Question

In problems 37-48 find either F(s) or f(t), as indicated.

$\mathit{L}\left\{e^{2-t}\mathcal{U}(t-2)\right\}$

Short answer

The f(t) of $\mathcal{L}\left\{e^{2-t}\mathcal{U}(t-2)\right\}$is obtained as$\mathcal{L}\left\{e^{2-t}\mathcal{U}(t-2)\right\}=\frac{e^{-2s}}{s+1}$

Step-by-step solution

Define the second Translation theorem

$\mathcal{L}\left\{f\right(t-a\left)u\right(t-a\left)\right\}=e^{as}F\left(s\right)\mathrm{Where}F\left(s\right)=\mathcal{L}\left(f\right(t\left)\right)$

Use Laplace Transform

The Laplace transform of the function given is

$\mathcal{L}\left\{e^{2-t}u(t-2)\right\}=\mathcal{L}\left\{e^{-(t-2)}u(t-2)\right\}$

Applying the theorem, we get

$\mathcal{L}\left\{e^{-(t-2)}u(t-2)\right\}=e^{-2s}\mathcal{L}\left\{e^{-t}\right\}$

Simplify

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

So, on further simplification, we get

$\mathcal{L}\left\{e^{2-t}\mathcal{U}(t-2)\right\}=\frac{e^{-2s}}{s+1}$

Thus, the solution is $\mathcal{L}\left\{e^{2-t}\mathcal{U}(t-2)\right\}=\frac{e^{-2s}}{s+1}$