Problem 9 In each of Problems 9 through 12 find the solution of the given initial value problem. Thenplot a graph of the solution.2 9. y" + 4y' = t;2 10. y4) +2y"+y = 3t + 4;2 11. у" — Зу" + 2y 3Dt+e';2 12. y() +2y" +y" +8y' – 12y = 12 sin t – e-';y"(0) = -1, y" (0) = 2У (0) %3D у (0) — 0, у"(0) —D 1У (0) %3D У (0) 3 0, у"(0) —D у" (0) — 1y(0) = 1, y'(0) =-, y"(0) = -y(0) = 3, y'(0) = 0,

Name: Problem 9 In each of Problems 9 through 12 find the solution of the gi
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Description: Problem 9 In each of Problems 9 through 12 find the solution of the given initial value problem. Thenplot a graph of the solution.2 9. y" + 4y' = t;2 10. y4) +2y"+y = 3t + 4;2 11. у" — Зу" + 2y 3Dt+e';2 12. y() +2y" +y" +8y' – 12y = 12 sin t – e-';y"

9 Given Initial value problem y'''+4y'=t ,y(0)=y'(0)=0, y''(0)=1 Put ddt=D,d2dt2=D2 & d3dt3=D3 , we get (D3+D)y=t A.E is D3+D=0 ⇒D(D2+1)=0⇒D=0,D=±i C.F is ae0t+bcost+csint i.e. C.F= a+bcost+csint P.I =1D(D2+1)t ==1D(1+D2)-1t =1D(1-D2+D4+...)t=1D(t-D2t+D4t+...)=1Dt=t22 Solution y= C.F+ P.I y= a+bcost+csint + t22 Now, using y(0)=0, we get 0=a+b⇒a=-b so, y= -b+bcost+csint+t22 Now, y'(t)=-bsint+ccost+t Using y'(0)=0 in above, we get 0=c i.e. c=0 so, y'(t)=-bsint+t Now, y''(t)=-bcost+1 Using y''(0)=1,we get, 1= -b+1 ⇒b=0 So, y(t)=t22 is the answer.

Answered: In each of Problems 9 through 12 find…

Math

Advanced Math

In each of Problems 9 through 12 find the solution of the given initial value problem. Then plot a graph of the solution. 9. у" + 4y 3 t; У (0) %3D У (0) — 0, у"(0) %3 1

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Problem 9

In each of Problems 9 through 12 find the solution of the given initial value problem. Then
plot a graph of the solution.
2 9. y" + 4y' = t;
2 10. y4) +2y"+y = 3t + 4;
2 11. у" — Зу" + 2y 3Dt+e';
2 12. y() +2y" +y" +8y' – 12y = 12 sin t – e-';
y"(0) = -1, y" (0) = 2
У (0) %3D у (0) — 0, у"(0) —D 1
У (0) %3D У (0) 3 0, у"(0) —D у" (0) — 1
y(0) = 1, y'(0) =-, y"(0) = -
y(0) = 3, y'(0) = 0,

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