Solution of the differential equation  $\cos x dy = y(\sin x - y ) dx$, $0<\mathrm{x}<\frac{\mathrm{\pi }}{2}$ is

cos x dy = y(sin x - y) dx
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ytan}\mathrm{x}-{\mathrm{y}}^2\sec \mathrm{x}\\ \frac{1}{{\mathrm{y}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{1}{\mathrm{y}}\tan \mathrm{x}=-\sec \mathrm{x}\\ \text{ Let }\frac{1}{\mathrm{y}}=\mathrm{t}\\ -\frac{1}{{\mathrm{y}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}\\ -\frac{dt}{\mathrm{dx}}-t(\tan \mathrm{x})=-\sec \mathrm{x}\Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}}+(\tan \mathrm{x})\mathrm{t}=\sec \mathrm{x}\end{array}$
I.F. = e ^{j tanx dx} = sec x
Solution is  $t(I.F) = \int (I. F) sec x dx\Rightarrow \frac{1}{y}secx=\int se{c}^{2}xdx$
$\frac{1}{\mathrm{y}}\sec \mathrm{x}=\tan \mathrm{x}+\mathrm{c}$