A First Course in Differential Equations with Modeling Applications Differential Equations with ...

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Complete Solutions Manual
A First Course in Differential Equations
with Modeling Applications
Ninth Edition
Dennis G. Zill
Loyola Marymount University
Differential Equations with
Boundary-Vary Problems
Seventh Edition
Dennis G. Zill
Loyola Marymount University
Michael R. Cullen
Late of Loyola Marymount University
By
Warren S. Wright
Loyola Marymount University
Carol D. Wright
* ; BROOKS/COLE
CENGAGE Learning-
Australia • Brazil - Japan - Korea • Mexico • Singapore • Spain • United Kingdom • United States
Table of Contents
1 Introduction to Differential Equations 1
2 First-Order Differential Equations 27
3 Modeling with First-Order Differential Equations 86
4 Higher-Order Differential Equations 137
5 Modeling with Higher-Order Differential Equations 231
6 Series Solutions of Linear Equations 274
7 The Laplace Transform 352
8 Systems of Linear First-Order Differential Equations 419
9 Numerical Solutions of Ordinary Differential Equations 478
10
Plane Autonomous Systems 506
11
Fourier Series 538
12 Boundary-Value Problems in Rectangular Coordinates 586
13 Boundary-Value Problems in Other Coordinate Systems 675
14 Integral Transforms 717
15 Numerical Solutions of Partial Differential Equations 761
Appendix 783
I Gamma function
II
Appendix Matrices 785
3.ROOKS/COLE
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1 3 4 5 67 11 10 09 08
1 Introduction to Differential Equations
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear bccausc of cos(r + u)
5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2
6. Second order: nonlinear bccausc of R~
7. Third order: linear
8. Second order; nonlinear because of x2
9. Writing the differential equation in the form x(dy/dx) -f y2 = 1. we sec that it is nonlinear in y
because of y2. However, writing it in the form (y2 — 1 )(dx/dy) + x = 0, we see that it is linear in x.
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu wc see that it is linear in
v. However, writing it in the form (v + uv — ueu)(du/dv) + u — 0, we see that it, is nonlinear in ■Ji­
ll.
From y = e-*/2 we obtain y' = — \e~x'2. Then 2y' + y = —e~X//2 + e-x/2 = 0.
12. From y = | — |e-20* we obtain dy/dt = 24e-20t, so that
% + 20y = 24e~m + 20 - |e_20t) = 24.
clt \ 'o 5 /
13. R'om y = eix cos 2x we obtain y1 = 3e^x cos 2x — 2e3* sin 2a? and y” = 5e3,x cos 2x — 12e3,x sin 2x, so
that y" — (k/ + l?>y = 0.
14. From y = — cos:r ln(sec;r + tanrc) we obtain y’ — — 1 + sin.Tln(secx + tana:) and
y" = tan x + cos x ln(sec x + tan a?). Then y" -f y = tan x.
15. The domain of the function, found by solving x + 2 > 0, is [—2, oo). From y’ = 1 + 2(x + 2)_1/2 we
1
Exercises 1.1 Definitions and Terminology
have
{y - x)y' = (y - ®)[i + (20 + 2)_1/2]
= y — x + 2(y - x)(x + 2)-1/2
= y-x + 2[x + 4(z + 2)1/2 - a;] (a: + 2)_1/2
= y — x + 8(ac + 2)1;/i(rr + 2)~1/2 = y — x + 8.
An interval of definition for the solution of the differential equation is (—2, oo) because y
defined at x = —2.
16. Since tan:r is not defined for x = 7r/2 + mr, n an integer, the domain of y = 5t£v.
{a; | 5x ^ tt/2 + 7i-7r} or {;r | x ^ tt/IO + mr/5}. From y' — 25sec2 §x we have
y' = 25(1 + tan2 5x) = 25 + 25 tan2 5a: = 25 + y2.
An interval of definition for the solution of the differential equation is (—7r/10,7T/10 . A:,
interval is (7r/10, 37t/10). and so on.
17. The domain of the function is {x \ 4 — x2 ^ 0} or {x\x ^ —2 or x ^ 2}. Prom y' — 2.:: -= -
we have
An interval of definition for the solution of the differential equation is (—2,2). Other
(—oc,—2) and (2, oo).
