Differential equations with boundary value problems 8th edition zill solutions manual

Prepared by Warren S. Wright

Carol D. Wright

Dennis Zill

Warren S. Wright

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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 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

Exponentiatingbothsidesoftheimplicitsolutionweobtain

Solving et 2=0weget t =ln2.Thus,thesolutionisdefinedon(−∞, ln2)oron(ln2, ∞). Thegraphofthesolutiondefinedon(−∞, ln2)isdashed,andthegraphofthesolutiondefined on(ln2, ∞)issolid.

20. Implicitlydifferentiatingthesolution,weobtain

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.

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

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.

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.

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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

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.

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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

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13. Fromtheinitialconditionsweobtain

e 1 c1 + ec2 =5

e 1 c1 ec2 = 5

Solving,wefind c1 =0and c2 =5e 1.Thesolutionoftheinitial-valueproblemis

y =5e 1 e x =5e 1 x .

14. Fromtheinitialconditionsweobtain

c1 + c2 =0

c1 c2 =0

Solving,wefind c1 = c2 =0.Thesolutionoftheinitial-valueproblemis y =0.

15. Twosolutionsare y =0and y = x3

16. Twosolutionsare y =0and y = x2 . A lso,anyconstantmultipleof x2 isasolution.

17. For f (x,y)= y2/3 wehaveThus,thedifferentialequationwillhaveauniquesolutioninany rectangularregionoftheplanewhere y =0.

18. For f (x,y)= √xy wehave ∂f/∂y = 1 2 x/y .Thus,thedifferentialequationwillhaveaunique solutioninanyregionwhere x> 0and y> 0orwhere x< 0and y< 0.

19. For f (x,y)= y x wehave ∂f ∂y = 1 x .Thus,thedifferentialequationwillhaveauniquesolution inanyregionwhere x> 0orwhere x< 0.

20. For f (x,y)= x + y wehave ∂f ∂y =1.Thus,thedifferentialequationwillhaveauniquesolution intheentireplane.

21. For f (x,y)= x2/(4 y2)wehave ∂f/∂y =2x2y/(4 y2)2.Thusthedifferentialequationwill haveauniquesolutioninanyregionwhere y< 2, 2 <y< 2,or y> 2.

22. For f (x,y)= x2 1+ y3 wehave ∂f ∂y = 3x2y2 (1+ y3)2 .Thus,thedifferentialequationwillhavea uniquesolutioninanyregionwhere y = 1.

23. For f (x,y)= y2 x2 + y2 wehave ∂f ∂y = 2x2y (x2 + y2)2 .Thus,thedifferentialequationwillhavea uniquesolutioninanyregionnotcontaining(0, 0).

24. For f (x,y)=(y + x)/(y x)wehave ∂f/∂y = 2x/(y x)2.Thusthedifferentialequation willhaveauniquesolutioninanyregionwhere y<x orwhere y>x.

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InProblems 25

28 weidentify f (x,y)= y2 9 and ∂f/∂y = y/ y2 9.Weseethat f and ∂f/∂y arebothcontinuousintheregionsoftheplanedeterminedby y< 3 and y> 3 withno restrictionson x

25. Since4 > 3,(1, 4)isintheregiondefinedby y> 3andthedifferentialequationhasaunique solutionthrough(1, 4).

26. Since(5, 3)isnotineitheroftheregionsdefinedby y< 3or y> 3,thereisnoguaranteeof auniquesolutionthrough(5, 3).

27. Since(2, 3)isnotineitheroftheregionsdefinedby y< 3or y> 3,thereisnoguarantee ofauniquesolutionthrough(2, 3).

28. Since( 1, 1)isnotineitheroftheregionsdefinedby y< 3or y> 3,thereisnoguarantee ofauniquesolutionthrough( 1, 1).

29. (a) Aone-parameterfamilyofsolutionsis y = cx.Since y = c, xy = xc = y and y(0)= c · 0=0.

(b) Writingtheequationintheform y = y/x,weseethat R cannotcontainanypointonthe y-axis.Thus,anyrectangularregiondisjointfromthe y-axisandcontaining(x0,y0)will determineanintervalaround x0 andauniquesolutionthrough(x0,y0).Since x0 =0in part(a),wearenotguaranteedauniquesolutionthrough(0, 0).

(c) Thepiecewise-definedfunctionwhichsatisfies y(0)=0isnotasolutionsinceitisnot differentiableat x =0.

30. (a) Since d dx tan(x + c)=sec2(x + c)=1+tan2(x + c),weseethat y =tan(x + c)satisfies thedifferentialequation.

(b) Solving y(0)=tan c =0weobtain c =0and y =tan x.Sincetan x isdiscontinuousat x = ±π/2,thesolutionisnotdefinedon( 2, 2)becauseitcontains ±π/2.

