<s>[INST] 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 ?
 [/INST] This is so close to an exact differential equation. But I'm going to make the substitutions $u=x+y$ and $v=x-y$ so that $x=\frac12(u+v)$ and $y=\frac12(u-v)$. Then
$$\left(\sqrt u+\sqrt v\right)\cdot\frac12(du+dv)+\left(\sqrt v-\sqrt u\right)\cdot\frac12(du-dv)=\sqrt v\,du+\sqrt u\,dv=0$$
This is now separable and
$$u^{-1/2}du=-v^{-1/2}dv$$
$$2u^{1/2}=-2v^{1/2}+C_1$$
$$u+v=\sqrt{x+y}+\sqrt{x-y}=\frac12C_1=C\tag{1}$$
Let's check this solution:
$$\left(\frac1{2\sqrt{x+y}}+\frac1{2\sqrt{x-y}}\right)dx+\left(\frac1{2\sqrt{x+y}}-\frac1{2\sqrt{x-y}}\right)dy=0$$
And on multiplying by $2\sqrt{x^2-y^2}$ which ends up being the reciprocal of the integrating factor that makes it exact we do in fact reproduce our original differential equation.  
EDIT: I was so busy congratulating myself on seeing the key to the problem that I didn't stop to see that there were several interesting subtleties in the solution. First off, I divided by $\sqrt{uv}$ without checking to see what happens if $u=0$ or $v=0$. In fact, these lead to the singular solutions to the differential equation, $x=y\ge0$ and $x=-y\ge0$. Then it's possible to simplify the general solution by squaring to
$$x+y+2\sqrt{x^2-y^2}+x-y=C^2$$
$$\sqrt{x^2-y^2}=\frac12C^2-x$$
$$x^2-y^2=\frac14C^4-C^2x+x^2$$
$$x=\frac{\frac14C^4+y^2}{C^2}\tag{2}$$
If $C=0$, that implies $y=0$ and so $x=0$ from eq. $(1)$, which was already covered under the singular solutions. Noting that eq. $(1)$ also implies $C\ge0$, we find that
$$\sqrt{x+y}=\frac{\left|\frac12C^2+y\right|}C$$
And
$$\sqrt{x-y}=\frac{\left|\frac12C^2-y\right|}C$$
Then for eq. $(1)$ to be satisfied,
$$-\frac12C^2\le y\le\frac12C^2$$
Thus if $x=f(y)$ we can only follow eq. $(2)$ forward until $y=\frac12C^2$, after which $x=y$. Similarly we can only follow eq. $(2)$ backwards until $y=-\frac12C^2$, before which $x=-y$. Here is a plot of some solution curves along with the two singular solutions.

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