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 Ap qy qy qy




plications, 12th Edition by De qy qy qy qy




nnis G. Zill qy qy




Complete Chapter Solutions Manual ar
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e 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 and Answer Guide qy qy qy




ZILL, DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS 2024,
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9780357760192; CHAPTER #1: INTRODUCTION TO DIFFERENTIAL EQUATIONS qy qy qy qy qy qy




TABLE OF CONTENTS QY QY




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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EXERCISES 1.1 QY




1. Second order; linear q y q y


4
2. Third order; nonlinear because of (dy/dx)
qy qy qy qy qy



3. Fourth order; linear qy qy



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


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

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



7. Third order; linear qy qy


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



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



10. First order; linear qy qy


2
11. Writing the differential equation in the form x(dy/dx) + y = 1, we see that it is nonli
qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy


near in y because of y . However, writing it in the form (y —
2 2
qy qy qy qy qy qy qy qy qy qy qy qy qy


1)(dx/dy) + x = 0, we see that it is linear in x.
qy qy qy qy qy qy qy qy qy qy qy qy qy


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


it is linear in v. However, writing it in the form (v + uv —
qy qy qy qy qy qy qy qy qy qy qy qy qy qy


ueu)(du/dv) + u = 0, we see that it is nonlinear in u.
qy qy qy qy qy qy qy qy qy qy qy qy qy



Fromqyyqy=qye− qyweqyobtainqyyjqy=qy—qy1qye− .qyThenqy2yjqy+qyyqy=qy—e− qy+qye− qy=qy0.
x/2 x/2 x/2 x/2
13. 2

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6 6 —
14. From y = qy qy — e we obtain dy/dt = 24e
qy qy qy qy , so that
qy qy

5 5
qyqy
dy −20t 6 6 qy

— −20t
5 qy

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


12e sin 2x, so that yjj — 6yj + 13y = 0.
3x
qy qy qy qy qy qy qy qy qy qy qy

j
16. From y = — qy qy qy = —1 + sin x ln(sec x + tan x) and
qy qy qy qy qy qy qy qy qy qy

cos x ln(sec x + tan x) we obtain y
qy qy qy qy qy qy qy qy qy qy

jj
y q y = tan x + cos x ln(sec x + tan x). Then y
qy qy qy qy qy qy qy qy qy qy qy qy q y + y = tan x.
qy qy qy qy



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


we have qy



j −
—x)y = (y — x)[1 + (2(x + 2) ]
q y qy qy qy qy qy qy qy




−1/2
= y — x + 2(y —
qy qy qy qy qy qy




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




= y — x + 8(x + 2)1/2
qy qy qy qy qy qy qy
−1/2q y =qyyq y — qyxqy+qy8.


An interval of definition for the solution of the differential equation is (—
qy qy qy qy qy qy qy qy qy qy qy qy


2, ∞) because yj is not defined at x = —2.
qy qy qy qy qy qy qy qy qy qy



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


{x q y q y 5x /= π/2 + nπ}
qy qy qy qy



or {x qy
q y
x /= π/10 + nπ/5}. From j y = 25 sec
qy qy
2 5x we haveqy qy qy qy q y qy qy q y qy qy




2 2 2
y .

An interval of definition for the solution of the differential equation is (—
qy qy qy qy qy qy qy qy qy qy qy qy


π/10, π/10). An- other interval is (π/10, 3π/10), and so on.
qy qy qy qy qy qy qy qy qy qy



19. The domain of the function is {x
qy 4— x qy qy qy qy qy qy qy /= 0} or {x
q y qy qy x /= —
q y q y


2 or x /= 2}. From y = 2x/(4 — x2)2 we have
qy qy q y q y qy qy q y qy qy qy qy qy


q y q y 1
yj = 2xqy qy q y = 2xy2. qy
2

4 — x2qy qy



An interval of definition for the solution of the differential equation is (—
qy qy qy qy qy qy qy qy qy qy qy qy


2, 2). Other inter- vals are (—∞, —2) and (2, ∞).
qy qy qy qy qy qy qy qy qy qy

√
20. The function is y = 1/ 1 — sin x , whose domain is obtained from 1 — sin x /= 0 or sin x /= 1.
qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy qy


Thus, the domain is {x x /= π/2 + 2nπ}. From y = 2— (1 — sin x) (— cos x) we have
qy qy qy qy q y qy qy qy qy qy qy qy qy q y qy qy qy qy qy qy qy




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




An interval of definition for the solution of the differential equation is (π/2, 5π/2). Anoth
qy qy qy qy qy qy qy qy qy qy qy qy qy qy

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