I need to find the inverse Laplace transform of the following function:
$$ F(s) = \frac{(s-2)e^{-s}}{s^2-4s+3} $$
I completed the square on the bottom and got the following:
$$ F(s) = (e^{-s}) \frac{(s-2)}{(s-2)^2+3} $$
Disregarding the $e^{-s}$ for the moment, I have a function of the following form:
$$ \frac{s-a}{(s-a)^2+b^2} = e^{at}\cos{bt}$$
That, combined with the fact that I have a function of the following form:
$$ e^{-cs}F(s) = u_c(t)f(t-c) \text{, where } u_c(t) \text{ is the unit step function}$$
Should give me the following:
$$ f(t) = u_1(t)e^{2(t-1)}\cos{(\sqrt3t-\sqrt3)} $$
But the answer in the back is
$$ f(t) = u_1(t)e^{2(t-1)}\cosh(t-1) $$
Are these the same thing, or did I do something wrong? What's the best way to go about figuring that out? Any help is appreciated. Thanks in advance!
Observe \begin{align*} \frac{(s-2)e^{-s}}{s^2-4s+3}&=\frac{(s-2)e^{-s}}{(s-2)^2-1^2}\\ \mathcal{L}^{-1}\left\{\frac{(s-2)e^{-s}}{s^2-4s+3}\right\}&=\mathcal{L}^{-1}\left\{\frac{(s-2)e^{-s}}{(s-2)^2-1^2}\right\}\\ \end{align*} Since $$\mathcal{L}^{-1}\left\{\frac{(s-2)}{(s-2)^2-1^2}\right\}= e^{2t}\cosh{t}$$ It follows $$\mathcal{L}^{-1}\left\{\frac{(s-2)e^{-s}}{s^2-4s+3}\right\}=\mathcal{U}_1(t)e^{2(t-1)}\cosh{(t-1)}$$