<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] The $x^2+y^2$ suggests using polar coordinates. Substitute $x=r\cos\theta$ and $y=r\sin\theta$. The difficult bit is the expression for $dy/dx$.
$$ \frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$$
Then you can substitute into your ODE.
$$\begin{align}
yy' + x &= \sqrt{x^2+y^2}\\
r\sin\theta\frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} + r\cos\theta &= r\\
\sin\theta(r'\sin\theta + r\cos\theta) + \cos\theta(r'\cos\theta - r\sin\theta) &= r'\cos\theta - r\sin\theta\\
r'\sin^2\theta + r\sin\theta\cos\theta + r'\cos^2\theta - r\sin\theta\cos\theta  &= r'\cos\theta - r\sin\theta\\
r' &= r'\cos\theta - r\sin\theta
\end{align}$$
It looks like $\dfrac{d}{d\theta}r\cos\theta$ might be useful on the right. Following this through led me to a solution that satisfied your ODE on substitution (which I will now add as the other answer provides the solution).
$$\begin{align}
r' &= \frac{d}{d\theta}r\cos\theta\\
r &= r\cos\theta + k\\
x^2 + y^2 &= x^2 + 2kx + y^2\\
y &= \sqrt{k^2 + 2kx}
\end{align}$$
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