# calc5 - rivas(kmr3394 Homework 05 staron(53595 This...

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rivas (kmr3394) – 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′′+ 36y= 01.k= 62.k= 36,363.k=364.k= 6,6correct5.k=66.k= 36Explanation:We will begin by solving fory′′.y= sinkty=kcoskty′′=k2sinkt.We computey′′+ 36y=k2sinkt+ 36 sinkt= (36k2) sinkt.Hencey′′+36y= 0 if and only ifk2= 36, i.e.,if and only ifk=±6. Thus,k= 6,6.00210.0pointsFind all nonzero values ofkfor which thefunctiony=Asinkt+Bcosktsatisfies thedifferential equationy′′+ 4y= 01.k=22.k= 4,43.k=44.k= 2,2correct5.k= 26.k= 4Explanation:We will begin by solving fory′′.y=Asinkt+Bcoskty=AkcosktBksinkty′′=Ak2sinktBk2coskt=k2(Asinkt+Bcoskt)=k2y.We computey′′+ 4y=k2y+ 4y= (4k2)y.Hencey′′+ 4y= 0 if and only ifk2= 4, i.e., ifand only ifk=±2. Hence,k= 2,2.00310.0pointsFind all values ofrfor which the functiony=ertsatisfies the differential equationy′′12y+ 32y= 0.1.r=8,42.r= 4,8correct
for all values ofAandB.
rivas (kmr3394) – Homework 05 – staron – (53595)23.r= 44.r=12,325.r= 126.r=32,12Explanation:We will begin by solving foryandy′′.y=erty=rerty′′=r2ert.We computey′′12y+ 32y=r2ert12rert+ 32ert= (r212r+ 32)ert= (r4)(r8)ert.Hencey′′12y+ 32y= 0 if and only ifk= 4,8. Thus,k= 4,8.00410.0pointsWhich of the following functions satisfy thedifferential equationy′′4y+ 4y= 0 ?1.y=e2t, te2tcorrect2.y=e4t, te2t3.y=e2t, te2t4.y=e2t, te4t5.y=e2t, te4t6.y=e2t, te2tExplanation: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=r2ert4rert+ 4ert= (r24r+ 4)ert= (r2)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)ert4(rt+ 1)ert+ 4tert= (r2t+ 2r4rt4 + 4t)ert=bracketleftbig(r24r+ 4)t+ 2r4bracketrightbigert=bracketleftbig(r2)2t+ 2(r2)bracketrightbigert= (r2) [(r2)t+ 2]ert.Hencey′′4y+ 4y=0ifandonlyif(r2) [(r2)t+ 2] = 0 for all values oft,so onlyr= 2 solves the differential equation.Thereforey=te2tis a solution.Since in both casesr= 2 solves the differ-ential equation, valid results for this problemincludey=e2tandy=te2t.
rivas (kmr3394) – Homework 05 – staron – (53595)300510.0pointsWhich of the following families of functions isthe solution to the differential equationy=8xy?1.y=Ce8x22.y=Cx183.y=Ce4x2correct4.y=x8+C5.y=Ce4xExplanation:This is a separable differential equation,