A First Course in Differential Equatio
l l l l l
nswithModeling Applications,12th
l l l l l
Editionby Dennis G. Zill l l l l
CompletelChapterlSolutionslManuall arelinclu
dedl(Chl1ltol9)
** Immediate Download
l l
** Swift Response
l l
** All Chapters included
l l l
,SolutionlandlAnswerlGuide:lZill,lDIFFERENTIALlEQUATIONSlWithlMODELINGlAPPLICATIONSl2024,l9780357760192;lChapterl#1:
Introductionl tol Differentiall Equations
SolutionandAnswerGuide l l l
ZILL,lDIFFERENTIALlEQUATIONSlWITHlMODELINGlAPPLICATIONSl2024,l 9780357760192;lCHAPTERl#1:lIN
TRODUCTIONlTOlDIFFERENTIALlEQUATIONS
TABLElOFlCONTENTS
Endl ofl Sectionl Solutions .......................................................................................................................................................................... 1
Exercisesl 1.1 ................................................................................................................................................................................................... 1
Exercisesl 1.2 ................................................................................................................................................................................................. 14
Exercisesl 1.3 ................................................................................................................................................................................................. 22
Chapterl1linlReviewlSolutions ................................................................................................................................................... 30
ENDlOFlSECTIONlSOLUTIONS
EXERCISESl 1.1
1. Secondl order;l linear
2. Thirdl order;l nonlinearl becausel ofl (dy/dx)4
3. Fourthl order;l linear
4. Secondl order;l nonlinear l becausel ofl cos(rl +lu)
√l
2ll
5. Secondl order;l nonlinearl becausel ofl (dy/dx) or 1l +l (dy/dx)2
6. Secondl order;l nonlinearl becausel ofl R2
7. Thirdl order;l linear
8. Secondl order;l nonlinearlbecausel ofl ẋl2
9. Firstl order;l nonlinearl becausel oflsinl(dy/dx)
10. Firstl order;l linear
11. Writingltheldifferentiall equationlinlthelforml x(dy/dx)l +l y2l =l 1,lwelseelthatlitlislnonlinearl inlylbecauselofly2.lHo
wever,lwritinglitlinlthelforml(y2l —l1)(dx/dy)l+lxl=l 0,lwelseelthatlitlisl linearl inl x.
12. Writingltheldifferentiallequationlinlthelformlu(dv/du)l+l(1l+lu)vl =l ueul welseelthatlitlisl linearlinlv.lHowever,l
writinglitlinlthelforml(vl+luvl—lueu)(du/dv)l+lul=l 0,lwelseelthatlitlisl nonlinearl inl u.
13. Fromlyl=le− x/2
welobtainlyjl =l—l1le− x/2
.lThenl2yjl +lyl =l—e− x/2
+le− x/2 =l0.
2
1
,SolutionlandlAnswerlGuide:lZill,lDIFFERENTIALlEQUATIONSlWithlMODELINGlAPPLICATIONSl2024,l9780357760192;lChapterl#1:
Introductionl tol Differentiall Equations
6 6
14. Froml yl = — e—20tlwelobtainldy/dtl=l24e−20tl,lsolthat
5 5
dyl+l20yl =l24e−20t 6 6l −20t
+l 20 —ll e =l 24.
dt 5 5
15. Fromlyl=le3xlcosl2xlwelobtainlyjl =l3e3xlcosl2x—2e3xlsinl2xlandlyjjl =l5e3xlcosl2x—12e3xlsinl2x,l sol thatl yjjl —
l6yjl +l13yl =l 0.
j
16. Fromlyl =l —lcoslxlln(seclxl+ltanlx)lwelobtainlyll =l—1l+lsinlxlln(seclxl+ltanlx)land
jj jj
yll =ltanlxl+lcoslxlln(seclxl+ltanlx).lThenlyll +lyl=ltanlx.
17. Thel domainl ofl thel function,l foundl byl solvingl x+2 l ≥l 0,l isl [—2,l∞).l Froml yjll =l 1+2(x+2)−1/2
wel have
j −
(yl —x)yl =l(yl—lx)[1l+l(2(xl+l2)ll 1/2l ]
=lyl—lxl+l2(yl—x)(xl+l2)−1/2
=lyl —lxl+l 2[xl+l 4(xl+l 2)1/2ll—x](xl +l 2)−1/2
=lyl—lxl+l8(xl+l2)1/2(xl+l2)−1/2l =l yl—lxl+l8.
