ODE $x'' + 2x' +5x = \sin3t$, $x(0)=1,\ x'(0)=-1$, Solve using Laplace Transform While solving the differential equation
$$x'' + 2 x' + 5 x = \sin3t, \quad x(0) = 1, \quad x'(0) = -1$$
by use of Laplace transform I got to 
$$X(s^2 +2s+5)=\frac{(3)}{s^2 +9} +s +1$$
$$X=\frac{\left(\:s^3+s^2+9s+12\right)}{\left(s^2+9\right)\left(s^2-2s+5^{\:}\right)}$$ 
Is it correct, if yes how do i find the inverse of it?
X=f(s)
 A: Notes: 


*

*The first part of this solution is for an equation that contained an error. It remains here as a demonstration. The corrcted equation is the second half of this solution. 

*The process of the proposer is correct in process but does need some guidance in finding a nice result. 



Consider the differential equation
$$x'' + 2 x' + 5 x = 0, \quad x(0) = 1, \quad x'(0) = -1$$
and make use of 
\begin{align}
\mathcal{L}\{x''(t)\} &= s^{2} f(s) - s f(0) - f'(0) = s^{2} f(s) - s + 1 \\
\mathcal{L}\{x'(t)\} &= s f(s) - f(0) = s f(s) - 1
\end{align}
then
\begin{align}
(s^2 + 2s + 5) f(s) = s + 1
\end{align}
such that
\begin{align}
f(s) = \frac{s+1}{(s+1)^{2} + 2^{2}}.
\end{align}
Using the known transform
\begin{align}
\mathcal{L}\{ e^{-at} \, \cos(b t)\} = \frac{s+a}{(s+a)^{2} + b^{2}}
\end{align}
then the solution to the differential equation is $x(t) = e^{-t} \, \cos(2 t)$. 

Amended equation
$$x'' + 2 x' + 5 x = \sin(3 t)$$
becomes, after the Laplace transform,
\begin{align}
((s+1)^{2} + 4) \, f(s) = s+1 + \frac{3}{s^{2} + 3^{2}}
\end{align}
for which
$$f(s) = \frac{s+1}{(s+1)^{2} + 2^{2}} + \frac{1}{2} \, \frac{3}{s^{2} + 3^{2}} \cdot \frac{2}{(s+1)^{2} + 2^{2}}.$$
By using
\begin{align}
\mathcal{L}\{ e^{-at} \, \sin(bt)\} &= \frac{b}{(s+a)^{2} + b^{2}} \\
\mathcal{L}\left\{\int_{0}^{t} F(t-u) \, G(u) \, du \right\} &= f(s) \cdot g(s)
\end{align}
then
\begin{align}
x(t) &= e^{-t} \, \cos(2t) + \frac{1}{2} \, \int_{0}^{t} e^{-u} \, \sin(2u) \, \sin(3t-3u) \, du \\
&= \frac{1}{52} \, e^{-t} \, \left( 9 \, \sin(2t) + 58 \, \cos(2t) \right) - \frac{1}{26} \, \left( 2 \sin(3t) + 3 \cos(3t) \right).
\end{align}