18. The function is y — l/y/l — sins. whose domain is obtained from 1 — sinx ^ 0 or . = 1 T
the domain is {z | x ^ tt/2 + 2?i7r}. From y' = —1(1 — sin x) 2 (— cos.x) we have
2y' = (1 — sin;r)_‘?/’2 cos# = [(1 — sin:r)~1//2]3cos:r - f/3cosx.
An interval of definition for the solution of the differential equation is (tt/2. 5tt/2 A:.. .
is (57r/2,97r/2) and so on.
19. Writing ln(2X — 1) — ln(X — 1) = t and differentiating implicitly we obtain
2 dX 1 dX
2X-1 dt X - l dt
2X - 2 - 2X + 1 dX _
(2X - 1)(X - 1) dt
IX
— = -C2X - 1)(X - 1) = (X - 1)(1 - 2X .
2
Exercises 1.1 Definitions and Terminology
Exponentiating both sides of the implicit solution we obtain x
2X-1
-----= el
X - l
2X-1 = Xel - ef
-4 -2
(e* - 1) = (e‘ - 2)X
-2
ef' — 1
X = -4
e*-2 '
Solving e* — 2 = 0 we get t = In2. Thus, the solution is defined on (—oc.ln2) or on (In2, oo).
The graph of the solution defined on (—oo,ln2) is dashed, and the graph of the solution defined on
(In 2. oc) is solid.
20. Implicitly differentiating the solution, we obtain y
—x2 dy — 2xy dx + y dy — 0
2 xy dx + (a;2 — y)dy = 0.
Using the quadratic formula to solve y2 — 2x2y — 1 = 0 for y, we get
y = (2x2 ± V4;c4 +4)/2 = a’2 ± vV1 + 1. Thus, two explicit
solutions are y\ = x2 + \A'4 + 1 and y-2 = x2 — V.x4 + 1. Both
solutions are defined on (—oo. oc). The graph of yj (x) is solid and
the graph of y is daalied.
-2
21. Differentiating P = c\?} } (l + cie^ we obtain
dP _ (l + cie*) cie* - cie* • cie* _ Cie« [(l + cie‘) - cie4]
eft (1 + cie*)" 1 + cief 1 + cie(
Ci CiC
= P( 1 - P).
1 + CI&- 1 + ci ef
2 PX ,2 ,2
22. Differentiating y = e~x / e: dt + c\ e~x we obtain
Jo
y / = e -*2e r2 2xe___ 2* /fr xX e J*22 dt — 2c\xe -r.2=1 2xe x 2 rx e +2dt — 2cixe —X
Jo Jo
Substituting into the differential equation, we have
y' + 2xy = l — 2xe x I e* dt — 2cixe x +2xe x [ e* dt 4- 2cio;e x = 1.
Jo Jo
3
Exercises 1.1 Definitions and Terminology
23. From y — ci e2x+c.2xe2x we obtain ^ - (2c\ +C2)e2x -r2c2xe2x and —| = (4cj + 4c;2)e2x + 4c2xe2j'.
so that
— 4 ^ + Ay = (4ci + 4co - 8ci - 4c2 + 4ci)e2x + (4c2 — Sc2 + 4e2)xe2x — 0.
nd.Tx. 2 dd:xr
24. From y — Cix-1 + c^x + c%x ]n x + 4a;2 we obtain
^ = —c\x 2 + C2 + c$ + C3 In x + 8rc,
dx
d2y
= 2cix,_3 + C3;r_1 -f 8.
dx2
and
= —6cix-4 - c3ar2,
so that
dx'3 +_r “2aJ/' 2 dx2 ~ XJ' dx + V ~ ^_6ci + 4ci + Cl + Cl^x 3 + ^_ °3 + 2cs ~~ 02 ~ C3 + C2^X
t (—C3 + cz)x In a; + (16 - 8 + 4)x2
= 12x2.
( —x2, x < 0 , f —2x, x < 0
25. From y = < ' we obtain y' = < ^ ^ „ so that - 2/y = 0.
t x . x > 0 {2x, x > 0
26. The function y(x) is not continuous at x = 0 sincc lim y(x) = 5 and lim y(x) = —5. Thus. y’(x)
x—>0“ x—>0+
does not exist at x = 0.
