,Solution manual for A First Course in Differential
Equations with Modeling Applications, 12th
Edition Dennis G. Zill
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,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 ′ = − 21 e−x/2 . Then 2y ′ + y = −e−x/2 + e−x/2 = 0.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 1
website, in whole or in part.
,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.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 2
website, in whole or in part.
,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
dy dy 4
−2x2 − 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 y 2 − 2x2 y − 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.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 3
website, in whole or in part.
,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
23. Differentiating P = c1 et / 1 + c1 et we obtain
dP 1 + c1 et c1 et − c1 et · c1 et c1 et 1 + c1 et − c1 et
= =
dt (1 + c1 et )2 1 + c1 et 1 + c1 et
c1 et c1 et
= 1− = P (1 − P ).
1 + c1 et 1 + c1 et
2 dy 2
24. Differentiating y = 2x2 − 1 + c1 e−2x we obtain = 4x − 4xc1 e−2x , so that
dx
dy 2 2
+ 4xy = 4x − 4xc1 e−2x + 8x3 − 4x + 4c1 xe−x = 8x3
dx
dy d2 y
25. From y = c1 e2x + c2 xe2x we obtain = (2c1 + c2 )e2x + 2c2 xe2x and = (4c1 + 4c2 )e2x +
dx dx2
4c2 xe2x , so that
d2 y dy
−4 + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
dx2 dx
26. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
dy
= −c1 x−2 + c2 + c3 + c3 ln x + 8x,
dx
d2 y
= 2c1 x−3 + c3 x−1 + 8,
dx2
and
d3 y
= −6c1 x−4 − c3 x−2 ,
dx3
so that
d3 y 2
2 d y dy
x3 + 2x −x + y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
dx3 dx2 dx
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2 = 12x2
In Problems 25–28, we use the Product Rule and the derivative of an integral ((12) of this section):
x
d
ˆ
g(t) dt = g(x).
dx a
x x
e−3t dy e−3t e−3t 3x
ˆ ˆ
27. Differentiating y = e3x dt we obtain = e3x dt + · e or
1 t dx 1 t x
x
dy e−3t 1
ˆ
= e3x dt + , so that
dx 1 t x
ˆ x −3t ˆ x −3t
dy 3x e 1 3x e
x − 3xy = x e dt + − 3x e dt
dx 1 t x 1 t
ˆ x −3t ˆ x −3t
3x e 3x e
= xe dt + 1 − 3xe dt = 1
1 t 1 t
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 4
website, in whole or in part.
,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
√ x x
cos t dy 1 cos t cos x √
ˆ ˆ
28. Differentiating y = x √ dt we obtain = √ √ dt + √ · x or
4 t dx 2 x 4 t x
x
dy 1 cos t
ˆ
= √ √ dt + cos x, so that
dx 2 x 4 t
ˆ x ˆ x
dy 1 cos t √ cos t
2x − y = 2x √ √ dt + cos x − x √ dt
dx 2 x 4 t 4 t
ˆ x ˆ x
√ cos t √ cos t
= x √ dt + 2x cos x − x √ dt = 2x cos x
4 t 4 t
5 10 x sin t dy 5 10 x sin t sin x 10
ˆ ˆ
29. Differentiating y = + dt we obtain =− 2 − 2 dt + · or
x x 1 t dx x x 1 t x x
dy 5 10 x sin t 10 sin x
ˆ
=− 2 − 2 dt + , so that
dx x x 1 t x2
2 dy 5 10 x sin t 10 sin x 5 10 x sin t
ˆ ˆ
2
x + xy = x − 2 − 2 dt + +x + dt
dx x x 1 t x2 x x 1 t
ˆ x ˆ x
sin t sin t
= −5 − 10 dt + 10 sin x + 5 + 10 dt = 10 sin x
1 t 1 t
x x
dy
ˆ ˆ
2 2 2 2 2 2 2
30. Differentiating y = e−x + e−x et dt we obtain = −2xe−x − 2xe−x et dt + ex ·
0 dx 0
2
e−x
ˆ x
dy −x2 −x2 2
or = −2xe − 2xe et dt + 1, so that
dx 0
ˆ x ˆ x
dy 2 2 2 2 2 2
+ 2xy = −2xe−x − 2xe−x et dt + 1 + 2x e−x + e−x et dt
dx 0 0
ˆ x ˆ x
2 2 2 2 2 2
= −2xe−x − 2xe−x et dt + 1 + 2xe−x + 2xe−x et dt = 1
0 0
31. From
−x2 , x < 0
y=
x2 , x≥0
we obtain
−2x, x < 0
′
y =
2x, x≥0
so that xy ′ − 2y = 0.
32. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus,
x→0− x→0+
y ′ (x) does not exist at x = 0.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 5
website, in whole or in part.
,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
33. Force the function y = emx into the equation y ′ + 2y = 0 to get
(emx )′ + 2(emx ) = 0
memx + 2emx = 0
emx (m + 2) = 0
Now since emx > 0 for all values of x, we must have m = −2 and so y = e−2x is a solution.
34. Force the function y = emx into the equation 3y ′ − 4y = 0 to get
3(emx )′ − 4(emx ) = 0
3memx − 4emx = 0
emx (3m − 4) = 0
Now since emx > 0 for all values of x, we must have m = 4/3 and so y = e4x/3 is a solution.
35. Force the function y = emx into the equation y ′′ − 5y ′ + 6y = 0 to get
(emx )′′ − 5(emx )′ + 6(emx ) = 0
m2 emx − 5memx + 6emx = 0
emx (m2 − 5m + 6) = 0
emx (m − 2)(m − 3) = 0
Now since emx > 0 for all values of x, we must have m = 2 and m = 3 therefore y = e2x and
y = e3x are solutions.
36. Force the function y = emx into the equation 2y ′′ + 9y ′ − 5y = 0 to get
2(emx )′′ + 9(emx )′ − 5(emx ) = 0
2m2 emx + 9memx − 5emx = 0
emx (2m2 + 9m − 5) = 0
emx (m + 5)(2m − 1) = 0
Now since emx > 0 for all values of x , we must have m = −5 and m = 1/2 therefore
y = e−5x and y = ex/2 are solutions.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 6
website, in whole or in part.
,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
37. Force the function y = xm into the equation xy ′′ + 2y ′ = 0 to get
x · (xm )′′ + 2(xm )′ = 0
x · m(m − 1)xm−2 + 2mxm−1 = 0
(m2 − m)xm−1 + 2mxm−1 = 0
xm−1 [m2 + m] = 0
xm−1 [m(m + 1)] = 0
The last line implies that m = 0 and m = −1 therefore y = x0 = 1 and y = x−1 are
solutions.
38. Force the function y = xm into the equation 4x2 y ′′ + y = 0 to get
4x2 (xm )′′ + (xm ) = 0
4x2 · m(m − 1)xm−2 + xm = 0
4(m2 − m)xm + xm = 0
xm [4m2 − 4m + 1] = 0
xm [(2m − 1)2 ] = 0
√
The last line implies that m = 1/2 therefore y = x1/2 = x is a solutions.
39. Force the function y = xm into the equation x2 y ′′ − 7xy ′ + 15y = 0 to get
x2 · (xm )′′ − 7x · (xm )′ + 15(xm ) = 0
x2 · m(m − 1)xm−2 − 7x · mxm−1 + 15xm = 0
(m2 − m)xm − 7mxm + 15xm = 0
xm [m2 − 8m + 15] = 0
xm [(m − 3)(m − 5)] = 0
The last line implies that m = 3 and m = 5 therefore y = x3 and y = x5 are solutions.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 7
website, in whole or in part.
, Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
40. Force the function y = xm into the equation x2 y ′′′ − 3xy ′′ + 3y ′ = 0 to get
x2 · (xm )′′′ − 3x · (xm )′′ + 3(xm )′ = 0
x2 · m(m − 1)(m − 2)xm−3 − 3x · m(m − 1)xm−2 + 3 · mxm−1 = 0
(m3 − 3m2 + 2m)xm−1 − 3(m2 − m)xm−1 + 3mxm−1 = 0
xm−1 [m3 − 6m2 + 8m] = 0
xm [m(m − 2)(m − 4)] = 0
The last line implies that m = 0, m = 2, and m = 4 therefore y = x0 = 1, y = x2 , and y = x4
are solutions.
