to y yields %
Thus 27y? = 3(¢ + y — xy).
Differ-
mt
= 3a + 327y? + T'(y) = 327y? + 8x -— 3 so Ty) = —-3y.
2.6.
THE GENERAL SOLUTION OF A LINEAR EQUATION
17
23. The equation is linear in y.
d
‘The equation in standard form is = + (a)y = 6. Thus the integrating factor is
exp ¢ /az)
= exp ar.
d
Multiplying by the integrating factor we have qe
} = be**.
Integrating and simplifying
yields ye®® = (b/a)e** + ¢ or y = b/a + ce7*™.
The equation is also separable.
Q
yo. / dz. Integrating both sides yields —2 In (b — ay) =
Separating the variables gives
x+e.
Thus y = b/a+ce~™*.
25, The equation is exact.
OF
Let a
b-ay
= siny — ysing.
.
.
.
‘
Integrating with respect to z we have f = rsiny + ycosx + T(y).
OF
Differentiating with respect to y yields —T(y) = 0. Thus xsiny + ycos¢ = ce.
aq. The equation is linear in y.
d
The equation in standard form is -
= rcosy + casx + T’(y)
by
— (2cot x)y = sin? x. Thus the integrating factor is
exp(-2 / cot de
= exp[~2In (sinz)| = sin7? «.
Multiplying by the integrating factor we have
plifying yields y(sin~?xz) = —cosxz +c.
~ cos (7/2) +eorc=1s0y
= cosx + xcosy so
dz
[y(sin™? x)|
= sing.
Integrating and sim-
Substituting initial conditions gives, 1/sin? (7/2) =
= 2sin?x sin? $a.
29. The equation is separable.
Separating the variables we have,
/
dz
/
Vi — 22
dy
fi —y
Integrating both sides yields, arcsinz + arcsin y = c, or a part of the ellipse x? + 2eyry + y? +
cf — 1 = 0; where c; = cose.
31. Substituting, arcsinO = — aresin /3/2+c¢ or 0 = —7/34+c. arcsina + arcsiny = RT, or that
arc of the ellipse x? + zy + y? = 3 that is indicated by a light solid line in Figure 2.4.
35. The equation is linear in x.
The equation in standard form is
dz
dy
(1/y)z = 2/y*. Thus the integrating factor is
exp (-[%)
=exp(—Iny) = y7?}.
CHAPTER 2.
18
d
Multiplying by the integrating factor we have qe)
EQUATIONS
OF ORDER
ONE
= 2y7?. Integrating and simplifying
yields cy7! = -y7? +¢ or zy = cy? — 1.
i.
37. The equation is linear in y.
d
The equation in standard form is = — (tanaz)y = cosz. Thus the integrating factor i
exp(- / tanz az)
= exp [In (cos x)] = cos x.
Multiplying by the integrating factor we have ay cosz) = cos’ x. Integrating and simplifying
yields ycosx = 3(cosz sing
+ x) +c or 2y = sinz + (x +c) seco.
39.
The equation is linear in y.
d
The equation in standard form is = + [2/(1 —2*)iy = (1 ~ 32? + 224)/(1 — 2).
Thus the
integrating factor is
exp (/ ; —2
az)
= exp[~jln(1 ~2*)) = (1- a2)" U2,
ge
d
Multiplying by the integrating factor we have
ve
— gy)
= (1
1 —
ar
2x?
— g2)i/2°
Integrating
and simplifying yields y(1 — £2)~!/? = aresina + 2/1 — x? — arcsinx + ¢ or
yea—a+c(1—2*)/2,
41. The equation is linear in y.
d
.
.
:
The equation in standard form is i +(tanx)y = secz. Thus the integrating factor is
exp (/ tan edz)
= exp [In (sec x)] = seca.
d
Multiplying by the integrating factor we have an
eps
sec x) = sec” x. Integrating and simplifying
yields ycos™' 2 = tanz + ¢ or y = sina + ccosz.
43. The equation is linear in y.
d
:
The equation in standard form is = +[(1/x) ~ tanazly = 1. Thus the integrating factor is
exp (fore) — tan 2] az) = exp (Inz ~— In (sec x)| = a cos =.
d
Multiplying by the integrating factor we have a
fying yields ry cosa = c + cosx + rsinz.
cosz) = xcosz.
Integrating and simpli-
2.6.
THE GENERAL SOLUTION OF A LINEAR EQUATION
19
45. The equation is linear in 2.
d
.
‘
.
‘The equation in standard form is a — (3/y)a = y* - y. Thus the integrating factor is
exp(-s [
)
= exp (—3Ilny) = y7?.
d
Multiplying by the integrating factor we have ayy)
zy)
=
Iny+y7!
+c.
= yl
Substituting initial conditions gives
-1
-—y7*.
Integrating yields
= 0-1+c
orc
= 0 so
z=y"[1+yln(~y)].
47. The equation is linear in z.
d
The equation in standard form is a + (1/y)2 = -1+1/y?.
exp (/ #1)
Thus the integrating factor is
= exp (Iny) = y.
d
Multiplying by the integrating factor we have ay
a yt
L/y.
Integrating yields ry = —y*/2+Iny +c. Substituting initial conditions gives -1 = -1/2+¢
orce= —1/2so0 y7+22y+1=2Iny,
49. The equation is linear in y.
d
The equation in standard form is = ~y = 3a. Thus the integrating factor is
exp(-
fa)
= exp (—2).
Multiplying by the integrating factor we have < [y(e~*)]
y(e~*) = 3{-re~* — e~*) + ¢.
orce=Osoy
= —-3(r+ 1).
= 3ae™*.
Integrating
yields
Substituting initial conditions gives 0 = 3(—(—1)e — e] +c
51. The equation is linear in y.
dy.
The equation in standard form is = + (2/(1 — 2*)]y = 1. Thus the integrating factor is
ex
ae
Multiplying
P
2 f
dx
pone)
= exp
OP
.
.
by the integrating factor we have
yields y (55)
= 2+ 2in(z —-1)+c.
(In
z+]
_@+1
Ta)
d
— { y
dx
od
g+i
z~-1
=
2
1+———.
g-l
.
Integrating
Substituting initial conditions gives 3/1 = 2+0+¢
t-—
orc =1 so (24+ l)y = («@ — 1){e+14+2In
(x — 1)}.
20
CHAPTER 2.
EQUATIONS OF ORDER
ONE
53. The equation is linear in y.
d
‘The equation in standard form is ~ + (3/x)y = 1/z°. Thus the integrating factor is
exp (3 / =)
=exp(3inz)
= 2°.
d
Multiplying by the integrating factor we have a (ye?) = 277,2
Integrating yields yr? =
—z~! +c, Substituting initial conditions gives 1 = —1+corc=2
so ry = 2¢—1.
55. The equation is linear in y.
The equation in standard form is a + (5)
y= a(e
1
3)" Thus the integrating factor
is
exp (
54)
= exp (f2+c5«)
== exp[2In
x + In (ax ~ 2)] = 2*(x
— 2).
Multiplying by the integrating factor we have £ (ya?(a — 2)] = 2.
Integrating yields yx?(xz — 2) = x*/2 +c.
Substituting initial conditions gives 1{1 — 2)2 =
(1/2) + ¢ or e = 3 so 2a?(a — 2)y = 2? ~ 5.
21
Chapter
3
Numerical
3.2
Euler’s
Methods
Method
1. The results are found in following table.
y= 2e7~z—-1.
The correct solution is computed
Table 3.1;
x
0.0
y
1.00
«a2t+y
100
dy
0.10
Correct
1.00
0.1
#116
1.20
6.12
Lil
0.2
0.3
0.4
0.6
0.6
0.7
08
0.9
1.0
1.22
1.36
1.53
41.72
1.94
2.19
248
281
3.18
142
166
1.93
2.22
2.54
289
3.28
3.71
06.14
0.17
0.19
0,22
0.25
0.29
033
O37
1.24
1.40
1.58
1.80
2.04
2.33
2.65
3.02
3.44
3. The approximate value is y(1) = 5.78.
The actual value is y(1) = 6.15,
5. The approximate value is y(3) - 1.19.
The actual value is y(1) = 1.44.
7. The approximate value is y(2) = 1.09.
9. The approximate value is y(2) = 1.03.
11. The approximate value is y(2) = 2.09.
13. The approximate value is y(2) ~ .61.
from the solution
22
CHAPTER 3.
3.3.
A Modification
of Euler’s
NUMERICAL METHODS
Method
1. The approximate value is y{1) ~ 3.42.
The actual value is y(1) = 3.44.
3. The approximate value is y(1) = 6.12.
The actual value is y(1) = 6.15.
5. The approximate value is y(3) = 1.42.
The actual value is y(1}) = 1.44.
7. The approximate value is y(2) = 1.03.
9. The approximate value is y(2) ~ .97.
11. The approximate value is y(2) = 3.21.
13. The approximate value is y(2} = .70.
3.4
1,
A Method
Yo
= 1;
of Successive Approximation
«
n(z)=1+ f (¢+1)dt=1+24 42°;
0
ze
(a) =1+ [ (E+ L4+¢+¢7/2)dt=1l+ata? + 42°:
0
x
wn(a) = 1+ f (t+1+t+e?+¢9/6)dt=l+a+a?t do%4 dat
0
3. w(e)=14
ferrari
i
[24
(e-ae= 142-1)
+ He
1
n(x) =1+ [ [2+ 3(t — 1) + (¢~1)?/2] dt = 1+ 2a —1) + 3(@- 1)? + F(a — 1);
1
ys(z) = 1 +f
£
(2+3(t-1)+ 3-1)? + 4-1)
] at
1
=142%2—1)+3((e- 1)? +3(e-18+ d(e-1)*.
3.5
An Improvement
1. yo(t) =1+a;
n(e)=1+ [ (t+14+t)dt=1+24+2".
02
+ go°.
nts) =1+ f [e+ 1+t+P|dt=leata?
z
w(e) =1+ [ ft+1+t+e 448 )dt=1+et+a7+ 409+ det
3.6.
THE USE OF TAYLOR’S THEOREM
23
3. yo(z) = 2a— 1;
x
n(a) =1+ |
yo(x) -1+/
1x
a
y3(x) = 1 +
(t+ 2 -1dt=2e¢~1+
3(@- 1).
[t+ 2t~1+ 3(¢—1)*| dt = 22-14 3(2- 1)? + (27-1).
(t+ 2¢-1+ 3(¢- 1)? + 4(¢- 1) Jat
=142(¢—- 1) + 8(z-1)? + F(x-1)3 + giz ~ 1).
3.6
The
Use of Taylor’s Theorem
1. We first find the required derivatives:
yourty,
yl=lt+y,
y” ay",
y WH ay,aH
Then evaluate them at (zo, yo).
to=0,
yo=l,
m=1,
ye=lt+l=2,
yg = yy" =2.
This gives the desired polynomial.
ys(z) =L+ac+2? 4+ b0° 4+ dat + go.
3. Use the same derivatives as in Exercise 1, then evaluate them at (zo, yo)
to=1,
yw=l,
y=2,
yo=l1t2=3,
yf’ =”
=3.
This gives the desired polynomial.
ys(z) = 1+ 2(2~-2)4+ 8 (x — 2)? 4 $(x 13+
a(x ~1j44+ H(z —1)§,
5. Use the same derivatives as in Exercise 1, then evaluate them at (ro, yo)
to=2,
y=
1,
yeh
y=l+1=2,
yg!
= yg" =2.
This gives the desired polynomial.
ys(t) = —1 + (a ~ 2) + (@ — 2)? + $(e — 2)° + (a — 2)4 + Ala - 2).
7. Use the same derivatives as in the example in the section, then:
yS)
=
Qyy)
- 8y'y"
y'S)
=
2yy)
+ 10y’y)
+ 6y"y”,
of 20y"y"",
yi?
=
2yy)
f 12y’y")
4 30y”y)
+ 20y'"y".
Then evaluate them at (zo, yo).
zo = 0,
yo = 1,
yo
= 1,
ys) = 20, yf) =96,
ye
=1l+1=2,
yf) = 552,
yo
= 4,
yl” = 3776.
This gives the desired polynomial.
yr (x)
—
=
l+ao+
x
v
+
2
gx5
+
5
eat
+
4
x5
+ 376
+ B38 7
24
CHAPTER 3.
3.7
The Runge-Kutta Method
1. The results are found in the following table.
Table 3.2:
z
0.20
y
195
Ky
152
e+$h | 0.25
y+ 5hK, | 1.33
Ka
171
y+$hK2 | 134
K3
1.73
t+h
|030
ythK, | 142
K4
1.93
K
1,72
3.8
<A
0.30
142
193
0.35
1.52
2.19
1.53
2.22
0.40
164
2.53
2.21
Continuing Method
1. yf? = 1.93; y?) = 1.94; yf? = 2.36; x? = 2.37.
0.40
1.64
2.53
0.45
1.77
2.93
1.79
3.00
0.50
1.94
3.51
2.98
0.50
1.94
NUMERICAL METHODS
Chapter
4
Elementary
4.3
Applications
Simple Chemical Conversion
1. From Section 4.1, ve = /2gR= ./2(.165)(.00609)(1080) = 1.5 miles/sec.
3, From
u(l)
Section 4.2, u(t) = 70 +ce™*#. At t = 0, u(0) = 70+¢ = 18 orc = -32. Att =1,
= 70 — 52e7* = 31 or e~* = 3/4 so k = .29. Thus when t = 5, u(5) = 70 -
52 exp [-0.29(5)] = 58°.
. From Section 4.2, u(t) = -10 + ce7**. At 1:00 let t = 0, u(0) = -10+¢ = 700rc=
80. At t = 2, u(2) = —10
+ 80e7%* = 26 or ew** = 45 s0 k = .4. Thus when t = 5,
u(5) = —10 + 80 exp [-0.4(5)] = 1°F.
Starting over at 1:05, u(t) = 70+ ce~-** Let ¢ = 0 again u{(0) = 70+ ¢ = 1 or ¢ = —69. Thus
at 1:09 when ¢ = 4, u(4) = 70 — 69 exp [—0.4(4)] = 56°F.
. From Section 4.2, u(t) = 20+-ce7**. At 2:00 let ¢ = 0 so u(0) = 20+¢ = 80 or c = 60, At 2:03,
t = 3 so u(3) = 20 + 60e~%* = 42 or e~9* = .366 so k = 1/3. If the thermometer is brought in
at to, it reads
Starting over
At 2:10, ¢ =
exp (i9/3) — 1
u(to) = 20 + 60 exp (—to/3).
at t = to, u(0) = 80+ c = 20+ 60exp(—tg/3) or c = —60[1 — exp (—ty/3)].
10 — tp, thus u(10 ~ tg) = 80 -- 60/1 -- exp (—to/3)]
exp [-(10 ~ t9)/3] = 71 or
= 4.2 so tg = 5. The thermometer was brought in at 2:05.
From Section 4.3, 2(t) = ape7**.
At t = 10, 29/4 = roe7}* so In(3/4) = —10k or k = .028.
One-tenth will remain when 29/10 = zge~-©?** or In(1/10) = ~(.028)¢ or t = 80 sec.
ii. The equation is e
= —k,/z. Separating the variables gives / a2
de =x —k / dt. Integrat-
ing yields 221/? = ~kt +c or x(t) = (—kt/2+c)?. At t = 0, 2(0) = 29 = c? or c= \/f. The
substance will disappear when 0 = (—kt/2 + \/£o)* or t = 2,/£0/k.
13. Ifb >a
za;
ifbsa,r-b.
15. From Section 4.3, x(t) = rpe~**. One-half of the material remains at t = 38 so x9/2 = xpe725*
or In(1/2) = —38k so k = .018. One-tenth of the material reamains when ro/10 = ape7
01
or In(1/10) = —(.018)t so ¢ = 126 hr.
dv
19. From Exercise 17, (w/z
= w—kv.
This equation is linear in v.
Putting it in standard
form gives S +(g/b}v= g. The integrating factor is exp (g / at) =: exp(gt/b). Multiplying
26
CHAPTER 4.
ELEMENTARY APPLICATIONS
d
by the integrating factor yields 7 (est °y) = ge9'!” oy et/by = bettlbteo Att = 0, vo = b+e
or ¢ = ug — bso v = b + (v9 — b) exp (—gt/b).
al. 1.9 sec and 10.5 ft/sec.
dy
23. The rope will; always be tangent to the curve, thus in
yields fe
fae wah
~y
:
Jag
fq?
2 - [wy
.
Separating the variables
ap
Integrating givesx = aln (+vE=2)
~ Yar —y+e.
At the start, r= 0 and y =aso0=aln(a/a)-O+corc=0.
fae yp
Thus ¢= ain SE VEO
far,
25. (a) s = 160 Ib; (b) ¢ = 64 min,
a7.
4.4
From Section 4.3, y(t) = 100¢!, At t =1, y(1) = 100e% = 100(1.0513) = $105.13.
(a) $105.13 — $100 = $5.13; (b) 5.13%.
Logistic Growth
and the Price of Commodities
1. We want to maximize the growth rate f(z) = a(b ~ ax). Taking the derivative and setting it
equal to zero gives f'(xz) = ~2en +b = 0 or x = b/2a.
In 9
. Taking the derivative and setting it equal to zero gives t = ---————
in9 —In4
* 2.7 hr.
dg
The equation is HoT
kg which is linear in g. In standard form it becomes dg +kg = c,
dt
The integrating factor is exp(« / at) = exp (kt). Multiplying by the integrating factor yields
(eg) = ce’.
a(t} > e/k.
Integrating gives eg
= (c/k)e*t + ¢ or g(t) = (c/k) + ce-*).
As t - on,
a. The first dose is given at t = 0, so y(t) = yoe™**, for 0 <t <T so y(T) = ye *?.
b. With y(T) as the new initial condition, y(t) = yo(1 + e7*’ )e~**, for T < t < 27 so
y(2T) = yo(e7*? + e7 7),
c. Continuing in this way,
y(nT)
= yo(e7*?
+ e7 hk?
;
wees
of e7knT)
;
— yo(e"*T
+ (e~*T
This is a geometric series with r = e~** < 1 so, y(nT) = yo
d. Taking the limit as n - 00, y —
wee
db (e7kT)"),
evkt _ (e7*ty(n+4)
Taek
4.4.
LOGISTIC GROWTH AND THE PRICE OF COMMODITIES
27
9. From Section 4.4, = = k(D-~S)=k[(C ~ dP) - (a+ 6P + qsin Gt)| which is linear in P. In
dP
;
standard form it is oe +k(d+6)P = k(c — a) — kqsin Gt. The integrating factor is
exp(«a +5) / at) = exp [k(d + B)4).
Multiplying by the integrating factor yields
5 (peds+0ie) = k(e — ae(+0t — fegeK (+90 sin Bt.
Integrating gives
© @ k(d-+oye _ kgelarb)t
Pe R(dtbye =_ Tube
or
[Ad + b)sin: Gt — B cos Bt] + c; where Q* 2 = k(d + b)* 24 + 9°.92
Let (8/Q) = cos (Gt + a), then k(d+ b)/Q = sina, and
Ca
P=
=
Ty5t
Let P(0) = Po, then c; = Py — Tt
Cg
=G
~k(d+b)t
[cos(Bt +a)] + ce
_ oe
so
+ ak cos (Bt + a) + cpe7 MAFF),
P(t)=~ cs
d+b°
Q
28
CHAPTER 5.
Chapter
ADDITIONAL
TOPICS
ON EQUATIONS
5
Additional Topics
on Equations of Order
5.1
OF ORDER ONE
One
Integrating Factors Found
by Inspection
1. The terms of the differential equation may be rearranged to give
Qty? dx+ydz~-xdy=0
Integration now yields 2? + 7 =e
or
or
2x dz+
y de
—~ x dy
P
= 0,
2(2yt+1) =cy.
3. The terms of the differential equation may be rearranged to give
ay(y de+azdy)+dz=0
or
-as\3
Integration now yields ey
+Inflez]=0
(ry)*(ydz+x dy) + =
or
= 0.
a?y? = ~3ln lez}.
5. The terms of the differential equation may be rearranged to give
2,
ai(y dx +x dy) +y7(x dy—ydz)=0
1
3
Integration now yields zy + 3 (2) =e
or
dy
—
(y de +2 dy) + GEV
or
Lo,
y(3e4+y*) = 3er’.
7. The terms of the differential equation may be rearranged to give
y(ydz+ady)+ydz~-xdy=0
Integration now yields ry + 7 =e
or
or
(ydx+a dy)+
ydzr—-xdy_
Pp
2(y?+1) = cy.
9. The terms of the differential equation may be rearranged to give
xy*(y de+ady)=ydz-axdy
Integration now yields
(zy)?
2
= In <| +Infel
or
xy(ydz+zxdy)= we
or
(zy)? = 2In
7
|
0.
5.1.
INTEGRATING FACTORS FOUND BY INSPECTION
29
11. The terms of the differential equation may be rearranged to give
z*y*(y dx +2 dy) = my dz—na dy
or
d
mo
zy d(zy) =
_ nit.
x
Integration now yields
(xy)?
2
y
eg™
= mlIn|z|—nin|y|+in{e|
or
(xy)? = 2In
yn
. The terms of the differential equation may be rearranged to give
y*{y dx
+ x2 dy) +2ary dz —x* dy=0
z2
Integration now yields xy + 7 =c
or
or
d(xy)+
Qaey dx ~ x
dy
0.
y?
.
a(e+y?)= cy.
15, The terms of the differential equation may be rearranged to give
3x2
ay
a(3a*y dx — 2° dy) + a(x dy — y dx) + y? dx =0 or
dxdz ——
B
x
3d:
Yay
xz dy—y dz
de
+
y?
x
3
Integration now yields 7
_ 7 +injezj=0
or
ylnjer| = x(1 — 2”).
17. The terms of the differential equation may be rearranged to give
x?y(x dy — y dz) + y(y dv +x dy) +227 dy=0
Integration
now yields 2yy
+ 4
g
pine
ye
or
yde-cdy
or
ydr+ady
3
ray?
2dy
ys
z’yty+2=cry?
YT
=
cry’.
19. The terms of the differential equation may be rearranged to give
a? (y de ~ a dy) +y?(x dy~ydz)+d(zy)=0
Integration now yields 2 + y_
y
£
i +e=z0
xZy
or d (=) +d (2) + See
2+
or
cay +y? = 1.
2i, The terms of the differential equation may be rearranged to give
(2?
+ y*)(y dx +a dy)+xudy-—ydzr=0
Integration now yields ry + arctan - =
or
ydr+zxdy+
cdy-ydxr_
== Q,
x?
+ y?
0.
0.
30
CHAPTER
§
ADDITIONAL
TOPICS
ON EQUATIONS OF ORDER
ONE
23. The terms of the differential equation may be rearranged to give
=
ze3 ey (y dx +a dy)+y(a dy—ydz)=0
Integration now yields e™¥ + 5 (2)
2
= ;
or
df; +2YY
e ZY d(cy)
(2)
or
.
=0.
Q27e"4 + y? = cx?,
25. The terms of the differential equation may be rearranged to give
(2? — y*)(a2 dx —y dy) ~2? de=0
or
Co —y*)y — 7
= ¢.
When « = 2, y = 0, so that c = $. Therefore 3(x? — y?)? = 4(x° + 4).
or
xy d(xy)+2d @
-
“ol
xyi(y dx +a dy) +a7(2y dx -2Qa dy)-y? dx =0
i
2
27. The terms of the differential equation may be rearranged to give
1
2
1
Integration now yields a y+ = +5
+-=c
or
ry? +427 +2y = 2cry. When zx = 1, y=,
so that c = 7/2. Therefore z4y3 + to? — Tey + 2y= 0.
29. The terms of the differential equation may be rearranged to give
z"y"(2 de+ydy)+a(ydr+ady)=0
or
(a dxrt+ydy)+
a{y dx +x dy)
aig
= 0,
Integration now yields:
2
lfn #1,
2
z +H
zy?
Ifn= 1, a +
5.2
1.
len
My”
fag
- 5 tan |ayl = 0 or
or
(nm ~ 1)(x? + y? — c)(xy)"~! = 2a;
2? +4+y?—c= —2aln|ryl.
The Determination of Integrating Factors
W (F
_
|
= =,
tv tl} dz + = =
tyr
Therefore exp
/ = az|
= =
is an integrating factor.
dy= 0 is exact. The solution is
y?
g-=-~—=+y=ec
x
or
2? -y+ay-l=cr
That is,
5.2.
THE DETERMINATION OF INTEGRATING FACTORS
3.
a
(3
= tt
- a)
=
=
2
dz —
eee
Therefore exp
/ =
ay|
= =
31
is an integrating factor.
That is,
dy = 0 is exact. The solution is
9
2
= cy.
+2 +2in[yl =e or 227 + zy + 2yln|y|
5, £
(OM _ aN ) = ~—1. Therefore exp / ~] az| = exp(—c) is an integrating factor. That
“N\
@y
62
is, y(y + 2a ~ Qje~* daz ~ 2(c + y)e7* dy = 0 is exact. The solution is
~2Qeye"* —y*e"* = —c or y(2x+y) = ce”.
‘dy
ON ) _
Or
2
Therefore exp | / =az|
= 2” is an integrating factor.
That
is,
Ri
|
1 (FS
x*y(8a — 9y) dx + 2x7(x ~ 3y) dy = 0 is exact. The solution is
Qaty ~ 3n3y? =e
9.
5 (F
~ =)
=: 22. Therefore exp
or x y(Qr-y) =e.
/ 22 az| =: exp (2*) is an integrating factor. That is,
exp (x7)y(2x27 ~ zy + 1) dz + exp (z”)(x — y) dy = 0 is exact. The solution is
2
exp (z*) (zy - *) = : or y(2x — y) = cexp(—z7).
il.
a
N\
om
dy
_ On
=
= Ox
3
Therefore exp
z
/ 3 dc|
z
=
x° is an integrating factor.
That
is,
Qa3(Qy? + Sry — 2y + 4) dx + x4(Qa + 2y — 1) dy = 0 is exact. The solution is
Qaoy + ty? — 2ty +224 =e or ah(y* + 2ry—y 42) =c.
13.
4
(=
=)
= i
Therefore exp | / : «| = © is an integrating factor. That is,
Zz
N\ dy
dr
(Qry? + 3a*y — Qny + 6x7) da + (2? + 2a*y — z*) dy = 0 is exact. The solution is
xy + 27y? —2*y +20 =e or 2(y* +2y—y +22) =e.
15. We consider the equation
_ Me
Ma+Ny
a+
N
Mxu+ Ny
= 0 and check to see if it is exact. If we
write the equation as P dx + Q dy = 0 then
ap
_ ag
dy
ox
_
(Ma + Ny)
OM
aN
(F-=)
OM
+2N>—
~aM
(Ma
+ Ny)?
NG
~
OM
OM
ON
tae) "CR
(Mz + Ny)?
ON
OM
a
ON
tN a
ON
~ Me
32
CHAPTER 5.
ADDITIONAL
TOPICS
OM
Now from Euler’s thecrem we have oa
i
~
ON EQUATIONS
aM
+ vay
OF ORDER
ON
= kM and ra
+ 5
ONE
= KN, so that
= 0. That is (Mz + Ny)~' is an integrating factor of the original equation.
17. Because
is homogeneous,
ecause the the equation
equa
eNeous, an integrating
grating factor is gly
£
the differential equation -—~
“t
5
x? + 2y?
dx + ——--—-
2y3
i
ty
yh = Dy
Th ius
dy = 0 is exact. The solution is
+In{y| =In|ef
or 4y? in|2| = g?,
1
19. Because the equation is homogeneous, an integrating factor is (Mu + Nv)7! = ae
2
the differential equation —
9
du+ we’)
y
“ine
=| + In|e] = 0
~55
.
;
Thus
2
dy = Q is exact. The solution is
cy?
or u? = Qu? In |— ae
_ dy
;
.
al. The general linear equation of order one is is + P(2)y = O(e " We may rewrite the equation
in the form (Q— Py) de — dy = M dx +N
is a function of x alone so that exp
tion exp
/ P az|
Ee +Py|
dy
= 0. Now
/ P(x) da|
=z: Qexp
OM _ON\ _-P
_
“Oy
is an integrating factor.
/ P as|
is an exact equation.
ax}
—i
Hence the equaIntegrating we get
vero | [ P ae] = [ Qexp| [ P ae| dz +e.
5.4
Bernoulli’s Equation
1. We set u = 3a — 2y and dy =
ne
and simplify to get (5u +11) dz ~ (u+ 3) du = 0.
The variables now separate and we integrate to get 252 — 5u—41n/5u + 11| = cy. Substituting
for u and simplifying yields the solution 5(a + y + ¢) — 2ln|15a¢ —- 10y +11] = 0.
ld
di
9 and
nem
to get ae = 4u* +9. The variables
dz
4dz
4
de
now separate and we integrate to get arctan — = 62 +c. Substituting for u and simplifying
yields the solution 2(9z + 4y + 1) = 3tan (6x + c).
We set u = 92 + 4y +1
. We seu
=a2+y
and
and —dy = tL
wv
a du
dz
~ 1 and simplify to get du
separate and we integrate to get tanu — ou
dz
=
yields the solution x + c= tan {x + y) — sec (x + y).
+e.
= 1+ sinw.
The variables now
Substituting for u and simplifying
5.4.
33
BERNOULLI'S EQUATION
We set u = siny and du = cosy dy and simplify to get (3u? ~ 52ru) dr + 227 du = 0. We solve
this homogeneous equation to get 23(u — x)* = cu*. Substituting for u and simplifying yields
the solution 23(sin y — x)? = esin? y.
ar
dz
du
an 1~ da and simplify to get de + (x — 3)(a +1) du = 0. The
We set z = u—v and
variables now separate and we integrate to get (z — 3) exp (4u) = c(x + 1). Substituting for x
and simplifying yields the solution (u — v ~ 3) exp (4u) = c(u~ v + 1).
11. We set y = e”” and dy = 2e”” dv and simplify to get (ky —u) du = (y+ ku) dy. We solve this
homogeneous equation to get 24 arctan = =In ic(u® + y*)|. Substituting for y and simplifying
yields the solution 2k arctan (ue~?”) = In |c(u? + e*”)].
13. We set u = ¢ + 2y and dx = du — 2 dy and simplify to get (u — 1) du ~ (3u— 7) dy = 0. The
variables now separate and we integrate to get 3u + 41n|3u—7| —9y = —c. Substituting for
u and simplifying yields the solution 32 ~ 3y + ¢ + 41n [38x + 6y —7| = 0.
. A Bernoulli equation. We set v = y'~" and y7"y! =
Solving this linear equation we have (k +n —
‘
1l~n
to get v' —
l-—n
vs (l—nja®o},
1)v = (1~n)z* + cz'~™. Substituting for v we
get the solution (k +n —1)y!-" = (1 —n)z* + cx)".
1%. We set u = 2 + 2y and dz = du — 2 dy and simplify to get (wu ~ 1) du—dy = 0. The variables
now separate and we integrate to get (u — 1)* = 2y +c. Substituting for u and simplifying
yields the solution (x + 2y — 1)? = 2y +c. The original equation is also exact.
3
1
19. A Bernoulli equation. We set u = yo? and ul = —2y7Sy' to get uw! + —u = “3
.
Solving this
ic
linear equation we have ux® = c~z.
Substituting for u we get the solution y?(e — @) = x.
al. Suppose n = 1 and k= 0. Then ay’ = 2y. The variables separate and we get y = cz”.
Suppose n =
1 and k # 0.
a* = kin |).
x
Suppose n # 1 and k+n
equation v’ —
Then zy’ = (x* + 1)y.
The variables separate and we get
= 1. The solution proceeds as in Exercise 15 down to the linear
—* v = (1—n)a*~!
which in this case becomes v’ ~ 1a,
= (1-2.
This linear differential equation has solution v = (1 — n)x'~" In |er|. Substituting for v gives
us yi” = (1 ~ nj?" In [ca].
23. We set u = 3a + y and y’ = u’ — 3 and simplify to get u! = 2(u? +1). The variables now
separate and we integrate to get arctanu = 2x +c. Substituting for u and simplifying yields
the solution arctan (32+ y) = 22 -+¢. When « = 0, y = 1 implies that c = w/4. Thus
4arctan (32 + y) = 84 +7.
25, A Bernoulli equation.
We set u = y~? and
linear equation we have 2?u = «+c.
= —ay*@
to get a + =u = =
Solving this
Substituting for u we get the solution 2? = y9(x +c).
When x = 2, y = 1 implies that c= 2. Hence x? = y*(x + 2).
34
CHAPTER 5.
ADDITIONAL
TOPICS
ON EQUATIONS OF ORDER ONE
‘
d
3
x
27. We set u = y” and du = 2y dy to get = “gus 5 Solving this linear equation we have
2
x2
us my +ex*. Substituting for u we get the solution y* = -> +cex®, When
=1, y=1
implies that c = 3/2. Thus 2y? = 2?(3a — 1).
5.5
Coefficients
Linear
in the Two
Variables
1. The lines y = 2 and z~-y-—1 = 0 intersect at the point (3, 2) so we set x = u+3 and y = v+2.
We get v du —(u—v) dv = 0. We solve this homogeneous equation and get u = —v In [cu].
Substituting for u and v gives us 2 ~ 3 = (2 — y}In|e(y — 2}.
. The lines y = 2z and 4x + y ~ 6 = O intersect at the point (1, 2) so we set c = u+ 1 and
y=u+2. We get (2u—v) du+(4u+v) dv = 0. We solve this homogeneous equation and get
(w+ vu)? = c(2u + v)?. Substituting for u and v gives us (rz + y — 3)? = e(2x + y — 4).
. We let u = 2+y and du = dx +dy and get (u—1) du+(u+2) dy = 0. The variables separate
and we have u+y+c=3ln|u+ 2). Substituting for u we obtain c+ 2y+c= 3ln|x+ y+ Qi.
The lines cg ~ 3y + 2 = 0 and x + 3y ~ 4 = 0 intersect at the point (1, 1) so we substitute
z=u+landy=v+1. We get (u —3v) du+ 3(u + 3v) dv = 0. We solve this homogeneous
equation and get In (u? + 9v?) —2 arctan a
Substituting for u and vu gives us the solution
In [(x ~ 1)? + 9(y — 1)*) ~ 2arctan
a
3(y-1) > *
. The lines 92 ~— dy + 4 = 0 and 22 — y + 1 = Q intersect at the point (0, 1) so we need only set
y= util. We get (92 —4u) dx - (22 —u) du = 0. We solve this homogeneous equation and get
u = 3(u— 3a) In |e(3x — u)|. Substituting for u gives us y-1 = 3(y—3a2 —1) In |c(3x ~ y + 1)].
11. The lines 2 + 2y —1 =O and 27 + y — 5 = 0 intersect at the point (3, -1) so we set 2 = u+3
and y= u-~
1. We get (u + 2v) du ~— (2u+ 4) dv = 0. We solve this homogeneous equation
and get (uv)? = c(v+u). Substituting for u and v gives us (x — y ~ 4)? = e(z + y — 2).
13. The lines 34 + 2y + 7 = 0 and 22 — y = 0 intersect at the point (—1, —2) so we set r= u-~1
and y = v— 2. We get (8u + 2v) du + (2u —v) dv = 0. We solve this homogeneous equation
and get 3u* + 4uv —v? = c;. Substituting for u and v gives us 3x2? + dary — y? +142 =e. The
original equation is exact, so an easier method of solution is available.
15. We let u = 22-y+3 and dy = 2 dr— du and get (24+2u) dr—udu
= 0. The variables separate
and we have 2c —-u+In|ju+1| = —c—3. Substituting for u we obtain y-+e = — In jar — y + 41.
i7. We let u = « — y+ 2 and dy = dx ~ du and get (u+ 3) dz ~3 du = 0. The variables separate
and we have z -+c=3lnju + 3|. Substituting for u we obtain 2 +c = 3in ja — y + 5).
19. The lines 22 — 3y +4
= 0 and x = 1 intersect at the point (1, 2) so we set c = u+1 and
y=u+2. We get (Qu - 3v) du + 3u dv = 0. We solve this homogeneous equation and get
2uln |u| + 3v = cu. Substituting for u and v gives us 2(r — 1) In |x — 1] + 3(y ~ 2) = e(@ — 1).
When c = —1, y = 2 implies c = 21n2, so that 3(y ~ 2) = ~2(2 —1)In
=z
a
SOLUTIONS INVOLVING NONELEMENTARY INTEGRALS
5.6.
36
zi. The lines + y ~ 4 = 0 and 3z — y — 4 = 0 intersect at the point (2, 2) so we set r= u+2
and y= ut+2. We get (u+v) du—(3u—v) dv = 0. We solve this homogeneous equation and
a4
= c. Substituting for u and v gives us in|y — 2] +
get In ju ~ uj + ——
x = 3, y = 7 implies c = 4 + In4. Thus y — 62 +8=2(y ~2)In(#
2(a — 2)
—2z
=c.
When
—£
).
4
ad. Equation (B) is a* — 5a + 6 = 0 with roots a; = 2 and a2 = 3. We substitute « = 2u + 3v
and y = u+v and get (v — 1) du + (—2u — 4) dv = 0. The variables separate and we get
(v — 1)? + c(u+ 2) = 0. Substituting for u and v we get (x ~ 2y ~ 1)? = c{x ~ 3y — 2).
25. Equation (B) is a? + 3a+2 = 0 with roots a; = —2 and ag = ~1. We substitute 2 = —2u—v
and y = u+v and get 3 du+(—u-+2) dv = 0. The variables separate and we get the solution
v+ce
3in|u — 2|. Substituting for u and v we get e+ 2y+e¢= 3ilnjx+y+2l.
27.
Equation (B) is a? — 3a +2 = 0 with roots a, = 1 and ag = 2. We substitute c = u + 2u
and y = u+v and get (—2u + 4) du+(u+3) dv = 0. The variables separate and we get the
solution (u + 3)? = e(2 — v). Substituting for u and v we get (2y — 2 + 3)? = cly — x +2).
29. We substitute z = aju+ $v and y = u+v. The coefficient of du will be given by the
expression 1 [ai(a1u + Gv) + b)(u + v) + ci] + [aa(aru + Bv) + bo(u + v) +c]. In order for
the resulting differential equation to be linear in u, the coefficient of u in this polynomial must
be zero. That coefficient is aja? + (a2 + b:)ay + by. Thus if a; is a root of the equation
007 + (ag + b,)o + bo = 0 the resulting equation will be linear. The condition that 8 4 a, is
required so that x 4 ayy.
31. Equation (B) is now a? +4a+4
y=utv
= 0. We take ay = ~2 and § = 0. We substitute « = ~2u and
d
1
and get ~ ~ - = ~~ 1, This linear equation has solution u +1 — ev = ~vIn |v}.
,
.
Substituting for u and v gives
9,
5 +1-
ale + y)
x
=
+O
ee
choice of § will give a different form for this solution.
:
Yin
2
Sa
¥|
Bach different
In this particular form, if we put
cy, =: —1 — In 2c we get the answer given in the text.
33. Equation (B) is now a? ~ 3a+2
= 0.
1
and y = u-+ vu and get =
a qt
We take a; = 1 and 6 = 0.
-1-
ory:
We substitute z = u
This linear equation has solution
(u + 2v +3)? = e(u +2). Substituting for u and v gives (2y — z +3)? = cly—x +2).
5.6
Solutions Involving Nonelementary
d
1. The variables separate to give oY
ae
in the form infey| =2~
[
0
¢
|1 — exp (—2*)]
1
exp (—6") df =x —~ 5
eo
Integrals
.
dx. Upon integrating we have the solution
wert x.
36
CHAPTER
5.
ADDITIONAL
TOPICS
ON EQUATIONS OF ORDER
3. The equation is linear with the integrating factor exp (f 4° dx) = exp(z*).
given by yexp (x*) = [
0
@
g
y = e* |
An integrating factor is e~* so that
o-B
F
x
The solution is
exp (6*) df +e.
d
5. The equation is linear and may be written = -y= -.
ye * = |
ONE
dG, where c > 0. The condition y = 0, x = 1 requires that c = 1. Therefore
en8
——
B
df.B
7. This linear equation may be written —- ~ 2zy = 2°
and has an integrating factor exp (—x?).
The solution may be written yexp(—z?) = fz*exp(~2z?) dx +c.
An integration by parts
gives
yexp (—a) =
> exp (~z) + ; [ exp (—§7) dB +e.
0
Vr
The initial condition yields c = 1. Thus 2y = 2exp (27) ~ 2+ “a exP (x*) erf x.
Miscellaneous
Exercises
d
1. We set u = y* ~ 3y, and du = (2y ~ 3) dy and simplify to get ~ +u= a. We solve this linear
equation and obtain u = x — 1+ ce~*.
y? — 3y-2+1=ce*,
Substituting for u and simplifying yields the solution
The lines 2 +3y—5 = 0 and x ~y--1 = 0 intersect at the point (2, 1) so we set c = u+2 and
y=util. We get (u+3v) du—(u—v) dv =0. We solve this homogeneous equation and get
2v = (u+v) In |e(u + v)|. Substituting for u and v gives us 2(y—1) = (x +y~—3) In |e(x + y — 3)}.
. This differential equation is exact as it stands. The standard procedure for solving exact
equations yields the solution 2x7 + 2ay — 3y? — 8x + 24y =e.
d
12
. We set u = 24, and du = 42° dx and simplify to get a + 7
dy
= 4y?.
We solve this linear
equation and obtain l5uy!? = 4y!> +¢. Substituting for u and simplifying yields the solution
L5aty!? = dy 4+
a
We set u = 27, and du = 2x dx and simplify to get i
equation and obtain u = yt + cy®.
6
~ =u = ~2y3.
We solve this linear
Substituting for u and simplifying yields the solution
x? = yi(1+ cy’).
11.
We set u = e”, and du = e” dy and simplify to get o
equation and obtain x°(z + u)* = c.
o(e + ev)? =o.
+ =u
Substituting for u and
= -3.
We solve this linear
simplifying yields the solution
5.6.
37
SOLUTIONS INVOLVING NONELEMENTARY INTEGRALS
18. The lines « — 3y +4 = 0 and x — y ~ 2 = O intersect at the point (5, 3) so we set z=
u+5
and y= u+3. We get (u ~ 3v) du + 2(u—v) du = 0. We solve this homogeneous equation
and get (u +v)4 = c(u—2v). Substituting for u and v gives us (x + y — 8)4 = e(z — 2y +1).
. A Bernoulli equation. We set u = 27! and i
to get e + ou = —y°, Solving this
= -2F
linear equation we have y(5u+y4) = c. Substituting for u we get the solution y(5+ay4) = cz.
17. A Bernoulli equation.
We set u = a7? and -
= ~20F
to get -
+ ou = 2. Solving this
linear equation we have y?(2y — 3u) = c. Substituting for u we get y*(20%y — 3) = ca.
19. We set u = 4x + 3y, and dy = wie
and simplify to get ~(u + 25) dx + (w+1) du = 0.
The variables now separate and we integrate to get u —- 2 + 3c = 241n{u + 25).
Substituting
for u and simplifying yields the solution z + y +¢ = 81n [4x + 3y + 25],
ai. We set u = x — y, and dx = du + dy and simplify to get (3u — 2) du+ (2u —3) dy = 0. The
variables now separate and we integrate to get ~Gu ~ 4y + 2c = 51n |2u — 3]. Substituting for
uw and simplifying yields the solution 2(y ~ 32 +c) = Sin |22 — 2y — 3).
23. The lines « ~ y~1 = 0 and y = 2 intersect at the point (3, 2) so we set z = u+3 and
y= ut2. We get (u—v) du-2v dv = 0. We solve this homogeneous equation and get
(u + v)?(u — 2v) = c. Substituting for u and v gives us (x+y — 5)*(a - 2y+1) =e.
. We set u = o+ 2y, and dr = du — 2 dy and simplify to get (Qu — 1) du —5(u - 1} dy = 0. The
variables now separate and we integrate to get In|u — 1| = 5y— 2u+c.
Substituting for u and
simplifying yields the solution Inia + 2y—-1) = y-2¢+e.
27.
1/d0M
W
ea
an
- a)
= 3. Therefore e° is an integrating factor. That is, the differential equation
e** (6ay ~ 3y? + 2y) dx + 2e**(x — y) dy = 0 is exact. The solution is y(2x — y) = ce~**.
29. We set u = ax + by +e
and du = a dz + 6 dy and simplify to get “
this linear equation and obtain bu = —a + ce".
—bu
=a.
We solve
Substituting for u and simplifying yields the
solution 6?y = cye’* — aba — a — cb.
d
ai
al. A Bernoulli equation. We set u = v~* and - = —Qy73 =
x
d
to get =
~- 2, = 1. Solving this
x
linear equation we have u = cz? — z. Substituting for u we get the solution v7(cer? ~ 2) = 1.
When x = 1, v = § implies that ¢ = 5. Therefore zv?(5a — 1) = 1.
33. We set u = x+y, and dr = du — dy and simplify to get (1+ u*) du— yu dy = 0. The variables
now separate and we integrate to get y? = u? + 2In|uj +c.
Substituting for u and simplifying
yields the solution y? = (x + y)? + 2In|z+yj+e.
35, This differential equation is exact as it stands.
The standard procedure for solving exact
equations yields the solution x? + 2 — 32y ~ y? + 4y =.
38
CHAPTER
5.
ADDITIONAL
37. This equation is homogeneous.
we get x(v? +1)
x=
= e(v? — 1).
TOPICS
ON EQUATIONS OF ORDER ONE
dz
du
We set y = ur and get 7 CWT
dv = 0.
Integrating
Substituting for vu gives x(y? + 2?) = e(y? — x).
When
1, y = 2 implies that c = 5/3. We finally have 3x(y? + 2*) = 5(y* — x?) which may be
written y?(5 ~ 32) = 27(5 + 32).
39, We set u = 2 — y+ 2, and u’ = 1 — y’ and simplify to get du + (u — 1) dx = 0. The variables
now separate and we integrate to get Inju~ 1] = ¢— a. Substituting for u and simplifying
yields the solution In |v —- y +1] = ¢— x.
39
Chapter
6
Linear Differential Equations
6.2
An Existence and Uniqueness
Theorem
1. All the coefficients and the function R(x) are continuous for all z. Moreover bo(z) = x — 1 is
zero for only x = 1. Hence the equation is normal for « > 1 and for x < 1.
3. All the coefficients are continuous for all x. R(x) = Inz is continuous for x > 0. bo{x) = x? is
not zero for z > 0. Hence the equation is normal for z > 0.
5, Since e* and e~* are solutions of the differential equation, y = c,e* + cge~* is a solution also.
For this solution y(0) = 4 and y’(0) = 2 implies that c, +c2 = 4 and c; —cg = 2. Hence c) = 3
and cz = 1. It follows that y = 3e” + e~* is the unique solution of the initial value problem.
7. Since e* and xe* are solutions of the differential equation, y = c,e* + core* is a solution also.
For this solution y(0) = 7 and y’(0) = 4 implies that c, = 7 and c; +c, = 4. Hence c; = 7 and
Cg = ~—3. It follows that y = 7e? — 3xre™ is the unique solution of the initial value problem.
d"y
.
:
i
9. Consider
the general homogeneous linear equation bolt) a
d™—ly
+ b(t) 55
++ ba(cjy
= 0
that is normal on the interval J. The function y{z) = 0 that has value zero for all 2 on the
interval J is clearly a solution of this differential equation.
y(zg) = y'(ap) = --- = y-) (a9) = 0 for any
Moreover
for this zero function
on the interval J. By Theorem 6.2 this zero
function is the unique solution that has these properties.
6.4
The
Wronskian
loz
1Wi)=|.
0
1
00
3. W(z)
=
e*
|e?
e*
x
veo
ght
Qe
ss
(n- Ux"?
2,
0:
cos x
-~sing
-—cosz
1b
|= OL LP Qt---(m—
(n~ 1)!
sing
cosa]
—sing
=|
e*
e™
2e"
cosz
—sing
0
sing
cosz|
0
= 2e*
cos
—sint
sinz
coszi
= 2e".
W (2) is never zero these three functions are linearly independent on any interval.
Since
40
CHAPTER
6.
LINEAR DIFFERENTIAL EQUATIONS
The trigonometric identity cos (wt — 8) = coswt cos 6 + sinwt sin § may be written in the form
1- cos (wt — 8) + (— cos 8) - coswt + (~sin 8) - sinwt = 0 for all t. Hence the three functions
are linearly dependent on any interval.
The trigonometric identity sin? z+cos? z = 1 may be written 1-1+( —1)-sin? 2+(~1)-cos? x = 0
for all x. Hence the three functions are linearly dependent on any interval.
we)=|f
f
f"
af
ass
afl +2fl
xf
ataraflelf
af" +Acfi+2f|
f
0
7p
[f" af!
aeefl flopa9] =27°C)
Aafl+2F
Since f(z) is never zero on the interval the Wronskian is never zero on the interval and the
functions are linearly independent by Theorem 6.3.
11. Since the equation is homogeneous the zero function, f(z) = 0 for ali x in the interval, is a
solution of the differential equation. Moreover the zero function satisfies the initial conditions
f(zo) = f’(ze) = 0. By the uniqueness theorem the zero function is the only function that
has these properties.
Therefore y = f.
13. Since y, and yg are solutions of the homogeneous equation (A), by Theorem 6.1 the linear
combination y(x) = €:y1(2) + G2y2(z) is also a solution. Moreover y(xo) and y'(x9) are both
zero. Hence by the result of Exercise 11, y(a) = 0 for all x in the interval.
6.8
The Fundamental Laws of Operation
1. From the definitions we have
(4D+ 1)(D~ Qy= (4D + 1)[(D- 2)y] = 4D+ ow ~ 2y)= (4D)(y' — 2y) + (y' — 2y)
= 4(y" — 2y')+ (y’ — 2y) = dy” — Ty! — 2y = (4D? — 7D ~ 2)y,
Hence (4D + 1)(D — 2) = 4D* ~ 7D - 2.
From the definitions we have
(D +2)(D? — 2D + 5)y= (D + 2)(y"— 2y/ + 5y) = Dy”— 2y' + Sy) + 2(y” — 2y’ + 5y)
= (y! — Qy” + By’) + 2(y”— 2y’ + Sy)
= yl +y' + 10y = (D? + D+ 10)y.
Hence (D + 2)(D? -2D +5) = D3 + D+ 10.
. 2D7+3D —~2=(D+2)(2D — 1).
. D'-2D* -5D +6 = (D—1)(D? — D—6) = (D—1)(D + 2)(D — 3).
. D'—4D?
= D*(D? ~ 4)= D?(D— 2)(D +2).
il.
D3 — 21D +20 = (D— 1)(D?
+ D — 20) = (D — 1)(D
— 4)(D +5).
13. 2D4 +11D3 + 18D? +4D —8 = (D+ 2)(2D° + 7D? + 4D — 4) = (D + 2)?(2D7 + 3D — 2) =
(D + 2)3(2D — 1).
41
SOME PROPERTIES OF DIFFERENTIAL OPERATORS
6.9.
15. D* 4 D3 — 2D? 4 4D — 24 = (D — 2)(D9 + 3D? + 4D + 12) = (D — 2)(D + 3)(D? +4).
17. From the definitions we have
(D -2)(D + x)y = (D~2)(y' + zy) = Dly' + zy) — aly + zy)
ay
+ay ty—ay -ay=y" t+ (L-2)y= (DP +(l-2*)y
Hence (D — x)(D +2) = D? +1~27,
19. D(xzD —1)y = D(zy' — y) = zy” = (2D*)y. Hence D(xD ~ 1) = xD*.
21. From the definitions we have
(wD + 2)(aD — L)y = (xD + 2)(zy' ~ y) = (eD)(zy' — y) + 2(zy’ — y)
= a2(xy") + Qay! — Qy = a? y" + Qay! — Qy = (47D? + xD — 2)y.
