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The Laplace Transform of $\frac{3}{(2s+5)^3}$ is given as $\frac{3 t^2}{16}e^{-\frac{5}{2}t}$ Can someone please walk through how this was obtained? Especially the $\frac{3}{16}$?

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The definition of Laplace Transform is $$ \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt = \hat{f}(s), $$ therefore $$ \mathcal{L}\{e^{-\alpha t} f(t)\} = \int_0^\infty e^{-(s+\alpha)t} f(t) dt = \hat{f}(s+\alpha) $$ Now, $$ \hat f \big(s + \tfrac{5}{2}\big)= \frac{3}{8}\mathcal{L}\{e^{-\frac{5}{2}t} f(t)\} $$ where $$ f(t) = \mathcal{L}^{-1} \left(\frac{1}{s^3}\right) $$ Finally $$ \mathcal{L}\{t^{n-1}\} = \frac{(n-1)!}{s^n} $$ Can you take it from here?

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Thank you!!!!!! – Shankar Kumar Nov 19 '12 at 4:21

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