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
,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