Hence (2D +2)(aD —1) = z°D? + 22D — 2.
6.9
Some Properties of Differential Operators
1. Given the equation
(D — 2)°y = 0 we multiply by the factor e~?*
shift e~?*(D — 2)8y = D'(e~**y) = 0.
and use the exponential
We now obtain e~?”y = cy + cox + cx" so that
y= (cy + cox + e3n")e**.
3. Given the equation (2D ~ 1)?y = 0 we multiply by the factor e~*/? and use the exponential
"shift e~*/2(2D — 1)2y = (2D)*(e~*/*y) = 0. We now obtain e~*/2y = c, + cyx so that
y = (cy + ega)e*/?,
5. Suppose the four functions are linearly dependent
on an interval J.
Then
there exist four
constants, not all zero, such that cpe73" + cgre73* + cgz%e73* + cyaFe73* = 0 for all x in I.
But e7* is never zero so we can divide both sides by e~5* and obtain cj +ce22-+c3z*+c4x? = 0.
It follows that 1, 2, z?, z° are linearly dependent on J. This contradicts the results obtained
in Exercise 1 of Section 6.4. Therefore the supposition that the four functions are linearly
dependent is false.
42
CHAPTER
Chapter
7.
LINEAR EQUATIONS
WITH CONSTANT COEFFICIENTS
7
Linear Equations
with
7.2
Constant
Coefficients
The Auxiliary Equation:
Distinct Roots
1. The auxiliary equation is m* + 2m — 3 = 0 and its roots are m = —3,
y= cpe73? + ege®,
1.
The solution is
3. The auxiliary equation is m? + m — 6 = 0 and its roots are m = —3,
2.
The solution is
y = cye"9* + ene”,
5, The auxiliary equation is m? + 3m? — 4m = 0 and its roots are m = 0, 1, ~4.
The solution
is y = cy + cge™ + cge7*.
7. The auxiliary equation is m3 + 6m* + 11m +6 = 0 and its roots are m = ~1, -2, —3. The
solution is y = cje~* + cge7?* + cge7**.
9. The auxiliary equation is 4m? ~ 7m +3 = 0 and its roots are m = 1, a1
~2.
The solution is
y = cje* + cg exp ($2) +3 exp (—32).
11. The auxiliary equation is m? + m? — 2m = 0 and its roots are m = 0, 1, —2. The solution is
y = C1 + cge’ + ge.
13. The auxiliary equation is 9m> — 7m + 2 = 0 and its roots are m = —1, 4, 3. The solution is
y= ce"? + cgexp (42) + €3 exp (22).
15. The auxiliary equation is m? ~ 14m +8 = 0 and its roots are m = —4,
2+ /2.
The solution
is y = cye~* + cy exp [((2 + V2)z] + cg exp [((2 — V2)z].
17. The auxiliary equation is 4m‘ — 8m —7m*+11m+6 = 0 and its roots arem = —1, 2, —4, 3.
The solution is y = cye7* + cge*” + c3 exp (—42) + e4 exp ($x).
19. The auxiliary equation is 4m* + 4m — 13m? —7m+6 = 0 and its roots are m = —1, —2, 3, 3.
The solution is y = ce~* + cge7** + cg exp ($2) + ca exp (32).
21. The auxiliary equation is m* - 4am + 3a? = 0 and its roots are m = 3a, a. The solution is
y = 68" + coe,
23. The auxiliary equation is m? —2m—3
= 0 and its roots are m = —1, 3. The general solution is
y = ee"
* +cge"* and y! = —c,e~*+3ce2e"*. But y(0) = ce, +e = Oand y/(0) = —e,+3e2 = —4,
so that c; = 1, cg = ~1. The particular solution is y = e7* — e**.
7.3.
THE AUXILIARY EQUATION: REPEATED ROOTS
43
20. The auxiliary equation is m* — 2m — 3 = 0 and its roots are m = 3, —1. The general solution
is y = cye** + cpe7* and y'! = 3cye** — cge™*. But y(0) =e, + cp = 4 and y’(0) = 3c; — cg = 0,"
so that cy = 1, co = 3. The particular solution is y = e** + 3e7*. Thus y(1) = e* + e7}.
a7. The auxiliary equation is m? —m —6 = 0 and its roots are m = 3, —2. The general solution is
y = c,e°* +972" and y! = 3c) e9* —2cge7**. But y(0) = ce, +¢2 = 3 and y’(0) = 3¢;-2cg = -1,
so that cy = 1, c, = 2. The particular solution is y = e9* +2e7?*. Thus y(1) = e? + 2e7?.
29. The auxiliary equation is m? ~ 2m? —-5m+6
= 0 and its roots are m
=
1, 3,
-2.
The
general solution is y = cye* + c2e5* + cge7**, so that y/ = cye” + 3cge°* — Qege~**, and
y” = cye™ + 9cge?® + dege~**. But y(0) = cr t+ eo +e3 = 1, y’(O) = cy + 3cq - 2c3 = —7,
and y"(0) = ¢, + 9c2 + 4cg = —1. Thus c; = 0, cg = ~1, cg = 2. The particular solution is
y = —e* + 2e7™* and y(1) = —e? + 2e7?.
7.3
The Auxiliary Equation:
Repeated
Roots
1. The auxiliary equation is m? — 6m +9 = 0 and its roots are m = 3, 3. The general solution
is y = (cy + cox)e*.
The auxiliary equation is 4m? + 4m? +m
solution
is Y=
cy
+
(cg
+
C3)
= 0 and its roots are m = 0, -},
~4.
The general
exp (—4z).
The auxiliary equation is m+ + 6m3 + 9m? = 0 and its roots are m = 0, 0, ~3, ~3.
The
general solution is y = cy + coz + (c3 + caz)e7 3".
The auxiliary equation is 4m3 — 3m +1
= 0 and its roots are m = —1,
4,
d
The general
solution is y = (c, + cox) exp ($2) + cge7*.
The auxiliary equation is m? + 3m? + 3m +1 = 0 and its roots arem = —1,
general solution is y = (c, + car +cgr7)e7.
11.
—1,
—1.
The
The auxiliary equation is m> —_m? = 0 and its roots are m = 0, 0, 0, 1, —1. The general
solution is y = c; + cox + ¢gx? + cge™ + cge7*.
13. The auxiliary equation is 4m‘ + 4m — 3m? — 2m+1 = 0 and its roots are m = —1, —1, 4, 4.
The general solution is y = (cy + conje~* + (eg + cgx) exp ($2).
15. The auxiliary equation is m4 + 3m — 6m? —28m—24
= 0 and its roots arem = —2, —2, —2, 3.
The general solution is y = (cy + cg@ + egz7)e7?* + ege**.
17. The auxiliary equation is 4m —23m3 ~33m?—1l7m—3 = 0 with roots m = —1, —1, 3, — $3 ~%.
The general solution is y = (ce; + cgr)e7* + cge?* + (cq + csr) exp (—§2).
19. The auxiliary equation is m* — 5m? — 6m — 2 = 0 and its roots are m
= 1+ /3, —1, —1. The
general solution is y = (c, + cgr)e~* + cg exp [(1 + V3)a] + cg exp [(1 — V3)z].
ai. The
auxiliary equation
is m? + 4m +4
= 0 and its roots are m
solution is y = (c; + cgz)e~** and y! = (—2e; + cg — 2cgr)e~**.
=
—2,
—2.
The general
But y(0) = c, = 1 and
y'(0) = —2c, +c, = —1, so that c; = cg = 1. The particular solution is y = (1 + z)e7?*.
CHAPTER
44
7.
LINEAR EQUATIONS
WITH CONSTANT COEFFICIENTS
23. The auxiliary equation is m? — 3m — 2 = 0 and its roots are m = —1,
—1, 2. The general
solution is y = (cy; + cox)e~* + cze?*, so that we have y! = (—c, + ¢2 — conje~* + Qoge**, and
yl! = (c, — 2eg + egx)e"* + dege**. But y(0) = ce, +3
= 0, y'(0) = —cy +c + 2c3 = 9, and
y’ (0) = ce, — 2co + 4e3 = 0, so that c) = —2, co = 3, and cg = 2. The particular solution is
y = (-2 4+ 3n)e"* + 2e7*.
25. From Exercise 24 we have y = ¢; + ¢2¢ + cge~® + cge**, so that y/ = cg — cge™* — 2cqe72*
y" = cge~* + 4ege~**, and y” = —cge~* ~ Bcqge7**. We have y(0) = cy +¢3 +4 = 0,
y'(0) = eg ~ ¢3 ~ 2c4 = 3, y"(0) = c3 + 4eq = —5, and y”"(0) = —cg ~ 8cq = 9. Solving
these equations
gives us ¢;
=
2,
cg
=
0, and cg
=
cy
=
—1.
The
particular solution is
y= 2~e7? — e7*,
27.
The auxiliary equation is 4m? ~ 4m +1 = 0 and its roots are m = a 4. The general
solution isi y = (c, +cgr) exp (42) and y! = (fer +e +c@)
exp ($2). But y(0)= c; = ~2 and
y'(0) = ger yor = 2,80 that c; = —2, cz = 3. The particular solution is y = (~2+3z) exp ($2)
and y(2) =
29.
The auxiliary equation is m3 + 5m? + 3m —9 = 0 and its roots are m = 1, ~3, ~3.
The
general solution is y = cye* + (c2 + cgz)e~** and y/ = cye® + (~3c2 + cg + cgz)e73*. But
y(0) = ¢) + co = 1 and y(1) = eye + (cg + cg}e7™? = 0. Moreover, Jimy =x 0 implies that
ce, = 0, so that cy = 0, cg = ~1, and cg = 1. The particular solution is y = (—1+ xr)e~*, and
y(2) = e~®
7.6
A Note on Hyperbolic
Functions
1. In what follows we will use equation (5) on page 114 of the text.
[(D ~ a)? + 07] y = (D ~a)?y + by
= ¢3(D — a)* [e** cos bz] + c4(D — a)? [e* sin bz]
+- b* [e3e9” cos ba + eye sin bz]
=: ¢3€* D? (cos ba] + cge**
D? [sin ba] + cgbe%* cos bx + c4b2e™ sin bx
= 0,
. The auxiliary equation is m? — 2m +2 = 0 and its roots are m = 1 +7. The general solution
is y = e*(c, cose + cgsin x).
. The auxiliary equation is m? — 9 = 0 and its roots are m
y = c, cosh 3z + cg sinh 3a.
=
+3.
The general solution is
The auxiliary equation is m* —4m+-7 = 0 and its roots are m = 2+ /3i. The general solution
is y = e?* (ce; cos /3z + cz sin V3z).
The auxiliary equation is m4 + 2m
+ 10m?
= 0 and its roots are m = 0, 0,
general solution is y = cy + cox + e7*(ci cos 3z + eg sin 3x).
-1 43%.
The
7.6.
11
A NOTE ON HYPERBOLIC FUNCTIONS
45
The auxiliary equation is m* + 18m? + 81 = 0 with roots m = 3i, 34, ~3i, —3i. The general
solution is y = (cy + c2x%) cos 3x + (cg + c4x) sin 32.
—i,
13. The auxiliary equation is m® + Om4 + 24m? +16 = 0 with roots m =i,
2i, 21, —2%,
—2i.
The general solution is y = cy cos@ + cg sinx + (c3 + c4x) cos 2x + (cs + Cgx) sin 2a.
15. The auxiliary equation is m? — 1 = 0 and its roots are m = +1. The general solution is
y = c, cosha + cysinha and y’ = ce sinhx + cy coshz. But y(0) = c, = yo and y’(0) = co = 0,
so that cy; = yo and cg = 0. The particular solution is y = yo cosh x.
17. Here m? + 7m? + 19m
y = cye7* + cge7** cos
(—2cg — 3c3)e7** sin 2x,
But y(0) = cy + c2 = 0,
+13
=
0 with roots m
=
—1,
-3+
2.
The
general solution
is
2z + cge~** sin Qa, Thus y’ = —cye7* + (~3e9 + 2c3)e73* cos 2x +
and y” = cye"? + (5c, - 12e3)e79* cos 2x + (12c2 + 5ceg)e7* sin 2z.
y’(0) = —cy — 3cg + 2e3 = 2, and y"(0) = c) + Seg ~ 12cg = ~—12, 80
that ¢) = cg = 0, and cg = 1. The particular solution is y = e73* sin Qa.
19. The
auxiliary equation is m? +k?
dx
.
x = c, cos kt + cgsin kt, a=
But «(0) = c; = 0 and
ys
= 0 with roots m = tki. Thus, the general solution is
d*x
.
~ke;, sin kt + keg cos kt, and He sz —k*e, coskt ~ k*egsin kt.
= (0) = keg = Ug, 80 that c¢, = 0, cz = =.
The particular solution is
(2) sin kt.
al. The auxiliary equation is m? + 2bm + k* = 0 with roots m = ~btiv k? — b?, where k > b > 0.
The general solution is x = cye~™' cos Vk? — Bt + coe7 cos Vk? — Ht. But x(0) = c, = 0
and z'(0) = —bey +
¥
Qs ae
Miscellaneous
Vk? ~ bce, so that c, = 0, c2 =
. The particular solution is
vo
Vk2 — 7
sin («/k? ~ 622).
Exercises
1. The auxiliary equation is m? + 3m = 0 and its roots are m = 0, —3. The general solution is
y = cy + cge"3*,
The auxiliary equation is m? + m — 6 = 0 and its roots are m = —3,
2. The general solution
is y = c,e73* + cpe**.
The auxiliary equation is m? ~ 3m? +4 = 0 and its roots are m = —1, 2, 2. The general
solution is y = c,e7* + (cg + c3x)e7*.
The auxiliary equation is 4m? — 3m +1 = 0 and its roots are m = —1, 4, 4. The general
solution is y = cye7* + (cg + c32) exp ($2).
. The auxiliary equation is m? + 3m? + 3m+1
general solution is y = (c; + coz + cgz*)e7*.
=0
and its roots are m = ~—1,
~—1,
—1.
The
46
CHAPTER
7.
LINEAR EQUATIONS
WITH CONSTANT
COEFFICIENTS
11. The auxiliary equation is 4m? — 7m +3 = 0 and its roots are m = 1, i, 3.
solution is y = cye” + cg exp (4x2) + cg exp (~ $x).
The general
13. The auxiliary equation is 8m° — 4m? ~ 2m +1 = 0 and its roots are m = 3, 3, ~}.
general solution is y = c, exp(—$2) + (ez + c3x) exp (42).
The
15. The auxiliary equation is m4 —2m3 + 5m? — 8m +4 = 0 and its roots are m = 1, 1, £24. The
general solution is y = (cy + cov)e” + cz cos 2z + cy sin 2a
17. The auxiliary equation is m+ + 5m? + 4 = 0 and its roots are m = +i, £2i.
The general
solution is y = c, cos z + cg sina + cq cos 2x + cysin 2a
19. The auxiliary equation is m4 — 11m + 36m? ~ 16m — 64 = 0 and m = ~4, —4, —4, —1. The
general solution is y = cye~* + (c2 + ega + c4z”)e~™.
21. The auxiliary equation is m* + 4m? + 2m? — 8m — 8 = 0 and its roots are m = —2, ~2, £,/9.
The general solution is y = (c; + cgr)e~?* + cgev = + cge™ V2,
23. The auxiliary equation is 4m* + 20m3 + 35m? + 25m +6 = 0 and m= -—1, —2, ~%, ~2. The
general solution is y = cye~* + cge** + cg exp (—}2x) + cq exp (—3z).
25. The auxiliary equation is m* + 5m? + 7m +3
= 0 and its roots are m = ~-1l,
—1,
~3.
The
general solution is y = (cy + corje~* + cge"**.
27. The auxiliary equation is m? ~ m? +m —1 = 0 and its roots are m = 1, +i. The general
solution is y = c,e" + cg cosx + cg sing.
29. The auxiliary equation is m4 — 13m? + 36 = 0 and its roots are m = +3, +2. The general
solution is y = e,e2* + cge73* + age? + ege7?.
31. The auxiliary equation is 4m? + 8m? ~ 11m + 3 = 0 and its roots are m = —3, 4, 4. The
general solution is y = cye~3* + (cg + cgx) exp (4a).
33. The auxiliary equation is m4 — m> — 3m? + m +2 = 0 and its roots are m = 1, 2, ~-i,
The general solution is y = cye™ + +cge** + (c3 + egx)e*.
-1.
35. The auxiliary equation is m> + m* — 6m3 = 0 and its roots are m = 0, 0, 0, ~3, 2. The
general solution is y = cy + cox + cgx* + cge7* + c5e**.
37. The auxiliary equation is 4m> + 12m? + 13m + 10 = 0 and its roots are m = —2, -4 +i. The
general solution is y = cye7?* + exp (~$2)(ca cos x + ¢3 sin 2).
39. The auxiliary equation is m® — 2m* — 2m* — 3m —2 = 0 and its roots are m = —1, ~1, 2, +i.
The general solution is y = (c; + cor)e~* + cge*® + egcosz + essing.
41. The auxiliary equation is m> ~ 15m3 + 10m? + 60m — 72 = 0 and m = 2, 2, 2, —3,
—3.
The
general solution is y = (cy + cov + cgxz”)e* + (cq + cgx)e7*”.
43. The auxiliary equation is m4+3m* —6m? —28m—24 = 0 and its roots arem = —2, —2, —2, 3.
The general solution is y = (c) + cot + e3z")e7** + cge**.
7.6.
A NOTE ON HYPERBOLIC FUNCTIONS
45. The auxiliary equation is 4m> ~ 23m’ — 33m? — 17m — 3 = 0 and m = 3, —1,
The general solution is y = cye?* + (ce, + egx)e~* + (cq + esx) exp (— $2).
AT
~1, ~4, —}.
47. The auxiliary equation is m4 + 6m? + 9m? = 0 and its roots are m = 0, 0, ~3, —3. The
general solution is y = c; + cga + (e3 + eqx)e73*, Hence, y! = co + (~3ce3 + cq ~ Scqxje7*,
and y” = (9c3 ~ 6c4 + 9cazje7**.
But y(0) = cr +3 = 0, y/(0) = co - 3c3 +4 = 0,
y"(0) = 9e3 — 6ceq = 6, and
tim
y’ = eg = 1. It follows that c; = 0, cz = 1, c3 = 0, and
cq = —1. The particular solution is y = z ~ xe79", so that y(1) = 1 - e7?.3
aa
49. The auxiliary equation is m> + m* — 9m3 — 13m? + 8m + 12 = 0, the roots of which are
m=1,
3, -1, ~2, —2. The general solution is y = cye™ + cge** + cge~* + (cg + epxje7**,
51. The auxiliary equation is m> + m4 ~ 7m? — 11m? — 8m — 12 = 0, the roots of which are
m= —2, ~2, 3, +i. The general solution is y = c,e5* + (eg + egz)e7** + ca cosz + cs sin z.
CHAPTER 8.
48
NONHOMOGENEOUS EQUATIONS:
UNDETERMINED
COEFFICIENTS
Chapter 8
Nonhomogeneous Equations:
Undetermined Coefficients
8.1
Construction of a Homogeneous Equation from a Specific
Solution
1. From y = 4e?* + 3e~* an auxiliary equation is (m—2)(m +1) = 0. Hence a
linear differential
equation is (D ~ 2)(D + 1)y = 0.
. From y = ~22 + ge an auxiliary equation is m?(m —~ 4) = 0. Hence a linear differential
equation is D?(D — 4)y = 0.
From y = 2e" cos 3a an auxiliary equation is (m — 1)? +9 = 0.
equation is [(D ~ 1)* + 9]y =0.
Hence a linear differential
. From y = ~-2e%" cosx an auxiliary equation is (m — 3)? + 1 = 0. Hence a linear differential
equation is [(D — 3)? + 1]y = 0.
From y = re7* sin 2z + 3e7* cos 2x an auxiliary equation is [(m + 1)? + 4)? = 0. Hence a
linear differential equation is [(D +1)? + 4] *y = 0.
11. From y = cos kx an auxiliary equation is m? +k? = 0. Hence a linear differential equation is
(D? + k*)y = 0.
13. From y = 4sinhz an auxiliary equation is m? — 1 = 0. Hence a linear differential equation is
(D* —1)y = 0.
15. From y = 3ze** an auxiliary equation is (m— 2)" = 0. Hence m = 2, 2.
17. From y = e~* cos 4z an auxiliary equation is (m +1)? + 4? = 0. Hence m = —14 43.
19. From y = a{e** + 4) an auxiliary equation is m?(m ~ 2)? = 0. Hence m = 0, 0, 2, 2.
2. From y = ze® an auxiliary equation is (m — 1)? = 0. Hence m=
1, 1.
23. From y = 4cos 2z an auxiliary equation is m? +4 = 0. Hence m = +2i.
20. From y = xcos 2x an auxiliary equation is (m? + 4)? = 0. Hence m = +2i, +21.
27. From y = xcos2z ~ 3sin 2z an auxiliary equation is (m? + 4)* = 0. Hence m = +2i, +2i.
8.3.
THE METHOD
29.
OF UNDETERMINED
From y = sin?s = 4(3sinz —sin3z),
COEFFICIEENTS
49
(m* +1)(m? +9) =0. Hence m = ti, £3i.
31. From y = 2? -z+e7*(x+cosz)an auxiliary equation is m3 [(m +1)? + 1](m+1)? = 0. Hence
m=,
0, 0, ~lti,
~1,
<1.
33. From y = x? 2 sinx +2 cosx an auxiliary equation is (m? + 1)3 = 0. Hence m = +i, -bi, +i.
8.3.
The
Method
of Undetermined
Coefficieents
1. From (D* + D)y = ~cosx we have m = 0, ~1 and m’ = +i. Hence y, = c, + cge~* and
Yp = Acosz + Bsinz. Substituting y, into the differential equation and simplifying we have
~A-~B=O,
B
Nie
A
Hi
-A+B=-1,
Bors
(-A+ B)cosz + (—A — B)sinag = —cosz, so that
The general solution is y = cy + cge7* + 4 cos x ~ $ sin @.
From (D*+3D+2)y = 122? we havem = —2, —1 and m’ = 0, 0, 0. Hence y. = cre7™” +c,e7*
and yp, = A+ Bet Cx*. Substituting y, into the differential equation and simplifying we have
(24+ 3B +2C) + (2B + 6C)z + 2Cz? = 122”, so that
9A+3B+2C =0,
2B +6C =0,
aC = 12,
A=21,
= —18,
C=6.
or
The general solution is y = ci e7 2% 4 oge7® +91 — 182 + 62".
From (D?+49)y = 5e™~ 1622 we have m = +31 and m’ = 1, 0, 0. Hence y, = cy cos 3x+cg sin 3x
and yp = Ae*+B+Cz.
Substituting y, into the differential equation and simplifying we have
10Ae* + 9B + 9Cax = 5e* ~ 1622, so that A=
is y = c, cos3z + cgsin3e + 3e" ~ 182.
4,
B=
0, and C = ~-18. The general solution
From y — dy’ — 4y = 30e7 we have m = 4, —1 and m’ = 1. Hence y, = cye4* + cge7* and
Yp = Ae®, Substituting yp into the differential equation and simplifying we have --6Ae* = 30e”,
so that A = —5. The general solution is y = cye* + cge7™® — 5e*.
From (D? — 4)y = e7 + 2 we have m = +2 and m' = 0, 2. Hence y, = cye”™ + cge7?* and
Yp = A+ Bre**. Substituting y, into the differential equation and simplifying we obtain
the equation 4Be* — 44 = e** +2, so that A = —-4, B = 4.
The general solution is
y = ce"* + cge~** — 3 + dze?*,
ii. From y” ~ 4y' + 3y = 20cosz we have m = 3, 1 and m'’ = +t. Hence y, = cye** + cge* and
Yp = Acosz + Bsinz. Substituting y, into the differential equation and simplifying we have
(2A ~ 4B) cosz + (4A +2B)sing = 20cosz, so that
2A ~-4B = 20,
4A+2B =0,
The general solution is y = ce?” + cge” + 2cosx ~ A4sina.
A= 2,
B= —4,
50
CHAPTER 8.
NONHOMOGENEOUS
EQUATIONS:
UNDETERMINED
COEFFICIENTS
13. From y” + 2y'+ y = 7+ 75sin2z we have m = —1, ~1 and m’ = 0, +27. It follows that
Ye = (C1 + gzje™* and yp = A+ Bcos2x + Csin2z. Substituting y, into the differential
equation and simplifying we have A+ (~3B +4C) cos 2x2 + (—4B — 3C) sin 2e = 7 + 75sin2z,
so that
A=7,
—-3B +4C = 0,
—~4B — 3C = 75,
A=7,
Bm 12,
C= —9.
The general solution is y = (c, + cax)e7* + 7 — 12 cos 2x — 9sin 22.
18. From (D? + 1)y = cosa we have m = +i and m’ = +1. Hence y, = c, cosz + cosine and
Yp = Axcost + Brsing. Substituting y, into the differential equation and simplifying we
obtain the equation —-2Asing + 2B cos = cosz, so that A=0,
is y = c, cos¢ + cg sin + dasin a.
B=
4. The general solution
17. (D? — ly = e7*(2sinz +4cosz) yields m= 1, —1 and m! = ~1+i. Hence y, = ce” +c9e77
and y, = Ae~* sin +Be~* cosz. Substituting yp into the differential equation and simplifying
we have (~A + 2B)e~* sing + (-2A — B)e~* cosz = 2e7* sinz + 4e7* cosa, so that
—-A+2B =2,
-2A-B=4,
= ~2,
B=0.
The general solution is y = c,e* + cge~* — 2e7* sin x.
19. From (D3 — D)y = x we have m = 0, +1 and m’ = 0, 0. It follows that y, = c, + c9e” +c3e7*
and yp = Ax + Bz*. Substituting y, into the differential equation and simplifying we obtain
~A~2Bx =z, so that A= 0, B = —}. The general solution is y = cy + cge* + cge7* — 427.
2l. From (D? + D? — 4D — 4)y = 3e7* — 4a — 6 we have m = 2, —2,
~1 and m’ = 0, 0, -1. It
follows that y, = cye** + cge7?* + cge~* and Yp = A+ Br + Cuxe~*.
Substituting y, into the
differential equation and simplifying we have ~-3Ce~* ~ 4Br + (--4A — 4B) = 3e7* — 4z ~ 6,
so that
—3C = 3,
C=
1,
—4B
= —4,
B=1,
-4A—4B = —6,
A=.
The general solution is y = cye?* + cpe~?" + ege7® + 5 +2 —sae7*,
23.
From the equation (D4 — 1)y = e7* we
obtain yo = cye™ + cge~* + egcosx +
differential equation and simplifying we
solution is y = cye™ + cge7* + ¢3 cosa +
26.
From (D* + 1)y = 12cos*z = 6 + 6cos2z we have m = ti and m’ = 0, +2i. It follows
have m = 1, ~1, i, —i and m' = —1. We therefore
cysing and y, = Azre™*. Substituting Yp into the
have ~4Ae7* = e7*, so that A = ~4. The general
cysinz — ize*,
that ye = c,cosz +cgsing and yp = A+ Bcos2r + Csin2z.
Substituting yp, into the
differential equation and simplifying we have A — 3B cos 2x2 — 3C’ sin 3a = 6 + 6cos 2z, so that
A=6, B= -~2, and C = 0. The general solution is y = ¢, cos + cysinz +6 — 2cos 22.
8.3.
THE METHOD
OF UNDETERMINED
COEFFICIEENTS
51
27. From y” ~ 3y' ~ 4y = 16x — 50cos2x we have m = 4, -1 and m’ = 0, 0, +£2i. Hence
Yo = cye** +cge7* and yp = A+ Bzu+Ccos2z+ Dsin 2x. Substituting y, into the differential
equation we get (—4A — 3B) ~4Bz + (--8C — 6D) cos 2x + (6C — 8D) sin 2x = 16x — 50 cos 2z,
so that
-4A—3B
—4B
~8C - 6D
6C — 8D
=0,
= 16,
= —50,
=0,
A=3,
B= ~—4,
C= 4,
D=3.
The general solution is y = ce** + cge7? +3 — 4a + 4cos 2x + 3sin 2c.
29. From the equation y” + 4y' + 3y = 15e*” + e~* we have m = —3, —1 and m’ = 2, ~—1. Hence
Ye = C1e7? + cpe~3" and y, = Ae** + Bare~*. Substituting yp into the differential equation
and simplifying we have 15Ae”* +2Bae~* = i5e”* + e~*, so that A= 1, B = 4. The general
solution is y = cye7* + cge7** + e?* + gze~*.
3l. From y" —y' ~2y = 62+6e7* we have m = 2, —land m! = 0, 0, —1. Hence y, = cye** +ene7*
and yp = A+ Br +Cxe~*. Substituting yp into the differential equation and simplifying we
have (~2.A — B) — 2Ba ~ 3Ce7* = Gz + Ge*, so that
—~2A-~B=0,
A= 3,
~2B = 6,
B= -3,
C= ~2.
—3C = 6,
The general solution is y = cye?® + cge7® + 3 ~ 30 — 2ne7*,
33. From (D* ~ 3D? + 4)y = 6+ 80cos2z we have m = —1,
2, 2 and m’ = 0, +2i.
Hence
Yo = €1e7* + (cg +c32)e*” and Yp = A+Bcos2z+C sin 2x. Substituting yp, into the differential
equation and simplifying we have 4A + (16B — 8C’) cos 2x + (8B + 16C) sin 22 = 6 + 80 cos 22,
so that
4A = 6,
16B — 8C = 80,
8B +16C = 0,
A= 3,
B =A,
C= ~2.
The general solution is y = cye™* + (eg + egx)e** + 3 + 4¢os 2x — 2sin 27.
35. From (D* + D* — 4D — 4)y = 84 +8 + be~* we have m = +2,
Yo = cye** + cge7** + cge7* and y, = A+ Bx +Cxre-*.
—1 and m’ = 0, 0, —1. Hence
Substituting yp into the differential
equation and simplifying we have (-4A — 48) - 4Br — 3Ce™* = 82 + 8 + Ge", so that
—~4A —4B = &,
~4B = 8,
—3C = 6,
A= 0,
B= -2,
C = —2,
The general solution is y = cye”” + coe~** + cge7* — 2a — Qre™*.
52
CHAPTER 8.
NONHOMOGENEOUS
EQUATIONS:
UNDETERMINED
COEFFICIENTS
37. From (D? — 4)y = 2— 8x we have m = 2, ~2 and m’ = 0, 0. Hence y, = cye*® + cge7?*
and y, = A+ Ba. Substituting this y, into the differential equation and simplifying we have
~4A-4Bzr = 2-82, so that A= ~}, B = 2. The general solution is y = cye””-+c2e77* 4422.
But
y(0) =a te.— 4 =0,
cq = 1,
y'(0) = 2c, ~ 2eg +2 = 5,
C2 =~.
The particular solution is y = e27 — deW%
_ $ + 2n..
39. From the differential equation (D? + 4D + 5)y = 10e73* we have m = —2-4i and m’ = —3.
Hence yo = e”?*(c; cosa + c2sinz) and yp = Ae~**. Substituting this y, into the differential
equation and simplifying we have 2Ae~%* = 10e~3", so that A = 5. The general solution is
y= e7**(e, cosz +cgsinz) + 5e74*, But
y(0) =e,
+5
= 4,
cq =~,
y'(0) = -2c, +p -15 = 0,
C2 = 13.
The desired solution is y = e~°7(— cosx + 13sinz) + 5e7*,
41, From ¢+4¢+52 = 8sint we have m = ~2+1 and m’ = +i. Hence x, = e~*#(c, cost
+ cg sint)
and z, = Acost + Bsint. Substituting z, into the differential equation and simplifying we
have (4A + 4B) cost + (—4A + 48) sint = 8sint, so that
4A +4B =0,
—4A4+4B =8,
= 1,
B=,
The general solution is x = e~**(c, cost + cg sint) ~ cost + sint. But
(0) =e, -1 = 0,
a(0) w ~Je,
+g
cy
= 1,
+1=0,
ecg= 1.
The desired solution is z = e~**(cost + sint) ~ cost + sint.
43,
From (D°? + 4D? +9D+10)y
= —24e7 we have m = ~2, ~1+2i and m’ = 1. Hence
Yo = cre** + e7 7 (cy cos 2x + cg sin 2z) and y, = Ae™.
equation and simplifying we have 24Ae”
Substituting yp into the differential
= ~24e7, so that A = ~—1.
The general solution is
y = cye7** + e7* (eg cos 2x + eg sin 2x) — e*. But
y(0) =e, +eg-1=0,
y'(0)
sz ~2¢y
— Co + 2cg3 —1
e = 2,
=
—4,
y"(0) = 4e, — 3e2 — deg — 1 = 10,
The particular solution is y = 2e77* ~ e~* cos 2x — e*.
tf =-—l,
cz = 0.
8.3.
45,
THE METHOD
OF UNDETERMINED
COEFFICIEENTS
53
From y” + 2y’ + y = © we have m = —1, -—l and m’ = 0, 0. Hence y. = (cy + coxje7*
and yp = A+ Bz. Substituting y, into the differential equation and simplifying we have
A+2B+4+Bz = 2,80 that A= —2, B = 1. The general solution is y = (¢: +eagz)e7* +2—2.But
y(0) =e; —2= -3,
cq =~l,
y(1) = (ce, + ee"? - 1 = -1,
c= 1.
The particular solution is y = (—1+xr)e7* + « — 2. It follows that y(2) = e~* and y'(2) = 1.
47. From
4y” + y = 2 we have m = +4i and m' = 0, Hence y, = cj cos$x + cgsin dz and
Yp = A. Substituting y, into the differential equation and simplifying we get A = 2. The
general solution is y = c, cos gx + €2 sin ax + 2.But
y(t) = cg +2 = 0,
cc; = —2,
tg = -2.
y'(m) = -3q =],
The particular solution is y = —2 cos 5—2 sin pe+2.
It follows that y(2) = ~2cos1—2sin1+2
and y'(2) = sin1 — cos 1.
49. From (D? + Djy = 2 +1 we have m = 0, —1 and m’ = 0, 0. Hence yo = cy + cge7?
and y, = Av + Bzr*. Substituting yp, into the differential equation and simplifying we get
(A+2B)+2Brc=2+1,sothatA=0,
But
y(0)0 == cy
+ c2
B= 4.
The general solution is y = c, + cge7* + 50",
ot
“e
i as
= 1,
=,
-
1
y(1) = ci tee"! +5 = 4,
The particular solution is y = i
~enl
—~e7l
a= Tet
1
+
L-~en}
1
e* + 5e
.
:
and y(4) = 8 - e7! ~ e7* — e739.
51. From (D? + 1)y = 2cos2 we have m = +i and m' = +i. Hence y, = c,cosx + cysinx and
Yp = Axcosz+ Basina. Substituting y, into the differential equation and simplifying we get
= 2cosz, so that A=
0, B = 1. The general solution of the differential
equation is y = cycosx +cosing + xsing.
~2Asing + 2Bcosz
But y(0) = c, = 0 and y(7) = —cy = 0, so that
¢; = 0. There are an unlimited number of solutions, y = cgsinz + asin 2.
53. From (D* ~ D)jy = 2 -— 2x we have m = 0, 1 and m’ = 0, 0. Hence y, = c, + ene”
and yp = Ar + Bx. Substituting yp into the differential equation and simplifying we get
~A+2B —2Bz = 2-22, so that A = 0, B = 1. The general solution of the differential
equation is y = c; + cge? + 27. We want y'(a) = cge° + 2a = 0 and y"(a) = cge* + 2 = 0, so
that a = 1. But
yQ) =e, + @e+1=0,
cy =1,
y'(1) = ce+2 = 0,
cg = ~2e7?.
The desired solution is y = 1 ~ 2e-! + z°.
54
CHAPTER 8
8.4
NONHOMOGENEOUS
EQUATIONS:
UNDETERMINED
COEFFICIENTS
Solution by Inspection
1. Ifo #4 a then
(D? + a?) [(a? — b?)~! sin ba] = —b?(a? — b?)~! sin ba + a(a? — 6*)~! sin bx
= sin ba.
Hence (D* + a”)y = sin bx has the particular solution y = (a? — 6?)~! sin bz if a # b.
ont
. Ify=3 then (D? + 4)y =4-3 = 12. Hence y = 3 is a particular solution.
. Ify = 2 then (D?+4D+4)y =4-2 = 8. Hence y = 2 is a particular solution.
. Ify = —f then (D® - 3D + 2)y = 2. (—2) = —7. Hence y = —2 is a particular solution.
. If y = 3x then (D? + 4D)y = 4-3 = 12. Hence y = 3z is a particular solution.
11. If y = 3x then (D? + 5D)y = 5-3 = 15. Hence y = 3z is a particular solution.
13. If y = —3x” then (D4 — 4D?)y = (—6) - (—4) = 24. Hence y = ~—3z? is a particular solution.
15. If y = —42° then (D® ~ D)y = 4! = 24. Hence y = —4z° is a particular solution.
17. If y = 2sinz then (D? + 4)y = —2sinz + 8sinz = 6sinz. Hence y = 2sinz is a particular
solution.
19. Ify
= 20+ 4 — 3e* then (D? + 4)y = —3e% + 84 +1 — 12e7
yo e+ ; — 3e* is a particular solution.
21.
If y = 3e7* then (D?+3D—4)y = 12e?*
+ 18e?*
= —15e" + 84 + 1. Hence
—12e** = 18e?*. Hence y = 3e?* is a particular
solution.
23, If y = je** then (D? — 1)y = $e** — 4e%* = 2e3*. Hence y = 4e** is a particular solution.
25, Ify = —}cos2z then (D? ~ 1)y = 3cos2x + }cos2r = cos2z. Hence y = —} cos
2z is a
particular solution.
27, Ify = $e” + 3a then (D? +1)y
= Je" + fe" + 3a = e* +3. Hence y = de" + 3z is a particular
solution.
29. If y = —} cos 2x — 2a then (D? + 1)y = 4cos 2x — § cos 2x — 2x = cos 2x ~ 2x. It follows that
y=
-} cos 2x ~- 2x is a particular solution.
31. Ify = —#sin4z then (D? + l)y = #2 sin dx — 2sin4r = 10sin4z. Hence y = ~—2 sin 4g is a
particular solution.
33. If y = 3e” then (D? + 2D +1)y = 3e + Ge? + Se” = 12e". Hence y = 3e” is a particular
solution.
35. Ify = 3e~* then (D? -2D+1)y = 3e7* +6e7* 4+3e7* = 12e7*. Hence y = 3e7* is a particular
solution.
8.4.
SOLUTION BY INSPECTION
55
37. Ify= —te* then (D? — 2D — 3)y = —}e” + 4e* + 2c” = e*. Hence y = —4e? is a particular
solution.
39. If y = —4sin x then (4D? +1)y = 16sinz-~-4sina = 12sinz. Hence y = ~4sinz is a particular
solution.
Al. Ify = 2e7 — 5 then (4D* + +4D + 1)y = 8e" + 8e + 2e7 —5 = 18e” — 5. Hence y = 2e* — 5 is
a particular solution.
43.
If y = —}e~* then (D3 ~1)y = ge7* + de7* = e~*, Hence y = —je™* is a particular solution.
45, If y = ge** then (D? ~ D)y u = 2e”* — Ze?” = e?*. Hence y = Ze”* is a particular solution.
47,
If y = #sin2c then (D*+4)y = #sin2c + Hsin
= 6sin2r. Hence y = 3 sin2x is a
particular solution.
49, Ify = $ cos 2x then (D* — D)y = 4sin2z+sin2z = 5sin 2x. Hence y = $ cos 2z is a particular
solution.
56
CHAPTER
Chapter
VARIATION OF PARAMETERS
9
Variation
9.2
9.
Reduction
of Parameters
of Order
1. One solution of the homogeneous
simplify to obtain
equation
is e*.
We
make
the substitution y =
ve™
and
vu" + Qu! = (x — le™*.
Multiplying by the integrating factor e** and integrating we get
e*y! = re™ — Qe* +,
Dividing by e?* and integrating again yields
vs
—ze7* + e7* + ce"?
4 ¢g,
y == ve™ = cge”*
+ cg3e7 ~2 +1.
3. One solution of the homogeneous equation is e?*. We make the substitution y = ve and
simplify to obtain v” = e7*. Integrating twice gives v = e~* + c; + cax. It follows that
y = ve" = (c, + cgn)e™* + e?.
5. One solution of the homogeneous equation is sina. We make the substitution y = vsinx and
simplify to obtain
v” + 2cot aru’ = secxcsex.
Multiplying by the integrating factor sin? z and integrating we get
sin? v’ = —In|cosa{ + cy.
Dividing by sin? « and integrating again yields
v = cotxzin|cosa| + x + c2 cot z + ¢g,
y = using = ¢3 sing + ¢2 cosxz + sing + coszln|cosz}.
7. One solution of the homogeneous equation is e~*. We make the substitution y = ve7* and
simplify to obtain v” = e*(e7 ~ 1)~*. Integrating twice gives v = c,z + cg ~ In|1 ~e7*|. It
follows that y = ve7* = e7* leyx +¢2~-Injl- e~*{].
9.2.
REDUCTION OF ORDER
57
9. One solution of the homogeneous equation is cosxz. We make the substitution y = ucosz and
simplify to obtain
vy
uv
,
— 2tanzu
~
= seczrese 7.
Multiplying by the integrating factor cos* x and integrating we get
cos*x v' = In|sinz| + cz.
Dividing by cos* x and integrating again yields
vs
—x+tangin|sina|+cotang +e),
y = ve” = c) cosx
+ cgsinz — xcoszr + sinzln{sina|.
11. One solution of the homogeneous equation is cosa. We make the substitution y = vcosa and
simplify to obtain
vo” — Qtanav’ = secr ese’ x.
Multiplying by the integrating factor cos” xz and integrating we get
1
cos? eu! = “3 cot?
x + ¢}.
Dividing by cos*x and integrating again yields
1
uo
9 Ste
+ C1 tanz
+ cs,
y=
voor
= 5 COS E cote + €, SIND + cg cos 2,
1
.
i
= Cy sing + ¢2 cos a7 + 3 ese x.
13. One solution of the homogeneous equation is 2. We make the substitution y = ux and simplify
to obtain
it
vo+
5
on"
—
i
=
0.
Multiplying by the integrating factor x°/? and integrating we get
ghltyl
= Cg.
Dividing by z°/* and integrating again yields
v= cge8! + ey,
Y=
url= ce + con
}/?,
58
CHAPTER
9.
VARIATION OF PARAMETERS
15. One solution of the homogeneous equation is x. We make the substitution y = va and simplify
to obtain
z(1 —x*)u" + (2 — 42*)y' = 0.
This may be written
Upon integrating we obtain
ay =x
21
e(i~-x?)”
=e
2
“ia?
+ i
"jon
--
1
T+al-
Integrating again yields
ees]
+ Ca,
y= VE = Cy [-2+2In
1
it]
Y= Cy fo
x
9.4
tin
l+z
l-z
+ Con.
Solution of y’ +y = f(z)
1. We consider
y = Ae* + Be~*,
y = Ae” — Be™* + A’e* + Blew®.
Setting A’e™ + B’e* = 0 gives
y = Ae* + Be~* + A’e® — Ble™*,
y -y= Ae —- Ble =e* +1.
We must solve the system
Ale* + Ble"? = 0,
Ale® ~ Ble * =e +1,
for A’ and B’ and then integrate to find A and B. We get
Ala de
fo. ~ge422
B=
ie
ge",
= fondo,
wx:
i p2t
_ lp®
B=—je
ge.
Thus
{i
L.-@)
Yp =— (Aa
— fe7*)
e* 2 + (—fe**1
2a
1.2\
— de”)
e
»72
9.4,
59
SOLUTION OF Y" + Y = F(X)
8. We consider
y = Acosz+ Bsinz,
y’ = ~Asinz + Bcosz + A’ cosa + B’ sing.
Setting A’ cosz + B’ sinz = 0 gives
y" = ~—Acosz — Bsing — A’ sing + B’ cosz,
y +y = —A’ sing + B’ cosx = esez.
We must solve the system
A’cosz + B’ sina = 0,
—A'sins + B’ cosx = ese,
for A’ and B’ and then integrate to find A and B. We get
Al=
~I,
A=
B' = cota,
Es
B = In{sing|.
‘Thus
Yp = ~zZcos£ +sinzlnjsinz|,
¥ = Ye — Ecosz +sinzln|sinz|.
5. We consider
y = Acosx+ Bsing,
y’ = ~Asinag + Boosa + A’cosx + B’ sina.
Setting A’ cosz + Blsina = 0 gives
y” = ~Acosz — Basing — A’sing + B' cosz,
y +y = —A'sing + B’ cosa = sec* x.
We must solve the system
A’cosz+ Bi sina = 0,
—A’ sina + B’ cosa = sec® zx,
for A’ and 8’ and then integrate to find A and B. We get
Al = ~sinxsec? z,
B! = sec? 2,
= —4 tan? 2,
G = tanz.
Thus
:
Yp == (—} (1 tan* tan?-x) cosx + tanzsing
mdig
= $secr —~d } cosa,
Y= Ye + § sec z.
60
CHAPTER 9.
VARIATION OF PARAMETERS
7. We consider
y = Acosx + Bsinag,
y = ~Asing + Bcosz + A’cosx + B'sinz.
Setting A’ cosz + B’sinxg = 0 gives
y” = —Acosz ~ Bsing — A’ sina + B’ cosz,
y +y=—A’sing + B’ cosa = tang.
We must solve the system
A’cosz + B’sinz = 0,
—A’ sing +B’ cosz = tang,
for A’ and B’ and then integrate to find A and B. We get
A’ = ~sinztanz,
A=sinz —In|seca + tan},
A’ = sing,
B= —cosaz.
Thus
Yp = (sinx — In| seca + tan z}) cosz — coszsinz
= —coszIn|seca + tanz|,
Y= Ye ~ cosxin|seca
+ tang].
. We consider
y= Acosx + Bsinz,
y = —Asine + Boosr + A’cosz + B’ sina.
Setting A’ cosa + B’sinx = 0 gives
y’ = —Acosz ~ Bsing — A’sinz + B’ cosa,
y” +y= —A’sine + B’ cosx = secrcsez.
We must solve the system
A’cosx + B’ sing = 0,
~A'sing + B' cosa = sec xcsc x,
for A’ and B’ and then integrate to find A and B. We get
Al = —secz,
B! =csen,
A=-—In|secx
+ tang],
B= -—In{escx
+ cota}.
Thus
Yp = ~cosz
in| seca + tanz| — sing In| csez + cot 2,
¥ = Yc — coszln|
seca + tanaz| — sinzln|cscx+ cota].
9.4,
SOLUTION OF Y"+Y = F(X)
61
ll. We consider
y = (A+ Ba)e*,
y’ = Ae* + Ble® + ze") + Ale™ + B're®.
Setting A’e* + B’re* = 0 gives
y" = Ae® + B(2e" + ze") + A’e® + Bi (e* + xe*),
y" ~ 2y +1 = Ble® =e (e? +1)7?.
We must solve the system
A’e® + B’xe™ =0,
Ble® = e**(e" + 1)7?,
for A’ and B’ and then integrate to find A and B. We get
A’ = ~xe*(e? +1)7?,
A=a(e"+1)71+In(1 +e7*),
Bl = e*(e7 +1)°?,
B= —(e*+1)7},
Thus
Yp = [x(e* + 1)7! + In(1 + e7*)] e® + [-(e* + 1)7*] ze*
=e" In(1 +e7*) = e* [In(1 +e”) —2],
y = ye + e7 In(1 + e”),
13.
We consider
y = Ae™ + Be**,
y’ = Ae” + 2Be** + A'e® + Ble®*,
Setting A’e™ + B’e?* = 0 gives
y=
Ae” + 4Be** 4+. A’e® + 2B’ e??,
y” — 3y' + 2y = A’e* + 2B’e** = cos(e~*).
We must solve the system
A’e® + B’e®® = 0,
A'e® + 2B’e** = cos (e~*),
for A’ and B’ and then integrate to find A and B. We get
A’ = —e7* cos (e7*),
A = sin (e7*),
B! = e~** cos(e~*),
B=
—e~* sin (e~*) — cos (e~*).
62
CHAPTER 9.
VARIATION OF PARAMETERS
Thus
Yp = e* sin (e~*) + [—e7* sin (e~*) — cos (e~*)] e*
= —e** cos (e~*),
Y = Yo — e** cos(e7*).
15. We consider
y = Ae* + Be,
y = Ae® ~ Be7* + A’e® + Ble™*.
Setting A’e* + B’e~* = 0 gives
y” = Ae* + Be~* + A’e® ~ Ble*,
y” —y=
Ale®
— Blev®
= et
sin (e~*),
We must solve the system
Ale? + Ble“? = 0,
A’e® — Ble™* = e~** sin (e~*),
for A’ and B’ and then integrate to find A and B. We get
A! = he~** sin (e7*),
A=
B! = —4e™* sin (e~*),
B =~} cos(e*).
ze
cos (e~*) ~ e~* sin (e~*) ~ cos(e~*),
Thus
Yp» = — sin (e~*) — e* cos (e7*)
Y = Ye ~ sin (e~7) — e” cos (e~”).
17. We consider
y= A+ Be® + Ce,
y = Be* — Ce™* + A’ + Ble® + Cle",
y"
=
Be*
+
Ce
he
Be
ae
Cle™*,
y" = Be® —Ce~* + B’e® + C'e*
In the first two differentiations we have let the terms involving A’, B’, and C’ equal zero. If
we now substitute back into the differential equation we have
yl"
an y!
=
B'e®
++
Ole
*
= 2,
A’ + Ble® + Cle* =0,
Ble® ~ Cle* = 0.
9.4, SOLUTION OF Y" +¥ = F(X)
63
Solving this system and integrating we obtain
A’
i
=
_
—Z,
ip
=~
Rr,
B! = kxre™*,
B= ~4xe7* — de®,
C' = ize®,
C =: —dre* — Let.
Thus
Up
wm
Lb mpe
gr"
+
appt
(—gae
_
Lat)
te~*)
e*
pe
Lypt
_ Lp®\
+ (4ae*
~ de*)e
p=
= ~4z? ~4,
—_
1,2
Y= Yo gx
1
19. We consider
y= Ac + Be*,
y = A+ Be* + A'x + Be*.
Setting A'z + B’e* = 0 gives
yl"
=
Be®
+
Al
+
B’e*,
(l—a)y"” +a2y-y=A'(1-2)4+ B(1~ax)e” = 2(2 — 1)*e7*.
We must solve the system
A's + B'e® =0,
A’ + Ble® = 21 — zje™*,
for A’ and B’ and then integrate to find A and B. We get
A=
2e7*,
A=
Bl = —2re™**,
~—2e7*,
B= xe" + Le~2,
Thus
Up = (~2e7*) a + (ze~** + 4e7**) e?
=-e-*(0~ 3),
Y= Yo— e7*(a — $).
21. We consider
y= Acosz + Bsing,
yi = ~Asinz + Boosa
+ A’cosx + B’ sinz.
64
CHAPTER 9.
VARIATION OF PARAMETERS
Setting A’ cosz + B’ sina = k gives
y” = ~Acosz — Bsinz — A'sina + B’ cosa,
y +y = —A’sinz + B’ cosxz = sec® x.
We must solve the system
A’ cosa + B’ sina = k,
—A'sina + Bi cosa = sec? x,
for A’ and B’ and then integrate to find A and B. We get
A’ = kcosa — sinasec® g,
A=ksing — } tan? z,
B! =sec*2 +ksinz,
B= tanz
~ keosex.
Note that when we form yp = Acosz + Bsinz the terms involving k will cancel leaving us
with the same solution as we had in Exercise 5.
