System of ODEs - from Cartesian to polar Given the system of ODEs,
$$\dot{x}=x^2+3y^2-1$$
$$\dot{y}=-2xy$$
How does one transform it into polar coordinates $(\rho, \theta)$?
Here's my line of reasoning: let $x=\rho \cos(\theta)$, $y=\rho \sin(\theta)$; then:
$$\dot{x}=-\rho\sin(\theta)\dot{\theta}+\dot{\rho}\cos(\theta)=\rho^2\cos^2(\theta)+3\rho^2\sin^2(\theta)-1$$
$$\dot{y}=\dot{\rho}\sin(\theta)+\rho\cos(\theta)\dot{\theta}=-2\rho^2\cos(\theta)\sin(\theta)$$
But I'm not quite sure how to then solve for $\dot{\rho}$ and $\dot{\theta}$. Is there a simpler method to do this transformation? Otherwise, I'd appreciate some help with solving for the time derivatives.
 A: $$\frac{dx}{dy}=\frac{x^2+3y^2-1}{-2xy}\quad\to\quad -4y^2xdx=(x^2+3y^2-1)2ydy$$
$Y=x^2\quad$ and $\quad X=y^2$
$$-2XdY=(Y+3X-1)dX$$
$$2X\frac{dY}{dX}+Y=-3X+1$$
This is a linear ODE easy to solve :
$$Y=\frac{c_1}{\sqrt{X}}-X+1$$
$$x^2=\frac{c_1}{y}-y^2+1$$
Solution expressed on implicit form (i.e.: the equation of the trajectory) :
$$y^3+(x^2-1)y-c_1=0$$
Solving for $y$ this cubic equation gives $y(x)$
Relationship with $t$ :
$$\frac{dy}{dt}=-2xy=\pm 2y\sqrt{\frac{c_1}{y}-y^2+1}$$
$$t=\pm\int\frac{dy}{2\sqrt{c_1y-y^4+y^2}}+c_2$$
This would involve feasible but arduous calculus.
If the initial condition (at $t=0$) was specified, the calculus might be simplified in some casses.
In polar coordinates, the equation of the trajectory can easily be derived :
$$(\rho^3-\rho)\sin(\theta)+c_1=0$$

A: Notice: $U \equiv x-\sqrt3 y $ satisfies $\dot U = U^2 -1$, which you can solve. If you substitute from $U$ the $x$ solution in the second equation, one gets: $\dot y =-2y(U+\sqrt 3 y)$, which is a Ricatti ODE. For this one, if you can guess a solution, you get the other. 
In short, your equations are nonlinear ODE, that's not textbook.