(c) Thelargestintervalonwhichthesolutioncanexistis( π/2,π/2).

31. (a) Since d dx 1 x + c = 1 (x + c)2 = y 2,weseethat y = 1 x + c isasolutionofthedifferential equation.

(b) Solving y(0)= 1/c =1weobtain c = 1and y =1/(1 x).Solving y(0)= 1/c = 1 weobtain c =1and y = 1/(1+ x).Beingsuretoinclude x =0,weseethattheinterval ofexistenceof y =1/(1 x)is(−∞, 1),whiletheintervalofexistenceof y = 1/(1+ x) is( 1, ∞).

(c) Byinspectionweseethat y =0isasolutionon(−∞, ∞).

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32. (a) Applying y(1)=1to y = 1/(x + c)gives

33.

(c) No,theyarenotthesamesolution.Theinterval I ofdefinitionforthesolutioninpart(a) is(−∞, 2);whereastheinterval I ofdefinitionforthesolutioninpart(b)is(2, ∞).See thefigure.

(b) Solving3x2 y2 =3for y weget

(c) Only y = φ3(x)satisfies y( 2)=3.

34. (a) Setting x =2and y = 4in3x2 y2 = c weget 12 16= 4= c,sotheexplicitsolutionis

= 3x2 +4 , −∞ <x< ∞.

(b) Setting c =0wehave y = √3x and y = √3x,both definedon(−∞, ∞)andbothpassingthroughthe origin.

weconsiderthepointsonthegraphswith x-coordinates x0 = 1, x0 =0,and x0 =1.Theslopesofthetangentlinesatthesepointsarecomparedwiththeslopesgivenby y (x0) in (a) through (f)

InProblems 35–38

35. Thegraphsatisfiestheconditionsin(b)and(f).

36. Thegraphsatisfiestheconditionsin(e).

37. Thegraphsatisfiestheconditionsin(c)and(d).

38. Thegraphsatisfiestheconditionsin(a).

InProblems 39-44 y = c1 cos2x + c2 sin2x isatwoparameterfamilyofsolutionsofthesecondorderdifferentialequation y +4y =0.Insomeoftheproblemswewillusethefactthat y = 2c1 sin2x +2c2 cos2x

39. Fromtheboundaryconditions y(0)=0and y π 4 =3weobtain

(0)= c1 =0

Thus, c1 =0, c2 =3,andthesolutionoftheboundary-valueproblemis y =3sin2x

40. Fromtheboundaryconditions y(0)=0and y(π)=0weobtain y(0)= c1 =0

(π)= c1 =0.

Thus, c1 =0, c2 isunrestricted,andthesolutionoftheboundary-valueproblemis y = c2 sin2x, where c2 isanyrealnumber.

41. Fromtheboundaryconditions y (0)=0and y π 6 =0weobtain

42. Fromtheboundaryconditions y(0)=1and y (π)=5weobtain

=5. Thus, c1 =1, c2 = 5 2 ,andthesolutionoftheboundary-valueproblemis y =cos2x + 5 2 sin2x 43. Fromtheboundaryconditions y(0)=0and y(π)=2weobtain

Since0 = 1,thisisnotpossibleandthereisnosolution.

CHAPTER1 INTRODUCTIONTODIFFERENTIALEQUATIONS

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DiscussionProblems

45. Integrating y =8e2x +6x weobtain

At x =1the y-coordinateofthepointoftangencyis y = 1+5=4.Thisgivestheinitial condition y(1)=4.Theslopeofthetangentlineat x =1is y (1)= 1.Fromtheinitial conditionsweobtain

47. When x =0and y = 1 2 , y = 1,sotheonlyplausiblesolutioncurveistheonewithnegative slopeat(0, 1 2 ),ortheredcurve.

48. Ifthesolutionistangenttothe x-axisat(x0, 0),then y =0when x = x0 and y =0. Substitutingthesevaluesinto y +2y =3x 6weget0+0=3x0 6or x0 =2.

49. Thetheoremguaranteesaunique (meaningsingle) solutionthroughanypoint.Thus,there cannotbetwodistinctsolutionsthroughanypoint.

50. When y = 1 16 x

, y =

(2)= 1 16 (16)=1.When

and y(2)= 1 16 (16)=1.Thetwodifferentsolutionsarethesameontheinterval(0, ∞),which isallthatisrequiredbyTheorem1 2 1.

51. At t =0, dP/dt =0.15P (0)+20=0.15(100)+20=35.Thus,thepopulationisincreasingat arateof3,500individualsperyear.Ifthepopulationis500attime t = T then

Thus,atthistime,thepopulationisincreasingatarateof9,500individualsperyear.