Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition Dennis G. Zill

A First Course in Differential
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Equations with Modeling
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Applications, 12th Edition by
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Dennis G. Zill fg fg fg




Complete Chapter Solutions Manual
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are included (Ch 1 to 9)
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** Immediate Download
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** Swift Response
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** All Chapters included
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,Solution fgandfgAns wer fgGuide: fg Zill, fgDIFFERENTIAL fgEQUATIONS fgWith fgMODELING fgAPPLICATIONS fg2024, fg9780357760192; fgChapter
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Introduction f g to f g Differential f g Equations


SolutionandAnswer Guide fg fg fg




ZILL, DIFFERENTIAL EQUATIONSW ITH MODELING APPLICATIONS 2024,
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9780357760192; CHAPTER #1: INTRODUCTION TO DIFFERENTIAL EQUATIONS
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TABLE OF CONTENTS fg g
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End of Section Solutions...................................................................................................................................... 1
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Exercises 1.1 .................................................................................................................................................................................... 1
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Exercises 1.2 ..................................................................................................................................................................................14
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Exercises 1.3 ..................................................................................................................................................................................22
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Chapter 1 in Review Solutions ....................................................................................................................... 30
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END OF SECTION SOLUTIONS
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f g
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EXERCISES 1.1 f g




1. Second f g order; f g linear
2. Third order; nonlinear because of (dy/dx)4
fg fg fg fg fg



3. Fourth order; linear fg fg



4. Second order; nonlinear because of cos(r + u)
fg fg fg fg fg fg fg

√ fg


5. Second order; nonlinear because of (dy/dx)2
fg fg fg fg fg
fg f g
or 1 + (dy/dx)2
f g f g


2
6. Second order; nonlinear because of R fg fg fg fg fg



7. Third order; linear fg fg



8. Second order; nonlinear because of ẋ 2 fg fg fg fg fg fg



9. First order; nonlinear because of sin (dy/dx)
fg fg fg fg fg fg



10. First order; linear fg fg



11. Writing the differential equation in the form x(dy/dx) + y2 = 1 , we see that it is
fg fg fg fg fg fg fg fg fg
f g
f g fg fg fg fg fg


nonlinear
fg in y because of y2 . However, writing it in the form (y2 — 1)(dx/dy) + x =
f g fg fg fg fg fg fg fg fg fg fg fg
fg
fg fg fg fg


0, we see that it is linear in x.
fg fg fg fg fg fg f g fg fg



12. Writing the differential equation in the form u(dv/du) + (1 + u)v = ue u we see
fg fg fg fg fg fg fg fg fg fg fg f g fg
fg
fg


that it is linear in v. However, writing it in the form (v + uv — ue u)(du/dv) + u = 0, we
fg fg fg f g fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg


see that it is nonlinear in u .
fg fg fg fg f g fg f g


1
13. From y = e− x/2 we obtain yj = —
fg
2
fg fg fg fg
f g
fg fg
x/2
. Then 2yj + y =
fg fg
fg
fg fg
x/2
+ x/2
= 0. fg


e− fg—e
fg
− e−
fg




1

,Solution fgandfgAns wer fgGuide: fg Zill, fgDIFFERENTIAL fgEQUATIONS fgWith fgMODELING fgAPPLICATIONS fg2024, fg9780357760192; fgChapter
fg#1:


Introduction f g to f g Differential f g Equations

— 6 6
14. From y = — e 20t we obtain dy/dt = 24e−20t , so that
fg f g
fg
fg fg fg fg
fg
fg fg

5 5
dy + 20y = 24e−20t
fg 6 6 f g
−20t
— e
fg fg fg
+ 20 f g fg f g = 24. fg
dt 5 5

15. From y = e 3x cos 2x we obtain yj = 3e3x cos 2x—2e3x sin 2x and yjj = 5e3x cos 2x—12e3x
fg fg fg
fg
fg fg fg fg
fg
fg
fg
fg
fg
fg fg fg
fg
fg
fg
fg


sin 2x, so that yjj — 6yj + 13y = 0 .
fg
gf f g fg fg
f g
fg
fg
fg fg fg

j
16. From y = — cos x ln(sec x + tan x) we obtain y
fg fg fg fg fg fg fg fg fg fg fg fg fg fg f g = —1 + sin x ln(sec x + tan x) and
fg fg fg fg fg fg fg fg fg fg

jj jj
y fg f g = tan x + cos x ln(sec x + tan x). Then y
fg + y = tan x.fg fg fg fg fg fg fg fg fg fg fg fg f g fg fg fg fg



17. The domain of the function, found by solving x+2 ≥ 0, is [—2, ∞). From yj
fg fg fg fg fg fg fg fg f g fg fg fg fg f g fg
fg f g
= 1+2(x+2)− 1/2
f g


we have fg



j − 1/2 f g
(y —x)y fg f g = (y — x)[1 + (2(x + 2)
fg fg fg fg fg fg fg fg f g ]

= y — x + 2(y —x)(x + 2)−1/2
fg fg fg g
f fg fg fg fg




= y — x + 2[x + 4(x + 2)1/2 —x](x + 2)−1/2
fg fg fg fg fg fg fg fg fg
fg fg
fg fg




= y — x + 8(x + 2)1/2(x + 2)−1/2 = y — x + 8.
fg fg fg fg fg fg fg fg fg
f g
fg fg fg fg fg




