Solution of the differential equation cos x d y=y(sin x-y) d x, 0 < x < π 2 is

$$\begin{aligned} & \cos x d y=y(\sin x-y) d x \\ & \frac{d y}{d x}=y \tan x-y^2 \sec x \\ & \frac{1}{y^2} \frac{d y}{d x}-\frac{1}{y} \tan x=-\sec x \\ & \text { Let } \frac{1}{y}=t \\ & -\frac{1}{y^2} \frac{d y}{d x}=\frac{d t}{d x} \\ & -\frac{d y}{d x}-t \tan x=-\sec x \Rightarrow \frac{d t}{d x}+(\tan x) t=\sec x . \\ & \text { I.F. }=e^{f \tan x d x}=\sec x \end{aligned}$$​cosxdy=y(sinx−y)dxdxdy​=ytanx−y2secxy21​dxdy​−y1​tanx=−secx Let y1​=t−y21​dxdy​=dxdt​−dxdy​−ttanx=−secx⇒dxdt​+(tanx)t=secx. I.F. =eftanxdx=secx​ Solution is $t\left(\right.$t( I.F) $\displaystyle =\int$=∫ (I.F) $\sec x d x$secxdx $\frac{1}{y} \sec x=\tan x+c$y1​secx=tanx+c