Answer:
Option A
Explanation:
We have,
$\frac{dy}{dx}=\frac{x+y}{x-y}$
Put y=vx
$\frac{dy}{dx}= v+x \frac{dv}{dx}$
$\therefore$ $v+x \frac{dv}{dx}= \frac{x+vx}{x-vx}=\frac{1+v}{1-v}$
$\Rightarrow$ $x \frac{dv}{dx}=\frac{1+v}{1-v}-v= \frac{1+v-v+v^{2}}{1-v}=\frac{1+v^{2}}{1-v}$
$\Rightarrow$ $\frac{1-v}{1+v^{2}}dv= \frac{dx}{x}$
$\Rightarrow$ $\int \frac{1}{1+v^{2}}dv-\frac{1}{2}\int \frac{2v}{1+v^{2}}dv=\int\frac{dx}{x}$
$\Rightarrow$ $\tan^{-1}v-\frac{1}{2} \log |1+v^{2}|=\log x+C$
$\Rightarrow$ $\tan^{-1}\frac{y}{x}= \log x+\frac{1}{2} \log \left(1+\frac{y^{2}}{x^{2}}\right)+C$
$\Rightarrow$ $\tan^{-1}\frac{y}{x}= \log \frac{x(\sqrt{x^{2}+y^{2}})}{x}+C$
$\Rightarrow$ $\tan^{-1}\frac{y}{x}= \log \sqrt{x^{2}+y^{2}}+C$
