Solving an integral using Laplace transform and inverse Laplace transform I want to solve this integral equation using Laplace:
$$ Y(t) + 3{\int\limits_0^t Y(t)}\operatorname  d\!t = 2cos(2t)$$
if $$ \mathcal{L}\{Y(t)\} = f(s)$$
then, $$ f(s) + 3 \frac{f(s)}{s} = \frac{2s}{s^2+4} $$
doing some operations I obtain
$$ f(s) = \frac{2s^2}{s^3+3s^2+4s+12} $$
and using Ruffini in the denominator
$$ f(s) = \frac{2s^2}{(s+3)(s^2+4)} $$
How can I calculate the inverse Laplace transform of the following function to get the solution?
 A: Using partial fractions, we can obtain
$$
\begin{align}
\frac{2s^2}{(s+3)(s^2+4)}&=\frac{8(s-3)}{13(s^2+4)}+\frac{18}{13(s+3)}\\
&=\frac1{13}\left(\frac{8s}{s^2+2^2}-\frac{12\cdot2}{s^2+2^2}+\frac{18}{s-(-3)}\right).
\end{align}
$$
Take the inverse we will obtain
$$
\frac1{13}\left(8\cos2t-12\sin2t+18\ e^{\large-3t}\right).
$$
A: Building on my comment, try the following:
$$f(s)=\frac{2s^2}{(s+3)(s^2+4)} = \frac{As+B}{s^2+4} + \frac{C}{s+3}$$
So $As(s+3) + B(s+3) + C(s^2+4) = 2s^2$, whence\
\begin{align*}
A+C &= 2\\
3A + B &= 0\\
3B+4C &= 0
\end{align*}
We can shortcut the solution a bit, as follows:
Plug in $s = -3$, we get: $13C = 18$, whence $C = \frac{18}{13}$.
Plug in $s = 0$, we get $3B+4C = 0$, so $B = -\frac{24}{13}$.
Then using the first equation, we get $A = \frac{8}{13}$.
So we have
\begin{align*}
f(s) &= \frac{\frac{8}{13}s}{s^2+4} - \frac{\frac{24}{13}}{s^2+4} + \frac{\frac{18}{13}}{s+3}\\
&= \frac{1}{13}\left(8\frac{s}{s^2+4} - 12\frac{2}{s^2+4} + \frac{18}{s+3}\right)
\end{align*}
And at this point I can refer you to a table of Laplace transforms (which implicitly gives the termwise inverse Laplace transform).
$$f(t) = \frac{1}{13}\left(8\cos(2t) - 12\sin(2t) + 18e^{-3t}\right)$$
