$\frac{x}{y}=e^x\left(x^2+x-4\right)+C$

$2 x y=e^x\left(x^2-2 x+4\right)+C$

$\left(x^2+2\right) y=e^x\left(x^2-2 x+4\right)+C$

$\left(x^2+2\right)^2 y=e^x\left(x^2+2 x-4\right)+C$

$x-2 y+\log |3 x-4 y+6|=C$

$5 x-15 y-4 \log |15 x-20 y-12|=C$

$5 x-15 y+14 \log |15 x-20 y-12|=C$

$8 y-4 x+\log |9 x-12 y+4|=C$

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

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

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

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

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

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

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

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