Anl intervall ofl definitionl forl thel solutionl ofl thel differentiall equationl isl (—2,l∞)l becausel yjl isl notl definedl atl xl =l —
2.
18. Sinceltanlxlislnotldefinedlforlxl =l π/2l +l nπ,lnl anlinteger,ltheldomainloflyll =l 5ltanl5xlis
{xll 5xl/=lπ/2l+lnπ}
orl{xll xl/=lπ/10l+lnπ/5}.lFromlyl j=l25ls ecl25xlwelhave
jl
y =l25(1l+ltan2l 5x)l=l25l+l25ltan2l 5xl=l25l+ly 2 .
Anlintervallofldefinitionlforlthelsolutionlofltheldifferentiallequationlisl(—π/10,lπ/10).lAn-
l otherlintervallisl(π/10,l3π/10),l andlsolon.
19. Theldomainl ofl thel functionlisl {xlll 4l —lx2 /=l 0}lorl{x xl /=l —2lorlxl /=l 2}.lFromly jl =
2x/(4l —lx2)2l wel have
1 2
=l 2xy2.
yjll=l 2x
4l—lx2
Anl intervall ofl definitionl forl thel solutionl ofl thel differentiall equationl isl (—2,l2).l Otherl inter-l valsl arel (—∞,l —
2)l andl (2,l ∞).l
√
20. Thelfunctionlisl yl =l 1/ 1l —lsinlxl,l whosel domainlisl obtainedl froml 1l —lsinlxl /=l 0l orl sinlxl /=l 1.
Thus,ltheldomainlisl{xll xl/=l π/2l+l2nπ}.lFromlyl j=l—l (11l—lsinlx)l −3/2l2(—lcoslx)lwelhave
2yjl =l(1l—lsinlx)−3/2lcoslxl=l[(1l—lsinlx)−1/2]3lcoslxl=ly3lcoslx.
Anl intervall ofl definitionl forl thel solutionl ofl thel differentiall equationl isl (π/2,l5π/2).l Anotherl onel isl (5π/2,l 9π/2),l a
ndl sol on.
2
, SolutionlandlAnswerlGuide:lZill,lDIFFERENTIALlEQUATIONSlWithlMODELINGlAPPLICATIONSl2024,l9780357760192;lChapterl#1:
Introductionl tol Differentiall Equations
21. Writinglln(2Xl —l 1)l —l ln(Xl —l 1)ll=ll tlandldifferentiating x
implicitlyl wel obtain 4
— =l 1 2
2Xl—l1l dt Xl—l1l dt
t
2 1 dXll –l4 –2 2 4
— =l 1
2Xl—l1 Xl—l1 dt
–2
–l4
dX
=l—(2Xl—l1)(Xl—l1)l=l(Xl—l1)(1l—l2X).
dtl
Exponentiatingl bothl sidesl ofl thel implicitl solutionl wel obtain
2Xl—
l1l Xl—l1
=letl
2Xl —l1l=lXetl —let
(etl—l1)l=l(etl—l2)X
et 1
Xl =l .
etl —l2l
Solvingletl —l2l =l 0lwelgetltl =l lnl2.l Thus,lthel solutionlisldefinedl onl(—
∞,llnl2)l orlonl(lnl2,l∞).l Thel graphl ofl thel solutionl definedl onl (—
∞,llnl2)l isl dashed,l andl thel graphl ofl thel solutionl definedl onl (lnl 2,l ∞)l isl solid.
22. Implicitlyl differentiatingl thel solution,l wel obtain y
2ll dy dy 4
—2xll —l4xyl+l2yl =l0
dxl dxl 2
—x2l dyl—l2xyldxl+lyldyl=l0
x
2xyldxl+l(x2l —ly)dyl=l0. –l4 –2 2 4
Usinglthelquadraticl formulaltolsolvely2ll —l 2x2yl —l 1ll=ll0 –2
√l √l
forly,lwelgetlyl = 2x2lll± 4x4l +l4ll /2l =l x2 4l
± x +l1l.
√ l –
Thus,ltwolexplicitlsolutionslarely1ll =l x2l + x4l +l1l and l4
√l
y2ll =l x2ll — x4l +l 1l.l Bothl solutionsl arel definedl onl (—∞,l∞).
Thel graphl ofl y1(x)l isl solidl andl thel graphl ofl y2ll isl dashed.
