Write your general solution as $ y_g(x) = y_c(x) + y_p(x) $, then use the initial conditions to find $c_1,c_2$. Note that, you have two constants to determine, so you need two equations in $c_1,c_2$. That means, you will use $y_g(x)$ and $y'_g(x)$. Once you find $c_1$ and $c_2$ plug them back in $y_g(x)$. The final answer is
$$ y(x) = -\frac{2}{5}\,{{\rm e}^{-3\,x}}+\frac{1}{5}\,{{\rm e}^{-2\,x}}+\frac{1}{5}\,\cos \left(x \right) + \frac{1}{5} \,\sin \left( x \right)$$
$$y_p = A\sin{kx} + B\cos{kx}$$
where k is the coefficient for the trig function radian measure.
$$y'p = -2A\sin{2x} + 2B\cos{2x}$$
$$y''p = -4A\cos{2x} - 4B\sin{2x}$$
A problem results here. When taking $k=2$, you should see
$$y_p = A \sin 2x + B \cos 2x$$
and then you get the derivatives
$$y_p' = 2A \cos 2x - 2B \sin 2x \;\;\;\;\; y_p'' = -4A \sin 2x - 4B \cos 2x$$
You seem to have forgotten to taken the derivative of the trigonometric functions themselves when getting $y_p'$.
This is mostly just an arithmetic error, though, and the overarching idea (and the complimentary solution) are correct.
Edit:
After an edit made to the OP, that error was fixed. The other error that results is finding $A,B$, the latter in particular.
$A=3/26$ is correct. However, an arithmetic error seems to have resulted in finding $B$. Using substitution into $-4A-6B=0$,
$$-4\left( \frac 3 {26} \right) - 6B = 0 \implies B = \frac{-1}{6} \cdot 4\left( \frac 3 {26} \right) = \frac{-12}{156} = \frac{-1}{13} \ne \frac{-12}{216} = \frac{-1}{18}$$
Best Answer
This problem can be easily solved directly. Multiplying both sides by $e^{2x}$ yields
$$e^{2x}y''+4e^{2x}y'+4e^{2x}y=(e^{2x}y)''=5x$$
Integrating twice and multiplying by $e^{-2x}$ gives the solution you've obtained.