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 App Qi Qi Qi




lications, 12th Edition by Den
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nis G. Zill Qi Qi




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 Qi Qi Qi




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




TABLE OF CONTENTS QI QI




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 QI




1. Second order; linear Q i Q i


4
2. Third order; nonlinear because of (dy/dx)
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3. Fourth order; linear Qi Qi



4. Second order; nonlinear because of cos(r + u)
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5. Second order; nonlinear because of (dy/dx)2 or
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2
6. Second order; nonlinear because of R Qi Qi Qi Qi Qi



7. Third order; linear Qi Qi


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



9. First order; nonlinear because of sin (dy/dx)
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10. First order; linear Qi Qi


2
11. Writing the differential equation in the form x(dy/dx) + y = 1, we see that it is nonlin
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ear in y because of y . However, writing it in the form (y —
2 2
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1)(dx/dy) + x = 0, we see that it is linear in x.
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u
12. Writing the differential equation in the form u(dv/du) + (1 + u)v = ue we see that it
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is linear in v. However, writing it in the form (v + uv —
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ueu)(du/dv) + u = 0, we see that it is nonlinear in u.
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FromQiyQi=Qie− QiweQiobtainQiyjQi=Qi—Qi1Qe −x/2.QiThenQi2yjQi+QiyQi=Qi—e−x/2Qi+Qie−x/2Qi=Qi0.
x/2 i
13. 2

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66 —
14. From y = Qi Qi — e we obtain dy/dt = 24e
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5 5
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dy −20t 6 6 Qi

— −20t
5 Qi

e
3x
15. From y = e cos 2x we obtain yj = 3e3x cos 2x—2e3x sin 2x and yjj = 5e3x cos 2x—
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12e sin 2x, so that yjj — 6yj + 13y = 0.
3x
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j
16. From y = — Qi Qi Qi = —1 + sin x ln(sec x + tan x) and
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cos x ln(sec x + tan x) we obtain y
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jj
y Q i = tan x + cos x ln(sec x + tan x). Then y
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17. The domain of the function, found by solving x+2 ≥ 0, is [—2, ∞). From yj = 1+2(x+2)−
1/2
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we have Qi



j −
—x)y = (y — x)[1 + (2(x + 2)
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−1/2
= y — x + 2(y —
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−1/2
= y — x + 2[x + 4(x + 2)1/2 —
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= y — x + 8(x + 2)1/2
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−1/2Q i =QiyQ i —QixQi+Qi8.


An interval of definition for the solution of the differential equation is (—
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2, ∞) because yj is not defined at x = —2.
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18. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
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{x Q i Q i 5x /= π/2 + nπ}Qi Qi Qi Qi



or {x Qi
Q i
x /= π/10 + nπ/5}. From jy = 25 s2ec 5x we have
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2 2 2
y .

An interval of definition for the solution of the differential equation is (—π/10, π/10). An-
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other interval is (π/10, 3π/10), and so on.
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19. The domain of the function is {x 4 — x /= 0} or {x
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Q i Q i


2 or x /= 2}. From y = 2x/(4 — x2)2 we have
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Q i Q i 1
yj = 2xQi Qi Q i = 2xy2.
Qi
2

4 — x2
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An interval of definition for the solution of the differential equation is (—
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2, 2). Other inter- vals are (—∞, —2) and (2, ∞).
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√
20. The function is y = 1/ 1 — sin x , whose domain is obtained from 1 — sin x /= 0 or sin x /= 1.
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Thus, the domain is {x x /= π/2 + 2nπ}. From y = —
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2
(1 — sin x) (— cos x) we have Qi Qi Q i Qi Qi Qi Qi Qi Qi Qi Qi Q i Qi Qi Qi Qi Qi Qi Qi




2yj = (1 — sin x)−3/2 cos x = [(1 — sin x)−1/2]3 cos x = y3 cos x.
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An interval of definition for the solution of the differential equation is (π/2, 5π/2). Anoth
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