Solution of ${\mathrm{y}}^2\mathrm{dx} = (\mathrm{xy} - {\mathrm{x}}^2 )\mathrm{dy}$ given that y = 1 when x = 1

$\begin{array}{l}Given\frac{dx}{dy}=\frac{xy-x^2}{y^2}\\ \Rightarrow \frac{dx}{dy}=\frac{x}{y}-{\left(\frac{x}{y}\right)}^2\\ Putx=vy\Rightarrow \frac{dx}{dy}=v+y.\frac{dv}{dy}\\ \Rightarrow v+y\frac{dv}{dy}=v-y^2\\ \Rightarrow y\frac{dv}{dx}=-v^2\\ \Rightarrow \int \frac{-1}{v^2}dv=\int \frac{dy}{y}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{v}=\log y+c\\ \Rightarrow \frac{y}{x}=\log y+c\\ Putx=1,y=1\\ \Rightarrow 1=\log 1+c\\ \Rightarrow c=1\\ \Rightarrow \frac{y}{x}=\log y+1\\ \Rightarrow y=x\log y+x\end{array}$