An interval of definition for the solution of the differential equation is (—2, ∞)
fg fg fg fg fg fg fg fg fg fg fg fg fg


because y j is not defined at x = —2.
fg fg
fg
f g f g f g f g fg f g



18. Since tan x is not defined for x =
fg fg fg fg fg fg fg f g f g π/2 + nπ, n an integer, the domain of y =
fg fg fg fg fg fg fg fg fg fg f g f g 5 tan 5x is
fg fg fg


{x fg f g 5x /
= π/2 + nπ} fg fg fg fg



or {x fg fg f g x /= π/10 + nπ/5}. From y j= 25 sec 25x we have
fg fg fg fg fg fg f g fg fg fg fg fg



j
= 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y2 .
fg
y fg fg fg
fg
fg fg fg fg fg
fg
fg fg fg fg




An interval of definition for the solution of the differential equation is (—π/10, π/10). An-
fg fg fg fg fg fg fg fg fg fg fg fg fg fg


other interval is (π/10, 3π/10) , and so on.
f g fg fg fg fg fg fg fg



19. The domain of the function is {x
fg fg fg fg fg fg fg fg f g 4 /
= 0} or {x x / = 2} . From y j =
= —2 or x / fg fg fg fg fg fg fg fg fg fg fg
f g



f g — fg x
2

2x/(4 — x2)2 we have fg fg
f g
fg
1 2
= 2xy2. fg


= 2x 4 —x
2
yj f g f g fg
gf fg




An interval of definition for the solution of the differential equation is (—2, 2).
fg fg fg fg fg fg fg fg fg fg fg fg fg


Other inter- vals are (—∞, —2) and (2, ∞).
√
fg fg f g fg fg fg fg fg fg fg



20. The function is y = fg fg fg f g 1 — sin x , whose domain is obtained from 1 — sin x /= 0 or sin x /= 1.
fg fg fg fg fg fg fg fg fg fg fg fg fg fg f g fg fg fg fg f g


fg 1/
Thus, the domain is {x fg fg fg fg fg f g x /= π/2 + 2nπ}. From y j= —
fg fg
2
fg fg fg fg f g fg f g (11 — sin x) fg fg fg f g
−3/2 f g (—fgcos fgx) fgwe fghave


2yj = (1 — sin x)− 3/2 cos x = [(1 — sin x)− 1/2] 3 cos x = y3 cos x.
fg
fg fg fg fg
fg
fg fg fg fg fg fg
fg
fg fg fg
fg
fg




An interval of definition for the solution of the differential equation is
fg fg fg fg fg fg fg fg fg fg fg f g (π/2, 5π/2). fg


Another
fg one is (5π/2, 9π/2), and so on. f g f g f g fg f g f g f g




2

, Solution fgandfgAns wer fgGuide: fg Zill, fgDIFFERENTIAL fgEQUATIONS fgWith fgMODELING fgAPPLICATIONS fg2024, fg9780357760192; fgChapter
fg#1:


Introduction f g to f g Differential f g Equations



21. Writing ln(2X fg f g — 1) — ln(X — 1) = t and differentiating
fg fg fg f g fg fg f g fg f g fg fg x

implicitly we obtain fg fg 4


— = 1 fg 2
2X — 1 fg fg f g dt X —1 fg fg f g dt
t
2 1 dX
— –fg4 –2
fg f g
= 1 2 4
2X — 1 X — 1 dt
f g
fg fg fg fg


–2


–fg4
dX
= —(2X — 1)(X — 1) = (X — 1)(1 — 2X).
fg fg fg fg fg fg fg fg fg fg fg
dt f g



Exponentiating both sides of the implicit solution we fg fg fg fg fg fg fg


obtain fg




2X — 1
= et
fg fg fg


X —1
fg
fg fg
fg




2X — 1 = Xe t — et
fg fg fg fg
fg
fg




(et — 1) = (et — 2)X
fg
fg fg fg
fg
fg




et 1
X = .
et — 2
fg f g
fg
fg fg



Solving e t — 2 = 0 we get t = ln 2. Thus, the solution is defined on (—∞, ln 2) or on
fg
fg
fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg fg


(ln 2, ∞). The graph of the solution defined on (—∞, ln 2) is dashed, and the
fg fg fg f g fg fg fg fg fg fg f g fg fg fg fg fg fg


graph of the solution defined on (ln 2, ∞) is solid.
fg fg fg fg f g fg fg fg fg fg fg




22. Implicitly differentiating the solution, we obtain
f g f g fg fg fg y

2 fg fg dy dy 4

—2x — 4xy + 2y = 0
fg f g fg fg fg fg fg
dx dx fg fg
2
—x2 dy — 2xy dx + y dy = 0
fg
fg fg fg fg fg fg fg fg



x
2xy dx + (x2 — y)dy = 0.
fg fg fg
fg
fg fg fg
–fg4 –2 2 4

–2
Using the quadratic formula to solve y2
fg fg fg fg fg fg — 2x2y — 1 = 0
fg f g
fg f g fg fg fg fg f g

√ fg √ fg

for y, we get y =
fg fg fg fg fg 2x2 fg f g
4x42 + 4 fg
fg fg f g /2 = ± x4 + 1 . f g
fg
fg fg
–
fg
± x
f g
fg4
√ fg

Thus, two explicit solutions are y1
fg fg fg fg fg fg f g = x4 + 1 and fg
fg f g

f g x2 + f g


√ fg fg

y2 = x 2 —
fg f g fg
fg f g
x4 + 1 . Both solutions are defined on (—∞, ∞).
f g
fg fg fg fg fg fg fg fg fg



The graph of y1 (x) is solid and the graph of y2 is dashed.
fg fg fg fg fg fg fg fg fg fg fg f g fg




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