Describe the non-trivial periodic orbit of a system of ode's Consider the system

$\frac{dx}{dt}=y+ax-x(x^2+y^2),\;x(0)=x_0$

$\frac{dy}{dt}=-x+ay-y(x^2+y^2),\;y(0)=y_0$

The question is describe for $a&gt; 0$ the non-trivial periodic orbit and determine its period.

My ideas:

Where, the fixed point is $(x,y)=(0,0)$ and using the polar coordinates, I obtained the following system of ode's

$\frac{dr}{dt}=ar-r^3$

$\frac{d\theta}{dt}=-1$.

If $\frac{dr}{dt}&lt;0$ then $r\geq \sqrt{a}$ and if $\frac{dr}{dt}&gt;0$ then $0&lt;r\leq \sqrt{a}$.

From here I don't know how to continue to find the orbit.

Thanks for your help.
Hint: You almost have it.  $dr/dt = 0$ if $r = \sqrt{a}$.
From

$$
\cases{
\frac{dr}{dt}=ar-r^3\\
\frac{d\theta}{dt}=-1
}
$$

with $\theta = t_0-t$ we have after integration

$$
\frac{\ln \left(r_1^2-a\right)-\ln \left(r_2^2-a\right)-2 \ln (r_1)+2 \ln (r_2)}{2
   a} = t_2-t_1 = -2\pi
$$

and solving for $r_1$

$$
r_1 = \frac{\sqrt{a} e^{2 \pi  a} r_2}{\sqrt{e^{4 \pi  a} r_2^2+a-r_2^2}}
$$

and $r_1 = r_2 = \sqrt a$

NOTE

From

$$
\cases{
x = r(t)\cos(\theta(t))\\
y = r(t)\sin(\theta(t))\\
\theta(t) = t_0 -t
}
$$

we have $T = -2\pi$ so along any orbit we have

$$
\frac{\ln \left(r_1^2-a\right)-\ln \left(r_2^2-a\right)-2 \ln (r_1)+2 \ln (r_2)}{2
   a} = t_2-t_1 = T
$$

and solving for $r_1$

$$
r_1 = \frac{\sqrt{a} r_2}{\sqrt{\left(a-r_2^2\right) e^{2 a T}+r_2^2}}
$$

so along a limit cycle, after a period $T$ we have $r_1 = r_2\Rightarrow r_1 = r_2 = \sqrt{a}$

characterizing a circle with radius $\sqrt a$ and period $T$