<s>[INST] How do I solve $yy'+x=\sqrt{x^2+y^2}$? I tried this:
$yy'+x=\sqrt{x^2+y^2}$
$y'=-\frac{x}{y}+\frac{1}{y}\sqrt{x^2+y^2}$
$y'=-\frac{x}{y}+\sqrt{(\frac{x}{y})^2+1}$
Substitution: $v=\frac{y}{x}$
$v'x+v=-\frac{1}{v}+\sqrt{(\frac{1}{v})^2+1}$
$v'x=-\frac{1-v^2}{v}+\sqrt{(\frac{1}{v})^2+1}$
$v'x=-\frac{1-v^2+ v\sqrt{(\frac{1}{v})^2+1}}{v}$
$v'x=-\frac{1-v^2+\sqrt{1+v^2}}{v}$
If I have to separate at this point, it's going to be pretty awkward to work with.
Is there a better way?
 [/INST] Once you set $z=x^2+y^2$, then things get extremely simple. Indeed:
$$
z'=2x+2yy'=2\sqrt{x^2+y^2}=2z^{1/2},
$$
and hence (assuming that $z>0$)
$$
z^{-1/2}z'=2,
$$
or
$$
\big(z^{1/2}\big)'=1,
$$
or
$$
z^{1/2}=x+c
$$
for some constant $c$, and finally
$$
z=(x+c)^2,
$$
equivalently
$$
x^2+y^2=(x+c)^2,
$$
and hence
$$
y^2=2cx+c^2.
$$
</s>