Find the solution of the differential equation $\frac{dy}{dx}=\frac{x+y}{x-y}$ with the initial condition $y(1)=0$.

Solution

1. Substitute $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$ into the equation to get $v+x\frac{dv}{dx}=\frac{1+v}{1-v}$.
2. Rearrange to $$\begin{array}{rl}x\frac{dv}{dx}& =\frac{1+v}{1-v}-v\\ & =\frac{1+v^2}{1-v}\end{array}$$, which separates into $\int \frac{1-v}{1+v^2}dv=\int \frac{1}{x}dx$.
3. Integrate both sides to obtain ${\tan }^{-1}(v)-\frac{1}{2}\ln (1+v^2)=\ln |x|+C$, which simplifies to ${\tan }^{-1}(\frac{y}{x})=\frac{1}{2}\ln (x^2+y^2)+C$.
4. Apply the initial condition $y(1)=0$ to find ${\tan }^{-1}(0)=\frac{1}{2}\ln (1)+C$, yielding $C=0$.
5. Hence the answer is (A).