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

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




Complete Chapter Solutions Manual
are included (Ch 1 to 9)




** Immediate Download
** Swift Response
** All Chapters included

,SolutionEandEAnswerEGuide:EZill,EDIFFERENTIALEEQUATIONSEWithEMODELINGEAPPLICATIONSE2024,E9780357760192;EChapte
rE#1:
IntroductionE toE DifferentialE Equations



Solution and Answer Guide E E E


ZILL,EDIFFERENTIALEEQUATIONSEWITHEMODELINGEAPPLICATIONSE2024,E 9780357
760192;ECHAPTERE#1:EINTRODUCTIONETOEDIFFERENTIALEEQUATIONS


TABLE OF CONTENTS E E




EndE ofE SectionE Solutions..............................................................................................................................................1
ExercisesE 1.1......................................................................................................................................................................... 1
ExercisesE 1.2....................................................................................................................................................................... 14
ExercisesE 1.3....................................................................................................................................................................... 22
ChapterE1EinEReviewESolutions ............................................................................................................................30




END OF SECTION SOLUTIONS
E E E




EXERCISES 1.1 E


1. SecondE order;E linear
2. ThirdE order;E nonlinearE becauseE ofE (dy/dx)4
3. FourthE order;E linear
4. SecondE order;E nonlinearE becauseE ofE cos(rE +Eu)
√E
5. SecondE order;E nonlinearE becauseE ofE (dy/dx)2EE or 1E +E (dy/dx)2
6. SecondE order;E nonlinearEbecauseE ofE R2
7. ThirdE order;E linear
8. SecondE order;E nonlinearEbecauseE ofE ẋE2
9. FirstE order;E nonlinearE becauseE ofEsinE(dy/dx)
10. FirstE order;E linear
11. WritingEtheE differentialE equationEinEtheE formE x(dy/dx)E +E y2E =E 1,E weEseeEthatEitEisEnonlin
earE inEyEbecauseEofEy2.EHowever,EwritingEitEinEtheEformE (y2E —
E1)(dx/dy)E+ExE=E 0,EweEseeEthatEitEisE linearE inE x.

12. WritingEtheEdifferentialEequationEinEtheEformEu(dv/du)E+E(1E+Eu)vE =E ueuE weEseeEthatEi
tEisE linearEinEv.EHowever,EwritingEitEinEtheEformE(vE+EuvE—
u
Eue )(du/dv)E+EuE= E 0,Ewe Esee EthatEitEisE nonlinear E inE u.

13. FromEyE=Ee x/2 weEobtainEyjE =E—2 x/2
.EThenE2yjE +EyE =E— x/2
+Ee x/2
=E0.
− 1E −
E e− e−




1

,SolutionEandEAnswerEGuide:EZill,EDIFFERENTIALEEQUATIONSEWithEMODELINGEAPPLICATIONSE2024,E9780357760192;EChapte
rE#1:
IntroductionE toE DifferentialE Equations

6 6 —
14. FromE yE = — e 20tEweEobtainEdy/dtE=E24e−20tE,EsoEthat
5 5
dyE+E20yE =E24e−20t 6 6E −20t
+E 20 —EE e =E 24.
dt 5 5

15. FromEyE=Ee3xEcosE2xEweEobtainEyjE =E3e3xEcosE2x—2e3xEsinE2xEandEyjjE =E5e3xEcosE2x—
12e3xEsinE2x,E soE thatE yjjE —E6yjE +E13yE=E 0.
j
16. FromEyE =E —EcosExEln(secExE+EtanEx)EweEobtainEyEE =E—1E+EsinExEln(secExE+EtanEx)Eand
jj jj
yEE =EtanExE+EcosExEln(secExE+EtanEx).EThenEyEE +EyE=E tanEx.
17. TheE domainE ofE theE function,E foundE byE solvingE x+2 E ≥E 0,E isE [—2,E∞).E From E yjEE =E 1+2(x+2)−1/2
weE have
j −
(yE —x)yE =E(yE—Ex)[1E+E(2(xE+E2)EE 1/2E ]

=EyE—ExE+E2(yE—x)(xE+E2)−1/2

=EyE —ExE+E2[xE+E4(xE+E2)1/2EE—x](xE +E2)−1/2

=EyE—ExE+E8(xE+E2)1/2(xE+E2)−1/2E =E yE—ExE+E8.

