The general solution of the differential equation \(\frac{d y}{d x}=\frac{3 x+y}{x-y}\) is (where \(C\) is a constant of integration.)

\(\frac{d y}{d x}=\frac{3 x+y}{x-y}\)
\(\begin{aligned} & \Rightarrow v=x \cdot \frac{d v}{d x}=\frac{3+v}{1-v} \quad[\text { let } y=v x] \\ & \Rightarrow x \cdot \frac{d v}{d x}=\frac{3+v}{1-v}-v \\ & \Rightarrow x \frac{d v}{d x}=\frac{3+v^2}{1-v} \\ & \Rightarrow \int \frac{1-v}{3+v^2} d v=\int \frac{d x}{x} \\ & \Rightarrow \int \frac{d v}{3+v^2}-\frac{1}{2} \int \frac{2 v d v}{3+v^2}=\int \frac{d x}{x} \\ & \Rightarrow \frac{1}{\sqrt{3}} \tan ^{-1} \frac{v}{\sqrt{3}}-\frac{1}{2} \log \left(3+v^2\right)=\log x+C\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{1}{\sqrt{3}} \tan ^{-1} \frac{y}{x \sqrt{3}}-\frac{1}{2} \log \left(\frac{3 x^2+y^2}{x^2}\right)=\log x+C \\ & \Rightarrow \frac{1}{\sqrt{3}} \tan ^{-1} \frac{y}{x \sqrt{3}}-\log \left(\frac{y^2+3 x^2}{x^2}\right)^{\frac{1}{2}}=\log x+C\end{aligned}\)