M408D SPRING 2015 HOMEWORK #5 - cheatham (sc36975) HW04 um...

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cheatham (sc36975) – HW04 – um – (53890)1Thisprint-outshouldhave14questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0 pointsFind all nonzero values ofkfor which thefunctiony=Asinkt+Bcosktsatisfies thedifferential equationy′′+ 25y= 0for all values ofAandB.1.k= 252.k=53.k=254.k= 25,255.k= 56.k= 5,5correctExplanation:We will begin by solving fory′′.y=Asinkt+Bcoskty=AkcosktBksinkty′′=Ak2sinktBk2coskt=k2(Asinkt+Bcoskt)=k2y.We computey′′+ 25y=k2y+ 25y= (25k2)y.Hencey′′+25y= 0 if and only ifk2= 25, i.e.,if and only ifk=±5. Hence,k= 5,5.00210.0 pointsFind all values ofrfor which the functiony=ertsatisfies the differential equationy′′2y8y= 0.1.r=4,22.r=2,4correct3.r=8,24.r= 2,85.r=26.r= 8Explanation:We will begin by solving foryandy′′.y=erty=rerty′′=r2ert.We computey′′2y8y=r2ert2rert8ert= (r22r8)ert= (r+ 2)(r4)ert.Hencey′′2y8y=0ifandonlyifk=2,4. Thus,k=2,4.00310.0 pointsThe family of solutions to the differentialequationy= 10xyisy=Ce5x2.Find the solution that satisfies the initialconditiony(3) = 2.1.y= 2e5(x2+9)2.y= 2e5(x29)correct
cheatham (sc36975) – HW04 – um – (53890)23.y= 2e5(x29)4.y= 2e5(x3)25.y=e5(x29)+ 1Explanation:This equation contains only one unknownconstant, so we will substitute the indicatedvalues ofxandyinto the equation to solveforCas follows:y(3) =Ce5(3)2=Ce45.This shows thaty(3) = 2 if and only ifC= 2e45. Thus,y(x) = 2e5x2e45.Hence,y(x) = 2e5(x29).00410.0 pointsThe family of solutions to the differentialequationy=3y2isy=13x+C.Find the solution that satisfies the initialconditiony(2) =5.1.y(x) =53(x+ 2)12.y(x) =515(x+ 2)1correct3.y(x) =13(x+ 2)54.y(x) =253(x+ 2)55.y(x) =115(x+ 2)5Explanation:This equation contains only one unknownconstant, so we will substitute the indicatedvalues ofxandyinto the equation to solveforCas follows:y(2) =13(2) +C=1C6.This shows thaty(2) =5 if and only ifC=15+ 6. Thusy(x) =13x15+ 6=13(x+ 2)15.Hence,y(x) =515(x

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