# Math Help - find the inverse of the laplace transform

1. ## find the inverse of the laplace transform

ok well here is the problem.. $\mathcal{L}^{-1}\left \{ \frac{1}{s^{2}+3s} \right \}$

$\mathcal{L}^{-1}\left \{ \frac{1}{s^{2}+3s} \right \}$

and here is my answer, $\frac{1}{3}+\frac{2}{3}e^{-3t}$

im not seeing where i went wrong, used partial fractions to break it down, i figured out A to be 1/3 and B to be 2/3...

any help would be appreciated, thanks in advance

2. Your partial fractions are wrong.

It should be A/s + B/(s + 3) = 1/(s^2 + 3s)

[A(s + 3) + Bs]/[s(s + 3)] = 1/[s(s + 3)]

A(s + 3) + Bs = 1

As + 3A + Bs = 1

(A + B)s + 3A = 0s + 1

A + B = 0 and 3A = 1

A = 1/3 and B = -1/3.

So 1/(s^2 + 3s) = 1/(3s) - 1/[3(s + 3)]. Now find the inverse Laplace transform.

3. nevermind i caught it, it was soooo simple. 0 = As + Bs LOL yup im a lil tired...