yet another simple Laplace transform

Question

what is $ℒ(t^2e^{3t})$

I have got this far so far:

$=\int_{0}^\infty (t^2e^{t(3-s)})$

Integration by parts using:

$u = t^2$ and $du = 2t$

$v = \frac{e^{t(3-2)}}{3-s}$ and $dv = e^{t(3-s)}$

Which I think yields: $0 - \int_{0}^\infty \frac{2t}{3-s} e^{t(3-2)}$

and now i'm stuck. if someone could do a step by step instructions, that would be so helpful!

Answer 1

Consider $$ \frac{\partial^2}{\partial s^2}\mathcal{L}\{e^{3t}\}, $$ Can you easily determine the Laplace transform of of $e^{3t}$? If so, what is the second derivative in $s$?

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\begin{align} \frac{\partial^2}{\partial s^2}\int_0^{\infty}e^{3t}e^{-st}dt &= \int_0^{\infty}\frac{\partial^2}{\partial s^2}e^{3t}e^{-st}dt\\ &= \int_0^{\infty}t^2e^{3t}e^{-st}dt\\ &= \mathcal{L}\{t^2e^{3t}\} \end{align}

Answer 2

In order to find the Laplace Transform of:

$$t^2e^{3t}$$

Your integral should be:

$$\int_{0}^{\infty} t^2e^{3t}e^{-st}dt = \int_{0}^{\infty} t^2e^{t(3-s)}dt$$

Using integration by parts, with $u = t^2$ and $dv = e^{t(3-s)}$

We then get:

$$\int_{0}^{\infty} t^2e^{t(3-s)}dt = \frac{t^2e^{t(3-s)}}{3-s} - \int \frac{e^{t(3-s)}2t}{3-s}dt$$

You would then perform integration by parts again, this time with: $u = 2x$ and $dv = \frac{e^{t(3-s)}}{3-s}$

Can you take it from here?