Solution of the differential equation $x \frac{d y}{d x}=y+\sqrt{x^2+y^2}$, is

$y+\sqrt{x^2+y^2}=C x^2$

Substituting $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$, we get

$v+x \frac{d v}{d x}=v+\sqrt{1+v^2} \Rightarrow \frac{1}{\sqrt{1+v^2}} d v=\frac{1}{x} d x$

On integrating, we get

$\log \left(v+\sqrt{v^2+1}\right)=\log x+\log C$

$\Rightarrow v+\sqrt{v^2+1}=C x \Rightarrow y+\sqrt{x^2+y^2}=C x^2$