23. If we let W
= yoy, ~ yiyg then the differential equation can be written W’ + pW
Multiplying by the integrating factor exp
exp
/ P az| gives us
P az| Wiz) =e,
W(2) = cexp - [? az],
25. We choose a = Zq in the solution given in equation (10). Thus
=
y=ecosz+onsine + [
f(8)
sin (x — 8) dg,
zo
y’ = ~c, sin re + ca cosz,
y(xo) = cy 008 Zo + cg sin zo = Yo,
y'(q) = —c1 sin Zp + €2 cos Zp = yp.
We must solve the last two equations for c; and cz. We obtain
C1 = Yo cOSXo — YQ sin Zo,
C2 = Yo COSXq + yo sin Zo.
Finally we have
y = [yo co8 29 — yo sin Zo] cos x + [yg cos rp + yo Sin 9] sinz + /
at
to
= yo cos (x — Zo) + usin (x ~ 20) + f f(8)sin
(x — B) df.
f(8) sin (2 — 8) dB
= 0.
9.4.
SOLUTION OF Y" +Y = F(X)
Miscellaneous
1. We
65
Exercises
consider
y= Ae” + Be,
yl = Ae® — Be™* + Ale® + Ble™.
Setting A’e* + B'e™* = 0 gives
y = Ae* + Be™* + A’e® — B'e™*,
y! —y = A’e® — Ble™® = 2e-*(1 +7 ™*)-?,
We must solve the system
Ale® + Ble™* = 0,
A’e® — Ble® = Qe"(1 +e ™*)-?,
for A’ and B’ and then integrate to find A and B. We get
A
t
=e
goat
(l+e
1/2
—2ny~2
)
;
—_
= Ty ends?
~ Bar
Bo=—(1L+
~2z
(+e)
“2
Ly
ants€ sannsetngn
+ ayer?
~ 2x
€
——_—,
B=~a-din(it+e7%)4
pert
Thus
Yo =
1/2
[es]
+
[-2-
gina
te
.
)
4
1/2ele 7. =
= —ze~* — fe7* In(1 +73") + de®,
Y * Yo — Te™* — ge" In(1 + e777),
3. We consider
y= Acose+ Bsinz,
y' 7 = —Asing + Bcosx + A’ cosa + B' sina.
Setting A’ cos¢ + B’ sina = 0 gives
"= —~Acosgx ~ Bsing ~ A’ sing + B’ cosz,
y +y = —A’sinaz + B’ cosx = sec’ xtanz.
We must solve the system
A’ cosa + B’ sing = 0,
~A'sina + B' cosx = sec* x tana,
1/2
L+e73"
66
CHAPTER
9.
VARIATION OF PARAMETERS
for A’ and B’ and then integrate to find A and B. We get
A! = —secxtan’ z,
B' = secetane,
A=~}secrtanx + $In|secz +tanal,
B= secs.
Thus
Hl
Hl
Yp = [—-4secxtanz + 4 In| seca + tan2z|] cosa +seczsinz
3 tanz + 3 coszln|secz
+ tanz|,
Yo + stanaz+ dcosxin|secx + tangzl.
5. We consider
y = Acosx + Bsina,
y = —Asinz + Bcosx+ A'cosz
+ B’ sina.
Setting A’cosx + B’ sina = 0 gives
y = —Acosz — Bsinag — A’ sina + B’ cosa,
y’ +y = —A’sing + B’ cosz = cota.
We must solve the system
A’ cosxz + B' sinx = 0,
~A' sina + B’ cosx = cot,
for A‘ and B’ and then integrate to find A and B. We get
A’ = — cosa,
A=~sing,
B! = cosxcotz,
B= cosz —In|cscxr
+ cota}.
Thus
Yp = ~ sina cosa + [cosa — In|csex + cota} sinz
= —singln|cscz + cota,
¥ = Yo — sina
In| ese x + cota,
7. We use equation (10) of Section 9.4 with a = 0 and f(z) = seca.
=
Y= Yo + [
sec Gsin (x — 8) dB
0
a
=uet [ (sinz
— cosztan @) dG
0
=
= Ye + [asin
+ cas
In| cos
= Ye + xsinaz + cosz in| cos zy}.
0
9.4. SOLUTION OF Y" + Y = F(X)
67
9. We consider
y = Ae® + Be,
yi = Ae® — Be~* + A’e® + Ble.
Setting A’e* + B’e-* = 0 gives
y" = Ae” + Be™* + A’e* — Ble,
y
n
Alot
~ Bre
= Ale”
—y
or
2
ae
"= eee
We must solve the system
A’e® + Ble™* = 0,
A’e*
~
Ble
a
2
et — e- =’
for A’ and B' and then integrate to find A and B. We get
Al =
~
a
et
ert
—é
~
x
B —= orl our
a
A=nm iinjl—e
7**|
2
[—en-2z?
—€
22
GEL]
=
Ad
22
B=ww» —3lnje*
— I.
Thus
Yp = [Ein[1 —e7?*|] e* + [—2 In Je* — 1] e~*
= ~xe7* + $(e* — e7*) In{l — et,
y= Yo ~ ze7™* + E(e® — e7*) In|1 — 7 7],
li. We consider
y= Ae? + Be™*,
y! = Ae® — Be™* + A’e® + Ble™*.
Setting A’e? + B'e~* = 0 gives
y = Ae® + Be~* + A’e® — Ble™*,
1
et _
y" a ~y =— Ale*
~ Ble *=
aT
We must solve the system
A’e? + Ble? = 0,
Ale?
— Ble?
=
°
°
1
er
+ 1"
;
,
68
CHAPTER 9.
VARIATION OF PARAMETERS
for A’ and BY’ and then integrate to find A and B. We get
ze
A =“ey
7
1
Te
367°
y = 2°ge7*
p@
“t+
— 4.
Tipent’
A==
—ip-t_1
—5e
5
lo-s
1
ge
L4es’
arctan
2
2
B = —5 arctan (e~*).
Thus
Yp = [~ge"* — 3 arctan (e~*)] e® + [—d arctan (e"*)] ev?
e* + e-*
_
a
arctane™*,
= -$
Y= Ye — 4 — cosh rarctane™*.
13. We consider
y = Ae~* + Be*,
y’ = ~Ae~* — 3Be™** + Alem? 4 Blew,
Setting A’e~* + Ble~3* = 0 gives
y” = Ae~* + 9Be™** — A’e~? — 3B’e7 5,
y” + 4y' + 3y = —A’e™* — 3B’e"* = sin (e*).
We must solve the system
Ale~* + Ble“ = 0,
~A’e~* — 3B’e5* = sin (e*),
for A’ and B’ and then integrate to find A and B. We get
A’ = $e” sin (e”),
A= ~} 0s (e*),
B’ = —1e** sin (e*),
B = he** cos (e") — e* sin (e*) — cos (e*).
Thus
Yp = ~~
sin (e*) — e~5* cos (e*),
Y¥ = Yo — €”** sin (e”) — e~ > cos (e*).
15. We consider
y= Ae™ + Be*,
y = Ae* ~ Be™* + A’e® + B’e™?,
(e~*),
9.4.
SOLUTION OF Y" +Y = F(X)
69
Setting A’e* + B’e~* = 0 gives
4 = Ae® + Be™* + A’e® — Ble,
yl! —~y = A’e® ~ Ble?
= et [3 tan (e*) + e* sec” (e*)].
We must solve the system
A’e® + Ble* =0,
A’e® ~ Ble~* = e** [3 tan (e*) + €* sec? (e*)],
for A’ and B’ and then integrate to find A and B. We get
A’ = $e" [3 tan (e”) + e* sec” (e*)],
A= ~In|cos(e*)| + ge” tan (e*),
B! = —}e**[3 tan (e*) + e* sec” (e*)],
B = —he** tan (e*).
Thus
Yp = ~e” In| cos (e*)| = e” In| sec (e”)},
y = yc +e” In| sec (e*)].
17. We consider
y= Acosz+ Bsing,
y = —Asinz+ Boosz + A’ cosz + B’sinz.
Setting A’ cosa + B’sinz = 0 gives
y = —Acosz ~ Bsinaz — A’sinz + B’ cosz,
y +y = ~A’ sing + B’ cosa = secz tan’ z.
We must solve the system
A’ cosa + B' sing = 0,
~A'sing + B’ cosa = secx tan® z,
for A’ and B’ and then integrate to find A and B. We get
A’ = —tan’x,
A= -—In|cosa|— tan? 2,
B’ = tan’ z,
B-stang-.
Thus
Y¥p = [— In| cos 2| — tan?x] cosa+ [tanz ~ 2|sinz
= $sinztanz — xsinx —coszln|coszl,
Y=Yet zsinztanz ~ zsing ~ coszin| cosa].
70
CHAPTER
Chapter
10.
APPLICATIONS
10
Applications
10.3.
Resonance
1. The spring is stretched 4 ft by the 5-lb weight so 5 = ak or k = 10 lb/ft. The initial value
problem is therefore
Sax"(t) + 10x(t) = 0; x(0) == $, 2/(0) = -6.
We therefore have
x(t) = cy sin 8t + c2 cos 8t,
2(0) =e, = 4,
x’ (t) = 8c; cos 8t — 8c sin 8t,
z'(0) = 8c, = —6,
co = 4,
= —¥
The position of the weight is given by
a(t) = } cos8t — 3 sin 8.
3. The initial value problem is
He"(t) + 162(t) = 0; 2(0) = 4, 2'(0) = 8.
We therefore have
z(t) = c, sin 16t + cg cos 16t,
x(0) = co = 3,
C2 = 4G,
z'(t) = 16c; cos 16¢ ~ 16c2 sin 16¢,
x'(0) = 16c; = 8,
c=.
The position of the weight is given by
a(t} = 3 sin 16t + } cos 16t.
5. The initial value problem is
gx"(t) + 82(t) = 3 cos8t; 2(0) = 0, 2'(0) = 4,
The differential equation
x(t) + 64an(t) = 4 cos 8t
10.3.
RESONANCE
71
a(t) = c; sin 8t + cg cos 8E + ftsin 8E,
a(t) = 8c; cos 8t — 8cq sin 8t + 2tcos8t+ ¢sin8t,
z(0) = c2 = 0,
2’(0) = 8c, = ~4,
3o a9S
fi
it
i =
toh
has the general solution
The position of the weight is given by
a(t) = Ftsin 8t — } sin 8.
. From Exercise 6 we have
a(t) =
=
=
=
4 cos8t — }sin8¢ + 3 sin4t
172(4V2
cos 8t — 4V2sin 8) + 4 sin
4
4\/2(cos8t cos 4a — sin 8tsin }r) + 4 sin4t
4V2cos (8t + 4m) + 4 sin
4t.
. From Exercise 8 we have x'(t} = —2sin 8t + 2ésin 8 = ~2(1 — t)sin8¢.
whenever 2’(t) = 0.
That will happen when t = an,
weight at each of these times is given by x = ~0.15,
dn, 1,
an (sec).
A stop will occur
The position of the
+0.05, +0.03, +0.04 (ft), respectively.
11. We must solve the initial value problem
ahx'(t) + 1282(t) = 360cos4t; x(0) = 5, 2'(0) = 4.
The differential equation becomes x(t) + 2562(t) = 720cos4¢ and its general solution can be
found to be
x(0) =e +3 = Z,
cy = ~§,
z’(t) = —16c; sin 16¢ + 16c2 cos 16f — 12sin 4t,
z'(0) = 16c2q = 4,
2 =
when
x(t) = c, cos 16¢ + cg sin 16t + 3cos4t,
The motion is described by
a(t} = -§ cos 16¢ + 4 sin 16¢ + 3.cos 4¢.
Finally, 2($7) = —3, and 2'(}4) = —8.
13. We need to solve the initial value problem
x(t) + 128x(t) = 0; 2(0) = 4, 2’(0) = -8.
The general solution and its derivative are given by
a(t) = c; cos 8V2t + cz sin 8V2t,
2(0) = c = 4,
ey = 9,
a! (t) = ~8V2c; sin 8V2t + 8V 2c, cos8V2t,
—-x'(0) = 8V2eg = -8,
cg = — 4 V2.
Hence the motion can be described as
x(t) = 5 cos 8x/2t — 4V2sin gv 2t.
72
CHAPTER
10.
APPLICATIONS
15. The initial value problem is
a0 (t) + 242(t) = 0; 2(0) = ~44, x'(0) = 8.
We therefore have
x(t) = cy sin y/ 422¢ + cg cos
1824,
2(0) =
a!(t) = / Bey
cos y/ M4t ~ 4/ Beco
sin / 22,
= —%,
a'(0)
= fa =8,
G=-—L
1 =8 ios.
The position of the weight is given by
a(t) = —§ cos 1/28t + 84/585 sin
1/4122.
The amplitude of this motion is ,/ 2 + 64-3; = 4/109 ft or 2/109 in. Because the weight
began to fall 14 in. above the equilibrium and ends its fall 2/109 in. below equilibrium, the
answer to the question is 14 + 2/109 in. or approximately 35 in.
17. We need to solve the initial value problem
Hx"(t) + 16z(t) = 0; (0) = ~2, 2’(0) = -15.
The general solution and its derivative are given by
z(t) = cy cos 8V/2t + co sin 8V2t,
z(0})=c¢, = —i,
cy = —4,
2'(t) = ~8V2c; sin8V2t + 8V/2cqcos8V2t,
—'(0) = 8V2ey = -15,
cp = V2.
Tt follows that
a(t) =
4V2sin 8/2t ~ 15 cos 8V/2t.
The first stop will occur when the weight reaches its highest point. The second stop will occur
half a period later, when the weight reaches its lowest point. The first stop occurs the first
time x(t) = 0. That will happen when tan(8/2t) = $2;
Since half of a period of the motion is yrv2,
t= arctan (2/2)
+ agrv2 = 0.4 (sec).
tan
that is, at t ston
=
(2/2
(gv)
the weight will reach its lowest point when
8/2
19. We need to solve the initial value problem
x’ (t) + 96x(t) = 0; 2(0) = 3, 2’(0) = —8.
The general solution and its derivative are given by
x(t) = c, cos 4V6t + co sin 4V6t,
x(0) = c, = 3,
C1 = 5,
z'(t) = —4/6c, sin 4V6t + 4V6c2 cos 4VEt,
x'(0) = 4V6cq = —8,
Cg = —i V6.
Hence the motion can be described as
z(t) = 5 cos 4/6t ~ ;
6sin 4V6t.
10.4.
DAMPED
73
VIBRATIONS
21. We need to solve the initial value problem
a" (t) + 64x(t) = 8sin8t; (0) = 4, x(0) = 0.
4
+
The method of undetermined coefficients may be used to obtain rp = —3t cos 8t as a particular
solution of this differential equation. The general solution and its derivative are given by
co = t,
z(0) = cg = 4,
a(t) = ¢ sin 8t + ca cos 8t — $t cos 8,
ax'(t) = 8c, cos 8t — 8c sin 8t + 4t sin 8t — Zcos8t,
x'(0) = 8c, — 4 =0,
i:
q=
Tt follows that
a(t) = —2sin 8t + 4tsin 8t = 4(¢ — }) sin 8t.
i
Stops will occur when ¢ =
nx/8,
n =
1,
2,---.
The first four stops in order are:
An, 4, 40, 3m (sec).
33. From Exercise 22 we have
a(t) = }(1 ~ 8) cos 16¢ + z sin 16¢,
z(t) = ~4(1 ~ 8t) sin 16¢.
An, An
veoes will occur when t = 3, ;knm. The first four stops will occur at t = §, at,
sec).
10.4
Damped
Vibrations
1. The auxiliary equation is m? + 2ym + 169 = 0 and its roots are —y + 1/7? ~ 169. Critical
damping will occur when y = 13. For this value of + we have m = —13, ~13 and the general
solution of the differential equation is
a(t) = (ce, + egt)e**,
x(0) = ec, = 0,
cy = 0,
z(t) = [-13(c, + e2t) + cgje7**,
x'(0) = —18e; + cg = 8,
Co = 8.
Thus the solution with critical damping is z(t) = 8te7!**.
If y = 12, the auxiliary equation is m? + 24m +169 = 0 with roots m = —12+4 5%. The general
solution of the differential equation is
x(0) = cg = 0,
cg = 0
x'(t) = e7!*[(5e, — 12c2) cos 5t + (—12e, ~ Sez) sin Bt],
2/(0) =5e: -12e2 = 8,
c= 8
i
a(t) = e7!**(c, sin 5t + c2 cos 5t),
‘Thus we have the damped oscillatory solution a(t) = 8e~!% sin 5¢.
If 7 = 14, the auxiliary equation is m? + 28m + 169 = 0 and m=
—7+ 373 = —8.8,
—19.2.
The general solution is
z(t) = eye 88! + eye 9%,
2(0) = c1 +e. =0,
cy = 0.77,
a'(t) = —8.8c,e785 — 19.2ege7 194,
x" (0) = -8.8c) - 19.2c9 = 8,
cg = 0.77.
The overdamped motion is described by x(t) = 0.77 (e788! — e194),
74
CHAPTER
10.
APPLICATIONS
3. The initial value problem is
4
x(t) + z(t
32
+ = 4(t) = 0; 2(0) = —4, 2(0) <5.
The differential equation may be written 2” (t) + 22‘(t) + 50a(¢) = 0 with auxiliary equation
m? + 2m +50 = 0. Here m = ~1+ 74. The general solution is
x(t) = e~*(e, sin 7t + eg cos 7t),
(0) = c2 =~,
cg = ~4,
x'(t) = e~*[ (Te, — c2) cos 7t + (—cy ~ Teq)sin7t],
2"(0) = Tey -—cp = 5,
a = Z,
The motion is described by a(é) = je~*(2sin 7t — cos 7¢).
5. The initial value problem is
4
,
3? () + 2'(t) + 10x(t) = 0; «(0) = 0, 2'(0) = —2.
The differential equation can be written x”(t) + 82'(t) + 80x(t) = 0 with auxiliary equation
m? + 8m + 80 = 0. Here m = ~4+8i. The general solution is
a(t) = e~**(c; cos 8t + cg sin 8t),
x(0) = ce, = 0,
z(t) =e" [(—8e, ~ deg) sin 8t + (—4er + 8eq) cos 8¢],
cy = 0,
x'(0)-= —4e, + 8eg = -2,
cg = ~2,
The motion is described by x(t) = —je~* sin 8¢.
7. The initial value problem is
Zo") + x'(t) + 4a(t) = 2sin8t; 2(0) = 4, 2/(0) = 0.
The differential equation can be written a(t) + 16x’(t) + 642(t) = 32sin 8t with auxiliary
equation m” + 16m +64
= 0.
Here m = —8,
~8.
The method of undetermined coefficients
can be used to find a particular solution y, = ~4 cos 8t. The general solution is
a(t) = (cy + cot)e* ~ } cos8t,
2(0) =e, ~ $= 4,
c =k,
a’ (t) = (—8c1 + cg — 8cat)e~** + 2sin 8t,
x'(0) = —8e; + cz = 0,
cg = 4,
The motion is described by x(t) = (4 + 4t)e~® — 3 cos 8.
9. Here
z(t) = e~*(0.50.cos 6.1t ~ 1.23sin 6,14),
a'(t) = e'(~-8 cos 6.1t ~ 1.82 sin 6.12).
~8
The motion will stop when 2’{t) = 0. This will occur when tan 6.1t = [3°
this equation, a negative solution, is t =
arctan (—;§5)
Si
_
=: —0.22.
One solution of
The period of the motion
10.4.
DAMPED
is a
75
VIBRATIONS
The stopping places will occur every half-period. The first three stopping places, with
positive values of t, and the corresponding values of x are
il.
2
ty = —0.22 + mat =0.3 (sec),
ty= 0.22 + =~ = 0.8 (sec),
xy = ~12 (in.),
tq =7
tg = —0.22 +
{in.),
= 1.3 (sec),
zx —4 {in,),
From Exercise 10 we have z(t) = 1.22e7°1¢sin9.8t.
The initial value of the damping
factor1.22e7°-1® is 1.22. We wish to determine the value of ¢ such that 1.22e~°
16 = 4 (1.22).
10 _ 14.4 (see).
In10
Solving this equation for ¢ gives us t = O16
13.
The initial value problem is
30 z(t) + .6x'(t) + 4x(t)= $sin82; x(0) = 4, 2'(0) =0
The differential equation can be written 2“(t) + 9.6a’(t) + 64r(t) = 4sin 8t with auxiliary
equation m* +9.6m+64
= 0. Here m
= —4.846.41.
The method of undetermined coefficients
can be used to find a particular solution yp = —0.05 cos 8t. The general solution is
a(t) = e~*®'(e, cos 6.4t + cg sin 6.4t) ~ 0.05 cos 8,
a’ (t) = e~***((—6.4c, — 4.8c2) sin 6.4t + (—4.8c, + 6.4e2) cos 6.44] + 0.4 sin 82.
Therefore «(0)
= c, — 0.05 = 0.25 and 2'(0)
= ~4.8¢, + 6.4c2 = 0, so that c, = 0.30 and
C2 = 0.22. The motion is described by x(t) = e7***(0.30 cos 6.4t + 0.22 sin 6.4t) — 0.05 cos 8t.
15, The initial value problem is
ao"(t) +2'(t) + 4a(t) = ¢sin 8t; x(0) = 3, 2'(0) = 0.
The differential equation can be written 2”(t) + 162’(t) + 64z(t)= 4sin 8t with auxiliary
equation m? + 16m + 64 = 0. Here m = ~8, ~8. The method of undetermined coefficients
can be used to find a particular solution y, = -#% cos 8t. The general solution is
x(t) = (c1 + egtje7™ — 2 cos 8t,
z(0)=q -4
=F,
a(t) = (-8cy + cg — 8cgt)e7* + 4 sin 8,
2'(0) = 8c, +e, = 0,
= #,
ca = 4.
The motion is described by x(t)= $(8t + 1)e7®' — 4 cos8¢. Note that when ¢ = 1, the first
term in the solution has value 84e~8= 0.001, so that x(t) = —g cos 8t for t > 1, to the nearest
0.01 ft.
17. The initial value problem is
Bet}+
4a'(t) + 24a(t)= 9sin dt; x(0)= 4, 2’(0) =
76
CHAPTER
10.
APPLICATIONS
The differential equation can be written 3x"(t) + 42z'(t) + 1922(t) = 72sin 4¢ with auxiliary
equation 3m? + 4m +192 = 0. Here m = -§24 8. Oi. The method of undetermined coefficients
can be used to find a particular solution y, = 3 sin 4t — a cos 4t. The general solution is
a(t) = e7 3 *(cy cos 8.0t + cg sin 8.0¢) + By4 sin
4t — qaq cos 4t,
z(t)=e" 3 *[(—#e1 + 8c) cos 8.0t + (—8e) — ¢2) sin 8.0¢]+ £1 cos4t ++ fy sin dt.
Therefore 2(0) = ce, ~ 727 = 4 and 2'(0) = ~3c; + 8c. + $1 = 0, so that c, = 0.30 and
C2 = ~0.22. Finally, 2(¢) = e7 34(0.30 cos 8.0¢ — 0.22 sin 8.0t) — 0.05 cos 4¢ 4- 0.49 sin 4¢.
19. We assume that the acceleration due to gravity is a constant g. Then Newton's law, F = ma,
d?
becomes
oc.
qg di2
= w for x < Ah. The initial value problem may be written
dz
Ga =H H(0) =0, 21(0) =0.
We get
a(t) = gg? +e: + Cat,
2(0) =e, = 0,
"(t) = gt +e2,
x'(0) =e) =0.
Finally, x(t) = }gt?.
ai,
From uv = gt + v9 we get t = —.
Thus
p= i @am)? , volv~ v0)
2g
g
Qgx = (v — v9)? + 2uo(v — ve) = (v — U9)(U + U9).
Finally, v? = vg + 2gz.
23.
aa
ott 4
(avg (avo — a)
g)(1 - a7
a
)
a
t
v
‘
= = + (= _ 5) (at— 407?
+ 2072? — date +---)
= £ t + (uot — dugat?+ duga7t? +--+.) + (-£ t + dt? — dgat® + pga + ~-)
= dgt? + vot — dat?(3uo + gt) + Hat? (4u0 + gt) +--+.
25. The initial value problem is
x(t) + 62’ (t) + 25a(t) = 0; (0) = 0, x/(0) = -12.
10.5.
THE SIMPLE PENDULUM
77
The auxiliary equation is m? + 6m + 25 = 0 with roots m = —3+4 44. The general solution is
x(t) = e~*'(e, sin 4t + c2 cos 4t),
x(0) = cg = 4,
co = 0,
g’(t) = e"*# [(—3e, — 4c) sin 4t + (4c, ~ 3c2) cos 4t],
x’(0) = 4c, — 3cg = -12,
cy = —3.
We have z(t) = —3e75'sin 4t. Stops will occur when 2'(t) = ~3e74'(4cos4t — 3sin 4t) = 0.
‘The first stop will be at t = arctan 4, Since the period is #/2, stops occur at intervals of length
w/4. Thus th = arctan 4 + inn,
for n= 0, 1, «-- . For these values of ¢
x(t) = —3 exp [~3t,] sin 4(arctan § + inn)
= ~3exp [—3¢,] sin 4(arctan $).
Thus
E(tnyi)
zen)
=
exp[—3tn4i]
exp [—3tn]
== €Xp [~3(tn41 _ tn)| = exp [—37/4].
This ratio is independent of n.
10.5
The Simple Pendulum
1. Using the linear approximation to the pendulum problem the motion is governed by the differential equation ag +648 = 0. Thus the period of the motion is 1/4 (sec). If the clack ticks
dt?
4
120
once every period, in 30 seconds it will tick 30- ai
38 times.
3. Here 6 = 0.1 cos 8t (radians). The maximum value of 6 will occur when 0”{t) = ~6.4cos 8t = 0.
That will first happen when ¢ = 7/16 = 0.2 (sec). The maximum angular speed is therefore
|’
(w /16)| = 0.8 (radians/sec).
5. The motion is described by @ =
6 cos 8t ~ q sin 8t.
This is a simple harmonic motion with
amplitude 4/ (75)? = ($)? = 0.16 (radians). Converting to degrees gives 9°.
7. The motion is simple harmonic with amplitude 1/62 + 8-?u2.
78
CHAPTER
Chapter
11.
LINEAR SYSTEMS OF EQUATIONS
11
Linear Systems of Equations
11.2
First-Order Systems
with Constant
Coefficients
1. If we let u = y’ then we get the system y/ = u, u’ = —8y +6ut+242,
3. If we let u = y’ then we get the system y’ = u, u! = —gy — pu + f(z).
5. If we let u=y’, v=’, and w = v’ then we get the system y’ =u,
u’ =v,
vu’ = w, w' =y
7. We can write the given system in the form
3u'
+ ww! = —2u + 6w + Se”,
4u' + w! = —2u + Bw + Se® + Qn — 3,
Solving for v’ and w’ yields the system
u' = Qw +22
~ 3,
uw’ = —Qu + Se” ~ 6x +9.
9. If we let u = y’ and w = v’ then we get the system
y' =u,
vo =w,
w=
y+ Qw +l,
wl = —y + 4u — Qu.
11.4
Some Matrix Algebra
1 2
2
1) +2(3
1. A428 = (
1 2\/2
3. AB+a=(
)
O\_
/f1 2\,f4
O'\_ fs 2
3)=(
+
»)=( i):
0
1 0\_ (4 ~2\, (2 0\_
(6 -2
4) +2(4
=
+6
= (
se-at=( ~3)-a(0 )=G a)-G 2)-( >):
1
+1
1
0
1 -l
2
0) f-1l
-1
i):
11.5.
FIRST-ORDER SYSTEMS REVISITED
fi
2\ fl
AC — B= (
» p
i) (;
-1
2
79
O\_ /3
>) ~ (
4)
3\ (2
~ ¢
4)
(
OO)
_ fi 3
4) ~ (
0):
Let d,; and e;; be the elements of D and & that are in the ith row and the jth column.
Then
the corresponding elements in D + FE and E + D are dj; + ei; and e,; + dij. Since these two
numbers are always equal, we have D+
E = E+ D.
Ll. Let aj; be the element of A that is in the ith row and the jth column, The corresponding
element in O is 0. But a,; + 0 = aj; for all values of i and 7. Hence A+ O= A,
13. Since k(a;j + bij) = kai; + kbi;, it follows that k(A+ B) = kA+kB.
15. Since det A = —3, the only solution of the system is the trivial solution, O.
17. Here the two equations 2x +y = 0 and ~4x ~ 2y = 0 have the same solutions. Hence y = —-2z
and (*)
= (_.%)
=
(2)
where z is arbitrary.
x
19. Since det A = 0, the only solution is ( ] = O.
z
ai. Here 3x + 2y = 0 so that if y = 3c then x = ~2c.
23.
=
(“)
Here x+y +2 = 0 so that if 2 = band z = —c then y = ~—b+c.
()-C)G)eQ)
x
solutions
b
fy}
=
1
[-6+ej=b
1
> -1]+ef
-l
25. X'= AX+Bwhere|1
bet
RO Of
(:
1
0}
t
_
4
If m = 4, (, 6
-~3
3)
ey)
()
We therefore have the
1).
1].
ef
sint
and B=
First-Order Systems
1. The characteristic equation is
(~3):
t
-ljandB=[
#1
pet
27. A’ = AX + B where
=o
0
1
2
2-1
11.5
That is (5)
8— 16
ly.
0
Revisited
_8 =
=m? —~16 = (m— 4)(m + 4) = 0.
_
:
:
3
= O, so that 4c; = 3cg. One such eigenvector is ( i):
80
CHAPTER
lim = ~4, (7
74)
16
(2)
—4
11.
LINEAR SYSTEMS OF EQUATIONS
= O, so that 12c,; = deg.
Cg
One such eigenvector is (3).
Two
solutions of the system are therefore (3) e* and (1) e~*t, The linear independence is checked
by computing W(0) =
‘
i = 8&0.
. The characteristic equation is
If m = 2, (i
3)
Ifm = ~2, e
3)
(=)
a 4
-4— 2
=m? — 4 = (m-2)(m+2)=0.
= O, so that 2c; + 3cg = 0. One such eigenvector is (3).
@)
= O, so that 6c; + 3c2 = 0. One such eigenvector is (2):
solutions of the system are therefore (3
checked by computing W(0) =
. The characteristic equation is
3
2
z
3~m
e** and
e~*t|
The linear independence is
=80,
3 a
aenn™
-1
(3
a
2m =_ m(m
— 2) == 0.
If m = 0, (3
3)
(3)
= O, so that 3ci + 3c, = 0. One such eigenvector is (i):
lfm = 2, (
3)
(&)
= O, so that ce; + 3c2g = 0. One such eigenvector is(_}).
solutions of the system are therefore
by computing W7(0) =
a
as
_
4
4)
4
Ifm = 6, (|
4
3) (3)
~6
cq
“)
(a)
-—-160
c3
= 220.
, {l2-m 4 =-4—-mi=™>
18g
35
8m + 12 = (m — 2}{m — 6) = 0.
= O, so that 10c; — 15c_ = 0. One such eigenvector is (3):
= O, so that 6c; — 15cg = 0. One such eigenvector is ().
solutions of the system are therefore ()
by computing W(0) =
Two
and (3) e*t, The linear independence is checked
. The characteristic equation is
If m = 2, (“4
Two
of| = 4 #0.
e*# and
(3)
Two
e®'. The linear independence is checked
11.6.
81
COMPLEX EIGENVALUES
11.6
Complex
Eigenvalues
1. The characteristic equation is
_
9;
If m = 22,
[4-28
_4
BY
—4- oi)
.
5} f 5
is (_ 4+ 2) = (_ :)
A =e}
(i)
lfm = —-i+i, (° ~s
= (3)
~
(
4—m
2
—5-i}
is (, _ a)
—13
-~6—m
= (*3)
+1 (5).
1("3)
~
vl
(isfby )
4 7
cos 2t — (2)
ON
-2
(4)
0
2
=0.
13\..
cost + ( :) sin| .
=m? —8m+20 =(m—4)?+4=0.
sin 2| + cet
(2)
0
2
cos 2t + (73)
sin 2| :
0
1—m
~2) = —(m —1)[(m—-1)? +4] =0.
2
il-m
fh
bo | = O so that 2b; ~ 2b; = 0 and
2
eigenvector is (-3
.
The general solution is therefore
3
0
sin 2
= Q, so that (8—2i}b; —17b2 = 0. One such eigenvector
l—m
12
cos 2f + (5)
=m? +%m+2=(m+1—i)(m+1+i)
OV,
~17\
1-8-2)
00
(2)
5) cost ~ (_{) sint| + cet
7. The characteristic equation is
Ifm=1,
+c
The general solution is therefore
13
8 —
X =ce#
sin 2
= O, so that (5 ~1)b, — 13bg = 0. One such eigenvector
\be
5. The characteristic equation is a
lfm = 4424,
foe
85
\
+4
= (m — 2i)(m
+ 27) = 0.
on
_
.
.
=: O, so that (4 - 2¢)b; + 5b2 = 0. One such eigenvector
cos 2t — ()
+7 (_!).
X=ce™'
fbr\
(i)
) (3)
2
*)
4
.{0
ws
+i (3). The general solution is therefore
3. The characteristic equation is
is (,
Be
ca-nl=™
4—m
with corresponding solution
(3)
2
et.
3b; + 2b2
= 0.
One
such
82
CHAPTER
11.
LINEAR SYSTEMS OF EQUATIONS
fbr
~2t
0
O}1
2
3
-2i
2
~—-2
-2i
be | = O, so that b; = 0 and —2ibg — 2b3 = 0. One such
ba
+4 (
0
).
If m = 1 + 23,
0
.
eigenvector is (
0
= (:)
~i
0
The general solution is therefore
~I
2
X = ce’ | —3 | + cet
2
0
1 ] cos 2t —
0
0
0 | sin 2t!
~1
0
0}
-1
+ cae!
0
{1
0
cos2t+
sin2t}.
i
9. We let u = y’ so that u! = y” = —2y —2u. Thus (%) =e e
.
equation is
| m
1
2
-o-nl
.
Ifm = ~1+i then
1-2
-2
.
1
1
is (, + ) = (i)
Y\
(?)
= Ce
ons
2)
(:). The characteristic
=m?+%m+2=(m41)?+1=0.
-1~i)
1
by
lo)
.
:
= O, so that (1—1)b; +52 = 0. One such eigenvector
0\.
Lo
+ (?) i. The general solution is
1
(1)
COS x
_f{9\..
—
0
(:) sing| + Cg€
I(:)
cos + (;
a
)
1\.
sinz|.
Note that y = e~*(c; cosz + cg sinz).
i
11. We let u = y’ so that u’ = y” = ~dy.
equation is a
If m
1
(.:)
1
si)
0o\.
+ ()
¥\
(:)
(2)
e
(2).
The characteristic
nl =m'?+4=0,
= 27 then
= (3)
Thus
( )
= QO, so that ~2ib; + bg = 0.
One such eigenvector is
Ls
i. The general solution is
= Cy 6
1).
cos 2x
Note that y = c, cos 2x + eg sin 2x.
_
{9\.
0
(3) sin 22| + Cg IC)
cos 24 + ¢
i\..
sin2z|
.
11.7.
REPEATED EIGENVALUES
83
13.
For m =p+qi,
(OP
c
For m = p — gi,
d-—p—qij
apn) (S)=%
\ca
(OP
,
c
peat) GQ) =
d-ptgi]
\c
so that (a — p ~ gije, + beg = 0.
so that (a ~ p+ gije, + beg = 0.
One such eigenvector is
One such eigenvector is
,
(,-a-a)=(p-e) “(Qt
(,-a+a) = (pa) * (9)
These eigenvectors are complex conjugates of each other.
b
15.
W[X1(0),
11.7
X2(0)|
=
Repeated
l
0
~
gq
|= ba #0.
Eigenvalues
lim
=
6,
@
»)
Xy = () e®,
:
*)
_4
=
O,
so that
1
We now seek Xz =
.
tions now requires that
One
=m?
B-mi7™
4—m
1. The characteristic equation is
1
&t
(Je
solution of this system
9
ca
+6(S)e
is (2)
=
1
6 == (m (m—6)2 —6)* == 0.
12m19m ++ 336
—2e, +c;
=
Ps,
te®t 4.
2
6t
4
=\|_4
(‘).
0.
One
e®
1
C4
Thus Xg
of the system
is
The system of differential equaC4
8
solution
e”™*
=
6t
(3)
-2
or
1
4
cy)
2
te® 4 (*)
1
ca}
e®.
Nay
Note
that
W(X), Xa] =|) {| =1 40.
i
0
2-m
3. The characteristic equation is
lfm
=
4,
Xy=
(2)
+
“a
>
(&
—1
| =m?
6—-m
4
=
QO, so that
now requires that
1
(2)
at
cg
One solution of this system is
W1X:(0), X2(0)] =|_5 21
4t
+a(S)e
0
-1
xn
os
_{]=-1 40.
~—2¢, — cg = 0.
tett +
a, We now seek Xz = (3
— 8m +16 = (m~ 4)? =0.
One
solution of the system
is
) e, The system of differential equations
=\4
_.
2
-1
cg
3
. Thus Xq =
elle
e*
At
3
or
-2
4
—-1
2}
te** 4. (3)
7
leq)
=\-2)
1
e**, Note that
CHAPTER
i-m
5. The characteristic equation is
0
—]
=
2,
2
—i1
Oo
by
|
LINEAR SYSTEMS OF EQUATIONS
2
—t
0
2-m
l-m
0
lfm
11.
=
O,
1] = —(m —~ 1)*(m — 2) = 0.
so that
—b; + 2bg ~ b3
=
0 and
—by +b,
=
O.
b3
1
One solution of the system is X,; =
[1]e*%.
Ifm=1,
1
0
2
{0
0
1
00
ba
1
bs | = O, so that
1
be
I
2bs ~ bg = 0 and bg = 0. Thus a second solution of the system is X2. = | 0]
e'. We now seek
0
1
a third solution of the form X3 =
[0]
by
te’ + | bg | ef.
0
1
now requires that | 0 ] e' +
by
fs
0
One solution of this system is
3
=
1
0
2
1
0
0
0
[1/2].
0
bg
I
The system of differential equations
by
Ps
02
eor/0
0
Thus X3 =
+1
0
0
1
[0 | teé+
0
1
by
bg]
1
bg
1
= | O].
0
0
[1/2 | et. The general
6
solution can now be written
0
X(t) = cy
e* +e
e +c
t+{1
et.
0
—m
. The characteristic equation is|
3
1
2-—m
~
OO Go
—1,
8
m=
ee
bret ed
fen
hm
RM
= 4,
gah
1
lim
3
—2
01
1
by
bg]
3-4
' (5)
= —(m+1)2(m~— 4) =0.
3
= O.
by
1
-m
One solution of the system is X,
i
= [| 1}
t)=(
1
by
1
bs | = O, so that bg + 3bg + bg = 0. One solution is | bs | =
b4
A second solution is (|
be
(*:
a
0
1}.
-—3
e**.
If
1
(
)
oO}.
Thus two solutions of the system of differential equations
85
THE PHASE PLANE
11.8.
1
|
~ 1
(
(
e~' and X_ =
0
)
—3
e~*. The general solution can now be written
0
font
are Xg =
X(t) = c (
7 + 9 (
—1
If m
= 2,
2
m=-2,)/2
2
-—3-—m
2
~l
-1
3)
fh
2
-—5
3
ba | = QO.
-1
3
~1
3/
~1
fb
1
-1
2
~2
2 -1
-1 3\
e +03 (:) et,
~3
—m
. The characteristic equation is}
1
')
3
3} = —(m + 2)?(m — 2) =0.
i-m
1
One solution of the system
[1
=
is A;
bs
If
e*
1
bg
3
bs | = O, sa that 2b4—05+3bg = 0. One solution is { b3 | =
Oo}.
\bs
be
—2
bg
1
A second solution is | 6s | = | 2]. Thus two solutions of the system of differential equations
bg
0
3
1
are Xq = ( | e~*t and X3 = (:) e~*, The general solution can now be written
2
0
1
3
X(t) = cy(:) et 4 c (
1
7
1
+¢3 (2)
~2
ent,
0
1
il. Each of the functions X; = (°)
0
et! Xg= (:)
0
eft, Xa = (°)
0
0
1
e
is a solution of the system.
The solutions are linearly independent since W [X1(0), X2(0), X3(0)] =
1
0
[0
1
0
=10.
001
11.8
The Phase Plane
1. The general solution is X = c (5) et 4+. co
i
ee,
Here m, = ~4 and mg = 4 so that
almost all of the solutions approach infinity as t ~» co. The exceptions being those for which
c; = 0, in which case the solution approaches the origin along the line y = 47.
The origin is
an unstable saddle.
The general solution is X(t) = c [(4)
cos 2t — () sin 2| + 9 I()
cos 2t + (3)
Here a = 0, and all trajectories move around the origin. The origin is a stable center.
sin 2| .
86
CHAPTER
11.
LINEAR SYSTEMS OF EQUATIONS
5. The general solution is
X=ce*
(6
cost — (3) sint|
+ cge™* (3)
cost + (75)
sin j .
Since a = ~1 the origin is a stable spiral.
«gs
.
Id
7. Here the characteristic equation is |
—m
4
-8 al
= (m+2)* = 0. Since the repeated roots
are negative the origin is a stable node.
15. The characteristic equation is
0
0
OF
0}
fh
le
-l-—m
0
-l1-ml
:
0
= (m+
.
1)? = 0.
= O, so that there are two linearly independent
The general solution is X = [es (3)
+2
()|
etx
.
If m
eigenvectors
= —1 we have
1
0
and
i}:
(c ) e~*. For any choice of c; and cg,
not both zero, the corresponding trajectory is a half line approaching the origin as t +
The origin is a stable node.
0
oo.
87
Chapter
12
Nonhomogeneous
12.1
Systems of Equations
Nonhomogeneous
Systems
1. As in Example 12.1
ai (t) = (2e' — 2)e7* = 2 ~ 2e7*,
a(t) = (2 ~ef)e"*! = 2e7** — e*,
so that a(t) = 2t + 2e7~' and ag(t) = —e~*¢ + e~*. The particular solution is
zy
({)
The
3. From
=_ err.
ty) (je
{¥\ oe ,+(-e7*
7 p~2t +e p-ty (Je
{1 \ o2t zs© yt
[2 +¢ (3)
¢f1\_ + (0):
fi
te (3)
8 general solution is X = a Wy l e +a 2\9 1 e* + tet
Exercise 5 in Section
11.5 we have X,
= a4 C:
2
+e
2
1
3\
+ a2 (.;)
the original system of equations forces us to take a},(t)(.;)
a,V(t) (t) ==
—£—3
5 5
and(t) ==
and a(t)
pe
_?
4
+ ag(t) (
+
1
0}.
ot
e*",
i)
—l
Substituting into
7! = (;).
(t+ 1)e7*# , 80 that a, (t} ___3t
=
r
3 and a2(t) _=
2
One particular solution is
X=
21
3t
2/\-1
WV (1,3
4
8/\-1)
3\
1-2
qte
Thus
et,
53-2
t?
~ 18-9
(-2
_ 1
2t74+1464+3)°
8\
A slightly simpler particular solution may be obtained by rewriting this particular solution as
1
Xp = 3 (
f—2t?-18t-6\
24? 4.14:)
-3
3/
1
(1)
.
.
and incorporating the last term as part of the homoge-
neous solution. The general solution then becomes
_
X=er (1)
1
3\
+e (te
2 , £
+3 (
(—2t?
— 18-6
24? + 14t) °
5. We follow Example 12.1 and have a(t) = (~3e')e~' = —3 and a4(t) = (3e')e~*! = 3e7* so
that a,(t) = —3t and a(t) = —3e7'. Finally X,(t) = —3te! @
— 3et ().
88
CHAPTER
12.2.
Arms
12.
NONHOMOGENEOUS
SYSTEMS OF EQUATIONS
Races
1. The systern to be solved is X’ =
must satisfy (
“
3)
X + (3).
The constant particular solution (5)
3) (;) + (3) = @) so that e = 1 and f = 2.
The characteristic equation of A is
-h—-m
2
4
_g—m| = MP 48+
T= (m+
m4+7) =0
giving eigenvalues —1 and ~7. These eigenvalues correspond to eigenvectors (3)
and (1):
Thus the general solution is
X(t) = ey () e+ eg (.:)
ety () ,
The initial conditions zg = 8, yo = 7 require
(:)+e(4)+()=(),
an equation that has solution ¢; = 4, cg = 3. Finally,
X(t) =4 () e' +3 (.:)
ey
(3) .
The solution will approach a stable solution x = 1 and y = 2.
2
3. The system to be solved is X'= | — 4
solution
(5)
must satisfy (i
i)
8
2) X+ ()
(5)
+ ()
, &(0) = (;). The constant particular
= (3)
so that e = —2 and f = —3.
The characteristic equation of A is
—-2—-m
4
giving eigenvalues 2 and —6.
4;
oma
og
Me
oe
+4m —12= (m+ 6)(m—2)=0
These eigenvalues correspond to eigenvectors (;)
Thus the general solution is
X(Q)=e (3) ot + oy (.;) e784 (<5)
The initial conditions 29 = 5, yg = 2 require
«()+e(2)-G)=@),
and
(1):
12.2.
89
ARMS RACES
an equation that has solution ¢, = 6, cz = 1. Finally,
X(t) =6 (3) em 4 (.;) e~8t _ (5)
There will be a runaway arms race.
2
2
(*)
must satisfy (f
so that e= land f = 1.
= (3)
+ (<2)
(5)
3)
The constant particular solution
5) X + (- 2 ).
5. The system to be solved is X’ = C 4
The homogeneous solution is the same as in Exercise 3. Thus the general solution is
X(t) = cy (;) e7! + eg (1)
et
4. @
.
The initial conditions require that
~1
fa\
_
[te-1
ca
~
Yo
1]?
1
an equation that has solution c, = 5 (to + yg — 2), cg =
ee
l\
1
(vo — yo). Finally,
bo]
1
X(t) = 5 (0 + yo — 2)e# (1) + 5(t0 ~ yo)e7* (.;) + (7) .
If xo + yo < 2 the arms expenditures for both countries will become and remain negative, a
condition of disarmament. If 29 + yo > 2 there will be a runaway arms race.
ae
4
7. The systern to be solved is X‘ =
must satisfy o
)
(;) se (=!)
The constant particular solution (5)
A+ ().
2)
so that e = -ar + 2s) and f = ~Z(2r +8),
The homogeneous solution is the same as in Exercise 3. Thus the general solution is
1
X(t)
=
Cy
@
1\
e*
+
ca
(_;)
.
é
Gt
1
fr+2
6
Ger :)
.
The initial conditions require that
(i) (2)= (3) +3 GP):
.
1
an equation for which ce, = 5 (x + Yo +
r+s
i").
.
Because of the factor e*! in the first term
of the general solution, that term will dictate the behavior of the solution as t + oo. Ife; > 0
-T-s
there will be a runaway arms race. This will occur if xg + yo >
2
90
CHAPTER
12.
NONHOMOGENEOUS
SYSTEMS OF EQUATIONS
9. Since pq — ab < 0 the two eigenvalues are real and have opposite signs. We set
u
- —Ptgt+
Veta
~ pg?ab)
~ eta) - Vip +a)? — Aq — a8)
2
2
We also note in passing that since pg — ab < 0 and q > 0,
+q+J7(p+a)?
pru<p- Prat Vera
_
<0,
(12.1)
The constant particular solution (5) must satisfy (
) (5) = (=) so that we have
rb
ex 4 18a and f = ptr
. Note that e and f are both negative. If we use the eigenvalue
pq —ab
pq - ab
m=
tu we must satisfy (? ~ b
‘) (&:)
~G{n-u
== QO so that one eigenvector is (
C2
‘):
If
p+u
we use the other eigenvalue m = v we get in the same manner an eigenvector
pt *):
Thus
the general solution of the system is
X(t) = cye™ (, + i) + ce” (, + :) + (7) .
Since u, a, and p+ wu are all positive and v is negative, the character of these solutions will
depend on whether c, is positive or negative.
If it is positive we will have a runaway arms
race. We now apply the initial conditions X(0) =
«
ptu)
ate
2).
(a8
Therefore ¢; = eo
een
vr’
2)
and get
\wo-f/pte) (8)\eo} ~~= (0-4)
Remembering that e, f, p+v, and v are negative
while xo, yo, u, and a are positive establishes that c; is positive.
arms
12.4
Tace,
Simple Networks
. This is Example 12.6.
3. The initial value problem is
I(t) + F(t) = # sin wt,
I(0) =0.
We therefore have a runaway
12.4.
91
SIMPLE NETWORKS
R
The linear differential equation has as an integrating factor exp ($+)
exp Gi
= z_| sinatexp (4)
and I{t}) = A (-wL coswt + Rsinwt) + ec; exp (74).
choose cy = =
5. The initial value problem
that Q(t)
dt
The initial condition forces us to
so that finally
I(t) = a
Q(t) exp (2504)
so that
=
-wL coswt + Rsinwt + wl exp (-#)|
d
is “
+ 25090
0.02 exp (250¢) + c;.
= 0.02 — 0.005exp(—250¢).
The
= 5,
Q(0)
= 0.015.
The general solution is
initial condition requires that c; =
Therefore I(t) = dQ
dt
J(0) = 1.25(amp).
7. Here I(t) = 3000texp(—1000¢).
.
—0.005,
= 1.25exp(—250t}(amp)
so
and
Thus J'(t) = (—3 - 10° + 3000)
exp (—1000¢) so that the
maximum current occurs at t = 0.001(sec). We have Imax = 3e7!{amp).
9. We have the problem RJ + LS
L
088
i ,
RO
pot
TS
+ aa = Bsinwt,
@Q(0) =0, £(0) = 0 which may be written
+Q= EC sinwt, Q(0) = 0, Q’(0) = 0. Here LCm? + RCm
~RC + J/R?C? — 420
R
=—o7
+
R
WT
1
~Fe=
+1 = 0 so that
~at Bi.
Thus Q, = e7**(c; cos Gt + cosin Gt). We seek a particular solution that has the form
Qp = Asinwt+Bcoswt. Substituting into the differential equation and determining the undeER
vE
ER
VE
.
.
.
Senne’ coefficients yields A = ~"3
and B=
“TRE
Thus Q, = — oH
sin wt— ow
cos wt
an
Q(t) == e~™ —at (ec, cos Bt + cz sin Bt)
a
VE
ER
—=;
ae sin wt ~ azz oS ut,
Q(t) = e~*!(—e1
8 sin Bt + cof cos Bt) — ae ~4te, cos Bt + cg sin Bt) —
ve
cos wt + ees sin wt.
92
CHAPTER
12.
NONHOMOGENEOUS SYSTEMS OF EQUATIONS
ER =
But Q(0) = 6 and Q’(0) = 0 so that ce, —
= 0 and cf — ac, — VE5 = 0.
5
R
wZ?
and cy = ae G + 1): Substituting c, and cg into I(t) = Q'(t) gives us
I(t)=e
(2
- a
VE
ER
Zz
Ze
~— +> coswt + —=
E
= evat E
11. We know that C = AL
~ fn
1) bin t+
{SS
+ a
(2
+1) } cos,
sinwt
a’R
VE
~ ap 7 3) sin Bt +2
Fz cos pt] + + E(Rsinwt
(-2
Fup
7
sinwt
wt ~—
G
ER
Thus ¢1 = “33
ycoswt) + ——Ee™
VE
80 that y = wh
R
1
By-y cos Gt
wo —_
a\
wl — Re
4tw
— ycosuit)
y+ + a2 J sinsi GE)
Gt .
. Th
a
Rt
~ aD
=
a
From this we get --—~ = =~
» = 2ay— Rw. Finally ~ay~-—— = ay— Rw. Substituting
this expression for the coefficient of t in the last term in the answer of Exercise 10 yields the
desired form.
