Instructor’s Resource Manual
Prepared by Warren S. Wright
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EIGHTH EDITION and A First Course in Differential Eqautions
EDITION
Carol D. Wright
Differential Equations with Boundary Value Problems
TENTH
Dennis Zill
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Warren S. Wright
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CONTENTS Chapter 1 Introduction To Differential Equations 1 Chapter 2 First-Order Differential Equations 30 Chapter 3 Modeling With First-Order Differential Equations 93 Chapter 4 Higher-Order Differential Equations 138 Chapter 5 Modeling With Higher-Order Differential Equations 256 Chapter 6 Series Solutions of Linear Equations 304 Chapter 7 The Laplace Transform 394 Chapter 8 Systems of Linear First-Order Differential Equations 472 Chapter 9 Numerical Solutions of Ordinary Differential Equations 531 Chapter 10 Plane autonomous systems Chapter 11 Orthogonal functions and Fourier series Chapter 12 Boundary-value Problems in ectangular Coordinates Chapter 13 Boundary-value Problems in Other Coordinate Systems Chapter 14 Integral Transform method Chapter 15 Numerical Solutions of iii Partial Differential Equations © 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 556 588 639 728 781 831 853 pp I 855 pp II Not For Sale © Cengage Learning. All rights reserved. No distribution allowed without express authorization.
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INTRODUCTIONTO DIFFERENTIALEQUATIONS 1
INTRODUCTIONTODIFFERENTIALEQUATIONS
1.1 DefinitionsandTerminology
1. Secondorder;linear
1.1 DefinitionsandTerminology
2. Thirdorder;nonlinearbecauseof(dy/dx)4
3. Fourthorder;linear
4. Secondorder;nonlinearbecauseofcos(r + u)
5.
6. Secondorder;nonlinearbecauseof R2
7. Thirdorder;linear
8. Secondorder;nonlinearbecauseof˙x2
9.
Chapter1
Secondorder;nonlinearbecauseof(dy/dx)2 or 1+(dy/dx)2
x.
u(dv/du)+(1+ u
ue
weseethatitislinear
v.However,writingitintheform(v + uv ueu)(du/dv)+ u =0,weseethatitisnonlinear in u. 11. From y = e x/2 weobtain y = 1 2 e x/2.Then2y + y = e x/2
e x/2 =0.
y = 6 5 6 5 e 20t weobtain dy/dt =24e 20t,sothat dy dt +20y =24e 20t +20 6 5 6 5 e 20t =24
y = e3x cos2x weobtain y =3e3x cos2x 2e3x sin2x and y =5e3x cos2x 12e3x sin2x, sothat y 6y +13y =0.
= cos
ln(sec
x)weobtain y = 1+sin x ln(sec x +tan x)and y =tan x +cos x ln(sec x +tan x).Then y + y =tan x © 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. Not For Sale © Cengage Learning. All rights reserved. No distribution allowed without express authorization.
Writingtheboundary-valueproblemintheform x(dy/dx)+ y2 =1,weseethatitisnonlinear in y becauseof y2.However,writingitintheform(y2 1)(dx/dy)+ x =0,weseethatitis linearin
10. Writingthedifferentialequationintheform
)v =
u
in
+
12. From
13. From
14. From y
x
x +tan
CHAPTER1 INTRODUCTIONTODIFFERENTIALEQUATIONS
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15. Thedomainofthefunction,foundbysolving x +2 ≥ 0,is[ 2, ∞).From y =1+2(x +2) 1/2 wehave
Anintervalofdefinitionforthesolutionofthedifferentialequationis( 2, ∞)because y is notdefinedat x = 2.
16. Sincetan x isnotdefinedfor x = π/2+ nπ, n aninteger,thedomainof y =5tan5x is {x 5x = π/2+ nπ} or {x x = π/10+ nπ/5}.From
17.
Anintervalofdefinitionforthesolutionofthedifferentialequationis( π/10,π/10).Another intervalis(π/10, 3π/10),andsoon.
Anintervalofdefinitionforthesolutionofthedifferentialequationis( 2, 2).Otherintervals are(−∞, 2)and(2, ∞).
