Question

Dealing with a Laplace Transform of a given function times a Heaviside function. Posted below in a moment.

Answer 1

\[L\left\{(t-2)^2e^{(t-2)}U(t-2)\right\}\]

Answer 2

Already in a convenient enough form for using it with the Unit Step Function;\[a=2, \ f(t)=t^3e^t\]

Answer 3

\[L(t^3e^t)=\int\limits_{0}^{\infty}t^3e^te^{-st}dt=\int\limits_{0}^{\infty}t^3e^{(1-s)t}dt\]

Answer 4

@Concentrationalizing Do you think I should really be finding the laplace this way, or is there something else I should be doing here?

Answer 5

Shouldnt be necessary. Your f(t) is \(t^{2}e^{t}\). So if you can find \[L \left\{ t^{2}e^{t} \right\}\] then you should be okay to just multiply that by \(e^{-2s}\) to get your final answer (the \(e^{-2s}\) coming from the unit step function part and the formula that requires \(e^{-as}F(s)\) )

Answer 6

Whoops, that's supposed to be cubed.

Answer 7

Oh! Nevermind, shifting.

Answer 8

Yeah, no sort of convolution needed, they're all regular formulas

Answer 9

\[L\left\{\vphantom{} t^3e^t\right\}=L\left\{\vphantom{}t^3\right\}\Bigg|_{s \rightarrow s-1}\]

Answer 10

\[L\left\{\vphantom{} t^3\right\}\Bigg|_{s \rightarrow s-1}=\left\{\vphantom{}\frac{3!}{s^4}\right\}\Bigg|_{s \rightarrow s-1}\]

Answer 11

Okay, so now you need the shift that \(e^{t}\) causes.

Answer 12

Alright, site it back for me, one moment.\[\left\{\frac{3!}{s^4}\right\}\Bigg|_{s \rightarrow s-1}=\frac{6}{(s-1)^4}\]

Answer 13

Moving from here,\[\frac{6e^{-2s}}{(s-1)^4}\]

Answer 14

Looks good to me. See, not that bad :)

Answer 15

Thank you!

Answer 16

Yep, no problem :)