Answer
$y=\ln (\sec x)+C\ln(\tan x + \sec x)+k$
Work Step by Step
We are given:
$y''-y'\tan x=1$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dx}=\frac{d^2y}{dx^2}$
Subsitute to the equation:
$\frac{du}{dx}-u\tan x=1$
Integrating factor:
$I=e^{\int -tan x dx}=e^{-(-\ln |\cos x|)}=\cos x$
The equation becomes:
$\frac{d}{dx}(\cos x)=\cos x$
Integrating the last equation:
$u \cos x=\sin x + C$
$u=\frac{\sin x +C}{\cos x}$
$\frac{dy}{dx}=\frac{\sin x+ C}{\cos x}$
$dy = \frac{\sin x +C}{\cos x}dx$
$dy=(\tan x +\frac{C}{\cos x})dx$
Integrating the last equation:
$y=-\ln |\cos x|+C \ln|\tan x +\sec x|+C_1$
$y=\ln (\sec x)+C\ln(\tan x + \sec x)+C_1$
The final solution is:
$y=\ln (\sec x)+C\ln(\tan x + \sec x)+k$
