Using the Laplace transform and showing the details solve qu

Using the Laplace transform and showing the details, solve
(question 21)

224 CHAP. 6 Laplace Transforms 31. Disch 18-27/ IVPs, SOME WITH DISCONTINUOUS Using the Laplace t 18, 4y\"-12y, + 9y = 0, y(0) = 2/3, y\'(0)-1 19, y\"-6y\' + 8y = e-t-e-4t, y(0)-1, y, (0) find in Fi INPUT the Laplace transform and showing the details, solve 4 32-34 Using the c 20. y\" + 10y, + 24y-144t2, y(0)-19/12. y\'(0) =-5 bry\" +4y = 4 cos t, if0

Solution

Laplace transform:
s^2Y(s)-sy(0)-y\'(0)+4Y(s)=4s/(s^2+1)
since y(0)=y\'(0)=0

Y(s)[s^2+4]=4s/(s^2+1)
Y(s)=s4/(s^2+1)^2
use partial fractions.
4s/(s^2+1)^2=(As+B)/(s^2+1) + (Cs+D)/(s^2+1)^2
multiply both sides by (s^2+4)^2
s=(As+B)(s^2+1)+Cs+D
s=As^3+4As+Bs^2+4B+Cs+D(I foiled)
coefficients of s^3 is 0:
0=A
coefficients of s^2 is 0:
0=B
coeffieicnets of s is 1:
1=4A+C
since A was 0. Then C=1.
coeffieicents of s^0 is 0:
0=4B+D
since B=0 then D=0 as well.
so plugging in what you know for A, B, C, and D.
Y(s)=1/(s^2+1)^2
use a covolution integral.
Y(s)=1/(s^2+1) * 1/(s^2+1)
multiply top and bottom of each by 2
Y(s)=(4/2)*2/(s^2+1) * (4/2)*2/(s^2+1)
Y(s)=2*cost * 2cost