In Problems 41–44, we substitute y = c into the differential equations and use y ′ = 0 and y ′′ = 0
41. Solving 5c = 10 we see that y = 2 is a constant solution.
42. Solving c2 + 2c − 3 = (c + 3)(c − 1) = 0 we see that y = −3 and y = 1 are constant solutions.
43. Since 1/(c − 1) = 0 has no solutions, the differential equation has no constant solutions.
44. Solving 6c = 10 we see that y = 5/3 is a constant solution.
45. Substituting y = (x + c1 )2 into differential equation yields,
2
dy
= 4y
dx
[2 (x + c1 )]2 = 4 (x + c1 )2
4 (x + c1 )2 = 4 (x + c1 )2
Both sides of the differential equation are zero when y = 0. No value of c1 in the family of
solutions gives y = 0 and thus the trivial solution y = 0 is a singular solution.
46. Substituting y = 3 sin (x + c1 ) into differential equation yields,
2
dy
= 9 − y2
dx
[3 cos (x + c1 )]2 = 9 − (3 sin (x + c1 ))2
9 cos2 (x + c1 ) = 9 − 9 sin2 (x + c1 )
9 cos2 (x + c1 ) = 9 − 9 1 − cos2 (x + c1 )
9 cos2 (x + c1 ) = 9 − 9 + 9 cos2 (x + c1 )
9 cos2 (x + c1 ) = 9 cos2 (x + c1 )
Both sides of the differential equation are zero when y = 3. No value of c1 in the family of
solutions gives y = 3 and thus the solution y = 3 is a singular solution.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 8
website, in whole or in part.
Equations with Modeling Applications, 12th
Edition Dennis G. Zill
Notes
1- All Chapters are step by step.
2- We have shown you 10 pages.
3- The file contains all Appendix and Excel
sheet if it exists.
4- We have all what you need, we make
update at every time. There are many
new editions waiting you.
5- If you think you purchased the wrong file
You can contact us at every time, we can
replace it with true one.
Our email:
Our website:
testbanks-store.com
,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 ′ = − 21 e−x/2 . Then 2y ′ + y = −e−x/2 + e−x/2 = 0.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 1
website, in whole or in part.
,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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website, in whole or in part.
,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
dy dy 4
−2x2 − 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 y 2 − 2x2 y − 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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website, in whole or in part.
,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
23. Differentiating P = c1 et / 1 + c1 et we obtain
dP 1 + c1 et c1 et − c1 et · c1 et c1 et 1 + c1 et − c1 et
= =
dt (1 + c1 et )2 1 + c1 et 1 + c1 et
c1 et c1 et
= 1− = P (1 − P ).
1 + c1 et 1 + c1 et
2 dy 2
24. Differentiating y = 2x2 − 1 + c1 e−2x we obtain = 4x − 4xc1 e−2x , so that
dx
dy 2 2
+ 4xy = 4x − 4xc1 e−2x + 8x3 − 4x + 4c1 xe−x = 8x3
dx
dy d2 y
25. From y = c1 e2x + c2 xe2x we obtain = (2c1 + c2 )e2x + 2c2 xe2x and = (4c1 + 4c2 )e2x +
dx dx2
4c2 xe2x , so that
d2 y dy
−4 + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
dx2 dx
26. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
dy
= −c1 x−2 + c2 + c3 + c3 ln x + 8x,
dx
d2 y
= 2c1 x−3 + c3 x−1 + 8,
dx2
and
d3 y
= −6c1 x−4 − c3 x−2 ,
dx3
so that
d3 y 2
2 d y dy
x3 + 2x −x + y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
dx3 dx2 dx
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2 = 12x2
In Problems 25–28, we use the Product Rule and the derivative of an integral ((12) of this section):
x
d
ˆ
g(t) dt = g(x).
dx a
x x
e−3t dy e−3t e−3t 3x
ˆ ˆ
27. Differentiating y = e3x dt we obtain = e3x dt + · e or
1 t dx 1 t x
x
dy e−3t 1
ˆ
= e3x dt + , so that
dx 1 t x
ˆ x −3t ˆ x −3t
dy 3x e 1 3x e
x − 3xy = x e dt + − 3x e dt
dx 1 t x 1 t
ˆ x −3t ˆ x −3t
3x e 3x e
= xe dt + 1 − 3xe dt = 1
1 t 1 t
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website, in whole or in part.