18. Thefunctionis y =1/√1 sin x ,whosedomainisobtainedfrom1 sin x =0orsin x =1. Thus,thedomainis
Anintervalofdefinitionforthesolutionofthedifferentialequationis(π/2, 5π/2).Another intervalis(5π/2, 9π/2)andsoon.
19. Writingln(2X 1) ln(X 1)= t anddifferentiatingimplicitlyweobtain
(y
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.
x)y =(y x)[1+(2(x +2) 1/
]
=25sec
y =25(1+tan
5x)=25+25tan2 5x =25+ y 2 .
y
2 5x wehave
2
Thedomainofthefunctionis {x 4 x2 =0} or {x x = 2or x =2}.From y =2x/(4 x2)2 wehave y =2x 1 4 x2 2 =2xy 2 .
{
= π/2+2nπ}.From y = 1 2 (1 sin x) 3/2( cos x)wehave 2y =(1 sin x) 3/2 cos x =[(1 sin x) 1/2]3 cos x = y 3 cos x.
x x
2 2X 1 dX dt 1 X 1 dX dt =1 2 2X 1 1 X 1 dX dt =1 2X 2 2X +1 (2X 1)(X 1) dX dt =1 dX dt = (2X 1)(X 1)=(X 1)(1 2X). 2 © 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
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Exponentiatingbothsidesoftheimplicitsolutionweobtain
Solving et 2=0weget t =ln2.Thus,thesolutionisdefinedon(−∞, ln2)oron(ln2, ∞). Thegraphofthesolutiondefinedon(−∞, ln2)isdashed,andthegraphofthesolutiondefined on(ln2, ∞)issolid.
20. Implicitlydifferentiatingthesolution,weobtain
1.1DefinitionsandTerminology
2
X 1 = e t 2
t e t e t 1=(e t 2)X X = et 1 et 2 . 3 2 1 1 2 3 2 1 1 2 3 4 y x
X 1
X 1= Xe
dx =0 x 2 dy 2xydx +
=0
xydx +(x 2 y)dy =0
Usingthequadraticformulatosolve y2 2x2y 1=0for y, weget y = 2x2 ± √4x4 +4 /2= x2 ± √x4 +1.Thus, 4 2 2 4 4 2 2 4 y x twoexplicitsolutionsare y1 = x2 + √x4 +1and y2 = x2 √x4 +1.Bothsolutionsare definedon(−∞, ∞).Thegraphof y1(x)issolidandthegraphof y2 isdashed. 21. Differentiating P = c1et/ 1+ c1et weobtain dP dt = 1+ c1et c1et c1et c1et (1+ c1et)2 = c1et 1+ c1et 1+ c1et c1et 1+ c1et = c1et 1+ c1et 1 c1et 1+ c1et = P (1 P ) 22. Differentiating y = e x2 x 0 e t2 dt + c1e x2 weobtain y = e x2 ex2 2xe x2 x 0 e t2 dt 2c1xe x2 =1 2xe x2 x 0 e t2 dt 2c1xe x2 Substitutingintothedifferentialequation,wehave y +2xy =1 2xe x2 x 0 e t2 dt 2c1xe x2 +2xe x2 x 0 e t2 dt +2c1xe x2 =1 23. From y = c1e2x + c2xe2x weobtain dy dx =(2c1 + c2)e 2x +2c2xe 2x and d2y dx2 =(4c1 +4c2)e 2x + 4c2xe 2x,sothat d2y dx2 4 dy dx +4y =(4c1 +4c2 8c1 4c2 +4c1)e 2x +(4c2 8c2 +4c2)xe 2x =0. 3 © 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. Not For Sale © Cengage Learning. All rights reserved. No distribution allowed without express authorization.
2x 2 dy dx 4xy +2y dy
ydy
2
.
CHAPTER1 INTRODUCTIONTODIFFERENTIALEQUATIONS
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26. Thefunction y(x)isnotcontinuousat x =0sincelim
27. From y = emx weobtain y = memx.Then y +2y =0implies
29.
2 emx 5memx +6emx =(m 2)(m 3)emx =0
Since emx > 0forall x, m =2and m =3.Thus y = e2x and y = e3x aresolutions.
