AFirst Course inDifferential Equatio
j j j j j
nswithModeling Applications,12thE
j j j j j
ditionby DennisG.Zill j j j j
CompleteChapterSolutionsManual are inclu
j j j j j
ded (Ch 1 to 9) j j j j
** Immediate Download
j j
** Swift Response
j j
** All Chapters included
j j j
,SolutionjandjAnswerjGuide:jZill,jDIFFERENTIALjEQUATIONSjWithjMODELINGjAPPLICATIONSj2024,j9780357760192;jChapterj#1:
Introductionj toj Differentialj Equations
SolutionandAnswerGuide j j j
ZILL,jDIFFERENTIALjEQUATIONSjWITHjMODELINGjAPPLICATIONSj2024,j 9780357760192;jCHAPTERj#1:jIN
TRODUCTIONjTOjDIFFERENTIALjEQUATIONS
TABLEOFCONTENTS j j
Endj ofj Sectionj Solutions........................................................................................................................................................................... 1
Exercisesj 1.1 ................................................................................................................................................................................................... 1
Exercisesj 1.2 ................................................................................................................................................................................................. 14
Exercisesj 1.3 ................................................................................................................................................................................................. 22
Chapterj1jinjReviewjSolutions .................................................................................................................................................... 30
ENDOFSECTIONSOLUTIONS
j j j
EXERCISES 1.1 j
1. Secondj order;j linear
2. Thirdj order;j nonlinearj becausej ofj (dy/dx)4
3. Fourthj order;j linear
4. Secondj order;j nonlinearj becausej ofj cos(rj +ju)
√j
5. Secondj order;j nonlinearj becausej ofj (dy/dx) or 2jj
1j +j (dy/dx)2
6. Secondj order;j nonlinearj becausej ofj R2
7. Thirdj order;j linear
8. Secondj order;j nonlinearj becausej ofj ẋj2
9. Firstj order;j nonlinearj becausej ofj sinj(dy/dx)
10. Firstj order;j linear
11. Writingjthejdifferentialj equationjinjthejformj x(dy/dx)j +j y2j =j 1,jwejseejthatjitjisjnonlinearj injyjbecausejofjy2.jHowev
er,jwritingjitjinjthejformj(y2j —j1)(dx/dy)j+jxj=j 0,jwejseejthatjitjisj linearj inj x.
12. Writingjthejdifferentialjequationjinjthejformju(dv/du)j+j(1j+ju)vj =j ueuj wejseejthatjitjisj linearjinjv.jHowever,jwrit
ingjitjinjthejformj(vj+juvj—jueu)(du/dv)j+juj=j 0,jwejseejthatjitjisj nonlinearj inj u.
13. Fromjyj=je− x/2
wejobtainjyjj =j—j1je− x/2
.jThenj2yjj +jyj =j—e− x/2
+je− x/2
=j0.
2
1
,SolutionjandjAnswerjGuide:jZill,jDIFFERENTIALjEQUATIONSjWithjMODELINGjAPPLICATIONSj2024,j9780357760192;jChapterj#1:
Introductionj toj Differentialj Equations
6 6 —
14. Fromj yj = — e 20tjwejobtainjdy/dtj=j24e−20tj,jsojthat
5 5
dyj+j20yj =j24e−20t 6 6j −20t
+j 20 —jj e =j 24.
dt 5 5
15. Fromjyj=je3xjcosj2xjwejobtainjyjj =j3e3xjcosj2x—2e3xjsinj2xjandjyjjj =j5e3xjcosj2x—12e3xjsinj2x,j soj thatj yjjj —
j6yjj +j 13yj =j 0.
j
16. Fromjyj =j —jcosjxjln(secjxj+jtanjx)jwejobtainjyjj =j—1j+jsinjxjln(secjxj+jtanjx)jand
jj jj
yjj =jtanjxj+jcosjxjln(secjxj+jtanjx).jThenjyjj +jyj=j tanjx.
17. Thej domainj ofj thej function,j foundj byj solvingj x+2 j ≥j 0,j isj [—2,j∞).j Fromj yjjj =j 1+2(x+2)−1/2
wej have
j −
(yj —x)yj =j(yj—jx)[1j+j(2(xj+j2)jj 1/2j ]
=jyj—jxj+j2(yj—x)(xj+j2)−1/2
=jyj —jxj+j 2[xj+j 4(xj+j 2)1/2jj—x](xj +j 2)−1/2
=jyj—jxj+j8(xj+j2)1/2(xj+j2)−1/2j =j yj—jxj+j8.
