Primedocs
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)
Pr
im
ed
oc
** Immediate Download
s
** Swift Response
** All Chapters included
Primedocs
,Primedocs
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
Pr
END OF SECTION SOLUTIONS
im
EXERCISES 1.1
ed
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
oc
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
s
√
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 y′ = − 12 e−x/2. Then 2y′ + y = −e−x/2 + e−x/2 = 0.
1
Primedocs
,Primedocs
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 6 6
+ 20y = 24e−20t + 20 − e
−20t
= 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
′ −1/2
(y − x)y = (y − x)[1 + (2(x + 2) ]
= y − x + 2(y − x)(x + 2)−1/2
Pr
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
im
= 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
ed
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
oc
{x 5x /= π/2 + nπ}
or {x x /= π/10 + nπ/5}. From y ′= 25 sec 25x we have
s
′
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 − x 2 /= 0} or {x x /= −2 or x /= 2}. From y ′=
2x/(4 − x2)2 we have
2
1
y′ = 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.
′ 1 −3/2
Thus, the domain is {x x /
= π/2 + 2nπ}. From y = − (1
2 − sin x) (− cos x) we have
2y′ = (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.
2
Primedocs
, Primedocs
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 dX1
− =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
Pr
= et
X −1
2X − 1 = Xet − et
im
(et − 1) = (et − 2)X
ed
et — 1
X= .
et − 2
oc
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
s
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
−x 2 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.
3
Primedocs
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)
Pr
im
ed
oc
** Immediate Download
s
** Swift Response
** All Chapters included
Primedocs
,Primedocs
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
Pr
END OF SECTION SOLUTIONS
im
EXERCISES 1.1
ed
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
oc
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
s
√
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 y′ = − 12 e−x/2. Then 2y′ + y = −e−x/2 + e−x/2 = 0.
1
Primedocs
,Primedocs
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 6 6
+ 20y = 24e−20t + 20 − e
−20t
= 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
′ −1/2
(y − x)y = (y − x)[1 + (2(x + 2) ]
= y − x + 2(y − x)(x + 2)−1/2
Pr
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
im
= 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
ed
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
oc
{x 5x /= π/2 + nπ}
or {x x /= π/10 + nπ/5}. From y ′= 25 sec 25x we have
s
′
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 − x 2 /= 0} or {x x /= −2 or x /= 2}. From y ′=
2x/(4 − x2)2 we have
2
1
y′ = 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.
′ 1 −3/2
Thus, the domain is {x x /
= π/2 + 2nπ}. From y = − (1
2 − sin x) (− cos x) we have
2y′ = (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.
2
Primedocs
, Primedocs
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 dX1
− =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
Pr
= et
X −1
2X − 1 = Xet − et
im
(et − 1) = (et − 2)X
ed
et — 1
X= .
et − 2
oc
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
s
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
−x 2 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.
3
Primedocs
