Almost done solving IVP using Laplace transform .. Need advice/guidance I have the following IVP:
$$\begin{cases}y''-2y'+5y = -8e^{-t},\\ \ \\  y(0) = 2\\ \ \\ y'(0) = 12\end{cases}$$
I will show my steps so far. 
After taking the transform of each term I get:
$$s^2Y(s) - 2sY(s) + 5Y(s) - 2s - 8 = \frac {-8}{s+1}$$
I simplified the L.H.S and got:
$$ Y(s)(s^2-2s+5) = \frac {-8}{s+1} + 2s+ 8 $$ 
Then the R.H.S becomes:
$$ Y(s)(s^2-2s+5) = \frac {2s^2+10s}{s+1}$$
I then get:
$$Y(s) =  \frac {2s^2+10s}{(s+1)(s^2-2s+5)}$$
I then used partial fraction decomposition and got $A = -1, B = 3$, and $C = 5$
I then have the following:
$$Y(s) = \frac {-1}{s+1} + \frac {3s+5}{s^2-2s+5}$$
I am now stuck finding the inverse Laplace transform of the second term. 
Please can someone check my work and let me know if I did this right and, if so, what is that inverse Laplace transform I mentioned at the last step? 
Help is greatly appreciated. Thank you. 

You have  $s^2-2s+5=(s-1)^2+4$. So the first term looks like 
$$
\frac {3s+5}{(s-1)^2+4}
=\frac {3(s-1)+8}{(s-1)^2+4}
=\frac {3(s-1)}{(s-1)^2+4}
+\frac {8}{(s-1)^2+4}.
$$
Since $\frac{s}{s^2+4}$ is the transform of $\cos 2t$ and $\frac2{s^2+4}$ of $\sin 2t$, with the shifting rule you get 
$$
\mathcal L^{-1}\left[\frac {3s+5}{s^2-2s+5}\right]
=\mathcal L^{-1}\left[\frac {3s+5}{(s-1)^2+4}\right]
=3e^t\cos2t+4e^t\sin 2t
$$
