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

$\tan^{-1}\frac{y}{x}= \frac{1}{2}\log_e(x^2+ y^2)+C$: C is an arbitary constant

The correct answer is Option (4) → $\tan^{-1}\frac{y}{x}= \frac{x}{2}\log_e(x^2+ y^2)+C$: C is an arbitary constant

Given differential equation: $ \frac{dy}{dx}=\frac{x+y}{x-y} $

Let $v=\frac{y}{x}$ so that $y=vx$ and $ \frac{dy}{dx}=v+x\frac{dv}{dx} $.

Substitute:

$ v+x\frac{dv}{dx}=\frac{1+v}{1-v} $

$ x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v^{2}}{1-v} $

$ \frac{1-v}{1+v^{2}}\,dv=\frac{dx}{x} $

Integrate both sides:

$ \displaystyle\int\frac{1}{1+v^{2}}\,dv-\int\frac{v}{1+v^{2}}\,dv=\int\frac{dx}{x} $

$ \tan^{-1}v-\frac{1}{2}\log(1+v^{2})=\log x + C $

Substitute $v=\frac{y}{x}$ and simplify the logarithmic term showing cancellation:

$ \tan^{-1}\!\left(\frac{y}{x}\right)-\frac{1}{2}\log\!\left(1+\frac{y^{2}}{x^{2}}\right)=\log x + C $

Substitute back:

$ \tan^{-1}\!\left(\frac{y}{x}\right) -\frac{1}{2}\log(x^{2}+y^{2}) + \log x = \log x + C $

The $+\log x$ terms on both sides cancel, leaving

$ \tan^{-1}\!\left(\frac{y}{x}\right) -\frac{1}{2}\log(x^{2}+y^{2}) = C $

Rename constant and write solution:

$\displaystyle \tan^{-1}\!\left(\frac{y}{x}\right)=\frac{1}{2}\log(x^{2}+y^{2})+C$.