Solution of ${(\mathrm{x}-\mathrm{y})}^2\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{a}}^2$

$\begin{array}{l}Given{(x-y)}^2\frac{dy}{dx}=a^2\\ {(x-y)}^2=a^2\frac{dx}{dy}\\ Putx-y=v\\ \Rightarrow \frac{dx}{dy}-1=\frac{dv}{dy}\\ \Rightarrow \frac{dx}{dy}=1+\frac{dv}{dy}\\ \Rightarrow 1+\frac{dv}{dy}=\frac{v^2}{a^2}\\ \Rightarrow \frac{dv}{dy}=\frac{v^2-a^2}{a^2}\end{array}$

$\begin{array}{l}\Rightarrow \int \frac{a^2}{v^2-a^2}dv=\int dy\\ \Rightarrow a^2\frac{1}{2a}\log \left(\frac{v-a}{v+a}\right)=y+c\\ \Rightarrow a\log \left(\frac{x-y-a}{x-y+a}\right)=2y+c\\ \Rightarrow 2y=c+a\log \left(\frac{x-y-a}{x-y+a}\right)\end{array}$