,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
√ x x
cos t dy 1 cos t cos x √
ˆ ˆ
28. Differentiating y = x √ dt we obtain = √ √ dt + √ · x or
4 t dx 2 x 4 t x
x
dy 1 cos t
ˆ
= √ √ dt + cos x, so that
dx 2 x 4 t
ˆ x ˆ x
dy 1 cos t √ cos t
2x − y = 2x √ √ dt + cos x − x √ dt
dx 2 x 4 t 4 t
ˆ x ˆ x
√ cos t √ cos t
= x √ dt + 2x cos x − x √ dt = 2x cos x
4 t 4 t
5 10 x sin t dy 5 10 x sin t sin x 10
ˆ ˆ
29. Differentiating y = + dt we obtain =− 2 − 2 dt + · or
x x 1 t dx x x 1 t x x
dy 5 10 x sin t 10 sin x
ˆ
=− 2 − 2 dt + , so that
dx x x 1 t x2
2 dy 5 10 x sin t 10 sin x 5 10 x sin t
ˆ ˆ
2
x + xy = x − 2 − 2 dt + +x + dt
dx x x 1 t x2 x x 1 t
ˆ x ˆ x
sin t sin t
= −5 − 10 dt + 10 sin x + 5 + 10 dt = 10 sin x
1 t 1 t
x x
dy
ˆ ˆ
2 2 2 2 2 2 2
30. Differentiating y = e−x + e−x et dt we obtain = −2xe−x − 2xe−x et dt + ex ·
0 dx 0
2
e−x
ˆ x
dy −x2 −x2 2
or = −2xe − 2xe et dt + 1, so that
dx 0
ˆ x ˆ x
dy 2 2 2 2 2 2
+ 2xy = −2xe−x − 2xe−x et dt + 1 + 2x e−x + e−x et dt
dx 0 0
ˆ x ˆ x
2 2 2 2 2 2
= −2xe−x − 2xe−x et dt + 1 + 2xe−x + 2xe−x et dt = 1
0 0
31. From
−x2 , x < 0
y=
x2 , x≥0
we obtain
−2x, x < 0
′
y =
2x, x≥0
so that xy ′ − 2y = 0.
32. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus,
x→0− x→0+
y ′ (x) does not exist at x = 0.
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website, in whole or in part.
,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
33. Force the function y = emx into the equation y ′ + 2y = 0 to get
(emx )′ + 2(emx ) = 0
memx + 2emx = 0
emx (m + 2) = 0
Now since emx > 0 for all values of x, we must have m = −2 and so y = e−2x is a solution.
34. Force the function y = emx into the equation 3y ′ − 4y = 0 to get
3(emx )′ − 4(emx ) = 0
3memx − 4emx = 0
emx (3m − 4) = 0
Now since emx > 0 for all values of x, we must have m = 4/3 and so y = e4x/3 is a solution.
35. Force the function y = emx into the equation y ′′ − 5y ′ + 6y = 0 to get
(emx )′′ − 5(emx )′ + 6(emx ) = 0
m2 emx − 5memx + 6emx = 0
emx (m2 − 5m + 6) = 0
emx (m − 2)(m − 3) = 0
Now since emx > 0 for all values of x, we must have m = 2 and m = 3 therefore y = e2x and
y = e3x are solutions.
36. Force the function y = emx into the equation 2y ′′ + 9y ′ − 5y = 0 to get
2(emx )′′ + 9(emx )′ − 5(emx ) = 0
2m2 emx + 9memx − 5emx = 0
emx (2m2 + 9m − 5) = 0
emx (m + 5)(2m − 1) = 0
Now since emx > 0 for all values of x , we must have m = −5 and m = 1/2 therefore
y = e−5x and y = ex/2 are solutions.
