Thank you for reporting, we will resolve it shortly
Q.
The general solution of the differential equation $\frac{d y}{d x}$ $=y \tan x-y^{2} \sec x$ is
Differential Equations
Solution:
We have
$\frac{d y}{d x}=y \tan x-y^{2} \sec x$
$\Rightarrow \frac{1}{y^{2}} \frac{d y}{d x}-\frac{1}{y} \tan x=-\sec x$
Putting $\frac{1}{y}=V \Rightarrow \frac{-1}{y^{2}} \frac{d y}{d x}=\frac{d V}{d x}$
we obtain $\frac{d V}{d x}+\tan x \cdot V=\sec x$, which is linear.
I.F. $=e^{\int \tan x d x}=e^{\log \sec x}=\sec x$
Hence, the solution is
$V \sec x-\int \sec ^{2} x d x+c $ or $ \frac{1}{y} \sec x-\tan x+c$
or $\sec x=y(c+\tan x)$