AnE intervalE ofE definitionE forE theE solutionE ofE theE differentialE equationE isE (—
2,E∞)E becauseE yjE isE notE definedE atE xE =E —2.
18. SinceEtanExEisEnotEdefinedEforExE =E π/2E +E nπ,EnE anEinteger,EtheEdomainEofEyEE =E 5EtanE5xEis
{xEE 5xE/
=Eπ/2E+Enπ}
=Eπ/10E+Enπ/5}.EFromEyE j=E25EsecE25xEweEhave
orE{xEE xE/
jE
y =E25(1E+Etan2E 5x)E=E25E+E25Etan2E 5xE=E25E+Ey 2 .

AnEintervalEofEdefinitionEforEtheEsolution EofEtheEdifferentialEequation EisE(—
π/10,Eπ/10).EAn-E otherEintervalEisE(π/10,E3π/10),E andEsoEon.
19. TheEdomainE ofE theE functionEisE {xEEE 4E — /
=E 0}EorE{x xE /=E —2EorExE /=E 2}.EFromEy jE =
Ex
2

2x/(4E —Ex2)2E weE have 1 2
=E 2xy2.
yjEE=E 2x 4E—Ex
2


AnE intervalE ofE definitionE forE theE solutionE ofE theE differentialE equationE isE (—
2,E2).E OtherE inter-E valsE areE (—∞,E —2)E andE (2,E ∞).E
√
20. TheEfunctionEisE yE =E 1 1E—EsinExE,E whoseE domainEisE obtainedE fromE 1E —EsinExE /=E 0E orE sinExE /=E 1.
/
=E π/2E+E2nπ}.EFromEyE j=E—E2(11E—EsinEx)E −3/2E (—EcosEx)EweEhave
Thus,EtheEdomainEisE{xEE xE/

2yjE =E(1E—EsinEx)−3/2EcosExE=E[(1E—EsinEx)−1/2]3EcosExE=Ey3EcosEx.

AnE intervalE ofE definitionE forE theE solutionE ofE theE differentialE equationE isE (π/2,E5π/2).E Anoth
erE oneE isE (5π/2,E 9π/2),E andE so E on.

2

, SolutionEandEAnswerEGuide:EZill,EDIFFERENTIALEEQUATIONSEWithEMODELINGEAPPLICATIONSE2024,E9780357760192;EChapte
rE#1:
IntroductionE toE DifferentialE Equations



21. WritingEln(2XE —E 1)E —E ln(XE —E 1)EE=EE tEandEdifferentiating x

implicitlyE weE obtain 4


— =E 1 2
2XE—E1E dt XE—E1E dt
t
2 1 dXEE
— =E 1 –E4 –2 2 4
2XE—E1 XE—E1 dt
–2


–E4
dX
=E—(2XE—E1)(XE—E1)E=E(XE—E1)(1E—E2X).
dtE
Exponentiating E bothE sidesE ofE theE implicitE solutionE we E obta
in

2XE—
E1E XE
=EetE
—E1
2XE—E1E=EXetE —Eet

(etE—E1)E=E(etE—E2)X
et 1
XE = E .
etE —E2E
SolvingEetE —E2E =E 0Ewe EgetEtE =E lnE2.E Thus,EtheE solutionEisEdefinedEonE(—
∞,ElnE2)EorEonE(lnE2,E∞).E TheE graphE ofE theE solutionE definedE onE (—
∞,ElnE2)EisE dashed,E andE theE graphE ofE theE solutionE definedE onE (lnE 2,E ∞)E isE solid.

22. ImplicitlyE differentiating E theE solution,E weE obtain y

2EE dy dy 4

—2xEE —E4xyE+E2yE =E0
dxE dxE 2
—x2E dyE—E2xyEdxE+EyEdyE=E0
x
2xyEdxE+E(x2E —Ey)dyE=E0. –E4 –2 2 4

–2
UsingEtheEquadraticE formulaEtoEsolveEy2EE —E 2x2yE —E 1EE=EE0
√E √E
forEy,Ewe EgetEyE = 4x 4E +E4EE /2E =E x x4E+E1E.
2 ± –
2x2EEE E4
±
√E
Thus,EtwoEexplicitEsolutionsEareEy1EE =E x2 x4E +E1E and
E
+
√EE
y2EE =E x2EE— x4E +E 1E.E BothE solutionsE areE definedE onE (—∞,E∞).

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