Find $\lim\limits_{t\to\infty}x(t)$ if $x'= (x-y)(1-x^2-y^2)$, $y' = (x+y)(1-x^2-y^2)$ Suppose $x_0,y_0$ are reals such that $x_0^2+y_0^2>0.$ Suppose $x(t)$ and $y(t)$ satisfy the following:


*

*$\frac{dx}{dt} = (x-y)(1-x^2-y^2),$

*$\frac{dy}{dt} = (x+y)(1-x^2-y^2),$

*$x(0) = x_0,$

*$y(0) = y_0.$


I am asked to find $\lim_{t\to\infty}x(t).$
Dividing the two differential equations, we have $$\frac{dy}{dx} = \frac{x+y}{x-y} = \frac{1 + \frac{y}{x}}{1-\frac{y}{x}}.$$
Let $v(x)=\frac{y}{x},$ so that $y = xv(x),$ and $$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$ This implies that $$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v^2}{1-v}.$$ Rearranging and integrating, we get $$\log(x) = \arctan(v)-\frac{1}{2}\log(1+v^2)+c,$$ so $$\log(x) = \arctan\left(\frac{y}{x}\right)-\frac{1}{2}\log\left(1+\left(\frac{y}{x}\right)^2\right)+c.$$ By writing $1+\left(\frac{y}{x}\right)^2$ as $\frac{x^2+y^2}{x^2},$ we can simplify the above equation to $$\frac{1}{2}\log(x^2+y^2) = \arctan\left(\frac{y}{x}\right) + c.$$
However, I'm not sure how to proceed from this implicit equation. Any help would be much appreciated!

Let $r^2 = x^2+y^2$, $x = rcos\theta$, and $y = rsin\theta$.
Then, you get the following equations:
$$\begin{align}\frac{dr}{dt} &= r(1-r^2) \\
\frac{d\theta}{dt} &= (1-r^2)
\end{align}$$
The solution to the first equation is: $r^2=\frac{ke^{2t}}{1+ke^{2t}}$, where $k=\frac{r^2_0}{1-r_0^2}$.
Then, the solution to the second equation is: $\theta-\theta_0=log\ \frac{r}{r_0}$,
where $r_0^2=x_0^2+y_0^2$ and $tan\ \theta_0= \frac{y_0}{x_0}$.
