Question with Solution

We have $\frac{dy}{dx}=y\tan x-y^2\sec x$

$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\tan x=-\sec x$

Putting $\frac{1}{y}=v\Rightarrow \frac{-1}{y^2}\frac{dy}{dx}=\frac{dv}{dx}$,

we obtain $\frac{dV}{dx}+\tan x.v=\sec x$

which is linear I. $F=e^{\int \tan xdx}=e^{\log \sec x}=\sec x$

∴ The solution is

$$\begin{array}{rlr}& v\sec x=\int {\sec }^{2}xdx+c\Rightarrow \frac{1}{y}\sec x=\tan x+c& \Rightarrow \sec x=y(c+\tan x)\end{array}$$