ODE help with initial conditions Initial conditions are $f(0)=1, f'(0)=0$
Applying the Laplace Transform to both sides and simplifying I reach the form
$$ L(y) = \frac{s^3 + 5s}{s^2 + 4}, $$ 
which I then further simplify in the following steps
$$ L(y) = \frac{s(s^2 + 5)}{(s^2 + 4)(s^2 + 9)} $$
$$ L(y) = \frac{s(s^2 + 4 + 1)}{(s^2 + 4)(s^2 + 9)}$$
$$ L(y) = \frac{s}{s^2 + 9} + \frac{s}{(s^2 + 4)(s^2 + 9)}. $$
I know the first term to be the Laplace of $\cos(3x)$ , but when I solve for the other term I reach a result of 
$$ L^{-1}\biggl[\frac{s}{(s^2 + 4)(s^2 + 9)}\biggr] = \frac{4}{5}\cos(3x) + \frac{1}{5}\cos(2x) $$
Giving a final result of 
$ y = \cos(3x) + \frac{4}{5}\cos(3x) + \frac{1}{5}\cos(2x) $
Which I know to be incorrect after approaching the problem with the method of undetermined coefficients. Any help would be appreciated.

$$y'' + 9y = \cos(2x)$$
$$\implies r=\pm3i$$
$$y(x)=c_1 \cos 3x + c_2 \sin 3x$$
Then for the particular solution
$$y_p=A\cos (2x) \implies A=\frac 15$$
The solution is therefore:
$$y(x)=c_1 \cos (3x) + c_2 \sin (3x) +\dfrac 15 \cos (2x)$$
Apply initial conditions:
$$c_2=0 \text { and } c_1=\frac 45$$
So that:
$$y(x)=\frac 45 \cos (3x)+\dfrac 15 \cos (2x)$$
Note that you have a little mistake in the fraction decomposition
$$ g(s)=\frac{s}{(s^2 + 4)(s^2 + 9)}=\frac s 5 \left(\frac{1}{(s^2 + 4)}-\frac{1}{(s^2 + 9)} \right)$$
$$\implies g(x)=\frac 1 5 (\cos (2x)-\cos(3x))$$
And you get the same answer:
$$y(x)=\frac 45 \cos (3x)+\dfrac 15 \cos (2x)$$
