Table Of ContentInstructor’s Resource Manual
Differential Equations with Boundary
Value Problems
EIGHTH EDITION
n.
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ati
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hori
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pr A First Course in
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wit Differential Eqautions
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allo
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stri TENTH EDITION
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d.
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v Dennis Zill
er
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All ri Warren S. Wright
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g Prepared by
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©
Warren S. Wright
Carol D. Wright
Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States
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1 2 3 4 5 6 7 16 15 14 13 12
CONTENTS
Chapter 1 Introduction To Differential Equations 1
Chapter 2 First-Order Differential Equations 30
Chapter 3 Modeling With First-Order Differential Equations 93
Chapter 4 Higher-Order Differential Equations 138
Chapter 5 Modeling With Higher-Order Differential Equations 256
n.
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zati Chapter 6 Series Solutions of Linear Equations 304
ori
h
ut Chapter 7 The Laplace Transform 394
a
s
s
e Chapter 8 Systems of Linear First-Order Differential Equations 472
pr
x
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ut Chapter 9 Numerical Solutions of Ordinary Differential Equations 531
o
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wit Chapter 10 Plane autonomous systems 556
d
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w Chapter 11 Orthogonal functions and Fourier series 588
o
all
n Chapter 12 Boundary-value Problems in R ectangular Coordinates 639
o
uti
strib Chapter 13 Boundary-value Problems in Other Coordinate Systems 728
di
o Chapter 14 Integral Transform method 781
N
ed. Chapter 15 Numerical Solutions of Partial Differential Equations 831
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ht A pp II 855
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©
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© 2013 Cengage LearnNing. All Rightso Reserved.t May no t be cFopied, scanoned, or duprlicated, in whSole or in parat, except forl use aes permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
Not For Sale
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©
1 INTRODUCTION TO
DIFFERENTIAL EQUATIONS
Chapter 1
INTROD1U.C1TIONDTeOfiDnIFiFtEioREnNsTIaALnEdQUTAeTIOrmNSinology
n. 1. Second order; linear
o
ati
z
ori 1.21. TDheirfidnoirtdieorn;snoannlidneaTrebremcaiunsoeloofg(ydy/dx)4
h
ut
a
s
s 3. Fourth order; linear
e
pr
x
e
ut 4. Second order; nonlinear because of cos(r+u)
o
h
wit (cid:112)
d 5. Second order; nonlinear because of (dy/dx)2 or 1+(dy/dx)2
e
w
o
all 6. Second order; nonlinear because of R2
n
o
uti
b 7. Third order; linear
stri
di
o 8. Second order; nonlinear because of x˙2
N
d.
e
erv 9. Writing the boundary-value problem in the form x(dy/dx)+y2 = 1, we see that it is nonlinear
s
s re in y because of y2. However, writing it in the form (y2 −1)(dx/dy)+x = 0, we see that it is
ht
g linear in x.
All ri
g. 10. Writing the differential equation in the form u(dv/du)+(1+u)v = ueu we see that it is linear
n
arni in v. However, writing it in the form (v+uv−ueu)(du/dv)+u = 0, we see that it is nonlinear
e
L
e in u.
g
a
g
en 11. From y = e−x/2 we obtain y(cid:48) = −1e−x/2. Then 2y(cid:48)+y = −e−x/2+e−x/2 = 0.
C 2
©
12. From y = 6 − 6e−20t we obtain dy/dt = 24e−20t, so that
5 5
(cid:18) (cid:19)
dy 6 6
+20y = 24e−20t+20 − e−20t = 24.
dt 5 5
13. From y = e3xcos2x we obtain y(cid:48) = 3e3xcos2x−2e3xsin2x and y(cid:48)(cid:48) = 5e3xcos2x−12e3xsin2x,
so that y(cid:48)(cid:48)−6y(cid:48)+13y = 0.
14. From y = −cosxln(secx+tanx) we obtain y(cid:48) = −1+sinxln(secx+tanx) and
y(cid:48)(cid:48) = tanx+cosxln(secx+tanx). Then y(cid:48)(cid:48)+y = tanx.
© 2013 Cengage LearnNing. All Rightso Reserved.t May no t be cFopied, scanoned, or duprlicated, in whSole or in parat, except forl use aes permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
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2 CHAPTER 1 INTRODUCTIONTODIFFERENTIALEQUATIONS
15. The domain of the function, found by solving x+2 ≥ 0, is [−2,∞). From y(cid:48) = 1+2(x+2)−1/2
we have
(y−x)y(cid:48) = (y−x)[1+(2(x+2)−1/2]
= y−x+2(y−x)(x+2)−1/2
= y−x+2[x+4(x+2)1/2−x](x+2)−1/2
= y−x+8(x+2)1/2(x+2)−1/2 = y−x+8.
