= Solution
The <equilibria> satisfy $y=0$ and $x-x^3=0$, giving $(0,0)$ and $(\pm1,0)$. At the origin,
$$
J_0=\begin{pmatrix}0&1\\1&-\Omega\end{pmatrix},\qquad\lambda_\pm=\frac{-\Omega\pm\sqrt{\Omega^2+4}}2.
$$
Their product is $-1$, so \b[the origin is a <saddle equilibrium> for every real $\Omega$].
At either nonzero <equilibrium>,
$$
J_\pm=\begin{pmatrix}0&1\\-2&-(\Omega+1)\end{pmatrix},\qquad\lambda^2+(\Omega+1)\lambda+2=0.
$$
The <determinant> is positive and the <trace> is $-(\Omega+1)$. Thus both <equilibria> are attracting for $\Omega>-1$, repelling for $\Omega<-1$, and have purely imaginary <eigenvalues> $\pm i\sqrt2$ at $\Omega=-1$. More precisely they are <stable focus>[stable foci] for $-1<\Omega<-1+2\sqrt2$ and <stable nodes> for $\Omega>-1+2\sqrt2$; they are un<stable focus>[stable foci] for $-1-2\sqrt2<\Omega<-1$ and un<stable nodes> for $\Omega<-1-2\sqrt2$. At the two discriminant-zero endpoints they are repeated-eigenvalue <degenerate nodes>, with a nondiagonalizable Jacobian. At $\Omega=-1$, <linearization> alone does not determine nonlinear stability or prove a nonlinear centre.
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