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Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill - All Chapters ( 1-9) Latest A++$12.99
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Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill - All Chapters ( 1-9) Latest A++
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DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS
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DIFFERENTIAL EQUATIONS WITH MODELING APPLICATIONS
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A First Course in Differential Equations with Modeling Applications
Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill - All Chapters ( 1-9) Latest A++
Solution Manual For A First Course in Differential Equations with Modeling Applications, 12th Edition Dennis G. Zill Completed ISBN:9780357760192 Newest Edition 2024 Instant Pdf Download
Solutions for A First Course in Differential Equations with Modeling Applications, 12th Edition Zill, All Chapters 1-9|Complete Guide A+
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
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 ′ = − 12 e−x/2 . Then 2y ′ + y = −e−x/2 + e−x/2 = 0.
1
,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.
2
,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
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.
3
, 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
4
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