Laplace transform using 2nd shifting Theorem

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$$\mathcal{L}(U(t-5)e^{t-5} )$$

I’ve been taught this method and im not sure of and whether it will be correct, so I came here to ask.

$\mathcal{L}(U(t-5)e^{t-5} = \mathcal{L} (U(t-a) f(t-a) ) = e^{-as}F(s) $

a = 5, Therefore, $f(t-5) = e^{t-5} $

since $\mathcal{L} (f(t)) = F(s) $

$f(t) = e^t $

and $\mathcal{L} (f(t)) = \frac{1}{s-1} $

Therefore the answer is - $e^{-5s} \frac{1}{s-1} $

Am I correct ?

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Yes, you have applied $$\mathcal{L}(U(t-5)e^{t-5} = \mathcal{L} (U(t-a) f(t-a) ) = e^{-as}F(s) = e^{-5s} \frac{1}{s-1}$$

correctly to get your answer $e^{-5s} \frac{1}{s-1}$