© 2023 Cengage. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ot a publicly accessible 6
website, in whole or in part.
,Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
37. Force the function y = xm into the equation xy ′′ + 2y ′ = 0 to get
x · (xm )′′ + 2(xm )′ = 0
x · m(m − 1)xm−2 + 2mxm−1 = 0
(m2 − m)xm−1 + 2mxm−1 = 0
xm−1 [m2 + m] = 0
xm−1 [m(m + 1)] = 0
The last line implies that m = 0 and m = −1 therefore y = x0 = 1 and y = x−1 are
solutions.
38. Force the function y = xm into the equation 4x2 y ′′ + y = 0 to get
4x2 (xm )′′ + (xm ) = 0
4x2 · m(m − 1)xm−2 + xm = 0
4(m2 − m)xm + xm = 0
xm [4m2 − 4m + 1] = 0
xm [(2m − 1)2 ] = 0
√
The last line implies that m = 1/2 therefore y = x1/2 = x is a solutions.
39. Force the function y = xm into the equation x2 y ′′ − 7xy ′ + 15y = 0 to get
x2 · (xm )′′ − 7x · (xm )′ + 15(xm ) = 0
x2 · m(m − 1)xm−2 − 7x · mxm−1 + 15xm = 0
(m2 − m)xm − 7mxm + 15xm = 0
xm [m2 − 8m + 15] = 0
xm [(m − 3)(m − 5)] = 0
The last line implies that m = 3 and m = 5 therefore y = x3 and y = x5 are solutions.
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website, in whole or in part.
, Solution and Answer Guide: Zill, DIFFERENTIAL EQUATIONS With MODELING APPLICATIONS 2024, 9780357760192; Chapter #1:
Introduction to Differential Equations
40. Force the function y = xm into the equation x2 y ′′′ − 3xy ′′ + 3y ′ = 0 to get
x2 · (xm )′′′ − 3x · (xm )′′ + 3(xm )′ = 0
x2 · m(m − 1)(m − 2)xm−3 − 3x · m(m − 1)xm−2 + 3 · mxm−1 = 0
(m3 − 3m2 + 2m)xm−1 − 3(m2 − m)xm−1 + 3mxm−1 = 0
xm−1 [m3 − 6m2 + 8m] = 0
xm [m(m − 2)(m − 4)] = 0
The last line implies that m = 0, m = 2, and m = 4 therefore y = x0 = 1, y = x2 , and y = x4
are solutions.
In Problems 41–44, we substitute y = c into the differential equations and use y ′ = 0 and y ′′ = 0
41. Solving 5c = 10 we see that y = 2 is a constant solution.
42. Solving c2 + 2c − 3 = (c + 3)(c − 1) = 0 we see that y = −3 and y = 1 are constant solutions.
43. Since 1/(c − 1) = 0 has no solutions, the differential equation has no constant solutions.
44. Solving 6c = 10 we see that y = 5/3 is a constant solution.
45. Substituting y = (x + c1 )2 into differential equation yields,
2
dy
= 4y
dx
[2 (x + c1 )]2 = 4 (x + c1 )2
4 (x + c1 )2 = 4 (x + c1 )2
Both sides of the differential equation are zero when y = 0. No value of c1 in the family of
solutions gives y = 0 and thus the trivial solution y = 0 is a singular solution.
46. Substituting y = 3 sin (x + c1 ) into differential equation yields,
2
dy
= 9 − y2
dx
[3 cos (x + c1 )]2 = 9 − (3 sin (x + c1 ))2
9 cos2 (x + c1 ) = 9 − 9 sin2 (x + c1 )
9 cos2 (x + c1 ) = 9 − 9 1 − cos2 (x + c1 )
9 cos2 (x + c1 ) = 9 − 9 + 9 cos2 (x + c1 )
9 cos2 (x + c1 ) = 9 cos2 (x + c1 )
Both sides of the differential equation are zero when y = 3. No value of c1 in the family of
solutions gives y = 3 and thus the solution y = 3 is a singular solution.
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