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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 $(r\cos(t)-r\sin(t)-r^3\cos(t)+r\cos(t)\sin(t),r\cos(t)+r\sin(t)-r^3\cos(t)-r\cos(t)^2)$.

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Two things. Your notation is obscure, as $t$ is function of the variable of integration, i.e. $\frac{d t}{d \xi} \equiv {t}'$, hence you are using the chain rule wrong; also, the cubic term on the second component of your substitution should read $-r^3 \sin t$. See my answer for details. –  Pragabhava Jan 30 '13 at 1:25
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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

$$\left ( \begin{array}\\ r' \\ t' \end{array} \right ) = \left ( \begin{array}\\ r-r^3 \\ 1 - \cos{t} \end{array} \right ) $$

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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}

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