Laplace Transform

Anybody feel like helping a guy practice for his differential equations final? I'm stuck on a Laplace problem.

y''+2y'+2y=h(t); y(0)=0, y'(0)=1
h(t)={0, 0≤t<π; 1, pi≤t<2π; 0, t≥2π}

I got to where I found:

y(t)=L^-1{e^(-πs)/(s[s^2+2s+2])}-L^-1{e^(-2πs)/(s[s^2+2s+2])}+L^-1{1/(s^2+2s+2)}

But I think I made a mistake somewhere because that polynomial is unfactorable and there's no inverse for it. (I thought of trying a variable change, but we never learned anything like that in class)

i think you made the problem more complex than it should be.

if you solve it right, you should just need to find the inverse laplace transorm of the function:

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

Y(s) is equal to 1/[(s+1)^2+1]... and if you look on a Laplace-1 table you can find:

exp(a*t)*sin(b*t) --> b/[(s-a)^2+b^2]

if you take b=1 and a = -1 this is the same form of the solution from Y(s), hence

the answer is exp(-t)*sin(t)


PROVE ME WRONG

Oh shit, you're right. I can't believe I forgot the first translation theorem, I was beating my head against that problem for so long (although I knew I only had to get the last one to get the other two).

Thank you, Anonymous!