Using polar co-ordinates to solve a dynamical system I have the system 
$x^\prime = x-y-x(x^{2}+y^{2})$
$y^\prime = x+y-y(4x^{2}+y^{2})$.
I want to show with polar coordinates that I eventually end up with
$r^\prime = r-r^{3}(1+ \frac{3}{4}$sin$^{2}(2 \theta))$.
I have used the following substitutions:
$x = r$cos$(\theta)$, $y=r$sin$(\theta)$, $x^{2} + y^{2} = r^{2}$ and $rr^\prime = xx^\prime + yy^\prime $.
I have made a few attempts but my recent one is as follows:
\begin{align}
rr^\prime &= xx^\prime + yy^\prime 
\\
&=x(x-y-x^{3} -xy^{2}) + y(x+y-4x^{2}y^{2}+y^{3})
\\
&=x^{2} -xy + x^{4} + x^{2}y^{2} + yx + y^{2} - 4x^{2}y^{2} + y^{4}
\\
&=r^{2} + (x^{4} + y^{4}) - 3x^{2}y^{2} .
\end{align}
I am not sure if I am heading in the right direction. I also have noted that $\sin(2\theta) = 2\sin(\theta)\cos(\theta) $.
Any help would be much appreciated.

Actually you are on the right track but you've just made a few slips with your algebra. 
You should have $$rr'=x(x-y-x^{3} -xy^{2}) + y(x+y-4x^{2}y^{\color{red}{1}}\color{red}{-}y^{3})$$
this will simplify to $$x^2+y^2-(x^4+y^4)-5x^2y^2$$
$$=r^2-(r^4-2x^2y^2)-5x^2y^2$$
$$=r^2-r^4-3x^2y^2$$
now use the double angle formula for $\sin2\theta$ and you are done.
I hope this helps.
