Laplace Transform of $e^{2t-12}u(t-6)$.

Question

I am trying to find the Laplace Transform of $e^{2t-12}u(t-6)$.

All I know is that $\mathcal{L}\{e^{-at}\} = \dfrac{1}{s+a}$ and that $\mathcal{L}\{u(t-a)\} = \dfrac{e^{-as}}{s}$.

I also know that $\mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}\mathcal{L}\{f(t)\}$

I tried dividing the exponent of $e$ by 2 to get $e^{t - 6}$, and letting $f(t) = e^{t - 6}$ so I could get $e^{-6s}\mathcal{L}\{e^{t-6}\}$ however I'm unsure how to calculate $\mathcal{L}\{e^{t-6}\}$. Any help would be appreciated.

Answer 1

Using the property \begin{align} \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}\mathcal{L}\{f(t)\} \end{align} then consider a function $f(t) = e^{p t}$ for which \begin{align} \mathcal{L}\{e^{p(t-a)} \, u(t-a)\} = e^{-as} \, \mathcal{L}\{e^{p t} \}. \end{align} Since $\mathcal{L}\{ e^{pt} \} = \frac{1}{s-p}$ then \begin{align} \mathcal{L}\{ e^{p(t-a)} \, u(t-a) \} = \frac{e^{-as}}{s-p}. \end{align} Now, for the case of the proposed problem, let $p=2$ and $a=6$ to obtain \begin{align} \mathcal{L}\{ e^{2t-12} \, u(t-6) \} = \frac{e^{-6s}}{s-2}. \end{align}