An interval of definition for the solution of the differential equation is (−2,∞) because y(cid:48) is
not defined at x = −2. n.
o
ati
z
16. Since tanx is not defined for x = π/2 + nπ, n an integer, the domain of y = 5tan5x is ori
{x (cid:12)(cid:12) 5x (cid:54)= π/2+nπ} or {x (cid:12)(cid:12) x (cid:54)= π/10+nπ/5}. From y(cid:48) = 25sec25x we have uth
a
s
s
e
y(cid:48) = 25(1+tan25x) = 25+25tan25x = 25+y2. xpr
e
ut
o
h
An interval of definition for the solution of the differential equation is (−π/10,π/10). Another wit
interval is (π/10,3π/10), and so on. ed
w
o
(cid:12) (cid:12) all
17. Thedomainofthefunctionis{x (cid:12) 4−x2 (cid:54)= 0}or{x (cid:12) x (cid:54)= −2orx (cid:54)= 2}. Fromy(cid:48) = 2x/(4−x2)2 n
o
we have uti
b
y(cid:48) = 2x(cid:18) 1 (cid:19)2 = 2xy2. distri
4−x2 o
N
An interval of definition for the solution of the differential equation is (−2,2). Other intervals ed.
v
are (−∞,−2) and (2,∞). ser
e
s r
√ ht
18. TThhuesf,utnhcetidoonmisaiyn=is 1{/x (cid:12)(cid:12)1x−(cid:54)=sπin/x2,+w2hnoπs}e.dFormomainy(cid:48)is=o−bt1a(i1ne−dsfirnoxm)−13/−2(s−incxos(cid:54)=x)0woerhsianvex (cid:54)= 1. All rig
2 g.
n
ni
2y(cid:48) = (1−sinx)−3/2cosx = [(1−sinx)−1/2]3cosx = y3cosx. ar
e
L
e
g
a
An interval of definition for the solution of the differential equation is (π/2,5π/2). Another g
n
e
C
interval is (5π/2,9π/2) and so on. ©
19. Writing ln(2X −1)−ln(X −1) = t and differentiating implicitly we obtain
2 dX 1 dX
− = 1
2X −1 dt X −1 dt
(cid:18) (cid:19)
2 1 dX
− = 1
2X −1 X −1 dt
2X −2−2X +1 dX
= 1
(2X −1)(X −1) dt
dX
= −(2X −1)(X −1) = (X −1)(1−2X).
dt
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
1.1 DefinitionsandTerminology 3
Exponentiating both sides of the implicit solution we obtain y
4
2X −1
= et 3
X −1
2
2X −1 = Xet−et
1
et−1 = (et−2)X
(cid:45)3 (cid:45)2 (cid:45)1 1 2 3 x
et−1 (cid:45)1
X = .
et−2 (cid:45)2
Solving et−2 = 0 we get t = ln2. Thus, the solution is defined on (−∞,ln2) or on (ln2,∞).
Thegraphofthesolutiondefinedon(−∞,ln2)isdashed, andthegraphofthesolutiondefined
n.
o on (ln2,∞) is solid.
ati
z
ori 20. Implicitly differentiating the solution, we obtain y
h
ut 4
a
s dy dy
es −2x2 −4xy+2y = 0
pr dx dx 2
x
ut e −x2dy−2xydx+ydy = 0
o
with 2xydx+(x2−y)dy = 0. (cid:45)4 (cid:45)2 2 4 x
ed (cid:45)2
w
o Using the quadratic formula to solve y2−2x2y−1 = 0 for y,
all √ √ (cid:45)4
on we get y = (cid:0)2x2± 4x4+4(cid:1)/2 = x2± x4+1. Thus,
uti √ √
strib two explicit solutions are y1 = x2 + x4+1 and y2 = x2 − x4+1. Both solutions are
di defined on (−∞,∞). The graph of y (x) is solid and the graph of y is dashed.
o 1 2
N
ved. 21. Differentiating P = c1et/(cid:0)1+c1et(cid:1) we obtain
er
s res dP = (cid:0)1+c1et(cid:1)c1et−c1et·c1et = c1et (cid:2)(cid:0)1+c1et(cid:1)−c1et(cid:3)
All right dt c1et (1(cid:20)+c1etc)12et (cid:21) 1+c1et 1+c1et
g. = 1+c et 1− 1+c et = P(1−P).
n 1 1
ni
ear (cid:90) x
ge L 22. Differentiating y = e−x2 et2dt+c1e−x2 we obtain
ga 0
n
Ce (cid:90) x (cid:90) x
© y(cid:48) = e−x2ex2 −2xe−x2 et2dt−2c1xe−x2 = 1−2xe−x2 et2dt−2c1xe−x2.
