How do I express this system of differential equations in polar coordinates? I'm supposed to express this system of differential equations in polar coordinates. $\begin{cases} \frac{dx}{dt}=\mu x-\omega y-x(x^2+y^2)\\\frac{dy}{dt}=\omega x+\mu y-y(x^2+y^2)\end{cases}$. I'm a bit confused as to what my expression for dx/dt and dy/dt should be but other than that my expression for the dx/dt equation is $\mu rcos(\theta)-\omega rsin(\theta)-rcos(\theta)(r^2)$ and similar for dy/dt.
 A: Recall that for polar coordinates we have 
$$
\begin{cases}
  x = r \cos \theta \\
  y = r \sin \theta,
\end{cases}
\implies 
\begin{cases}
 \frac{\partial x}{\partial r} = \cos \theta, &
  \frac{\partial x}{\partial \theta} = -r \sin \theta,\\
 \frac{\partial y}{\partial r} = \sin \theta, &
  \frac{\partial y}{\partial \theta} = r \cos \theta.
\end{cases}
$$
Now,
$$
\begin{cases}
 x(t) = x\big(r(t),\theta(t)\big) = r(t) \cos\big(\theta(t)\big) \\
 y(t) = y\big(r(t),\theta(t)\big) = r(t) \sin\big(\theta(t)\big)
\end{cases}
\Rightarrow
\begin{cases}
 \dfrac{d x}{d t} = \dfrac{\partial x}{\partial r} \dfrac{dr}{dt} +   
  \dfrac{\partial x}{\partial \theta} \dfrac{d\theta}{dt} = 
  \cos\theta\,\dfrac{d r}{dt} - r\sin\theta\,\dfrac{d\theta}{dt},\\
 \dfrac{d x}{d t} = \dfrac{\partial x}{\partial r} \dfrac{dr}{dt} +   
  \dfrac{\partial x}{\partial \theta} \dfrac{d\theta}{dt} = 
  \sin\theta\,\dfrac{d r}{dt} + r\cos\theta\,\dfrac{d\theta}{dt}.\\
\end{cases}
$$
Then your system of equations look like
$$
\begin{cases}
\cos\theta\,\dfrac{d r}{dt} - r\sin\theta\,\dfrac{d\theta}{dt}=
\mu r \cos \theta -\omega r \sin \theta -r^3\cos\theta \\
\sin\theta\,\dfrac{d r}{dt} + r\cos\theta\,\dfrac{d\theta}{dt}=
\omega r \cos \theta + \mu r \sin \theta -r^3\sin\theta 
\end{cases}
$$
After some simplifications you will be able to get the system of the form
$$
\begin{cases}
 \dfrac{dr}{dt} = \dots \\
 \dfrac{d\theta}{dt} = \dots \\
\end{cases}
$$
where $r=r(t)$ and $\theta = \theta(t)$.
