Solution Manual for A First Course in Differential Equations with Modeling Applications, 12th Edition by Dennis G. Zill

A First Course in Differential Equati
A A A A A




ons with Modeling Applications, 12t
A A A A




h Edition by Dennis G. Zill
A A A A A




Complete Chapter Solutions Manual are incl
A A A A A



uded (Ch 1 to 9) A A A A




** Immediate Download
A A




** Swift Response
A A




** All Chapters included
A A A

,SolutionAandAAnswerAGuide:AZill,ADIFFERENTIALAEQUATIONS AWithAMODELINGAAPPLICATIONS A2024,A9780357760192;AChapterA
#1:
IntroductionA toA DifferentialA Equations



Solution and Answer Guide A A A



ZILL,ADIFFERENTIALAEQUATIONSAWITHAMODELINGAAPPLICATIONSA2024,A 9780357760192;ACHAPTERA#1
:AINTRODUCTIONATOADIFFERENTIALAEQUATIONS


TABLE OF CONTENTSA A




EndA ofA SectionA Solutions ........................................................................................................................................................................ 1
ExercisesA 1.1................................................................................................................................................................................................... 1
ExercisesA 1.2................................................................................................................................................................................................. 14
ExercisesA 1.3................................................................................................................................................................................................. 22
ChapterA1AinAReviewASolutions................................................................................................................................................. 30




END OF SECTION SOLUTIONS
A A A




EXERCISES 1.1 A



1. SecondA order;A linear
2. ThirdA order;A nonlinear A because A ofA (dy/dx)4
3. FourthA order;A linear
4. SecondA order;A nonlinear A becauseA ofA cos(rA +Au)
√A
5. SecondA order;A nonlinear A becauseA ofA (dy/dx) or 2AA
1A +A (dy/dx)2
6. SecondA order;A nonlinearA becauseA ofA R2
7. ThirdA order;A linear
8. SecondA order;A nonlinearAbecauseA ofA ẋA2
9. FirstA order;A nonlinear A because A ofAsinA(dy/dx)
10. FirstA order;A linear
11. WritingAtheA differentialA equationAinAtheA formA x(dy/dx)A +A y2A =A 1,A weAseeAthatAitAisAnonlinearA inAyAbecauseAo
fAy2.AHowever,AwritingAitAinAtheAformA (y2A —A1)(dx/dy)A+AxA=A 0,AweAseeAthatAitAisA linearA inA x.
12. WritingAtheAdifferentialAequationAinAtheAformAu(dv/du)A+A(1A+Au)vA =A ueuA weAseeAthatAitAisA linearAinAv.AH
owever,AwritingAitAinAtheAformA(vA+AuvA—Aueu)(du/dv)A+AuA=A 0,AweAseeAthatAitAisA nonlinearA inA u.
13. FromAyA=Ae− x/2
weAobtainAyjA =A—A1Ae− x/2
.AThenA2yjA +AyA =A—e− x/2
+Ae− x/2 =A0.
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,SolutionAandAAnswerAGuide:AZill,ADIFFERENTIALAEQUATIONS AWithAMODELINGAAPPLICATIONS A2024,A9780357760192;AChapterA
#1:
IntroductionA toA DifferentialA Equations

6 6
14. FromA yA = — e—20tAweAobtainAdy/dtA=A24e−20tA,AsoAthat
5 5
dyA+A20yA =A24e−20t 6 6A −20t
+A 20 —AA e =A 24.
dt 5 5

15. FromAyA=Ae3xAcosA2xAweAobtainAyjA =A3e3xAcosA2x—2e3xAsinA2xAandAyjjA =A5e3xAcosA2x—
12e3xAsinA2x,A soA thatA yjjA —A6yjA +A13yA =A 0.
j
16. FromAyA =A —AcosAxAln(secAxA+AtanAx)AweAobtainAyAA =A—1A+AsinAxAln(secAxA+AtanAx)Aand
jj jj
yAA =AtanAxA+AcosAxAln(secAxA+AtanAx).AThenAyAA +AyA=AtanAx.
17. TheA domainA ofA theA function, A foundA byA solvingA x+2 A ≥A 0,A isA [—2,A∞).A FromA yjAA =A 1+2(x+2)−1/2
weA have
j −
(yA —x)yA =A(yA—Ax)[1A+A(2(xA+A2)AA 1/2A]

=AyA—AxA+A2(yA—x)(xA+A2)−1/2

=AyA —AxA+A 2[xA+A 4(xA+A 2)1/2AA—x](xA +A 2)−1/2

=AyA—AxA+A8(xA+A2)1/2(xA+A2)−1/2A =A yA—AxA+A8.

