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?

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
$$
y^{2} = k^{2} + 2kx \implies y = \sqrt{k^{2} + 2kx}
$$
