# How to integrate $xy' = \sqrt{x^2-y^2}+y$

I have to integrate:

$$xy' = \sqrt{x^2-y^2}+y$$

It is supposed to be an homogeneous differential equation, but I can't see the variable change to make it happens. I tried:

$$y' = \sqrt{1-(y/x)^2} + y/x$$

And then $$z = y/x$$, but I don't think that that is the path.

• Unless I misread something, my edit should provide the correct rewrite when dividing through by $x$... Mar 19 at 14:34
• You're almost there ["I think" - because of my next comment]. Moving the $y$ over to the LHS gives you $xy'-y$, which is the numerator of your $z'=(y/x)'$, so... Mar 19 at 14:48
• [Proceeding that way does not give you a homogeneous equation, I think, but certainly one that is separable.] Mar 19 at 14:49
• looks ok to me, carry on with the sub Mar 19 at 15:04

If $$z=y/x$$, then $$z'= {xy'-y\over x^2}.$$ Rearranging the original equation, and dividing by $$x^2$$, one gets
$${xy'-y\over x^2}= {1\over x}\sqrt{1-\left({y \over x}\right)^2}$$ Therefore $$z'= {1\over x}\sqrt{1-z^2},$$ which is doable (separable).
$$xy' = \sqrt{x^2-y^2}+y$$ $$xdy= (\sqrt{x^2-y^2}+y)dx$$ $$f(x,y)dy=g(x,y)dx$$ is homogeneous if $$f$$ and $$g$$ are homogeneous functions of the same degree: $$f(tx,ty)=tx=tf(x,y)$$ $$g(tx,ty)=\sqrt{t^2x^2-t^2y^2}+ty=t (\sqrt{x^2-y^2}+y)=tg(x,y)$$ The two functions are homogeneous ones of degree $$1$$ and the substitution $$y=tx$$ gives a separable DE.