Particular solution of $y''+4y=x\cos x$ 
Find both the general solution $y_h$ and a particular solution $y_p$ for
  $$y''+4y=x\cos x$$

So far I've got $y_h$ from factoring the characteristic polynomial:
$$y_h=C_1\sin2x+C_2\cos2x$$
But the $y_p$ part troubles me, any help?

For $y_p$ use method of variation of parameter
$$
\begin{aligned}
& W=\left|\begin{array}{cc}
y_{1} & y_{2} \\
y_{1}^{\prime} & y_{2}^{\prime}
\end{array}\right| \\
& W=\left|\begin{array}{cc}
\sin 2 x & \cos 2 x \\
2 \cos 2 x & -2 \sin 2 x
\end{array}\right|=-2 \\
\therefore y_{p} &=-y_{1} \int \frac{y_{2} x}{w} d x+y_{2} \frac{y_{1} x}{w} d x .
\end{aligned}
$$
$y_{1}=\sin 2 x\\
y_{2}=\cos 2 x
$
[Solution by complement function.]
$x=x \cos x$ [Forcing function]
Original image
See the steps for finding particular integral.
Try using variation of parameter.
