$\frac {dx}{dt}=y+x(1-x^2-y^2)$, $x(0)=x_{0}$
$\frac {dy}{dt}=-x+y(1-x^2-y^2)$, $y(0)=y_{0}$
For $(x_{0},y_{0}) \neq (0,0)$, I need to determine the asymptotic behavior of the solution as $t$ goes to infinity.
First, I am not sure how to change above equations into matrix forms. Since it has square terms $-x^2-y^2$, it looks like there are three solutions by letting $z=x^2+y^2$. Anyone can help me find matrix forms of above differential equations?
If I can find them, I think I will be able to determine how solutions behave as t goes to infinity.
Thanks.
I guess you have gotten the hint to look at the function $z(t) =x(t)^2+ y(t)^2$.
We find that $$z' = 2 (x x' + y y') = z (1-z)$$ with $z(0) =x_0^2+y_0^2$. This is a simple uncoupled differential equation for $z(t)$. You will have no problems to find the solution, e.g., by separation. You will notice that, irrespective of the initial conditions, the solution will approach $z=1$.
Once the solution $z(t)$ is found, you can aim at determining $x(t)$ and $y(t)$. For this, observe that $$ x' = y + (1-z)x,\qquad y' = -x + (1-z)y\;.$$
In the asymptotic limit, we have $z\approx 1$. If we plug $z=1$ into the equations above, we need to solve the linear set of ODE $$ x' =y \qquad y' =-x$$ valid for asymptotically long times.
I am sure, you can take it from here.