27. From y = emx we obtain y' = mernx. Then yf + 2y — 0 implies
rnemx + 2emx = (m + 2)emx = 0.
Since emx > 0 for all x} m = —2. Thus y = e~2x is a solution.
28. From y = emx we obtain y1 = mernx. Then by' — 2y implies
brriemx = 2e"lx or m =
5
Thus y = e2:c/5 > 0 is a solution.
29. From y = emx we obtain y' = memx and y" = rn2emx. Then y" — 5y' + Qy = 0 implies
m2emx - 5rnemx + 6emx = (rn - 2)(m - 3)emx = 0.
Since ema! > 0 for all x, rn = 2 and m = 3. Thus y = e2x and y = e3:r are solutions.
30. 0
From y = emx we obtain y1 = rnemx an<l y" = rn2emx. Then 2y" + 7y/ — 4y = implies
2m2emx + 7rnemx - 4ema: = (2m - l)(m + 4)ema' = 0.
Exercises 1.1 Definitions and Terminology
Since emx > 0 for all x, rn ~ | and m = —4. Thus y — ex/2 and y = e ^ are solutions.
31. From y = xm we obtain y' = mxm~1 and y" = m(m — l)xm~2. Then xy" + 2y' = 0 implies
xm(m, — l)xm~2 + 2mxm~l = [m(rn -1)4- = (m2 + m)xm_1
- rn(m + l).xm_1 = 0.
Since a:"'-1 > 0 for ;r > 0. m = 0 and m = — 1. Thus y = 1 and y — x~l are solutions.
32. From y = xm we obtain y' = mxm~1 and y" = m(m — l)xm~2. Then x2y" — 7xy' +15y — 0 implies
[m(m
x2rn{rn — l)xrn~2 — lxmxm~A + 15:em = — 1) — 7m + 15]xm
= (ro2 - 8m + 15)a,m = (m - 3) (to - 5)xm = 0.
Since xm > 0 for x > 0. m = 3 and m = 5. Thus y — x^ and y = xa are solutions.
In Problems 33-86 we substitute y = c into the differential equations and use y' — 0 and y" — 0
33. Solving 5c = 10 we see that y ~ 2 is a constant solution.
34. Solving c2 + 2c — 3 = (c + 3)(c — 1) = 0 we see that, y = —3 and y = 1 are constant solutions.
35. Since l/(c — 1) = 0 has no solutions, the differential equation has no constant solutions.
36. Solving 6c = 10 we see that y = 5/3 is a constant solution.
37. From x — e~2t + 3ec< and y — —e~2t + 5ew we obtain
^ = —2e~2t + 18e6* and = 2e~2t + 30e6*.
dt dt
Then
x- + 3y = (e~2t + 3e6t) + 3(-e'2* + oe6t) = -2e"2* + 18e6t = ^
\Jub
and
5:r + 3y = 5(e~2* + 3eet) + 3(-e~2* + 5e6') = 2e~2t + 30e6* = ^ .
at
38. From x = cos 21 + sin 21 + and y — — cos 21 — sin 21 — we obtain
— = —2 sin 2t -f 2 cos 22 + and ^ = 2 sin 22 — 2 cos 2t — -e*
d.t 5 dt 5
and
d2:r , „ . . ^ 1 ^ , ^2V , 1 /
= —4 cos 2t — 4 sm 22 + re and -r-^- = 4 cos 2t + 4 sin 22--e .
dt2 Id d22 5
Then
cPx
1 1
4 y + et = 4(— cos 21 — sin 21 — pef) + el — —4 cos 21 — 4 sin 22 + -el = -7-^
0 o dt
and
5
Exercises 1.1 Definitions and Terminology
4x — ef = 4(cos 21 + sin 21 -I- ^e*) — e* = 4 cos 2£ + 4 sin 2t — \ef —
39. (t/)2 + 1 = 0 has no real solutions becausc {y')2 + 1 is positive for all functions y = 4>(x).
