The general solution of \(\frac{d y}{d x}=\frac{x+y}{x-y}\) is

\(\begin{aligned} & \frac{d y}{d x}=\frac{x+y}{x-y} \\ & \text { Put } y=v x \\ & \therefore \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \therefore v+x \frac{d v}{d x}=\frac{x+v x}{x-v x}=\frac{1+v}{1-v} \\ & \therefore x \frac{d v}{d x}=\frac{1+v}{1-v}-v=\frac{1+v-v+v^2}{1-v}=\frac{1+v^2}{1-v} \\ & \therefore \int \frac{(1-v)}{1+v^2} d v=\int \frac{d x}{x} \\ & \therefore \int \frac{1}{1+v^2} d v-\frac{1}{2} \int \frac{2 v}{1+v^2} d v=\int \frac{d x}{x} \\ & \therefore \tan ^{-1}(v)-\frac{1}{2} \log \left|1+v^2\right|=\log |x|+c \\ & \therefore \tan ^{-1}\left(\frac{y}{x}\right)-\frac{1}{2} \log \left|1+\frac{y^2}{x^2}\right|-\frac{1}{2} \log \left|x^2\right|=c\end{aligned}\)
\(\begin{aligned} & \tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\frac{1}{2}\left[\log \left|\frac{\mathrm{x}^2+\mathrm{y}^2}{\mathrm{x}^2}\right|+\log \left|\mathrm{x}^2\right|\right]=\mathrm{c} \\ & \tan ^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)-\frac{1}{2} \log \left|\mathrm{x}^2+\mathrm{y}^2\right|=\mathrm{c}\end{aligned}\)