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The Laplace Transform I'm having trouble with is:

$$f(t) = 6te^{-9t}\sin(6t)$$

I'm not sure what the protocol is for multiplying t into it.

The Laplace Transform for $f(t) = 6e^{-9t}\sin(6t)$ is $\dfrac 6{(s+9)^2 - 36}$.

Can't figure out how to add in the $t$.

Thanks in advance for your help.

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If $\displaystyle F(s) = L[f(t)](s) = \int_0^\infty f(t)e^{-st} \ dt $

and

$ \displaystyle G(s) = L[tf(t)](s) = \int_0^\infty tf(t)e^{-st} \ dt$

then

$\displaystyle {dF \over ds} = {d\ \over ds} \int_0^\infty f(t)e^{-st} \ dt = \int_0^\infty {\partial\ \over \partial s} f(t)e^{-st} \ dt = \int_0^\infty -tf(t)e^{-st} \ dt = - \int_0^\infty tf(t)e^{-st} \ dt = -G(s)$

Hence $\displaystyle G(s) = -{dF \over ds}$.

Thus $\displaystyle L[ 6te^{-9t}\sin(6t)](s) = -{d\ \over ds}\left(\dfrac 6{(s+9)^2 - 36}\right) = \ ...$

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Here is a partial solution: \begin{align} \mathcal{L}\{f(t)\}&=\mathcal{L}\{e^{-9t}[6t\sin(6t)]\} \\ &=\mathcal{L}\{6t \sin(6t)\} \vert_{s \to s+9} & \text{First Translation Theorem}\\ &= 6 \mathcal{L}\{t\sin(6t)\} \vert_{s \to s+9} & \text{linearity} \\ &=-6\frac d{ds} \mathcal{L}\{\sin(6t)\} \vert_{s \to s+9} & \text{transform derivative principle} \\ &= -6 \left. \frac d{ds} \frac {6}{s^2+36} \right\vert_{s \to s+9} \end{align} Now evaluate $\frac d{ds} \frac {6}{s^2+36}$, then replace all the $s$ in your derivative with $s+9$.

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