0 0
Substituting into the differential equation, we have
(cid:90) x (cid:90) x
y(cid:48)+2xy = 1−2xe−x2 et2dt−2c xe−x2 +2xe−x2 et2dt+2c xe−x2 = 1.
1 1
0 0
dy d2y
23. From y = c e2x+c xe2x we obtain = (2c +c )e2x+2c xe2x and = (4c +4c )e2x+
1 2 dx 1 2 2 dx2 1 2
4c xe2x, so that
2
d2y dy
−4 +4y = (4c +4c −8c −4c +4c )e2x+(4c −8c +4c )xe2x = 0.
dx2 dx 1 2 1 2 1 2 2 2
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with a certain product or service or otherwise on a password-protected website for classroom use.
Not For Sale
4 CHAPTER 1 INTRODUCTIONTODIFFERENTIALEQUATIONS
24. From y = c x−1+c x+c xlnx+4x2 we obtain
1 2 3
dy
= −c x−2+c +c +c lnx+8x,
1 2 3 3
dx
d2y
= 2c x−3+c x−1+8,
dx2 1 3
and
d3y
= −6c x−4−c x−2,
dx3 1 3
so that
x3 ddx3y3 +2x2 ddx2y2 −x ddxy +y = (−6c1+4c1+c1+c1)x−1+(−c3+2c3−c2−c3+c2)x ation.
z
+(−c3+c3)xlnx+(16−8+4)x2 ori
h
= 12x2. aut
s
s
e
pr
(cid:40) (cid:40) ex
25. From y = −x2, x < 0 we obtain y(cid:48) = −2x, x < 0 so that xy(cid:48)−2y = 0. out
x2, x ≥ 0 2x, x ≥ 0 with
d
e
w
26. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus, o
x→0− x→0+ all
y(cid:48)(x) does not exist at x = 0. n
o
uti
b
27. From y = emx we obtain y(cid:48) = memx. Then y(cid:48)+2y = 0 implies stri
di
o
N
memx+2emx = (m+2)emx = 0. d.
e
v
er
s
Since emx > 0 for all x, m = −2. Thus y = e−2x is a solution. s re
ht
g
28. From y = emx we obtain y(cid:48) = memx. Then 5y(cid:48) = 2y implies All ri
g.
2 n
5memx = 2emx or m = . ni
5 ear
L
e
g
Thus y = e2x/5 > 0 is a solution. a
g
n
e
C
29. From y = emx we obtain y(cid:48) = memx and y(cid:48)(cid:48) = m2emx. Then y(cid:48)(cid:48)−5y(cid:48)+6y = 0 implies ©
m2emx−5memx+6emx = (m−2)(m−3)emx = 0.
Since emx > 0 for all x, m = 2 and m = 3. Thus y = e2x and y = e3x are solutions.
30. From y = emx we obtain y(cid:48) = memx and y(cid:48)(cid:48) = m2emx. Then 2y(cid:48)(cid:48)+7y(cid:48)−4y = 0 implies
2m2emx+7memx−4emx = (2m−1)(m+4)emx = 0.
Since emx > 0 for all x, m = 1 and m = −4. Thus y = ex/2 and y = e−4x are solutions.
2
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
1.1 DefinitionsandTerminology 5
31. From y = xm we obtain y(cid:48) = mxm−1 and y(cid:48)(cid:48) = m(m−1)xm−2. Then xy(cid:48)(cid:48)+2y(cid:48) = 0 implies
xm(m−1)xm−2+2mxm−1 = [m(m−1)+2m]xm−1 = (m2+m)xm−1
= m(m+1)xm−1 = 0.
Since xm−1 > 0 for x > 0, m = 0 and m = −1. Thus y = 1 and y = x−1 are solutions.
32. From y = xm we obtain y(cid:48) = mxm−1 and y(cid:48)(cid:48) = m(m−1)xm−2. Then x2y(cid:48)(cid:48) −7xy(cid:48) +15y = 0
implies
x2m(m−1)xm−2−7xmxm−1+15xm = [m(m−1)−7m+15]xm
n. = (m2−8m+15)xm = (m−3)(m−5)xm = 0.
o
ati
oriz Since xm > 0 for x > 0, m = 3 and m = 5. Thus y = x3 and y = x5 are solutions.
h
ut
s a In Problems 33–36 we substitute y = c into the differential equations and use y(cid:48) = 0 and y(cid:48)(cid:48) = 0.
s
e
pr
ex 33. Solving 5c = 10 we see that y = 2 is a constant solution.
ut
o
with 34. Solving c2+2c−3 = (c+3)(c−1) = 0 we see that y = −3 and y = 1 are constant solutions.
