= Solution
Differentiate the <Friedmann equation> and use the <cosmological continuity equation> to obtain $\dot H+H^2=-(4\pi G/3)(\rho+3P/c^2)$. The relation extends to turning points by continuity. Radiation obeys $\rho_R=\rho_{R0}a^{-4}$, while <dark energy> with $P_\Lambda=-\rho_\Lambda c^2$ has constant density. Substitution at $a(t_0)=1$ gives $kc^2=\beta H_0^2$, and at arbitrary $a$ gives
$$
H^2=H_0^2\frac{a^4-\beta a^2+\beta}{a^4},\qquad
\dot H=-\beta H_0^2\frac{2-a^2}{a^4}.
$$
Put $x=a^2$. For $\beta>4$, the upward-opening polynomial $P(x)=x^2-\beta x+\beta$ has roots $x_\pm=(\beta\pm\sqrt{\beta(\beta-4)})/2$, with $1<x_-<2<x_+$. The expanding <Big Bang> branch reaches $x_-$ in finite time. There $H=0$ but $\dot H<0$, so expansion turns to contraction; it returns to $a=0$ in finite time, a <Big Crunch>. A sketch of $a^4H^2/H_0^2=P(x)$ has the forbidden negative interval between the roots, while $a^4\dot H/H_0^2=\beta(x-2)$ is a line crossing zero at two.
For $\beta=4$, $P(x)=(x-2)^2$. The branch starting at zero has $\dot x=2H_0(2-x)$, giving
$$
\boxed{a(t)=\sqrt{2(1-e^{-2H_0t})}.}
$$
At small $t$, $a\sim2\sqrt{H_0t}$, the radiation-dominated <scale factor>. At large $t$, $a=\sqrt2[1-\tfrac12e^{-2H_0t}+O(e^{-4H_0t})]$. It approaches an <Einstein static universe> with equal radiation and dark-energy densities, rather than recollapsing or entering unbounded expansion.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ii/paper-1-radiation-lambda.png]
{title=Radiation–Lambda turning points and the critical scale factor}
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