13. We have from Kirchhoff’s laws
101; + 2000Q2 = 60,
101; + 2013 = 60,
@o(0) = 0.03,
h=h+Idy.
From the second equation we have I; = 3 — 41 1. Substituting into the last equation yields
di
= 3 lp +2,
Thus the differential equation may be written oe
+ 300Q2 = 6. The general
solution of this equation is Q2 = ¢ +c1e79%. The initial condition yields c; = 1/100 so that
Op = be 4 1g- soot
50°
100
Tz = ~3e7 3008,
Ty = 2 — 2e7 0,
Ig == 2 + @7 3008,
15. Application of Kirchhoff’s laws yields the system
dl.
Ail, + Rela + a
= BE,
Ril; + Roly + 5-Qs
3
= E
fy =Ig4+
dy.
12.4,
93
SIMPLE NETWORKS
dl.
The first of these equations can be written wet
R
~—h
at
E
+-—.
- fey,
La
Ly
second equation and using the third equation to eliminate J, yields
al,
Riz
+ Rt=
aly
di,
Riv
Replacing
I
a
+ oh
diy
+ Bs (Ge -
Ly"
Differentiating the
=0,
1
+e-x- (I) ~ In)
= 0.
by its equivalent gives us
(Ry + Rs)
(Ry
+ Rg)
di
R
- zt ~Ril; ~ Relg+ E) + a (h- fn) = 0,
dl,
_
ct
=
(
RRs
|
Tp
1
a)
he (-
RRs
Lo
1
+z)
I+
RE
The system in J, and Iz can now be written
_ (- cent;
CaRyRatL
_
~obiteteey \
alte
Z
Re ~ L
—
-R
I
(Lh), TaRi +R)
RE
£
Ty
a
*
The nature of the solutions of this system depend upon the roots of the characteristic equation
_ GaRyRstle
9
Cebgthi the)
__OsRgR
R
~ GbR
— #2
—m)
2
which may be written
m+
=|
CsRiRg + Lo
C3L2(R; + Rs)
Ro(CgRiR3+L2)
C3E2(Ri + Ra)
— L2)
Ri(CsR2Rs
C3L3(Ry + Rs)
== Q,
or
C3Lo(Ry
+ Rg)m?
+
[(C3(RiR, + Rohs
+ Rg Ry) + L|m
+ hy + Rg =0.
Note that the answer given in the book has a typographic error. The last term should be Re,
not Rg.
94
CHAPTER
Chapter
The
13.
THE EXISTENCE AND UNIQUENESS OF SOLUTIONS
13
Existence
and
Uniqueness
of
Solutions
13.2
An Existence and Uniqueness
Theorem
1, Here f(x, y) = x, xo = 2, and yo(x) = 1. The sequence defined by (2) becomes
yo(z) = 1,
n(a) = 1+ f
=
t?
215
2
x
ua(e) =1+ [
x
tam i+ [5 | =—-—1,
2
tdt=--—1.
2
2
2
It follows that B00
lim y,(z) = =2 — 1.
3. Here f(z, y) = 2y, ro = 0, and yo{x) = 1. The sequence defined by (2) becomes
yo(x) = 1,
we
w(z) =1+ /
2 dt = 1+ [2t]) = 1+ (22),
wala) =1 +f
[2+ 2(2t)] dt=1+ le
ape eee opeoy
Let us suppose that y,(a@) = 1 + (22) + (ay
By induction we have y,,(z
4g ) kd |,
42
k+l
2
>
Then
SF 2
ad k | dt =1+4 [cy + Qty
*
Yeti (Z) = 1+ f Prams
= 1+ (2) + ey
+ orl- 1+ (2a) + Gey
+
+e
(ay
1
. It follows that Jim
Ya(z) = 5
k=O
(2k)*
a
e%,
95
Chapter
14
The Laplace Transform
14.3.
Transforms
of Elementary
Functions
1. From elementary calculus we obtain
L{cos kt} = [
Ox
e **(—scoskt + ksin =) °°
e~* coskt dt =
Sie
9
For positive s, e~** —
Therefore we have
0 as t —
oo.
Furthermore sinkt and coskt are bounded
L{cos kt} = =H
From Euler’s formula we get sin kt =
ikt __ »—tkt
5;
. Formally using the result of Example 14.1
k
1
1 f
nut,
ly,
atty
le
.
Usin kt} == ite} —— —L{e*}
Me) == —5 Snag a ae
saik| | ~ae ae,
PER
Let?
4t}
== L{8}
{tth=[a-ata
Lf}+ +4L(t}=2-244
L{t?}—- 10}
{t° —— 2°#2 ++ 4t}
80.0
. Using the result of Example 14.1 we have
. 4ty
.-2t) = 3L{e*}
2)
L{3e* dt — e~**}
— L{e7“*}
11.
L
in?
{sin’ kt}
ee
1 ~ cos 2kt
2
|
SOT
1
bet Le L{1 —~ cos* Co8* kt} Se
57
13. L{sin®tn? kt}=
15.
L{e~*
_ et}
~
1
sta
3
1
28 +160
sod _ 3337 = GrDGTD’
eeyalee * PLR
p=ee 5eesnk
ninasiassamnmnssteonit
uf
1
s+b_
1fi
2 5
fe
oo.
s>o,
we have
‘ kt} = 5 Lfe ee
- Lisinh
e€
as t ~
8
a? wd
sce
3? + 2k?
ageteneegi
S(sh
+ aR
FE
b-@
,
(s+ a)(s +6)
3 > kl.
2k?
s(s? + 4k?)’
ceemnertteememeemmn
2k?
~ meee
Sl?
+ aR’
s>4.
$>0
s>0 .
83> max(—a,
—0).
.
96
CHAPTER
14.
THE LAPLACE
TRANSFORM
17. From the definition of the Laplace transform we have
Le}
= f
2
20
en st a+ [
te~*t dt=
S —st | 72 +|=# ~§ tie
_ §
e728
=
$
1
e728
“bo
+ —EO
8
Jo
§
Gt, TOO
82
3
s>0.
&
19. From the definition of the Laplace transform we have
_
st gi
Tt
_
14.6
Functions
of Class
x
0
_ 2(1-e7**)
st4+4d
~ ast
‘
[e7*{—ssin
2t — 2cos 2t)
¢
a> 0.
A
1. If F(t) and Fo(t) are of exponential order as t + oo then there exist constants M,, Mz, and
b,, be and fixed numbers ¢,, te such that
LFi(t)| < Mie", fort >t,
and
[Fo(t)| < Mae", for t > tg.
If we take tg =
max(t1,
tg), we have for ¢ > ts
LF i (t)| - [Fe(t}| < My Mo exp [(b, + b2)e]
or
[Fi (t) - Fa(t)| < Mi Mz exp [(b1 + ba)é}.
Thus F,(£) - Fo{(t) is of exponential order as t + oo.
Similarly, if we take Mg =
max(M],,
M2),
b3 =
max(b),
62), and tg =
write
[Fi (t)| < Mie’! < Mse*,
for t > ta,
|FQ(t)} < Moe"
for t > ty.
and
< Mge®s*,
Thus
[Fy (t) + Fe()| < [FA(O)| + [Fa@)| < 2Mge"", for t > ty.
That is, F(t) + Fo(é) is of exponential order as ¢ + 00.
max(t:,
tg) we may
14.6.
FUNCTIONS OF CLASS A
Consider im ett?
= Jim
_
ot OO
97
tx
—.
Tfa < 0 this limit is zero and ¢* = Oe’)
as t -
oo.
If
E
0< 2 < 1, we use the tools of elementary calculus to evaluate the indeterminate form
£z
gtt~}
lim —- =
foo
= 0.
lim
2
t-+00
=O
Again we see that t? = O(e'). Finally, if0 <n <2 <n+1,
n+1 times to obtain
x
ne
lim ~ = lim ae
too
€
1
ene
EOP
we
this argument may be repeated
am
Weenie”
)(ea=n+ Ut
t-400
et
= too
tim TLD
~
nee
eae
et
~
Z—-n—-L
Log,
Thus
t? = O(e').
The function cos kt is continuous and bounded by unity for all t, hence | cos kt| < 2e® for all
t. Thus cos kt is of exponential order as t ~» oo. Hence cos kt is of class A.
The function sinh kt is continuous for all ¢ and if k > 0 then
kt
| sinh kt} =
em €
5
kt
e kt
< 7
for all t > 0.
lfk < 0 then
| sinh kt| = | — sinh [(—K)t]| = {sinh [(-&)¢]] < ot
for t > 0.
Hence sinh kt is of exponential order as t —+ oo. Therefore sinh kt is of class A.
For t > 1,
sin kt
sin At
sin kt
.
< |sinkt| < e®. Hence
is continuous for t > 0.
sin kt .
.
is of exponential order as t + oo. Moreover,
.
. sinkt
It is also true that lim
.
= lim keoskt
= k.
is sectionally continuous over every finite interval in the range ¢ > 0. Thus
Therefore
sin kt
is of
class A.
il. By Exercise 8, t” is of class A. Also e** is of class A. Hence by Exercise 2, t"e* is of class A.
13. By Exercise 8 and Exercise 5, t” and cos kt are of class A. Hence by Exercise 2 their product
is of class A.
15. By Exercise 8, t” is of class A, and by Exercise 6, cosh kt is also of class A. It follows from
Exercise 2 that ¢” cosh kt is of class A.
98
CHAPTER
14.
THE LAPLACE TRANSFORM
cost ~ cosht
17, We let F(t) = ——__——-.
Then for t > 0, F(t) is a continuous function. Moreover,
lim Fé) = lim (— sint ~cosht) = ~1, Hence, F(t) is sectionally continuous for t > 0. For
cost — cosht
;
t > 1,
< |cost — cosht|,
But by Exercises 5, 6, and 2, cost — cosht is of
exponential order as t -+ oo; that is, there exist constants Mand
b and a fixed tp such that
|cost—cosht| < Me" for t > tg. It follows from the above inequalities that for t > max(1, tg)
cost — cosht
t
< Me®,
That is, F(t) is of exponential order as t — oo. Therefore, F(t} is of
class A.
14.10
Periodic
Functions
1. By equation (6), page 269,
17} = FEM),
1T(1/2)
— 9 87
by Theorem 14.10,
= 55572
by equation (7), page 269,
= 2 (2),
for
s > 0.
ivi
3. sin kt is the unique solution of the initial value problem
y+ k?y = 0, y(0) =0, y/(0) =k.
Using the Laplace transform on this problem gives
s*L {sin kt} — k + k*L{sin kt} = 0.
Thus
L{sin kt}
k
=wok
5. If P(t) = sin kt then F’(t) = kcoskt. For this F(t), Theorem 14.5 states that
Lik cos kt} = sL{sin kt} — 0
or
kL{cos kt} = sL{sin kt}.
14.10.
PERIODIC FUNCTIONS
99
7. Using Theorem 14.9 we have
|
k
|
L{t? sin kt} = L{(—t)* sin kt} = a
ds? | 3? +k?
_ ~2k(s? + k?)? + 2ks-2(s?-+h*)-28 — 2k(3s? ~ k*)
~
Cea
“Gh yp Rays
FO
50
9. From the definition of the Laplace transform we obtain
L{F(t)} = [ * ot 41) dt-+ [ ” ge78" at
_
~—e7
(t+ 1)
~
From Theorem
_ est
8
s2
.
st]?
e728
1
8
9
s
gs
Q
14.5 we obtain
_yeice™
1
L{P'(t)} = s oo 3
—3e7
+ otp
;
3g?
s>0.
A second derivation of this formula can be obtained by using the definition of the Laplace
transform directly on the function F/(t)=1, O<t<2; P(t)=0, t> 2.
11. Using Theorem 14.12 we obtain
L{T(t, oj} = Bee
1 — ees
9B _ Se Be"? a8 + J" (20 ~ Bye™*? a8
1 ~~ er ees
c
= _
i
1 ~ en tes
_ 1
.
ZB
8
1l-2e* em
= Ql
98
owe 1
s?
oes
1
=
0
2c
+
=e
9p
8
+
B
8
-sB
+
1
s?
(1 — e789)?
oer
13. Using Theorem 14.12 we have
E{| sin kt} =
n kB — kcoskp)|"/*
sinkG dp ~ [e~*9(1(—ssi
o e~*
— e~87/k)(52 + £2)
1 — ev sa/k
(k) +k
_= (= e7S/K
en sx/k){ 52 +k?)
k
8
sre Oth a
_
=
ok
PLR
14 ecst/h
,
Ser
ss
e
100
CHAPTER 14.
15. From Theorem
THE LAPLACE TRANSFORM
14.12 we have
L{G(t)} =
fo en*%e8 dB
fe elt
l-e"s
_
ag
Cs ee
el #+)8
© _
1
- aaa], ~~ 3=1
17. From Theorem
1I-—exp ([c(1
—s)]
T—exp(=es)
,
>
14.12 we obtain
1
2n fw
FO} = peas
1
=
f
i
1 ~ e- 9ra/w
a
t
e~ 8 E(B)
8 =
wt fw
Samara:
|
ev ams/w
sinw? — wees)"
gs? + we
0
w
=
dB = [
[ro
dt df= [ [ro
a.
ee
“E
=1 {FO},
1
site
19. Using an argument similar to the proof given in the text for Theorem
[
sb
e * sinwB dg
‘ 1 — en re/w
14.8, we have
dB dt
101
Chapter
Inverse
15.1
15
Transforms
Definition
1. Using Theorem
r7}
of an Inverse
Transform
15.2 we obtain
1
ee
(zener
1
cee
lorie}
= fi
bee pv tpl
ev
1
{ass}
1
ge
tent
=
gi
Se.
sinst
3. Using Theorem 15.2 we obtain
Dkk.
38
ee
lores}
hee
we
=e
_1 POE [ 3(s +2) —-6
gr tn
pratt
_op-l
‘oorss}
E
dard
33 ~—6
oe
§
{ava}
TV 49
3
2b
\wre}|
= e~""(3. cos 3t — 2sin 3t).
5, Using Theorem
15.2 we obtain
-1
L
1
pel
\zrurapnt
I
—pvatp-l
(arms
fil
L
_
ae
aerate”
7. Using Theorem 15.2 we obtain
L
wij
2s —3
\aoare}
ba
_
88S
L
pid
2s—3
een
typo}
evL
won}
s+]
{aa}
pit
9. Using Theorem 15.2 we obtain
L7 {ets
11. Using Theorem
po {26495 SI a endte-l (B5-] =e M(at — 82),
15.2 we obtain
L
~1
1
{t payer \
= e at
eel
yal
1
(a
\
=
le
e*' (2 cos t + § sin 2t).
tre
nm
102
CHAPTER
15.
INVERSE TRANSFORMS
13, Using Theorem 15.2 we obtain
1
~1
.
1
pratr~
1
mat
os
I opareef ote" {arp}
= jectsinee
Ox
15. Consider L {F (;) \ ss [
e AF (é)
0
df. We change the variable of integration by setting
(= at and df = a dt to get
L {F (¢) \ =a [ * e=* F(t) dt = af (as).
That is L“{ f(as)} = =F(£).
15.2
Partial
1. £7}
|
e4p2fl__
s* + a8
2
8
Fractions
a
ro
s3 + 3? — 2s
4s +4
_ Eo}
5 o
{es}
.
-t
aL
9, Lo
15.3
_~
=
8
fet
L
fe+
8
3
Lo}
53-2
eee
aUEHEcD
S
(s? + a)(s? + 6?)
Initial Value
a
d
~
s—-1l
3.
yg}
g+2
not
ee.
2
=~ - ) = 3e7 — 3 - Qt.
8
a}
se
8
3
EL
=+(1—e-*).,
$+a
-1f
2
1
i Co
=
1
~1
2 +a
zp1
=n?
tite
}
~ 3 ; af”
at—B
t . ~2t .
+e
(cos bt — cos at)
‘
Problems
1. y’ = e*; y(0) = 2. Applying the Laplace transform to both sides of the differential equation
and using Theorem 14.6 allows us to write
1
gu(s)
~2=
pel?
Bo ait
us()=“ s(s-1)
8
1
1
1
s—1'
The inverse transform now gives y{t) = 1 +e. Verification of this solution is straightforward.
15.3,
103
INITIAL VALUE PROBLEMS
3. y/+y =e; y(0) = 0. Applying the Laplace transform to both sides of the differential equation
and using Theorem 14.6 allows us to write
1
su(s) + u{s) = s23
1
3
u(s) = (s—2)\(s+1)
3
s-2
s+1
The inverse transform now gives y(t) = }e** — $e7*. Verification of this solution is straightforward.
.y” +a7y = 0; y(0) = 1, y/(0) = 0. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
s*u(s) — 8 +a7u(s) = 0,
8
us) = aya
The inverse transform now gives y(t) = coset. Verification of this solution is straightforward.
. yo” — By’ + 2y = e%*; y(0) = 0, y’(0) = 0. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
s*u(s) — 3su(s) + 2u(s) = oy,
1
1
1
2
1
u(s) = (gs — 1}(s ~ 2)(s ~ 3) slA-Stsl
1
The inverse transform now gives y(t) = 5 [et — 2e%* + e**].
Verification of this solution is
straightforward.
y” ~ 2y' = —4; y(0) = 0, y(0) = 4. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
4
s’u(s) — 4 —2su(s) = “>
_-4(i-s)
u(s) =
(sa)
1
21,2
* 32
ard
The inverse transform now gives y(t) = e”* — 1+ 2t. Verification of this solution is straightforward.
11. a” — 4a’ + 4a = 4e7#; 2(0) = —1, 2'(0) = ~4.
Applying the Laplace transform to both sides
of the differential equation and using Theorem 14.6 allows us to write
su(s) + 8 +4 —4su(s) — 4+ 4u(s) = a
u{s) =
—s?4+Qs+4
(s-2))
_
4
(s—2)8
_
9
1
(s—2)?
s-2
104
CHAPTER
15.
The inverse transform now gives z(t) = e**(2t? — 2t — 1).
straightforward.
INVERSE TRANSFORMS
Verification of this solution is
13. y" ~y
= 4cost; (0) = 0, y'(0) = 1. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
4s
7u(s) — 1 —u(s) =
i
4s
u(s) =
+
1
(s?~1)(s?+1)
_
28
_
2s
s?~1 °° s?-1
i"
1
5241 ° 521°
The inverse transform now gives y(t) = 2cosht + sinht — 2cost.
is straightforward.
Verification of this solution
15. x" +47 =t+4; 2(0) = 1, 2/(0) = 0. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
1
$ 2 u(s) — s + du(s) ==
4
5 +
3°
1
4
8
u(s) = s*(s? + 4) + ero
The
inverse transform
now
gives z(t)
=
1
l
+34
1
qt _ 5 on 2é+
1
~ 4s?
1.
1
A(s? +4) as
Verification of this solution is
straightforward.
17. x" +2 = det; (0) = 1, 2'(0) = 3. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
2
s“u(s)
—
—s
—_
—-3+u(s)
u(s) =
4
eed
scT?
4
(s—1)(s?4+1)
a
(8? +1)
a
ae
+=
s?+1
°° s—1 °° s?4+1
The inverse transform now gives z(t) = 2e' + sint — cost.
5
92477
Verification of this solution is
straightforward.
19, y” + 9y = 40e"; y(0) = 5, y’(0) = -2. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
40
s*u(s) — 58 +24 9u(s) = yo
u(s) =
40
(s ~ 1)(s? +9)
5s —2
+ aT
4
=
a-l
8
6
+359" TO
The inverse transform now gives y(x) = 4e” + cos 3x — 2sin 3x. Verification of this solution is
straightforward.
15.3.
108
INITIAL VALUE PROBLEMS
ai. gl +a! +22 = 4t?; x(0) = 0, z’(0) = 0. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
s’u(s) + 3su(s) + 2u(s) = a
we
s8(s+1)(s+2)53
5%
8
s+1
8s4+2
The inverse transform now gives z(t) = 2t? -6t+7-—8e7'+e7*'. Verification of this solution
is straightforward.
23. y" +y = ~cosz; y(0) = a, y'(0) = b. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
s*u(s) ~ sa~b+ su(s)-a=—aAs,
~8
as+atb
u(s) = s(s + 1)(s? +1) +
The inverse transform now gives y(z)
solution is straightforward.
a+b
b+3
8
stl
s(s +1)
$8
5
s@4+1
5241
= c; + cge7* + 4 cos x _ $ sin az.
Verification of this
25. yf! + 3y' + 2y = 122";
y(0) = a, y’(0) = 6. Applying the Laplace transform to both sides of
the differential equation and using Theorem 14.6 allows us to write
s’u(s) ~— sa ~ 6 + 3su(s) -a + 2u(s) = x
u(s) =
24
+
si(s+1)(s+2)
_b-24
a-6+3
s+1
as+a+
(s+1)(s +2)
s+2
12
18,
ssf
21
gs!
The inverse transform now gives y(z) = cye~” + cge7** + 6a? ~ 18a + 21. Verification of this
solution is straightforward.
27,
y’ + 4y' + 5y = 50x + 13e9"; y(0) = a, y/(0) = 6. Applying the Laplace transform to both
sides of the differential equation and using Theorem 14.6 allows us to write
a
s*u(s)
—_ sa ——~hb+d
b + 4su(s) ~oo a + 5u(s)
50
13
ep cee
et fe ert
pty
u(s)=
50
5
s*(s?+48+5)
_ (a+ Fs +2)
(s +2)? +1
+
+
13
as+a+b
3
+ 3
(s-3}(s?+45+5)
57 44845
b-a+}
(s+2)?+1
10 8
=
8?
3
33-3
The inverse transform now gives y(z) = e~**(c cosx + cgsinz) +102 —-8+ de°* . Verification
of this solution is straightforward.
106
CHAPTER 15.
INVERSE TRANSFORMS
29. yl” +y" — dy! — dy = 3e7* — dor ~ 6; y(0) = a, y/(0) = 6, y"(0) = c. Applying the Laplace
transform to both sides of the differential equation and using Theorem 14.6 allows us to write
s°u(s)
— s*a — sb -—c+8° u(s)
- sa—-6—- dsu(s) + 4a
du(s) = ——— -
u(s) = 388"(s4s+ +1) — Gs(s +1) | s?a + o(a +b) +b
1)?(s — 2)(s + 2)
(s + 1)(s— 2)(s
+ 2)
_ W(-3+2a+3b4+0)
=
3.2
F(1+2a+b-c)
~
5 ~~,
+e~4a
1 + 4.1
&
F(-9+8a-2)
+2
s+1
~ (s+1)2
"3?
The inverse transform now gives y(ax) = cye?* + cge7™ + cge7* — we7® ba + 4. Verification
of this solution is straightforward.
3i. yf" —~y = e*; y(0) =a, y/(0)= b, y"(0)=, y'"(0)= d. Applying the Laplace transform
to both sides of the differential equation and using Theorem 14.6 allows us to write
s4u(s) — s%a ~ s?b— se—d—u(s) = ay
u(s) =
1
ey
ca
s-1
The
inverse transform
now
sta +s"b+sc+d
375
+
3
(s — 1)(s + 1)2(s? +1)
(s—1)(s + 1)(s? +1)
;
3H
c3s
“Grip?
gives y(z)
=
C4
erie
cye* + ege7* +
s+)"
c3cos@ + cqsing — te~
x
, where
cy = 4(1 + 2a + 2b + 2c + 2d), C2 = 3 (-3
+ 2a — 2b + 2c
+ 6d), c3 = 4(-1+
2a + 4b — 2e— 4d),
and cq = i(-—1+ 2b ~ 2d). Verification of this solution is straightforward.
33. y” ~ dy = 2— 8x; y(0) = 0, y'(0) = 5. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
2
u(s) = s(s?— 4)
8
5
?(s?
— 4) + Ia
The inverse transform now gives y(x) = e* — be? + 22 — ‘.
1
$
5237
2
state
[rom
2
8
2
s“uls)
~5 — 4u(s)= ~ — 32
3
Verification of this solution is
straightforward.
35. y” + 4y’ + 5y = 10e~9*; y(0) = 4, y/(0) = 0. Applying the Laplace transform to both sides of
the differential equation and using Theorem
14.6 allows us to write
10
2a fe\ —— 4s + 4su(s) —_ 16 + 5u(s) — 0
s*u(s}
sa3?
NS)
(3) =
10
(s+ 3)(s2 +4845)
13
,
+2
48 +16
s?bde 45
5
“(erat (asi! ses
15.3.
INITIAL VALUE PROBLEMS
107
The inverse transform now gives y(z) = e7**(13sinz — cosz) + 5e73*.
solution is straightforward.
Verification of this
37, €+4¢+52
= 8sint; 2(0) = 0, 2(0) = 0. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
s*u(s) + 4su(s) + 5u(s) = a
UNS)
0
(s) = ——-—2.
(32 + 4s + 5)(s2 +1)
a+2
1
8
~ (so? 41° (ra? s1°
1
el”
The inverse transform now gives z(t) = e~*‘(cost — sint) ~ cost + sint.
FT
Verification of this
solution is straightforward.
39. yf’! + dy" + Oy’ + 10y = —24e"; y(0) = 0, y’(0) = —4, y”(0) = 10. Applying the Laplace
transform to both sides of the differential equation and using Theorem 14.6 allows us to write
s?u(s) + 4s — 10 + 48?u(s) + 16 + 9su(s) + 10u(s) = a
ds? ~ 23 - 18
uls) = poe
(s ~ 1)(s3 + 4s? + 9s + 10)
2
sti
1
st+2. (st+l)*+4
8-1
The inverse transform now gives y(x) = 2e7°* — e~* cos 2z — e”. Verification of this solution
is straightforward.
41. y+ 2y' + y = 2; y(0) = —3, y’(0) = a. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
1
s*u(s) + 38 —a+ 2su(s) ~6+u(s) = a
u(s) =
~38° + (a + 6)s?
s*(s +1)?
72,1
ss?
1
, a+10
g4+1
The inverse transform now gives y(z) = —2+ a4 —-e7? + (a +10)ze™*.
y(1) = ~1—e7' + (a+ 10)e7! = ~1, so that a = ~9 and
ya) = -24+a—-e°% 4+ xe",
y'(x) = 1+ 2e7* — xe™”.
It follows that y(2) = e~? and y'(2) = 1.
(s+1)?"
From this we get
108
.
CHAPTER 15.
INVERSE TRANSFORMS
43. x" — 4a’ + 4x = e**; x(0) = 6, x'(0) = 0. Applying the Laplace transform to both sides of the
differential equation and using Theorem 14.6 allows us to write
s"u(s) ~ bs — 4su(s) + 4b + 4u(s) =
3
b
2b
“= Gigs tsa BaF
The inverse transform now gives the fnetion z(t) = 1 2g + be* — 2bte**. From this we get
al} = $e 2 4 be? —2be? = 0, so that b = , and z(t) = a(1 ~t)*e*. Verification of this solution
is straightforward.
15.4
A Step Function
9. We first write the function F(t) in terms of the a@ function and then use Theorem
obtain the Laplace transform.
15.3 to
P(t) = 4 —4a(t — 2) + (2t ~ l)a(t ~ 2) = 4 - aft - 2) + 2(¢ — 2)alt — 2),
HR W)= 5-4
e728
e728
11. We first write the function F(z) in terms of the a function and then use Theorem
obtain the Laplace transform.
15.3 to
F(t) = ¢? — a(t — 1) + 8a(t — 1) — 3a(t — 2)
= , ~ (t — 1)?a(t ~ 1) — 2(t — Na(t — 1) + 2a(t - 1) - 3a(t — 2),
De _ Bem2# me 0
2
*) _ 3e7*s
= ee Qe“ _ De®
,~8 (:
IFO}=
3
5?
+
8
3
33 re
g*
3
13. We first write the function F(t) in terms of the a function and then use Theorem
8
15.3 to
obtain the Laplace transform.
F(t) = e~* — e~fa(t — 2) = et — e~2e7
1
HPO}
=
e74e74s
4
St]
1
Malt — 2),
e7 28-2
6st+l.)6os+i.
15. We first write the function F(¢) in terms of the a function and then use Theorem
obtain the Laplace transform.
F(t) = sin 3¢ — sin 3¢ a(t — 7) = sin 3¢ + sin 3(t — 7) a(t — 7),
3e7™
L{F(t
=
sagt
3249)
15.3 to
15.4.
109
A STEP FUNCTION
i
|
Hl
ta
bebe
ane
oh!
no
By
he
BS
boteR
Me
i
i
seoeecana,
ed
%
be
omen
17. We have
Fert
(t — 4)? exp [~2(t — 4)] a(t — 4).
19. We have
si
= t— 4(t — 2) a(t
gt
52
— 2) + 3(t — 4) a(t — 4).
Therefore F(1) = 1, F(3) = 3-—4= —1, and F(5)=5~-4-.34+3 = —4.
ai. We have
oD
oo
F(t) = sint v(t, 7) = 53 (-1)" sint a(t — nr) = > sin (t — nm) a(t — nn),
n=0
L{F(t)}
_
3
© SX
ev hns
41
_
s
ned
n=)
(e78*)"
sée+l
_
1
1
s?4+1
1-—e79"
23. We first rewrite H(t) and find its transform.
H(t) = 3-3 a(t — 4) + (2¢ ~ 5) a(t — 4)
=3~—3 a(t ~ 4) + 2(t- 4) a(t - 4) +3 a(t ~ 4)
= 3+ 2(t — 4) a(t — 4)
3.
L{H(t)} =~ +
Qen48
a
Now applying the Laplace transform to the differential equation we get
#*L{a(t)} — s+ L{2()} =~ + ee
?
8
3
2e~4
L{e(t)} = erie s(s? +1) + 2(s? + 1)
_
Qs
3
-g5ti+(S
1
1
wri)
The inverse transform now yields
a(t) = 3—2cost + 2[(t — 4) — sin (¢ — 4)] a(t — 4).
~dAg
‘
110
CHAPTER
15.
INVERSE TRANSFORMS
25. x” + 42 = M(t); 2(0) = 0, z'(0) = 0. The Laplace transform of M (t) can be determined to
1
Ons
be L{M (t)} = yi
SoTT
We now apply the transform to the differential equation to
obtain
2
1
ae
euens
ule) + 4u(s) = Sy eT
1
en ana
u(s) = (s?+1)(s?+4)
~5
3
7 (8? + 1)(s? + 4)
i
i
3
1i—2xs
a
“ssl
a&
+4
etl
1-296
ge
+t ya?
a(t) = gsint — }sin2t — + sin (¢ ~ 2%) a(t
~ 27) + 2 sin
2(t — 2x) a(t
~ 2x)
= gsint ~ qsin2s ~ g sing a(t 2n) + g sin2t a(t ~ 2)
= §[1 — a(t ~ 2n)](2sint ~ sin 2t).
27. v(t) + 22’(t) + x(t)= 2+ (t~3) a(t ~3); 2(0)= 2, x'(0)= 1. We use the Laplace transform
on both sides of the differential equation and get
s°u(s) ~ 2s — 1+ 2su(s) —4+ u(s) = 32, + e438
3
e738
s*(s +1)?
u(s} =
i
a
2,1,
“8
2s? +58+2
s(s +1)?
2
etsy
1
2
(s +1)?
We obtain x(t) = [-2 + (t —3) + 2e7F-9) 4. (¢ - 3)e~t-3)]
s'
1
Hit
a(t -3)+2+te7'.
a(1) =2+e7) and 2(4) = [-2+142e71 + e~1]) +24 4e74 =1+43e714 de~4,
15.5
A Convolution
Theorem
t
LL if (t ~ B) sin 38as} = L{t} - L{sin3t} = ery:
3. 1f ['a- aye ap} = L{t3}-L{e (=a
5. L
af
i f[e
-[
twra}-
~26
di ag.
=
|—t--28|
5°
Ge
_ 1-2
a1
5e
.
We now have
15.6.
111
SPECIAL INTEGRAL EQUATIONS
7. We use the convolution theorem to obtain
L
a
t
1
: si} = [ sin sin (t ~ 8) dB = -5
WEED
1
t
1
ee
[cost — cos (26 ~ t)] dB
i
1
=-5 cost [B]y — 7 [sin (28 — t)]y = ~ zt cost
1,
.
~ qisint +sint)
= -5(t cost — sint).
9. We use the Laplace transform to solve the initial value problem
y — Ky = H(t); y(0) =0, y(0) = 0.
s*u(s) — k*u(s) = L{A(t)},
us) =
L{H(t)} =_
i L{H()} . L {sinh kt}.
t
An inverse transform yields y{t) = : [
6
A(t —BP)sinhké dg.
11. We use the Laplace transform to solve the initial problem
w" + 6a’ +92 = F(t); x(0) = A, 2'(0) = B.
s*u(s) — sA ~ B + 6su(s) — 6A + 9u(s) = L{F(t)},
u(s) = L{F(t)}+sA+6A+B
(s + 3)?
_ L{F(t)}
A(s+3)
(s+3)? © (+3)?
3A+B
9 (s+3)?"
t
An inverse transform yields x(t) = e~% [A+(3A+ B)t} + [
15.6
0
Be
9 F(t — B) dB..
Special Integral Equations
t
1. F(t} =14+2 [
F(t ~ B)e~*? dB. We apply the Laplace transform to the integral equation,
0
solve for u(s), and find the inverse transform.
_i
2u(s)
u(s) = 8 + s+2°
1
2
u(s)= = +a:
F(t)
= 1+ 2t.
Verification of this solution is straightforward.
,
112
CHAPTER
15.
INVERSE TRANSFORMS
t
3. F(t) = 1+ [
F(t— @)e~* df. We apply the Laplace transform to the integral equation, solve
for u(s), and find the inverse transform.
0
u(s
us) = 3+
yt
50,
F(t) = se +t.
Verification of this solution is straightforward and tedious.
t
. F(t)=t8+ [
F°() sin (t — 8) dB. We apply the Laplace transform to the integral equation,
solve for u(s), and find the inverse transform.
a
6
u(s)
us) = a+
ayy
6
6
u(s) = wt
3,
Fit)=t
ee
+ 55°
Verification of this solution is straightforward and tedious.
t
. F(t) =t?-2 [ F(t— G) sinh 2 d@. We apply the Laplace transform to the integral equation,
solve for u(s), and find the inverse transform.
= 5-3)
u(s)
=
Fz
s
1
F(t)=t? ~~
Verification of this solution is straightforward and tedious.
t
. H(t) = 9e"*~2 | H(t—@)cos@ df. We apply the Laplace transform to the integral equation,
0
solve for u(s), and find the inverse transform.
M8) =
u(s) =
9
2su(s)
S97 P41
9s* +9
_
5
= (-Distly s-2
H(t) = 5e* + 4e~* — 6te~*.
+
4
s4l
Verification of this solution is straightforward and tedious.
_6
(s +1)?’
15.6.
SPECIAL INTEGRAL EQUATIONS
113
ae
ll. g(z) =e"
7-2 [
g(@) cos (x ~ 8) dB. We apply the Laplace transform to the integral equa0
tion, solve for u({s), and find the inverse transform.
_
us) =
i
4
2su(s)
st? +1’
u(s) =
s?+1
1
2
2
(+i
sai
(ti
an
g(x) = e7* — Qae~* + w*e"* = e (1 — 2)”.
Verification of this solution is straightforward and tedious.
t
13. F’(t) = t+ /
F(t~ 8)cos @ dG, F(Q) = 4. We apply the Laplace transform to the equation,
0
solve for u(s), and find the inverse transform.
sul) -4=
1
suls
a 4
et,
_ (8? +148?
+1)
u(s) =
38
F(t) = 4+
Bt?
>
4,5
1
=ctate
et
+ 5y:
Verification of this solution is straightforward and tedious.
t
15. Given F(t) = t+ [
Q
F(t ~ B)e~* dB, we multiply both sides of the equation by e* to obtain
t
' ef F(t) = te? +f
0
F(t — pye’* ag.
In the integral we let t- G = y to get
e' F(t) = te’ — |
0
t
F(pje" du = tet +f
t
0
e® F(R) df.
Differentiating both sides of this equation with respect to t yields
eF'(t) +e F(t) =te'+e' + e' F(t),
Pt) =t+1,
F(t) = st +tt+e
But F(0) = 0, so that c = 0, and finally F(t) = t + 42?.
114
CHAPTER
15.8
The
Deflection
15.
INVERSE TRANSFORMS
of Beams
1, This problem is the same as the problem in Example 15.22 except for the second pair of
boundary conditions. Here y(2c) = 0 and y’(2c) = 0. Equations (11) and (12), page 309, are
still valid here. We substitute < = 2c into each of these equations to obtain
= 5 A(4e?) + = B88) + + ze [Be(16e4}— 320° + 5],
i502
0 = A(2c)
+ =Bsc?) + 7, de(8e*) — 16c8 +e'}.
Solving these equations for A and 8 yields A = Fe wc?
and B=
~Zwoe.
Equation (11),
page 309, now becomes
2%
5
3
Ely(x) = Jeg Woe F 2 ~ Gowoct 54
5
[Sex
~az° +(x —c)® a(x —c)}.
3. The boundary value problem to be solved is
EIT4 = wolt ~ a(x - 0]; u(0) = 0, ¥(0) =0, y(2c) =0, y"(2e) =0
di
To solve this problem we use the same notations as in Example 15.22. Applying the Laplace
transform to the differential equation and solving for u(s) yields
stu(s) ~ 68-0
1
6?-0- 9: A~ B= u9(2a
ue)
=
eS
; )
8B
$+
1
5
+m
ee
(5-<).
It follows that
Ely(a) = 5 An? + - Be" + ar [x* ~ ( ~c)4 a(x —c)],
Ely'(x) = Ax + 5B
+ = > [2° — (x — 0)? a(x ~ e)],
Ely'(r) = A+ Bart > [2* — (x -—c)? a(x -c)].
Using the boundary conditions at x = 2c yields
0 = 2Ac* + 5Be
2c
tw
[25
cA
- 5)
2
O=A+2Be+ wp [2e?- $].
9
Solving this system for A and B gives us A =
2
Wor
and B= —
Ely(x) = gg9 woeang
2" ~ 19
Fag woos * + ixy uo[x" _ tn
(x
57wac
. Finally we have
—c)* a(x -e)].
15.9.
SYSTEMS OF EQUATIONS
15.9
115
Systems of Equations
1. Applying the Laplace transform directly gives
:
14
s*u(s) — 3su(s) ~ su(s) + 6.5 + 2u(s) = 3 + <
su(s} — 3u(s) + sv(s} - 6.5 = 7
or
(s? — 3s)u(s) - (s — 2)u(s) = = +=3 - 65,
(s — 3)u(s) + sv(s) = : + 6.5.
Solving for u({s) and v(s) gives us
a
—9+12
2
1721
*7 = 3(s—ij(s—3)(s+2)8
v(s) =
8-1
6.539 +7.587-3s-14
5
Sis-l1)\(sed)
so?
st+2
7
8-3’
1,
5/2
sol
342"
The inverse transform now yields
a(t) = 2— =e! ~ et
se
5
y(t) =5+7t~e' + 3°
3. Applying the Laplace transform directly gives
su(s) ~ 3 ~ 2u(s) ~ sv(s) ~ o(s) = - : 5
2su(s) — 6 — 3u(s) + sv(s) — 3v(s) = —
or
(3 ~2)u(s) ~ (s+ 1)o(s) = E—4,
(25 ~ 3)u(s) + (s— 3)u(s) = E=2).
Solving for u(s) and v(s) gives us
3s* — 6s ~1
us)= asa)
_
—~3s¢+5
v(s) = (s—1)?(s—3)
1
2
2
1
i
sai t Gaye tye
ol
s—1
(s—-1)? s—3°
116
CHAPTER
15.
INVERSE TRANSFORMS
The inverse transform now yields
z(t) = e' + Qte’ + 2e**,
y(t) = e' — tet — e*.
. Applying the Laplace transform directly gives
s"u(s) — u(s) + 5su(s) ~ 5 = =
s*u(s) — s ~ 4u(s) — 2su(s) = -*,
or
2
5s? +1
(s* ~ 1)u(s)} + 5sv(s) =
3
2
~Qsu(s) + (s* — 4)u(s) = . 5 2
Solving for u(s) and v(s) gives us
us) =
pet
8B,
8/8
~ s%(s24+1)(s? +4)
82
g2@ 41° 9244"
v(s) =
st +797 +4
ai,
Zs _ 2s
~ s(s?+1)(s2+4)
s
s?+1
52447
The inverse transform now yields
5,
4.
a(t) = ~-t —~ 3 sint + 3 sin 2t,
2
2
y@) = 14+ 3 cost - 3 00s 2t.
. Applying the Laplace transform directly gives
s’u(s) — 1+ su(s) — v(s) = 0,
2su(s) ~ u(s) + sw(s) ~— w(s} = 0,
su(s) + 3u(s) + sv(s) — 4v(s) + 3w(s) = 0,
or
su(s) + (s — 1)v(s) = 1,
(2s — 1lju(s) + (9 — 1)w(s) = 0,
(s + 3)u(s) + (3 — 4)u(s) + 3w(s) = 0.
15.9.
SYSTEMS OF EQUATIONS
117
Solving for u(s), v(s), and w(s) gives us
i
i
us)= Go
u(s)
=
(
= 7 tec
a
1 ~ 28
i
1
1
The inverse transform now yields
z(t)=—-1+e’,
y(t)
=
—te’,
z(t) = 1 ~e' — te’.
9. We have
a
_ c18+e2 , sf(s)+g(s}
oe
rr
_ c28+e1
v(s) = I]
oe
sg(s) + f(s)
2op
+
The inverse transform now yields
£
x(t) = c, cosht + cg sinht + [
Q
{cosh BF(t — 8) + sinh BG(t — B)| df,
t
y(t) = e, sinht + eg cosht + [ [cosh GG(t — 8) + sinh BF(t — B)| af.
Q
11. Applying the Laplace transform directly gives
su(s) ~ 1 — 2u(s) = f(s),
svu(s) + 2u(s) = gfs),
or
su(s) — 2u(s) = f(s) +1,
2u(s) + su(s) = g{s).
Solving for u(s) and vfs) gives us
_ 8f(s)+s+29(s)
u(s) =
u(s)
8
sf(s)
gs +4
~ g2 44° g2 44°
st +4
sttd4 0 gt+4
_ sg(s)—2-2f(8)_ 2,
sys)
| 2g(s)
5244’
2f(s)
52 44°
CHAPTER
118
15.
INVERSE TRANSFORMS
The inverse transform now yields
t
a(t) = cos 2t + [
0
[cos 2GF(é — 8) + sin 26G(t — 8)| af,
t
y(t) = —sin 2¢+ [ [cos 28G(t — 8) — sin 26F(t — 8)] df.
6
13. Applying the Laplace transform directly gives
su(s) —c, = au(s) + bvo(s)+ E{f(t)}},
svu(s) ~ cg = cu(s) + du(s) + L{g(t)},
or
(s — a)u(s) — bu(s) = cr + L{F(t)},
—cu(s) + (s — dju(s) = cg + L{g(t)}.
Solving for u(s) and v(s) gives us
_
atLi{ft)}
—-b
gq
8b
Lift}
jeat+L{glt)}
s-dl
f2
s-—d
L{g(t)}}
u(s) = 3?—(a+d)s+ad~be
s-a
-c
@—(atdjstad—be* 8? ~(a+d)s + ad — be’
atL{f(t)}
¢2+L{g(t}}
$S-a
—c
v(s) = s?—(a+d)s+ad—be
s-—d
ey
ty
So (ads bad
s~a
~c
be *
L{f(t)}
L{g(t)}
(at ds +ad—be
We note that the first term on the right side of each equation depends on c, and ce, but is
independent of L { f(t)} and L {g(t)}. Also the second term on the right side of each equation
is independent of c, and co, but depends on L {f(t)} and L {g(t)}}. It follows that
a(t) = x(t) + p(t),
y(t) = ye(t) + p(t),
where z,(t) and y(t) depend on c; and cg whereas xp(t) and yp(t) depend on f(t) and g(t).
15. Kirchhoff’s laws require that the following system of equations be satisfied:
dl.
Ryly + Role + laF = EB,
1
Ruf + Rglg + Ges =&,
3
iy
=
dQs
I
+ ts,
“a 7
15.9.
SYSTEMS OF EQUATIONS
119
An application of the Laplace transform gives
Ryi; + Rete + Lesig
.
= E/s,
1
Ryi; + Rgizg + —-93 = E/s,
C3
4
Multiplying the second equation
allows us to write
by sC3
Ry,
=
tg + a3,
$q3
> dg.
and substituting for sg3 from the fourth equation
+ (Re + Dys)ig = E/s,
CzRy si, + (CeRas + 1)ig = Cok,
4) —ig —ig = 0.
The nature of the solutions of this system will depend on the value of the determinant
Ry
Ro t+ Les
C3Rys
0
~1
1
0
C3Rgs4+1)
om
R,
=
Ry + Ro+
(CyRy,s
1
Las
Ry
Cahis
0
Cs(Ry + Rs)s+1
0
Los + Ri + Re
C3his
= Cgle(Ry
Ry
C3(Ry
+ R3)s? + [Cs(R, Rg + RoRg + Ag Rhy) + Ly]s +R,
+ R3)s +1
+ Re.
120
CHAPTER
Chapter
16.
NONLINEAR EQUATIONS
16
Nonlinear Equations
16.2.
Factoring the Left Member
1. x*p? — y? = (xp - y)(xp + y) = 0. It follows that
ip=y
or
ap = —y,
or
Pp
pi
y 2
y= Cpe
=
_
y
x
y = C2 /a.
or
3. 2°p? ~ 5ryp + by? = (xp — 2y)(xp — 3y) = 0. It follows that
zp
= 2y
p_
2
or
y =
or
y= cx"
or
zp = dy,
3
P= ~,
yo
x
y = cor.
5. ap? + (1 —2z*y)p — zy = (xp + 1)(p — zy) = 0. It follows that
rps
—1
p=--
1
Lf
y = — In [ego
or
Peal,
or
z= In|epyl.
¥
9. (x +y)*p? —y? = [(c+y)p— y] [(z + y)p+y] = 0. It follows that
(z+ y)p-y=0
or
(g+y)pt+y =0.
16.2.
FACTORING
THE LEFT MEMBER
Each of these equations is homogeneous,
variables. The resulting equations are
de,
tide
x
v
121
so we substitute y = ux in order to separate the
_
dz
0
(w+ Ide
or
x
vinjcjzv| = 1
or
In |? (v? + 2v)] = 2In fal,
we +Qu
0,
& == yln jevy|
or
y(2z + y) = 2p.
ll. p*? — ay(z+y)p+ zy) = (p — x*y)(p ~ zy?) = 0. It follows that
p= ay
or
Ply
or
¥
2° = 3ln |egy|
p= xy",
£ = £,
¥
or
y{x? +c) = —2.
13. (a — y)*p? ~ y® = [(a - y)p— y] [(c@ — y)p + y] =O. It follows that
(c-y)p-y=0
Each
or
of these equations is homogeneous,
variables. The resulting equations are
d
x
_
"3
1
1—
vu
, Ua vde
(t~y)pt+y =0.
so we substitute y = vz in order to separate the
_
or
In Jeqyru|
2 =—ylnlery|
dz
x
(l
|
vjdv
_ 6,
2u — v2
‘
or
x?(2u ~ vu") = ca,
or
y(2e — y) = eg.
15. (2? + y?)?p? — dar?y® = [(2? + y?)p — 2ay] [(a? + y?)p + 2xy| = 0. It follows that
(x? + y*)p — 2ay = 0
Each of these equations is homogeneous,
variables. The resulting equations are
or
so we substitute y = vx in order to separate the
dx + (L+v?)du _
z
wy
(2? + y?)p + 2xy = 0.
dx | (itv)dv
°
°F
z
In |a(v? ~ 1)| = In fan
Pay
7%
or
In [z3(u3 + 3u)| = In [b},
z(v? -1)=cu
or
z3(v? + 3v) = ea,
y? — a? = oy
or
y(32? + y*) = cg.
17. xy(x? +y?)p? — (at + 27y? + y4)p — ay(a? +y*) = [(eyp — (a? + y)] [(2? +y?)pt zy] =O. It
follows that
xryp — (a? oo y*) = 0
or
(x? + y*\p + zy = 0.
CHAPTER
122
16.
NONLINEAR EQUATIONS
Each of these equations is homogeneous, so we substitute y = va in order to separate the
variables. The resulting equations are
dx
Tt dy
1
~ In lege] + 5
=0
y? = 227 In legal
or
dx
zt
(l+v")dv
wae
or
xiy?(y? +2) = ¢1,
or
y?(y? + 227) = cy.
0,
19. In order for the solution to pass through the point (1, 1) the solution must have the form
y= 4/-2(2~—c,) and 1 = /-2(1~c¢,). Therefore c, = 3/2 and y = J/3
— 2a, valid for
x < 3/2.
al, Let
F(z) = /2— 22, forz <1,
and
F'(z) =
F(z) = —Inz, for
and
F'(z) = “,
>1,
Te
for 2 <1
for z > 1,
Note that F(x) is defined for all x, but F’(x) is not defined for c = 1. We have for x < 1,
yp’ +(e+y)p+1=
af2 — 2x _ at Ve~ 2x
2-22
V2-—- 2a
+1=0.
And for ¢ > 1
typ? +(x+y)p+1=—-
zing
2-Ing
+1
= 0.
Therefore F(z)is a solution of the differential equation so long as x # 1.
23. Let
G(a) = V3 — 2a, for z <1,
and
G' (zr) = a
G(x) = 1—Ingz, for x > 1,
and
G'(z) = =,
~ 22
Note that G(x) and G’(x) are defined for all x.
for x <1,
for x > 1.
Substituting y = G and p = G’ into the
differential equation leads us to
2
_ivs—2
stWoe
v8— ee +1=0,
zyp’+(c+y)p+i=
3 oe
=
zyp? +(e+y)pti== (i-Inz)
forr <<1,
z+i-ing 4+1=0, forg>1.
x
Therefore G(x) is a solution of the differential equation for all x.
16.5.
THE P-DISCRIMINANT EQUATION
123
26. From equation (5) we seek a solution y = \/—2(x —c1) that passes through (~4, 2). We
must have 2 = ,/—2(—3 — 1) or ¢; = 3/2. Thus y = 3 — 2g, valid for x < 3/2. From the
equation (6) we seek a solution y = — In |c22| that passes through (~3, 2). This requires that
2 = —In| — 4e| or —4c2 = e~*, so that co = —2e~* andy = — In| — 2e~*x| = 2~In 2—In (—z),
valid for x <0. Each of these solutions is valid on the interval ~1 <2 < ~—j.
27. The function y = 3 ~ 2x is the only solution of the equation that is valid for x <
passes through the point (~4, 2). In order to define a function that is a solution
interval -1 <x < 2 we must find a function of the form y = ~In|cga| that passes
the point (1, 1). We require that 1 = — In |eq|; that is, cp = e~! and y = —In|e!z| =
3/2 and
on the
through
1-—Inz
for 2 > 1. We now define the function
G(z) = ¥3—22,
=l-—Inz,
forz <1,
forz
> 1.
This is the function of Figure 16.1
16.5
The p-Discriminant Equation
1. f = Ap?+
Bp+C
= 0,
ee
= 2Ap+ B=
0.
If we multiply the first equation by 2, the
second equation by p, and subtract the second from the first, we obtain the pair of equations
2Ap+B=0, Bp+2C = 0. Multiplying the first of these equations by B, the second by 2A,
and subtracting yields B? — 4AC = 0.
p? + Ap? + B = 0, —of = 3p? +2Ap = 0. If we multiply the first equation by 3, the
second by p, and subtract we obtain the pair of equations 3p* + 2Ap = 0, Ap? + 3B = 0.
f=
>
ap
Elimination of p* from these equations yields 2A°p
= 9B,
giving us the pair of equations
4A7p? — 81B? = 0, Ap? + 3B = 0. Elimination of p* from these equations gives the desired
result B(4.4° + 27B) = 0.
