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′=Akcoskt−Bksinkty′′=−Ak2sinkt−Bk2coskt=−k2(Asinkt+Bcoskt)=−k2y.We computey′′+ 25y=−k2y+ 25y= (25−k2)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′′−2y′−8y= 0.1.r=−4,22.r=−2,4correct3.r=−8,−24.r= 2,85.r=−26.r= 8Explanation:We will begin by solving fory′andy′′.y=erty′=rerty′′=r2ert.We computey′′−2y′−8y=r2ert−2rert−8ert= (r2−2r−8)ert= (r+ 2)(r−4)ert.Hencey′′−2y′−8y=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(x2−9)correct