Math 22B, Homework #8 - UC Davis Mathematics

1 math 22B, Homework #81 #1We find a particular solution of the ODEy 5y + 6y= 2etusing the method of variation of parameters and then verify the solution using themethod of undetermined we solve the homogeneous equation using the characteristic equationr2 5r+ 6 = 0 which has rootsr= 3,2. Thus a fundamental set of solutions for thehomogeneous equation isy1=e3tand y2=e2tThus we assume that a particular solution has the formyp(t) =u1(t)e3t+u2(t)e2tdifferentiating we get thaty p= 3u1e3t+ 2u2e2ty p= 3(u 1+ 3u1)e3t+ 2(u 2+ 2u2)e2tThe form ofy pcomes from the constituent equationu 1e3t+u 2e2t= 0(1)Pluggingypand its derivatives back into the ODE we get a second equation3u 1e3t+ 2u 2e2t= 2et(2)Putting equations (1) and (2)

2 Together we get the system[e3te2t3e3t2e2t][u 1u 2]=[02et]Solving the system we obtain[u 1u 2]=[2e 2t 2e t]Thus after integrating we get:u1= e 2tand u2= 2e tReturning to the original form we getyp(t) =u1(t)e3t+u2(t)e2t= e 2te3t+ 2e te2t=etMath 22B, Homework #82 UCNow we can check the answer above by the following: Assume thatyp=AetwhereAis a constant. Then we getAet 5 Aet+ 6 Aet= 2et 2A= 2 ThusA= 1 andyp=etProblem #5We use Variation of Parameters to solve the ODEy +y= tant0< t < 2yh(t)The characteristic equation isr2+1 = 0 which has rootsr= i.

3 Thus the generalsolution to the homogeneous equation isyh=Acost+BsintVOPWe assume thatyp(t) =u1(t) cost+u2(t) sintThen the derivatives ofyphave the formy p= u1sint+u2costy p= (u 2 u1) cost (u 1+u2) sintThe form ofy pcomes from the constituent equationu 1cost+u 2sint= 0(1)Pluggingypand its derivatives back into the ODE we get another equation u 1sint+u 2cost= tant(2)Putting equations (1) and (2) together we get the system[costsint sintcost][u 1u 2]=[0tant]Which after solving yields[u 1u 2]=[ sect+ costsint]u1(t)Now we haveu 1(t) = sect+ costthusu1(t) = ( sect+ cost)dt= sectdt+ costdtThe second integral is easy, but to do the first we writesect=1cost=1cost costcost=cost1 sin2tMath 22B, Homework #83So sectdt= (cost1 sin2t)dtNow, letu= sintanddu= costdt, then (cost1 sin2t)dt= du1 u2=12[ du1 +u+ du1 u]=12ln 1 +u1 u So finally we get sectdt= (cost1 sin2t)dt=12ln 1 + sint1 sint Thusu1(t)

4 = sint 12ln 1 + sint1 sint u2(t)Thankfully, the expression foru 2is much easier. We haveu 2= sintthusu2(t) = costyp(t)Now we assumed thatyp(t) =u1(t) cost+u2(t) sintThusyp(t) =[sint 12ln 1 + sint1 sint ]cost+ [ cost] sintSoyp(t) = 12ln 1 + sint1 sint costy(t)Now for the general solutiony(t) =yh(t) +yp(t), thusy(t) =Acost+Bsint 12ln 1 + sint1 sint costProblem #16 Given thaty1=etandy2=tform a fundamental set of solutions for(1 t)y +ty y= 0 we find a particular solution for(1 t)y +ty y= 2(t 1)2e tAssume thatyp(t)

5 =u1et+u2tThen the derivatives ofypbecomey p=u1et+u2y p= (u1+u 1)et+u 2with the constituent equationu 1et+u 2t= 0(1) math 22B, Homework #84 Pluggingypand its derivatives into the ODE we getu 1et+u 2= 2(1 t)e t(2)Putting (1) and (2) together we have[ettet1][u 1u 2]=[02(1 t)e t]Inverting gives[u 1u 2]=[ 2te 2t2e t]Thus, after integrating, we getu1(t) =te 2t+12e 2tand u2(t) = 2e tThusyp(t) =u1et+u2t=(te 2t+12e 2t)et+ ( 2e t)t=12e t te tProblem #21 Lety(t) be any solution to the IVPL[y] =y +p(t)y +q(t)y=g(t) ;y(t0) =y0;y (t0) =y 0By Theorem there exists a unique functionu(t) which is a solution to the IVPL[u] =u +p(t)u +q(t)u= 0 ;u(t0) =y0;u (t0) =y 0 Then letv(t) =y(t) u(t).

