<s>[INST] 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. 
 [/INST] 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:
$$\arctan\left( \frac{\sqrt{x^2 + y^2}}{\sqrt{a^2 - x^2 - y^2}} \right)  = \arctan\left(\frac{y}{x} \right) + C$$
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