Solving the given differential equation. We need to solve : $$ ( \sqrt{x+y} + \sqrt{x-y}) \,dx + ( \sqrt{x-y} - \sqrt{x+y})\,dy=0$$
I tried as follows : 
$$ \frac{dy}{dx} = \frac{ \sqrt{x+y} + \sqrt{x-y}}{\sqrt{x-y} - \sqrt{x+y}}$$
And hence letting $ y = vx$ yields : 
$$ v + x \: \frac{dv}{dx} = \frac{\sqrt{1+v} + \sqrt{1-v}}{\sqrt{1-v} - \sqrt{1+v}}$$
Continuing from here yields a very complicated integral , is there a simpler way to proceed ?

Here is an alternate approach:
\begin{equation}
( \sqrt{x+y} + \sqrt{x-y}) \,dx + ( \sqrt{x-y} - \sqrt{x+y})\,dy=0
\end{equation}
Before making the substitution $y=xv$ one should first multiply  by the conjugate of the second term and simplify the result to obtain
\begin{equation}
( x + \sqrt{x^2-y^2}) \,dx -y\,dy=0
\end{equation}
Then the $y=vx$ substitution gives, after simplification
\begin{equation}
( x-xv^2+x\sqrt{1-v^2}) \,dx -vx^2\,dv=0
\end{equation}
dividing by $x$ and separating variables and doing a bit of rationalizing of the denominator yields an expression solved by elementary integrals. 
\begin{equation}
\frac{1}{x}\,dx+\left(\frac{1}{v}-\frac{1}{v\sqrt{1-v^2}} \right)\,dv=0
\end{equation}