Anj intervalj ofj definitionj forj thej solutionj ofj thej differentialj equationj isj (—2,j∞)j becausej yjj isj notj definedj atj xj =j —2.
18. Sincejtanjxjisjnotjdefinedjforjxj =j π/2j +j nπ,jnj anjinteger,jthejdomainjofjyjj =j 5jtanj5xjis
{xjj 5xj/=jπ/2j+jnπ}
orj{xjj xj/=jπ/10j+jnπ/5}.jFromjyj j=j25jsecj25xjw ejhave
jj
y =j25(1j+jtan2j 5x)j=j25j+j25jtan2j 5xj=j25j+jy 2 .
Anjintervaljofjdefinitionjforjthejsolutionjofjthejdifferentialjequationjisj(—π/10,jπ/10).jAn-
j otherjinterval jisj(π/10,j3π/10),j and jsojon.
19. Thejdomainj ofj thej functionjisj {xjjj 4j —jx2 /=j 0}jorj{x xj /=j —2jorjxj /=j 2}.jFromjy jj =
2x/(4j —jx2)2j wej have
1 2
=j 2xy2.
yjjj=j 2x
4j—jx2
Anj intervalj ofj definitionj forj thej solutionj ofj thej differentialj equationj isj (—2,j2).j Otherj inter-j valsj arej (—∞,j —
2)j andj (2,j ∞).j
√
20. Thejfunctionjisj yj =j 1/ 1j —jsinjxj,j whosej domainjisj obtainedj fromj 1j —jsinjxj /=j 0j orj sinjxj /=j 1.
Thus,jthejdomainjisj{xjj xj/=j π/2j+j2nπ}.jFromjyj j=j—j (11j—jsinjx)j −3/2j (—2jcosjx)jwejhave
2yjj =j(1j—jsinjx)−3/2j cosjxj=j[(1j—jsinjx)−1/2]3jcosjxj=jy3j cosjx.
Anj intervalj ofj definitionj forj thej solutionj ofj thej differentialj equationj isj (π/2,j5π/2).j Anotherj onej isj (5π/2,j 9π/2),j andj
soj on.
2
, SolutionjandjAnswerjGuide:jZill,jDIFFERENTIALjEQUATIONSjWithjMODELINGjAPPLICATIONSj2024,j9780357760192;jChapterj#1:
Introductionj toj Differentialj Equations
21. Writingjln(2Xj —j 1)j —j ln(Xj —j 1)jj=jj tjandjdifferentiating x
implicitlyj wej obtain 4
— =j 1 2
2Xj—j1j dt Xj—j1j dt
t
2 1 dXjj –j4 –2 2 4
— =j 1
2Xj—j1 Xj—j1 dt
–2
–j4
dX
=j—(2Xj—j1)(Xj—j1)j=j(Xj—j1)(1j—j2X).
dtj
Exponentiatingj bothj sidesj ofj thej implicitj solutionj wej obtain
2Xj—
j1j Xj—j1
=jetj
2Xj —j1j=jXetj —jet
(etj—j1)j=j(etj—j2)X
et 1
Xj =j .
etj —j2j
Solvingjetj —j2j =j 0jwejgetjtj =j lnj2.j Thus,jthej solutionjisjdefinedj onj(—
∞,jlnj2)j orjonj(lnj2,j∞).j Thej graphj ofj thej solutionj definedj onj (—
∞,jlnj2)j isj dashed,j andj thej graphj ofj thej solutionj definedj onj (lnj 2,j ∞)j isj solid.
22. Implicitlyj differentiatingj thej solution,j wej obtain y
2jj dy dy 4
—2xjj —j4xyj+j2yj =j0
dxj dxj 2
2
—x dyj—j2xyjdxj+jyjdyj=j0
j
x
2xyjdxj+j(x2j —jy)dyj=j0. –j4 –2 2 4
Usingjthejquadraticj formulajtojsolvejy2jj —j 2x2yj —j 1jj=jj0 –2
√j √j
forjy,jwejgetjyj = 2x2jjj± 4x4j +j4jj /2j =j x2 ± x4j+j1j.
√j –j4
Thus,jtwojexplicitjsolutionsjarejy1jj =j x2j + x4j +j1j and
√j
y2jj =j x2jj — x4j +j 1j.j Bothj solutionsj arej definedj onj (—∞,j∞).
Thej graphj ofj y1(x)j isj solidj andj thej graphj ofj y2jj isj dashed.
