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

A First Course in Differential E
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quations with Modeling Appli v v v




cations, 12th Edition by Denni
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s G. Zill v v




Complete Chapter Solutions Manual are
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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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,SolutionvandvAnswervGuide:vZill,vDIFFERENTIALvEQUATIONSvWithvMODELINGvAPPLICATIONSv2024,v9780357760192;vChapterv#1:
IntroductionvtovDifferentialvEquations




Solution and Answer Guide v v v



ZILL,vDIFFERENTIALvEQUATIONSvWITHvMODELINGvAPPLICATIONSv2024,v9
780357760192;vCHAPTERv#1:vINTRODUCTIONvTOvDIFFERENTIALvEQUATIONS


TABLE OF CONTENTSV V




Endv ofv Sectionv Solutions ....................................................................................................................................... 1
Exercisesv 1.1 ......................................................................................................................................................... 1
Exercisesv 1.2 .......................................................................................................................................................14
Exercisesv 1.3 .......................................................................................................................................................22
Chapterv1vinvReviewvSolutions ........................................................................................................................ 30




END OF SECTION SOLUTIONS
V V V




EXERCISES 1.1 V




1. Secondv order;v linear
2. Thirdvorder;vnonlinearvbecausevofv(dy/dx)4
3. Fourthvorder;v linear
4. Secondvorder;vnonlinearvbecausevofvcos(rv+vu)
√
5. Secondvorder;vnonlinearvbecausevofv(dy/dx)2v or 1v+v(dy/dx)2
6. Secondvorder;vnonlinearvbecausevofvR2
7. Thirdvorder;vlinear
8. Secondvorder;vnonlinearvbecausevofvẋ 2
9. Firstvorder;vnonlinearvbecausevofvsinv(dy/dx)
10. Firstvorder;vlinear
11. Writingvthevdifferentialvequationvinvthevformvx(dy/dx)v+vy2v =v 1,vwevseevthatvitvisvnonlinearvi
nvyvbecausevofvy2.vHowever,vwritingvitvinvthevformv(y2v—
v1)(dx/dy)v+vxv=v0,vwevseevthatvitvisvlinearvinvx.


12. Writingvthevdifferentialvequationvinvthevformvu(dv/du)v+v(1v+vu)vv =vueuvwevseevthatvitvisvl
inearvinvv.vHowever,vwritingvitvinvthevformv(vv+vuvv—
u
vue )(du/dv)v+vuv =v 0,vwevseevthatvitvisvnonlinearvinvu.


13. Fromvyv=ve−x/2vwevobtainvyjv=v—v12ve−x/2.vThenv2yjv+vyv=v—e−x/2v+ve−x/2v=v0.




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,SolutionvandvAnswervGuide:vZill,vDIFFERENTIALvEQUATIONSvWithvMODELINGvAPPLICATIONSv2024,v9780357760192;vChapterv#1:
IntroductionvtovDifferentialvEquations


6 6 —
14. Fromv yv = — e 20t wevobtainvdy/dtv =v24e −20t ,vsovthat
5 5
vv
dy
+v20yv =v24e −20t +v20 6 —v 6 e−20t =v 24.
v


dt 5v 5v


15. Fromvyv =ve3xvcosv2xvwevobtainvyjv =v3e3xvcosv2x—2e3xvsinv2xvandvyjjv =v5e3xvcosv2x—
12e3xvsinv2x,vsovthatvyjjv—v6yjv+v13yv=v0.
j
16. Fromvyv =v —vcosvxvln(secvxv+vtanvx)vwevobtainvy =v—1v+vsinvxvln(secvxv+vtanvx)vand
jj jj
yv =vtanvxv+vcosvxvln(secvxv+vtanvx).vThenvyv +vyv=vtanvx.
17. Thevdomainvofvthevfunction,vfoundvbyvsolvingvx+2 v ≥v 0,visv[—2,v∞).v Fromvyjv =v 1+2(x+2)−1/2
wevhave
j −1/2
(yv —x)yv =v(yv—vx)[1v+v(2(xv+v2) ]

=vyv—vxv+v2(yv— x)(xv+v2) −1/2

=vyv—vxv+v2[xv+v4(xv+v2)1/2v— x](xv+v2) −1/2

=vyv—vxv+v8(xv+v2)1/2 (xv+v2) −1/2v =vyv —vxv+v8.

