ugurel (eu2243) – Homework 05 – staron – (53550)1Thisprint-outshouldhave18questions.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′′+ 9y= 01.k= 9,−92.k=−33.k= 3,−3correct4.k=−95.k= 96.k= 3Explanation:We will begin by solving fory′′.y= sinkty′=kcoskty′′=−k2sinkt.We computey′′+ 9y=−k2sinkt+ 9 sinkt= (9−k2) sinkt.Hencey′′+ 9y= 0 if and only ifk2= 9, i.e., ifand only ifk=±3. Thus,k= 3,−3.00210.0pointsFind all nonzero values ofkfor which thefunctiony=Asinkt+Bcosktsatisfies thedifferential equationy′′+ 36y= 01.k= 362.k= 36,−363.k= 6,−6correct4.k=−65.k=−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′′−4y′−12y= 0.1.r= 122.r=−6,2

for all values ofAandB.

ugurel (eu2243) – Homework 05 – staron – (53550)23.r=−12,−44.r= 4,125.r=−26.r=−2,6correctExplanation:We will begin by solving fory′andy′′.y=erty′=rerty′′=r2ert.We computey′′−4y′−12y=r2ert−4rert−12ert= (r2−4r−12)ert= (r+ 2)(r−6)ert.Hencey′′−4y′−12y= 0 if and only ifk=−2,6. Thus,k=−2,6.00410.0pointsWhich of the following functions satisfy thedifferential equationy′′+ 10y′+ 25y= 0 ?1.y=e5t, te5t2.y=e−5t, te5t3.y=e−5t, te−5tcorrect4.y=e−5t, te−10t5.y=e5t, te10t6.y=e−10t, te−5tExplanation:All answer choices above are of the formertandtert, so let us solve for potential valuesofrin each of those cases. We will begin byassumingy=ertand findingy′andy′′.y=erty′=rerty′′=r2ert.We computey′′+ 10y′+ 25y=r2ert+ 10rert+ 25ert= (r2+ 10r+ 25)ert= (r+ 5)2ertHencey′′+10y′+25y= 0 if and only ifr=−5.This meanse−5twill satisfy this equation.Now we will assumey=tertand findy′andy′′.y=terty′=rtert+ert= (rt+ 1)erty′′=rert+r2tert+rert=r2tert+ 2rert= (r2t+ 2r)ert.Now we computey′′+ 10y′+ 25y= (r2t+ 2r)ert+ 10(rt+ 1)ert+ 25tert= (r2t+ 2r+ 10rt+ 10 + 25t)ert=bracketleftbig(r2+ 10r+ 25)t+ 2r+ 10bracketrightbigert=bracketleftbig(r+ 5)2t+ 2(r+ 5)bracketrightbigert= (r+ 5) [(r+ 5)t+ 2]ert.Hencey′′+ 10y′+ 25y= 0 if and only if(r+ 5) [(r+ 5)t+ 2] = 0 for all values oft, soonlyr=−5 solves the differential equation.Thereforey=te−5tis a solution.Since in both casesr=−5 solves the differ-ential equation, valid results for this problemincludey=e−5tandy=te−5t.

ugurel (eu2243) – Homework 05 – staron – (53550)300510.0pointsWhich of the following families of functions isthe solution to the differential equationy′= 8xy?1.y=Cx182.y=Ce4x3.y=Ce4x2correct4.y=x8+C5.y=Ce8x2Explanation:This is a separable differential equation,