SolutionandAnswerGuide:Zill,DIFFERENTIALEQUATIONSWi thBOUNDARY-VALUEPROBLEMS2024,9780357760451;Chapter# 1:
IntroductiontoDifferentialEquations
SolutionandAnswerGuide
ZILL,DIFFERENTIALEQUATIONSW ITHBOUNDARY-VALUEPROBLEMS2024,
9780357760451;C HAPTER#1:INTRODUCTIONTODIFFERENTIALEQUATIONS
TABLEOFCONTENTS
EndofSectionSolutions.............................. ................................................ 1
Exercises1.1....................................... ................................................... ... 1
Exercises1.2....................................... ................................................... ... 14
Exercises1.3....................................... ................................................... ... 22
Chapter1inReviewSolutions.......................... ............................................. 30
ENDOFSECTIONSOLUTIONS
EXERCISES1.1
1.Secondorder;linear
2.Thirdorder;nonlinearbecauseof (dy/dx)4
3.Fourthorder;linear
4.Secondorder;nonlinearbecauseof cos(r+u)
5.Secondorder;nonlinearbecauseof (dy/dx)2or/radicalbig
1+(dy/dx)2
6.Secondorder;nonlinearbecauseof R2
7.Thirdorder;linear
8.Secondorder;nonlinearbecauseof ˙x2
9.Firstorder;nonlinearbecauseof sin(dy/dx)
10.Firstorder;linear
11.Writingthedifferentialequationintheform x(dy/dx) +y2= 1,weseethatitisnonlinear
inybecauseof y2.However,writingitintheform (y2−1)(dx/dy) +x= 0,weseethatitis
linearin x.
12.Writingthedifferentialequationintheform u(dv/du) + (1 + u)v=ueuweseethatitis
linearin v.However,writingitintheform (v+uv−ueu)(du/dv) +u= 0,weseethatitis
nonlinearin u.
13.Fromy=e−x/2weobtain y′=−1
2e−x/2.Then2y′+y=−e−x/2+e−x/2= 0.
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IntroductiontoDifferentialEquations
14.Fromy=6
5−6
5e−20tweobtain dy/dt= 24e−20t,sothat
dy
dt+20y= 24e−20t+20/parenleftbigg6
5−6
5e−20t/parenrightbigg
= 24.
15.Fromy=e3xcos2xweobtain y′= 3e3xcos2x−2e3xsin2xandy′′= 5e3xcos2x−12e3xsin2x,
sothaty′′−6y′+13y= 0.
16.Fromy=−cosxln(secx+tanx)weobtain y′=−1+sinxln(secx+tanx)and
y′′= tanx+cosxln(secx+tanx).Theny′′+y= tanx.
17.Thedomainofthefunction,foundbysolving x+2≥0,is[−2,∞).Fromy′= 1+2(x+2)−1/2
wehave
(y−x)y′= (y−x)[1+(2(x+2)−1/2]
=y−x+2(y−x)(x+2)−1/2
=y−x+2[x+4(x+2)1/2−x](x+2)−1/2
=y−x+8(x+2)1/2(x+2)−1/2=y−x+8.
Anintervalofdefinitionforthesolutionofthedifferential equationis (−2,∞)because y′is
notdefinedat x=−2.
18.Sincetanxisnotdefinedfor x=π/2 +nπ,naninteger,thedomainof y= 5tan5 xis
{x/vextendsingle/vextendsingle5x/ne}ationslash=π/2+nπ}
or{x/vextendsingle/vextendsinglex/ne}ationslash=π/10+nπ/5}.Fromy′= 25sec25xwehave
y′= 25(1+tan25x) = 25+25tan25x= 25+y2.
Anintervalofdefinitionforthesolutionofthedifferential equationis (−π/10,π/10).An-
otherintervalis (π/10,3π/10),andsoon.
19.Thedomainofthefunctionis {x/vextendsingle/vextendsingle4−x2/ne}ationslash= 0}or{x/vextendsingle/vextendsinglex/ne}ationslash=−2orx/ne}ationslash= 2}.Fromy′=
2x/(4−x2)2wehave
y′= 2x/parenleftbigg1
4−x2/parenrightbigg2
= 2xy2.
Anintervalofdefinitionforthesolutionofthedifferential equationis (−2,2).Otherinter-
valsare(−∞,−2)and(2,∞).
20.Thefunctionis y= 1/√
1−sinx,whosedomainisobtainedfrom 1−sinx/ne}ationslash= 0orsinx/ne}ationslash= 1.
Thus,thedomainis {x/vextendsingle/vextendsinglex/ne}ationslash=π/2+2nπ}.Fromy′=−1
2(1−sinx)−3/2(−cosx)wehave
2y′= (1−sinx)−3/2cosx= [(1−sinx)−1/2]3cosx=y3cosx.
