The solution of the equation $\frac{dy}{dx}=\frac{x+y}{x-y}$, is

We have,

$\frac{dy}{dx}=\frac{x+y}{x-y}$

Substituting $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$ we get

$\begin{array}{r}v+x\frac{dv}{dx}=\frac{1+v}{1-v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1-v}\\ \Rightarrow \frac{1-v}{1+v^2}dv=\frac{dx}{x}\end{array}$

$\begin{array}{l}\Rightarrow {\tan }^{-1}v-\frac{1}{2}\log \left(1+v^2\right)=\log x+\log C\text{ [On integrating] }\\ \Rightarrow {\tan }^{-1}v=\log \left\{Cx\sqrt{1+v^2}\right\}\\ \Rightarrow C{\left(x^2+y^2\right)}^{1/2}=e^{{\tan }^{-1}y/x}\end{array}$