AnA intervalA ofA definitionA forA theA solutionA ofA theA differential A equationA isA (—
2,A∞)A becauseA yjA isA notA definedA atA xA =A —2.
18. SinceAtanAxAisAnotAdefinedAforAxA =A π/2A +A nπ,AnA anAinteger,AtheAdomainAofAyAA =A 5AtanA5xAis
{xAA 5xA/=Aπ/2A+Anπ}
orA{xAA xA/=Aπ/10A+Anπ/5}.AFromAyA j=A25AsecA25xAweAhave
jA
y =A25(1A+Atan2A 5x)A=A25A+A25Atan2A 5xA=A25A+Ay 2 .

AnAintervalAofAdefinitionAforAtheAsolutionAofAtheAdifferentialAequationAisA(—π/10,Aπ/10).AAn-
A otherAintervalAisA(π/10,A3π/10),A and A soAon.


19. TheAdomainA ofA theA functionAisA {xAAA 4A —Ax2 /=A 0}AorA{x xA /=A —2AorAxA /=A 2}.AFromAy jA =
2x/(4A —Ax2)2A weA have
1 2
=A 2xy2.
yjAA=A 2x
4A—Ax2
AnA intervalA ofA definitionA forA theA solutionA ofA theA differentialA equationA isA (—2,A2).A OtherA inter-A valsA areA (—
∞,A —2)A andA (2,A ∞).A
√
20. TheAfunctionAisA yA =A 1/ 1A —AsinAxA,A whoseA domainAisA obtainedA fromA 1A —AsinAxA /=A 0A orA sinAxA /=A 1.
Thus,AtheAdomainAisA{xAA xA/=A π/2A+A2nπ}.AFromAyA j=A—A (11A—AsinAx)2
A
−3/2A (—AcosAx)AweAhave


2yjA =A(1A—AsinAx)−3/2AcosAxA=A[(1A—AsinAx)−1/2]3AcosAxA=Ay3AcosAx.

AnA intervalA ofA definitionA forA theA solutionA ofA theA differentialA equationA isA (π/2,A5π/2).A AnotherA oneA isA (5π/2,A 9π
/2),A andA soA on.




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, SolutionAandAAnswerAGuide:AZill,ADIFFERENTIALAEQUATIONS AWithAMODELINGAAPPLICATIONS A2024,A9780357760192;AChapterA
#1:
IntroductionA toA DifferentialA Equations



21. WritingAln(2XA —A 1)A —A ln(XA —A 1)AA =AA tAandAdifferentiating x

implicitlyA weA obtain 4


— =A 1 2
2XA—A1A dt XA—A1A dt
t
2 1 dXAA –A4 –2 2 4
— =A 1
2XA—A1 XA—A1 dt
–2


–A4
dX
=A—(2XA—A1)(XA—A1)A=A(XA—A1)(1A—A2X).
dtA
ExponentiatingA bothA sidesA ofA theA implicitA solutionA weA obtain

2XA—
A1A XA —
=AetA
A1



2XA —A1A=AXetA —Aet

(etA —A1)A=A(etA —A2)X
et 1
XA =A .
etA —A2A
SolvingAetA —A2A =A 0AweAgetAtA =A lnA2.A Thus, AtheA solutionAisAdefinedA onA(—
∞,AlnA2)A orAonA(lnA2,A∞).A TheA graphA ofA theA solutionA definedA onA (—
∞,AlnA2)A isA dashed,A andA theA graphA ofA theA solutionA definedA on A (lnA 2,A ∞)A isA solid.

22. ImplicitlyA differentiatingA theA solution,A weA obtain y

2AA dy dy 4

—2xAA —A4xyA+A2yA =A0
dxA dxA 2
2
—x dyA—A2xyAdxA+AyAdyA=A0
A


x
2xyAdxA+A(x2A —Ay)dyA=A0. –A4 –2 2 4


UsingAtheAquadraticA formulaA toAsolveAy2AA —A 2x2yA —A 1AA=AA 0 –2
√A √A
forAy,AweAgetAyA = 4x4A +A4AA /2A =A x2 ± x4A+A1A.
2AAA –
2x A4
±
√A
Thus,AtwoAexplicitAsolutionsAareAy1AA =A x2A + x4A +A1A and
√AA
y2AA =A x2AA — x4A +A 1A.A BothA solutionsA areA definedA onA (—∞,A∞).
TheA graphA ofA y1(x)A isA solidA andA theA graphA ofA y2AA isA dashed.



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