Solution Manual for A First Course in Differential Equations with Modeling Applications 12th Edition by Dennis Zill , ISBN: 9780357760192 Chapter 1-9 |All Chapters Verified| Guide A+

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A First Course in Differential
Equations with Modeling
Applications, 12th Edition by
Dennis G. Zill
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Complete Chapter Solutions Manual
are included (Ch 1 to 9)
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FD
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** Immediate Download
** Swift Response
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** All Chapters included

,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
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Exercises 1.3............................................................................................................................................... 22
Chapter 1 in Review Solutions ............................................................................................................ 30




END OF SECTION SOLUTIONS
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EXERCISES 1.1
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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)
√
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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
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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) + 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′ = − 12e−x/2. Then 2y′ + y = −e−x/2 + e−x/2 = 0.




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


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14. From y = − e−20t 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
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′ −
(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
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= 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
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not defined at x = −2.
18. Sinc. e tan x is not defined for x = π/2 + nπ , n an integer, the domain of y = 5 tan 5x is
{x . 5x /= π/2 + nπ}
.
.
or {x x /= π/10 + nπ/5}. From y′ = 25 sec2 5x we have
′
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y = 25(1 + tan 2 5x) = 25 + 25 tan 2 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.
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. .
19. The domain of the function is {x . 4 − x2 /= 0} or {x . x /= −2 or x /= 2}. From′ y =
2x/(4 − x2 )2 we have
2
′ 1 2
y = 2x = 2xy .
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}. Fromy = − 2 (1 − 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.



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, 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
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Exponentiating both sides of the implicit solution we obtain

2X − 1
= et
X −1
2X − 1 = Xet − et
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(et − 1) = (et − 2)X
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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.
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22. Implicitly differentiating the solution, we obtain y

dy dy
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4
−2x2 − 4xy + 2y =0
dx dx
−x2 dy − 2xy dx + y dy = 0 2

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.




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