. xyp? +(2+y)p+1=0.
Solving for p we have
_ a~tyt V(etypP day
ps
Qry
-(e@+y)tJS(z-y)?
~
Qry
‘
Thus the condition that the equation have equal roots in p is (x — y)? = 0. But for y = x we
have p = 1 so that ryp? + (x+y)p+1=
27 4+22+10.
differential equation. There are no singular solutions.
Hence y = 2 is not a solution of the
The function has as its graph the left half of the semicircle y = a? — (x +-2a)*] V2 and the right
half of the semicircle y = (a? ~(£— 2a)2]*/?,
connected by the line segment y = a between
x = —2a and x = 2a. Note that at the points where these arcs are joined, the slopes of the
arcs are both zero, so that the given function is a solution for the entire interval ~3a < xz < 3a.
CHAPTER
124
16.
NONLINEAR EQUATIONS
3a4p? — ap —y = 0. From Exercise 1 the p-discriminant equation is
B? —4AC = 2? + 12a4y = 27(1 4 1227y) = 0.
~1
1
Substituting this function into the differential
with p= és
Consider the function y = 1x2
3a
ik
1
equation gives 3a4p? —xp—y = me -é3 + jo
= 0. That is, one solution of the differential
equation is given by 1227y = —1.
il. p2 -~zp+y
= 0. The p-discriminant equation is B? ~ 4AC = 2* — 4y = 0. For y = 27/4,
p= 2/2 and p? —zp+y=
ge
gt
777 + 7
equation.
0. Thus y = z*/4 is a solution of the differential
°
13. 4y3p? — dap + y = 0. The p-discriminant equation is B? — 4AC = 16(a — y*)(x + y?) = 0.
Rather than testing each of the four parabolic arcs y = JZ, y = —J/Z, y = /—a, and
y = —/—2 individually to see if they represent solutions of the differential equation, we note
that yt = 22 implies that 2yp = x. Multiplying the differential equation by y* and checking
4y®p? — day?p + y' = 2? ~ 2x? + 2? = 0, we see that each of the four functions represents a
solution of the differential equation.
15.
y2p? — 6xp + 2y = 0. In order to use the formula given in Exercise 2, we need to divide first
by y?. We must therefore check to see if y = 0 is itself a solution, and indeed it is. Then the
ifferential equation
becomes p” ~
ye + yn
108
4A° 427B = ye
ow the p-discriminant equation is given by
— 2x)(yt + Qxy? + 427) = 0.
For the parabola y? = 2x we have yp = 1. If we multiply the differential equation by y we
may then write for the parabola y5p? — 6xyp + 2y? = 2x — 62 + 4z = 0. Thus both y = V2z
and y = —\/2r are solutions of the differential equation.
17. The quadratic equation c? + exy + 4a = 0 has equal roots when xy? ~ 16x = 0. That is, when
zy” = 16. This the same as the singular solution of the differential equation.
16.7
Clairaut’s Equation
1. Here we have
t= —f'(a)
dx = ~ f(a) da
It follows that p =
and
and
y= fla) ~af'(a),
dy = —af" (a)
da.
dy ne and
y — pr — f(p) = f(a) - af'(a) + af'(a) ~ f(a) = 0.
That is, the original equations are the parametric form of a solution of the differential equation
y = px + f(p).
16.7.
CLAIRAUT’S EQUATION
125
. p? + 23p — 2a*y = 0. Dividing by c* and differentiating with respect to z eliminates y and
.
yields x?(2p + 2)
.
d
= px(2p + 2°). Thus ze
=p or p= ~—z7/2. We solve this differential
equation and obtain p = cx. Substituting p = cx into the original differential equation yields
2y = c* +cx*, Substituting p = —c3/2 into the original differential equation yields 8y = —24.
. 2ap? — 6yp? + 24 = 0. Dividing by p* and differentiating with respect to z eliminates y and
d
yields x(p* — oe
= 2p(p? — x*). Thus « P= 2p or p = xz. We solve this differential
dx
equation and obtain p = cx*. Substituting p = cr? into the original differential equation yields
2chx* = 1 ~ 6c?y, Substituting p = x into the original differential equation yields 2y = 2”.
.y = px + kp*. This is a Clairaut equation with f(p) = kp* and f'(p) = 2kp. The general
solution is y = cx + ke*, and the singular solution is given parametrically by 2 = ~2¢p,
= —kp*. Eliminating p from these equations gives us x? = —4ky.
. t4p? + 223yp — 4 = 0. Dividing by xy and differentiating with respect to x eliminates y and
yields —o(x'p? +4)
= 3p(z4p* +4). Since x4p?+4 cannot be zero we have of +3p =0. We
solve this differential equation and obtain p = k/x*. Substituting p = k/x* into the original
differential equation yields 1?(1 + cy) = c?, where c = ~k/2. There is no singular solution.
il. 3x4p? — xp —y
= 0.
Differentiating both sides
ad
with respect to x eliminates y and results in
2(62°p— ye +2p(62%p—1) = 0. Thus o +2p = Oor p = 1/(6xr°). We solve this differential
equation and obtain p = c/x?. Substituting p = c/x® into the original differential equation
yields cy = c(3cr — 1). Substituting p = 1/(62°) into the original differential equation yields
122?y = ~1.
12. p(zp-y+k)+a
= 0. We can write this equation in the Clairaut form y = xp +k + a/p.
The general solution is y = cr + k +a/c or cece
~-y+k)+a=0. Here f(p) = k+a/p and
f'(p) = —a/p*. Parametric equations for the singular solution are x = a/p?, y = k + 2a/p. It
follows that (y — k)* = daz.
15.
= g°p3 — xp,
Differentiating
equation 2(32°p? — N=
with respect to xz eliminates y and gives us the differential
+ 2p(3a°p? —1) = 0. Thus aoe
solve this differential equation and obtain p = c/x?.
differential equation gives us cy = c(c*x—1).
+ 2p = 0 or p? = 1/(325).
We
Substituting p = c/a? into the original
The original differential equation may be replaced
by y? = 21p® — 2x7pt + x*p*. Substituting p = 1/(325) into this differential equation yields
2723y? = 4,
17. zp® — yp* +1
= 0.
We can write this equation in the Clairaut form y = pr + +1/p”.
The
general solution is y = cx + 1/c? or c*y = clx +1. Parametric equations for the singular
1
solution are z = -—5, y= ~5 + — = —. It follows that 4,9 = 272°.
p
19. p? —~xp—-y = 0.
pp
po
Differentiating both sides with respect to x eliminates y and gives us the
differential equation (2p — a)
d
= 2p. We solve this homogeneous differential equation and
CHAPTER
126
16.
NONLINEAR EQUATIONS
obtain 32 = 2p+kp'-!/2), Substituting this value for 3z into the original differential equation
yields a second equation 3y = p? — kp!/?, The last two equations give the general solution in
terms of the parameter p.
al. 2p? + xp — 2y = 0.
Differentiating
the differential equation dp
both sides with respect to x eliminates y and gives us
l, =z 4, We solve this linear differential equation and obtain
xz = 4pln|pcl. Substituting this value for z into the original differential equation yields a
second equation y = p* (1 +2in [pel]. The last two equations give the general solution in terms
of the parameter p.
23. 4rp* — 3yp +3 = 0. Dividing by p and differentiating both sides with respect to z eliminates y
and gives us the differential equation é +-£= 3.
We solve this linear differential equation
and obtain 22 = 3p7* + ep74. Substituting this value for 2x into the original differential
equation yields a second equation 3y = 9p"! + 2ep~3. The last two equations give the general
solution in terms of the parameter p.
25. 5p? + 6ap — 2y = 0. Differentiating both sides with respect to x eliminates y and gives us the
3
5
differential equation o + ap" = 5
We solve this linear differential equation and obtain
x = —p+kp'-3/2)_
The form of this solution may be changed to get p*(x + p)? = c, which,
when taken with the original differential equation, gives parametric equations for the solution.
27. 5p? + 32p -y = 0. Differentiating both sides with respect to 2 eliminates y and gives us the
differential equation
o + =
«x —§.
We solve this linear differential equation and obtain
a = —2p + kp(-3/2), The form of this solution may be changed to get p(x + 2p)? = c, which,
when taken with the original differential equation, gives parametric equations for the solution.
29. y = xp + x%p*. Differentiating both sides of this differential equation eliminates y and yields
+22%p +1 = 0. Substituting v = x? changes this equation into ob + = = - 3p.
aap
2
We solve this linear differential equation and obtain v = 2? = ep'~4/5) — 2p~1, which, when
taken with the original differential equation, gives parametric equations for the solution.
16.9
Independent
Variable Missing
1. y’ = z(y’)°. Since y does not appear explicitly we set p = y’ and get a differential equation
we must solve for p. Then an integration yields y. We get
ap
aoe
yea
and
adz
= + aresin (=)
3
5
and
de
= fy
ha
#1
xz = cy sin (y + cg).
16.9.
INDEPENDENT
VARIABLE MISSING
127
Since x does not appear explicitly we set p = y/ and yf = P=dp to get a
. yy” + (y’)? = 0.
differential equation we must. solve for p. Then an integration yields y. We get
dp | dy
+
Py
=0,
z= cy— In fegyl.
an
dy
=e
Pde
d
y
ey- 1?
_—
. 2ay”+(y’)? = 0. Since y does not appear explicitly we set p = y' and get a differential equation
we must solve for p. Then an integration yields y. We get
dp
_
205 +dzx = 0,
and
y = t2J/a/z— 1 +c,
and
_dy
da
+a
Jena’
(y — ¢2)* = 4a(x — ey).
" s= Qy(y')?. Since z does not appear explicitly we set p = y! and yf? = pe
differential equation we must solve for p. Then an integration yields y. We get
dy
~3
=
2ydy,
and
to get a
1
de
Pte?
y° = 3(cg - x ~ cry).
. yy” + (y')? = 0. Since x does not appear explicitly we set p = y! and y= =edp to get a
differential equation we must solve for p. Then an integration yields y. We get
zt
Poy
=),
ay
and
— =
1
PS de ~ inleayl’
f= ec, +yln leoy| —
it. a3y" — x?y' = 3 — x? Since y does not appear explicitly we set p=y' and get a differential
equation we must solve for p. Then an integration yields y. We get
po~
=
y=
a!
y=
+e+e27
and
pa
a
a=
~e
+l teas,
+c.
13. y” = e*(y')*. Since y does not appear explicitly we set p = y' and get a differential equation
we must solve for p. Then an integration yields y. We get
dp,
Es
p*
ede
1
y= ~< In|ee™* — 1] ~ eg,
1
an
d
and
= dy
dx
1
cy~e®
e*
ae
cen — 1’
cy
+ eg = —Injeye7?
— 1].
CHAPTER
128
16.
NONLINEAR EQUATIONS
15. y” =14(y’)*. Since y does not appear explicitly we set p = y’ and get a differential equation
we must solve for p. Then an integration yields y. We get
dp
p=”dy = tan(z+c1),
and
itp
e” cos (x + c1) = €.
and
y == ~ In| cos (x + ¢1)| + Incsg,
17. (1+ y7)y” + (y’')8 + y’ = 0. Since x does not appear explicitly we set p = y' and y” = pe
get a differential equation we must solve for p. Then an integration yields y. We get
y+e1
dy
dy
dp
m1
pe+1
= 0,
yeti
and
az
dr
to
’
cyy-i
zg = eg +ey — (1 +c?) nly + eal.
19. zy” = y'(2— 3zy’). Since y does not appear explicitly we set p = y/ and get a differential
equation we must solve for p. Then an integration yields y. We get
x2
dy SEED armenia
.
2
2p
dp
Bre
3
P= ee eentone
and
oe omen
anne
yp” (Bernoulli)
ine
3y = In {a? + c1| + €2.
ai. (y”)? — ay" + y' = 0. We set w = y’, w’ = p, and get w = px — p*, a Clairaut equation. The
general solution is w = cix — c?, so that y = $cyx? — cfx +3 or 2y = oz? — 2c?x + cg. In
the Clairaut equation f(p) = —p? and f’(p) = —2p. The parametric equations of the singular
solution are « = 2p and w = ~p? + 2p’ = p*. Eliminating the parameter p gives us w = 52°.
Tt follows that y = Be? +4 or 12y = 23 +k is a family of singular solutions.
23. 3yy'y" = (y’)8 — 1. Since 2 does not appear explicitly we set p = y’ and y=
differential equation we must solve for p. Then an integration yields y. We get
dp _EE dy
ap?ee
_
and
p-lsy
dy
Pin
(y+ eo) 1/8
als
d
Poe
to get a
,
2Tex(y + 1)? = B(x + e2)?.
25. z?y" + (y')? —22y’ = 0; y(2) = 5, y'(2) = —4. Since y does not appear explicitly we set p = y’
to get a differential equation we must solve for p. We then use an initial condition to obtain
¢;. Then an integration yields y. Finally, we determine the value of cp. We have
z2
2p
dp
i
z
ec
p(2)Q)=
—4
p ed
a
.
an
(Bernoulli),
gi
us
gives
c, =a=
~3
~3,
+ 32 + 91n |x — 3| +2,
+ 3a + 9lnix
- 3] —3.
d
so
but
that
p=
voces
_ fy
=3,=
y{2) = 5,
a
'
;
z=
sothat
p43
¢
cg = —3;
oe:3
16.9.
INDEPENDENT VARIABLE MISSING
27, ay” =y'+2°;
129
y(1) = 4, y(1) = 1. Since y does not appear explicitly we set p = y' to get a
differential equation we must solve for p. We then use an initial condition to obtain c,. Then
an integration yields y. Finally, we determine the value of cg. We have
PP
apd
d
oat,
c= 7:3
p(l}=1 gives us
teas;
and
pa
so that
papa
yest
cS
but
te,
3x?
+ RD
+
oS 24
:
i
2
mG
3x73
«®
dy
1
C2 = 53
so that
y(l) = 5,
24y = o® + O27 + 2.
io
;
ta}
29. y" if + Bry
= 0. Since
x does not appear explicitly
we set p = y’ f and y” a = Pedp to get a
xe
differential equation we must solve for p. Then an integration yields y. We get
di
— 62y?,
z= = +4/c2
pdp + By dy = 0,
and
p=
arcsin By
and
y = ¢, sin O(@ + co).
cg
= D(x +),
31. y” = x(y’)?; y(0) = 1, y’(0) = 4. Since y does not appear explicitly we set p = y' to get a
differential equation we must solve for p. We then use an initial condition to obtain c;. Then
an integration yields y. Finally, we determine the value of cp. We have
op = ede,
cy = —4,
»(0) = 5
gives us
y=5in
ete ton
and
Po ao
so that
pay
but
y(0)
= 1,
- 28,
= 74
so that
c.=1;
ate.
y=t+5in
33. y” = ~—e7*¥; y(3) = 0, y(3) = —1. Since x does not appear explicitly we set p = y’ and
yf = Poe to get a differential equation we must solve for p. We then use an initia! condition
to obtain c,. Then an integration yields y. Finally, we determine the value of co. We have
pdp=
p(3)
—e7 2
=]
and
dy,
gives
us
¢) = 0,
so that
ex eo
ra
_
+04;
env:
sothat
ev = —r +e,
but
y(3) =—-1,
ev =4d—ax,
or
y= iIn(4~ 2).
cp = 4;
CHAPTER
130
35. 2y" = sin2y;
y= PT
y(0) = —2/2,
16.
NONLINEAR EQUATIONS
y’(0) = 1. Since 2 does not appear explicitly we set p = y’ and
to get a differential equation we must solve for p. We then use an initial condition
to obtain c;. Then an integration yields y. Finally, we determine the value of cp. We have
4pdp = 2sin 2y dy,
p(0) = 1 and y(0) = —7/2
gives us
c; = 1,
—Injescy+coty| = —z +e,
and
Qp* = — cos 2y + €4;
so that
p=~siny;
but
y(0) = —72/2,
so that c2 = 0;
a = In(— cscy — cot y).
37. 2y” = (y')' sin 2z;y(0) = 1, y/(0) = 1. Since y does not appear explicitly we set p = y’ to
get a differential equation we must solve for p. We then use an initial condition to obtain ¢.
Then an integration yields y, Finally, we determine the value of cg. We have
24p
Pp
= sin 2x dz,
oe
p(0) =1
and
‘
1
gives us
—ps
h
c, = at
y = In] seca + tanz| + ce,
2
008 2
2
Tiwi
so that
p=
but
y(0) =1,
+01;
2
=— sec’ 2 x;ns
so that
c.=1;
y= 1+ In(secx
+ tan 2).
39. gy” = Qy
d:
yl! = pe
[1
~ y' siny — yy’ cos y}.
Since z does not appear explicitly we set p = y' and
P
.
.
.
to get a differential equation we must solve for p.
Then an integration yields y. We
get
dpdz _ P_aa
ay
y
y
(Bernoulli),
and
p= 42
—¥__.,
dx
-ysiny
+ ¢
x= cj In|legy| — cosy.
41. gy” = y'(22 —y’); y(-1) = 5, y’(-1) = 1. Since y does not appear explicitly we set p = y!
to get a differential equation we must solve for p. We then use an initial condition to obtain
cy. Then an integration yields y. Finally, we determine the value of cz. We have
dp
ne
2p
Ee
oo
2p"e
.
(Bernoulli)
an d
p(-1)
1) ==11 gives
gives us cy ==2 2,
so
2
y= S~ 2a t4ln|a +2] ter,
but
y= x?
yet
Alnie +2 +5,5
that
a?
p = tia ;
P=ogo,
Tyo"
pho
y(—1) =5,
so that cp = 5/2;
(Qy — 1) = (a — 2) 2 + 8 in|
+ 2].
43. (y")? — 2y” + (y’)? — 2ay’ + a = 0; y(0) = §, y’(0) = 1. Since y does not appear explicitly
we set p = y/ and get (p’)? — 2p’ + p? — 2zp +z? = 0. This equation may be rewritten as
16.9.
INDEPENDENT
VARIABLE MISSING
131
(p' - 1)” +(p—a)* = 1. Here we observe that (p — x)'= p'—1. This observation suggests
that we let g = p— zx. This substitution gives us q! = 41/1 ~ q? and q = sin(x +c ,). Thus
p= x+sin(z+c,). The condition p(0)= 1 now forces us to choose c) = a/2, so that
2
p=z+cosz, and y = > +sing + ey. The condition y(0) = 3 now requires that co = i, and
finally 2y = 14+ 2° + 2sinz.
Miscellaneous
Exercises
1. 23p? + 2%yp+4 = 0. Dividing both sides of this equation by xp and differentiating both sides
with respect to x eliminates y and gives x(2°p* — ae + 2p(a3p? ~ 4)= 0. Thus oe +2p=0
or z°p* = 4. We solve this differential equation and obtain p = c/2?. Substituting p= efx?
into the original differential equation gives c? + czy + 4z = 0. Substituting x3p? = 4 into the
original differential equation gives p = Fay" Eliminating p from the last two equations gives
us the singular solution ry? = 16.
3. 9p?+3ay4pty? = 0. Dividing both sides of this equation by 3y*p and differentiating both sides
with respect to x eliminates x and gives 3y*(y® ~ gp?) <2 — 12y%p(y° — 9p?) = 0. Remember
dy
that o
=
2.
obtain p = ky
Thus 3y ‘e
— 12y*p = 0 or Gp* = y>. We solve this differential equation and
Substituting p = ky’ into the original differential equation gives ceyi(x—c) = 1,
where c = —3k.
Substituting 9p?= y° into the original differential equation
Eliminating p from the last two equations gives us the singular solution 221°
5. 5p? — Qap~ dy = 0. ap
y and gives a(z°p — ye
gives p = ~.
= 4.
eatins both sides of this equation with respect to x eliminates
+ 3p(z°p ~ 1) = 0. Thus oo
2e
differential equation and obtain p = a
+ 3p =0orp=
2e
Substituting p = “3
275.
We solve this
into the original differential
equation yields x*(y ~ c*) = e. Substituting p = 2~® into the original differential equation
yields 4aty = —1.
7. 5p? +-6xp - 2y = 0. Differentiating both sides of this equation with respect to x eliminates y
and gives (10p + 60) 2 +4p = 0, We solve this homogeneous differential equation and obtain
« = cp~4/? — p, Substituting into the original differential equation yields 2y = 6ep~1/? — p?
The
last two equations
are parametric
equations of the solution of the original differential
equation.
9. 4z°p? + 1224yp + 9 = 0.
Dividing both sides of this equation by z‘p and differentiating
both sides with respect to x eliminates y and gives x(4a°p? ~— ns
Thus of
+4p
= 0 or 4x°p*
= 9.
+ 4p(4a°p? — 9) = 0.
We solve this differential equation and obtain p =
k
af
CHAPTER
132
16.
NONLINEAR EQUATIONS
k
Substituting p = yA into the original differential equation yields 2°(2cy — 1) = c*, where
c= ~2k/3. Substituting 42°p* = 9 into the original differential equation gives us p = - 55
Eliminating p from the last two equations yields xy? = 1.
il. p'+ap—3y = 0. Differentiating both sides of this equation with respect to x eliminates y and
gives z
_
= 2p”. We solve this linear differential equation and obtain 5x = 4p? + cp!/?.
Substituting into the original differential equation yields 15y = 9p + cp/?.
The last two
equations are parametric equations of the solution of the original differential equation.
13. zp) — Qryp? + yp +1 = 0. Dividing by p and completing the square allows us to rewrite
this equation as (xp — y)* + — = 0. We note that p must be negative, so that we can factor
1
the left hand side to get (2p —yt+ = (2
V~P
1
y = px ~ —z==.
v~P
.
i
=)
V-P
= Q. Thus y = pr+
:
1
or
vp
:
‘
Hach of these equations is of Clairaut type so that their general solutions
i
are y = cx + Ta and y = cz — —==.
—c
ye
These two solutions can be combined to give a single
equation in the form z7c3 — 2ryet +yc+1=
f=
-—yr-
(—p)~*/? and f'(p) = 5(-p)
0. For the first Clairaut differential equation
2 so that equations for the singular solution are given
53
parametrically by 2 = ~3(-P) ~3/2 and y = =(—p)7¥/ *| Eliminating the parameter p from
1
these equations gives 27x = —4y?.
16.
zp? ~ (x? + 1)p +a = 0. Here it is possible to factor the equation into (xp — 1)(p — x) = 0.
The first. factor yields the solution y = In jcgz|, the second y = a7
+ cy.
17. p® ~2x2p~-y = 0. Differentiating both sides of this equation with respect to x eliminates y and
2x
gives ap + 3p == p. We solve this linear differential equation and obtain 8% = 3p? + ep
2/3,
Substituting into the original differential equation yields 4y = p>—cp'/3. The last two equations
are parametric equations of the solution of the original differential equation.
19. xp? —
as (zp
y = px
are y =
(Qry + 1)p+ y? +1
= 0.
Completing the square allows us to rewrite this equation
~ y)* = p—1. We note that p ~ 1 must be positive, so that y = px + \/p~1 or
~— \/p—1. Each of these equations is of Clairaut type so that their general solutions
cxt+ Vc — land y = cxn—vVc— 1. These two solutions can be combined to give a single
equation in the form zc? ~ (2Qay+1)c+y* +1 = 0. For the first Clairaut differential equation
f(p) = (p— 1)? and f'(p) = 5° — 1)~'/?, so that equations for the singular solution are
1
given parametrically by x = -5( — 17M? and y = (p— 1)? — ah(p ~1)7'/?,
the parameter p from these equations gives 4x7 — day — 1 = 0.
Eliminating
16.9.
INDEPENDENT VARIABLE MISSING
133
21. t*p* = (c—y)*. The equation factors and yields two differential equations that may be written
dy
1
- +o¥
dy
1
= 1 and a. z=
;
2
~1. The solution of the first of these linear equations is xy = x +e
which may be rewritten as z(x ~ 2y) = c,. The second differential equation has the solution
y = —xIn leer.
23. p? — p* + ap —y = 0, This equation may be rewritten y = px + p* — p? and recognized as
a Clairaut equation.
The general solution is y = cx +c? —c?.
Here f(p) = p3 — p? and
f'(p) = 3p? — 2p, so that we obtain 2 = —3p? + 2p and y = —2p3 + p’ as parametric equations
for the singular solution.
25. yp? —(x+y)p+y
Dividing both sides of this equation by p and differentiating both
d
sides with respect to z eliminates y and gives y(p — 1)(p + Na + p*(p ~ 1) = 0. Remember
that e
= re
= 0.
d,
Thus y({p + Na
+p?
= 0 or p = 1.
We solve this differential equation
and obtain py = cexp(—1/p). This equation taken with the original differential equation
px = y(p? ~ p+ 1) gives the solution of the original equation in parametric form. Substituting
p = 1 into the original differential equation gives the singular solution y = zx.
27.
ap? — 2yp? + 4a? = 0. Dividing both sides of this equation by p* and differentiating both
sides with respect to x eliminates y and gives x (p* — 82) e = p(p* — 8x). Thus oe = por
p = 22/3, We solve this differential equation and obtain p= kz. Substituting p = kr into the
original differential equation yields x? = 4c(y~ 8c”), where c = 1/(2k). Substituting p = 2a!/3
into the original differential equation yields 8y? = 2727.
134
CHAPTER
Chapter
Power
17.5
17.
POWER SERIES SOLUTIONS
17
Series
Solutions
Solutions Near an Ordinary Point
1. y” +y
= 0. The elementary solution is y = c; cosz + sing. Substituting y = s- nx",
we
n=O
oO
oO
obtain Ss; n(n ~ l)aga”~? + S. a,2" = §. Replacing n by n + 2 in the first series gives us
ned
oo
xO
ne
> _(n +2){n+ lenzex” + s a,x" = 0. Combining these series and equating each coefficient
nD
n=O
to zero yields the recurrence relation an42 = weHeey’
and n=
for n > 0. For n = 0,2,--- ,2k
1,3,---,2k
— 1 we have
==
0
on od
4"
Gop
ok
4-3
=
mt?
1
3.2
pn
a
2k(2k — 1)
\k
It follows that ag, =“aa
2 HL
3
~]
and
@apa1 =
B4
ee
ak= 1
2k+1 ™ (Qk + 1k)
k
eo
for k > 1 with ao and a, arbitrary
constants. The original series becomes
)Fgthtt
yay
ee
[+a
[erp
Geer
oO
3. y + 3xy’ + 3y = 0.
ie. 4]
s; n(n — Lanz"? +3
2
oo
So [(n + 2)(n + Lange + 3(n + Lap] a” = 0.
neat
| =socore + ysns
oO
> NynZ” +3 S| Gn2” = 0,
nee
neo
17.5.
SOLUTIONS NEAR AN ORDINARY POINT
The recurrence relation is a,,2 = apa”
Go
=
135
for n > 0.
—~3ag
aa
3=
=
—3a;
as
=
—3ag
2
_
—3ag
a4 =
nn
a
doe = ~dagn.2
2k
3\Fx 2k
+s
Py
(- sya Qh+1
Ok 41)
x
5. (1 — 4r*)y" + 8y = 0.
(~1)*3*a,
+
2k kl
oa]
ae
(~
8 etk=1
ake = SE
doeei =
aa, = (~1)*3*ag
y¥ = ao pay
3
B..-(2k4+1)°
; valid for all finite z.
oo
oo
Sonn ~ Lanz”? — > 4n(n - lana” + s- 8a,x"
Texd
oo
n=Q
n=O
[(n + 2)(n + l)@nge — 4(n — 2)(n + Dey] 2" = 0. The recurrence relation is
n~ 2) Qan
mln
forn
> 0.
On+42 = a nah
forn>0
ag =
—4ap
=
=
PO
hand
+ by
k=nQ
2
g2k
We
_ 4F(-1)-1-3++-(2k
~ Bay
math
ay, = 0 for k > 2
3
— 42°)
a5
a
Qo, = Ofork
>2
y = ag{1
4s . (Yar
Che
4(2k — 3jaay—
Q2k+1 = ACK — S)azi—a
ie
2
aa, = O for k > 2
>
-— 42°) +a)
=
4-la
aq =0
y = ap{1
ag
Q2k41 = Wri
=F fork > 1.
Atk tk+h
ET
ak+t
1L
3-5-.: (2k +1)
. Replace a; with —b, to get
1
; valid for |z| < ry
=: Q,
136
CHAPTER
17.
os
7. (14 27)y" + l0zy! + 20y = 0.
POWER SERIES SOLUTIONS
oO
S> n(n — lana"? + S
TZ
[n(n —1)+10n + 20ja,2” = 0,
neB
oO
Solr +2)(n + Danze + (n+ 5)(n + 4an|c" = 0. The recurrence relation is
n=l
(n+ 5)(n+ dan
On+2 TER meee
(nb 2\(n $1)
a
forn
>> 0 0.
_ -5 + 4ao
ae
a
a
8
_ 7+ 6ag
a
48
=
—(2k + 3)(2k + 2)aon—2
2k(2k ~ 1)
—
(~1)*4-5-++ (2k + 3)aq
ak
1-2-+- (2k)
aon =
w=
on
ak
lt
a
54
_
—(2k + 4)(2k + 3)aan-1
a+r
(2k + 1)(2k)
om
= (~1)*5 -6+-- (2k + 4)ay
aR
(-1)*(k + 1)(2k + 1)(2k + 3)aq
3
ea
2
_ 78+ Tag
23+
(Qk + 1)
A2k+1 =
+ 1)(2k + 1)\(2k + 3)a07* =
k=l
yn = 22S 6 pant(1+ I)(k+2)(26-+3)0%™4 = 26
Cop he
aon
_ 76 - 5a;
a8
oO
(~1)*(k + 1)(k + 2)(2k + 3)ay
6
s(t
.
+ 1)(2k + 1)(2k + 3)x?*.
kad
(—1)*(k+ 1) (k+2)(2k+ 3)x7*+1, valid
k=O
for jz] <1.
9. (a* - 9)y" + 32y’-3y=0.
oO
oo
n=2
n=
~—9 So nln ~ lage"? + S> [n(n ~ 1) + 3n — 3fanz” = 0,
Om
‘> {[-9(n + 2)(n + Langa + (n + 3)(n — 1)an|
2” = 0. The recurrence relation is
n=O
(n+ 3)(n — Lan
ng = METI
O80 tor n> 0.
Qnt2 = “Gin 4 indi) "=
3+ (—L)ag
OST R1
SE
Gg =
eterna
5-1
~~ 9-4-3
_ (2k + 1)(2k ~ 3)a2%-2
9(2k)(2k — 1)
Ook ~
“3
=
ag = 0
on4¢i1 == 0 0 f for k >
L.
17.5.
SOLUTIONS NEAR AN ORDINARY POINT
a
ak
y=
137
_ 3-5+--(2k+1)-(-1)-1---(2k—3)ap
—[8-5--- (2k
+ IJay
ORQERI- 1-3. (Qk — 1)
aq
.
[3-5--(2k + 1)] 27
ll — »
TB)F(ak- Da
~
ame (2k
— 1k
;
++ @,@; valid for iz| <3.
oD
lL. (x? + 4)y"+6ary'+4y=0.
oO
4 S- n(n — Lanz”? + S- [n(n ~1)+6n+ 4ja,2” = 0,
oo
n=2
ned
s- [4(n + 2)(n + Lange + (n+ 4)(n + i)a,|c” = 0. The recurrence relation is
n=aQ
—(n+4)a,
An+2 = ee
din +2)
a
=
for
for nn >> 00.
—4ag
aa
29-2
—6ag
“4
on
4-4
2k
doc
4(2k)
=
4k/2.4..-(2k)]
4-38
x beh + 3)Q2k=1
oat TS
G2k+1
42k + 1)
=
TRTQ_ e
Ufem
aA
4*[3-5-.-(2k
+ 1)]
_ (EK 2k + 8)ar
Cok) Sk
1
CO
|
OO
13, ¥ y” + 2?y = 0.
er
1)" (2k
3)a 2k+1
ee
; vali
for [a]d
< 2.
oO
n(n ~ 1)an2"-? +
read
Sin +2}(n+lengex” +
S*a,a"*? = 0,
pet}
oO
n=
Fag
(—1)*[5.7--- (2k + 3)}a1
en
_ (EIR
+ Lay
rn)
14S
=
(—1)*[4-6--- (2k + 2)]ao
Q2k =
y = do
~B04
rer:
aon az Leth + 2)o2n-2
dak
=
a
ox
> Qn—g0" = 0, 2ag+6agzrt+ s: I(n +2)(n+ lange +dn—2ja” = 0.
n=2
nonQ
138
CHAPTER
Hence, a2 = a3 = 0 and any. =GED
Q4k+Q
=
0
and
7
=
Oy,
=
age
=
ED
0,
and
— 220
oS
_ 74
“8 37
dag
=
“oo
~Gdk—4
a
ak ™ 4k(4k — 1)
“8
=
(—1)*ao
a
RIB
7+ (4k — D]
.
POWER SERIES SOLUTIONS
for n > 2. It follows that ag = aig = «++ =
“eo
dae
17.
= ft
54
_
ag
9-8
we
dk 3
Akt ™ (4k + 1)(4k)
EES
_
(~ ijFa 4k
y = ag
1+) ag
(—1)Fay
[5-9
(4k + 1) aE RL
(- 1)Fz ak
$07. (dk
1)
+ ay
2+)" amg
;
5.9. (4k
4 1)
.
; valid for all
finite x.
oO
15. (1+ 22)y” + 3ry' ~ 3y = 0.
oC
J nln ~lanz™? + S > [2n(n ~1)+3n ~ 3]an2" = 0,
n=2
rea
oO
Ss; [(n + 2)(n + Lany2 + (2n + 3)(n ~ 1)an|c” = 0. The recurrence relation is
neo=Q
Qnt2 =
~(2n + 3)(n — lay
er
te
Ein pty
>
rr 20
0.
—3-(—l)a
a=
aie
ag
=
0
ada
=
0
_=
0
~T-la
wok
ag
=
ag,
=
are
_ (4k — 2k(2k
1)(2k ~ — i3)aax—2
Ook.
for
k
>a
L.
_ (-1)*[8 7-- (4k - D][(-1)-1--- (2k —-3)]ao — (— 147 [3-7--- (4k — 1) a0
y = ao
21 -3--- (2k — 1)|
oe
1+)
(-1)**13.7.--(4k — 1)x?
arRaE— 1)
-
D
otk
2Fkl(2k — 1)
.
+ a2; valid for jz}
< 1/72.
17, y”14 xy! + By= x, We f first solve the homogeneous equation.
S n(n — Lanz? + s(n + 3)anz” = 0,
3 [(n + 2)(n + lense + (n+ 3)an]z” = 0.
need
n=)
n=O
17.5.
SOLUTIONS NEAR AN ORDINARY POINT
139
~(n+ dan
.
;
The recurrence
relation
isj gn4,2 = mmnennutnsetnnn
int d(n an+ 1)nae for 2 >> 0.
a
—3ag
2=
a
2-1
__ 78a
an
“as 43
~(2k + L)aop—2
Oak =
am
2k
ao
= (~1)*3-5--.
(2k + Dao
a
_ (-1)F(2k + Lao
ang
CORRE 1 3-- (2k = 1)
14S
~
~ Gag
54
Oak+l = ~ (RF 1)(2k)
C re
6.--(2 be Pay
MAT BBs (2k
+ 1) [2-4-- (2k)]
(18k + Day
AALS 3.5. (Qk +1)
OO + _4\k
=
a
3-2
=
~(2k + 2)aax—1
TOR (ak — 1)
Qk!
Ye
—~4a;
4=
2k
CUE)
Qk!
k
0
Sy
tht)
hoes
. We
(2k +1)
now
seek
Yp = A+Br+Cex*, asolution to the nonhomogeneous equation. We need 2C + Bu+2Cx? +3A+
3Ba + 3Cx? = 27, or 5Cz? + 4Bx + (2C +3A) = 2?. It follows that A=-—2%, B=0,
C=}.
Thus yp = —~% + dx, Finally, y = yc + yp: valid for all finite x.
oO
19. y” + Say’ + 7y = 0.
OG
» n(n — lanz™”? + S_(3n + Tanne” = 0,
tes?
nm
S| [( + 2)(2 + lange + (30 + 7)a,] 2” = 0. The recurrence relation is
—(3n
+ Tan
= ee
for n > 2.
ont
Te dDintl)
"=
ao
=
ae
Qa
=
—7ag
aa
OTT
=
ae
~L3ag
ae
=
ae "4 3
ao.
=
ae
a
y = ao
a
OK(2k
— 1)
a
(2k)!
(—1)*7-13---
valid for all finite z.
(2k)!
_
okt
= (~1)*7-13--- (6k + lag
1
3-2
—l6a3
er
—(6k+ laop_2
ak
—10a,;
+a;
+ 4)don—y
(2k
+ 1) (2k)
_ (-1)F10-16--- (6k + day
okt h
(6k
+ 1)x?*
{6K
(2k + 1)!
ri
k=l
1
.
1)*
10-16---
(6k
+ 4)x
eff
(e+ ij!
2h+1
;
140
CHAPTER
oD
21. (a2? + 4)y" + ry’ - Oy = 0.
17.
POWER SERIES SOLUTIONS
oO
3 4n(n ~ lage"? + S [n(n —1) +n —- 9janz” = 0,
n=?
n=O
oO
» [4(n + 2)(n + 1)any2 + (n — 3)(n + 3)an] x” = 0. The recurrence relation is
n=f)
Ont2 =FE
—(n+3)(n - 3a,
ttre tener
>
yon
pty rn 2o
0.
_ -3(-3)ao
_ ~4(-2)ay
BST a1
13
9-3-3
—i(~L)a
.
Ook
—(2k + 1)(2k ~ 5)aon—2
Oak =
_
ac
Agn41
Ak) QR-1)
_ 8(-1)F3.5+-(2k + Dao
> 2
~O3RKI (Dk ~ 1)(Qk — 3)
(~1)* 3-5 ++ (2k + 1)?
od
= 0 fork
1.31. ot
a
iy GR ay | tole + a2"): valid for [al <2.
y=ao|1 +3). Saker
OO
23. (1+ 92")y"—18y=0.
OO
So n(n—1an2z"-? +S [9n(n - 1) — 18]an2" = 0,
n=?
n=O
oS
> [(n + 2)(n + Lande + 9(n — 2)(n + 1)a,|x” = 0. The recurrence relation is
n=Q
an+2
~9(n — Za,
Sa ete
a,
for n >> 0
0.
a3
_= ~%(-2)ao
3
a=
_ ~X-1ar
3
a4
=0
as
—9-
2+ = OE
OO
c+ >>
k=l
Daa
aang =
ag, = 0 for k > 2
y = ao{1 +927] + ay
lag
teen
5
(—1)§t1g2hg2k+2
Wri
TGR FI)
; valid for |x| < 4,
17.5.
SOLUTIONS NEAR AN ORDINARY POINT
141
ce
25. (1 + 2x*)y” + lizy’ + Sy = 0.
ox
S- n(n — Lanz? + s~ [2n(n ~ 1) + lin + 9]anz” = 0,
oo
n=
a=d
> [(n + 2)(n + Lange + (2n + 3)(n + 3)an |x” «=: 0, The recurrence relation is
n=l
—(2n + 3)(n + 3)ay
SE
for n >
Qnt2=
“Ty bym py
N29
0.
_ —3+3a9
Oe
a
FI
a=
—7
- Sag
a
aa 8
aon
*k
= —(4k — 1)(2k + L)aon—2
a
(2k)(2k — 1)
_ 7D: day
se
oe
ake
32
~—9 + Gag
54
_ 7(4k + 1)(2k + 2)aap-1
(2k
+ 1)(2k)
(-1)*(2k + 1)3-7-+-(4k — Lao
a2k
=
does
ORE!
’
= (—1)*(k +1)5-9--- (4k
+ Lay
okt
3-5---(Qk +1)
oO
= ay h + a
(~1)*(2k +1)3-7--- (4k — 1)x?*
sil
vm [ery
—1)F(k+1)5-9-. (4k + Lxteth
y 1
| ; valid for (z| < 1/./2.
+-(2k +1)
a
27. y” +(x — 2)y= 0. We set z = x ~ 2 and obtain a5
ds y + zy = 0.
oO
Oo
3 n(n — Lanz"? + x Qnz?t} = 0,
sin + 2)(n + Danzgez” + So an-12” = 0,
ns2
n=O
n=d
nel
oO
2aq + S
[tn + 2)(n + l)any2 + Qn—i]z " == 0. Hence, ag = 0 and an49 =
ot]
n>
Oy
(nF d(n+i)
1.
ag eee
= —2
6
_
743
65
—Agk~3
a3k = (Gk)(3k—1)
a4 =4-3
OT
O3k+1
=
=
a,
2G
7-6
—3k-2
(3k + 1)(3k)
= 0
ag = 0
a3h42
=0
142
CHAPTER
____(-I)Fag
Osk = BR Rl D-5---(3k—1)
-
sx
oe,
[8441
__(-1)K(e
~ 2)%
=
POWER SERIES SOLUTIONS
oie
Be Ria 7... (BR 41)
y= 0014 > seaTa-5 Geo | 79 |
all finite x.
17.
_ a) 4 So ME w= 2H]
)+ 2 aaaR EA] | Vl for
143
Chapter
18
Solutions Near Regular Singular Points
18.1
Regular
Singular Points
1
4
- p(t) = =i az) = Bei)
19.
. No singular points in the finite plane.
. pz) = za
23.
i
~24
R.S.P.
R.S.P. at @ = 2; 15.P. at a = 0.
25.
4
at 2 =i,
13,
27.
LS.P. at 2 = 0.
29.
17. p(x) =-— (Gat i)e
18.4
—I
ara)"
POT(2)
= Qa
—=———x;
—
+ tle—3)) (2)=Pee
R.S.P. at z = —4, 3.
6x
+422)?
3
p(x) =aa 25): q(x) =
R.S.P. at «= 0, 3, —-3.
me) = ya 1) = Trae
p(x) =
1+ 22
ia a)e
—9
R.S.P. at g = Ri, gl.
15.
~i.
R.S.P. atc = 0; 15.P. at e = 1.
R.S.P. at ¢ = ~2; LS.P. at a = 0.
6a
6
p(x) = 0; g(x) =
- P(t) = 33 a(2) = ward)
il. p(z) = 0; g(a) = a
—2i.
p(z) = Ty gn a) =7yaF
» P(x) = 53 q(z) = ed)
4
g(x) = rt44°
21. No singular points in the finite plane.
LS.P, at x = 0.
1
3
ad
R.S.P. at 2 = 21,
R.S.P. at ¢ = 1; L5.P. at x = 0.
1
—62
p(x) =
LS.P.
at
z=
LS.P. at z==
zi,
—1—5.
— $i.
31.
1
4
=
0:
a
R.S.P. at 2 = $i,
nrc
~$i.
Difference of Roots Nonintegral
oO
1. 2a(a + Dy” + 3(a + Dy’ — y = 0. We substitute y =
> a,z"** and obtain
nal
oS
Ss; [2(nte)(n+e—1)+3(n+c) — Llane???+ s [2(n+e)(n+e-1)+3(n+c)lana™*e"! = 0,
n=l
n=d
144
CHAPTER
18.
oO
SOLUTIONS NEAR REGULAR SINGULAR POINTS
oO
SS (an + 2e~—1)(n+e¢4 l)agz™t? + soln +¢)(2n + 2+ lagz® te"! = 0,
n=l
n=l
Ox
e(2c + 1)agx®7! + s- [(n + )(8n + 2e + lay + (n + c)(Qn + 2c — 3)an-1]a"**!
n=1
—(2
ee
= 0. The
_
indicial equation is c(2c +1) = 0 and the recurrence relation is a, = wnt
e~ 3)an—1
3
2n + 2e+1
n>
1. Taking c= ~5 yields a, = —(R = 2)enn
for
so that a; = ap and a, = 0 for n > 2. Thus
one solution is y, = 27'/? + 2/2. Taking ¢ = 0, we have an =n
=
=3)@n-1
an+1
_ ~(-1)a0
3
a=
ag
_ —(jar
5
=
a, = MMH) 1+ On = 3)ao _ (-1)**a9
7
3-5---(2n +1)
.
.
_
oD
We therefore obtain a second solution yg = 1+ 2
an?
—7
(-1)"t1z?
in? ni
0G
3. 4a*y"+4ry’—(4a?+1)y = 0.
&
s> [A(nte)(nte~1)+4(n+e)—lJa,a™t*— 5° da,x™ tet? — 9,
n=0
oo
Sn
n=)
oa
+ 2e— 1)(2n + 2¢4 ljage™t? ~ 3 4dn22"** = 0, (2c — 1)(2e + L)agx®
n=O
oo
+ (2c +1) (2c + BJaya°t? +
nese
[(Qn + 2c ~ 1)(2n + 2c + Ian — 4an—2]z7** = 0. The indicial
n=2
4an—2
equation is (2c—1)(2c+3) = 0, and the recurrence relation is a, = Gn42eiGnb
deh 1)
We take a, = 0, and therefore aa.4, = 0 for k > 0. Forc= 4 we have a, = ne
n(n +1)
70.
2-2-3
ag = 22
4-5
_O2k-1
Oak = Dkk +1)
= — 0
(Qk+ 1)!
for
k
> 0.
18.4.
DIFFERENCE OF ROOTS NONINTEGRAL
SO
Therefore one solution is y, =
145
ptkt/2 — ginhs
1
an—2
——s=~. For ¢ = —3, we have ay = (a—1n’
-a(QF ij
il
wo
al
we
= 18
il
®
Me
£
oO
VE
G2k-2
Ao, = Gk- DGB
Therefore a second solution is y
“3
2
=
y gk?
(2k)!
a
™ oni
j
_ cosh
Vr
=
for k > 0.
.
5. 227(1 — xy" — 2(1 + Tx)y +y = 0.
Ss; [2(n-+e)(n+e-1)—(n+c)+lla,a"t — >, [2(in+e)(n+e—-1)+7(n+e)lana°tet! = 0,
neo
n=O
+ c)(2n +2e+5)ane™?et! = 0,
5 (an + 2¢—1)(n +¢—1)anz"* Sn
nsal}
n=O
° = 0. The
(2c— 1){e — Lag + 3 [(2n +2c—1)(n+e~—1)a, —(n+e—1)(2n+ 2c4+3)a,_1]2"*
res]
indicial equation is (ce — 1)(2c - 1) = 0, and the recurrence relation is a, = (2n + 20+ 3)an-2
3
5
for n > 1. For e=1, an = CUT DMM
an+1
a
2n+2c-—1
foe
> 1,
Tao
m3
a
9a;
a
on=
5
2
Ant
5an—
ont5)an-a
_
2
Cn + Bao = 4 (2n + 3)(2n
+ 5)ao.
Qn 41)
146
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
Oo
Hence one solution is y, = «+ + Son
+ 3)(2n + 5ja?t?,
n=l
For c= 3,
n+
2)an..
an = it 2)Om=
porn > 1,
n
a
3ag
my
a
_ 4a
ae
"e
3-4---(n
+ 2)ag
Qn _= (n+ 2)an-1
:
=
= 3(n+1)(n
2)ag
+
Thus a second solution is yo = 2!/?+
ee
oO
2 ntI)(n+2)artY?,
OG
7, 8x7y"” + 10ry’ ~ (1+2r)y =0.
xO
Sagar tett = 0,
n=O
> [8(n + ¢)(n + ¢~ 1) + 10(n +c) — 1Ja,a"*°
oO
n=0
oO
S$ o(4n + de ~ 1)(2n + 2c+ lana*t? — S~ an_iz"** = 0,
n=O
oO
(de ~ 1)(2c+ L)aga® + S: [(4n + 4e—1)(2n + 2¢+ lan — an—1]2"** = 0. The indicial equation
res]
:
is (4c —~— 1)(2c +1) == 0 and a,
=
an-1
2n(4n + 3)
fi
SET
Oni
1
a eeeennnenenentenennennrstenminnanernmeretantn
(in
bdo~ ijn
db bed)sett forn >> 1. 1, For c == 7, we have
21.
orn2t
a,=—2
21-7
9-2-1
Gy =
On 1
an(4n +3)
=
ag
2" nl [7-11---(4n43))
18.4.
DIFFERENCE OF ROOTS NONINTEGRAL
147
grt
Thus
one
solution
is
nh
gil/4
+
y 42
nl
[7-11--(4n +3)]
For
c=
~%
we
take
Bpm1
al. for n > 1.
on = 2n(4n~ 3)
or
a
ag
abe
=
a
ae 9-2-5
ay—]
an =
ao
2n(4n— 3) ~ 2° nl fl -8+++(4n ~3)]
n- 1/2
T hus
soluti is y. == x gi?
a second d solution
or——
fi-5.(in
5]
9. 2a(z + 3)y" — 3(a + Dy! +2y= 0.
s
2 (n+e)(n+e-1)-3 (n+e)+2Jaa"4 5° | [6(n +¢)(n+e—1)-3(n+e)fa,z"te! = 0,
rm
n=O
oO
om
> (2n +2ce~1)(n+e~ 2)agz"t? + s 3(n + ¢)(2n + 2e ~ 3)aya” te"? = 0,
n=O
n=O
oO
oO
>> (an 4+ 2e-~ 3)(n +e = 3)anyatort + Ss; 3(n + c)(2n + 2c — Bjaya™ te? = 0,
ned
n=0
oo
=O.
3e(2c ~ 3)agzo"! + 3 [3(n +e)(2n + 2¢ — 3)ay + (Qn + 2c ~ 3)in+e—- 3)an—1jarte-!
n=l
The indicial equation is c(2c ~ 3) = 0 and a, = (n+ o~ 3)dn—1
a(n +c)
a = wen= Bana
"
forn > 1. Forc=
z, we have
3(2n +3)
_ —(—1)ao
ae
ao
=
a
"
s-5
—i
ay
"3-7
_ (-1*[(-1) 1 (@n—3)]
3°[5-7---(2n+3)]
oO
One solution is yy = 23/7 + s- a
nw
On
= a(n ~3)dn—1
Thus
a,
=
|
(-1)"+ag
3" 1(2n — 1)(2n + 1)(2n +3)"
(—1)ttignts/2
1(2n ~ 1)(2n + 1)(2n +3)"
~(—9
={o?)00
n
A second solution is yo = 1+ 42+ 42?
_
245,
ag
=
For c = 0 we get
—{—-]
1
ba,
Gn
=
0 for
n
> 3.
148
11.
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
«(4 — ayy"
+ (2-—a)y' + 4y= 0.
5
[din tojinte- 1)+2(n+e)lanc™ ten? + x [-(n+e}(n+c—1)~(n+c)+4]anz”*? = 0,
n=o
n=O
s 2(n + c)(2n + 2c — Ljaga™te7} — srr +e+2)(n+c¢-2ayr™** = 0,
OO
2e(2c — Lag + J [2(n + e)(2n + 2c — lag — (N+ e+1)(n+e¢—3)en—i]2"**! = 0. The
n=l
ae
.
(nteo+1)(n+e—-3)aq-1
indicial equation is c(2c ~ 1) = 0 and a, = “Sin teGn tient)
For c = 4 we have
_ (2n+3)(2n ~ 5)an—1
an
8n(2n+ 1)
5+ {-3aa
a=
2
‘
8-1-3
_ i (-Yar
8-2-5
_ (8n+3)(Qn—Sjan-,
an =
8n(2n + 1)
~
[5+ 7++-(2n + 3)] [(—3)(—1)--- (2n — 5)]a0
2" nl 3-5---(an
+1)
— 5)] a0
---(2n
_ (2n+3)[(-3
)(-1)
938n nl
4.
~
— 5) ert?
f 2n n—5)]x
3)|(-3)(-1
+
(Qn n+3)[(Ke
One solution is y, = 2!/? + y
n=
For c == 0 we get an
to
— 3)an-1
1)(n
(n+ Sane)
ns forn >> 1.1
ie te
a, =
a{—2
(
3(—1)a
ag
209
zits
= AoDe
=
i
300
Qn = O for n > 3.