3
l l l l l
nswithModeling Applications,12th
l l l l l
Editionby Dennis G. Zill l l l l
CompletelChapterlSolutionslManuall arelinclu
dedl(Chl1ltol9)
** Immediate Download
l l
** Swift Response
l l
** All Chapters included
l l l
,SolutionlandlAnswerlGuide:lZill,lDIFFERENTIALlEQUATIONSlWithlMODELINGlAPPLICATIONSl2024,l9780357760192;lChapterl#1:
Introductionl tol Differentiall Equations
SolutionandAnswerGuide l l l
ZILL,lDIFFERENTIALlEQUATIONSlWITHlMODELINGlAPPLICATIONSl2024,l 9780357760192;lCHAPTERl#1:lIN
TRODUCTIONlTOlDIFFERENTIALlEQUATIONS
TABLElOFlCONTENTS
Endl ofl Sectionl Solutions .......................................................................................................................................................................... 1
Exercisesl 1.1 ................................................................................................................................................................................................... 1
Exercisesl 1.2 ................................................................................................................................................................................................. 14
Exercisesl 1.3 ................................................................................................................................................................................................. 22
Chapterl1linlReviewlSolutions ................................................................................................................................................... 30
ENDlOFlSECTIONlSOLUTIONS
EXERCISESl 1.1
1. Secondl order;l linear
2. Thirdl order;l nonlinearl becausel ofl (dy/dx)4
3. Fourthl order;l linear
4. Secondl order;l nonlinear l becausel ofl cos(rl +lu)
√l
2ll
5. Secondl order;l nonlinearl becausel ofl (dy/dx) or 1l +l (dy/dx)2
6. Secondl order;l nonlinearl becausel ofl R2
7. Thirdl order;l linear
8. Secondl order;l nonlinearlbecausel ofl ẋl2
9. Firstl order;l nonlinearl becausel oflsinl(dy/dx)
10. Firstl order;l linear
11. Writingltheldifferentiall equationlinlthelforml x(dy/dx)l +l y2l =l 1,lwelseelthatlitlislnonlinearl inlylbecauselofly2.lHo
wever,lwritinglitlinlthelforml(y2l —l1)(dx/dy)l+lxl=l 0,lwelseelthatlitlisl linearl inl x.
12. Writingltheldifferentiallequationlinlthelformlu(dv/du)l+l(1l+lu)vl =l ueul welseelthatlitlisl linearlinlv.lHowever,l
writinglitlinlthelforml(vl+luvl—lueu)(du/dv)l+lul=l 0,lwelseelthatlitlisl nonlinearl inl u.
13. Fromlyl=le− x/2
welobtainlyjl =l—l1le− x/2
.lThenl2yjl +lyl =l—e− x/2
+le− x/2 =l0.
2
1
,SolutionlandlAnswerlGuide:lZill,lDIFFERENTIALlEQUATIONSlWithlMODELINGlAPPLICATIONSl2024,l9780357760192;lChapterl#1:
Introductionl tol Differentiall Equations
6 6
14. Froml yl = — e—20tlwelobtainldy/dtl=l24e−20tl,lsolthat
5 5
dyl+l20yl =l24e−20t 6 6l −20t
+l 20 —ll e =l 24.
dt 5 5
15. Fromlyl=le3xlcosl2xlwelobtainlyjl =l3e3xlcosl2x—2e3xlsinl2xlandlyjjl =l5e3xlcosl2x—12e3xlsinl2x,l sol thatl yjjl —
l6yjl +l13yl =l 0.
j
16. Fromlyl =l —lcoslxlln(seclxl+ltanlx)lwelobtainlyll =l—1l+lsinlxlln(seclxl+ltanlx)land
jj jj
yll =ltanlxl+lcoslxlln(seclxl+ltanlx).lThenlyll +lyl=ltanlx.
17. Thel domainl ofl thel function,l foundl byl solvingl x+2 l ≥l 0,l isl [—2,l∞).l Froml yjll =l 1+2(x+2)−1/2
wel have
j −
(yl —x)yl =l(yl—lx)[1l+l(2(xl+l2)ll 1/2l ]
=lyl—lxl+l2(yl—x)(xl+l2)−1/2
=lyl —lxl+l 2[xl+l 4(xl+l 2)1/2ll—x](xl +l 2)−1/2
=lyl—lxl+l8(xl+l2)1/2(xl+l2)−1/2l =l yl—lxl+l8.
