Question

In Problems 3 - 24 fill in the blanks or answer true or false.

$\mathbf{22}\mathbf{.}\mathbf{I}\mathbf{f}\mathbf{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{,}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{L}\left\{t{e}^{8t}f\left(t\right)\right\}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Short answer

$\mathcal{L}\left\{t{e}^{8t}f\left(t\right)\right\}=-\frac{d}{ds}\left[F\right(s-8\left)\right]$

Step-by-step solution

The first translation equations

If$\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$then$\mathcal{L}\left\{t{e}^{8t}f\left(t\right)\right\}$since, we know from First Translation Theorem$\mathcal{L}\left\{e^{at}f\left(t\right)\right\}=F(s-a)$

And from differential form of Laplace transform

$\mathcal{L}\left\{t^{n}f\left(t\right)\right\}=(-1{)}^n\frac{d^n}{d{s}^n}\left(F\right(s\left)\right)$

Thus,

$\mathcal{L}\left\{t{e}^{8t}f\left(t\right)\right\}=-\frac{d}{ds}\left[\mathcal{L}\left\{e^{8t}f\left(t\right)\right\}\right]=-\frac{d}{ds}\left[{\left.\mathcal{L}\left\{f\right(t\left)\right\}\right|}_{s\to s-8}\right]$

$=-\frac{d}{ds}\left[F\right(s-8\left)\right]$

Final proof

$\mathcal{L}\left\{t{e}^{8t}f\left(t\right)\right\}=-\frac{d}{ds}\left[F\right(s-8\left)\right]$