A second solution is yz = 1 — 2¢ + $2?
13. ay"
+ (1+ 2x}y’ + 4y= 0.
s [Ain +c)(n+e-1)4+(n+c)jana™t) + 5
na)
[21 n+e)+4la,z™*° = 0,
n=
ox
Soin + c)(2n + 2¢— agate? 4 s> 2(n + e+ Nan_ye"te} = 0,
n=
n=l
c(2e ~ 1)agxe! + S[m
ne]
+ ¢)(2n + 2c - Ian + 2(n +04
Van-i]2"to"
= 0. The indicial
18.4,
DIFFERENCE OF ROOTS NONINTEGRAL
149
2
Lean
(nte+
equation is c(2c - 1) = 0 and a, = htoQn+
On
_
Van-1
26-1)
>2
forn
1.
Fore
=
$ we have
—~(2n + 3)an—1
(an +1)
a
a
_
75ao
pe
7-3
_
aia,
ae
O85
_ —(2n+3)an-1 — (—1)"[5-7---(2n+3)]a0
nm
nant 1)
onl [8-5--(Qn +1}
OO
fp
tan iise ay,
One solution
y, az= a°/* of + 2.
a,
=
yn
(1)? (2n + 3)zx
no
—2-
ae
n+1/2
3 nl
—2
4 1)
n
Gn—1
For c _= 0 we get ay, _= ani)
1)
2a
PV
m=
a,
(=1)"(2n
+ 3)ag_
~2+ 3a;
98
= (-2)(n+ l)an-y — (—1)"2"(n + Lag
a"
n(Qn-1)
xD
a
.
A second solution is y2 = 1 +d
7
1-8---(Qn—1)°
(—1)"2"(n + 1)a”
13° (n-1)~
15, Qe22y" — 32(1 ~ zy’ + Qy = 0.
oO
oO
s [a(n + ¢)(n+o—1) ~ 3(n +c) + QMana™*? +S” 3(n + c)ana™***! = 0,
n=O
n=l
oO
den
oO
+e~-U(nt+e— ane"? +
n=O
> 3(n +e - lana"?
= 0,
res]
ox
(2c — 1)(e ~ 2)aga® + s; [(2n + 2e~ 1)(n +e
nee]
2)ay + 3(n+e—1an_i]2"t° = 0. The indicial
~3(n +e ~- Tans
equation is (2c - 1)(e — 2) = 0 and a, =(nd 2-1)inbe~2) forn
> 1. For
c = 2 we
150
CHAPTER
have
ay
~3
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
1)an..
VYan-1
n(2n
+ 3)
= ~3(n +
a
= —~3 +2 ag
eT 5
a=
—3
ae
dy,
ates
One solution
a=
_
is yy, = 2
2
ay
~3(n +1) an-1 _= (-1)" 3°%(n +1) ag
=
0
3
ONT
n(2n + 3)
5-7---(2n+
3)
(-1)" 37¥(n
+ Ia™t 2
+S
Gay
. For c=
4
_ 78(2n
~ 1) an-1
3 we get a, =
ofan —3)
—3:-1 ag
te 2-1(-1)
Qo
=
—3:-3
a,
eee 2
a, a wom = Vann _ (-1)"3"[1- 3-2 = 1)] ag _ (-1)"#49"(2n
- 1) a
7"
9n(in-3)
ar nl [(—1) -1-+-(2n — 3)]
.
ss
A second solution is y, =__ 21/?ml /2 +
oO
>
(~))
n+132(2n
~
27 nl
.
_ Latt/2
an nl
n=l
17. Qry — (1 + 2x*)y’ — xy = 0.
oO
oO
>> 2mtel(nte-1)-(n+o)]a™*e) — $° [ain t+c) + 1]a"tet? = 0,
nee)
re)
moO
oo
IG + ¢)(2n + 2c — 3)a,z7ter! — > (2n +2c+ lagc*et! = 0,
n=O
n=
oO
c(2c ~ 3)apz°"! + (e + 1)(2e — Nayr* + So (2n + 2c ~ 3}[(m + clan — dn—g]a"to"! = 0. We
need
Gn —2
take c(2c — 3) = 0, a1 = 0, and ay = ="
for n > 2. The choice c = 0 gives us a, =
aQ=—
aq
=
fak =
a
2
ag
4
a3
=
0
5
=
0
a5
O2k~2
“pe
ao
= oF ET
Gand, = 0.
On—2
na 2
18.4.
DIFFERENCE OF ROOTS NONINTEGRAL
2k
Thus one solution is y, = 1+ YH
ea Te
a=
151
(427). A second solution comes from ¢ = 3 and
20n—2
ne an+3
ag =
2aq7
a4
203
=
ei
2024.2
O2% = 4k+3
w-
2* ap
a TTT ae
7-11---(4k
+3).
gk gtk+3/2
8/2
Hence yo= a+ > 7-11: (4k-+ 3).
oO
19. 2x*y" + ay’ — y = 0. >. [2(n +e)(n+e-1)+(n+c) —1anr"** = 0,
oo
m0
So (ant+2c+1)(n+e~1anr”** = Q,
oo
(2c+1)(c—1)aoz®+ >" (2n+2c+1)(n+ce—])ayz™*
n=d
= 0.
nel
We can make the first term vanish by choosing ce = 1 or ¢ = -3 while leaving ag arbitrary.
We are also forced to take a, = 0 for n > 1. For c = 1 we get the solution y; = a. Force = ~4
we get the solution yg = «71/2.
21. 9n7y" +2y = 0.
oO
Ss; [9(n+e)(n+e— 1) +2]a,z"t° = 3 (an+ 3e— 2)(38n
+ 3e— lagz”™*
n=O oo
n=
(3e~ 2)(3¢— L)agx® + WED + 3c —2)(8n+ 3c—1)anz"** = 0. We choose ¢ = z orc = % and
n=l
aq arbitrary. We are forced to take a, = 0 for n > 1. For c = 2 we get the solution yy = 27/8.
For c = 4 we get the solution y2 = 21/3,
23. If y is a function of x and we let t = Inw for
aydz
dt dy
il agg PY
xdt
dtdz
2 0 then
vat
dydt _ on (Pydt? tydt
dt? dz
25. Qa2y" + ay’ a y = 0. We set t = Ina and use the results of Exercise 23 to get the differential
d
d
equation 2 <a ~ e
+ —~ —y =0 or (2D? — D— Ly = (D —-1)(2D + 1)y = 0. It follows
that y = cre’ + cge(—/* or y = yr + egal V2),
2
27, On*y"
i
.
*
+ 2y = 0. Letting ¢ = Ing we obtain 9
d*y
We
dy
he
+ 2y = (9D*
3
— 9D
+ 2)y =
(3D — 2)(3D — 1)y = 0. It follows that y = cye(?/9* + cge(/9)* = ey x?/8 + ega'/3,
152
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
dy
“)
2,7 + Qay’ a— 12y == 0. Substituting
‘tuti
29. a*y”
¢ == Inz we obtaini ( one
Gt eteGe
dyat oe 12=[Doe
+2 mete
(D? + D ~12)y = (D — 3)(D + 4)y = 0. Thus y = cye** + ege™®* = eyr8 + cga4.
Py
dy\
31, x*y” — 3zy' + 4y= 0. Letting
t = Ina we get (A
-#)
~
dy
at
w=
(D* — 4D + 4)y = 0. The general solution is y = (c, + cgt)e”* = x?(cy + cgln zr).
2
33. z*y” + 52y’ + 5y= 0. Substituting ¢ = Inx we get (5
- 4
+ sf + 5y =
(D? + 4D +5)y = 6. The roots of the auxiliary equation are ~2 + i and —2 —i. The general
solution is y = e~**(c; cost + cz sint) = 27" [c, cos (Inz) + cz sin (In).
18.6
Equal Roots
oS
L. a®y” —a2(1+a)y' +y = 0.
oa
[(n+e)(n+e—1)—(n+e)+1la,2"F* - Si (nteja,ztert =
n=O
oO
n=0
~
KG
+e-
1)?a,2"*° ~ So(n +6¢—1)an-:2"'* = 0. We therefore have
neeQ
n=l
(e—1)*agx° +
> [(n+ce—-1)an(n+c-1)an—1]z"t° = 0. The indicial equation is (e— 1)? = 0
n=1
and
ay
=
On—1
Peni
>
ern el
1.
ao
aq=e
c
a2
Qe
=
=
"
a
i
On-1
nte-1
=
&o
cet+])---(c+n-1)
Thus ye = 2" ‘Laeryaery
gtte
BYc
Be
gente
=welnst yo ele+l) (e#n=1)
yields n =2+
1
|e
1
e4l
c
1
ohne
. Substituting c = 1
got
A, oth
= 77 = we® and yo = yr Ing — $" 2 —
ns]
3. x?y" +2(c—-3)y! +4y= 0.
x (n+e¢}(n+e~-1)— B(nte)+-4Jana™*457( (n+e)anatet) =
oo
4
nO
oO
Son +e-~2)Panz"*? + yin +¢~l)an-12"** = 0. We therefore have
ned
nel
18.6.
EQUAL ROOTS
153
oO
(c-2)%agz°+ > [(n-+e~ 2)7ant+(n+e~1)@n-1 ]a"*¢ = 0. The indicial equation is (e~ 2)? = 0
fas]
_
~(n+e-
and dy =
_
~C
ag
“ (e=1?
—(e+1)
a= ——
a
a=
a,
~(n+e~1)
"
1) ani
(n+e¢—- 2)?
a1
(~1)"e(e+1)---(n+e-1) ap |
(n+e~ 22
(e-~1)%e?---(n+e-2)?
OO
Therefore y, =mx 2 pt >
>
roe
(-1)"(n+e-1) ate
(e=i)ee+
1)
(n bend)
(-1)"{n +¢- 1)x"*¢
1
(c—1)*ele+1)---(nte-2)
|n+e-1
2
Ye
and <#e
Fe
1
e-1
oO 7 oy\n
(—1)"(n+1)a7*? [
(-1)"+! [n+ (n+
oO
Yi a>
1
m+o-2]°
an
n
1
_
e4+1
n+2
hee
oO
:
Yelna+
1
ec
Substituting ¢ = 2 we have y, = 27+ Ceti
s;
_
(-1)"(n
+ c¢—1) ag
~~ (e-1)%e(e
+ 1)--- (n +e — 2)
1
_
1)H,|2"+ 2
ml
.
oO
§. a(14+cr)y"
+ (1+52)y! + 3y = 0.
oO
So [(n+o(n+e-1) + (n+e)]anz*o“} +S
3
co
na
e~1)4+5(n +0) + 3)ana"** = So(n + cane”??? 4 Son
nal
[(ntelnt
tet l)(n+e+ 3)agz™t =
nel
eo
aga) + 3 [(n+c)*an + (n+e)(n+0+2)an—1]2"*o"! = 0. The indicial equation is c? = 0
nom]
and a, =
—(n+e4+2) an-1
for
i
>
1.
nee
ay
_ ~(e+3) ao
~
e+]
_ 7le+4) a
ay
—
an
~
e+2
_ a(nt+e+2)an-1
c+n
~
(-1)*(c+3)---(c+n4+2) a9
(~-1)"(n+c4+1D(n+e42) ag
(e+ Ile+2)fe+n)
(c+ 1)(c + 2)
CHAPTER
154
SOLUTIONS NEAR REGULAR SINGULAR POINTS
(-1I)"(ntetI(ntet2)art
oO
Therefore y. == 72 +h
Be
18.
{e+ 1)(e + 2)
ane
yeinz +
tn
tet Minter 2ar**
n=l
{e+ 1)(e + 2)
!
Ban s
ntet+l
nt+et+2
—
tt
e+)
c+2)°
ye
Lo,
“
nlng +5)
=
wing
1)"(n
~
an
I
So(-p"(n + 1)(n 4+ 2)" and
need
+ 1)(n
+ 2)z
5(m 1) +5 DD
3
Bop
oO
Substituting c = 0 we have y; = 1+
=
(Qn +3)
wt
all
1
at
in
1
wo Lam) |
2®,
~~
ay” + a(e - iy + (L- ay
= 0
SO[nt+ (n+
e- 1) - (n+) + lana? +
re)
ce 27
oO
oD
Sia +e—Tage™ter! = Soin +e~1)ana"t? + So (n +6 —Qag—12"*? = (e—1)*agx® +
nod
n=O
n=l
CO
S[@
n=1
+c0-1)*a, + (n+ ec - 2)an-1|x"t° = 0. The indicial equation is (e — 1)? = 0 and
—(n+e-—2) ay-1
merrenrenennnemnenne
>1
An SEE eater (nben
ip?
forn>1
ay = Ea
a
=
—C
@y
2” (e+ 1)?
dn = _ (n+e¢-2) — 2) ani
dn
(n+e-1)?
There is a problem here since we eventually wish to substitute both n = 1 and ¢ = 1 into
our recurrence relation. We therefore keep the a, term separate from the infinite series. Thus
_ —(c—1) ag
_ (-1)"(e- De-+-(n+¢—
2) any
a=
So
(e—
we oe
Ye =
Bye
——
Oc
and a, =
1actt
ce
Betis
20
for n > 2. We have
(-1)"(c — L)xtte
<> [ele +1)---(n+e-2)|(n+e-1)?
e~1 | 1
“|
= y, ng
— —z— | ——— =]
¥
(nfteni)
e
je-1
¢
*
2%}
a
Ler)
and
(-1)"(c
~ 1)z™t¢
tena
reriy
18.6.
EQUAL ROOTS
1
[ty
1
155
1
ee i
Uae
a2
_
Yo = ying —s2
1
ee
When
9°
2
(—1)?artt
+o
2
sl |)
i
90
Gainne =ylng
Substituti
ubstituting ¢ =} we have y; =so x and
(-1)"
nm
«
pati
+ })-—___.
n=?
n=l
dy
d?y
:
9. x(a —2)y"”nt +2(a—1)y’ t —2y = 0. We set z = 2-2 and obtain x{2+2) 75 +2(z+1)—- —2y = 0,
ox
where z = 0 is a regular singular point. Setting y =
oo
oo
> Gnz"** we have Ss; [(n-+e)(n+e~-1)+
n=l
wo
n=0
2(n+c)—2] Onzrte+ 5 > [2(n+e)(n+e—1)+2(n+c)]anz"te7} = So (n+e+2)(n-+e~Lanz"te+
xo
n=O
wo
> An+e)Pagz tor! = WePagze) +
nad
a=
> [(n+e+1)(n+¢-2) @p-1 +2(n+c)?an] grte-l — Q,
rel
—(n+e+1)(n+e— 2) an
1 for n > 1. Here we
2(n +c)?
must be careful to keep the n = 1 term separate because when c = 0, a, = 0 for n > 2.
The indicial equation is c? = 0 and a, =
_ ~f{e+2){e~ 1) ao
ms
2
Bet 12
_ -(e+3)ea
Oe+ 2)?
- —(nteti)(n+e-—2)a,-1
on =
2(e+n)?
Th
US Yo == gf2
Ye _
Oc
(e+2)(e—1) et?
1
1
2e+1)?
le+2
e-1
gern
a(ce+ij(nte-iint+e)
1) 2"
2. etiinte-iyn+a
(c+2)(e-1)[
5p ute= Den +e+t)
44
SX (-1)"(e- De(n +4
Be +1)?
= Yelnz
— (-1)"(e- Ie(n+e+1) ag
~ 2%(c+1)\(n+e-I(nt+e)"
2 |
e4+1
ett
ut
fy
yi,
i
[e~1
¢
mtetl
1
e+1l
nte-1
1
nate]!
Substituting geo=
c = 0 we have y;
z and yoyo =i
= Inz+ [}-1-g]2$0 CMe
w== 1+
4
2.
any
Thus yx = 1+ (a2)
5
= (-1)"(n
+ 1)(2 — 2)"
and ya = ya In(a — 2) ~ 5(2— 2) ~ J
156
CHAPTER
11. ay” + y' +2y =0.
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
oO
oo
reed}
n=xQ
So [in +e)(n+o-1)+ (n+ Canto! + 5" anarter = 0.
3
io8)
So(n +c)Panertend + >> Gnga" te! =
n=
nd
oO
ctagx®! + (c+ 1)?a,2* + S; [(n + c)?an + Qn—2]a"te"! = 0
need
We take c? = 0, a; = 0, and a, = mage
om
ag =
2”
ore?
(e+2 2)?
ag =
4
(e+ 4)?
aot +
wo
on
(—1)* ao
ae
(e+ 2k)?
(e+ 2)?--- (e+ 2k)?
kp tk-+e
So,
Oe L
uee® +d (e+ 2)? (e+ 2k)?’ Bc
|;
2
ren
2
war
,
‘ Substituting
Hi
Yo
0
kpak-te
(=)
yen +) (e+ 2)?---(c+
2k) *
c=0
.
yields
oo (—1)*a2*
let 2 ‘2k (Rye
y=
and
1)*x?4
=n
1
13. Ho
1
1
~ 5 Hk = (Gran +aatE)Oo
v2 of Exercise 12 becomes yo = y; na + D
oO
15. 427 y"+8a(2+1)y'+y = 0.
oo
1
1
1
5 (145 ++: i)
1-3-.-(Qk ~1)|
a
<a
Hop
— $H,
eae
[4(n+e)(n+e-1)+8(n+c)+1]an2"*°+ 5
n=O
2
9
m+ 2c
+1)"
2
a,r
nad
nee
nm+e~ lan
n+e
=
n=l
(2c + 1)" agx* + > [(Qn + 2c + 1)an + 8(n +e— l)an—1]a"** = 0.
n=l
"+ 37: The
«] 2
OO
n=O
8
+
1
at
B(nte)anzr tet! =
18.6.
EQUAL ROOTS
157
The indicial equation is (2c + 1)*= 0 and ay, =
—8(n+c¢-1) an-1
f
> 1.
(2n + 2c +1)?
Orne
—8e ag
a,
=
(2c + 3)?
= ~—8(¢+1) ay
a
a
(26 + 5)?
= —~8(n+e-~Wdp-1
"
Hence
er
Ye =r
Pe -
pl
(-1)"8"e(c+1) +--+ (mn +e—1) ag
(Qn+2e4+1)?
— (2e 4+ 3)2(2¢ + 5)? --- (e+ In + 1)?"
os (—1)"8"c(e+ 1) +++
(n +e—1) arte
+h
(2c + 3)2(2c + 5)? ---(2e+
2n + 1)?
and
(-1)"8"c(e+ 1)---(n +o-—1) Pte
yelnz + > (e+ 3)2(2c+5)2---(2e+ antl)”
Loy
bt
4
m+e-1
+3
ce
For ¢ =-ly
qr?
»
wy
~1)8" [(-4)(4)-(2ony?- Pl ar’?
2
R42
= £ -1/2 + ys
“1/2
(—1)"(—1)-1-++(2n
— 3) 27/7
(ni)?
(- y*(—~1)-1+--(2n— 3) a*-¥/2
(ni)?
a ing + >»
_
sand yo = yilnag+
2
—O49424...42~%
ator gt
2
4 34
7
tee
aT
(nl)?
oo
oo
yee
_
(—1)"(—1) ---(2n— 3)(2Han—2 — Hn-1 — 2Hy — 2) 2-1/7?
17. zy" +(1-2x)y'-y = 0
c)taga™te~?
2e+2n+1]°
nal
a4 | ox
on
.
oO
So [(nte)(nte-1)4(nt+-c)Jane™*2~S net lana”? = So (n+
n=O
_ So (nt
|
oo
ean"
we
= eagx®!
an—
The indicial equation is c? = 0 and a, = ~
a
+ >
nes
[(nt
n=]
; for n > 1.
fo
e+l1
}
y=
_
ay
e+2
an-1
“etn
n=0
c)*an _ (n+c)an—y]27te—}
ag
(e+1le+2)---(e+n)
= 0.
158
CHAPTER 18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
ghte
Hence y, == fe
&
Oye
Be
=
yy
weet
to
Grier
aera)
and
eniynnstnramsenmmmmnamanananntienimsieiettieniteteaninnatnisitsi
gree
DG cree
1
erm
1
1
|oed 7 o42
i,
cin
. We have the
%
solutions y; = 1 +e ~~ cee” and yg = y, Ing— se a
nai”
18.7
no]
Equal Roots, an Alternative
3. L(y) = zy" + x(x — 3)y’ + 4y. We previously determined the nonlogarithmic solution of
+ Dah
L(y) =0 to be yy = SOU
ni
. We now seek a second solution of the form
n=0
OO
ya = ying + > bya?t?
fisxl
oO
yo = y, Ina t+ vi tT S (n+ 2)b,2"* 1
nel
oo
7 ~ By
vy = yf ina
+ “BE
(n+ 2)(n + 1)bn2”,
n=l
so that
Oo
co
L(y2) = L(ys) ina + 2zy; +(e —4)y + (n+ 2)(n t+ L)baa™t? + (n+
YS
baat —
n=l
oo
nyse]
oO
3 So(n + 2)baa™t? 44 > baa”t?
nel
oo
n=]
oo
= Qaey} + (@—4)yi + Ss; n?b,art? 4 Sin + 1bp—1a**?
2s.OO ! ¢ _4\n
1)"(n
n=}
+ 2)(n+ lx
ned
re+2
nl
n=2
tyoO
1)"
int
Ltt
je
aye
—1)"(
ia nt
Latt2
x
+
ne=Q
n=O
oS
bx? +
I [n7bn + (n+ 1)bp1j2"?
ne?
n
=>
(-}) ate +1)x
tee
=
(by
—
3)x°
+ 3
n+2
+d
1*(n +2
peur
nly
(~ 4 Soa
ptt?
oo
+b2° + So [n?bp + (n+ 1bn-a]a"?
n=2
+ .
+ (n+
1)b
4
x
a
+2.
ned
ayn
In order to make L(y2) = 0 we must take by = 3 and nb, + (n+ 1)bn—1 + a
for n > 2.
=
18.7.
EQUAL ROOTS, AN ALTERNATIVE
159
5. We let L(y) = a(1+a)y" + (1+ oy
+ 3y. We previously determined the nonlogarithmic
solution of L(y)= 0 te be y, = 1+-5LS
~1)"(n+1)(n+2)z". We now seek a second solution
of the form
.
=1
oO
yo=yiInat+ >> bax”,
nok
Yh = yi Ing + YtHS
ngs
,
nel
yy = yf (Ing +
yjWo
BS
n(n
1)b,2"~?,
tes]
so that
L(y2) = L(y) Ina + 2(@ + yi +41 + DO n(n — 1)bae™) + Yonin ~ 1baz"+
ot}
> nbyz™ 1 + Yo Ss"
+ Yates"
nel
= 2e+l)yy + 4yi + 3 nbat™) + s(n
nl
= at
+ 1}(n + 3)b,2"
nl
OO
ly, + yr. +b: + 5 [nbn + n(n + 2)bni]2"-
OG
n=a
oO
= SU(-1)? a(n + 1)(n t 2a" + $7 (-1)"n(n + 1)(n + 2)! + 44
n=l
ns]
oa
250 (-)"(n + 1)(n + 2)0” + br +S” [n7b" + n(n + 2ba—a]a"?
n=}
ne?
= (by ~ 2) + $0(-1)"(n + 1)(n + 2)?"+
oo
ne]
So [(-D"a(n + In + 2) + bq + n(n + Wby—i]2™
n=?
oo
= (by ~2)+ $0 (-1)"" n(n + 12a
Oo
n=2
¥, [(-U"n(n +1)(n +2) +n7b, + n(n t 2)b, ~ije"?
n=?
oO
= (b) — 2) + > [(-1)"n(n + 1) + 07d, + n(n + 2)bn-1]2"71.
nad
In order to make L(y2) = 0 we must take b; = 2 and nb, + (n+ 2)b,_1 + (—1)"(n +1) = 0
n> 2.
for
160
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
7. We let L(y) = 27y” + x(x — ly’! + (1 — z)y. We previously determined the nonlogarithmic
solution of L(y) = 0 to be y, = x. We now seek a second solution of the form
oo
ye=ning+ >> ba",
n=l
;
Yoa= Yhiy net
UL
oO
+ Dnt)
b
nk mt
yi wy Ayn(n+ 1)baa"}
uf = vin + ewe
tone,
n=l
so that
OO
oO
E(y2) = L(y) lng + Qey, + (c — 2)yi + > n(n +1)b,2"*} + Son + 1)bya"*?
nos]
OO
KG
nel
m=
+ 1)b,2"*}
oo
+ s> n*b,2"*!
n=l
oO
+ >. baw™t}
n=l,
_ s. batt?
nes]
ne]
oO
+ \, nb, xz? t?
ri]
oO
= (by +1)2? + S$ > [n?bn + (nm — 1)dp_aje™*.
n=2
In order to make L(y2) = 0 we must take 6; = ~1 and nb, + (n — l)bp_1 = 0 for n > 2.
In this exercise the two-term recurrence relation can be solved in the usual way to obtain the
oO
(-1)"2*t!
solution yg = 41 Ine + S|
nl
n=l
18.8
Nonlogarithmic
Case
1. Set L(y) = x?y” + Qx(x — 2)y’ + 2(2 — 3x)y. Then
oO
Ly) = Dom +e)(n+e—Vana*? +250 (n+oo clanartort—
oe
os
nag,
4 Yn
n=)
+ c)agx"*? +4 >
n=O
n=0
a,rrte -6
: anv tert
neo
oo
=<
= s> [(n + c)* wes 5(n +e) + dana" t¢ +2 So (n +e-
B)azzr ter!
ne
n=)
oo
Si(nte-4)a,i2"*°
=) \(nte-I(nt+e-4)anz"** +2 nel
naQ)
= (c— 1)(e ~ 4)agz* + y [(n+e-1)(n+e— 4)an + 2(n +0 -A)ag_i]2"*°.
n=]
18.8.
NONLOGARITHMIC CASE
161
We choose the smaller of the two roots of the indicial equation, namely c = 1, and consider
n(n ~ 3ja, +2(n - B)an-1 = 0,
forn >].
1{—2)a;
+2(-2)ao
=,
a, = —2a9,
2(—L)ag
+2(—L)ay
= 0,
ag = ~a4,
3(O)az
+2(O)ag
=Q,
ag arbitrary,
Oy = 22 Guat
n
for n > 4,
—2ag
Q4
=
4
4
a a (Bas _ 6-2)" as
nS 45 -n
el
6(——2)" 3x Fed
Thus y = a [z — 22? + 22°] + a3
x “uy
i
nm
3. Set L(y) = z*y" + 2(2 + 32)y’ — 2y. Then
oO
OS
L(y) = > [(n+e)(n+e—1) + 2(n +c) —Qlane"** + Ss; 3(n + c)a,z™ter!
n=O
n=Q
= dom +e+2)}(nt+e-Dagr™t? +
n=l
> 3(n +=
Lan_ya"*?
ne]
= (e+ 2)(e ~ l)aga* + Yin te-1)[(n+e+ 2a, + 3a,_i]z"**.
nse]
We choose the smaller of the two roots of the indicial equation, namely c = ~2, and consider
n(n ~ 3)an +3(0 -3jan-1
U-2a,
+3(—2)ay
= 0,
=0,
forn> 1.
a, = ~3ag,
2(—l)ag
+3(—-l)ai
=0,
a2 = Bap,
3(0)a3
4-3(D}ag
=0,
ag arbitrary,
On = =3 Gn-1 for n > 4.
n
a=
—3a3
|
(~1)?-83"-3
ay
=
ag
4.5.0.7
2(—1)"-33"-2
~
nl
(—1}"- wt
Thus
y = ao [2
2
3¢7 +g]
+0 fee
Teme Be
a
162
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
5. Set L(y) = x(14+a)y" + (w+ 5)y’ — dy. Then
+ So [(m +o)(n-+e~1) +(n+e) — dana"?
L(y) = So [(n +e)(n+e-1)+5(n+ cane!
n=l
n=O
= Sim +e)(n+e+ agate!
+ do(a +e+2)(n+c-2)anz"t¢
nesQ
need
= c(e + 4)apx°) + S| [(n+c)(n+e+4)an+(ntce+ 1j(nte- 3)an—ija*te?,
nel
We choose the smaller of the two roots of the indicial equation, namely c = —4, and conside:
n(n ~ 4a, +(n —-3)(n—- Tan)
=0,
forn>1.
1(-3)a,
=Q,
a, = 4a,
+(—2)(—6)ag
=0,
a2 = 5a,
3(—lag
4(O)a4
2(—2)a2
+ (0) (—4) ag
+(1)(—3)as
=O0,
=Q,
a3 = 0,
30,
5(1)as
+(2)(—2)a4
= 0,
a5 = fas,
6(2)ag
7(3)ay
+(3)(—1)as
+(4)(O)ag
=Q,
=0,
dg = ga4,
a7 =0.
All
+(—1)(—5)ay
subsequent
a’s
are
proportional
to ay
4
a arbitrary,
and
1
y = ag [o~4 + 49 + 5a] + 14 . + get 3
are therefore
zero.
Finally,
we
have
.
7. Set L(y) = z*y" + x7y! — Qy. Then
oO
L{y) = S
oO
[a +e)(n+e—-1) - 2lage™t* + Son t+eja,a"tet!
n=o
oem
= So(n teot+i)(n+e-2)anz™t* + Si(n +¢-l)ap-i2"**
n=
co
n=l
= (e+ 1}(c — 2)agx® + > [((n+c+1)(n+e—2an+(n+e —l)ag_;]2"t?.
n=]
We choose the smaller of the two roots of the indicial equation, namely c = —1, and consider
n(n —~3)a, + (n—-2)an-1=0,
forn>1.
\(—2)a1
=0,
a, =—~§ao,
+ (—lL)ao
2(—1L)ag
+
(Ojai
mz 0,
ag = Q,
3(O)a3
+ (lag
=Q,
a9=0,
ag arbitrary,
18.8.
NONLOGARITHMIC CASE
dg
=
163
oe)?
for
n{n~ 3)
Qa =
—2 ag
as =
—3
n> 4,
4-1
a4
5-2
ne’
Thus y =— a9 [2
_
=
(—1)"~$[2.3...
(n — 2)}
=
ag
[a-5--nj[l-2--(n—-3))
1
if 5] +402 2 +
oo
6(—1)"-3
ag
n(n—1)-(m-3y
for n > 4.
6(—1)? 3g"?
At
natneaay
.
Sb +38dy ~ 2y = 0.
9. We set z = x — 1 and obtain L(y) = x24
oO
oo
Ly) = So [(n +c)(m+e-1)- 2lanz"*¢ + S- [(n +e)(n+te-1)+3(n4+ c)la,z"te"!
oot
n=O
8
= Si(n +et+1)}(n+e— Q)anz"** + Si(n te\(ntco+2anz"*te-!
n=
n=O
= e(e + 2)agz°! + WG
+e)(n+e+2)an + (n+ e)(n+e-3)ani[27ter?.
ns]
We choose the smaller of the two roots of the indicial equation, namely c = —2, and consider
n{n —2)an + (n~- 2)(n—-5)e,-1 =0,
forn> 1.
U(-l)a,
+ (—1)(—-4)ao
=0,
4a; = dag,
2(0)ag
+ (0)(—3)a,
=0,
a, arbitrary,
We have an = =(n = 8) anon for n > 3, so that ag = 2a9, a4 = fag, and a, = 0 for n > 5.
n
Thus y = ao [(a ~1)7? + 4(2 ~ 1)7*] + de [ + “(¢ -1)+
(a ~ 1)
.
11. Set L(y) = xy” — 2(a + 2)x’ + dy. Then
oO
L(y) = 7[(n t+ Qn +e-1) -A(n + e)Jane™*! ~ 2"(n +6 — Qanz"*?
as
cas so(n
n=
oo
ns?
+ e)(n +e-
5)ana™ ten}
~2 Sm
+en
an
e"ten}
n=l
n=
= e(e — Sjaga*-? + s- [(n +e)(n +e-5)ay
nel
- 2(n+e-
3)an—1Ja7te-},
164
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
We choose the smaller of the two roots of the indicial equation, namely c = 0, and consider
n(n - 5jan ~2(n — 3)an-1 = 0, for n > 1.
1(—4)ay
2(-3)ag
—2(—2)ao
~2(—1)ay
= 0, a; = ao,
= 0, a2 = $a,
3(—2)ag
—2(O)ag
= 0, a3 = 0,
4{—l)az
—2(Lja3
=
5(O)as
—2(2)a4
=0,aq=0,
0,
a(n — 3) ay-
ay = eee
oo
wT
=
_
2°3
as
_
2-4
a6
==
6-1
T9
aq
= 0,
ag arbitrary,
for n > 6
an-5/3.4---(n~
3)] as
ep ala (n—5)]
_
1»
Thus y = ao [ites
3
5.
| + a5 F
60
- 2-5 a,
Go Slaln— Ina
=
60. gr~5yn
+h
(n—5)l n(n
iin)
28
>
6.
|"
13. Set L(y) = x(@ + Dy” ~ Oy! — 6y. Then
Ly)
y [(n
=
+ c)(n +ce-
1) _ 6lanx™te
4 5 [3(n
fe e)(n
om 1)
_ O(n
+e)lanz"te-!
nxt
nel
= Son +e-3)(n+e+ 2)ayz™t? + s; 3(n+e)(n+e~4)a,za™te!
nexO
nol
= 3e(e — 4)aga®* + $7 [3(n + c)(n+0-4)an + (n+e-4)(n+e+ VagJarte?.
a]
We choose the smaller of the two roots of the indicial equation, namely c = 0, and consider
3n(n ~ Jan +(n -4)(n+ Tan)
3:
1(~3)a,
+(—3)
. 289
3: 2{—2)a2
+(—2)
‘ 301
3: 3(—l)as
+(-1)
4a
3-4(O)aqg
+(0)-5az
=0,
forn>1.
a
0,
a
=
~- ao,
=
0,
ag
=
4%,
~ gh ao,
= 0,
a3 =
=,
a4 arbitrary.
18.9.
LOGARITHMIC CASE
_
—(n+
On =
1) @n-1
3n
_
for
n > 5.
76 a4
05
an
165
3-5
= —~7T as
6
"3-6
ag (144+) a
NAG Tt)
ow
"
3°-4(5-6---n)
Finally, y = ag ft _ a
+ 5
for n > 5.
5. gna
-4z s] +04 G eh ~1)"(n
+ 1)z*
5 - 3n-4
ned
15. Set L(y) = zy” + (x5 — 1)y’ + 2?y. Then
oo
Ly) = So [nt od(nte-1) -(nte)jage™te-} + y(n ++ la,z”ter?
ma
oO
n=Q
x
= Sin +e)(n+e~2)anz™*1 +S (nte- ange"!
n=0
= ¢c(e ~ 2)agx°™!
n=}
+ (e+ 1)(e— l)aye® + (e+ 2)cagr“t?
ao
+ Yin +e~2)[(n+c)an tan—g]z"*?~
naeg
We choose the smaller of the two roots of the indicial equation, namely c = 0 and are forced
to choose a, = 0. However, ag can be left arbitrary and a, = ——°—
a=
= _%3
ag =
aq
= 43
03,
Finall
inally,
18.9
3k
(—1)* ao
= pe
a.
3kK!
Logarithmic
ag =
a7 = 0
a3 =
_ 2%—8
a3n41 == 0
—aF
y = 0 [i+ 5
= 2%5
0
_
_ —A3K~3
Ask =
—=
1x3
(-))'s"*
yp
Case
1. Set L(y) = cy” +y. Then
Sk+1
| +
Ak+2 == —vak=1
E>
= 9
|e 45
for n > 3.
a:
ah+a
=
(~1)Fg 3442
E ©
(—1)* ag
[5-8--- (3k + 2)]
ETD] |
166
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
oo
oo
Ly) = Soin +e)(n+e~lagz?*e1 + > anzte
ned
nel
ox
oO
= S"(nte)(n+c-l)anz™*o? +
a=0
Gy yx tend
nel
ox
= e(e ~ ljagz®! + > [(n teln+e- lan +anifx*tet,
ne]
Oni
The indicial equation is c(c ~ 1) = 0 and a, =
ay =
1
(n-+e}(n+e-1)
forn2
1
(e+ Ie
Qo =
** (e+ 2)(e
+1)
a. =
~An-1
*
(e+ny(etn=1)
Ye = ao tS
nal
.
(~1)” a
(e+1)+:-(c+n)ele+1).-(e+n—1)
Cy
(e+1)---
(c+ nje(e+1)---(c+n—-1)
We wish to take ag = c, so we write
goth
Ye
= Oe ad
‘lear
Me ine pat Bg
Ge S¥elnate * Crip
S
(-1)Pz"te
‘(c+ n)le+1)---(e+n—1)’
(carats
Der
ermMeroerne”
Substituting c = 0 we obtain
Lea"
n=-24
eo er1!
n=2
_
w=nmnetite-S)
nm?
3. Set L(y) = 2ry” + 6y' +y. Then
(-1)" (Hp + H,~1)2"
nl (n= 1)!
.
18.9.
LOGARITHMIC CASE
167
L{y) = x [An te)(nte-1)+6(n+c)lanz™*! + s, ante
n=}
nmQ
oO
Oo
== » Qn + c)(n+e+
ayz tte} + s> Gin yt?
n=O
nal
= Qe(e + 2)aga°! +
IEG t+e)(n+e+2)an tan-1ja7te!.
nw
eae
:
‘
The indicial equation is c(e+ 2) == 0 and a,
ay
ST
hy 1
Bntonte+d)
nremeunenareaatenmenei
eee
tne
>
forn>1
.
= ———
1 (e+ Ile+ 3)
ag =
7” Q(e+ 2)(c+ 4)
On
"
=
anol
=
AUc+nfle+n+2)
.
(=1)” ao
Wle+1)---(c+n)(c+3)---(e+n4+2)
£4 2%(64+1)--(e+ n)le+3)--- (e+ n+ 2) |
We need to take ag = ¢ + 2, 50 we write
_
:
(e+
Q)ae+1
get?
Ye = (c+ 2)a* ~ FT aleas) * Fler ied ders)
So
+d
Oye _
2
(e+2)2e+4
Be = velna
+ ah — 2(c + 1)(e + 3)
(~1)"a"+e
mer Dlera) vernleray
1
es
ger?
1
“etl
1
3]
1
1
+ Ret
et ayer 4) [ty ors
y
need
ern aay
1
al
(~1)"a"t¢
ar(e+ 1)[(e + 3)---(c+n)|
{(c+ 3)--- (c+ n+ 2)]
1
e+1
e
fy,
ty fst
e+3
c+n
e+3
cHen+24]°
168
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
Substituting c = —2 we obtain
SS
(-1)tttan-2
“3+ omyi) ‘Gn 3 nl~ 24 2% nl (n= 2)!"
yo =ying+a77+
=o
+h
2 (-a)n(- 1+ Ayog + H,,)x"~?
3 nl (n 2)
nes
=ying+y
+2
~Y
+52
aly
(— 1)" (Hn
1
Oe
+
(n
Hy.
3)a
9)
ns
5. Set L(y) = x*y” ~ 2(6 + x)y’ + 10y. Then
oO
CO
L(y) = s [(n+e)(n+c-1) - 6(n+¢) + 10Jana"** — So(n + c)ayzte+}
n=G
n=}
ax
=
oO
KG +e-—2)(n+e-5)anz"?? — IC +e-lay—i2"t?
nal
nel
= (e— 2)(e ~ B)aga® + S$” [(n +e ~ 2)(n +e — 5)an ~ (n+
— 1)an—1]2"**.
neel
The indicial equation is (c — 2}(c — 5) = 0 and a, =
(e-Nle-4)
_t)a
_
e{c ~ 3)
(e+2)ag
3 (e+ Dle-2)
ce
forn 2 1.
c ag
oO
on
(n+e-~-1) Gy
(n+e-2)(n+ce-5)
_
{e+ 1) ag
(e- 1)(e — 4)(e — 3)
(e+ 2) a
(e- Nle—4)(e— 3)(c— 2)
(nte-Vane
in+e-2)(nte-5)
cagz**}
(c+n~-1)
ao
(e-l\(e-4(c—3)(e2) (e+ n—5)
(c+ lagz*t?
Yom aot’ + —TELH * He
(c+ 2)agaet?
Alers) * Elie ale~ Ben]
(c+n— lagx™*¢
+)
Grier
ale~
le
Nee
ATH
18.9,
LOGARITHMIC CASE
169
We need to take ag = c ~ 2, so we write
_
«,
e-2)at!
(e+ 1)(e- 2)a°t?
{c+ 2)x°t8
=(e~ 2)0* + Erayen a * (om Nle=Hle=3) * C= Nle= Hle=F
(c+n—1)a"te
‘eo
Bye _
«,
ele~
2)actl
Be ~Yelnzte +e
fi
(e+ 1)(c — 2)a°t?
* G21
ee
1
1
I
+c
1
1
‘(e+n—5)]’
1
1
1
A(e~-3) Esareseeestrereres]
(e+ 2)act8
1
1
1
1
(e~ INe~ 4)(e ~- 3) le
saosl
(c+n—i)z"te
‘acne
1
4){c~ 3)[(e-1)---(ct+n—- By)
1
lepwca
1
sca
1
1
ecac ca
1
fea}.
Substituting c= 2 we obtain
_
(n+ 1ant?
(n+ 1)27*?
al +3nixed Gebe
“Uns
“Bn
ear3)”
3
3
=
‘
Eea+ a 5t a +e
gea=ying+as
1
== y, Ing+ 5Y! +a?
+ »
( n+ ijort?
Ans)!
a3, 45] [1
—
| were
1 oe oe1
|n+i
ae1
itgti
~ Hiya]
—(n+1)Hn—s] 2" +2
a(n— 3)!
need
7. Set L(y) = 2(1 — zy” +201 — 2)y’ + 2y. Then
ox
Ly) = >, [(n+ec)(n+e-1)+2(n+o)lanz7te"!
n=
_ s(n +e)(n+e-1) +2(n +e) — 2lanz™t?
=O
oo
oO
me Son +e)(n+e+ lane}
read
— dln +e4+2)(n+e-l)anz"*?
n=O
= ¢(e+ Lapa?! + y [(nte\(nt+et lan -(nt+e+1(nte—2an_y]arter?.
n=]
170
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
The indicial equation is c(c+1) = 0 and a, =
(n+¢— 2) ag-1
whe
for
n > 1.
_ (e-1) ag
a=
e+1
n=
Ca,
{e— le ag
ae e+2.
(e+ 1)(e + 2)
_ (n+e-2)an-1
7”
ctn
(e-les-(e+n—2)
ag _
~
fetl)(etn)
(c~ le ag
~~ (e+ n—letn)
We need to take ag = c+ 1, so we have
(ce - Dext? +d
= (c+ La? + (e~ 1a
Oye _
Go
c.
Yen
+a
+ +
cet
(ct+n—Ij(e+n) ’
le teatt®
tant
tat
(c+ 2)
$57 ea Beles Darts
(c~ Ie(e+ 1a*te
e-1 +e
1
1
4 (c+n—I)(c+n)
e432
1
1
1
c+l
ec+n-1
cH+n]
Substituting c = —1 we obtain
y= —2+ 2z,
Qar-1
ox
yo=
=
v1 +1
ying +a27°°
~
Bet
OLD
—eeencammracnemerencaneneateetie.
need
9. (1 —z)y” + 2(1—a)y’ + 2y. Let z= 2 —1 to get L(y)= 2(z+
oO
fy + act
~ 2y = 0. Then
ag
L{y) = S> [(n+e)(n+e-1) +2(n +e) —2lagz™*¢ + Son +e)(n+e—lagz"te-!
n=O
ne
oo
= WC
Oo
+e+2)(n+e—lanz™t? + Yn
n=)
+e)(n+e-lapz*te?
n=
= e(e ~ 1)agz** + y [(n+e)(n+e—- Lan + (ntet1)(n+e-2Qa,_yJ27ter}.
n=l]
18.9.
LOGARITHMIC CASE
171
The indicial equation is c(c - 1) = 0 and a, =
a=
order)
_
(@tem+e-i)
ani
re)
~({c+2)(e-1) ap
(e+e
_ =(e+3)ca,
2
—(n+e+(n+e-2)
— (e~ 1)(e+3) a9
(et+ip
(nt+e+Dnte~2) any
(n+e)(n+e-1)
_ (-1)"(et+ nt 1(e—1) ao
~
(e+1(e+n-1)
*
(1
[e+ 2)--- (et n+)] [le-1)--- (c+
-
—2)] ao
[(e+1)---(c+n)][e---(c+n—1)]
We need to take ap = c, so we have
Ye = C2
c
(c+ 2)(c~ Wz"
e+1
+e
dye
e (e+ 2)(e- yt
de = yolnz
+ 2° —
e+1
VS
—~
(e+ 1)(e+n-1)
(-1)"(c+ n+ 1) (ce - Lez"
(e+ 1)(e+n—-1)
,
1
1
1
chat enl ofl
|
1 4
1 ail
1
c+n+1
e-1
ce e+1
Substituting c = 0 we obtain
yi = 22 = 2(e— 1),
- 1) 41-3
te = In(w
1) +5 Ue
Men"
ne!
ll. Set L(y) = x*y” — 5ary’ + (8+ 52z)y. Then
oO
oo
L(y) = > [(n + e)(n +e — 1) —5(n +c) + 8]a,2"** +
res)
> 5a,x" tert
n=O
oO
Oo
= Sin +e-2)(n+e~- AayatT? + S- Bane"
n=O
= (ce — 2}(c — 4Jagx® +
nol
> [(n +ce—2)(n +c
n=1
4)an + 5an—i]2"**.
1
|
ct+n-l
172
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
The indicial equation is (¢ — 2}(c — 4) = 0 and a, =
_
“1
—~§
Gg
~ba
5? ag
2 = o(e-2)
_
We need
for n> 1.
= Ne=3)
_
n=
ied
cle-1)(e~3)(e— 2)
—5
Qn-1
_
(n+e-2)(n+e-4)
(—1)"5"
ag
— [(e-1)---(e+n—2)][(e-3)---(e+n-4)]
to take ag = c - 2, so we have
_
,
5(c
_ 2)xct}
azet2
= (€~ 2) ~ Taye
ad) * Aen inen3)
(—1)"5% ante
+
[(e— le---(e+n— 2)] (ce ~ 3)[(e~1)--- (etn —-4)}]"
ac
Ing +a" — (ce ~ 1}{e - 3)
5get?
5 ~e-1
1
“de-e-9
1
]
5(e — 2jact}
¢
Oye
1
- 4
1
|-i--4-
oy)
(-1)"5%a7te
1
(e- le---(e+n—2)(e-3)[(C-1)-(e+n—4)]
1
1
{ett
1
tac}
1
a
1
ton
“
sraza}|
Substituting c = 2 we obtain
Ba
wk
+)<<
1 nt i Bi gnt2
nl (n-2)!
oo
w=
aeyk
1)rt
Br gent?
erent
~1)"5"
nine +2? +5054 50!
)
(
na?
oO oy
=yinet+y +2? +5254 50!
1
A,
+
Hy~2
_
,
ijant?
nl (n — 2)!
)"5"(H,
nd
+
Hp—2)2"*?
nl (n — 2)!
13. Set L(y) = 9x"y" — lizy’ + 7(1+2)y. Then
oO
L(y) = $7 [(9(n+e)(nt+e-1) —15(n +c) + Tlana™t? +
n=Q
oO
> Ta,x"tetl
n=l
18.9,
LOGARITHMIC CASE
= 5 (an + 3c —
173
7)(3n + 3c — Dana™t? + > Tan—yx"*?
n=Q
n=l
oO
= (3c — 7)(3e ~ 1japa® + $ > [(8n + 3e ~ 7)(3n + 3c — Lan + 7ap_a]z™**.
n=]
The indicial equation is (3¢ — 7)(3¢c — 1) = 0 and a, =
_
~7F
_
~T ay
~7
Ay]
—_———
(Gnt3e-7)\(In45en1)
P=!>
ag
= (3c — 4) (30 4 2)
“2
Qn =
Ge-1)(3e+5)
_
49 ap
(8e—4)(3e — 1)(3e + 2)(3c+ 5)
~T an-1
_
(-1)"7" ap
(8n
+ 3c ~ 7)(8n+3e~1) ~~ [(3e ~ 4)--(Be + 8n — 7)] [(8e + 2)-+- (3c + 8n — 1]
We need to take ag = 3c — 1, so we have
= (3¢~ lat ~
7(3¢ — 1)a¢t!
Dae dD)
>
(—1)"77gnte
&
(8c
~ 4) [(3c
+ 2)-+- (Be
+ 3n — 7)] [(8c+ 2)-+- (e+ 3n -1)]’
3
3
3
1(8e- ett?
2
Oye _
Be = Yelna
+ 32° ~ Bed 3) Soc
+,
na2
~ 3e-4
at
(—1)"tl7rgnte
(8c
~ 4) [(3e
+ 2) --- (8c-+
3n ~ 7)| [(Sc+ 2)---(Se+ 3n—1)]
Sof
3c ~ 4
8
3c +2
yg 8
3c + 3n
P88
3c +2
— 7
3
$e4+38n-—1)
Substituting ¢ = 4 we obtain
& 1 yrzagnt1/3
oo
“- n=Z
Loy
—3)[3---(3n— 6)][3---Gn)]
7
yo = yi ina
+ 3a
1/3 + 32 4/3 yn
new’
(—1)*th yng
th/3
2, FT a (n— 2)!"
ne Z
(—7)" (Hy + Ayo ~ La2rtt/s
327-1 nl (n— 2)
ih
1.
174
CHAPTER
18.10
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
Solution for Large x
1. 23(2 —1)y” + (x ~ 1)y’ + 4zy = 0. Using the substitutions suggested in the text we get
1
fi
g@’y
wy? dy
1
w? dy
als) (# “aut *™ qw)*\o
-wih) + Sue %
wi ~u) St¥4(2- 2w ~ w +9)
i
0.
Since w = 0 is a R.S.P. of this equation the point at infinity is a R.S.P. of the original equation.
3. y+ acy = 0. Again the change of variable w =
gives
d?y
dy
1
w!
wa tu wit
ion +a 2i 0,
welt
ww
——5
+ 2w*
aay
ty
0.
w= 0 is an 1.S.P. of this equation. It follows that + = 0 is an I.S.P. of the original equation.
1
5. xty” +y = 0. Setting w= = yields
3 (w'
ad?y
a
dw
dy
w5 4 + 2b
HV + y= 0,
+wy = 0.
w= 0 is a R.S.P. of this equation. It follows that 2 = 0 is a R.S.P. of the original equation.
7. cty”
dy
+ 2(1 + 227)y' + by= 0. We set w= 1/ and get Ly) = £4 -wH + 5y= 0. w = 0 is
an ordinary point so we put y = 3 aw".
eQ
oO
oO
8]
L(y) = dona ~ l)a,w"~? — s- nayw” + Ss; Sanw"
wed
n=]
oO
nel
oO
= Yon +2)(n + Danpow” + Sag nex0
iG ~ 5)a,w™
nem]
oo
== Sag + 2a + > [(n + 2)(n + Langa — (n — 5)an|
w
n=l
= > [(n +2)(n + Danse — (n- 5)an| we”
noel
18.10.
SOLUTION FOR LARGE X
175
(n ~ 5) ap
The recurrence relation is dn42 = heiin+d)
a
_
75
for n > 0.
ag
a
eT 2
ag =
ae
a2
=!
