Ansätze to solve 2-dimensional nonlinear integral curve I try to solve the initial value problem analytically $X(t_0) = X_0, X \in \mathbb{R}^2$ for the following integral curve:
$x'(t) = -y(t)+x(t) (x(t)^2+y(t)^2-1) \\
y'(t) = x(t)+y(t) (x(t)^2+y(t)^2-1) $
According to the Picard–Lindelöf theorem there should be a solution for at least some period of time since the system is locally Lipschitz continuous. All differential equations I solved analytically so far were either one dimensional (seperation of variables, using $e^\lambda$) or linear (variation of parameters).
So far I figured out, that trajectories will converge towards the unit circle. But I still can not figure out a method how to calcluate an analytical solution. 
Is there a known method to solve these systems? I would be gratefull just for the hint to a method I could use. Or is it completely hopeless to find a closed solution for this problem.

Try the following. We expand both equations by $y$ (first equation) and by $x$ (second equation).
$$x'y=-y^2+xy(x^2+y^2-1)$$
$$y'x=x^2+xy(x^2+y^2-1)$$
Now subtract both equations
$$y'x-x'y=x^2+y^2$$
After deviding by $x^2$, we obtain:
$$\frac{y'x-yx'}{x^2}=1+\frac{y^2}{x^2}$$
Notice that we have the derivative of $y/x$ on the left side:
$$\frac{d}{dt}\left(\frac{y}{x}\right)=1+\frac{y^2}{x^2}$$
Substitute $u(t)=\frac{y}{x}$ to get 
$$\frac{d u}{dt}=1+u^2$$
This DE has $u(t)=\tan(t)$ as solution.
Now you can substitute $y(t)=x(t)\tan(t)$ into the second equation. Which will be in $x$ and $t$ only.
