Question regarding $\mathcal{L}^{-1}\{\frac{s}{s^2+4s+5}\}$

Question

The book asks for: $\mathcal{L}^{-1}\{\frac{s}{s^2+4s+5}\}$

So I can see: $\frac{s}{s^2+4s+4+1} = \frac{s}{(s+2)^2+1}$

From the properties of the inverse Laplace transform: $\mathcal{L}^{-1}\{F(s)\}|_{s->s-a} = e^{at}f(t)$

$\mathcal{L}\{\cos (kt)\} = \frac{s}{s^2+k^2}$, so I believe the answer should be $e^{-2t}\cos{t}$, but the book shows:

$e^{-2t}\cos{t}-2e^{-2t}\sin{t}$

I can't seem to figure out why this is. Any help would be greatly appreciated.

Answer 1

When using $\mathcal{L}^{-1}\left\{{f(s-a)}\right\}= e^{at}\mathcal{L}^{-1}\left\{{f(s)}\right\}$, You have replaced $s-a$ with $s$ in the denominator of your transform, but not the numerator.

Add $+2-2$ to the numerator and use linearity of $\mathcal{L}^{-1}$ to split the transform and you should get the book's answer.