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mehany (wm6948) – Homework 05 – staron – (53595)1Thisprint-outshouldhave20questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsFind all values ofkthat don’t result in a zerofunction for which the functiony= sinktsatisfies the differential equationy′′+ 25y= 01.k= 52.k= 253.k= 25,254.k=55.k=256.k= 5,5correctExplanation:We will begin by solving fory′′.y= sinkty=kcoskty′′=k2sinkt.We computey′′+ 25y=k2sinkt+ 25 sinkt= (25k2) sinkt.Hencey′′+25y= 0 if and only ifk2= 25, i.e.,if and only ifk=±5. Thus,k= 5,5.00210.0pointsFind all nonzero values ofkfor which thefunctiony=Asinkt+Bcosktsatisfies thedifferential equationy′′+ 49y= 01.k=72.k= 7,7correct3.k=494.k= 495.k= 49,496.k= 7Explanation:We will begin by solving fory′′.y=Asinkt+Bcoskty=AkcosktBksinkty′′=Ak2sinktBk2coskt=k2(Asinkt+Bcoskt)=k2y.We computey′′+ 49y=k2y+ 49y= (49k2)y.Hencey′′+49y= 0 if and only ifk2= 49, i.e.,if and only ifk=±7. Hence,k= 7,7.00310.0pointsFind all values ofrfor which the functiony=ertsatisfies the differential equationy′′+ 4y21y= 0.1.r=21,42.r=4,21
for all values ofAandB.
mehany (wm6948) – Homework 05 – staron – (53595)23.r= 214.r=7,3correct5.r=3,76.r=7Explanation:We will begin by solving foryandy′′.y=erty=rerty′′=r2ert.We computey′′+ 4y21y=r2ert+ 4rert21ert= (r2+ 4r21)ert= (r+ 7)(r3)ert.Hencey′′+ 4y21y= 0 if and only ifk=7,3. Thus,k=7,3.00410.0pointsWhich of the following functions satisfy thedifferential equationy′′+ 4y+ 4y= 0 ?1.y=e2t, te2t2.y=e2t, te2t3.y=e4t, te2t4.y=e2t, te4t5.y=e2t, te4t6.y=e2t, te2tcorrectExplanation:All answer choices above are of the formertandtert, so let us solve for potential valuesofrin each of those cases. We will begin byassumingy=ertand findingyandy′′.y=erty=rerty′′=r2ert.We computey′′+ 4y+ 4y=r2ert+ 4rert+ 4ert= (r2+ 4r+ 4)ert= (r+ 2)2ertHencey′′+ 4y+ 4y= 0 if and only ifr=2.This meanse2twill satisfy this equation.Now we will assumey=tertand findyandy′′.y=terty=rtert+ert= (rt+ 1)erty′′=rert+r2tert+rert=r2tert+ 2rert= (r2t+ 2r)ert.Now we computey′′+ 4y+ 4y= (r2t+ 2r)ert+ 4(rt+ 1)ert+ 4tert= (r2t+ 2r+ 4rt+ 4 + 4t)ert=bracketleftbig(r2+ 4r+ 4)t+ 2r+ 4bracketrightbigert=bracketleftbig(r+ 2)2t+ 2(r+ 2)bracketrightbigert= (r+ 2) [(r+ 2)t+ 2]ert.Hencey′′+ 4y+ 4y=0ifandonlyif(r+ 2) [(r+ 2)t+ 2] = 0 for all values oft,so onlyr=2 solves the differential equa-tion. Thereforey=te2tis a solution.Since in both casesr=2 solves the differ-ential equation, valid results for this problemincludey=e2tandy=te2t.
mehany (wm6948) – Homework 05 – staron – (53595)300510.0pointsWhich of the following families of functions isthe solution to the differential equationy=6xy?1.y=Cx162.y=Ce3x2correct3.y=Ce6x24.y=Ce3x5.y=x6+CExplanation:This is a separable differential equation,