Solve the Differential equations y'' + 4y = 4 tan^{2} x

We need to solve following differential equation
$y"+4y=4{\tan }^2\hspace{0.25em}x$
Let us first solve the homogeneous part.
$y"+4y=0$
Let $y=e^{m}x$ be the trial solution. Then we get the auxiliary equation
$m^2+4=0$
This gives us
$m=\pm 2i$
Thus the solution of the homogeneous part of the differential equation be
$y_h(x)=c_1\cos (2x)+c_2\sin (2x)$
We use variation of parameter to calculate a
particular solution y_h. Let us take
$y_1=\cos 2x,y_2=\sin 2x$
Then we get
$y{‘}_1=-2\sin 2x,y{‘}_2=2\cos 2x$
The wro
ian of $y_{1}and{y}_2$ is
$W=y_{1}y{‘}_2-y{‘}_1{y}_2$
$=(\cos 2x)(2\cos 2x)-(-2\sin 2x)(\sin 2x)=2\ne 0$
In this case it is given that
$f(x)=4\tan (2x)$
Then by the formula we obtain
$u_1=\int \frac{-y_{2}f}{W}dx$
$=\int \frac{-(\sin 2x)(4\tan 2x)}{2}dx$
$=-2\int \frac{{\sin }^2(2x)}{\cos (2x)}dx$
$=-2\int (\frac{1-{\cos }^{2}2x}{\cos 2x})dx$
$=2\int (\cos 2x-\sec 2x)dx$
$=2\int (\cos 2x-\sec 2x)dx$
$=\sin 2x-\ln |\sec 2x+\tan 2x|$
Similarly
$u_2=\int \frac{-y_{1}f}{W}dx$
$=\int \frac{(\cos 2x)(4\tan 2x)}{2}dx$
$=2\int \sin (2x)dx$
$=-\cos (2x)$
Thus by variation of parameters the patricular solution is
$y_p=u_1{y}_1+u_2{y}_2$
$=(\sin 2x-\ln |\sec 2x+\tan 2x|)(\cos 2x)+(-\cos 2x)(\sin 2x)$
$y_p=-\cos 2x\ln |\sec 2x+\tan 2x|$
Hence the general solution is
$y=y_c+y_p=c_1\cos 2x+c_2\sin 2x-\cos 2x\ln |\sec 2x+\tan 2x|$