# ODE Review

1. May 8, 2014

### 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!

2. May 10, 2014

### 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.

3. May 10, 2014

### 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
$$\mathcal{L}^{-1} \{ \mathrm{F}(s) \} = t \, \mathcal{L}^{-1} \left\{ \int_s^\infty \! \mathrm{F}(u) \, \mathrm{d}u \right\}$$

4. May 11, 2014

### metdave

Thanks for the help!