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. 
 A: 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
$$ \frac{dz}{ds} = \sqrt{a^2-z^2}.$$
A: This question is a bit of a nightmare for an entrance exam. Anyway, first observe that $$d(x^2 + y^2) = 2xdx + 2ydy$$ This suggests that using a variable $u = x^2 + y^2$ is useful, as we also have the RHS (right hand side)
$$RHS = \sqrt{\frac{a^2 - u}{u}}$$
The denominator on the LHS (left hand side) is slightly tricker; the minus sign suggests a derivative of $1/x$. Let's try $v = y/x$, then $$dv = \frac{1}{x} dy - \frac{y}{x^2} dx$$ 
Hence
$x^2 dv = xdy - ydx$ and we can write the LHS 
$$LHS = \frac{x dx + y dy}{x dy - y dx} = \frac{1}{2x^2} \frac{du}{dv}$$
If we can write the $x^2$ terms of $u$ and $v$ we will have an ODE in just those variables: $$ \frac{1}{v^2 + 1} = \frac{x^2}{x^2 + y^2} = \frac{x^2}{u} \ \ \hbox{ hence } \  x^2 = \frac{u}{v^2 + 1}$$ Thus we can write
$$LHS = \frac{v^2 + 1}{2u} \frac{du}{dv} = RHS = \sqrt{\frac{a^2 - u}{u}}$$
or
$$\frac{du}{dv} = 2\sqrt{u(a^2 - u)} . \frac{1}{v^2 + 1}$$
This equation is separable 
$$\int \frac{du}{\sqrt{u(a^2 - u)}} = 2\int \frac{dv}{v^2 + 1}$$
...after a bit of work,
$$\arctan\left( \frac{\sqrt{u}}{\sqrt{a^2 - u}} \right)  = \arctan v + C$$
or back in the original variables:
$$\arctan\left( \frac{\sqrt{x^2 + y^2}}{\sqrt{a^2 - x^2 - y^2}} \right)  = \arctan\left(\frac{y}{x} \right) + C$$
