ds.dynamical systems - Linearization at equilibrium points

As Victor says:

I am confident that the system

$$x' = -y + ( x^2 + y^2) x,$$
$$y' = x + ( x^2 + y^2) y$$
has the origin repelling nearby trajectories, while
$$x' = -y - ( x^2 + y^2) x,$$
$$y' = x - ( x^2 + y^2) y$$
has the origin attracting nearby trajectories, and
$$x' = -y$$
$$y' = x$$
has just periodic orbits near the origin. But all three
linearize to the same thing at the origin,
$$left( begin{array}{rr} 0 & -1 \ 1 & 0 end{array} right) .$$
with eigenvalues $pm i.$

EDIT: Indeed, given a constant real number $lambda$ and system

$$x' = -y + lambda ( x^2 + y^2) x,$$
$$y' = x + lambda ( x^2 + y^2) y ,$$
we find that
$$frac{d}{dt} ; (x^2 + y^2) = 4 lambda (x^2 + y^2)^2.$$

EDIT some more: so, for the nonconstant paths, if we set time to $0$ when the trajectory crosses the unit circle, we get

$$x^2 + y^2 = frac{1}{1 - 4 lambda t}$$
showing that when $lambda > 0$ the path reaches infinite radius in finite time, while with
$lambda < 0$ the path spirals in to the origin, as expected.
Then, if we set $$x = r cos theta, ; y = r sin theta$$
as usual, the rate of change of $theta$ does not depend on $lambda$ and $forall lambda,t$ we have
$$frac{d theta}{d t} = 1.$$

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