Oak =
SEE
an =z
Bd
2,
8
1g
aw '4s-5) 15
(2k
— 7) aax-2
eenmanttinsunneionramntenniuiens
(2k — 1k)
~5L—~3).-a
5)(=8)
«(2h = 7) ao
"T.3--:(2k—1) 2° kl
2
~rim_
a2 3 OO
ang, == O fork
>> 3
—
_
19 ao
(2k —5)(2k —3)(Qk—
1) WF Al
~15w*
lt follows that
—15a~2*
w= » 2 kl (2k — 5)(2k — 3)(2k — 1) “yo 2 kl (2k —5)(2k — 3)(2k — 1)’
Ye = w
2qu
1
be
+ Rv
za -i_
223
L —5
37
+ TRF
9. 2(1 — x)y" — 3y' + 2y = 0. We set w = 1/z and get
L(y)= (w8 - ut)
ee (5w? — uw) ady + 2y= 0.
oO
w= 0 is a regular singular point so we put y = Ss; anw"t,
n=
tet}
(nt+e}(ntco—-1+5(n+ c)ja,w
d=
- Qlanw™t?
[(n +e)(n+¢-1)+2(n4+ ce)
-
ned
oo
= son +e)(n+e+4a,wtet? — Sn +e+2)(n+c~la,gwt?
nf}
nol
= —(¢ + 2)(e— l)apw®
+
> [(n+e-1)(n+e+ 3)an_1 -(n+04+2)(n+¢- l)an}w™*’.
n=l
176
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
The indicial equation is (¢ + 2)(c ~ 1) = 0. Choosing the smaller root, namely c = ~2, we
must satisfy the recurrence relation n(n — 3) a, = (n + 1)(n — 3) ayp_y for n > 1.
1(-2)a, = 2(—2)ao,
a = ag,
2(—1)ag = 3(—1)ai,
@2 = 3ao,
3(O}ag = 4(0)aa,
Ay
ag arbitrary
= (m +3) dn
for
n
a=
5 as
as
6
7
>
4
te"
=
eS
ay,
=
a4
§-6---(n9+
1) ay-1
=
q-5e on
n+1
q
%
y = ag(w~? + 2w7) +3) +43
|wt 3 NAT nn
nad
oo
+ Ss.
Ps ag (x? +22 +3) +43 le
4
ie
ned
1
= ao(z* + 22+ 3)+ rh
oo
iG + 4)a7P8-},
til}
11. 227(x2 — 1)y" + 2(5a — 3)y’ + (c+ ly = 0. We set w = 1/x and get
L{y) = 2w*(1 ~ Ww)
d
d
~ w(w + N=
+(w+1)y = 0.
oO
w = 0 is a regular singular point so we put y = s a,wt,
n=O
oO
L(y) = —- s [2(n + ¢)(n+e~1)+(n+c)—lagw™*et}
aol)
"00
+ s; [2(n +e)(n+e-1)—(n+c)+1lanw™t
=0
“xo
=
ox
5 (2n + 2e+1)(n+e—lagw?*et! + 5" (an +2e-1(n+e—-l)anw™t?
n=O
n=O
= (2c ~ 1)(e— Dagw*
+ y [(2n + 2c — 1)(n +6
ru]
1)an — (2n + 2c —1)(n+¢-2an_s]w"t?,
18.10.
SOLUTION FOR LARGE X
177
The indicial equation is (2c - 1)(c ~ 1) = 0 and a, =
we have a,
=
(n —1) ana
a
for n >
1.
(n+e¢—- 2)
Thus a; = 0 and a,
2n — 3)
yi = w= 27), For c= 1/2 we have a, = Ce
dy.
Qn]
(n+e-1)
forn>1.
= 0 for alln
>
tor n 21.
_ (-1)1 ao
a=
a
=
1- a,
a3
a
_ (-1)-1--- (an— 3) a9
ne
4-36-(Qn=-1) 0
wrtt/2
Thus ya =
1 an -{t
ao
an 1
xn 2
-y4 an —
n=l
13. 227(1 — x)y” — 5a(1 + a)y! + (5 -x)y = 0. We set w = 1/z and get
L(y) = 2w?(w — Se
+ w(Sw + no
+ (Sw — Dy= 0.
oo
w == 0 is a regular singular point so we put y = > anw*?,
n=
oo
Ly) = S> [2(n+e)(n+e-1) +9(n+c) +5]anw™ ter
woe)
°
oS
- S- [2(n +c)(n+e-1) -(n+0) + llanw"t?
n=0
oO
OO
= > “(an + 2+ 5)(n+e+ Lagwitet! — > (2n + 2¢-1)(n+¢-l)ayw"**
n=0
nmQ
= (2c — Ife — Dagw*
+ x [(2n + 2c — 1)(n +e — lay — (2n + 2e + 3)(n + c)an—a]w"*e.
n=l
Forc=1
1.
That
:
is,
178
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
evn
ts
_
l=
da.=
The indicial equation is (2c -~1)(e ~ 1) = 0 and a,
(2n + 5)(n+1) an-1
c= 1 we have a, =
a
n(2n +1)
forn
> 1.
_ 1-2 ag
re
a
(2n + 2c+ 3)(n +c) an- 1
> 1.
(Ont de~1inte~t)
for n > 1. For
T3
_
9-3 a1
ae
eB
= [7-9---(Q2n+5) (n+1)!] ap — (n+1)(2n+3)(2n
+5) ao
nl 3-5---(Qn+1)
~
15
noo
.
ao
Thus y, = w+ rd (n+ 1)(2n+3)(2n+ 5)w"tt=
° nat (n+ 2}(2n +1) dn-1
ce = 1/2 we have
a, =
ae
a
_
ae
n(n
3-3
ag
4:5
a
TT
1)
3c +1)(2n+3)(2n+5)a7"7?, For
5
for
n >= 1
0
1.
"2-3
= [3-4---(n+2)][3-5-+-(Q2n+1)] ag (n+1)(n+2)(2n
+1) ap
nl1-3---(Q2n—1)
~
2
—_
im
oS
Thus yo= w'/? + =Lyn +1)(n + 2)(2n + lw tl? =
2 n+1)(n+2)(2n
4 1j)e7" 7/2,
2 onl
15. a(1+ax)y” + (1+ 5az)y’ + 3y = 0. We set w = 1/a and get
Ly) = ww
+4 LY
dw
+ wiw —3)$Y + ay = 0.
oO
w == 0 is a regular singular point so we put y =
> anw"™*?,
ne
OO
L(y) = > [(n+e)(n+e-1)+(n+ec)]a,w*tet?
=0
.
oo
+
> [(n+c)(n+e-1)-3(n+e)+3]anw"*
nd
oO
x
= S "(n+ caw tt! +S "(n+e-1)(n +c -3)aqu™*
nod
n=O
18.10.
SOLUTION FOR LARGE X
179
= (e— 1)(c = 3)agw® + 3 [(n+e-1)(n+e¢-3)an + (n+e—1)%ag_\}w™*?.
n=]
The indicial equation is (c ~ 1}(c - 3) = 0 and a, =
a=
te
—(n+e-1) a,(wie
3)
for n > 1.
“CC ag
e-2
ap = —(e+1) a;
e-1
a=
no
{(-1)"c(e+1):--(c+n—D ag
(e~2)(e-1)---(e+n—-3)
(-1)"(n+ce-2)(n+e—1) ap
—
(ce ~ 2)(e—1)
.
We need to take ap = ¢ ~ 1, so we have
aye. Ce- Dutt!
Yo
=
Oye =
dc
(€
lw
c—?
A
(-I)"(n+e-2)(n+e- lw
+d
e-~2
,
1 ~ 1
y]
e_ e(c ~ 1)wet? 2[1
eRwtw
e~2
ctenl
end
a
eel
Poy
perl
e-2
m+e-2
“5:
n+e-1l
e-2
Substituting ec = 1 we obtain
= nla
= S-1)n(a arm
n=?
n=?
=
+w+w"w +d
y2 = ylnw
yiln
1
1
—1)"t
a(n — lw wt! 11 4totes
= +
ypr'n(n
oO
= In(i/e)+a7' +a? + S$ (-1ynF tn
Darth
n=2
17. a(1 — x)y” + (1 — 4z)y' — 2y = 0. We set w = 1/2 and get
Ly) (y) ==
w*(w
w"(w
d?y
Fae
-1)55
dy
+ wlw + 2)5
a
ox
w = QO is a regular singular point so we put y = s
n=O
oo
Ly) = So [(nte)(nte-1)+(n+eJanw teh
nm
anw"t,
2y ==0.0
~ 2y
180
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
~ 3 [(n +e)(n +e- 1) — An +c) + Qlanw™te
n=
OO
oO
= WC +c)aywtet? — wG
ned
te~1)(n+e-2)anwt?
feed
= (e~ 1)(e— 2)aqw® + y [(n+e-I(nt+e~2)an — (n+e-1)?ap_y]w*.
n=l
.
(n+c-—1) @n-1
.
wae
The indicial equation is (c — 1)(c — 2) = 0 and On =
op
fore.
a=
c ag
en l
a = eta
c
a
_ ee+1)--(c+n~Ijag
ne
(e=—Dew-(e+n-2)
(n+e-—1)
ao
e~1
We need to take ag = c — 1, so we have
OO
Yo = (c— lw + Yo(n +e~ lw",
m=]
Be _ y. Inw + ue + Sous,
ac
.
a
Substituting c = 1 we obtain
oo
w= So
nel
nw"?
oo
= Sone-"},
n=l
xO
oO
Ye = yi inw+w+ Sow
= yy In (1/z) + Soa
n=]
n=O
1
ey
19. vty” +2c%y'+4y = 0. The substitution w = z and a bit of simplification gives us ao? +dy = 0.
It follows that
y = ¢, cos 2w + cg sin 2w
= ¢, cos (2271) + eg sin (227*).
18.11.
MANY-TERM RECURRENCE RELATIONS
18.11
Many-Term
181
Recurrence Relations
1. Set L(y) = x?y" + 8ay' + (L+2+2°%)y. Then
oo
co
°°
Ly) = So [(n+e)(nte-1)+3(n+e) + onz"*? + So ane™tt! +S" anarters
nse)
n=O
oO
oO
= Yon
+e+
1)7a,2"** + s; Qn ya"?
n=O
neD
OO
+ S,
n=l
An gu"?
n=3
= (c+1)’agx® + [(e + 2)?ay + ag]x°*! + [(c + 3)?a2 + a ]2°t?
oO
+ Ss, [(n +e+1)*an + an. + Qn—3|a"*°.
reed
The repeated root c = ~—1 of the indicial equation will be used to obtain one solution with
ay = 1. Equating each coefficient to zero yields
a1 + ag = 0,
fag + a, = 0,
Naa +On-1 +Gn-3 = 0, forn > 3.
oO
Thus y. = So ana"?
n> 3.
in which ag = 1, ay = —1,
ag = i and a, =
—(On—1
7a
+ On—3)
for
n=O
‘The second solution will be of the form
OO
yo=yiinet+ >” bye??,
n=l
oo
yy =y{ Ine +a7ly + So (n- baa’,
nel
oO
yo = yf Ina t+ Qe ty, ~ Fy, + Si(n —1)(n — 2)byz"74.
n=l
Substituting into the differential equation and simplifying gives us
oO
OO
oOo
oO
So anage”* + So bye) + So baz") + S- bn-3z*7? = 0,
n=l
n=]
nD
nimd
oo
(2a, + by) + (4a + dbo + 8;)x + (Gaz + 9b3 + bg)z? + Ss; [2nan + n7by + ba—1 + bp—s]z"~! = 0.
nod
182
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
Equating each coefficient to zero leads to
b, =
—2a;
= 2,
bo 2 x Dita
_= 84’
4
b
_
oe
by =
~bg-6a3
19
108?
—Uncit bnns)
nn
_ 2m
Tt
torn
>
4
OO
3. Set L(y) = cy” +y' +20. +a2)y. We put y = s; a,xte,
n=)
oC
oo
oo
L(y) = s; [(n+e)(nte-1) + (n+e)]anz™*o! + So agar tett + » ane” ter?
n=O
oo
= So(n
n=O
+ cPPana™te-!
nos}
= aga!
n=O
oo
20
+
> Ongar ten! + Sangre
nm?
nex3
+ (e+ 1)%a,2° + [(e+ 2)"a2 + ag|a°t?
oO
+ ang]aPtet.
+5" [(n +c) an + Qn—2
ned
oo
We choose c = 0 and obtain one solution of the form y, = s
Qn2", where we arbitrarily take
nad
ag = 1, and then must take a, = 0, aj = -4,
and a, = —
Gn—2 + On—3
na
for n > 3,
A second solution will be be of the form
oO
ye = yy inet So bax”,
mah
yy =yinzta ty
oO
+ > nbac™,
rem]
oO
yh = yf Ine + Qe7ly, — Py
+ »- n(n — 1)b_z"~?,
n=?
Substituting into the differential equation and simplifying gives us
OO
OG
oO
OO
> n*b,27} + s ba-gz™) + >; bygt” } = ~2y) = ~2 S Na@,z"1,
fend
nag
reed
nol
oO
(by + 2a,) + (4bo + dag)a + (9bs + by + Gag)z? + 5” [n2b, + ba—2 + dns + 2nay] 2") = 0.
nag
18.11.
MANY-TERM RECURRENCE RELATIONS
183
Equating each coefficient to zero leads to
by
=
~2a, a=
0,
1
by = a2 = 7,
1
6
2
bs = — Gh - 593 = 55,
by = _bn=2 +n
2an forn > 4
ni
5B y= Yo anc”,
Pr
n>
ao arbitrary, a; = 2a9,
az arbitrary, and a, = — (2+ Van-1 = 6an-2
n{n— 2)
for
3.
_
4ag~60,
3
ay =
“4
4322
=
_ _ Sag — Gag
19 |
8
12°77
Gaq — Gaz
7
a5 >
15
os =
Tag
~ 6a
‘=
24
+ 4ag,
=
5
9°
13
gaa
+
_ 53
5 A
_ 83
©6772 *
60
Thus y = ag(x7} +2442? — 8$23 + 321 — 8 x4. -) +a2(2 ~ 42? +22 3 _ ZA + 82° tree).
7. We note that ag =—~ 1, a) =—
i a3 —=
~1, ag =_ 5°
@1
1
aq = 94°
and a
_(n
+ Lan—1
n(n +1) + On—§
for n > 5. Indeed
ag = Oe too
5
30.
ae = etm
oe
a2
w=s
a? +
24°
_ 31
«1008
1,
52°
14,15
+ aq”
at
+ To087
aoe,
x —
[CPR + 2) an + G24-<
ae
1
31,
Note that
ang
2k
x — | PRA Mak + 02-5]
2k(2k + 1)
Observe that each ay, for k = 0,
1,
og
ak+d
(2k + 1)(2k + 2)
2, 3 is positive while each ao,4;
for k = 0,
1, 2,
3
is negative. From the recurrence relation for a9, we see that each successive az, will be the
negative of the sum of two negative numbers; that is, a2, will always be positive. Similarly,
@2%41 Will always be the negative of the sum of two positive numbers, so that a9,4., will always
be negative.
184
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
2
9. a(x—2)?y” —2(2—2)y'+2y = 0. We set z = x—2 and get L(y) =
(242)
9,4
42 rae ty = 0,
OO
where z = 0 is a regular singular point. Setting y =
> an2z"** we obtain
net
L{y) = s(n +e)(n+e-lagz™tert + 5 [2n +c)(n+e-—1) ~2(n+c) + 2la,2"**
n=O
neQ
oO
xO
= Soin +e)(n+e-—lagz™tet! + S 2(n + e—1)?an2z”*°
n=
nad
oo
= 2(¢ — 1)?agz* + S- [2(n +e - 1)?an + (nt+e~1)(n+¢—2)an-i]2"** = 0.
n=l
The indicial equation is (c — 1)? = 0 and ay, =
—~(n+c~2) ani
2(n+e—1)
forn > 1.
a, = =e s
_
“CA
02 = B(e+ 1)
a
= (-1)"({e— l)e---(e+n-2) a9
Wele+i--(e+n—1)
oO
a
Yes? +h
(-1)"(c
(-1)"({e-1) a
~
Wle+n—1)
’
—_ Ajz™te
sate nanaatnaste niainioniineemamessnowene
#(n+c-1)
OY
(-1)"(e— Dero f
dec = Yeln aos
Wile+n—-1)
1
1
le-1
c+n-1]/
Substituting c = 1 we obtain two solutions
Ye
eT
— 2;
nant
a= mine + OE
me
bi
= yy in (a — 7+
n+i
ore
naz]
11. Let L(y) = 2ay" + (1 — xy’ — (14+ 2)y. Then L(e”) = [2a +-2)-+
x) |e” == 0; that
is, e* is a solution of the differential equation Z{y) = 0. Moreover, the general solution is
y = C1y1 + Caye where y; and yg were obtained in Exercise 10. It follows that the solution e*
must be a linear combination of y, and ye; e* = cry, + cey2. Differentiation of this equation
18.11.
MANY-TERM RECURRENCE RELATIONS
185
yields e* = cyy} + cays. We look at the first few terms of the series for y1, yo, yj, yh and have
=
ty [et
+
i
50?
1
e”
=
Cy
+
|
+
Ca
[ber
1
gata]
€
[gotta
Gata
...|
+eg([l+a+---].
We examine the behavior of these series at « = 0. From the first equation we see that cz = 1.
From the second equation we must take ¢, = 0. Hence y, = e7. With this solution of the
differential equation we use the method of reduction of order to produce the general solution.
We set y = ve® and obtain 22u" + (82 + 1)v' = 0. Separating the variables and integrating
(~3/2)
yields the first-order equation vu’ = a
Wg = UE =e
so that finally
o
as another solution of the differential equation.
z
3
obtain ys; = 2e7 [
0
e2”
df.
——=—
VY
dy,
In this integral we substitute y = @? and
It follows that ys = kyy, + kaye. As before we can examine
the first few terms of the power series expansion for both sides of this last. equation for x = 0
and see that we must have ky = 0 and ky = 2. That is, y, = e” [
Miscellaneous
VE
0
7 8/28 ga.
Exercises
1. Set L(y) = cy” —(2+ax)y’ —y. Then
Ly) =o [(n + o)(nte-1) -2n + Jaga” — "lint 0) + Yaga?
n=O
n=O
= s(n +e)(n+e~ 3)anz™te-} ~ yin + c)an—ya"te!
ned
= e(e — 3)apx®* +
nal
> [(n+e)(n+e-3)an — (n+ c)an—s]2"te",
noel
186
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
The indicial equation is c(e— 3) = 0 and a, =
‘
relation we have an
c
(ec
By(e=
exett
Ye = c8 +
ag
Ne
ty
(e4n
eget
3)
for n > 1. Solving the recurrence
e now take ag = c and get
pets
9 + GIVER
* C-DE-D
oO
n--e
+3°
=
;
£~ (e-2)(e1) (e+ 1)---(c +n ~ 3)]
Oe
ac
Ing +2° +
tty
e-2)
cart
(e-2(e-1l)
1
1
oO
e-2
e-1
&
“ettsl est)
e-2|e
etd
(e~2Me-1) |,
1
1
1
e-2
e-1
ctl
v=
ying
+1—
,
le
,
c-2
,
c-l
Seaeaicery
ne
(c~ 2)(e~ 1) [(c+1)---(e+n — 3)]
1
c+n~3]-
1
1
52+
52° + 7 4D
wom
ai [5
~ Hal
3. Set L(y) = 27y” + 2x7y’ — 2y. Then
5s 2(n + c)ana™ tet?
‘ [(m+e)(n+e~1) — Qana?t? + ned
L(y) = ned
is]
OO
= Som +e-2(n+e+
Dagz™t? + s 2(n+¢~l)ag—ya"t?
ned
a=l
= (c— 2)(e + l)aga® + wc
+e-Din+e+ lan + 2(n+e-an-y]2"**.
nol
The indicial equation is (c— 2)(c +1) = 0 and (n+e—-2}(n+e4 ljan t+ 2(n+e—1)an-1 =0
for n > 1. Using the smaller root of the indicial equation, ec = -1, we have
(-2) ‘la,
(—1)-2
+2:
(~1) ag = 0,
ag+2- (0) a, =9,
(0)-3ag3+2-(1) a2 =0,
ay,
=
(n= 3)n
a, = ~a9,
dg = 0,
ag arbitrary,
for n > 4.
18.11.
MANY-TERM RECURRENCE RELATIONS
Solving this recurrence relation we get a,, =
187
6(—1)"-32"-4(n — 2) ag
al
. We therefore have the
solutions
M1
=a}
-1,
©
maa+65~!
nod
¢_y\n—-3on-3 7,
1)
2
(n
Th !
2)x
m1
.
5. Set L(y) = 27(1 + x?)y” + 22(3 + 2?)y’ + 6y. Then
oO
L(y) = > [(n+e)(nte-1) + 6(n +c) + 6Janz"*
nad
> [(n +c}(n+e~
+
1) + 2(n + c)ja,a"te+?
no=O
co
~
= s(n
+¢e+ 2)(n +¢e+ Zagat?
+ Sin
m=O
te)inte+
La," ter?
n=
= (e+ 2)(c+ 3)agz® + (c+ 3)(e+ Dayx*t
oO
+ s [(m+e+2)(n+e43)an +(nt+e-2)(n+e- 1)an—2|a"t?.
nes?
The indicial equation is (c+-2)(c+-3) = 0 and (n+c+2)(n+c+3)an+(n+e—2)(n+c—L)an_2 = 0
for n 2 2. Using the smaller root of the indicial equation, c = —3, we have can leave both ag
and a; arbitrary and a, =
~-(n ~ 5)(n - 4) ana
for n > 4, Thus a, = 0 for n > 4 and we
n(n — 1)
have the solutions y; = 27? — 327! and yz = 7? — 1,
7. Set L(y) = Qay” + (1+ 2z)y' ~ 3y. Then
oo
95
Ly) = 5° [2(n+e)(n+e-1) + (nt+c)janaz™to} + ¥- [a(n + ¢) - Janet
m0
nod
oo
oo
= IG + ¢}(2n + 2¢ — Lagz” te"! + S_(2n + 2¢ — 5)ay ye? te-4
n=O
nel
oO
= e(2¢ — L)agr®} + Ss; [(n + ¢)(2n + 2c — 1)an + (2n + 2c — 5)an—y]a?*e-?.
aml
The indicial equation is c(2c — 1) = 0 and a, =
c= 0 and solving the recurrence relation yields
_
on
~(2n + 2¢ — 5) Gy}
>
> 1.
(n+ ¢}(2n + 2c — 1) for n
3(~1)” ao
(2n — 1}(2n ~ 3)’
_
(—1)"2”
n=l +30 nl (Qn— 1)(2n— 3)"
OH
.
4
Choosing
188
CHAPTER
18.
Choosing c = 1/2 gives us ay =
SOLUTIONS NEAR REGULAR SINGULAR POINTS
>
— +2) 1)Gn—1
—2(nn(an
> 1. Thus a; == 2§ ag and a, == 0 for
wae forn 21.
anny, ares
n > 2. Finally, yo = 2'/? + 209/?,
9. Set L(y) = x(1 ~ 2*)y” — (7 +27)y' + 4zy. Then
oO
Ly) = S- [(n+e)(n+e-1)~7(n+c)lagz™te*
n=O
- Solin tein +e~1) + (n+0) ~ dana
nel)
=
+e)(n-+e~ 8)anzte"! — yin
0
+¢-2)(nt+e+2Manater
n=O
nad
= e(e ~ 8)agz*! + (e+ 1)(e — Taiz®
+ c)an_g]a"te?.
+ S [(n + ¢){n +¢~8)ay — (n+e-4)(n
n=?
The indicial equation is c(c ~ 8) = 0. Using the smaller root of this equation, c = 0, we also
need a, = 0 and (n — 8)a, = (n ~ 4)an—, for n > 2.
(—1)
(-3)
ag =
(2)
a4 = 0,
ao,
ag =
(2)
=
ag
= 0,
a5
= 0.
as =0,
(—1) a6 = (1) aa,
(0)
3 Go,
G2
a6,
=
We can therefore take ag arbitrary and ao, = (k
ao
—
R=fb. 3\fk 2
9°0 92
Finally, y = a0 [1+ $27] +a5 [a8 + k=5
Q\a2k
_
= Da — 2) ag for k > 8.
11. Set L(y) = 4x*y" — 2z(24+ x)y' + (34+ a)y. Then
L(y) = > [4(n + c)(n +e-1) —4(n +c) + 3]aga”t? — s [2(n +c) ~ ljagz™*ott
nO
n=O
oo
oo
a 5 (2n + 2e—1)(2n + 2c — 3)aga"** ~ So (an + 2¢ — 3)an_yx"*°
n=O
n=l
s= (2c — 1)(2c — 3)aga* + So(2n + 2¢ — 3)[(2n + 2c — ja, — ay_y]a"**,
nel
The indicial equation is (2c—1)(2c—3) = 0. Choosing the smaller root of the indicial equation,
c= i yields 2(n — 1)(2na, ~ dp1) = 0 for n > 1. This leaves both ao and ay arbitrary and
18.11.
MANY-TERM RECURRENCE RELATIONS
an
an = >
for n > 2.
189
tg
.
.
.
One solution is yy = z'/?. Solving the recurrence relation yields
ay
ghti/2
Qn = ga-1 yl
Thus a second solution is yg = 3/2 4. v5 5anl
gnt3/2
= 3/2 4 Yer PD
Note that the answer in the text is correct, but perhaps misleading.
a, = —1/2 in the answer in the text we get the solution y = 2'/?.
If we put ag = 1 and
13. Set L(y) = 2ry” + y' + y. Then
oO
oO
L(y) = ¥> [2in+e)(n+e-1) + (n+e)]ane te! + So ana
n=0
nod
mo
ox
= Sin + c)(2n + 2c — la,e? te?! + > Any xrter}
n=O
ress
OO
= e(2c ~ Ljaga®! +
> [(n + c)(2n + 2e— Lan + Qn—1ja"te?,
neexl
~an—1
The indicial equation is c(2c — 1) = 0 and an = Giontde2i)
for n > 1. Choosing
c= 0 and solving the recurrence relation yields
a. =
(=1)" ao
° nit 8 --Qn = 1))
"a"
nat
Choosing c = 1/2 and solving the recurrence relation gives us
ay,
=
(—1)" ao
nl (3-5---(2n+1)]’
=
wes
pif?
a0
_
(-1)"x
n+1/2
thon (3-5. (Qn+1))
15. Set L(y) = 4a7y” — a*y' + y. Then
oO
L{y) = > [4(n +e)(nt+e—1)+1Ja,2"t ~Sr(n
+c)a,a"tet!
nel
n=0
00
oS
= do (2n + 2e~1)Pa,a™t? — So(n +e~-1)an—y2"*°
n=l
n=O
Oo
= (2e~1)?apx® + S$” [(2n + 2¢ —1)?an — (n+¢- Van—1]2"*°.
re]
190
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
(n+¢-1) an-1
for n > 1. Solving the
(2n
+ 2c — 1)?
e(e+1)-+-(e+n—1) ap
recurrence relation we have a, =
. Therefore
(2c+ 1)?(2c
+ 3)?---(2e + 2n — 1)?
The indicial equation is (2c — 1)? = 0 and a, =
e-(c+1)---(c+n—1)a"te
arti
Ger aye. ++ (2c -+ 2n ~ 1)?’
ose +)
AeAepen
_ ne
e-(c+1)---(c+n—1)ar
eigen
1
4
1
c
2
c+n—-1
oe
_
2%x+1
1)?
2
2e+2n~1)
Setting c = 1/2 gives the solutions
on
BREE
2
3:
:
1-3---(Qn—1)a"t/2
Q
tary 7
2x*y” — (1 + 2z)y! + (14+ 32z)y. Then
SO
L(y) =
2
itgt
23" (nl)?
et
17. Set L(y) =
[2
oO
> [2(n + e)(n + ¢-—1)—(n+c) + llagz™*? — S| [2(n +c) — 3Jana™ter!
nel)
n=l
oO
oO
= S>(2n +2¢—1)(n+¢- 1)anzr"t? — So (an + 2¢ ~ B)an—iz"F°
n=Q
nol
= (2c ~ 1)(e — ljapx® + 5 [len + 2¢-1)(n +e ~ lan — (2n + 2c — B)ag_y]2"*°,
n=]
The
indicial equation
is
(2c ~
Choosing c = 1/2 and a
=
1)(e ~
2
1)
2)
eo
=
0 and
On—
(2n + 2¢ — S)an_1
a,
= (n+
2e-1inte—-1)
forn > 1.
we have a, = ~2a9 and a, = 0 for n > 2.
Therefore one solution is y, = x'/? — 223/?, Choosing c = 1 and solving the recurrence relation
gives us
On = nl (n—1)(2n +1)’
oo
a
or
grt
90
(Qn — 1)(2n +1) = boa
grt
(2n — 1)(2n + 1)"
18.11.
MANY-TERM RECURRENCE RELATIONS
191
19. Set L(y) = 4a*y" + 322y' + (1+ 3z)y. Then
oO
[4(n +e)\(n+e-1) + 1Ja,2"*° + 3
Ly) = >
n=O
3(n+e+l)agz"tt!
n=O
oO
o>
= So (2n +2¢—1)*ann"t? + s- 3(n + chan ya"**
n=O
nol
oo
= (2c — 1)?agz* + > [(2n + 2c — 1)?an +3(n + c)an—ija"**,
nel
The indicial equation
is (2c ~ 1)? = 0 and a,
=
~3(n +c) Qn—1
Gn 4 2e— 1?
for n
>
1.
Solving this
recurrence relation we get
ay =
(-1)"3% (c+ 1)---(c+n) ag
(2¢ + ye
-(2e+2n—1)? ’
_
x
os
ye
(—1)"3% (e+ 1)---(c+n) atte
(2c +1)*---(2e4+2n~—1)?
’
(-1)"3"%(c+1)---(c+n) ate
= yeina+
(2c4 1)2-- (e+
2n —1)2
Do,
e+l
dt 4%W+l
ec+n
2c+2n-—1]°
Substituting ¢ = 1/2 gives the solutions
yaar y y sat
nal
yo=yilng
>
aan
(=1)"3" [3 ae
=; Inz + »» (~1)"3"
mins
gntl/2
2)
2
i: bed
a
-
1)]a"*/2 E bok
—
_ n,
1)]artt/2
|
192
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
21. Set L(y) = 4x7y” + 227’ ~ (x + 3)y. Then
lJanartet}
oO
+ -1) — BJana"*? = = S72[2(n +c) —
L(y) = S> [A(n +0)(n
n=0
n=0
oO
oo
= >" (an + 2¢4 1)(2n + 2c ~ 3)anz™** + > (an + 2¢— 3)an_ya"t?
nol
n=l
oO
= (2e + 1)(2c — 3)aga* + >, [(2n + 2¢ + 1)(2n + 2c — 3)an + (2n + 2e — B)an—y]a"t*.
naz]
The indicial equation is (2c-+1)(2c—3) = 0. Choosing the smaller root of the indicial equation,
c= ~1/2, and n(n — 2)ay + (nm ~ 2)an_1 = 0 for n > 1, we have
2-1-(-1)
ay + (-1)-a9 =0, a ==,
2-2-Q0a2+0-a,
=0,
(-1)"-?
and for
n > 3, On = FF
ag
arbitrary,
ag
974
Thus
~1/2
1/2)
2
os (— 187 a
need
Note that if we take ay = 1 and aj = 3 we have the solution
yaa?
+e
b~ 3 + a
1)"2"
1a)
(s/so(-2/2),
a gl)
That is, 2'~1/2) e(-#/2) is also a solution of the differential equation.
23. Set L(y) = x2y” + 2(3+a)y + (1+ 2x)y. Then
oo
oO
L(y) = s- [(n+c)(n+e-—1)+3(n+c)+1a,e™*? + S(n +e+2agzter!
n=O
n=0
oO
oO
= Sota +e+1)apz™t? + Si(n +e+lay—i2"**
n=l
n=l
Om
= (e+ 1)*agz° + > [(n+c+1)*an+(n+e+I)an—y]2"*°.
mo]
18.11.
MANY-TERM RECURRENCE RELATIONS
193
— Ont
The indicial equation is (c + 1)? = 0 and a, = ntet+l
for n > 1. Solving this recurrence
relation we get
(-1)" ao
~ (e+ 2)--(etn +l)’
_
Yes
oe
of
+
(-1)*
ery
OYe
Bm
arte
ern Fy’
(-1)"
uma+
2?te
oh+2).- sees
| aa
cael
Substituting c = —1 gives the solutions
peatSe att
ye cota
rn
reed
1)"+1
Y=
whet
5
7,
gn
1
CU.
25. Set L(y) = a(1 — 2x)y” ~ 2(2 + x)y’ + 18y. Then
L(y) = Ss, [(n+e)(n+e-1)—4(n+0e)]a,2"te"!
n=O
- 5 [2n +c)(n+¢e—-1)+2(n +c) - 18]an2"**
n=O
= y(n
+e)(n+e-5)a,z™te-? — s An+e-3})(n+et+ 3)agz™t?
re=D
n=)
= e(e — 5)agz®! + 3 [(n +e)(n +e - 5a, — 2(n+e~- 4)\(n+e+ 2)an—sJarte“},
n=l
Choosing c = 0 and n(n — 5)ay = 2(n — 4)(n + 2)a,_1 for > 1,
L-(—4) a, = 2-(-3)-3 ao,
a; = 3 ap,
2+(—3)
a2 = 12 ag,
ag = 2+ (~2)-4
a),
3-(—2) ag = 2+(—1)-5 ag,
ag
= 20 ag,
4-({~1) ag = 2-(0)-6 a3,
5- (0) a5 = 2-(1)-7 ag,
aq
= 0,
ag arbitrary.
194
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
2"-5(n— 4)(n
+ 1)(n + 2) as
For n > 6 we solve the recurrence relation and get ay, =
|x?+
. Finally,
2-5
(n — 4) (n+1)(n + 2)2”
2
oo.
3 + a5
y _= ao (1+ Sa + 12z°2 + 202°]
607
5
ne
= 500 [2 +90 + 24x 2 + 402°]3 + 3
2 nm (n— 4)(n + 1)(n
+ 2)z"
ai
nm=$
27. Set L(y) = a7y” — 3xy' + 4(1+2)y. Then
oO
L(y) =
oo
> [(n + e)(n +e-1) — 3nt+e) + 4Janc?** + s 4a,ertert
ned
n=l
x
oO
= Sin +e~2)Pagx™t? + s- day xt
n=)
n=]
oO
= (c — 2)agr* + S; [(n+¢-2)"an + 4anr]a"*e.
nee]
ery
tds
*
.
_9\2
=
=
The indicial equation is (c - 2)* = 0 and ay
—4 an
Tt
>
mien?
Fea
for n > 1. Solving this recurrence
relation we get
_
7
(—1)"4" ag
(e-1)P
‘(e+n~ 2)?’
(- 1)F4"
ene “+>
ax
OY.
(
_
gre
(e~ 1)?--- (c+
— 2)?’
=
ante
(-1)"4"
See
2
e-1
~~
2
c+n~2
Substituting c = 2 gives the solutions
1=z
epee
~1
nan
ghee
head
am
n=}
1)"4"H, ot?
yo =
Ui Ing
-—2 s a
nel
—1)"4”"
ght?
18.11.
MANY-TERM RECURRENCE RELATIONS
195
29. Set L(y) = 4x?y" + a(x — 4)y’! + (5 ~ 3x)y. Then
L(y) = So [al + o)(n +e~1)—8(n+0)+5]ana™* +S [an +e) — sJagatte
n=O
n=O
oO
oo
= 5 (2n + 2c ~— 1)(2n + 2¢ — SJana™*? + S-(an + 2e ~ 5)ay—y2"t°
n=
nol
oO
== (2c — 1)(2c — 5)agr® + > [(2n + 2c - 5)(2n + 2c — lay + (2n + 2c — 5)@n— Jat.
nel
Choosing c = 1/2 and 2n(n — 2)an + (n ~ 2)an_1 = 0 for n > 1.
2-1-(~1)- a, + (-1)
+a = 0,
a, = —4 ao,
2-2-(0)-ag+(0)
-a, =0,
ag arbitrary.
—1)\"
For n > 3 we solve the recurrence relation and get a, = es,
oo
y = ao [2¥? > 42/2]
(-
+a2,)>
nax2
oo
=
dy
[a2
-
30°"
+
2
ay
n=O
Finally,
1)Pamti/2
gn-3 ny!
(-1)"a"t8/2
2e-) (n+ 2)
31. Set L(y) = a(1 — x)y” ~ (44+ 2)y' + 4y. Then
[(n+e)(n+e-1)-4(n+ c)|ana" te"?
L(y) = >
neal
_ Jolla +e)(n+ce—1)+(n+e) —4]agz™t*
nal
oo
oo
=} \(nte)(n+c-5)ana%te} —
reed
KG +e-~2)(n+e4 2)ayz™t?
ned
= c(e — Saga?7} +
oO
> [(n+c)(n+¢-5)ay -(n+e—3)(n+e4 Vania? te},
nse}
Choosing c = 0 and n(n — S)a, — (n ~ 3)(n + Lay, = 0 for n > 1 we have
1+ (4) ay ~ (-2)-2
ap =0,
a; = ao,
2-(-3)
= 0,
a3
=
ag = 0,
a3
= 0,
3-(—2)
ag ~(-1)-3a,
a3 — (0)-4
4-(-1) aq — (1)-5 ag = 0,
5+ (0) ag — (2)-6 ag = 0,
a9,
ag
= 0,
ag arbitrary.
196
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
We solve the recurrence relation a, =ene
Gn ~~
=
Motte
and get
for n > 6. Finally,
y= ag [1 +a2+ hx7] + ae s(n ~ 4)(n —3)(n + Dz".
riseh
2
33. 2(1—z)y"—(4+a)y'+4y = 0. We let z = 2-1 and get L(y) = (241) 4+ (245)
dy = Q.
Then
L(y) = y [(n+e)(nte-1) +(n +0) —4A]anz™*°
+ Y[tn +e)(nte-1) +5(nte)janzte!
to)(n+e+4)anzte!
7 ye Fem a)(n+et Danz0s + Yn
oO
= O(¢ + A)agz* + », [(n+e)(n+e+ dan + (n+e-3)(n+e4lanaj2rtet.
n=l]
Choosing ec = —4 and n(n ~ 4)a, + (n — 7)(n — 3)an-1 = 0 for n > 1 we have
1+ (~3) ay +
2+ (—2) ag +
3-(~1) a3
4-(0) ag
(—6) - (—2)
(—5)-(-1)
+ (—4)
- (0)
+ (—3)- (1)
ag
a,
ag
ag
= 0,
= 0,
= 0,
= 0,
a,
a2
a3
a4
= 4 do,
= 5 ao,
= 0,
arbitrary,
5. (1) a5 +(-2)-(@) a4 =0,
as = $ 04,
= 0,
(3) as
6 - (2) ag + (~—1)-
ag = £ Ga,
7+ (3) az + (0) - (4) ag = 0,
n > 7.
ay, = 0 for
Finally,
y = ag (274 + 4279 4 5277] + ag[1 + dz + b27]
= ag[(x ~ 1)~4 + 4(@— 1)79 + 5(@ - 1)7?] + aa [1 + f(@ - 1) + 3 (@-1)].
1811,
MANY-TERM RECURRENCE RELATIONS
197
35. Set L(y) = a(1 ~a)y” + (1 — 4ax)y’ — 2y. Then
*te!
L(y) = s [(n+e)(n+e-1)+(n+oc)jayz
n=O
oO
ee Sln
+
c)(n
+o
1) +
d(n
+ c)
“fe 2|anz"*¢
n=O
OO
oO
= o(n+cPanz™*e! — S"(n¢e+l(ntet 2)ay2"*¢
n=l
nal
oo
= Papa? + S7[(c+n)Pan — (e+ n)(c+n+ Iaaij2"t?.
nad
1)
aa.
The indicial equation is c? = 0 and ay, = (c+ n+ 1) Qncn
for n > 1. Solving this recurrence
relation we get.
_ {e+n+]) ao
an
~
c+1
oO
Ye
=
=X c +
RCA
OYe me ye In
mee
Oc
,
(c+n4+1) att
onl
,
ALERT SAORI
oS. (¢-+ +1) atte
yen
+ J
1
e+l
1
ns
ct+n+1
;
cl
Substituting c = 0 gives the solutions
oO
xD
w=1+So(nti1) a" =So(n4+)) 2%,
nal
nee
oO
1
oO
yo = yi Ina+ Sola +1) 2” sn
n=
n
_ | = ying ~ Sona”.
+
nol
37. Set L(y) = zy” + (1 — x)y’ + 3y. Then
oO
Ly) =
,
oO
| [(m+e)(n+e-—1)+(n+c)lana"te! — Sin +e-3)a,27*¢
nol)
n=O
OO
= (n+ cana!
nal
oO
— Son te
Aan
arte}
n=l
Oop
= Cagr® t+ s- [(e+n)an — (c+
no]
—4)an—a]a?ter?.
CHAPTER
198
18.
The indicial equation is c*= 0 and a
(c— 3) ao
relation we get a, =“edt?
ag =
SOLUTIONS NEAR REGULAR SINGULAR POINTS
=€ +is ar 2)
(c(er— 3)(c
— 2) ag
ipier a2”
forn > 1. Solving this recurrence
a3 =
(c— 3)(c
— 2)(c — 1) ao
(oF 1)
4 2)%e4 a2? nd for
e— 3}(e ~ 2)(c— Le ao
ned
a=
[e+ SFA
2,
ora
(e-Sjatt! = (e-3)(e—2)a°t?
Ye = e+
1)(c +n)
(ec — 3)(c — 2)(e — Ia**4
ye
+ eriie+a * (er ier 2)%e+ 3)
[(e+1)---
(c+ n)](c+n—3)(c+n—-2)(etn—-Il(e+n)
+ y
(ce ~ 3}(c
~ 2)(e — Yc aot”
&
[Aaa
TOS
Be sume
2
1
e~ 3jact!
Oye
(c—3)(e—2)a°t? [ 1
+ (e+ 1)*(c + 2)? [3
(c-3c-2(e-ae*
(e+
(e+
(e+ 3)?
1
ety-
fl
2
9
1,
iol
latstea
1
2
an
2
2
-aa
(e — 3)(c — 2)(c ~ 1)eae**
“hy (e+1)-
1
i
a=3te=2 ateni i+
ae
ctn-l
fetnlerne
3)(c+n—2)(c+n—-1)(e+n) © [P(e) ~ QCe)],
and Q(c) = cti’
+o4n'ctn-3
edna
+ orn
Substituting c = 0 gives the solutions
_
w=
32+
32
52
13
er
x
zt
aa
235
U 2
_ 23
pen
ya = ying
+ 72x
72 + 757 “6a n(n — 1)(n — 2)(n — 3)
39. Set L(y) = Qx(l — xy” + (1 — Qx)y’ + By. Then
L{y) = + [2(n +e)(n+e-1)+(n+c)]anz™te"?
ned
~
> [2(m + c)(n +e-~1) +2(n+c) —8]a,2"*°
nod
= Sr(n + ¢)(2n + 2¢ — lage” te"! — > An +c-2)(n+ e+ 2a z"*°
n=O
n=O
oO
(2c~ l)agr®! +
IG + e)(2n + 2e~ Ian — 2(n +e 3)(n+c4 1)ay—y]a"te".
nel
18.11.
MANY-TERM RECURRENCE RELATIONS
199
a(n +¢~3)(n +e-+1) ane
The indicial equation is c(2c ~ 1) = 0 and a, =
(ntcj(Qntac-1)
1
rr >2}
2(n ~ 3)(n +1) ana
we have a, = —8a, a2 = 8a,, and a, = 0
n(2n —1)
for n > 3. Therefore one solution is y, = 1 — 82 + 8x7. Choosing c = 1/2 and solving the
Choosing c = 0 and a, =
recurrence relation gives us
_
3[5-7---(2n+3)] a9
oF al Ga Gn = Hen
_
ee
i/2
1)’
(Qn + 3)antt/2
+8 aannl 5
3)(Qn —1)(Qn+ 1)
41. Set L(y) = xy” — 32(1 + z)y’ + 4(1 —ax)y. Then
oO
OO
L{y) = S| [(n +¢)(n+¢-1) -3(n +0) +4]anz**? — S= [3(n +c) +4]a,a"tet!
n=Q
ye}
oO
oo
= Yo(n +c-2)*a,a"t* — >_(3n + 3¢4+ Lanz"?
n=0
ni
oO
= (e — 2)Papz® +
: [m+e- 2)*a, — (8n + 8c+ Iaa—ija**.
n=]
The indicial equation is (c — 2)* = 0 and a.
_ (3nr +3¢4+1) an-i
(nte_ oy
for n > 1.
recurrence relation we get
an
= (3c + 4)--- (30+ 3n +1) a9
e
(e~1)?
(e+ n~2)?
oO
me
Ye=a°+ D0
’
(3¢ + 4)---(3e+
38n4+ 1) a™t¢
(c-1)?:-(etn—-2p
nel
ee
= Yelna +e
3
(8¢ + 4)- Get ant) gnte
(6-1)? (e4n—2)
0,
38
3c +4
2
3e+3n+1
e-1
c+n-2|°
Substituting c = 2 gives the solutions
naar had
=
wee (8n+7) x n+2
(nl)
oo
neulng+
2
7
10-+-(8n+7) 2™*?
ye
[3
jo tt
3
ayy
7 2A
Solving this
1.
200
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
43. Set L(y) = 2xy” + (3 —a2)y’ — 3y. Then
oo
co
L(y) = > [2(n + e}(n +¢~1) + 3(n + c)janz™te7} —
n=}
KG +c+3)ayz"t¢
nod
x
oOo
= Son + c)(2n + 2c + Lagx™t? — Si(n +e+2)an,ye"te-t
n=O
n=}
OO
= (2c+ 1)agz®! + 3 [(n + c)(2n + 2e+ lan — (n+ 64 2)an—-y]a"te?.
reed
The
indicial equation
is c(2c + 1) =
0 and
a,
(n+ e+ 2) any
= (ato)(an + 2eF1)
for n > 1.
e = 0 and solving the recurrence relation we have
On
"™
_= (n+
1)(n +2) a
SS DT
2[3- 5
@n+0)]
1) te Tse.
+ 2)a"
si+ )°A({n “:A
ie
(Qn+0]"
Choosing c = —1/2 and solving the recurrence relation gives us
7
fan)
ao
on = On nli-3---(Qn—1)
_
a7
1/2
(2n +1)(2n
+ 3) ag
3-27]
(2n +1)(2n
+ 3)z"-1/?
+ ene ral
,
;
hE
45, Set L(y) = xy” + 3y’ —y. Then
L(y) = 3 [(n+e)(n+e-1) +3(n+c)]ana”te-} — s ayn”te
ne
n=O
On
OO
= Sa +e)(n+e+ 2)agz™ten! — s a," ten}
Tx}
nol
oo
= ¢(¢ + 2)agzr°! + SS, [(m+e)(n ++ 2)an —an—ya7ten,
nem]
Choosing
18.11.
MANY-TERM RECURRENCE RELATIONS
201
ani
The indicial equation is c(c
+ 2} = 0 and a, = hio(ntedd
1 (et Ie +3)
=
ag
a
a
EO
ee
forn> 1.
2 (e+ I(e+ Het 3)(c + 4)
ay, =
oo
(c+ 1)(c + 2)[(e+ 3)---(e+n)] [(e + 3)---(c+n+2)]
for
n > 3.
We now take a9 = c+ 2 and get
_
.
(e+ 2)ac+}
Ye = (e+ 2a +
get
eds) * ebiler
ayers)
oo
gre
+d
Galery
Oy.
2,
Ge =Yelne ts
emerson
(e+2)c°*}
1
era ray
1
1
+ CPI (e+5) Estresieresd
e+?
_
‘ermeraners |
1
1
1
aa
gite
‘Dames
_
1
_
e+l1
1
“(e+n)|[(e+3)-(ctnt2)]
feed
e+3
1
_
c+n
1
4
4
e+3
1
etnt2}}-
Setting c = —2 gives the solutions
1
n=—3~
yi
oO
gt
2
Tee
V(n — 21
k
wd
va =yiing +a?
=
l
oO
ving +o
~o ama
era
1
W(l-F
+] “ys
_ gig
1)
Gl
7 -y!
n=]
n
Hy,
—2)a"~?
2
n! (n— 2)!
- An — Any2)2"
n+2
nl (n +2)!
202
CHAPTER
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
47. Set L(y) = 32y” + 2(1 — x)y’ — 2y. Then
L(y)
= S> [Bin + e)(n +e~ 1) + 2(n +c) ann” 1 S°(0n 4 204 2)an ante
n=O
n=O
oO
ox
= Soin + ¢)(8n + 3¢ — Dage™*te! — > 2(n + c)an—y2"ter!
n=O
spe]
oO
= (3¢— l)apz®? + S> [(n + €)(8n + 3e ~ Lan — 2(n + c)an—s]atO"",
net
2(n + ¢) dn-1
en
nt(Gn +en3c—1)
er for n >> 1. 1, Choosing
;
The indicial equation is c(8c — 1) = 0 and a, = (nt
c= 0 and solving the recurrence relation we have
_
2”
ay
~ 2-5++-(8n—
1)
Choosing c = 1/3 and solving the recurrence relation gives us
2"
2" a9
ay
=
3"
gn nl
¥
_. pl fS
peat
On
mee
9
3
aMSexp
(3 2).
“3 nl
49. Set L(y) = xy” + (3 — 2x)y’ + 4y. Then
xO
oo
L(y) = > [(nte)(n+e—1)+3(n+c)janz™te"! — > (2n + 2¢~ 4)anz™*?
aos)
nO
00
oO
= Sim +e)(n+e+2agz™te! — > 2(n +e —- Qanz™t?
n=O
n=)
oo
= e(e+ 2)agz*' + So [(nte)(ntet an — 2n+e-3)an—afate
new]
18.11.
MANY-TERM RECURRENCE RELATIONS
203
The indicial equation is c(c +2)= 0 and a,
wee
=
a=
forn>1
2{e — 2} ao
+ (e+ Dle+3)
_
27(¢
~ 2)(e — 1) ap
2 = (e+ 1(e+ 2)(e + 3)(e 4 4)
3
_
{ce — 2)(e — 1c ag
(e+ let Dle+3)%{e+
4 (c+5)
_
24(c — 2}le— 1c ag
“a= (c+ 2)(c + 3)?(c + 4)2(¢ + 5)(c + 6)
2" (e~ 2)(e— Le ag
“(e+n— 2)(e+n— 1)(e+n)(e+ 3)(e4+ 4)--- (c+ 242)
for n> 5.
We now take ag = c+ 2 and get
We-2N(e+ 2x!
(e+ 1)(e+3)
8(e — 2)(c ~ 1ex*t3
4(e— 2(e — 1) z°t?
(e+ I}(e+ 3)(c+ 4)
16(¢ ~ 2)(c — 1)er*4
Ye = (C+ Q)a* +
(CF NE+ Hl
+ >
nnd
lC+ NlE+ le+ NEF
EF 6)
2"(c — 2)}(e ~ Lele + 2a"t°
(ct+n—2)(c+n—-DYl(et+ net 3)(c+4)---(c+n+
2)’
Bye _
Oc
C+N EFS) "(CF
2(e—2)(e+2)aet! f
yelng + 2° +
{e+ i}(e+ 3)
A(e~2)(e—1)x°t? [
1
4
1
c~2
e+2
1
1
2
1
e+1
1
ec4+3
1
fe+1)(e+3)(c+4) [e-2
ec-1
ec+1
ct+3
e+4
8(e - 2)(e — 1jer*t?
1 1 1 gil
1 2
1
(e+ 1}(e+ 3)*(c+ 4)(e+5) l[e-2
c-1
¢
e+1
+3
ec+4
16(c — 2)(e — 1)ex*t4
1
1 afm1 oe 2 _ 2 wn 1
(+ 3)%(e + 4%(0-+ S)(0-+ 8)
c-~l1
6
ec+3
c+4
ec+5
ye
2"(c — Ble — lele + Barre
.
