- #1
metdave
- 5
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I'm out of college and am brushing up on Laplace Transforms. I have a problem I've solved, but I believe the solution I got is wrong and can't find my error.
The problem is 2x''-x'=t*sin(t) x(0)=5,x'(0)=3
My solution...
Take the Laplace Transform
2(s^2x-5s-3)-(sx-5)=2s/(s^2+1)^2
Rearranging, I get
x(2s^2-s)-10s-1=2s/(s^2+1)^2
Solve for x
x=(10s+1)/(2s^2-s)+2/((2s-1)(s^2+1)^2
Then, doing a PFD on the first term, I get -1/s+8/(2s-1)
Doing an inverse Laplace Transform, I get x(t)=-1+8e^(t/2)+Integral((sin(y)-ycos(y)(e^(1/2)((t-y))dy,0,y)
I used the convolution theorem on the second term on the RHS. That doesn't look right because the initial conditions aren't satisfied. Can anyone point me in the right direction?
Thanks!
The problem is 2x''-x'=t*sin(t) x(0)=5,x'(0)=3
My solution...
Take the Laplace Transform
2(s^2x-5s-3)-(sx-5)=2s/(s^2+1)^2
Rearranging, I get
x(2s^2-s)-10s-1=2s/(s^2+1)^2
Solve for x
x=(10s+1)/(2s^2-s)+2/((2s-1)(s^2+1)^2
Then, doing a PFD on the first term, I get -1/s+8/(2s-1)
Doing an inverse Laplace Transform, I get x(t)=-1+8e^(t/2)+Integral((sin(y)-ycos(y)(e^(1/2)((t-y))dy,0,y)
I used the convolution theorem on the second term on the RHS. That doesn't look right because the initial conditions aren't satisfied. Can anyone point me in the right direction?
Thanks!