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

A First Course in Differential
V V V V V


Equations with Modeling Ap
V V V


plications, 12th Edition by De
V V V V


nnis G. Zill V V




CompleteVChapterVSolutionsVManual
VareVincludedV(ChV1VtoV9)




** Immediate Download
V V


** Swift Response
V V


** All Chapters included
V V V

,SolutionVandVAnswerVGuide:V Zill,VDIFFERENTIALVEQUATIONSVWithVMODELINGVAPPLICATIONSV2024,V9780357760192;VChapt
erV#1:
IntroductionVtoVDifferentialVEquations


Solution and Answer Guide V V V

ZILL,VDIFFERENTIALVEQUATIONSVWITHVMODELINGVAPPLICATIONSV2024,
9780357760192;VCHAPTERV#1:VINTRODUCTIONVTOVDIFFERENTIALVEQUATIONS


TABLEVOFVCONTENTS
EndV ofV SectionV Solutions .........................................................................................................................1
ExercisesV1.1 .......................................................................................................................... 1
ExercisesV1.2 ........................................................................................................................ 14
ExercisesV1.3 ........................................................................................................................ 22
ChapterV 1V inV ReviewV Solutions ........................................................................................................... 30




ENDV OFV SECTIONV SOLUTIONS
EXERCISESV 1.1
1. SecondVorder;Vlinear
2. ThirdVorder;VnonlinearVbecauseVofV(dy/dx)4
3. FourthVorder;Vlinear
4. SecondVorder;VnonlinearVbecauseVofVcos(rV +Vu)
√
5. SecondVorder;VnonlinearVbecauseV ofV(dy/dx)2V or 1V +V (dy/dx)2
6. SecondVorder;VnonlinearVbecauseVofVR2
7. ThirdVorder;Vlinear
8. SecondVorder;VnonlinearVbecauseVofVx˙V2
9. FirstVorder;VnonlinearVbecauseVofVsinV(dy/dx)
10. FirstVorder;Vlinear
11. WritingVtheVdifferentialVequationVinVtheVformVx(dy/dx)V+Vy2V =V 1,VweVseeVthatVitVisVno
nlinearVinVyVbecauseVofVy2.VHowever,VwritingVitVinVtheVformV(y2V−V1)(dx/dy)V+VxV =V 0,V
weVseeVthatVitVisVlinearVinVx.
12. WritingVtheVdifferentialVequationVinVtheVformVu(dv/du)V+V(1V+Vu)vV =V ueuVweVseeVt
hatVitVisVlinearVinVv.VHowever,VwritingVitVinVtheVformV(vV+VuvV−Vueu)(du/dv)V+VuV=V
0,VweVseeVthatVitVisVnonlinearVinVu.
13. FromVyV=Ve−x/2V weVobtainVy′V2=V−V1e−x/2.VThenV2y′V +VyV =V−e−x/2V +Ve−x/2V =V0.




1

,SolutionVandVAnswerVGuide:V Zill,VDIFFERENTIALVEQUATIONSVWithVMODELINGVAPPLICATIONSV2024,V9780357760192;VChapt
erV#1:
IntroductionVtoVDifferentialVEquations
6V 6
14. FromVyV =V −V —e20t weV obtainV dy/dtV =V 24e
−20t ,VsoVthat
5 5
dyV
+V20yV=V24e−20tV + 6V 6V − =V 24.
−V
V20 20t
e
dt 5 5V

15. FromVyV=Ve3xVcosV2xVweVobtainVy′V =V3e3xVcosV2x−2e3xVsinV2xVandVy′′V =V5e3xVcosV2x−12
e3xVsinV2x,VsoVthatVy′′V−V6y′V+V13yV=V0.
′
16. FromVyV =V −VcosVxVln(secVxV+VtanVx)V weVobtainVy =V−1V+VsinVxVln(secVxV+V tanVx)V and
′′ ′′
yV =VtanVxV+VcosVxVln(secVxV+VtanVx).VThenVyV +VyV=VtanVx.
17. TheVdomainV ofV theVfunction,VfoundVbyVsolvingVx+2V ≥V 0,VisV[−2,V∞).V FromV y′V =V 1+2(x+2)−1/2
weVhave
′ −
(yV − x)yV =V(yV−Vx)[1V+V(2(xV+V2)
1/2
]

=VyV−VxV+V2(yV−V V x)(xV+V2)−1/2

=VyV−VxV+V2[xV+V4(xV+V2)1/2V−V V x](xV+V2)−1/2

=VyV−VxV+V8(xV+V2)1/2V (xV+V2)−1/2V =VyV−VxV+V8.

AnVintervalVofVdefinitionVforVtheVsolutionVofVtheVdifferentialVequationVisV(−2,V∞)Vbecaus
eVy′V isVnotVdefinedVatVxV=V−2.

18. SinceVtanVxV isVnotVdefinedVforVxV =V π/2V +V nπ,VnV anVinteger,VtheVdomainVofVyV =V 5VtanV5xV is
{xV 5xV/=Vπ/2V+Vnπ}
orV{xV xV/=Vπ/10V+Vnπ/5}.VFromVyV′=V25VsecV25xVweVhave
′
2 2 2
yV =V25(1V+VtanV 5x)V=V25V+V25VtanV 5xV=V25V+VyV .

AnVintervalVofVdefinitionVforVtheVsolutionVofVtheVdifferentialVequationVisV(−π/10,Vπ/10).VA
n-VotherVintervalVisV(π/10,V3π/10),VandVsoVon.
19. TheVdomainVofVtheVfunctionVisV{xV4V /=V 0}VorV{x xV/=V −2VorVxV/=V 2}.VFromVyV′=
−VxV 2
2x/(4V −Vx2)2V weV have VV 1 V2
=V 2xy2.
4V−Vx2
y′V =V2x
AnVintervalVofVdefinitionVforVtheVsolutionVofVtheVdifferentialVequationVisV(−2,V2).VOther
Vinter-VvalsVareV(−∞,V−2)VandV(2,V∞).
√
20. TheVfunctionVisVyV =V 1 −VsinVxV,VwhoseVdomainV′ isVobtained
1
VfromV1V−VsinVxV /=V 0VorVsinVxV /=V 1.
−3/2
1/
Thus,VtheVdomainVisV{xV xV/=V π/2V+V2nπ}.VFromVyV =V−V (21V−VsinVx) (−VcosVx)VweVhave
2y′V =V(1V−VsinVx)−3/2V cosVxV=V[(1V−VsinVx)−1/2]3VcosVxV=V y3VcosVx.

AnVintervalVofVdefinitionVforVtheVsolutionVofVtheVdifferentialVequationVisV(π/2,V5π/2).VAno
2

, SolutionVandVAnswerVGuide:V Zill,VDIFFERENTIALVEQUATIONSVWithVMODELINGVAPPLICATIONSV2024,V9780357760192;VChapt
erV#1: therVoneVisV(5π/2,V9π/2),VandVsoVon.
IntroductionVtoVDifferentialVEquations




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