Question

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

$\mathbf{23}\mathbf{.}\mathbf{I}\mathbf{f}\mathbf{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{k}\mathbf{>}\mathbf{0}\mathbf{,}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{L}\left\{e^{at}f(t-k)U(t-k)\right\}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Short answer

$\mathcal{L}\left\{e^{at}f(t-k)U(t-k)\right\}=e^{-k(s-a)}F(s-a)$

Step-by-step solution

The Laplace equations

$\mathcal{L}\left\{e^{at}{e}^{ak}{e}^{-ak}f(t-k)U(t-k)\right\}$

$=e^{ak}\mathcal{L}\left\{e^{a(t-k)}f(t-k)U(t-k)\stackrel{(*)}{=}\right\}{e}^{-k(s-a)}F(s-a)$

Thus,

$\mathcal{L}\left\{e^{at}f(t-k)U(t-k)\right\}=e^{-k(s-a)}F(s-a)$

If$F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}$and$a>0$, then$\mathcal{L}\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

If $\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$and is any real number, then$\mathcal{L}\left\{e^{at}f\left(t\right)\right\}=F(s-a)$

Final proof

$\mathcal{L}\left\{e^{at}f(t-k)U(t-k)\right\}=e^{-k(s-a)}F(s-a)$