Hardcore Differentials!

The given equation can be expanded as, $(ydx-xdy)\quad +\quad \sqrt { xy } (x+y)[xdx\quad +ydy]\quad =0$

on dividing the equation by ${ y }^{ 2 }$

$\frac { ydx-xdy }{ { y }^{ 2 } } +\frac { \sqrt { xy } }{ y } \frac { (x+y) }{ y } [xdx+ydy]\quad =\quad 0$

we can observe that the first term is actually , $d(\frac { x }{ y } )$

$\Rightarrow \quad d(\frac { x }{ y } )+\sqrt { \frac { x }{ y } } (\frac { x }{ y } +1)[xdx+ydy]\quad =\quad 0$

$\Rightarrow \quad d(\frac { x }{ y } )+\sqrt { \frac { x }{ y } } ({ (\sqrt { \frac { x }{ y } } ) }^{ 2 }+1)[xdx+ydy]\quad =\quad 0$

$\Rightarrow \quad d(\frac { x }{ y } )\quad \quad \quad =\quad -\sqrt { \frac { x }{ y } } ({ (\sqrt { \frac { x }{ y } } ) }^{ 2 }+1)[xdx+ydy]$

$\Rightarrow \quad \int { \frac { d(\frac { x }{ y } ) }{ \sqrt { \frac { x }{ y } } (({ \sqrt { \frac { x }{ y } } ) }^{ 2 }+1) } } =\quad -\quad \{ \int { xdx } +\int { ydy\} }$

on integrating both sides,

$\Rightarrow 2\tan ^{ -1 }{ \sqrt { \frac { x }{ y } } } \quad +\quad C\quad =\quad -\quad \{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ 2 } \} \quad$

$\Rightarrow \quad \frac { { x }^{ 2 }+{ y }^{ 2 } }{ 2 } \quad +\quad 2\tan ^{ -1 }{ \sqrt { \frac { x }{ y } } } \quad +\quad C\quad =\quad 0$

hence r+s+t+u = $\boxed { 8 }$

Hope you guys understood it :)

If you got an easier solution , don't hesitate to put it up!