40. The only solution of (?/)2 + y2 = 0 is y = 0, since if y ^ 0, y2 > 0 and (i/)2 + y2 > y2 > 0.
41. The first derivative of f(x) = ex is eT. The first derivative of f{x) = ekx is kekx. The differential
equations are y' — y and y' = k.y, respectively.
42. Any function of the form y = cex or y = ce~x is its own sccond derivative. The corresponding
differential equation is y" — y = 0. Functions of the form y = c sin x or y — c cos x have sccond
derivatives that are the negatives of themselves. The differential equation is y" -+- y = 0.
43. We first note that yjl — y2 = \/l — sin2 x = Vcos2 x = | cos.-r|. This prompts us to consider values
of x for which cos x < 0, such as x = tt. In this case
i {sklx)
%
= cosxl^,. = COS7T = —1.
dx
X=7T
but
\/l - y2\x=7r = V1 - sin2 7r = vT = 1.
Thus, y = sin re will only be a solution of y' - y l — y2 when cos x > 0. An interval of definition is
then (—tt/2, tt/2). Other intervals are (3tt/2, 5tt/2), (77t/2, 9tt/2). and so on.
44. Since the first and second derivatives of sint and cos t involve sint and cos t, it is plausible that a
linear combination of these functions, Asint+B cos t. could be a solution of the differential equation.
Using y' — A cos t — B sin t and y" = —A sin t — B cos t and substituting into the differential equation
we get
y" + 2y' + 4y = —A sin t — B cos t + 2A cos t — 2B sin t + 4A sin t + 4B cos t
= (3A — 2B) sin t + (2A + 3B) cos t = 5 sin t.
Thus 3A — 2B = 5 and 2A + 3B = 0. Solving these simultaneous equa+t ions TwT7e« ^fiITnidrl AA =-- -j-#- and
13
B = — . A particular solution is y = sint — ^ cost.
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 (—oo, 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, oo), is the part of the graph in the fourth quadrant.
6
Exercises 1.1 Definitions and Terminology
47. Differentiating (V1 + y^)/xy = 3c we obtain
xy(3x2 + 3y2y') - (a?3 + y*)(xi/ + y)
= 0
x?y2
3 x3y + 3 xy^y' — x'^y' — x% — xy^y’ — yA — 0
(3:ry3 - xA - xyz}i/ = -3x3y + xiy + y4
, = y4 - 2x3y _ y(y[i - 2x3)
^ 2.ry3 — x4 rt:(2y3 — a:3)
48. A tangent line will be vertical where y' is undefined, or in this case, where :r(2y3 — x3) = 0. This
gives x = 0 and 2y3 = a:3. Substituting y?J — a;3/2 into ;r3 + y3 = 3xy we get
x + h = x { w x)
3 3 3
-x3 = — r2
2 2V3a
a:3 = 22/V
z2(.x - 22/3) = 0.
Thus, there are vertical tangent lines at x = 0 and x = 22/3, or at (0,0) and (22/3,21'/3). Since
22/3 ~ 1.59. the estimates of the domains in Problem 46 were close.
49. The derivatives of the functions are ^(.x) — —xf a/25 — x2 and ^{x) = x/\/25 — x2, neither of
which is defined at x = ±5.
50. To determine if a solution curve passes through (0,3) we let 2 = 0 and P = 3 in the equation
P = c-ie1/(1 + eye*). This gives 3 = cj/(l + ci) or c\ = — |. Thus, the solution curve
(—3/2)e* = —3e*
1 - (3/2)eL 2 - 3e{
passes through the point, (0,3). Similarly, letting 2 = 0 and P = 1 in the equation for the one-
parameter family of solutions gives 1 = ct/(l + ci) or ci = 1 + c-|. Since this equation has no
solution, no solution curve passes through (0.1).
51. For the first-order differential equation integrate f(x). For the second-order differential equation
integrate twice. In the latter case we get y = f(f f(x)dx)dx + cja: + C -
2
52. Solving for y’ using the quadratic formula we obtain the two differential equations
y> = — ^2 + 2\J 1 + 3ar®^ and y1 = — ^2 — 2y 1 4-3a?^^ ,
so the differential equation cannot be put in the form dy/dx = f(x,y).
7