Hi I have this problem where I need to take the convolution of functions and I am not sure if I got the right answer or something close so any advice or help would be very appreciated. So here is the problem. $$t^{2} * e^{-2t}= L^{-1}(L(t^{2} * e^{-2t}))$$ $$=L^{-1}(L(t^{2}) \cdot L(e^{-2t}))$$ Which is: $$=L^{-1}((\frac{2}{s^{3}})(\frac{1}{s+2}))$$ $$=L^{-1}(\frac{2}{(s^{3})(s+2)})$$ Then I did partial fractions to solve this: $$\frac{2}{(s^{3})(s+2)}=\frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s^{3}} + \frac{D}{s+2}$$ Then I solved for the unknown coefficients and got this: $$\frac{2}{(s^{3})(s+2)}=\frac{1}{s} - \frac{1}{s+2}$$ Then we take the inverse Laplace transform of this: $$L^{-1}(\frac{1}{s} - \frac{1}{s+2})= 1- e^{-2t}$$ So my answer would be that $$t^{2} * e^{-2t}= 1- e^{-2t}$$ Now I am not sure if this is the correct answer or the most efficient way of solving a problem like this so any advice/help would be greatly appreciated.
Thanks!
Your approach is good. Using Laplace Transforms followed by Partial Fractions is probably the best way to solve this problem. (The next easiest way would be to evaluate $\int_{0}^{t}(t-\tau)^2e^{-2\tau}\,d\tau$).
When doing the partial fractions, you should have gotten:
$\dfrac{2}{s^3(s+2)} = \dfrac{\tfrac{1}{4}}{s} - \dfrac{\tfrac{1}{2}}{s^2} + \dfrac{1}{s^3} - \dfrac{\tfrac{1}{4}}{s+2}$.
Check your work on that step.