First Course in Differential Equations 12th Edition Zill – Complete Solutions Guide for Chapters 1-9

A First Course in Differential E
. . . . .




quations with Modeling Appli
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cations,12thEditionby Denni
. . . .




s G. Zill . .




CompleteChapterSolutionsManual a
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re included (Ch 1 to 9)
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**Immediate Download
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** Swift Response
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** All Chapters included
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,Solution.and.Answer.Guide:. Zill,.DIFFERENTIAL.EQUATIONS.With.MODELING.APPLICATIONS.2024,.9780357760192;.Chapter.#1:
Introduction.to.Differential.Equations




SolutionandAnswerGuide . . .


ZILL,.DIFFERENTIAL.EQUATIONS.WITH.MODELING.APPLICATIONS.2024,
9780357760192;.CHAPTER.#1:.INTRODUCTION.TO.DIFFERENTIAL.EQUATIONS


TABLEOFCONTENTS . .




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
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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)
√
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.x˙.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.b
ecause.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.−.ue )(du/dv).+.u.=. 0,.we.see.that.it.is.nonlinear.in.
u

u.
13. From.y.=.e−x/2. we.obtain.y′. =.−1. e−2 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


6. 6 —20t
14. From.y. =. −. e we. obtain. dy/dt. =. 24e −20t ,.so.that
5 5
dy.
+.20y.=.24e−20t. +.20 6. 6.
−. 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.th
at.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

=.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.d
efined.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./=.π/2.+.nπ}
or.{x. x./=. π/10.+.nπ/5}..From.y.′=.25.sec.25x.we.have
′
2 2 2
y. =.25(1.+.tan . 5x).=.25.+.25.tan. 5x.=.25.+.y. .



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
′ 1
.from.1.−.sin.x. /=. 0. or.sin.x. /=. 1.
−3/2
Thus,.the.domain.is.{x. x./=. π/2.+.2nπ}..From.y. =.−. (21.−.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
2X.−.2.−.2X.+.1. dX –2
. . =.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. t.X
=.e
.−.1



2X. −.1. =.Xet.−.et

(et.−.1).=.(et.−.2)X
et.—.1
X. =. t. .
e −.2.

Solving.et.−.2.=. 0.we.get.t. =.ln.2..Thus,.the.solution.is.defined.on.(−∞,.ln.2).or.on.(ln.2,.∞)..The.gra
ph.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

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.f or m ula .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. =. x . + 2
x4. +.1. and
√
y2. =. x2. −. x4. +.1...Both.solutions.are.defined.on.(−∞,.∞)..The.gr
aph.of.y1(x).is.solid.and.the.graph.of.y2. is.dashed.




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