Write the system
$x'=-y-xy$
$y'=x+x^2 $
in polar coordinates, use
$rr'=xx'+yy'$
$r^2\theta'=xy'-yx'$
Then for above example, have
$r'=0$ and $\theta'=1+x>0$, for $x>-1$.
$r(t)$ is constante and $\theta(t)$ increases without bound, when $t\to \infty$.
We have the point $(0.0)$ and the straight $x=-1$ of equilibrium points.
How can I sketch the phase portrait for this system? I understood that in a neighborhood of the origin has a center, but I am in doubt how the phase portrait looks for different solution curves. Any site where I can plot this system?
Hint.
Making
$$ \cases{ x \dot x = -x y-x^2 y\\ y \dot y = x y + x^2 y } $$
after addition we have
$$ \frac 12 \frac{d}{dt}(x^2+y^2) = 0\Rightarrow x^2+y^2= C $$
are the orbits. Follows a plot showing the stream plot in light blue and the orbit $x^2+y^2=0.5$ in red.
NOTE
This stream plot has a line $(x = -1)$ with null flux. Thus for $r \lt 1$ the orbits are given by
$$ x^2+y^2 = r^2 $$
while for $r \gt 1$ we have the orbits composed by two circle segments. For example, for $r = 1+\delta,\ \ \delta > 0$ we have
$$ (1+\delta)(\cos\alpha,\sin\alpha),\ \ \ \text{for}\ \ \ \arctan(-1,-\delta(\delta+2))\le \alpha\le \arctan(-1,\delta(\delta+2)) $$
and
$$ (1+\delta)(\cos(-\alpha),\sin(-\alpha)),\ \ \ \text{for}\ \ \ \arctan(-1,\delta(\delta+2))\le \alpha\le \arctan(-1,-\delta(\delta+2)) $$