Find $\mathcal{L}^{-1} \frac{9}{(s+3)^3} $

Question

Find $$\mathcal{L}^{-1}\left[ \frac{9}{(s+3)^3}\right].$$

How do I go about with the fraction inside? There is no fixed formula for this expression. I did a partial fraction of the repeated linear factors as $$9 = As^2 + 6As + 9A + Bs + 3B + C$$ This is just only going to tell me that $ 9A + 3B + C = 9$

Then I am stuck here.

Answer 1

Hint: If $\mathcal{L}(f(t)) = F(s)$ then $$\mathcal{L}\left(f(t)e^{at}\right) = F(s-a),$$ and $$\mathcal{L}\left(t^n\right) = \frac{n!}{s^{n+1}}.$$

Answer 2

Note that $L^{-1}\{F(s)\}=f(t)$, then $L^{-1}\{F(s-a)\}=e^{at}f(t)$

Therefore, we have $$L^{-1}\left\{\dfrac{9}{(s+3)^3}\right\}=e^{-3t}L^{-1}\left\{\dfrac{9}{s^3}\right\}$$ Now find $L^{-1}\left\{\dfrac{9}{s^3}\right\}$ $$L^{-1}\left\{\dfrac{9}{s^3}\right\}=L^{-1}\left\{\dfrac92\cdot\dfrac{2}{s^3}\right\}=\dfrac92L^{-1}\left\{\dfrac{2}{s^3}\right\}=\dfrac{9t^2}{2}$$

Therefore, $L^{-1}\left\{\dfrac{9}{(s+3)^3}\right\}=\dfrac{9e^{-3t}t^2}{2}$