The solution of the differential equation $(x+y)dy-(x-y)dx=0$ is
(where $C$ is integration constant)

The correct option is B $|y^2+2xy-x^2|=C$
$\frac{dy}{dx}=\frac{x-y}{x+y}$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+\frac{xdv}{dx}=\frac{1-v}{1+v}$
$\Rightarrow \frac{xdv}{dx}=\frac{1-2v-v^2}{1+v}$
$\Rightarrow \int \frac{(1+v)dv}{v^2+2v-1}=-\int \frac{dx}{x}$
$\Rightarrow \frac{1}{2}\ln |v^2+2v-1|=-\ln |x|+\ln |k|$
$\Rightarrow \ln |(v^2+2v-1)\cdot x^2|=\ln k^2$

Putting $v=\frac{y}{x}$
$\Rightarrow |y^2+2xy-x^2|=C$, where $k^2=C$