I am studying for a exam and one of model questions is solve a DOE system using polar coordinates. I've research and didn't find any reference about this subject.
System in question is
$$ \left\{\begin{matrix} x'= (1-x^2-y^2)x - y & \\ y'= x+(1-x^2-y^2)y& \end{matrix}\right. $$
I know that I can do $1-x^2-y^2 = 1-(x^2+y^2)=1-r^2$. I know, too, that solutions are as $c_{1}X_{1}e^{\lambda_{1} x}+c_{2}X_{2}e^{\lambda_{2} x}+...$, where $X_{n}$ are eigenvectors and $\lambda_{n}$ are eigenvalues. So:
$$ p(x)= det\begin{bmatrix} 1-r^2-\lambda & -1 \\ 1 & 1-r^2-\lambda \end{bmatrix} =(1-r^2-\lambda)^{2}+1 =0 $$
I will have a biquadratic equation with many lambda equations. Am I right or I got lost?
$$ \left\{\begin{matrix} x=\rho \cos(\theta) \rightarrow x' =\rho'\cos(\theta) -\rho \sin(\theta) \theta' & \\ y=\rho \cos(\theta) \rightarrow y' =\rho'\sin(\theta) +\rho \cos(\theta) \theta' & \end{matrix}\right. $$ $$ \left\{\begin{matrix} x'= (1-x^2-y^2)x - y = \rho'\cos(\theta) -\rho \sin(\theta) \theta' = (1-\rho^2)\rho \cos(\theta) - \rho \sin(\theta) & \\ y'= (1-x^2-y^2)y + x = \rho'\sin(\theta) +\rho \cos(\theta) \theta' = (1-\rho^2)\rho \sin(\theta) + \rho \cos(\theta) & \end{matrix}\right. $$ $$ \left\{\begin{matrix} \rho'\cos(\theta) -\rho \sin(\theta) \theta' = (1-\rho^2)\rho \cos(\theta) - \rho \sin(\theta) & \\ \rho'\sin(\theta) +\rho \cos(\theta) \theta' = (1-\rho^2)\rho \sin(\theta) + \rho \cos(\theta) & \end{matrix}\right. $$ First equation multiplied by $\cos(\theta)$ + second equation multiplied by $\sin(\theta)$
Second equation multiplied by $\cos(\theta)$ - first equation multiplied by $\sin(\theta)$
$$ \left\{\begin{matrix} \rho' = (1-\rho^2)\rho & \\ \theta' = 1 & \end{matrix}\right. $$ I let you continue and find $\rho(t)$ and $\theta(t)$.