30. From y = emx weobtain y = memx and y = m2emx.Then2y +7y 4y =0implies 2m 2 emx +7memx 4emx =(2m 1)(m +4)emx =0
Since emx > 0forall x, m = 1 2 and m = 4.Thus y = ex/2 and y = e 4x aresolutions.
y = c
x
c
x + c3x ln x +4x2 weobtain dy dx = c1x 2 + c2 + c3 + c3 ln x +8x, d2y dx2 =2c1x 3 + c3x 1 +8, and d3y dx3 = 6c1x 4 c3x 2 , sothat x 3 d3y dx3 +2x 2 d2y dx2 x dy dx + y =( 6c1 +4c1 + c1 + c1)x 1 +( c3 +2c3 c2 c3 + c2)x +( c3 + c3)x ln x +(16 8+4)x 2 =12x 2 .
From y = x
x2
weobtain y = 2x,x< 0 2x,x ≥ 0 sothat xy 2y =0.
24. From
1
1 +
2
25.
2,x< 0
,x ≥ 0
x→0
x)=5andlim x→0+ y
x
y(
(
)= 5.Thus, y (x)doesnotexistat x =0.
me
Since
2.Thus
mx +2emx =(m +2)emx =0.
emx > 0forall x, m =
y = e 2x isasolution.
5memx =2emx or m = 2 5 .
28. From y = emx weobtain y = memx.Then5y =2y implies
Thus y = e2x/5 > 0isasolution.
m
From y = emx weobtain y = memx and y = m2emx.Then y 5y +6y =0implies
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31. From y = x
32.
InProblems 33–36 wesubstitute y = c intothedifferentialequationsanduse y =0 and y =0.
33. Solving5c =10weseethat y =2isaconstantsolution.
34. Solving c2 +2c 3=(c +3)(c 1)=0weseethat y = 3and y =1areconstantsolutions.
35. Since1/(c 1)=0hasnosolutions,thedifferentialequationhasnoconstantsolutions.
36. Solving6c =10weseethat y =5/3isaconstantsolution.
From
1.1DefinitionsandTerminology
mxm 1 and y = m(m 1)xm 2.Then xy
xm(m 1)xm 2 +2mxm 1 =[m(m 1)+2m]xm 1 =(m 2 + m)xm 1 = m(m +1)xm 1 =0.
m weobtain y =
+2y =0implies
Since xm 1 > 0for x> 0, m =0and m = 1.Thus y =1and y = x 1 aresolutions.
mxm 1 and y = m(m 1)xm 2.Then x2y 7xy +15y =0
x 2 m(m 1)xm 2 7xmxm 1 +15xm =[m(m 1) 7m +15]xm =(m 2 8m +15)xm =(m 3)(m 5)xm =0.
From y = xm weobtain y =
implies
m =5.Thus
Since xm > 0for x> 0, m =3and
y = x3 and y = x5 aresolutions.
e 2t +3e6t and y = e 2t +5e6t weobtain dx dt = 2e 2t +18e 6t and dy dt =2e 2t +30e 6t Then x +3y =(e 2t +3e 6t)+3( e 2t +5e 6t)= 2e 2t +18e 6t = dx dt and 5x +3y =5(e 2t +3e 6t)+3( e 2t +5e 6t)=2e 2t +30e 6t = dy dt . 38. From x =cos2t +sin2t + 1 5 et and y = cos2t sin2t 1 5 et weobtain dx dt = 2sin2t +2cos2t + 1 5 e t or dy dt =2sin2t 2cos2t 1 5 e t and d2x dt2 = 4cos2t 4sin2t + 1 5 e t or d2y dt2 =4cos2t +4sin2t 1 5 e t . Then 4y + e t =4( cos2t sin2t 1 5 e t)+ e t = 4cos2t 4sin2t + 1 5 e t = d2x dt2 and 4x e t =4(cos2t +sin2t + 1 5 e t) e t =4cos2t +4sin2t 1 5 e t = d2y dt2 . 5 © 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. Not For Sale © Cengage Learning. All rights reserved. No distribution allowed without express authorization.
