<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] One should have $2yy'=(y^2)'$ burned into one's memory. Now notice
$$\color{Blue}{x}+(\color{Red}{y^2/2})'=\sqrt{\color{Blue}{x^2}+\color{Red}{y^2}}$$
The red observation tells us we could use the substitution $v=y^2$. The blue observation tells us we can go further, since $2x=(x^2)'$ (which is just our original observation a second time). Now:
$$\frac{d}{dx}\left(\frac{x^2+y^2}{2}\right)=\sqrt{x^2+y^2}\iff \frac{dr^2}{dx}=2r.$$
Simplifyig, $dr/dx=1\implies r=x+c\iff y^2=(x+c)^2-x^2=c(2x+c)$.
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