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

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 s s s




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


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




TABLE OF CONTENTS S S




End of Section Solutions ...................................................................................................................................... 1
s s s



Exercises 1.1 ......................................................................................................................................................... 1
s



Exercises 1.2 .......................................................................................................................................................14
s



Exercises 1.3 .......................................................................................................................................................22
s



Chapter 1 in Review Solutions........................................................................................................................ 30
s s s s




END OF SECTION SOLUTIONS
S S S




EXERCISES 1.1 S




1. Second order; linear s s


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



3. Fourth order; linear s s



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


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

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



7. Third order; linear s s


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



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



10. First order; linear s s


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


in y because of y . However, writing it in the form (y — 1)(dx/dy) + x = 0, we see that it is
2 2
s s s s s s s s s s s s s s s s s s s s s s s


linear in x.
s s s


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


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


nonlinear in u.
s s s



From sy s= se− swe sobtain syj s= s—s1se− . sThen s2yj s+ sy s= s—e− s+ se− s= s0.
x/2 x/2 x/2 x/2
13. 2

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



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

5 5
s s
dy −20t 6 6 s

— e−20t
5 5

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


so that yjj — 6yj + 13y = 0.
s s s s s s s s s

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

jj
y s = tan x + cos x ln(sec x + tan x). Then y
s s s s s s s s s s s s s + y = tan x.
s s s s



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


we have s



j −
—x)y = (y — x)[1 + (2(x + 2) s s s s s s s s ]
−1/2
= y — x + 2(y —
s s s s s s




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




= y — x + 8(x + 2)1/2
s s s s s s s
−1/2 s = sy s — sx s+ s8.


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


not defined at x = —2.
s s s s s s



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


{x s s 5x /= π/2 + nπ} s s s s



or {x s
s
x /= π/10 + nπ/5}. From y j = 25 sec 2 5x we have
s s s s s s s s s s s s




2 2 2
y .

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


other interval is (π/10, 3π/10), and so on.
s s s s s s s s



19. The domain of the function is {x 4 — x /= 0} or {x
s s s s s s s s s s s x /= —2 or x /= 2}. From y
s s s s s s s s


= 2x/(4 — x2)2 we have
s s s s s s


s s 1
yj = 2x s s 2 = 2xy2. s
s


4 — x2
s s



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


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

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


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




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




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


one is (5π/2, 9π/2), and so on.
s s s s s s s

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




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

implicitly we obtain s s 4

2 dX 1 dX
— s =1 s 2
2X — 1 dt s s s s X — 1 dt s s s s


s s s

2 1 dX t
—
s s s s
s =1 s –4 –2 2 4
2X — 1 X — 1 s s s s dt
–2
2X — 2 — 2X + 1 dX s s s s s s s s
=1 s

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




Exponentiating both sides of the implicit solution we obtain s s s s s s s s




2X — 1 s
= et
s s
s
X —1
s
s s




2X — 1 = Xet — et s s s s s s




(et — 1) = (et — 2)X
s s s s s s




et 1
X= .
et — 2
s s
s
s s




Solving e — 2 = 0 we get t = ln 2. Thus, the solution is defined on (—∞, ln 2) or on (ln 2, ∞).
t
s s s s s s s s s s s s s s s s s s s s s s s s s


The graph of the solution defined on (—∞, ln 2) is dashed, and the graph of the solution
s s s s s s s s s s s s s s s s s s


defined on (ln 2, ∞) is solid.
s s s s s s s




22. Implicitly differentiating the solution, we obtain
s s s s s y

2 s
dy dy 4
—2x s — 4xy + 2y =0 s s s s s
dx dx s s

2
—x2 dy — 2xy dx + y dy = 0 s s s s s s s s s



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

–2
Using the quadratic formula to solve y — 2x y — 1 = 0
2 2
√
s
√
s s s s s s s s s s s


for y, we get y =
s s 4x4 + 4 /2 = x2
s x4 + 1 . s s 2x2 s
s s s s
s
s s
–4
2
√
Thus, two explicit solutions are y1 = x + x4 + 1 and
s s s s s s s s s
s
s s

√
y2 = x2 — x4 + 1 . Both solutions are defined on (—∞, ∞).
s s s s
s
s s s s s s s s s



The graph of y1(x) is solid and the graph of y2 is dashed.
s s s s s s s s s s s s