I am having trouble trying to convert a first order ODE into polar coordinates.
My ODE system is as follows:
$$\frac{dx}{dt} =y$$
$$\frac{dy}{dt} = -p(x^2+y^2 -1)y - y$$
where $p$ is a parameter such that $p>0$. I know that $r^2=x^2+y^2$, and $\tan \theta=x/y$, but I don't exactly know how to manipulate this ODE to get it to be in the forms of $dr\over dt$ and $d\theta \over dt$.
Any help would be appreciated!
First DE gives us $$x'=y \implies xx'=yx$$ And from the second DE: $$\frac{dy}{dt} = -p(x^2+y^2 -1)y - y$$ $$y'y = -p(x^2+y^2 -1)y^2 - y^2$$ Then we have $$rr'=xx'+yy'$$ $$ \begin{align} rr'=&xy-p(x^2+y^2 -1)y^2 - y^2 \\ rr'=&r^2\cos \theta \sin \theta-p(r^2 -1)r^2 \sin ^2 \theta - r^2 \sin ^2 \theta \\ r'=&r\cos \theta \sin \theta-p(r^2 -1)r \sin ^2 \theta - r \sin ^2 \theta \\ r'=&r\cos \theta \sin \theta- r \sin ^2 \theta(p(r^2 -1) +1) \end{align} $$ Do the same for $\theta '$ with $$r^2 \theta '= y'x-x'y$$