<s>[INST] Solve the ODE $2y'' - y' = 2\cos(3t)$ using laplace transform 
$2y'' - y' = 2\cos(3t)$

\begin{align*}
2\mathcal{L}[y''] - \mathcal{L}[y'] &= 2\mathcal{L}[\cos3t] \\
2\bigg[-y(0) - sy(0) + s^2Y(s) \bigg] + y(0) - sY(s) &= \frac{2s}{s^2+3^2} \\
\end{align*}
This works out to $$Y(s)(s^2-s) = \frac{2s}{s^2+9} + 3 - 2s$$ and $$Y(s) = \frac{2s^3 + 3s^2-16s+27}{(2s^2-s)(s^2+9)} $$
I get the following partial fraction decomposition, but here is where I think I am messing up and getting the wrong coefficients
\begin{align*}
\frac{2s^3 + 3s^2-16s+27}{(2s^2-s)(s^2+9)} &= \frac{A}{s} + \frac{B}{2s-1} + \frac{Cs + D}{s^2+9} \\
2s^3+3s^2-16s + 27 &= 2As^3 + 18As - As^2 - 9A + Bs^3 + 9Bs + 2Cs^3 - Cs^2 + 2Ds^2 - Ds \\
2s^3+3s^2-16s + 27 &= s^3(2A+B+2C) + s^2(-A-C+2D) + s(18A+9B-D) - 9A
\end{align*}
Here is the corresponding system of linear equations 
\begin{align*}
2 &= 2A + B + 2C \\
3 &= -A - C + 2D \\
-16 &= 18A + 9B - D \\
27 &= -9A
\end{align*}
I got $A = -3, B = \frac{136}{35}, C = \frac{212}{35}, D = \frac{106}{35}$ which seems way off because the correct answer with coefficients is $y(t) = \frac{-4}{37}\cos3t - \frac{2}{111}\sin(3t) - 4 + \frac{152}{37}e^{t/2}$ 
I think something might be off even before partial fractions because $A=-3$ would imply that the corresponding laplace transform is $-3 = \frac{-3}{s}$, which doesn't even appear in the answer.
edit: Sorry I forgot the initial conditions! $y(0) = -1, y'(0) = =1$
edit: additional work on partial fractions
\begin{align*}
\frac{4s+2}{(s^2+9)(2s-1)} &=\color{blue}{2(2s+1)} \left ( \frac{A}{s^2+9} + \frac{B}{4s^2-1} \right ) \\
&=-\dfrac {2(2s+1)}{37}\left ( \dfrac{1}{s^2+9} - \dfrac{4}{4s^2-1} \right) \\
1= 4As^2-A + Bs^2 + 9B \\ \vdots \\ 
&=-\dfrac {2(2s+1)}{37}\left ( \dfrac{1}{s^2+9} - \dfrac{4}{4s^2-1} \right)
\end{align*}
I get following system of linear equations to solve for A and B and it isn't clear how the final result is reached...
\begin{align*}
0 &= 4A + B\\
1 &= -A + 9B
\end{align*}
\begin{align*}
A &= -\dfrac  1 {37}\\
B &= \dfrac 4 {37}
\end{align*}
 [/INST] You should get:
$$Y(s)(2s^2-s)+y(0)(1-2s))-2y'(0)=\dfrac {2s}{s^2+9}$$
Maybe you are given initial conditions ?
You wrote :
$$Y(s)(\color{red}{2s^2}-s) = \frac{2s}{s^2+9} + 3 - 2s$$
You should keep it this way for simplicity:
$$Y(s) = \frac{2}{(s^2+9)(2s-1)} + \dfrac { 3 - 2s}{s({2s}-1)}$$
It's easier to decompose in simple fractions.
$$Y(s) = \frac{2}{(s^2+9)(2s-1)} - \dfrac { 3}{s} + \dfrac {4}{{2s}-1}$$
$$Y(s) =-\dfrac {2(2s+1)}{37}\left ( \dfrac{1}{s^2+9} - \dfrac{4}{4s^2-1} \right) - \dfrac { 3}{s} + \dfrac {4}{{2s}-1}$$
$$Y(s) =-\dfrac {2}{37}\left ( \dfrac{(2s+1)}{s^2+9} \right)+ \dfrac{156}{37(2s-1)}  - \dfrac { 3}{s} $$
Apply inverse Laplace Transform.
$$\boxed {y(t)=-\dfrac {4}{37} \cos (3t)-\dfrac {2}{111} \sin(3t) -3 +\dfrac {78}{37}e^{t/2}}$$
</s>