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

Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:

,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:



Solution and Answer Guide r1 r1 r1




ZILL, DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS 2024,
r1 r1 r1 r1 r1 r1


9780357760192; CHAPTER #1: INTRODUCTION TO DIFFERENTIAL
r1 r1 r1 r1 r1 r1


EQUATIONS
r1




TABLE OF CONTENTS R1 R1




End of Section Solutions ................................................................................................................................... 1
r1 r1 r1



Exercises 1.1 ........................................................................................................................................................ 1
r1



Exercises 1.2 ......................................................................................................................................................14
r1



Exercises 1.3 ......................................................................................................................................................22
r1



Chapter 1 in Review Solutions...................................................................................................................... 30
r1 r1 r1 r1




END OF SECTION SOLUTIONS
R1 R1 R1




EXERCISES 1.1 R 1




1. Second r 1 order; r 1 linear
4
2. Third order; nonlinear because of (dy/dx)
r1 r1 r1 r1 r1



3. Fourth order; linear r1 r1



4. Second order; nonlinear because of cos(r + u)
r1 r1 r1 r1 r1 r1 r1


5. Second order; nonlinear because of (dy/dx) or 1 + (dy/dx)2 2
r1 r1 r1 r1 r1 r1 r1 r1

2
6. Second order; nonlinear because of Rr1 r1 r1 r1 r1



7. Third order; linear r1 r1


2
8. Second order; nonlinear because of ẋ
r1 r1 r1 r1 r1



9. First order; nonlinear because of sin (dy/dx)
r1 r1 r1 r1 r1 r1



10. First order; linear r1 r1


2
11. Writing the differential equation in the form x(dy/dx) + y = 1, we see that it is
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


nonlinear in y because of y . However, writing it in the form (y — 1)(dx/dy) + x = 0,
2 2
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


we see that it is linear in x.
r1 r1 r1 r1 r1 r1 r1 r1


u
12. Writing the differential equation in the form u(dv/du) + (1 + u)v = ue we see that
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


it is linear in v. However, writing it in the form (v + uv — ue )(du/dv) + u = 0, we
u
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


see that it is nonlinear in u.
r1 r1 r1 r1 r1 r1 r1


1 −
we obtain yj = — . Then 2yj + y = —e−
x/2 x/2 x/2
13. From y = e− r1 r1 r1 r1
2 e r1 r1 r1 r1 r1
r1
r1 r1 r1 r1 r1 r1 + e−x/2 = 0.
r1 r1 r1 r1

,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:



6 6 —
14. From y = — e r1 r1 we obtain dy/dt = 24e , so that
r1 r1 r1 r1 r1 r1

5 5
r1 r1
dy −20t 6 6 r1

— e−20t
5 5

3x 3x
15. From y = e cos 2x we obtain yj = 3e
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 cos 2x—2e3x sin 2x and yjj = 5e3x cos 2x—12e3x
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


sin 2x, so that yjj — 6yj + 13y = 0.
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1

j
16. From y = — cos x ln(sec x + tan x) we
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 = —1 + sin x ln(sec x + tan x) and
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1

obtain y
r1 r1

jj
y r 1 = tan x + cos x ln(sec x + tan x). Then y
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r 1 + y = tan x.
r1 r1 r1 r1



17. The domain of the function, found by solving x+2 ≥ 0, is [—2, ∞). From yj = 1+2(x+2)−
1/2
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


we have r1



j −
—x)y r 1 = (y — x)[1 + (2(x + 2) ]
r1 r1 r1 r1 r1 r1 r1




−1/2
= y — x + 2(y —
r1 r1 r1 r1 r1 r1




−1/2
= y — x + 2[x + 4(x + 2)1/2 —
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1




= y — x + 8(x + 2)1/2
r1 r1 r1 r1 r1 r1 r1
−1/2 r 1 = r1y r 1 — r1x r1+ r18.


An interval of definition for the solution of the differential equation is (—2, ∞) because
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


yj is not defined at x = —2.
r1 r1 r1 r1 r1 r1 r1 r1



18. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y
r1 r1 r1 r1 r1 r1 r1 r1 r 1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r 1 = 5 tan 5x is
r 1 r1 r1 r1


{x r 1 r 1 5x /= π/2 + nπ}
r1 r1 r1 r1



or {x r1
r 1
x /= π/10 + nπ/5}. Fromj y
r1 r1 r1 r1 r1 r1 r 1 = 252 sec
r1 r1 r 1 5x we haver1 r1




2 2 2
y .

An interval of definition for the solution of the differential equation is (—π/10, π/10).
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


An- other interval is (π/10, 3π/10), and so on.
r1 r1 r1 r1 r1 r1 r1 r1 r1



19. The domain of the function is {x
r1 4 — x r1 r1 r1 r1 r1 r1 r1 /= 0} or {x r 1 r1 r1 x /= r 1 r 1 —2 or r1


x /= 2}. From y
r1 r 1 = 2x/(4 — x2)2 we have
r 1 r1 r1 r 1 r1 r1 r1 r1 r1


r 1 r 1 1
yj = 2x r1 r1 = 2xy2. r1
2
r 1


4 — x2r1 r1



An interval of definition for the solution of the differential equation is (—2, 2). Other
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


inter- vals are (—∞, —2) and (2, ∞).
r1 r1 r1 r1 r1 r1 r1 r1

√
20. The function is y = 1/ 1 — sin x , whose domain is obtained from 1 — sin x /= 0 or sin x /= 1.
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1


Thus, the domain is {x x /= π/2 + 2nπ}. From y =2— (1 — sin x) (— cos x) we have
r1 r1 r1 r1 r 1 r1 r1 r1 r1 r1 r1 r1 r1 r 1 r1 r1 r1 r1 r1 r1 r1




2yj = (1 — sin x)−3/2 cos x = [(1 — sin x)−1/2]3 cos x = y3 cos x.
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1




An interval of definition for the solution of the differential equation is (π/2, 5π/2).
r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1 r1

, Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
r1 Another one is (5π/2, 9π/2), and so on.
r1 r1 r1 r1 r1 r1 r1