Homework 4-solutions.pdf - atchison(bma862 Homework 4...

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atchison (bma862) – Homework 4 – staron – (53940)1Thisprint-outshouldhave13questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsFind all nonzero values ofkfor which thefunctiony= sinktsatisfies the differentialequationy′′+ 16y= 0.1.k= 162.k= 4,-4correct3.k= 44.k= 16,-165.k=-166.k=-4Explanation:We will begin by solving fory′′.y= sinkty=kcoskty′′=-k2sinkt.We computey′′+ 16y=-k2sinkt+ 16 sinkt= (16-k2) sinkt.Hencey′′+16y= 0 if and only ifk2= 16, i.e.,if and only ifk=±4. Hence,k= 4,-4.00210.0pointsFind all nonzero values ofkfor which thefunctiony=Asinkt+Bcosktsatisfies thedifferential equationy′′+ 36y= 01.k= 362.k= 63.k= 6,-6correct4.k=-365.k= 36,-366.k=-6Explanation:We will begin by solving fory′′.y=Asinkt+Bcoskty=Akcoskt-Bksinkty′′=-Ak2sinkt-Bk2coskt=-k2(Asinkt+Bcoskt)=-k2y.We computey′′+ 36y=-k2y+ 36y= (36-k2)y.Hencey′′+36y= 0 if and only ifk2= 36, i.e.,if and only ifk=±6. Hence,k= 6,-6.00310.0pointsFind all values ofrfor which the functiony=ertsatisfies the differential equationy′′+y-6y= 0.1.r=-32.r=-2,33.r=-1,
for all values ofAandB.
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atchison (bma862) – Homework 4 – staron – (53940)24.r= 65.r=-6,16.r=-3,2correctExplanation:We will begin by solving foryandy′′.y=erty=rerty′′=r2ert.We computey′′+y-6y=r2ert+rert-6ert= (r2+r-6)ert= (r+ 3)(r-2)ert.Hencey′′+y-6y=0ifandonlyifk=-3,2. Thus,k=-3,2.00410.0pointsWhich of the following functions satisfy thedifferential equationy′′+ 4y+ 4y= 0 ?1.y=e2t, te2tcorrect2.y=e2t, te4t3.y=e2t, te2t4.y=e4t, te2t5.y=e2t, te2t6.y=e2t, te4tAll 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=Explanation:

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