Take the homogeneous part and find the roots to the characteristic equation:
\(y''+4y=0\implies r^2+4=0\implies r=\pm2i\)
This means the characteristic solution is \(y_c=C_1\cos2x+C_2\sin2x\).
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form \(y_p=ax\cos2x+bx\sin2x\). Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With \(y_1=\cos2x\) and \(y_2=\sin2x\), you're looking for a particular solution of the form \(y_p=u_1y_1+u_2y_2\). The functions \(u_i\) satisfy
\(u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx\)
\(u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx\)
where \(W(y_1,y_2)\) is the Wronskian determinant of the two characteristic solutions.
\(W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2\)
So you have
\(u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx\)
\(u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x\)
\(u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx\)
\(u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x\)
So you end up with a solution
\(u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x\)
but since \(\cos2x\) is already accounted for in the characteristic solution, the particular solution is then
\(y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x\)
so that the general solution is
\(y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x\)
