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$[(x+y)^{0.5} + (x-y)^{0.5}]dx + [(x-y)^{0.5} - (x+y)^{0.5}]dy = 0$

This is a homogeneous differential equation but am having difficulty solving it by using $y=vx$ transformation.

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    Is the problem written correctly? With one sign change it is also exact. – 2017-02-08
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    Seeing these type of question, I think of homogeneous differential equation. So I would prefer if I could solve it via that method otherwise if you know some other method, I will be ok with that aswell – 2017-02-08

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Dividing both top and bottom $\sqrt{x}$ we get $\frac{dy}{dx}=-\frac{\sqrt{1-\frac{y}{x}}+\sqrt{1+\frac{y}{x}}}{\sqrt{1-\frac{y}{x}}-\sqrt{1+\frac{y}{x}}}$ then it appears to be homogenous so making the obvious substitution $v'x+v=\frac{\sqrt{1+v}+\sqrt{1-v}}{\sqrt{1+v}-\sqrt{1-v}}$ this is sill terrible so we are going to rationalize the denominator of the right hand side and we get $\frac{2+2\sqrt{1-v^2}}{2v}=\frac{1+\sqrt{1-v^2}}{v}\implies$ $x\frac{dv}{dx}=\frac{1-v^2+\sqrt{1-v^2}}{v}$ this is separable $\int\frac{v}{\sqrt{1-v^2}(1+\sqrt{1-v^2})}\text{d}v=\int\frac1x\text{d}x$ the left hand side you can make the substitution $u=1-v^2\to \text{d}u=-2v\text{d}v\implies$ $-\int\frac{1}{2\sqrt{u}(1+\sqrt u)}\text{d}u=-\ln(1+\sqrt{u})$ $=-\ln(1+\sqrt{1-v^2})$ combining this with the right side of our equation we get $-\ln(\sqrt{1-v^2}+1)=\ln x+C$ through some simplification and undoing our substitution $\boxed{\frac{\sqrt{x^2-y^2}+x}{x}=Ce^{1/x}}$