37.
x =
CHAPTER1 INTRODUCTIONTODIFFERENTIALEQUATIONS
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DiscussionProblems
39. (y )2 +1=0hasnorealsolutionsbecause(y )2 +1ispositiveforallfunctions y = φ(x).
40. Theonlysolutionof(y )2 + y2 =0is y =0,since,if y =0, y2 > 0and(y )2 + y2 ≥ y2 > 0.
41. Thefirstderivativeof f (x)= ex is ex.Thefirstderivativeof f (x)= ekx is f (x)= kekx.The differentialequationsare y = y and y = ky,respectively.
42. Anyfunctionoftheform y = cex or y = ce x isitsownsecondderivative.Thecorresponding differentialequationis y y =0.Functionsoftheform y = c sin x or y = c cos x havesecond derivativesthatarethenegativesofthemselves.Thedifferentialequationis y + y =0.
43. Wefirstnotethat 1 y2 = 1 sin2 x = √cos2 x = | cos x|.Thispromptsustoconsider valuesof x forwhichcos x< 0,suchas x = π.Inthiscase
Thus, y =sin x willonlybeasolutionof y = 1 y2 whencos x> 0.Anintervalofdefinition isthen( π/2,π/2).Otherintervalsare(3π/2, 5π/2),(7π/2, 9π/2),andsoon.
44. Sincethefirstandsecondderivativesofsin t andcos t involvesin t andcos t,itisplausiblethat alinearcombinationofthesefunctions, A sin t + B cos t,couldbeasolutionofthedifferential equation.Using y = A cos t B sin t and y = A sin t B cos t andsubstitutingintothe differentialequationweget y +2y +4y = A sin t B cos t +2A cos t 2B sin t +4A sin t +4B cos t =(3A 2B)sin t +(2A +3B)cos t =5sin t.
Thus3A 2B =5and2A +3B =0.Solvingthesesimultaneousequationswefind A = 15 13 and B = 10 13 .Aparticularsolutionis y = 15 13 sin t 10 13 cos t.
45. Onesolutionisgivenbytheupperportionofthegraphwithdomainapproximately(0, 2 6). Theothersolutionisgivenbythelowerportionofthegraph,alsowithdomainapproximately (0, 2.6).
46. Onesolution,withdomainapproximately(−∞, 1.6)istheportionofthegraphinthesecond quadranttogetherwiththelowerpartofthegraphinthefirstquadrant.Asecondsolution, withdomainapproximately(0, 1 6)istheupperpartofthegraphinthefirstquadrant.The thirdsolution,withdomain(0, ∞),isthepartofthegraphinthefourthquadrant.
dy dx x=π = d dx (sin x) x=π =cos x x=π =cos π = 1, but 1 y2 x=π = 1 sin2 π = √1=1
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59,theestimatesofthedomainsinProblem46wereclose.
50. Todetermineifasolutioncurvepassesthrough(0, 3)welet t =0and P =3intheequation
passesthroughthepoint(0, 3).Similarly,letting t =0and P =1intheequationforthe one-parameterfamilyofsolutionsgives1= c1/(1+ c1)or c1 =1+ c1.Sincethisequationhas nosolution,nosolutioncurvepassesthrough(0, 1).
51. Forthefirst-orderdifferentialequationintegrate f (x).Forthesecond-orderdifferentialequationintegratetwice.Inthelattercaseweget y = ( f (x)dx)dx + c1x + c2
52. Solvingfor y usingthequadraticformulaweobtainthetwodifferentialequations y = 1 x 2+2 1+3x6 and y = 1 x 2 2 1+3x6 , sothedifferentialequationcannotbeputintheform dy/dx = f (x,y).
53. Thedifferentialequation yy xy =0hasnormalform dy/dx = x.Thesearenotequivalent because y =0isasolutionofthefirstdifferentialequationbutnotasolutionofthesecond.
