= Solution
The off-axis equilibria exist for $|a|\leq1$ and coalesce with an axial equilibrium at each endpoint. Thus the stationary bifurcations occur at
$$
\boxed{\ a=\pm1.\ }
$$
For the positive bifurcation, $a_0=y_0=1$. Put
$$
Y=y-1,
\qquad \mu=a-1,
$$
so that
$$
\dot x=2x(Y-\mu),
\qquad
\dot Y=-x^2-2Y-Y^2,
\qquad
\dot\mu=0.
$$
Write the extended <center manifold> as $Y=h(x,\mu)$. Its invariance equation is
$$
2x(h-\mu)h_x=-x^2-2h-h^2.
$$
Under the scaling $\mu=O(x^2)$, comparison of the terms of order $x^2$ gives
$$
\boxed{\ Y=h(x,\mu)=-\frac{x^2}{2}+O(x^4).\ }
$$
Substitution into the $x$ equation gives
$$
\boxed{\ \dot x=-2\mu x-x^3+O(x^5).\ }
$$
With the reversed parameter $\nu=-2\mu=2(1-a)$, this is the <pitchfork bifurcation> normal form $\dot x=\nu x-x^3$. Therefore the positive stationary bifurcation is a \b[supercritical pitchfork in the parameter $\nu=2(1-a)$], or equivalently the stable daughter equilibria exist on the $a<1$ side.
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