Solve the differential equation $xy'=y +\sqrt{y^2 -x^2}$ I have to find the solution for this equation $xy'=y +\sqrt{y^2 -x^2}$, but I'm in a dead spot. I divided the equation by $x$, but I'm not sure what to do in the next step.
 A: To make explicit the method put forth by projectilemotion, with the substitution $v=\frac{y}{x}$, the equation becomes $$ y' = v + \sqrt{v^2-1} $$ Now, by the chain rule we have $$\frac{\text{d}v}{\text{d}x} = \frac{\text{d}}{\text{d}x} \frac{y}{x} = \frac{y'x - y}{x^2} \Rightarrow \frac{y' - v}{x}  = v' \Rightarrow y' = v'x + v$$ Subbing this into the diffeq gives $$ v'x + v = v + \sqrt{v^2 - 1} $$ and then $$ \frac{\text{d}v}{\text{d}x} = \frac{\sqrt{v^2-1}}{x} $$ Which is a seperable differntial equation. Solving it in the regular way, we get  $$\frac{1}{\sqrt{v^2-1}} \text{d}v = \frac1x \text{d}x \Rightarrow \ln x + c  = \ln \left(v + \sqrt{v^2-1}\right) $$ where I've just ignored the absolute value to get a solution for positive $x$ only, which usually suffices. You could workout the casework for the other case too. This gives that $$ cx = v + \sqrt{v^2-1}$$ which gives $$(cx-v)^2 = v^2-1$$ so that $$ v = \frac{1}{2}\left(cx + \frac{1}{cx}\right)$$ subbing backing in our definition of $v$ gives us the family of solutions $$ \boxed{y = \frac{1}{2}cx^2 + \frac{1}{2c}}$$ where $c$ is an arbitrary constant.
