<s>[INST] 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.
 [/INST] Another way:
$$yy^\prime+x=\sqrt{x^2+y^2}$$
$$ydy+xdx=\sqrt{x^2+y^2}dx$$
$$\frac 12 (dy^2+dx^2)=\sqrt{x^2+y^2}dx$$
$$\frac 12 \frac {d(x^2+y^2)}{\sqrt{x^2+y^2}}=dx$$
Integrate.
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