Solving a Laplace Transform Problem: Where Am I Going Wrong?

[metdave]

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!

 

[cryora]

To get inverse laplace of 1/(2s-1) I would rewrite as (1/2)/(s-1/2) which becomes (1/2)e^(1/2t). It appears you did not include the 1/2 factor for two of you terms.

 

[lurflurf]

Two things jump out
1)in 8/(2s-1) the 8 should be 332/25
2)The convolution should involve trigonometric functions not exponents
This rule is also useful here
$$\mathcal{L}^{-1} \{ \mathrm{F}(s) \} = t \, \mathcal{L}^{-1} \left\{ \int_s^\infty \! \mathrm{F}(u) \, \mathrm{d}u \right\}$$

 

[metdave]

Thanks for the help!