3
j j j j j
nswithModeling Applications,12thE
j j j j j
ditionby DennisG.Zill j j j j
CompleteChapterSolutionsManual are inclu
j j j j j
ded (Ch 1 to 9) j j j j
** Immediate Download
j j
** Swift Response
j j
** All Chapters included
j j j
,SolutionjandjAnswerjGuide:jZill,jDIFFERENTIALjEQUATIONSjWithjMODELINGjAPPLICATIONSj2024,j9780357760192;jChapterj#1:
Introductionj toj Differentialj Equations
SolutionandAnswerGuide j j j
ZILL,jDIFFERENTIALjEQUATIONSjWITHjMODELINGjAPPLICATIONSj2024,j 9780357760192;jCHAPTERj#1:jIN
TRODUCTIONjTOjDIFFERENTIALjEQUATIONS
TABLEOFCONTENTS j j
Endj ofj Sectionj Solutions........................................................................................................................................................................... 1
Exercisesj 1.1 ................................................................................................................................................................................................... 1
Exercisesj 1.2 ................................................................................................................................................................................................. 14
Exercisesj 1.3 ................................................................................................................................................................................................. 22
Chapterj1jinjReviewjSolutions .................................................................................................................................................... 30
ENDOFSECTIONSOLUTIONS
j j j
EXERCISES 1.1 j
1. Secondj order;j linear
2. Thirdj order;j nonlinearj becausej ofj (dy/dx)4
3. Fourthj order;j linear
4. Secondj order;j nonlinearj becausej ofj cos(rj +ju)
√j
5. Secondj order;j nonlinearj becausej ofj (dy/dx) or 2jj
1j +j (dy/dx)2
6. Secondj order;j nonlinearj becausej ofj R2
7. Thirdj order;j linear
8. Secondj order;j nonlinearj becausej ofj ẋj2
9. Firstj order;j nonlinearj becausej ofj sinj(dy/dx)
10. Firstj order;j linear
11. Writingjthejdifferentialj equationjinjthejformj x(dy/dx)j +j y2j =j 1,jwejseejthatjitjisjnonlinearj injyjbecausejofjy2.jHowev
er,jwritingjitjinjthejformj(y2j —j1)(dx/dy)j+jxj=j 0,jwejseejthatjitjisj linearj inj x.
12. Writingjthejdifferentialjequationjinjthejformju(dv/du)j+j(1j+ju)vj =j ueuj wejseejthatjitjisj linearjinjv.jHowever,jwrit
ingjitjinjthejformj(vj+juvj—jueu)(du/dv)j+juj=j 0,jwejseejthatjitjisj nonlinearj inj u.
13. Fromjyj=je− x/2
wejobtainjyjj =j—j1je− x/2
.jThenj2yjj +jyj =j—e− x/2
+je− x/2
=j0.
2
1
,SolutionjandjAnswerjGuide:jZill,jDIFFERENTIALjEQUATIONSjWithjMODELINGjAPPLICATIONSj2024,j9780357760192;jChapterj#1:
Introductionj toj Differentialj Equations
6 6 —
14. Fromj yj = — e 20tjwejobtainjdy/dtj=j24e−20tj,jsojthat
5 5
dyj+j20yj =j24e−20t 6 6j −20t
+j 20 —jj e =j 24.
dt 5 5
15. Fromjyj=je3xjcosj2xjwejobtainjyjj =j3e3xjcosj2x—2e3xjsinj2xjandjyjjj =j5e3xjcosj2x—12e3xjsinj2x,j soj thatj yjjj —
j6yjj +j 13yj =j 0.
j
16. Fromjyj =j —jcosjxjln(secjxj+jtanjx)jwejobtainjyjj =j—1j+jsinjxjln(secjxj+jtanjx)jand
jj jj
yjj =jtanjxj+jcosjxjln(secjxj+jtanjx).jThenjyjj +jyj=j tanjx.
17. Thej domainj ofj thej function,j foundj byj solvingj x+2 j ≥j 0,j isj [—2,j∞).j Fromj yjjj =j 1+2(x+2)−1/2
wej have
j −
(yj —x)yj =j(yj—jx)[1j+j(2(xj+j2)jj 1/2j ]
=jyj—jxj+j2(yj—x)(xj+j2)−1/2
=jyj —jxj+j 2[xj+j 4(xj+j 2)1/2jj—x](xj +j 2)−1/2
=jyj—jxj+j8(xj+j2)1/2(xj+j2)−1/2j =j yj—jxj+j8.
