$(1+\sin x) y=n \cos x+C$

$(1+\cos x) y=x \sin x+C$

$(\sec x+\tan x) y=x \sec x+C$

$(\sec x+\tan x) y=x+C$

$$ \begin{aligned} & (\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x} -4(\sin 2 x+\cos 2 x) y=0 \end{aligned} $$

$$ \begin{aligned} & (\cos 2 x+\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x} -4(\sin 2 x-\cos 2 x) y=0 \end{aligned} $$

$$ \begin{aligned} & (\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+(4 \sin 2 x) \frac{d y}{d x} +4(\sin 2 x+\cos 2 x) y=0 \end{aligned} $$

$$ \begin{aligned} & (\sin 2 x-\cos 2 x) \frac{d^2 y}{d x^2}-(4 \sin 2 x) \frac{d y}{d x} -4(\sin 2 x+\cos 2 x) y=0 \end{aligned} $$

$\log \left|\frac{\tan \frac{(x-y)}{2}+1}{\tan \frac{(x-y)}{2}}\right|=x+C$

$\log \left|\frac{\tan \frac{(x-y)}{2}-1}{\tan \frac{(x-y)}{2}}\right|=x+C$

$\log \left|\frac{\tan (x-y)-1}{\tan (x-y)}\right|=x+C$

$\log \left|\frac{\sin (x-y)+\cos (x-y)}{\cos (x-y)}\right|=x+C$

$y^2=3 x^2 \log (C x)$

$y^2=\log x+C$

$y \log x=x+C y$

$y \log x=x^2+C$