Anl intervall ofl definitionl forl thel solutionl ofl thel differentiall equationl isl (—2,l∞)l becausel yjl isl notl definedl atl xl =l —
2.
18. Sinceltanlxlislnotldefinedlforlxl =l π/2l +l nπ,lnl anlinteger,ltheldomainloflyll =l 5ltanl5xlis
{xll 5xl/=lπ/2l+lnπ}
orl{xll xl/=lπ/10l+lnπ/5}.lFromlyl j=l25ls ecl25xlwelhave
jl
y =l25(1l+ltan2l 5x)l=l25l+l25ltan2l 5xl=l25l+ly 2 .
Anlintervallofldefinitionlforlthelsolutionlofltheldifferentiallequationlisl(—π/10,lπ/10).lAn-
l otherlintervallisl(π/10,l3π/10),l andlsolon.
19. Theldomainl ofl thel functionlisl {xlll 4l —lx2 /=l 0}lorl{x xl /=l —2lorlxl /=l 2}.lFromly jl =
2x/(4l —lx2)2l wel have
1 2
=l 2xy2.
yjll=l 2x
4l—lx2
Anl intervall ofl definitionl forl thel solutionl ofl thel differentiall equationl isl (—2,l2).l Otherl inter-l valsl arel (—∞,l —
2)l andl (2,l ∞).l
√
20. Thelfunctionlisl yl =l 1/ 1l —lsinlxl,l whosel domainlisl obtainedl froml 1l —lsinlxl /=l 0l orl sinlxl /=l 1.
Thus,ltheldomainlisl{xll xl/=l π/2l+l2nπ}.lFromlyl j=l—l (11l—lsinlx)l −3/2l2(—lcoslx)lwelhave
2yjl =l(1l—lsinlx)−3/2lcoslxl=l[(1l—lsinlx)−1/2]3lcoslxl=ly3lcoslx.
Anl intervall ofl definitionl forl thel solutionl ofl thel differentiall equationl isl (π/2,l5π/2).l Anotherl onel isl (5π/2,l 9π/2),l a
ndl sol on.
2
, SolutionlandlAnswerlGuide:lZill,lDIFFERENTIALlEQUATIONSlWithlMODELINGlAPPLICATIONSl2024,l9780357760192;lChapterl#1:
Introductionl tol Differentiall Equations
21. Writinglln(2Xl —l 1)l —l ln(Xl —l 1)ll=ll tlandldifferentiating x
implicitlyl wel obtain 4
— =l 1 2
2Xl—l1l dt Xl—l1l dt
t
2 1 dXll –l4 –2 2 4
— =l 1
2Xl—l1 Xl—l1 dt
–2
–l4
dX
=l—(2Xl—l1)(Xl—l1)l=l(Xl—l1)(1l—l2X).
dtl
Exponentiatingl bothl sidesl ofl thel implicitl solutionl wel obtain
2Xl—
l1l Xl—l1
=letl
2Xl —l1l=lXetl —let
(etl—l1)l=l(etl—l2)X
et 1
Xl =l .
etl —l2l
Solvingletl —l2l =l 0lwelgetltl =l lnl2.l Thus,lthel solutionlisldefinedl onl(—
∞,llnl2)l orlonl(lnl2,l∞).l Thel graphl ofl thel solutionl definedl onl (—
∞,llnl2)l isl dashed,l andl thel graphl ofl thel solutionl definedl onl (lnl 2,l ∞)l isl solid.
22. Implicitlyl differentiatingl thel solution,l wel obtain y
2ll dy dy 4
—2xll —l4xyl+l2yl =l0
dxl dxl 2
—x2l dyl—l2xyldxl+lyldyl=l0
x
2xyldxl+l(x2l —ly)dyl=l0. –l4 –2 2 4
Usinglthelquadraticl formulaltolsolvely2ll —l 2x2yl —l 1ll=ll0 –2
√l √l
forly,lwelgetlyl = 2x2lll± 4x4l +l4ll /2l =l x2 4l
± x +l1l.
√ l –
Thus,ltwolexplicitlsolutionslarely1ll =l x2l + x4l +l1l and l4
√l
y2ll =l x2ll — x4l +l 1l.l Bothl solutionsl arel definedl onl (—∞,l∞).
Thel graphl ofl y1(x)l isl solidl andl thel graphl ofl y2ll isl dashed.
3
