I'm not sure it's as hard as you're making it out to be. First, let's find an integrating factor. We want
$py''-2py'\tan x=(py')'$
$-2p\tan x=p'$
$\frac p{p'}=-2\frac{\sin x}{\cos x}$
$\ln p=2\ln(\cos x)=\ln(\cos^2x)$
$p=\cos^2x$
Multiplying through by our integrating factor, we get
$y''\cos^2x-2y'\sin x\cos x=(y'\cos^2x)'=\sec x$
$y'\cos^2x=\ln(\sec x+\tan x)+C$
$y'=\sec^2x\ln(\sec x+\tan x)+C\sec^2x$
$y=\int\sec^2x\ln(\sec x+\tan x)dx+C\int\sec^2xdx$
The second part is simply $C\tan x$. For the first integral, we'll use integration by parts. Obviously, we want that logarithm to go away, so that's the part we'll take the derivative of.
$u=\ln(\sec x+\tan x),du=\sec x$
$dv=\sec^2xdx,v=\tan x$
$\int\sec^2x\ln(\sec x+\tan x)dx=\tan x\ln(\sec x +\tan x)-\int\sec x\tan xdx=$
$\tan x\ln(\sec x+\tan x)-\sec x$
Putting everything together, we have
$y=\tan x\ln(\sec x+\tan x)-\sec x+k_1\tan x+k_2$