# 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.

• This post is visible from outer space$\ldots$ Nov 10 '14 at 14:04
• It looks like the imaginary part of (x+iy)(dy-idx) which is just z and -idz multiplied together in the numerator and the real part of that same expression in the denominator along with (x+iy)(x+iy)* in the radical. Nov 10 '14 at 15:43
• $sin^{-1}(\frac{\sqrt{x^2+y^2}}{a})=tan^{-1}(\frac{y}{x})+c$ @Kainui solution Nov 10 '14 at 16:05

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:

$$\arctan\left( \frac{\sqrt{x^2 + y^2}}{\sqrt{a^2 - x^2 - y^2}} \right) = \arctan\left(\frac{y}{x} \right) + C$$

• how come the answer be $sin^{-1}(\frac{\sqrt{x^2+y^2}}{a})=tan^{-1}(\frac{y}{x})+c$ Nov 10 '14 at 16:23
• These answers are equivalent. Draw a right triangle with hypotenuse $a$ and one side $\sqrt{x^2 + y^2}$. Then the other side is $\sqrt{a^2 - x^2 - y^2}$ and you can see that $$\arcsin\left(\frac{\sqrt{x^2 + y^2}}{a}\right) = \arctan\sqrt{\frac{a^2 - x^2 -y^2}{x^2 + y^2}}$$ Nov 10 '14 at 16:29
• ...and thanks for torturing us (me) by not giving us the answer up front! Nov 10 '14 at 16:30
• Sorry for the turmoil, I rather got this answer later, else I would have posted it earlier itself. Nov 10 '14 at 16:37
• In my argument above $\arcsin(stuff) = \arctan(other stuff)$ above, I've written the reciprocal of $otherstuff$. Nov 10 '14 at 22:57

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 $$\frac{dz}{ds} = \sqrt{a^2-z^2}.$$