Solve $yy^\prime+x=\sqrt{x^2+y^2}$ 
Solve $yy^\prime+x=\sqrt{x^2+y^2}$

Tried dividing by $y$ to get $$y^\prime+\frac{x}{y}=\frac{\sqrt{x^2+y^2}}{y}$$
$$(y^\prime)^2+2\frac{x}{y}+\frac{x^2}{y^2}=\frac{x^2+y^2}{y^2}$$
$$(y^\prime)^2+2\frac{x}{y}=1$$
$$y^\prime=\sqrt{1-2\frac{x}{y}}$$
Tried using $v=\frac{y}{x}$
$$y^\prime=v^\prime x+v$$
$$v^\prime x + v=\sqrt{1-2v^{-1}}$$
Not sure what to do now or if I've even done something right until now.

Let $r=\sqrt{x^2+y^2}$, then the equation becomes
$$
rr'=r
$$
which means that $r'=1$; that is, $r=x+c$ which is the horizontal parabola
$$
y^2=2cx+c^2
$$