Anj intervalj ofj definitionj forj thej solutionj ofj thej differentialj equationj isj (—2,j∞)j becausej yjj isj notj definedj atj xj =j —2.
18. Sincejtanjxjisjnotjdefinedjforjxj =j π/2j +j nπ,jnj anjinteger,jthejdomainjofjyjj =j 5jtanj5xjis
{xjj 5xj/=jπ/2j+jnπ}
orj{xjj xj/=jπ/10j+jnπ/5}.jFromjyj j=j25jsecj25xjw ejhave
jj
y =j25(1j+jtan2j 5x)j=j25j+j25jtan2j 5xj=j25j+jy 2 .
Anjintervaljofjdefinitionjforjthejsolutionjofjthejdifferentialjequationjisj(—π/10,jπ/10).jAn-
j otherjinterval jisj(π/10,j3π/10),j and jsojon.
19. Thejdomainj ofj thej functionjisj {xjjj 4j —jx2 /=j 0}jorj{x xj /=j —2jorjxj /=j 2}.jFromjy jj =
2x/(4j —jx2)2j wej have
1 2
=j 2xy2.
yjjj=j 2x
4j—jx2
Anj intervalj ofj definitionj forj thej solutionj ofj thej differentialj equationj isj (—2,j2).j Otherj inter-j valsj arej (—∞,j —
2)j andj (2,j ∞).j
√
20. Thejfunctionjisj yj =j 1/ 1j —jsinjxj,j whosej domainjisj obtainedj fromj 1j —jsinjxj /=j 0j orj sinjxj /=j 1.
Thus,jthejdomainjisj{xjj xj/=j π/2j+j2nπ}.jFromjyj j=j—j (11j—jsinjx)j −3/2j (—2jcosjx)jwejhave
2yjj =j(1j—jsinjx)−3/2j cosjxj=j[(1j—jsinjx)−1/2]3jcosjxj=jy3j cosjx.
Anj intervalj ofj definitionj forj thej solutionj ofj thej differentialj equationj isj (π/2,j5π/2).j Anotherj onej isj (5π/2,j 9π/2),j andj
soj on.
2
, SolutionjandjAnswerjGuide:jZill,jDIFFERENTIALjEQUATIONSjWithjMODELINGjAPPLICATIONSj2024,j9780357760192;jChapterj#1:
Introductionj toj Differentialj Equations
21. Writingjln(2Xj —j 1)j —j ln(Xj —j 1)jj=jj tjandjdifferentiating x
implicitlyj wej obtain 4
— =j 1 2
2Xj—j1j dt Xj—j1j dt
t
2 1 dXjj –j4 –2 2 4
— =j 1
2Xj—j1 Xj—j1 dt
–2
–j4
dX
=j—(2Xj—j1)(Xj—j1)j=j(Xj—j1)(1j—j2X).
dtj
Exponentiatingj bothj sidesj ofj thej implicitj solutionj wej obtain
2Xj—
j1j Xj—j1
=jetj
2Xj —j1j=jXetj —jet
(etj—j1)j=j(etj—j2)X
et 1
Xj =j .
etj —j2j
Solvingjetj —j2j =j 0jwejgetjtj =j lnj2.j Thus,jthej solutionjisjdefinedj onj(—
∞,jlnj2)j orjonj(lnj2,j∞).j Thej graphj ofj thej solutionj definedj onj (—
∞,jlnj2)j isj dashed,j andj thej graphj ofj thej solutionj definedj onj (lnj 2,j ∞)j isj solid.
22. Implicitlyj differentiatingj thej solution,j wej obtain y
2jj dy dy 4
—2xjj —j4xyj+j2yj =j0
dxj dxj 2
2
—x dyj—j2xyjdxj+jyjdyj=j0
j
x
2xyjdxj+j(x2j —jy)dyj=j0. –j4 –2 2 4
Usingjthejquadraticj formulajtojsolvejy2jj —j 2x2yj —j 1jj=jj0 –2
√j √j
forjy,jwejgetjyj = 2x2jjj± 4x4j +j4jj /2j =j x2 ± x4j+j1j.
√j –j4
Thus,jtwojexplicitjsolutionsjarejy1jj =j x2j + x4j +j1j and
√j
y2jj =j x2jj — x4j +j 1j.j Bothj solutionsj arej definedj onj (—∞,j∞).
Thej graphj ofj y1(x)j isj solidj andj thej graphj ofj y2jj isj dashed.
3
