Laplace transform of $u(t-5)t^2e^{-2t}$

Question

If t were equal to (t-5) I would use $\frac{n!}{(s-a)^{n+1}}$ which n is power of t and a is the power of $e^t$. However it is not. What should I do, can you give me a hint please?

Answer 1

I will use that the general form $u(t-c) g(t)$ has a Laplace transform of $e^{-cs} \mathcal{L}[g(t + c)]$. Then, in your case $g(t) = t^2 e^{-2t}$ and $c = 5$, so we simply need to find the Laplace transform of $$ (t + 5)^2 e^{-2(t+5)} = e^{-10} (t + 5)^2 e^{-2t}. $$ The constant $e^{-10}$ can be brought outside of the Laplace transform since it is a linear operator. Thus, we need to compute the following: $$ \mathcal{L}[(t + 5)^2 e^{-2t}] = \mathcal{L}[t^2 e^{-2t}] + 10 \mathcal{L}[t e^{-2t}] + 25 \mathcal{L}[e^{-2t}] $$ Each of these terms are common Laplace transforms that can be found on any Laplace transform table which gives us our answer of: $$ \mathcal{L}[u(t-5) t^2 e^{-2t}] = e^{-5s} \cdot e^{-10} \bigg( \frac{2!}{(s+2)^3} + \frac{10 \cdot 1!}{(s + 2)^2} + \frac{25}{s+2} \bigg) $$ After some simplification, we get $$ \mathcal{L}[u(t-5) t^2 e^{-2t}] =\frac{25s^2 + 110s + 122}{(s+2)^3} e^{-5(s + 2)} $$