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.
[Math] solve$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$
ordinary differential equations
Related Solutions
By dividing both side to $y^2$ we get $$ydx-xdy+3x^{ 2 }y^{ 2 }e^{ x^{ 2 } }dx=0\\ \frac { ydx-xdy }{ { y }^{ 2 } } +3x^{ 2 }e^{ x^{ 2 } }=0\\ d\left( \frac { x }{ y } \right) =3x^{ 2 }e^{ x^{ 2 } }\\ \int { d\left( \frac { x }{ y } \right) =\int { 3x^{ 2 }e^{ x^{ 2 } }dx } } =\frac { 3 }{ 2 } \int { x } d{ e }^{ { x }^{ 2 } }=\frac { 3 }{ 2 } \left( x{ e }^{ { x }^{ 2 } }-\int { { e }^{ { x }^{ 2 } }dx } \right) \\ \frac { x }{ y } =\frac { 3 }{ 2 } \left( x{ e }^{ { x }^{ 2 } }-\int { { e }^{ { x }^{ 2 } }dx } \right) +C\\ y=\frac { 2x }{ 3\left( \left( x{ e }^{ { x }^{ 2 } }-\int { { e }^{ { x }^{ 2 } }dx } \right) +C \right) } $$
The solution (below) is obtained on the form of an implicit equation : $$(y^2-x^2+1)^5-c\:(y^2+x^2-3)=0$$ where $c$ is a constant.
Since it's a quintic polynomial equation, $y^2$ cannot be expressed as a function of $x^2$ with a finite number of elementary functions.
Theoretically, $y(x)$ could be expressed on a closed form thanks to special functions (in terms of Jacobi theta functions), but this would be very arduous : http://mathworld.wolfram.com/QuinticEquation.html
Best Answer
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:
$$\arctan\left( \frac{\sqrt{x^2 + y^2}}{\sqrt{a^2 - x^2 - y^2}} \right) = \arctan\left(\frac{y}{x} \right) + C$$