Auxillary equation is $D^2+4 = 0 \\ D^2 = -4 = 4i^2 \\ D = ±2i \\ Y_c = C_1 \cos 2x + C_2 \sin 2x \\ \text{Let} y_1 = \cos 2x \space \text{and} \space y_2 = \sin 2x \\$ Now $$w=\begin{vmatrix} y_1 & y_2\ y_1' & y_2'\ \end {vmatrix}$$

$$=\begin{vmatrix} \cos 2x & \sin 2x\\ -\sin 2x.2 & \cos 2x.2\\ \end {vmatrix}$$

$=2\cos^22x+2\sin^22x\\ =2 \\ \mu=-\int\dfrac {y_2x}2 dx \\ =-\int \dfrac {\sin2x}2\tan2x \space dx\\ =-\dfrac 12\int \dfrac {\sin^22x}{\cos 2x}dx \\ =-\dfrac 12\int \dfrac {1-\cos^22x}{\cos 2x}dx \\ =-\dfrac 12\int (\sec 2x-\cos 2x)dx \\ = -\dfrac 12\Bigg[\dfrac {\log\sec 2x+\tan 2x}2-\dfrac {\sin 2x}2\Bigg] \\ =\dfrac 14[\sin 2x-\log (\sec 2x+\tan 2x)]$

And $v=\int\dfrac {y_1x}wdx \\ v=\int\dfrac {\cos 2x}2\tan 2x \space dx\\ =\dfrac 12 \int \cos 2x\dfrac {\sin 2x}{\cos 2x}dx \\ =\dfrac 12\int \sin 2x\space dx \\ =\dfrac 12\Bigg[\dfrac {-\cos 2x}2\Bigg] \\ =\dfrac {-\cos 2x}4$

Complete solution is $y_p + y_c \\ y= C_1 \cos 2x + C_2 \sin 2x +\dfrac 14\cos 2x \log (\sec 2x+\tan 2x)$