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?
 A: 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
$$
y^2=2cx+c^2.
$$
A: 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, $dr/dx=1\implies r=x+c\iff y^2=(x+c)^2-x^2=c(2x+c)$.
A: 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).
$$\begin{align}
r' &= \frac{d}{d\theta}r\cos\theta\\
r &= r\cos\theta + k\\
x^2 + y^2 &= x^2 + 2kx + y^2\\
y &= \sqrt{k^2 + 2kx}
\end{align}$$
A: I like all the answers gone before me. However, just to add another different (if not longer method)
We start by dividing through by x first we find
$$
\frac{y}{x}y^{'} + 1 = \sqrt{1 + \left(\frac{y}{x}\right)^{2}}
$$
then subbing in $v = \frac{y}{x}$
we obtain
$$
v\left(xv^{'} + v\right) + 1 = \sqrt{1 + v^{2}}
$$
re-arrange we obtain
$$
\int \frac{v}{1+v^{2}-\sqrt{1+v^{2}}}dv = -\int\frac{1}{x}dx = -\ln{x}
$$
turning out attention to the right hand side we can use the substitution $u^{2} = v^{2} + 1$
this reduces the integral to
$$
\int \frac{v}{1+v^{2}-\sqrt{1+v^{2}}}dv  = \int \frac{1}{u-1}du = \ln{\left(u-1\right)} 
$$
combining the two sides 
$$
\ln{\left(u-1\right)}  = -\ln{x} + C_1
$$
or 
$$
u - 1 = \sqrt{v^{2}+1} - 1= \frac{k}{x} 
$$
where $k = \mathrm{e}^{C_1}$
therefore
$$
v^{2} + 1 -2\sqrt{v^{2}+1} + 1 = \frac{k^{2}}{x^{2}}\implies\\
v^{2} + 2 -2\left(\frac{k}{x} +1\right) = \frac{k^{2}}{x^{2}},\\
v^{2} - 2\frac{k}{x} = \left(\frac{y}{x}\right)^{2} - 2\frac{k}{x} = \frac{k^{2}}{x^{2}}
$$
or
$$
y^{2} = k^{2} + 2kx \implies y = \sqrt{k^{2} + 2kx}
$$