Anintervalofdefinitionforthesolutionofthedifferential equationis (π/2,5π/2).Another
oneis(5π/2,9π/2),andsoon.
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website,inwholeorinpart.2 SolutionandAnswerGuide:Zill,DIFFERENTIALEQUATIONSWi thBOUNDARY-VALUEPROBLEMS2024,9780357760451;Chapter# 1:
IntroductiontoDifferentialEquations
21.Writingln(2X−1)−ln(X−1) =tanddifferentiating
implicitlyweobtain
2
2X−1dX
dt−1
X−1dX
dt= 1
/parenleftbigg2
2X−1−1
X−1/parenrightbiggdX
dt= 1
2X−2−2X+1
(2X−1)(X−1)dX
dt= 1
dX
dt=−(2X−1)(X−1) = (X−1)(1−2X).– 4 –2 2 4t
– 4–224x
Exponentiatingbothsidesoftheimplicitsolutionweobtai n
2X−1
X−1=et
2X−1 =Xet−et
(et−1) = (et−2)X
X=et−1
et−2.
Solvinget−2 = 0wegett= ln2.Thus,thesolutionisdefinedon (−∞,ln2)oron(ln2,∞).
Thegraphofthesolutiondefinedon (−∞,ln2)isdashed,andthegraphofthesolution
definedon (ln2,∞)issolid.
22.Implicitlydifferentiatingthesolution,weobtain
−2x2dy
dx−4xy+2ydy
dx= 0
−x2dy−2xydx+ydy= 0
2xydx+(x2−y)dy= 0.
Usingthequadraticformulatosolve y2−2x2y−1 = 0
fory,wegety=/parenleftbig
2x2±√
4x4+4/parenrightbig
/2 =x2±√
x4+1.
Thus,twoexplicitsolutionsare y1=x2+√
x4+1and
y2=x2−√
x4+1.Bothsolutionsaredefinedon (−∞,∞).
Thegraphof y1(x)issolidandthegraphof y2isdashed.– 4 –2 2 4x
– 4–224y
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website,inwholeorinpart.3 SolutionandAnswerGuide:Zill,DIFFERENTIALEQUATIONSWi thBOUNDARY-VALUEPROBLEMS2024,9780357760451;Chapter# 1:
IntroductiontoDifferentialEquations
23.Differentiating P=c1et//parenleftbig
1+c1et/parenrightbig
weobtain
dP
dt=/parenleftbig
1+c1et/parenrightbig
c1et−c1et·c1et
(1+c1et)2=c1et
1+c1et/bracketleftbig/parenleftbig
1+c1et/parenrightbig
−c1et/bracketrightbig
1+c1et
=c1et
1+c1et/bracketleftbigg
1−c1et
1+c1et/bracketrightbigg
=P(1−P).
24.Differentiating y= 2x2−1+c1e−2x2weobtaindy
dx= 4x−4xc1e−2x2,sothat
dy
dx+4xy= 4x−4xc1e−2x2+8x3−4x+4c1xe−x2= 8x3
25.Fromy=c1e2x+c2xe2xweobtaindy
dx= (2c1+c2)e2x+2c2xe2xandd2y
dx2= (4c1+4c2)e2x+
4c2xe2x,sothat
d2y
dx2−4dy
dx+4y= (4c1+4c2−8c1−4c2+4c1)e2x+(4c2−8c2+4c2)xe2x= 0.
26.Fromy=c1x−1+c2x+c3xlnx+4x2weobtain
dy
dx=−c1x−2+c2+c3+c3lnx+8x,
d2y
dx2= 2c1x−3+c3x−1+8,
and
d3y
dx3=−6c1x−4−c3x−2,
sothat
x3d3y
dx3+2x2d2y
dx2−xdy
dx+y= (−6c1+4c1+c1+c1)x−1+(−c3+2c3−c2−c3+c2)x
+(−c3+c3)xlnx+(16−8+4)x2= 12x2
InProblems 25–28,weusetheProductRuleandthederivativeofanintegral(( 12)ofthissection):
d
dxˆx
ag(t)dt=g(x).
27.Differentiating y=e3xˆx
1e−3t
tdtweobtaindy
dx=e3xˆx
1e−3t
tdt+e−3t
x·e3xor
dy
dx=e3xˆx
1e−3t
tdt+1
x,sothat
xdy
dx−3xy=x/parenleftbigg
e3xˆx
1e−3t
tdt+1
x/parenrightbigg
−3x/parenleftbigg
e3xˆx
1e−3t
tdt/parenrightbigg
=xe3xˆx
1e−3t
tdt+1−3xe3xˆx
1e−3t
tdt= 1
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