Solution Manual for A First Course in Differential Equations with Modeling Applications, 12th Edition – Dennis Zill | Verified Solutions | 2025/2026

,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)
√
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 ẋ 2
9. First order; nonlinear because of sin (dy/dx)
10. First order; linear
11. Writing the differential equation in the form x(dy/dx) + y2 = 1, we see that it is nonlinear
in y because of y2. However, writing it in the form (y2 — 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 yj = — 12 e−x/2. Then 2yj + 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 —
14. From y = — e 20t we obtain dy/dt = 24e−20t , so that
5 5
dy 6 6 −20t
+ 20y = 24e−20t + 20 — e = 24.
dt 5 5

15. From y = e3x cos 2x we obtain yj = 3e3x cos 2x—2e3x sin 2x and yjj = 5e3x cos 2x—12e3x sin 2x,
so that yjj — 6yj + 13y = 0.
j
16. From y = — cos x ln(sec x + tan x) we obtain y = —1 + sin x ln(sec x + tan x) and
jj jj
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 yj = 1+2(x+2)−1/2
we have
j −1/2
(y —x)y = (y — x)[1 + (2(x + 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 yj 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 /= π/2 + nπ}
or {x x /= π/10 + nπ/5}. From y j= 25 sec 25x we have
j
y = 25(1 + tan 2 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 /= 0} or {x x /= —2 or x /= 2}. From y j =
2x/(4 — x2)2 we have
2
1
yj = 2x = 2xy2.
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 /= 0 or sin x /= 1.
Thus, the domain is {x x /= π/2 + 2nπ}. From y j= — (11
2
— sin x) −3/2 (— cos x) we have

2yj = (1 — sin x)−3/2 cos x = [(1 — sin x)−1/2]3 cos x = y3 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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, Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations




21. Writing ln(2X — 1) — ln(X — 1) = t and differentiating x

implicitly we obtain 4

2 dX 1 dX
— =1 2
2X — 1 dt X — 1 dt
2 1 dX t
— =1 –4 –2 2 4
2X — 1 X — 1 dt
–2
2X — 2 — 2X + 1 dX
=1
(2X — 1) (X — 1) dt
–4
dX
= —(2X — 1)(X — 1) = (X — 1)(1 — 2X).
dt

Exponentiating both sides of the implicit solution we obtain

2X — 1
= et
X —1
2X — 1 = Xet — et

(et — 1) = (et — 2)X
et — 1
X= .
et — 2

Solving et — 2 = 0 we get t = ln 2. Thus, the solution is defined on (—∞, ln 2) or on (ln 2, ∞).
The graph of the solution defined on (—∞, ln 2) is dashed, and the graph of the solution
defined on (ln 2, ∞) is solid.

22. Implicitly differentiating the solution, we obtain y

2 dy dy 4
—2x — 4xy + 2y =0
dx dx
2
—x2 dy — 2xy dx + y dy = 0
x
2xy dx + (x2 — y)dy = 0. –4 –2 2 4

–2
Using the quadratic formula to solve y2 — 2x2y — 1 = 0
√ √
for y, we get y = 2x2 ±
4x4 + 4 /2 = x2 ± x4 + 1 . –4
√
Thus, two explicit solutions are y1 = x2 + x4 + 1 and
√
y2 = x2 — x4 + 1 . Both solutions are defined on (—∞, ∞).
The graph of y1(x) is solid and the graph of y2 is dashed.




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