d
e
w 35. Since 1/(c−1) = 0 has no solutions, the differential equation has no constant solutions.
o
all
on 36. Solving 6c = 10 we see that y = 5/3 is a constant solution.
uti
b
stri 37. From x = e−2t+3e6t and y = −e−2t+5e6t we obtain
di
d. No ddxt = −2e−2t+18e6t and ddyt = 2e−2t+30e6t.
e
v
ser Then
e
ghts r x+3y = (e−2t+3e6t)+3(−e−2t+5e6t) = −2e−2t+18e6t = dx
All ri and dt
g. dy
nin 5x+3y = 5(e−2t+3e6t)+3(−e−2t+5e6t) = 2e−2t+30e6t = .
ar dt
e
L
ge 38. From x = cos2t+sin2t+ 1et and y = −cos2t−sin2t− 1et we obtain
ga 5 5
n
e
C dx 1 dy 1
© = −2sin2t+2cos2t+ et or = 2sin2t−2cos2t− et
dt 5 dt 5
and
d2x 1 d2y 1
= −4cos2t−4sin2t+ et or = 4cos2t+4sin2t− et.
dt2 5 dt2 5
Then
1 1 d2x
4y+et = 4(−cos2t−sin2t− et)+et = −4cos2t−4sin2t+ et =
5 5 dt2
and
1 1 d2y
4x−et = 4(cos2t+sin2t+ et)−et = 4cos2t+4sin2t− et = .
5 5 dt2
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with a certain product or service or otherwise on a password-protected website for classroom use.
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6 CHAPTER 1 INTRODUCTIONTODIFFERENTIALEQUATIONS
Discussion Problems
39. (y(cid:48))2+1 = 0 has no real solutions because (y(cid:48))2+1 is positive for all functions y = φ(x).
40. The only solution of (y(cid:48))2+y2 = 0 is y = 0, since, if y (cid:54)= 0, y2 > 0 and (y(cid:48))2+y2 ≥ y2 > 0.
41. The first derivative of f(x) = ex is ex. The first derivative of f(x) = ekx is f(cid:48)(x) = kekx. The
differential equations are y(cid:48) = y and y(cid:48) = ky, respectively.
42. Any function of the form y = cex or y = ce−x is its own second derivative. The corresponding
differential equation is y(cid:48)(cid:48)−y = 0. Functions of the form y = csinx or y = ccosx have second
n.
derivatives that are the negatives of themselves. The differential equation is y(cid:48)(cid:48)+y = 0. o
ati
z
43. We first note that (cid:112)1−y2 = (cid:112)1−sin2x = √cos2x = |cosx|. This prompts us to consider hori
ut
a
values of x for which cosx < 0, such as x = π. In this case s
s
e
pr
(cid:12) (cid:12) x
ddxy(cid:12)(cid:12)(cid:12) = ddx(sinx)(cid:12)(cid:12)(cid:12) = cosx(cid:12)(cid:12)x=π = cosπ = −1, hout e
(cid:12)x=π (cid:12)x=π wit
d
e
but w
(cid:112)1−y2(cid:12)(cid:12)(cid:12) = (cid:112)1−sin2π = √1 = 1. n allo
x=π o
Thus,y = sinxwillonlybeasolutionofy(cid:48) = (cid:112)1−y2 whencosx > 0. Anintervalofdefinition stributi
is then (−π/2,π/2). Other intervals are (3π/2,5π/2), (7π/2,9π/2), and so on. o di
N
d.
44. Since the first and second derivatives of sint and cost involve sint and cost, it is plausible that ve
er
a linear combination of these functions, Asint+Bcost, could be a solution of the differential es
s r
equation. Using y(cid:48) = Acost − Bsint and y(cid:48)(cid:48) = −Asint − Bcost and substituting into the ht
g
differential equation we get All ri
g.
n
y(cid:48)(cid:48)+2y(cid:48)+4y = −Asint−Bcost+2Acost−2Bsint+4Asint+4Bcost arni
e
L
= (3A−2B)sint+(2A+3B)cost = 5sint. e
g
a
g
n
e
Thus 3A−2B = 5 and 2A+3B = 0. Solving these simultaneous equations we find A = 15 C
13 ©
and B = −10 . A particular solution is y = 15 sint− 10 cost.
13 13 13
45. One solution is given by the upper portion of the graph with domain approximately (0,2.6).
The other solution is given by the lower portion of the graph, also with domain approximately
(0,2.6).
46. One solution, with domain approximately (−∞, 1.6) is the portion of the graph in the second
quadrant together with the lower part of the graph in the first quadrant. A second solution,
with domain approximately (0, 1.6) is the upper part of the graph in the first quadrant. The
third solution, with domain (0, ∞), is the part of the graph in the fourth quadrant.
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