6 We haveL[v] =L[y u] =L[y] L[u] =g(t) 0 =g(t)andv(t0) =y(t0) u(t0) y0 y0= 0 ;v (t0) =y (t0) u (t0) =y 0 y 0= 0 Thusy(t) =v(t) +u(t) is a solution to the original IVP. #5(a) FindL[t]L[t] = 0te stdt=[ 1ste st 1s2e st] 0=1s2(b) FindL[t2]L[t2] = 0t2e stdt=[ 1st2e st 2s2e st 2s3e st] 0=2s3 math 22B, Homework #85(c) FindL[tn].L[tn] = 0tne stdt=n!sn+1 The reason is that all the terms in the integral are of the formtke stk6= 0 exceptfor the last term. The coefficient of the last term is n!sn+ #9 FindL[eatcoshbt]L[eatcoshbt] =L[eat(ebt+e bt2)]=12(L[e(a+b)t] +L[e(a b)t])12(L[e(a+b)t] +L[e(a b)t]) =12[1s (a+b)+1s (a b)]=s a(s a)2 b2 ThusL[eatcoshbt] =s a(s a)2 b2 Problem #13 FindL[eatsinbt]L[eatsinbt] =12i[L[e(a+ib)t] L[e(a ib)t]]=12i[1s (a+ib) 1s (a ib)]=b(s a)2+b2 ThusL[eatsinbt] =b(s a)2+b2 Problem #18 FindL[tneat]L[tneat] = 0tneate stdt= 0tne(a s)tdtLetS=s a.

7 Then 0tne(a s)tdt= 0tne Stdt=n!Sn+1=n!(s a)n+1 ThusL[tneat] =n!(s a)n+1 #11 Solve the IVPy y 6y= 0 ;y(0) = 1 ;y (0) = 1using a Laplace [y y 6y] = (s2Y(s) sy(0) y (0)) (sY(s) y(0)) 6Y(s) = 0= Y(s) =s 2s2 s 6=15 1s 3+45 1s+ 2 Thusy(t) =15e3t+45e 2tMath 22B, Homework #86 Problem #14 Solve the IVPy 4y + 4y= 0 ;y(0) = 1 ;y (0) = 1using a Laplace [y 4y + 4y] = (s2Y(s) sy(0) y (0)) 4(sY(s) y(0)) + 4Y(s) = 0= Y(s) =s 3(s 2)2=s 2(s 2)2 1(s 2)2= Y(s) =1s 2 1(s 2)2 Thusy(t) =e2t te2t= (1 t)e2tProblem #16 Solve the IVPy + 2y + 5y= 0 ;y(0) = 2.

8 Y (0) = 1using a Laplace [y + 2y + 5y] = (s2Y sy(0) y (0)) + 2(sY y(0)) + 5Y= 0= Y(s) =2s+ 3s2+ 2s+ 5=(2s+ 2) + 1(s2+ 2s+ 1) + 4= 2 s+ 1(s+ 1)2+ 4+12 2(s+ 1)2+ 4 Thusy(t) = 2e tcos 2t+12e tsin 2tProblem #18 Solve the IVPyiv y= 0 ;y(0) = 1 ;y (0) = 0 ;y (0) = 1 ;y (0) = 0using a Laplace [yiv y] = (s4Y s3y(0) s2y (0) sy (0) y (0)) Y= 0= Y(s) =s3+ss4 1=ss2 1 Thusy(t) = coshtProblem #21 Solve the IVPy 2y+ 2y= cost;y(0) = 1 ;y (0) = 0using a Laplace [y 2y + 2y] = (s2Y sy(0) y (0)) 2(sY y(0)) + 2Y=ss2+ 1=L[cost] math 22B, Homework #87= Y(s) =(s 2s2 2s+ 2) (1)+(s(s2+ 1) (s2 2s+ 2)) (2)We can deal with the two parts (1) and (2) seperately sinceL 1is a linear (1) corresponds to the homogeneous equation and (2) to a particularsolution to the given ODE.

9 (1) We haveYh(s) =s 2s2 2s+ 2=s 2(s 1)2+ 1=s 1(s 1)2+ 1 1(s 1)2+ 1 Thusyh(t) =etcost etsint(2) Here we haveYp(s) =s(s2+ 1) (s2 2s+ 2)Now we assume thats(s2+ 1) (s2 2s+ 2)=As+Bs2 2s+ 2+Cs+Ds2+ 1 Solving this we getA= 15;B=45;C=15;D= 25. ThusYp(s) =15[ s+ 4s2 2s+ 2+s 2s2+ 1]=15[ (s 1) 3(s 1)2+ 1+ss2+ 1 2 1s2+ 1]And finallyYp(t) = 15 (s 1)(s 1)2+ 1+35 1(s 1)2+ 1+15 ss2+ 1 25 1s2+ 1 Henceyp(t) = 15etcost+35etsint+15cost 25sintFinally if we put (1) and (2) together we gety(t) =yh(t) +yp(t) =45etcost 25etsint+15cost 25sintProble #23 Solve the IVPy + 2y +y= 4e t;y(0) = 2.

10 Y (0) = 1using a Laplace [y + 2y +y] = (s2Y sy(0) y (0)) + 2(sY y(0)) +Y=4s+ 1=L[4e 4t] math 22B, Homework #88= Y(s) =2s+ 3s2+ 2s+ 1+4(s+ 1)(s2+ 2s+ 1)=2(s+ 1) + 1(s+ 1)2+4(s+ 1)3 ThusY(s) =s 1s+ 1+1(s+ 1)2+ 2 2(s+ 1)3 And finallyy(t) = 2e t+te t+ 2t2e t