Laplace transform with heaviside step function

Hi Everyone,

I'm a little confused with this problem so I just wanted to check my solution. I have to find the Laplace Transform of the diff eq:

d^2y/dt^2+2dy/dt+2y=x(t)

where x(t) = 1, 0≤t≤1;

2, t≥2;

0 elsewhere

y(0)=0, y'(0)=1

I got the following:

Y=1/(s^2+2s+2)=1/(s+1)^2

Y=te^-t

The heaviside function is:

e^-s/s +2e^-2s/s

so would the final solution be:

te^-t=e^-s/s +2e^-2s/s??

Thanks for any help!!

Re:Laplace transform with heaviside step function

Well I seem to have a similar transform of the RHS as you but the LHS differs. The LHS seems incomplete? Is that considered the full solution to the differential equation?