The solution of the differential equation  $\frac{dy}{dx}-2y\tan 2x=e^x\sec 2x$ is 

$\frac{dy}{dx}+\left(-2Tan2x\right)y=e^x\sec 2x$

$I.F=e^{\int p\text{ dx}}$

$\begin{array}{l}=e^{-2\int Tan2x\text{ dx}}\\ =e^{\cancel{2}\frac{\log \left|\cos 2x\right|}{\cancel{2}}}\\ =\cos 2x\end{array}$

y.I.F $=\int Q.I.Fdx$

$y\text{ x}\cos 2x=\int e^x\sec 2x\cos 2xdx$

$=\int e^{x}dx$

$=e^x+c$