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

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

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




Complete Chapter Solutions Manual
are included (Ch 1 to 9)




** Immediate Download
** Swift Response
** All Chapters included

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




Solution and Answer Guide
ZILL, DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS 2024,
9780357760192; CHAPTER #1: INTRODUCTION TO DIFFERENTIAL EQUATIONS


TABLE OF CONTENTS
End of Section Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Exercises 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Exercises 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
Exercises 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
Chapter 1 in Review Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30



END OF SECTION SOLUTIONS
EXERCISES 1.1
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
p
5. Second order; nonlinear because of (dy/dx)2 or 1 + (dy/dx)2
6. Second order; nonlinear because of R2
7. Third order; linear
8. Second order; nonlinear because of ẋ2
9. First order; nonlinear because of sin (dy/dx)
10. First order; linear
11. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear
in y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is
linear in x.
12. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is
linear in v . However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is
nonlinear in u.
13. From y = e−x/2 we obtain y ′ = − 12 e−x/2 . Then 2y ′ + y = −e−x/2 + e−x/2 = 0.




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


6 6 −20t
14. From y = − e we obtain dy/dt = 24e−20t , so that
5 5
 
dy −20t 6 6 −20t
+ 20y = 24e + 20 − e = 24.
dt 5 5

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

(y − x)y ′ = (y − x)[1 + (2(x + 2)−1/2 ]

= y − x + 2(y − x)(x + 2)−1/2

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

= y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8.

An interval of definition for the solution of the differential equation is (−2, ∞) because y ′ is
not defined at x = −2.
18. Since tan x is not defined for x = π/2 + nπ , n an integer, the domain of y = 5 tan 5x is
{x 5x 6= π/2 + nπ}
or {x x 6= π/10 + nπ/5}. From y ′ = 25 sec2 5x we have

y ′ = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 .

An interval of definition for the solution of the differential equation is (−π/10, π/10). An-
other interval is (π/10, 3π/10), and so on.
19. The domain of the function is {x 4 − x2 6= 0} or {x x 6= −2 or x 6= 2}. From y ′ =
2x/(4 − x2 )2 we have
 2
1
′
y = 2x = 2xy 2 .
4 − x2
An interval of definition for the solution of the differential equation is (−2, 2). Other inter-
vals are (−∞, −2) and (2, ∞).
√
20. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x 6= 0 or sin x 6= 1.
Thus, the domain is {x x =6 π/2 + 2nπ}. From y ′ = − 12 (1 − sin x)−3/2 (− cos x) we have

2y ′ = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x.

An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another
one is (5π/2, 9π/2), and so on.


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