# Laplace transform stuff

• Feb 13th 2006, 07:24 AM
diff'eq_austin
Laplace transform stuff
the problem is: 4y'' + 4y' + 17y = g(t); y(0)=0 , y'(0)=0

I get Y(s)=G(s)/(4s^2 + 4s + 17) ... not sure how break up the bottom... i tried doing something like (2s+1)^2 + 4^2 ... but then i dont know how to account for the 2 in front of the "s" to make it similar to the form b/((s-a)^2 + b^2) ... any thoughts?
• Feb 13th 2006, 11:39 AM
topsquark
Quote:

Originally Posted by diff'eq_austin
the problem is: 4y'' + 4y' + 17y = g(t); y(0)=0 , y'(0)=0

I get Y(s)=G(s)/(4s^2 + 4s + 17) ... not sure how break up the bottom... i tried doing something like (2s+1)^2 + 4^2 ... but then i dont know how to account for the 2 in front of the "s" to make it similar to the form b/((s-a)^2 + b^2) ... any thoughts?

I would factor the 4 out of the denominator first. Then you have to work with \$\displaystyle 1/(s^2+s+17/4)\$. Now do a partial fraction decomposition. Set \$\displaystyle 1/(s^2+s+17/4) = 1/(s+a) + 1/(s+b)\$. Once you do that, you should be able to do the inverse transform.

-Dan