solve$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$ solve the differential equation. 
$$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$
The question is from IIT entrance exam paper. I have tried substituting $x^2=t\ and \ y^2=u$ but was not a worth try. 
Thanks in advance. 

This equation could have been solved easily in polar coordinates.
\begin{align}
x &= z \cos{s} \\
y &= z \sin{s} 
\end{align}
Then 
\begin{align}
dx = \cos(s) \, dz - z\,\sin(s)\, ds \\ 
dy = \sin(s) \, dz + z\,\cos(s)\, ds  
\end{align}
as well as
$$x^2+y^2 = z^2\big(\cos^2(s) + \sin^2(s)\big) = z^2$$ which means that
$$z = \sqrt{x^2 + y^2}$$
Now
\begin{align}
y\,dx =   z \sin(s) \cos(s) \, dz - z^2\,\sin^2(s)\, ds \\ 
x\,dy = z \cos(s) \sin(s) \, dz + z^2\,\cos^2(s)\, ds  
\end{align}
which leads to the difference
\begin{align}
x\,dy - y\,dx &= z \cos(s) \sin(s) \, dz + z^2\,\cos^2(s)\, ds\\
& -   z \sin(s) \cosh(s) \, dz + z^2\,\sinh^2(s)\, ds \\
&=   z^2\,\cos^2(s)\, ds +  z^2\,\sin^2(s)\, ds \\ 
&= z^2\big(\cos^2(s) + \sin^2(s)\big) \, ds\\
&= z^2\, ds
\end{align}
Moreover, 
\begin{align}
x\,dx + y\,dy &= \frac{1}{2}d \left(x^2+y^2\right)\\ 
&= \frac{1}{2} d(z^2)\\ 
&= z\,dz
\end{align}
Consequently
\begin{align}\frac{x\,dx + y\,dy}{ x\,dy - y\,dx } &= \frac{z\, dz}{z^2 \, ds}\\ &= \frac{1}{z} \frac{dz}{ds}\end{align}
and finaly the equations becomes
\begin{align}
 \frac{1}{z} \frac{dz}{ds} = \frac{x\,dx + y\,dy}{ x\,dy - y\,dx } = \sqrt{\frac{a^2 - x^2 - y^2}{x^2+y^2}} = \sqrt{\frac{a^2 - z^2}{z^2}} = \frac{\sqrt{a^2-z^2}}{z}
\end{align} 
multiply both sides by $z$ and obtain the simple differential equation
$$ \frac{dz}{ds} = \sqrt{a^2-z^2}.$$