1
ct+5
1
—
c+6
((e+n—2)(e+n—I)(e+n)[(e+3)---(ct+n+
2)
1
c~-2
i
ec~1
a4
1
1
e+2
c+n-2
_
1
1
1
c#n—-1l
cin
e+3
4
1
ec+n+2)]
Setting c = —2 gives the solutions
yy = ~24 + 322 — 82",
-2 + 8x7! + 26 _ e r+ 0 2 7)
yo =— yilng
+ a7?
gngn-2
snl(n~A(n—3)(n— 2)
204
CHAPTER
51. Set L(y)=
18.
SOLUTIONS NEAR REGULAR SINGULAR POINTS
2x2y” + 3ay’ -— (14+ ac)y. Then
gntetl
L(y) = $= [2(n+e)(nt+e~-1)+3(nt+e)-1] anr"t © Sam
n=0
n=0
oO
OO
a= So(n +e+1)(2n + 2c — lagnz™*? — 3 An—1z"t?
reael)
nol]
= (¢+1)(2¢ — 1)ag2® + yilln ++
1)(2n + 2¢— lan ~— a_—i]a"*°.
n=l
The indicial equation is (c + 1)(2c - 1) = 0 and a, = WietiGeteeD
Choosing c = ~1 and solving the recurrence relation we have
ay =
#0
ni (—1)-1---(2n
— 3)’
=
was
yl
oo
+a
x
n-1
(<i):
1 (Qn 3)"
Choosing c = 1/2 and solving the recurrence relation gives us
ay,
=
——_—e
_
nid. 7-.-(2n 43)’
= yi?
ya=e
es
gnti/2
“y
nb? (an+3).
forn > 1,
205
Chapter
20
Partial Differential Equations
20.3
1
Method of Separation of Variables
Ou _ pS
at? ~~"
We put u(x, t) = X(x)T(¢), separate the variables, and arrive at the equations
Aa?"
TH
xX”
apt
r=
If c = —f?
< 0 then T” + a?6°T
= 0 and X” + a287X
= 0, so that
u(x, t) = (A; cos aft
+ Ag sin aft)(B, cosafx
+ Bz sinaGz). Ifc = 0 then T” = 0 and X” = 0,
so that u(x, t) = (A3+A4t)(B3+B4r). Ife = G? > Othen T’-a?Z?T = 0 and X"—a?g2X =0,
so that u(x, t) = (As cosh aft + Ag sinh aft)(B; cosh ax + Bg sinhafz).
au
AE
+
du
2b
ft
=
if
ured
ver
@
2 Pu
aa
We set u(xt)
= X(x)T(t),
separate the variables, and arrive at
a
= -
= ¢, If ¢ = —k* we have T” + 267" + k’a®T = 0 and X" +42X
= 0. The
auxiliary equation m? + 2bm+ k*a? = 0 has roots m = —b+ Vb?
— atk. If b? — a?k? = 4? > 0
then T = e~*t(AyeT + Age) so that
u(x, t) = e~* (Aye + Age”) (B, coska + Bysin kz).
If b? ~ a?k? = —6? < 0 then T = e~ (Ag cos dt + Ag sin dt), so that
u(x, t) = e~°'( Ag cos dt + Aqsinét) (By coske + Bosin ka).
If b? — a?k? = 0 then T = e~**(As + Agt), so that
u(x, t) = e~ (As + Agt) (By coskz + Bg sin kz).
Ifc = 0 we have T” + 207" = 0 and X” = 0. The auxiliary equation m* + 26m = 0 has roots
m=
0, —2b, so that
u(z, t) = (Ar+ Age”) (Bs + Byz).
If c = k* > 0 we have T” + 267" — k?a?T = 0 and X" —k?.X = 0. Here the auxiliary equation
m? + 2bm — k?a? = 0 has real roots —b + /6? + k%a? = —b+ yp. It follows that
u(x, t) = 7
(Age
+ Aipe™#*) (Bge** + Bge~**).
206
CHAPTER 20.
ow
5. tr
~
ow
= w + Vay"
f
PARTIAL DIFFERENTIAL EQUATIONS
.
:
:
We put w = X(x)¥(y), separate the variables, and obtain the equations
4
= 1+ ue = k. Thus 2X’ ~-kX = 0 or y¥’ —(k -1)Y = 0. These equations have
solutions X = Bark and Y = Cy*—!, so that w = Ax*y*~},
Au
dui
du
ae
aai +4>——
aadt + b=
Oe
_
.
= 0. We make the substitution u(x, ¢) = X(z)T(t) and get the equation
X"(x)T(t) + 4X" (2)T(t) + 5X(2)T"(t) = 0. Dividing by the product X(x)T(t) yields the
x?
equation
-
xX
fr
+ 4
TH
FP + a
= 0, but the second term keeps us from separating the
f
variables in this equation unless either *
eo
is independent of x or i is independent of t. One
suggestion would be to set T(t) = e** so that z = k, Then the variables will separate.
Exercise 8.
= 0.
ff
frees
(on
=
*
See
We separate the variables by setting u = f{ax)g(y) and obtain
Byv
a. =.
If c= —4k? then f” + 4xf’ + 4k?
f = 0 and g” — 4k?g= 0. Thus
u(z, y) = [Are7*¥ + Age*¥] [Bi fi(x) + Bofo(z)],
where f; and fy are any linearly independent solutions of f” +42 f’+4k?f = 0. These solutions
ao
oO
may be found by setting f(z) = S> azz” to obtain s [(n+2)(nt+an42 +4(n+k*)an| a2” = 0.
nal
n=
The recurrence relation can be solved to get
__ (~4)™h2(k? +2) +++ (k? + 2m ~ 2) ag
A2m =
G2m+1
(2m)!
and
+ 3) - + (k? + 2m — 1) ay
__ (-4)™(k? +. 1)(k? (im+i)I
=
so that
— 2)?
+ 2mm-2)%m
++ (k?
+ 2)+at
(—4)™k?(k?
yk"
fi(z) “14
OO
f(z) =2+ 50
(4
4
mK?
44
k? 43)---
) [eet
m —— 1)a2m+1 1)e2™+1
k?
4+. 89m
msl
If c= 4k? then f” + 4zf! ~ 4k?f = 0 and g" + 4k?g = 0. Thus
ulz, y) = [As cos 2ky + Aqsin 2ky| [Bs fa(x) + Ba fa(x)],
20.3.
METHOD
OF SEPARATION OF VARIABLES
207
where fs and f, are any linearly independent solutions of f"+4a f'—4k? f = 0. These solutions
may be found by replacing k* by ~k? in fy and fg. We get
fala) =14 s (—4)™(—k?)(2 — k?)(4 — k®)--- (2m — 2 — k?)x?™
(2m)i
reel
OS
and
we
Am
—
be
2
4
nee
Bye
2m4}
If c= 0 then f” + 42f’ = 0 and g” = 0. These differential equations can be solved to obtain
u(z, y) = [As + Asy] E + Be / exp (—2z7) as| .
See Section 5.6 in the text for possible ways to treat the nonelementary integral in this set of
solutions.
11. u= fi(z — at) + fo(a + at). We let y = x — at and z= 2+ at. Then
Ou _ di Oy | df, 0z _ dfi | df
Pu Pf
@x
ax? dy? dc
dy Ox
dz Ox
dy + dz’
Gu_ fi Oy , Ufo dz _ di
Wa
at
a2’
dy Ot
dz Ot
“ty
Gu ag
OE
dy
Pfrdz CA
df
dz® ax ‘dy? + dz’
fy,
fo
@
Oz _ weft La 2d? fo
dy? at
"ae ot
dy?
‘dz
au = got
We see that —~
3
13.
Ag
Ow — df dz
df
Ow
_ af dz
df
w== (2x +9").2 We set z= == 22 +y". a Then a- == on
7 = = 25° and =By ~ dz dy ~= Mde’
follows that Ow
Oy
= Ow
Yon"
It
208
CHAPTER 21.
Chapter
21
Orthogonal
21.6
Other
ORTHOGONAL SETS OF FUNCTIONS
Sets of Functions
Orthogonal
Sets
1, From equation (3) in the text we have
D{ze™*Li(x)| +ne“*La(z) =0
and
D[xe~*Li,(x)] + me~*Lm(x) = 0.
Multiplying the first of these equations by L,,(2), the second by Z,(x), and subtracting we
obtain
(m — nje“* Lm(2)Ln(2) = Lm(2)D[xe™*
Li, (x)] — Lp(x)D|xe~*
Li, (x)].
(21.1)
We also note that
D{xe~* Lyn (x)Li,(2)] = L,(2)D|ze~*
Li, (x)| + ze™*
Li (x) Li, (2),
D{ze~* Lm(x)L,,(x)] = Lm(z)D[ze~* Li,(x)] + 2e7*
L,, (x) Li, (2).
It follows that
Lm{x)D[xe~*
Li, (2)] - La(x)D[xe*
Ly, (z)| = D[xe~*Lm(x)L},(2)| — D[we~*Ln(x)
Li, (2)].
Equation (21.1) can now be written
(m —nje"*
Ly (x) L,(z) = D [ze~* {Ln
(x) Lt, (2) - Ly(%) Li, (z)}] .
An integration gives us
(m—n) [ * e7* Lin (2) Ln (x) dx = [we~* {Lm(x)L},(x) - Ln(x)
Li, (2)}]>.
As x — oo the function on the right approaches zero because L,,(x) and L,(z) are polynomials.
Moreover xe~* is zero for = 0. Hence we have for m # n,
[
0
7 Lm (2) Ly (2) dx = 0.
The Laguerre polynomials are orthogonal with respect to the weight function e~* on the
interval 0 < @ < o<.
21.6.
OTHER ORTHOGONAL
SETS
209
3. From equation (6) in the text we have
D{exp (—x”)
Hi (x)] + 2nexp(—a?)H,(z) = 0,
Dlexp (—x?)
Hi, (x)| + 2mexp (~a*) Hy, (x) = 0.
Multiplying the first of these equations by H,,(x), the second by H,,(z), and subtracting we
obtain
2(m — n) exp (—2”)Hm(x)
Hn (2) = Hm(x)D[exp (—2)
Hi} (x)| — Hp(x)D [exp (~2?) Hi, (x)).
(21:2)
We
also note that
Dlexp (—x?)
Hn (2) Hp, (2)] = Ha(z)
D [exp (—2”)
A, (x)] + exp (—2)
H, (2) H,, (2),
D{exp (-2?)Hm(x)H,(2)] = Hm(2)D [exp (-2”)H,, («)] + exp (—2”)
H;, (x), (2).
It follows that
Hn (x) D [exp (—2”) H,(2)] - Ha(x)D[exp (—2”) H,, (2)]
= Dlexp (~2?)
Hm (2) Hi,(2)] — D[exp (-2?)
Hp (2), (2)].
Equation (21.2) can now be written
2(m — n) exp (—z)
Hm (2) H,(x) = D [exp (~2*) (Hn
(2) Hi, (2) — H,(x)H},(x)}} .
An integration gives us
2m —n) [exp (—2*)Hm()Fla() de = (exp (—2") {Hy (2)Hi (2) — Ha)
Hs (2)}]
OK
As x ~ oo and as z — —oo the function on the right approaches zero because H,,(x) and
H,,(z} are polynomials.
Hence we have for m 4 n,
[-
exp (~27)Hm(x)Hy(z) dx = 0.
The Hermite polynomials are orthogonal with respect to the weight function exp (—a?) on the
interval —co < x < 00.
210
CHAPTER 22,
Chapter
Fourier
22.3.
FOURIER SERIES
22
Series
Numerical
Examples
of Fourier Series
1. Using equations (10), (11), and (12) from the text we have
ay = + f (e~2) de=§,
i
m=
=f
f(z)~
4 +
f°
ee
OS
ly,
ae =
3 ft- (-1)"]
c
(c—2)sin “=
li
RET
n= 4 f (e~ aco
dt = —.
[fa ~({-1)
n
RAL
_ RAT
+ nm sin“)
} cos ~—
.
3. Using equations (10), (11), and (12) from the text we have
c
ay = =f
2
2 de
che
=,
&
an ==
f
edi.
—
cose
c
dp = 20(=D"
nig?
e
t=
=f
ce
~e
a? sin
de
c
dx =0
SA (-1)"
fe)~ ot ae
Lap
one
5. Using equations (10), (11), and (12) from the text we have
1
ao == |
c
dx = 1,
on = 2
“00s 2
dr = 0,
Clo
1
Moe
f° . nae
mies
1
f(a) ~ phon
Boc"l5
n
i)"].
Tyr
°
22.3.
NUMERICAL EXAMPLES
OF FOURIER SERIES
211
7. Using equations (10), (11), and (12) from the text we have
0
Fa
ay == f
(8r+22) de += f
rT
~ tT
i
7°
Qn = —
®
(x + 2x) dx = 4n,
Jo
1
(0 +l cose de
bn = Lf
/*
|
(3x + 2x) sinnz dz + —-t |
~~ ®
—?
bae-1 = 0,
w
0
(7 + 2x) cosnz dx = 0,
~—
(7+ 22) sinna dx = = [1 ~ (—1)"].
oO
ba =a,
fle) ~ ar — 25°
sin 2ka
k=l
9. Using equations (10), (11), and (12) from the text we have
weiner
=
f
fae
e+ 1yeos™%
ae +} [- cos =~ dz = apy")
att ~
nen?
kz]
ee
wera
2
Jo
nO
2
gp
2-1)"
1)"
nT
fiz) ~ = +5aoa = (0 _ (—1)"} cos “= +nn(—1)"*! sin =
11. Using equations (10), (11), and (12) from the text we have
Ls
ay ==
®
f
2
T
2? de =,
Jo
an == f
3
RT
b he
ee
do
fa)~ e+ s MOA
niet
dx = 2!
uy"
Jo
1)"44n?4?
ene 4 <
— 24 2(-1)"]
‘
5 [(-1)"44n?n? — 2+ 2(-1)"] sine.
nai
n=l
1)
2? cosne
™
13. Using equations (10), (11), and (12) from the text we have
M0
ms -[ cos” dx = 4
eG
Ljf*
«.
tn =
= [cos
sinne de
= - ff cos = cosna dt = 4p"
ne
(2
2
= 0,
~ 9(4n? — 1)’
4
f(a)~= +=)
(-1)"+! cosnz
4n?2 —1
°
212
CHAPTER 22,
15. Using equations (10), (11), and (12) from the text we have
_—
Lf
2 sinh
FOURIER SERIES
1 f°
sinhe
| e? cos EEc gg = 2(—ct I"+ n2q
fet acm 2mche, ga?cJ_.
a=}cJ_.
1
f°
,
h=cfe
sin
*
sinhe
f(a) ~
dexv=
2(-1)" sinhe
age
(MT).
vane sinh c[ecos(nax/c) — nz sin (nra/c)|
= >>
By nial
17. Using equations (10), (11), and (12) from the text we have
ag =~
1
¢
f°
1
dx = 5,
Jose
€
Qn =~
1
©
f°
NAE
—l . ner
cosa—— dy = —— sin
Jo/2
ec
ne
2°
if
nNE
_
nw
cos —~> )
bp nt =~ [ a in —— dz dt=— (cos nT —— cos
+ (cos nn — cos 2 | sin =|
cl
~ [sin ee2 cosc
F(z) ~ Lisl
wt
19. Let us put ¢ = 4 in Exercise 17. Then if f(z) is the function of Exercise 17, 1 — f(z) is the
function of this exercise. Thus for the function of this exercise
1
lel
torre
fiz)~
nH
T.
eEi glee
+2 bs = [sin 5
cos "= +
ne
lomo)
(cosrem — cos)
ae
DAT
sin“).
21. Using equations (10), (11), and (12) from the text we have
0
wml
(c+ 2) dx = £,
an =f
(e+ 2) cos “S* de =
5g(l- cos),
~¢
1
—C
ees
bn = of
¢cE
f(z)
©
1)" cos BToT — ae© sin
sin BT
—
bee{1 ~ (—1)"} cos
iF
n
23. a7 ~ ns
+ =
—o
Ure
. Since the function of the series is continuous at + = 0 we
‘ai
ce
have 0 = —-
(-1)"
yr
oy
oes
. It follows that yo
a
=F
22.4.
FOURIER SINE SERIES
25. cin
—a
8e4
3
213
1)"(n
3 a
_
6
nee
cos.
nan
Since the function of the series is continuous at
c
oO
1
z = ¢ we have c* = Sat
n=l
48
1
the use of the answer to Exercise 24. We have “i a na
=
}
that Dor
97.
=
x!
x
=~] + ; + 5
F
= =
-
90°
lim e—sinhe_= lim
1—coshe
= lim
—sinhe
’ o40 2c sinhe c0 2¢2 coshe + 4csinhe~
~~ 2c? sinhe + 8ccoshe at 4sinhe
_
~ cosh
2
= lim 2c? coshe + lWesinhe+ 12coshe
22.4
It follows
ci"
5
oF
neq? |
(-1y"t)
ox?
ne
12°
Hence y
Fourier Sine Series
1. From equations (1) and (2) in the text we obtain
2
f° . naz
2[1 - (-1)"]
bn =< = f sin ——— dx dos ot!
>
f(z)
4
bop bapas = (kis
ba,ob == 0, 0,
in tok + Lara
ty
5K +1"
7 hemo
C
3. From equations (1) and (2) in the text we obtain
c
4
_p\n+t
b, =ee egin ee de = 22 aS
C Jo
c
nm
an
nw 2c? y
[a ial _ 20
_ a]
nix
_
_
{14%
nix
|
sini a.
e
. The odd periodic extension of this function is the same as the odd periodic extension of the
function in op xercise 4, Hence the two functions have ths same Fourier sine series. Thus
(yn
sin
BIe ~ sin ——.
7. From equations (1) and (2) in the text we obtain
2
eh
f(x) ~
f°
_ ANE
ee
aE
RSI
+1)8
4c? (1— (-1)"|
in tok + Lynx
c
bor = 0,
bok+t
8c?
= OEP aaas
214
CHAPTER 22.
FOURIER SERIES
9. From equations (1) and (2) in the text we obtain
n=
.
mnt
2 [* sin
2
nit
ae = = [1 - cos
21
fla) ~ 2°
= (1
mato \
cos
.
vet
sin ——.
nol
11. From equations (1) and (2) in the text we obtain
1
by = af
(a — 4)sinnaz dz= -— cosna—
1/2
nr
=
_ sin
nx?
f(z) ~ 3 ee
nw
n=l
sin naz.
13. From equations (1) and (2) in the text we obtain
2
f*
by = —
WT
,
2/71.
cos2rsinna dz,
by = —
Jo
®
~ sin4da dx = 0.
Jo
2
Wr
Ifn # 2 then 6, = : [
{sin (n + 2)a +sin(n — 2)z] dx.
Q
(2k +1)
2k+1
4
=
From this integral it follows that
sin [(2k + 1)z]
Pak =O and Paet1 = 7 OR ayae pa) We set I=) ~ syk=O k= 1)(2k+3)
15. From equations (1) and (2) in the text we obtain
2
[% 2,
=ife
sin
nee,
— Inw[1—(—-1)"e~4]
—~
dz =
c? + n2n?
»
“
fz)~
>
ix
Qnn (1 —(—1)"e~*]
c+ nix?
17. From equations (1) and (2) in the text we obtain
_2
by = =f
oO
»
f(z)
f*
_ mmc ,
cosh kx sin
7
dz =
2nx[i — (-1)" cosh ke}
hy
nint
;
Qnx(L—(—1)"coshke] . naz
sin ——.
Re a nine
nak
19. From equations (1) and (2) in the text we obtain
c
n= = f
x4 sin
dx = 2
[icon {2 -
5 - Fa} + ead,
er
0
24
24.1.
Pin
{2n1 — 22nigt2 Eb
f(a) ~ 92Id So= (ays
nog
n=l
nex
sin
nwa
——.
22.5.
FOURIER COSINE SERIES
215
21. From equations (1) and (2) in the text we obtain
i
by"h
=
2
(z ~1)* sinnaz dr = 35 [n?n? —2+42(-1)"],
~2+2(-1)"
aoe
22.5
Fourier
Cosine
sin ng.
Series
1. From equations (1) and (2) in the text we obtain
1
a=
[
0
i
a=
2
cart
[
[ @-2)de=1,
l
nTeT
& COS ~~ az + [
2
naz,
(2 ~ 2) cos ~~ dz =
4
nT
5
5 [20s
1—(-1)
n
|.
oO
— 1 ~ (- 1)"| cos “=
[2 cos
fe)~ 5+ aa
3. From equations (1) and (2) in the text we obtain
ay =2 | * {e-1) 2 dx = 3,2 an
f(a) ~
i + 4
3
=2 |
,
(x — 1)
2
4
cosnma da = 55,
> cosnme
on? &
n@
5. From equations (1) and (2) in the text we obtain
wo == ['(e~2) dx =,
C
é
an = 2 f(e-2) cos
C
0
c
4c
G2 = 0,
0
a
4e—
fo.
fi
nex?
”
ce
Aenq1 = (2k + 1)?n?”
de = =)
1
F(z) ~ 9° x2 2. @e+ip
(2k+ lax
——
7. From equations (1) and (2) in the text we obtain
ao = hn (x - 3)$
=A
8
~s
<5
.
n=l
= >
ap = 2
x=; (cosmn ~ COS oy)
“ne
2
1
i
2
an
~ x)cosnmz dx = “a5 (cosa ~ COS >) :
cos nTs
,
9. From equations (1) and (2) in the text we obtain
2/7
a == [
Jo
cos 22 dz
= 0G,
2"
a == f
w
do
if
cos 22 cos
nx dx = -|
Rt
Tr
cos
22 cosnx da
216
CHAPTER 22.
FOURIER SERIES
But on the interval —7 < x < 7, cos2z is orthogonal to cosna for n 4 2. Hence a, = 0 for
n 32, Moreover ag = ‘f° cos? 2 dx = 1, so that f(a) ~ cos 2z.
11. From equations (1) and (2) in the text we obtain
ay ==
A
/2
x dz =,
Ha) ~ $+
as
a, == [
[nm sin =
/2
woos“ de = <5
— 2 (1
[nm sin S* ~2(1—cos ="),
cos =) | cos “E,
13. From equations (1) and (2) in the text we obtain
ao ==
f(z)
~
2
|
f*
2 sinh ke
coshkx dz =
sinhke
ke
2
ike
“.
+ sinh ke J
-=/
2ke(~1)”
Pe+nin
°°
f°
cosh kz cos
nex
2
dr =
sinh ke « 2ke(—1)"
Baan
ne
7.
15. From equations (1) and (2) in the text we obtain
gf
i
f(z)
ce
Jo
c
ode= ©,
2
T Sy
n=l
[S iy"
med
Cc
0
1)
2 eos
5
So
c
ae = 5S |f 1)
cos “TE.
424s
fon
n
(~1) |;
217
Chapter
23
Boundary
23.1
The
Value Problems
One-Dimensional
Heat Equation
1. The boundary value problem to be solved is
bu
aS
Ast—0*,
Asa
forO<a<c,
O< 1%;
u— ug,
for0O<a<e,
0t, u— 0,
Asr-e,
for
0 <¢;
uO,
for
0 < t.
The standard procedure for separating the variables will be used.
T
form u = X(z)T(t).
Thus we require that ——ap
We seek solutions of the
xX"
= —-
= k.
We now seek nontrivial (not
identically zero} solutions of the system
X"—-kX =0,
X(0)=X(c)=0.
Ifk
= 0 then
X = 6; + bow.
X(0) =0
implies that b; = 0.
X(c) = 0 implies that 6, + boc = 0.
Hence 6; = bp = 0 and X
= 0.
That is there are no nontrivial solutions in the case where
k = 0,
If k = 67 > 0 then X = c; sinh Bz + cy cosh Bz.
X(0) = 0 implies that cg = 0.
X{c) = 0 implies that c, sinh Be + cg cosh Bc = 0.
Hence ¢; = cz = 0 and there are no nontrivial solutions in the case where k is positive.
If k = —8? <0 then X = c; sin Gz + cg cos Bz.
(0) =0
X(c) =0
implies that cg = 0.
implies that c; sin Be + cg cos Be = 0.
218
CHAPTER 23.
BOUNDARY
VALUE PROBLEMS
We can now obtain nontrivial solutions by choosing 8c = na or @ = na/ec. These solutions are
. Wax
X(x) = ¢, sin —
Bape
The choice of the separation constant k = —§? =
having been determined by the
boundary conditions in z, we must now find functions that satisfy the differential equation
hen2r2
j
T+
9 = 0. These functions are T(t) = b, exp (—h?n?47t/c?).
We now have an infinite set of product functions
_ are
u(x, t) = A, sin —
—hn? att
exp (==)
,
each of which satisfies the required boundary conditions in x, but none satisfying the initial
condition in t. We consider the infinite series with undetermined coefficients A,
oo
u(x, t)=
A, sin —
—h2n2a?
exp (“3=)
(23.1)
ne]
and attempt to choose the A, in a manner that will satisfy the initial value condition. Setting
t= 0, we see that we want
oO
to =
_ mre
7) An sin —,
forO<a2<e.
n=l
That is we want the A, to be the Fourier sine coefficients for the constant function ug. Thus
e
An == |
clo
It follows that Ay, = 0 and Aggy:
ug sin
c
ge = 201
nt
Aug
= aQk 41)"
— cosnn).
Using these coefficients we obtain the final
solution
s
1
u(x, t) = Aug
—2v fg
SY ———
2k +1
_ (2k +1)rx
sin —
- exp
oe
+ we
a
|
. If u is independent of ¢ the heat equation reduces to u” = 0. Thus u(x) = a; + agz.
The
conditions u(0) = A and u(c) = 0 force us to choose a; and a2 so that a; = A and a; +age = 0.
Thus aj = A and a2 = —A/e. It follows that u(x) = A(c — z)/e.
5. The solution proceeds as in Exercise 4. We want a solution of the form
u(x, t) = Ale—2)/o+ 3 byexp - (my
n=l
sin o.
23.1.
THE ONE-DIMENSIONAL HEAT EQUATION
219
Each term of the right hand side is a solution of the heat equation. Moreover the two boundary
conditions are satisfied by each term. We need to choose 6, so that
f(zj)= A= 2) 4
by sin HE,
That is the constants 6, must be the coefficients of the Fourier sine series expansion of the
function f(2) — A(e — «)/c on the interval 0 << 2 <c.
7. The problem to be solved is
ou
$e
for 0 <a
< 20, O< f;
Ast —~+ 0+, u—» 120,
for 0< 2 < 15,
Ast
for 15 <a
< 20;
0+,
u— 30,
Asz—-0T, u— 30,
for
0 < t;
Asxz— 207,
for
0 < ft.
u- 30,
We choose a new variable v = u — 30. The new problem becomes
oy
Ae
for O< ¢ < 20, O <1;
Ast
0+, v— 90,
for 0 <2
< 15,
As t—0t, v0,
for 15 < x < 20;
Asx—0*,
for
0 <t;
v0,
As a 207,
v-»+ 0,
for
0 < é.
This problem is much like the problem in Exercise 1. We must replace h? with “, ¢ with
20, and the constant initial temperature function with a step function.
Exercise 1 becomes
net
v{a, =A,
(Ge)
Equation (23.1) in
RIL
307
n=l
The initial condition now requires that f(r)~ y
of this problem.
Q
b,= 20
bn sin —— = where f(x) is the step function
Thus
Jp
fe
f(z) sin
0
= de=
1
10
,
TNL
180 |,
de =
[1 20
Ssin—y
snr
4
=
it follows that
_
180 «=~ [1 — cos (3n7/4)]
u(x, t) = 30+ v(2, t)
= 30+ oy
~
n
—n7t\
| nwa
eXP | 7990 | SB
q-
220
CHAPTER 23.
BOUNDARY VALUE PROBLEMS
_1\k
9, We have
for t = 0,
yas
3 BE
= A,
But sin 2°" = 0 and sin OF + De
3
3
_ ())'v3
1) v3
2
_
sin CE+2)"
Thus
3
—1)*(2k + 1)
It follows that » «Gk
Ge + yk
+2)
=
(3k + 1)(3k +2)
8
£4) BRF1)8 ” (Bk+2)8)
3
n-
2k +1
(-1)*3V3 | (-1)'3v3 =); (—1)*oV3
3
3 1 nd OT
an
a
11. The problem can be interpreted as that of heat conduction in a slab of width c. The face at
ax = cis held at temperature zero while the face at x = 0 is insulated. The initial temperature
in the slab is given by f(z). We let u(z, t) = X(x)T(t) in the heat equation and obtain
f
a
tt
= x" = 4. The differential equation in X is accompanied by conditions X‘(0) = 0 and
X(e) = 0. As in Exercise 1 the cases where a = 0 and a > 0 have only the trivial solutions
X =0. When a = —a* the differential equation has the solutions X = Acosaz + Bsinaz,
so that X’ = —Aasinaz + Bacosaz. The condition X'(0) = 0 implies that B = 0. The
condition that X(e) = 0 forces us to choose a so that cosac = 0.
for k = 0, 1, 2,---,
2
T' + mek
(2k A
+1)
and
2g?
—T
2k+1
X = A, cos (2k + 1)nx
That is a = Cee e
Now the differential equation in T’ becomes
~h? (2k + 1)?x?
= 0, so that T = Byexp
AS].
We therefore have a
solution of the form
u(x, t) = yox box
Fears
| _p2h*(2k
+ Qed
1)*x
cos (2k + Uma
4c?
2c
The initial condition forces us to choose b, so that f(z} ~
a
(2k + l)rzr
by, cos ———.
k=0
23.4
Heat Conduction in a Sphere
1. We first solve the problem of equations (8) through (11). Separation of the variables can be
accomplished by substituting u(p, t) = Q(p)T(t) to obtain an
= <
« k, The boundary
value problem in @ becomes
Q” —kQ=0,
as p—+R,
Q-0,
and as p07,
> a limit
If k = 0 then Q = c, + cep and the boundary conditions force us to satisfy c; + cgR = 0 and
as p—» 0, - + cq has a limit. These equations can be satisfied only if cy = cg = 0.
23.5.
THE SIMPLE WAVE EQUATION
221
If k = a* > 0 we then obtain Q = cg cosh ap + cq sinh ap and the boundary conditions require
that cg cosha.R + cysinhaR = 0 and that seme tee
second condition forces us to choose cz = 0 and now
have a limit as p — 0. This
the first equation forces cq = 0.
If k? = —8? the solution of the differential equation takes the form Q = cs cos Bp + cg sin fip.
The condition as p — 0 again forces the choice c, = 0. Finally Q(2) = 0 requires that 6 = —
2,,2q2
and we get @ = c, sin ore
QQ?
T= a, exp | a)
The equation
in 7’ is now 7” + h ao
= 0 with solutions
This gives us
u(p, th= 5 bn exp
=
—h?n2 nt |
nip
R2
“R
nel
oO
We now force the condition in equation (11) and need pf(p) = s b, sin
be the coefficients of the Fourier sine series for pf(p). That is
2
bn = af
f®
_
Thus 6, must
need
np
pf (a) sin —- dp.
Finally with these values for b,
_t
u(p, t) = O91
ue
23.5
—h'n 2
neo |e
a
24
.
|sin
The Simple Wave Equation
1. The problem to be solved is
ay
3a
=
Ast
2 O*y
Ay?
for O<a<e,0<4;
0+, y— f(z),
for0<2<c;
Astor,
Yo,
for 0< 2 <¢;
Asx2—0T,
y—-0,
for
0 < ¢;
Asz-e,
y 70,
for
0 < t.
If we seek product solutions of the differential equation of the form y{x, t) = X(z)T(), we
ff
ue
have aan = x = k.
aT
solutions of the system
The two boundary conditions now require that we find nontrivial
X"—-kX=0,
X(0)=X(c)=
222
CHAPTER 23.
BOUNDARY
VALUE PROBLEMS
The only nontrivial solutions occur when k = —n?1?/c? and are of the form sinnwz/c. That
particular choice of the separation constant & requires us to find solutions of the system
2
T+" nm vr=0,
These solutions are of the form cos nat ve
NTL
nrat
Ay sin —~—- cos ---—-, We
c
¢
T0)=
Thus the desired product solutions are of the form
are therefore led to consider the series
NTE
->
oe eA
Any sin ——
c
In order to satisfy the remaining initial sondition we need to choose A,, in such a way that
f(z) = s An sin —,
for O< a<e.
(23.2)
n=l
That is, we require that
Cc
An=2 C [ Jo f(z)sin—c de.
(23.3)
For the particular f(x) of this problem we have
Q
rel?
An == [
NUT
de
|
rsin“— de + ~ ff, (e-=x) sin —— da = 2,3 in
It follows
OUOWS that
that Ag, Ag, == 0 and A
Qk+1 ==
(Sk4e(—1)"
+ 1)2x?"
(-1)*
WeV finally y have
(2k+1)ra
yz, t) = SL wea
nn
a
(2k + 1)xat
3. From equations (22.1) and (22.2) of Exercise 1 we have
ef4
An == [
¢ Jo
20.
2c
<=
de + = f
c
ae
_
30/4
sin "=
4
sino
=
y= 72 Seer a
si
ef4
2
< sin ——~ dx +
4
&
€
3e/4
(e~2)sin “= de
¢
4}
4. gin 3O™
m
ne
4
+ sin
3n7T
ecettceetes
4
COS
nant
| naz
sin
semcraeanaone
c
5. From equations (22.1) and (22.2) of Exercise 1 we have
274.
we
nxx
2
c/2
ee
.
ala
2 sin
.
nr
ane
(4¢- x) sin —
of4
Cc
nant
n
NE
~ sin --- | cos -———— sin.
2
c
2e
c
nh
nT
dt = — 5 (2sin*4 ~ sin)
2
rw
'
23.6.
LAPLACE’S EQUATION IN TWO DIMENSIONS
223
7. This is like Exercise 1 except that the system in T becomes
a.
2722
Tea
T+ +—~37T
nrat
which has solutions sin
so that y(z,t) =>
..
..
the remaining condition at t = 0, namely
@(z) ast — OF.
= 0,
T(0)
=
0,
nA
nirat
A, sin —— sin
By
Bt = s
nraA,
——
oS
Now we need to satisfy
. NNT
nat
sin > cos
~ nnaA,
AL
Thus we need ¢(x) = s- ~ sin =.
must approach
This can be accomplished by
n=1
choosing
_
¢
"na
2 / ae/ am sin
d
CJey3
2ugc
nT
an? 72
3
cos ant
3
Thus
y(z, th = S- Bn sin —
x
nrat
si
oO
Tis
Oy _ 3
at
We want d(x
->
By
nta
ne]
c
e
c
. 7
8 sin ~~; that is we must choose
nTrG
23.6
Bp nwa sin NUE cos nrat
Bn os ef
Laplace’s Equation in Two
o(z) sin
dz,
Dimensions
1. This is the problem of equations (2) to (6) in the text. We seek product solutions of the form
ut
u(z, y)= X(z)¥(y). Separation of the variables leads to ~ = ——
(4) together with this differential equation in X require that
a
= k, Equations (3) and
X" kX =0, X(0)=X(a) =
The only nontrivial solutions of this system come from taking k = ~—n?x?/a?. These solutions
are of the form sin (nra/a).
224
CHAPTER 23,
BOUNDARY VALUE PROBLEMS
If we use these same valus for the separation constant k, we must solve the following system
in the variable Y:
nen?
_
Y ~"Y= 0,
¥(0)=
The solutions of this system are sinh (nwy/a}. We therefore have an infinite set of product
solutions, sin (nw2/a)sinh (nry/a), each of which satisfies equations (2) to (5), but none of
which satisfies equation (6).
We now consider the infinite series
oO
NtZ
uz, y) = > Cp Sin =
nw
sinh =,
nod
and attempt to determine the constants c, in order to satisfy the remaining condition (6). We
therefore require that
nr bs NE
f(z) = Sey sinh 22
in.
n=l
This may be accomplished by choosing
a
cn sinh “Et
a
2 f f(z) sin
a Jo
=
a
dz.
. This is the same problem as Exercise 1 with f(x) = 1 for 0 < 2 < da, f(z) =0forda<a<a,
We need
.
|
sin
—_ 2(1 — cos (nm/2)|
~ asinh (nxb/a) Jy
a
~ parsinh (nrb/a) ’
2
[1—cos(n/2)]
ney , naz
OST » nsinh (nxb/a) sinh a
a
x?
-y"
Separation of the variables as in Exercise 1 gives us y=
boundary value problem in X is
X"—cX =0,
Nontrivial
solutions can occur only if e =
Ip as
This time however the
X'(0) = X(a) =
—8?
<
0, X
=
e, cos Gx + cesin Bx,
and
also
A’ = ~f sin Sr + cf cos Bx. The condition X’(0) = 0 causes us to choose cg = 0 and then
X(a)
= 0 requires that cosa
= 0.
This can happen only if 6 = Gh sue
choices of 8 we have X = A, cos(es De
yr"
—
(2k + 1)?”
4a2
and for these
We now have for the equation in Y
Y=0,
¥(0)=0.
23.6.
LAPLACE’S EQUATION IN TWO DIMENSIONS
Thus Y = 5, sinh Gee,
225
This leads to the series of products
OO
_
4
ula, yy = S
C,, sinh
(2k+ lary
5a
(2k +1)0x
ar a
k=O
Now we require that u(x, 8) = 1, so that
i= > Chs
inCE
(CF +
oo , (ak + i)re
oa
It follows that we must take
_
2
o
(ak+1jyez
4(—1)*
Qa
~ asinh ((2k + 1)nb/(2a)| [ cos
d ™ (2k + 1) sinh (2k + 1)nb/(2a)]
Finally we have
ulz, y) = sme
(-1)*
(2k+1)y
(2k +1)ex
sinh [(2k + 1)nb/(2a)] nag
88
ag
7. The solution of this problem is the same as that of Exercise 5 except that here we require
Y‘(0) = 0 instead of Y(0) = 0. That change will result in each sinh function being replaced
by a cosh function. All other details remain the same. We get
uz, y) = +
(—1)*
ETH aii [Ok + 1)nb/(ay C8
(Qk+1)ry
5g
88
(2k +1)rex
9. Let the temperature at the center of the given square plate be the constant u,. Now consider
four separate plates, each having a different face held at temperature unity while the other
three faces are held at temperature zero. For each such plate the temperature at the center
will be u,. Now if the four plates are superimposed we obtain a plate with all four boundaries
held at temperature unity and the temperature at the center will be 4u,. But such a plate will
have in the steady state a temperature of one at its center. Thus u, = 4.
8, with b = a, we get
2 5! (—1)* sech[(2k + 1)x/2 ai
7k
2k+1
~ 4"
Tt follows that
s
k=O
(—1)* sech[(2k + 1)2/2]
2k+1
ao
~ 8
Now from Exercise
226
CHAPTER 23.
BOUNDARY
VALUE PROBLEMS
11. Because of the symmetry of the boundary conditions the temperature in the plate will be a
function of r only. Hence the problem to be solved is
fu Lda
=Q
fora<r<b.
dr? or dr
As r-68",u- 8,
As
r-at,u- A.
The general solution of the differential equation is u = c, Inv + cg. The boundary conditions
require that B = c; Inb-+ cg and A = c, Ina + ce, so that
(B - A)lnr + Alnb- Blna — Bln(r/a) ~ Aln(r/b)
In (b/a)
In(b/a)
In (b/a)
,
13. The problem to be solved is
uj 10u, 1 ou
Gr?
or Or
=Q
—r® OG?
As
@—0t,
u-
As
@-
8B", u- 9,
As
r>
Ro,
As
r—0*,
for0<r<R,O<@<8.
90,
u-i,
u has a
limit.
a2
ut
We separate the variables by letting u(r, 8) = F(r)G(6). We get ae
f
re
= aol =.G
must now satisfy
G"+cG=0,
G(0)=0, G(s) =0.
ws
:
:
nen?
Nontrivial solutions occur only if c =
er
_ meré
and they are of the form G = A, sin a
F must
&
now satisfy
rR" 4rF
nxBed
rm
F=0,
and F(0) exists.
The general solution of this Cauchy type equation is F(r) = cyr"*/9 + egr-"*/F, To satisfy
the condition at r = 0, we must take cp = 0. Thus F(r) = B,r”*/*. It follows that
=
nne
u(r, 0) = S > Car®*/9 sin a"
n=)
oO
nro
Now we require that 1 = s- C, Rm! 8 sin a
so that
n=l)
C=
2
BR*/8
8
fo
nw
aR
dO =
2(1
— cos nz)
naRerfe
°
23.6.
LAPLACE’S EQUATION IN TWO DIMENSIONS
4
RORTTTETE
It follows that Cz = 0 and Cres = GE
227
Finally we have
u(r, 6) = Lyn (LE) OMe? sin 2k2k+ +11) 80/0),
* pao
228
CHAPTER 24.
Chapter
ADDITIONAL PROPERTIES OF THE LAPLACE TRANSFORM
24
Additional Properties of the Laplace
Transform
24.1
Power
Series and
Inverse
Transforms
1. We use equations (4) and (7) and obtain
sinkt)
1
fos (-1)"h? "4129 |
t
}=2{5
Om
(2n + 1)!
(—1)"k?"+}
oe 2
SA (-1) 7k?" 41 (20)!
\-¥
(2n + 1)! s@rt1
k
Ont
= arctan —,
3>0.
3. We use equations (6) and (10) and obtain
L sinh kt
“>
Gee
kantl yan
q (2n + 1)!
= YS arieer
kent
- oe
k’ntl(dn)!
4 (2n +1)!
+1)! s2ntt
s2ntl
=iin 1+ (k/s) = iin stk
« (n+ 1)s?e+t ~ 2° (k/s)
1
s—k’
s>k>0.
5. We use equation (1) and obtain
F(t) sz E77
(aaem}
OO
>
= D7?
ans
}-4 1 S(t
n=O
~ 2n)Pa(t
— an)
2
F(5) = 5 [5?a(5) + 3?a(3) + 1?a(1)] = 17.5.
7. We use equation (1) and obtain
“0-0 aaeaaa} 2 remem} = remy}
2
Ge 3967
Gen
-o' fy ere}
ned
$(10) = 7a(7) + 1a(1) = 344.
ae “foss-£
nes
e7{6nt3)s
sr}
oO
= Set
n=
9alt—on—2).
24.1.
POWER SERIES AND INVERSE TRANSFORMS
229
9. We use equation (1) and obtain
L~* {f(s) tanh (cs)} = £7 ¢ f(s
~ 208
pes baet {so j-
as [D7 (70)
pa
al}
1)" exp (-— mesa) |
nse)
= L7'¢
f(s) f +2 y(-
1)"*! exp (—2(n + ves
n=D
“efi
1
p +2 y-
wera
\
= F(t)+2 s(-1"Ft ~ Inc)a{t — ne).
n=l
ll. There is a notational problem here. We use Q(t) for the charge on the capacitor in the circuit
and Q(t, c) for the square-wave function. Note also that C is the capacitance and c is half the
period of the square-wave function. If we let i(t) = Q’(t) then the problem to be solved is
RQ! ()+ 20) = EQ(t, ¢), Q(0)=
Using the Laplace transform on this equation together with equation (16) from Section 14.10
we obtain
1
E
cs
Rsq(s) + aus) => tanh (S$),
E/R
cs
1
1
19) = sep apRey om (F) = CF E a “TR
To obtain the inverse transform we make use of the result in Exercise 9. We get
Q(t) =CE p 42 s7(-"alt _ vo
The current in the circuit can now be found by a differentiation. We have
i=5 5 |e
(Fe)
2
1yrexp {
29
cs
tanh (>),
bate - ro)
230
CHAPTER 24.
24.2
The
Error
ADDITIONAL PROPERTIES OF THE LAPLACE
TRANSFORM
Function
1. Consider the graph of y = erf x. From equation (4) in the text we see that the slope of the
graph is always positive. A second derivative of erf x shows that for x > 0 the graph is concave
up. There is an inflection point at z = 0. From equation (7) we also conclude for x > 0 that
the graph of erf x is asymptotic to the line y = 1. A similar argument for « < 0 yields the
result that the graph is asymptotic to the line y = —1. Thus the entire graph lies between the
lines y = land y = —~1.
3. Using PHopital’s rule we have
5. Since jerfe z| < 2 it follows that for m < 0, im, z™ erfc x = 0. The indeterminate form occurs
ifm > 0. In this case
lim 2” erfe
c=
x—tOO
lim
goo
.
fi
oe
gm
ilim inert
soo
| gpm
(~2/,/mjexp(—2*)
= in, geet
m+1
lm, gn
2
exp (2).
But we know from Chapter 14 that polynomials are of expomential order so that this last limit
is zero.
7. Using the suggestions made in the text we have
etre} (bith}-'
1
ts
l+s
1
l+s
&
s/l+s
1
a3
heatatah
1
1
ips?
l+s
Now from equation (3) and the equation preceding equation (2) we have
L
afoot
\ie}-
or ef =
Lt erf( Vt) + Tae =
erfe(/t).
gece} -¥{de-gian} ¥en) em
te
jt
9. From Exercise 8 we have
Now we use equation (1) from Section 14.6 to obtain =
7
follows that L~! {aa}
= a
— e! erfe(v't).
~Lo} {
|
\ = e! erfe(Vt). It
24.3.
BESSEL FUNCTIONS
231
11. We are given that “
=
1
erf (;).
Li{d(t)}=
~1)"
Fe y Z Foe,
nb a sinvi:
o(t)
3
From equation (5) in the text we get the series
:
;
:
:
Appl
Applying the inverse transform to both sides gives us
1)"4?"
=
=
es
(-1)" Ag
a
It follows that
é(vt)
=
Vai
7a &
“(Qn +1)
al’
Therefore
neeQ
(—1)”
2
fi
HOWD)= Fed
ad 4 Gn : ar att = ad (an 1)! (Vaynr? ~ Yas (=a):
13. The derivation of equation (29) that is given in the text gives us a clue as to how to proceed
here. We have
_
2
_
esch (x¥/s) = exp (2/8) —exp (~aV8)
= Qexp(—zy/s) Sex
_2exp(—zy/3)
~ y= exp (~22-V)
( ~2neyi) =2)° exp [—(2n
+ las}.
n=d
It follows that
yt {sev
asvun{t exp | ~(2n-+ 1)eva }.
nsxQ}
Now we use equation (24) from the text to obtain
Lo
24.3.
Bessel
{ osch (2's)
eve) \ = 2)
et ae *| .
Functions
1. We expand the exponential function into a power series to get
Lo
-
1
-
ian exp (2)}- L {3 aie}
ki
CO
=
pk gktn
t
meray
(k+n)! = (£)
a ae
a
on
aan}
(Vat)"*
Sera”
t
(z)
nf/2
I, (2va#)
232.
CHAPTER
25.
Chapter
PARTIAL DIFFERENTIAL EQUATIONS:
TRANSFORM METHODS
25
Partial Differential Equations:
Transform Methods
25.1
Boundary
Value Problems
1. We apply the Laplace transform in the variable t, letting L {y(z, t)} = u(x, s). We get
Ws
is
tsu =
_oe
ze-Q,
u- -s.
33
.
:
.
.
~2
This linear differential equation has general solution u = F +e,(s)e~**”. As x — 0 we need
4
—~2
to take 373
.
+ci(s).
6
~
That is c(s) = =
We therefore have u = =
+ Bente,
An
inverse transform now yields the solution y(x, t) = —¢? + 3(t ~ 4x) a(t ~ 4z).
3. We apply the Laplace transform in the variable t, letting L {y(x, t)}} = u(2z, 3). We get
du
~8
Gz t 4su- 4a =,
z—Q,
dz
du
Gy tf fsu=4t~
8
5,
uns
g£-0,
u-
This linear differential equation has general solution u = =
4
i
x — 0 we need to take =
have u =
= —
= Ti
-qa-F
i
2
y(a, th =a - zt —
stip
6
2
n
+c(s).
e74*® . An
i
8
.
4
a
s*
~ 2 + e;(sje"*#*,
6
1
That is c(s) = 3 + qt"
We therefore
transform now yields
i nverse transf
ields the
the solution
soluti
4 Ee — dar)? + ic ~ 4s)| a(t — 42).
5. We apply the Laplace transform in the variable t, letting L {y(z, t)} = u(z, 3). We get
2
oe
drt _ 16(s2u +2),
dy
3
9
7 16s*u = 392,
2-0,
r~+0,
u- ~z,
Ss
u-
As
1
a
.
im,
lim
2-700
exists,
exists,
25.2.
THE WAVE EQUATION
233
_
2
This linear differential equation has general solution u(x, 8) = c;(s)e#** + e2(s)e~ 48" — =z: In
]
order for lim, _.o. to exist we must take ci(s) = 0. As 2 — 0 we need to take 2
That is co(s) = s
We therefore have u(z, 4) = Senses - 2.
2
co(s) ~- Zz
An inverse transform now
yields the solution y(a, t) = 3(¢ — 4x)a(t — 4a) — 2t.
25.2
The Wave
Equation
1. Direct application of the Laplace transform gives us the transformed system
2
s*u
~ s(x ~ 27)=
The
linear equation
u
dx?’
z0t,
u70,
r 417, uO.
has as its general solution u =
zc
gg
2
~— — —
— 3
+ cye7** + cge®*.
8
condition z ~+ 0*, u — 0 implies that c; + cz = 2/s°. The condition z+
that e~*c, + e°co = 2/s3.
The
17, u—+0 implies
From these two equations we obtain
~ 38(1+e-*)
5 ~ 38axe
nex
y*
.
“na,
and cz
=
2e7*
si(l+e*) 33=
n=O
n
Wyre
~(n+l)s
Therefore
u(x, é) = - _ =
~ 5
+a2s
(-1)1)” etl
1)"e —(n+2)8 +32
n==Q
-2)s
re==0
The inverse transform now yields
y(z, t)= oat
24
S(-1)"[(t —n—2alt—n—2)
+ (¢-n—-142)alt—n—142)],
ned
25.5
Diffusion
in a Slab of Finite Width
1, Direct application of the Laplace transform gives us the transformed system
sw ~[=
d*w
F
g—0t,
w+0,2-417,
dw
—7
0.
The linear equation has as its general solution w = : + ¢, sinh (v/s) + cgcosh(x/s).
condition 2 — 0+, w — 0 implies that cg = ~1/s. Thus
oe = Vacs cosh (2/8) ~ % sinh (2/3).
The
234
CHAPTER 25.
PARTIAL DIFFERENTIAL EQUATIONS:
TRANSFORM METHODS
The condition z + 17, dw/da — 0 implies that c, = (tanh /s)/s. Thus
1 + tanh fssinh(x/s)
3
8
cosh (ry/s) Py_ 1
8
8
w(a, 3) = —
cosh [(z ~ 1)/3]
scosh \/s
,
The argument on pages 475 and 476 in the text that produced equation (29) can be slightly
altered ta give the result
h (tJ)
pt_ {sey
scosh ./s
rn
1—~a2+2n
= S*(-1)"
jerfe
tet)
2
)
aft
l+a+2n
4 ente
(AEE).
me
at
Using this inverse transform with z replaced by x — 1 allows us to write
u(x, t) = 1- you)"
ent
(Ae)
+ erfe (5)|
.
ne=Q
Diffusion in a Quarter-Infinite Solid
25.6
1. v=
erf (=) erf (4)
= 2 i
exp (|
iL"
exp (-a" aa
. Since the inte-
grand of each of these integrals is a positive decreasing function, the integrals themselves are
positive monotone increasing functions of x and y respectively for any fixed t > 0. Moreover
ox
T
the number [ exp (~ (7) d@ =ve is an upper bound for each integral. Hence for any fixed
0
z > 0, t > 0 each of these integrals has value less than fm /2. It follows that 0<u <1.
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