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I'm supposed to evaluate: $L\{t^{2}e^{7t}\sinh(3t)\}$.

I know that this can be broken using $(-1)^n \frac{d^n}{ds^n}F(s)$ so I end up with $(-1)^2 \frac{d^2}{ds^2}L\{e^{7t}\sinh(3t)\}$. I'm not sure how to proceed from here, as there are now two functions inside of the Laplace transform. Do I use some sort of unit step function? I saw in my notes that $L\{e^{-as}F(s)\}=f(t-a)U(t-a)$ but I'm not sure how to use this, or if I'm supposed to use it as all.

Any help would be greatly appreciated, thank you.

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up vote 2 down vote accepted

First, use

$$\sinh{x} = \frac{1}{2} (e^x - e^{-x})$$

as well as

$$\int_0^{\infty} dt \: t^2 e^{-a t} = \frac{2}{a^3}$$

Note that

$$\int_0^{\infty} dt \: t^2 \, e^{7 t} \sinh{3 t} e^{-s t} = \frac12 \int_0^{\infty} dt \: t^2 e^{-(s-10) t} - \frac12 \int_0^{\infty} dt \: t^2 e^{-(s-4) t} $$

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So from that, I get $\frac{1}{2}(-1)^2\frac{d^2}{ds^2}L\{e^{10t}\}-L\{e^{4t}\}$ – Manu Juneja Apr 23 '13 at 20:14
Which reduces to the answer, $\frac{1}{2}(-1)^2\frac{d^2}{ds^2}(\frac{1}{s-10}-\frac{1}{s-4})$? – Manu Juneja Apr 23 '13 at 20:16
@user1066886: it's even simpler than that. Forget the derivatives; use the hints I gave you. – Ron Gordon Apr 23 '13 at 20:30
So with your method, I'm resulting in $$\frac{1}{2}\int_0^{\infty} dt \: t^2 e^{10t}-t^2e^{4t}$$ – Manu Juneja Apr 23 '13 at 20:46
And by splitting it up, I result in $$\frac{1}{10^3}-\frac{1}{4^3}$$, is that correct? – Manu Juneja Apr 23 '13 at 20:47

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