1.1DefinitionsandTerminology 47. Differentiating(x3 + y3)/xy =3c weobtain xy(3x2 +3y2y ) (x3 + y3)(xy + y) x2y2 =0 3x 3 y +3xy 3 y x 4 y x 3 y xy 3 y y 4 =0 (3xy 3 x 4 xy 3)y = 3x 3 y + x 3 y + y 4 y = y4 2x3y 2xy3 x4 = y(y3 2x3) x(2y3 x3) . 48. Atangentlinewillbeverticalwhere y isundefined,orinthiscase,where x(2y3 x3)=0. Thisgives x =0and2y3 = x3.Substituting y3 = x3/2into x3 + y3 =3xy weget x 3 + 1 2 x 3 =3x 1 21/3 x 3 2 x 3 = 3 21/3 x 2 x 3 =22/3 x 2 x 2(x 22/3)=0 Thus,thereareverticaltangentlinesat x =0and x =22/3,orat(0, 0)and(22/3 , 21/3).Since 22/3 ≈ 1.
49.
φ1(x)= x/√25 x2 and φ2(x)= x/√25 x2,neither
Thederivativesofthefunctionsare
ofwhichisdefinedat x = ±5.
P = c1et/(1+ c1e
).Thisgives3= c
c1
3 2
P = ( 3/2)et 1 (3/2)et = 3et 2 3et
t
1/(1+ c1)or
=
.Thus,thesolutioncurve
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55. (a) Since e x2 ispositiveforallvaluesof x, dy/dx> 0forall x,andasolution, y(x),ofthe differentialequationmustbeincreasingonanyinterval.
(b)
dy dx approaches0as x approaches −∞ and ∞,thesolutioncurvehashorizontalasymptotestotheleftandto theright.
(c) Totestconcavityweconsiderthesecondderivative
Sincethesecondderivativeispositivefor x< 0andnegativefor x> 0,thesolutioncurve isconcaveupon(−∞, 0)andconcavedownon(0, ∞).x
56. (a) Thederivativeofaconstantsolution y = c is0,sosolving5 c =0weseethat c =5and so y =5isaconstantsolution.
(b) Asolutionisincreasingwhere dy/dx =5 y> 0or y< 5.Asolutionisdecreasingwhere dy/dx =5 y< 0or y> 5.
57. (a) Thederivativeofaconstantsolutionis0,sosolving y(a by)=0weseethat y =0and y = a/b areconstantsolutions.
(b) Asolutionisincreasingwhere dy/dx = y(a by)= by(a/b y) > 0or0 <y<a/b.A solutionisdecreasingwhere dy/dx = by(a/b y) < 0or y< 0or y>a/b.
(c) Usingimplicitdifferentiationwecompute
y = c1x + c2x2 weget y = c1 +2c2x and y =2c2.Then c2 = 1 2 y and c1 = y xy ,so y = c1x + c2x 2 =(y xy )x + 1 2 y x 2 = xy 1 2 x 2 y
1 2 x2y xy + y =0or x2y 2xy +2y =0.
CHAPTER1 INTRODUCTIONTODIFFERENTIALEQUATIONS 54. Differentiating
Thedifferentialequationis
2 =0andlim x→∞ dy dx =lim x→∞ e x2 =0.Since
lim x→−∞ dy dx =lim x→−∞ e x
d2y dx2 = d dx dy dx = d dx e x2 = 2xe x2
(d) 4 2 2 4 1 0.5 y x
d2y dx2
)+
a
)=
a 2by) Solving d2y/dx2 =0weobtain y = a/2b.Since d2y/dx2 > 0for0 <y<a/2b and d2y/dx2 < 0for a/2b<y<a/b,thegraphof y = φ(x)hasapointofinflectionat y = a/2b 8 © 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.
= y( by
y (
by
y (
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58. (a) If y = c isaconstantsolutionthen y =0,but c2 +4isnever0foranyrealvalueof c
(b) Since y = y2 +4 > 0forall x whereasolution y = φ(x)isdefined,anysolutionmust beincreasingonanyintervalonwhichitisdefined.Thusitcannothaveanyrelative extrema.
(c) Usingimplicitdifferentiationwecompute d2y/dx2 =2yy =2y(y2 +4).Setting d2y/dx2 = 0weseethat y =0correspondstotheonlypossiblepointofinflection.Since d2y/dx2 < 0 for y< 0and d2y/dx2 > 0for y> 0,thereisapointofinflectionwhere y =0.
Theoutputwillshow y(x)= e5xx cos2x,whichverifiesthatthecorrectfunctionwasentered, and0,whichverifiesthatthisfunctionisasolutionofthedifferentialequation.
