Question

Show that the solution of $\frac{dy}{dx}+y\tan x-\sec x=0$ is $y\sec x=\tan x+c.$

Answer 1

$\frac{dy}{dx}+\tan xy=\sec x$

This is a linear differential equation

Here, $P=\tan x$, $Q=\sec x$
$IF=e^{\int \tan x.dx}$
$=e^{ln|\sec x|}$
$=\sec x$.
Hence, the solution is given by
$\sec x.y=\int se{c}^{2}xdx$
$\sec x.y=\tan x+C$.