Mathematics - Differential Equation Question with Solution | TestHub

$$\begin{array}{rl}& \cos xdy=y(\sin x-y)dx\\ & \frac{dy}{dx}=y\tan x-y^2\sec x\\ & \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\tan x=-\sec x\\ & \text{ Let }\frac{1}{y}=t\\ & -\frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}\\ & -\frac{dy}{dx}-t\tan x=-\sec x\Rightarrow \frac{dt}{dx}+(\tan x)t=\sec x.\\ & \text{ I.F. }=e^{f\tan xdx}=\sec x\end{array}$$ Solution is $t($ I.F) $=\int$ (I.F) $\sec xdx$ $\frac{1}{y}\sec x=\tan x+c$