Anvintervalvofvdefinitionvforvthevsolutionvofvthevdifferentialvequationvisv(—
2,v∞)vbecausevyjvisvnotvdefinedvatvxv=v—2.
18. Sincevtanvxvisvnotvdefinedvforvxv =v π/2v+vnπ,vnvanvinteger,vthevdomainvofvyv =v 5vtanv5xvis
{xv vv 5xv/=vπ/2v+vnπ}
orv{xv xv/=vπ/10v+vnπ/5}.vFromvyv =j v25vsecv 5x
2 vwevhave

j
y =v25(1v+vtan 2 5x)v=v25v+v25vtan 2 5xv=v25v+vy 2 .

Anvintervalvofvdefinitionvforvthevsolutionvofvthevdifferentialvequationvisv(—π/10,vπ/10).vAn-
vother vintervalvisv(π/10,v3π/10),vandvsovon.


19. Thevdomainvofvthevfunctionvisv{x v 4v—vx2 /=v 0}vorv{x v xv /=v — j

2vorvxv /=v 2}.vFromvyv =v2x/(4v—vx2)2vwevhave
v v
2
1
v v v
yjv =v2x =v 2xy2.
4v—vx2
Anvintervalvofvdefinitionvforvthevsolutionvofvthevdifferentialvequationvisv(—2,v2).vOthervinter-
vvalsv arev (—∞,v—2)vandv (2,v∞).
√
20. Thevfunctionvisvyv =v 1/ 1v—vsinvxv,vwhosevdomainvisvobtainedvfromv1v—vsinvxv /=v 0vorvsinvxv /=v 1.
Thus,vthevdomainvisv{xv xv/=vπ/2v+v2nπ}.vFromvyv =j v—v (1
1
2
v—vsinvx)
−3/2 (—vcosvx)vwevhave


2yjv=v(1v—vsinvx)−3/2vcosvxv=v[(1v—vsinvx)−1/2]3vcosvxv=vy3vcosvx.

Anvintervalvofvdefinitionvforvthevsolutionvofvthevdifferentialvequationvisv(π/2,v5π/2).vAnothervon
evisv(5π/2,v9π/2),vandvsovon.


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, SolutionvandvAnswervGuide:vZill,vDIFFERENTIALvEQUATIONSvWithvMODELINGvAPPLICATIONSv2024,v9780357760192;vChapterv#1:
IntroductionvtovDifferentialvEquations




21. Writingvln(2Xv —v 1)v —v ln(Xv —v 1)v =v tvandvdifferentiating x

implicitlyvwevobtain 4

2 dX 1 dX
—v =v1 2
2Xv—v1v dtv Xv—v1v dtv
v v v
2 v vv1 dXv t
—v =v1 –v4 –2 2 4
2Xv—v1 Xv—v1 dt
–2
2Xv—v2v—v2Xv+v1vdXv
=v1
(2Xv—v1)v(Xv—v1)v dt –v4
dX
=v—(2Xv—v1)(Xv—v1)v=v(Xv—v1)(1v—v2X).
dtv

Exponentiatingvbothvsidesvofvthevimplicitvsolutionvwevobtain
2Xv—
v1v Xv—
=vetv
v1



2Xv—v1v=vXetv—vet

(etv—v1)v=v(etv—v2)X
et — 1
X v =v .
etv —v2v

Solvingvetv—v2v=v0vwevgetvtv=vlnv2.vThus,vthevsolutionvisvdefinedvonv(—
∞,vlnv2)vorvonv(lnv2,v∞).vThevgraphvofvthevsolutionvdefinedvonv(—
∞,vlnv2)visvdashed,vandvthevgraphvofvthevsolutionvdefinedvonv(lnv2,v∞)visvsolid.

22. Implicitlyv differentiatingv thev solution,v wev obtain y

2v dy dy 4
—2xv —v4xyv+v2yv =v0
dxv dxv
2
—x2v dyv—v2xyvdxv+vyvdyv=v0
x
2xyvdxv+v(x2v—vy)dyv=v0. –v4 –2 2 4

–2
Usingvthevquadraticvformulavtovsolvevy2v —v 2x2yv —v 1v =v 0
√ √
forvy,vwevgetvyv = 2x2 ± 4x4v +v4v /2v =v x2 ± x4v +v1v. –v4
√
Thus,vtwovexplicitvsolutionsvarevy1v =v x2v+v x4v+v1v and
√
y2v =v x2v —v x4v +v1v.v Bothvsolutionsvarevdefinedvonv(—∞,v∞).
Thevgraphvofvy1(x)visvsolidvandvthevgraphvofvy2v isvdashed.




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