Diff eq. transformation polar coordinates I have $(x',y')=(x-y-x(x^2+y^2)+\frac{xy}{\sqrt{x^2+y^2}},x+y-y(x^2+y^2)-\frac{x^2}{\sqrt{x^2+y^2}} )$
Now I want to use polar coordinates $(x,y)=(r\cos(t),r\sin(t))$ to get $(r',t')=(r(1-r^2),2\sin(\frac{t}{2})^2)$
I do not see this relation. When I put $x=\cos t$, $y=\sin t$ into the system of differential equations, I only get $(r\cos(t)-r\sin(t)-r^3\cos(t)+r\cos(t)\sin(t),r\cos(t)+r\sin(t)-r^3\cos(t)-r\cos(t)^2)$.
 A: Using $x=r \cos{t}$, $y=r \sin{t}$:
$$x'=(\cos{t}) r' - r (\sin{t}) \, t'$$
$$y'=(\sin{t}) r' + r (\cos{t}) \, t'$$
So we get
$$\left ( \begin{array}\\ \cos{t} & -r \sin{t} \\ \sin{t} & r \cos{t} \end{array} \right ) \left ( \begin{array}\\ r' \\ t' \end{array} \right ) = \left ( \begin{array}\\ r \cos{t} - r \sin{t} - r^3 \cos{t} + \sin{t} \cos{t} \\ r \cos{t} + r \sin{t} - r^3 \sin{t} -  \cos^2{t} \end{array} \right )  $$
Multiply both sides by the matrix inverse to get
$$\left ( \begin{array}\\ r' \\ t' \end{array} \right ) = \frac{1}{r} \left ( \begin{array}\\ r\cos{t} &  r\sin{t} \\ - \sin{t} &  \cos{t} \end{array} \right ) \left ( \begin{array}\\ r \cos{t} - r \sin{t} - r^3 \cos{t} + \sin{t} \cos{t} \\ r \cos{t} + r \sin{t} - r^3 \sin{t} -  \cos^2{t} \end{array} \right ) $$
Just do out the multiplication.  It is messy, but there is a lot of cancellation and we get
$$\left ( \begin{array}\\ r' \\ t' \end{array} \right ) = \left ( \begin{array}\\ r-r^3 \\ 1 -  \cos{t} \end{array} \right ) $$
A: Your notation is obscure and is getting in the way of the problem. What you have is
\begin{align}
x\color{red}{(t)} &= r\color{red}{(t)} \cos \theta \color{red}{(t)}\\
y\color{red}{(t)} &= r\color{red}{(t)} \sin \theta \color{red}{(t)}
\end{align}
so
\begin{align}
x'(t) &= r' \cos \theta - r \theta' \sin \theta \\
y'(t) &= r' \sin \theta + r \theta' \cos \theta \\
\end{align}
In vectorial form
\begin{equation}
\vec{X}(t) = r(t) \hat{r}(t), \mbox{ where } \hat{r} = \pmatrix{\cos \theta\\ \sin \theta}
\end{equation}
and
\begin{equation}
\vec{X}'(t) = r'\hat{r} + r \theta' \hat{\theta}
\end{equation}
Then
\begin{equation}
r'\hat{r} + r \theta' \hat{\theta} = r \hat{r} + r \hat{\theta} - r^3\hat{r} - r \cos\theta \,\hat{\theta}
\end{equation}
Due orthogonality
\begin{align}
r' &= r(1-r^2) \\
\theta' &= 1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)
\end{align}
