Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition Dennis G. Zill/ All chapters 1-9 fully covered

A First Course in Differential
b1 b1 b1 b1 b1




Equations with Modeling Ap b1 b1 b1




plications, 12th Edition by De b1 b1 b1 b1




nnis G. Zill b1 b1




Complete Chapter Solutions Manual ar
b1 b1 b1 b1




e included (Ch 1 to 9)
b1 b1 b1 b1 b1




** Immediate Download
b1 b1




** Swift Response
b1 b1




** All Chapters included
b1 b1 b1

,Solutionb1andb1Answerb1Guide:b1Zill,b1DIFFERENTIALb1EQUATIONSb1Withb1MODELINGb1APPLICATIONSb12024,b19780357760192; b1Chapter
b1#1:




Solution and Answer Guide b1 b1 b1




ZILL, DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS 2024,
b1 b1 b1 b1 b1 b1 b1


9780357760192; CHAPTER #1: INTRODUCTION TO DIFFERENTIAL EQUATIONS b1 b1 b1 b1 b1 b1




TABLE OF CONTENTS B1 B1




End of Section Solutions .................................................................................................................................... 1
b1 b1 b1



Exercises 1.1 ........................................................................................................................................................ 1
b1



Exercises 1.2 ......................................................................................................................................................14
b1



Exercises 1.3 ......................................................................................................................................................22
b1



Chapter 1 in Review Solutions ..................................................................................................................... 30
b1 b1 b1 b1




END OF SECTION SOLUTIONS
B1 B1 B1




EXERCISES 1.1 B 1




1. Second order; linear b 1 b 1


4
2. Third order; nonlinear because of (dy/dx)
b1 b1 b1 b1 b1



3. Fourth order; linear b1 b1



4. Second order; nonlinear because of cos(r + u)
b1 b1 b1 b1 b1 b1 b1


5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2
b1 b1 b1 b1 b1 b1 b1 b1

2
6. Second order; nonlinear because of R
b1 b1 b1 b1 b1



7. Third order; linear b1 b1


2
8. Second order; nonlinear because of ẋ
b1 b1 b1 b1 b1



9. First order; nonlinear because of sin (dy/dx)
b1 b1 b1 b1 b1 b1



10. First order; linear b1 b1


2
11. Writing the differential equation in the form x(dy/dx) + y = 1, we see that it is nonl
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


inear in y because of y . However, writing it in the form (y —
2 2
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


1)(dx/dy) + x = 0, we see that it is linear in x.
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


u
12. Writing the differential equation in the form u(dv/du) + (1 + u)v = ue we see that
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


it is linear in v. However, writing it in the form (v + uv —
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


ueu)(du/dv) + u = 0, we see that it is nonlinear in u.
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1



Fromb1yb1=b1e− b1web1obtainb1yjb1=b1—b11b1e− .b1Thenb12yjb1+b1yb1=b1—e− b1+b1e− b1=b10.
x/2 x/2 x/2 x/2
13. 2

,Solutionb1andb1Answerb1Guide:b1Zill,b1DIFFERENTIALb1EQUATIONSb1Withb1MODELINGb1APPLICATIONSb12024,b19780357760192; b1Chapter
b1#1:




6 6 —
14. From y = b1 b1 — e we obtain dy/dt = 24e , so that
b1 b1 b1 b1 b1 b1

5 5
b1b1
dy −20t 6 6 b1

— −20t
5 b1

e
3x
15. From y = e b1 b1cos 2x we obtain yj = 3e3x cos 2x—2e3x sin 2x and yjj = 5e3x cos 2x—
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


12e 3x
sin 2x, so that yjj — 6yj + 13y = 0.
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1

j
16. From y = — b1 b1 b1 = —1 + sin x ln(sec x + tan x) and
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1

cos x ln(sec x + tan x) we obtain y
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1

jj
y b 1 = tan x + cos x ln(sec x + tan x). Then y
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b 1 + y = tan x.
b1 b1 b1 b1



17. The domain of the function, found by solving x+2 ≥ 0, is [—2, ∞). From yj = 1+2(x+2)−
1/2
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


we have b1



j −
—x)y = (y — x)[1 + (2(x + 2) ]
b 1 b1 b1 b1 b1 b1 b1 b1




−1/2
= y — x + 2(y —
b1 b1 b1 b1 b1 b1




−1/2
= y — x + 2[x + 4(x + 2)1/2 —
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1




= y — x + 8(x + 2)1/2
b1 b1 b1 b1 b1 b1 b1
−1/2b 1 =b1yb 1 — b1xb1+b18.


An interval of definition for the solution of the differential equation is (—
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


2, ∞) because yj is not defined at x = —2.
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1



18. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
b1 b1 b1 b1 b1 b1 b1 b1 b 1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b 1 b 1 b1 b1 b1


{x b 1 b 1 5x /= π/2 + nπ}
b1 b1 b1 b1



or {x b1
b 1
x /= π/10 + nπ/5}. Fromj y = 252sec
b1 b1 b1 b1 b1 b1 b 1 b1 b1 b 1 5x we have b1 b1




2 2 2
y .

An interval of definition for the solution of the differential equation is (—
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


π/10, π/10). An- other interval is (π/10, 3π/10), and so on.
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1



19. The domain of the function is {x
b1 4 — x b1 b1 b1 b1 b1 b1 b1 /= 0} or {x
b 1 b1 b1 x /= —
b 1 b 1


2 or x /= 2}. From y = 2x/(4 — x2)2 we have
b1 b1 b 1 b 1 b1 b1 b 1 b1 b1 b1 b1 b1


b 1 b 1 1
yj = 2x b1 b1 b 1 = 2xy2.
b1
2

4 — x2b1 b1



An interval of definition for the solution of the differential equation is (—
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


2, 2). Other inter- vals are (—∞, —2) and (2, ∞).
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1

√
20. The function is y = 1/ 1 — sin x , whose domain is obtained from 1 — sin x /= 0 or sin x /= 1.
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1


Thus, the domain is {x x /= π/2 + 2nπ}. From y =2— (1 — sin x) (— cos x) we have
b1 b1 b1 b1 b 1 b1 b1 b1 b1 b1 b1 b1 b1 b 1 b1 b1 b1 b1 b1 b1 b1




2yj = (1 — sin x)−3/2 cos x = [(1 — sin x)−1/2]3 cos x = y3 cos x.
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1




An interval of definition for the solution of the differential equation is (π/2, 5π/2). Anot
b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1 b1

, Solutionb1andb1Answerb1Guide:b1Zill,b1DIFFERENTIALb1EQUATIONSb1Withb1MODELINGb1APPLICATIONSb12024,b19780357760192; b1Chapter
b1#1:
herb1oneb1isb1(5π/2,b19π/2),b1andb1sob1on.