Find the inverse Laplace transform of F(s) = (2s^2 + 20s + 36)/(s^3 + 6s^2 + 9s). I partially decomposed the fractions, and I don't really know where to go from there?

Question

anonymous · Answer

What did you get from the partiall decomposition?

anonymous · Answer

one sec, I'm typing it

anonymous · Answer

$$F(s) = (2s ^{2} + 20s + 36) [1/(9s) - 1/([9(s+3)]) - 1/[3(s+3)^2]$$

anonymous · Answer

so?

anonymous · Answer

I'm lost. I'm assuming I then distribute the first polynomial and then Laplace each thing, but they don't seem to fit any basic transformations.

anonymous · Answer

mmm wait

anonymous · Answer

I think you can simplify the expression before the partial fractions.

anonymous · Answer

To something like this:
$$\frac{2}{s} \frac{s^2+6s+9+14s+27}{s^2+6s+9}$$

anonymous · Answer

or$$\frac{2}{s}\frac{14s+27}{s^2+6s+9}$$

anonymous · Answer

hmm... that seemed to help. thanks!

anonymous · Answer

You're welcome