= Solution
For $\beta=4$,
$$
H^2=\frac{H_0^2}{a^4}(a^2-2)^2.
$$
The branch expanding from $a=0$ has
$$
H=\frac{H_0(2-a^2)}{a^2}
$$
until it approaches $a^2=2$. With $x=a^2$,
$$
\dot x=2a\dot a=2a^2H=2H_0(2-x),
\qquad x(0)=0.
$$
Solving this linear equation gives the <critical closed radiation--cosmological-constant solution>
$$
\boxed{a(t)=\sqrt{2(1-e^{-2H_0t})}}.
$$
For $H_0t\ll1$,
$$
1-e^{-2H_0t}\sim2H_0t,
\qquad
\boxed{a(t)\sim2\sqrt{H_0t}}.
$$
For $H_0t\gg1$,
$$
a(t)=\sqrt2\left(1-\frac12e^{-2H_0t}
+O(e^{-4H_0t})\right),
$$
so
$$
\boxed{a(t)\longrightarrow\sqrt2}
$$
exponentially from below.
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