Inverse Laplace transform

[tony873004]

L^-1(1/(s+2)³)
I don't see this one in the table. How do I solve the inverse Laplace transform?

I know from class notes that the answer is (1/2) t²e^-2t
But I don't know to get it.
Thanks!

 

[gabbagabbahey]

I assume that $$\mathcal{L}^{-1}\left[ \frac{1}{s^3} \right]$$ is in your table?

If so, just begin by applying the frequency shift rule:

$$\mathcal{L}^{-1}\left[ f(s+a) \right]=e^{-at}\mathcal{L}^{-1}\left[ f(s) \right]$$

 

[tony873004]

ok, thanks. That gives me
$$\frac{1}{(e^{-2t}L^{-1}(s))^3}$$

And the table gives for $$L^{-1}\frac{1}{s^3}$$ as $$\frac{t^2}{2}$$

What do I do from here?

 

[gabbagabbahey]

ok, thanks. That gives me
$$\frac{1}{(e^{-2t}L^{-1}(s))^3}$$

How do you get that ugly expression?

You should get

$$\mathcal{L}^{-1}\left[ \frac{1}{(s+2)^3} \right]=e^{-2t}\mathcal{L}^{-1}\left[ \frac{1}{s^3} \right]$$