1.1DefinitionsandTerminology (d) 2 1 1 2 4 2 2 4 6 y x
(d) 1 1 2 2 y x
Clear[y] y[x ]:=xExp[5x]Cos[2x] y[x] y [x]-20y [x]+158y [x]-580y [x]+841y[x]//Simplify
ComputerLabAssignments 59. In Mathematica use
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60. In Mathematica use Clear[y]
y[x ]:=20Cos[5Log[x]]/x-3Sin[5Log[x]]/x y[x]
x ∧ 3y [x]+2x ∧ 2y [x]+20xy [x]-78y[x]//Simplify
Theoutputwillshow y(x)= 20cos(5ln x) x 3sin(5ln x) x ,whichverifiesthatthecorrect functionwasentered,and0,whichverifiesthatthisfunctionisasolutionofthedifferential equation.
1.2 Initial-ValueProblems
1.2 Initial-ValueProblems
1.
2.
3. Letting x =2andsolving1/3=1/(4+ c)weget c = 1.Thesolutionis y =1/(x2 1).This solutionisdefinedontheinterval(1, ∞).
4. Letting x = 2andsolving1/2=1/(4+ c)weget c = 2.Thesolutionis y =1/(x2 2). Thissolutionisdefinedontheinterval(−∞, √2).
5. Letting x =0andsolving1=1/c weget c =1.Thesolutionis y =1/(x2 +1).Thissolution isdefinedontheinterval(−∞, ∞).
6. Letting x =1/2andsolving 4=1/(1/4+ c)weget c = 1/2.Thesolutionis y = 1/(x2 1/2)=2/(2x2 1).Thissolutionisdefinedontheinterval( 1/√2 , 1/√2).
InProblems 7–10 weuse x = c1 cos t + c2 sin t and x = c1 sin t + c2 cos t toobtainasystemoftwo equationsinthetwounknowns c1 and c2
7. Fromtheinitialconditionsweobtainthesystem c1 = 1 c2 =8.
Thesolutionoftheinitial-valueproblemis x = cos t +8sin t
CHAPTER1 INTRODUCTIONTODIFFERENTIALEQUATIONS
Solving 1/3=1/(1+ c1)weget c1 = 4.Thesolutionis y =1/(1 4e x).
Solving2=1/(1+ c1e)weget c1 = (1/2)e 1.Thesolutionis y =2/(2 e (x+1)).
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1.2Initial-ValueProblems 8. Fromtheinitialconditionsweobtainthesystem c2 =0 c1 =1. Thesolutionoftheinitial-valueproblemis x = cos t. 9. Fromtheinitialconditionsweobtain √3 2 c1 + 1 2 c2 = 1 2 1 2 c1 + √3 2 c2 =0. Solving,wefind c1 = √3/4and c2 =1/4.Thesolutionoftheinitial-valueproblemis x =(√3/4)cos t +(1/4)sin t. 10. Fromtheinitialconditionsweobtain √2 2 c1 + √2 2 c2 = √2 √2 2 c1 + √2 2 c2 =2√2 Solving,wefind c1 = 1and c2 =3.Thesolutionoftheinitial-valueproblemis x = cos t +3sin t. InProblems 11–14 weuse y = c1ex + c2e x and y = c1ex c2e x toobtainasystemoftwo equationsinthetwounknowns c1 and c2 11. Fromtheinitialconditionsweobtain c1 + c2 =1 c1 c2 =2 Solving,wefind c1 = 3 2 and c2 = 1 2 .Thesolutionoftheinitial-valueproblemis y = 3 2 ex 1 2 e x . 12. Fromtheinitialconditionsweobtain ec1 + e 1 c2 =0 ec1 e 1 c2 = e. Solving,wefind c1 = 1 2 and c2 = 1 2 e2.Thesolutionoftheinitial-valueproblemis y = 1 2 ex 1 2 e 2 e x = 1 2 ex 1 2 e 2 x . 11 © 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. Not For Sale © Cengage Learning. All rights reserved. No distribution allowed without express authorization. Download all pages and all chapters at: TestBankTip.com
