= Solution
Set $y=a^2$. Multiplication of the <Friedmann equation> by $4a^4$ gives
$$
\dot y^2=4\Gamma-4y+\frac{4\Lambda}{3}y^2.
$$
For the expanding branch emerging from $y(0)=0$, \b[$\dot y(0)=2\sqrt\Gamma$]. Differentiating where $\dot y\ne0$ gives $\ddot y=(4\Lambda/3)y-2$, and continuity extends it through a regular turning point. Put $\kappa=\sqrt{4\Lambda/3}$. Solving this linear equation with both initial data gives
$$
\boxed{a^2(t)=\frac3{2\Lambda}\left[1-\cosh\kappa t+\lambda\sinh\kappa t\right],\qquad
\lambda=2\sqrt{\frac{\Lambda\Gamma}{3}}}.
$$
Initially $a(t)\sim(2\sqrt\Gamma\,t)^{1/2}$. If $\lambda>1$, $y$ stays positive and strictly increasing; at late times $a\sim[3(\lambda-1)/(4\Lambda)]^{1/2}e^{\kappa t/2}$. This universe expands forever.
If $0<\lambda<1$, expansion stops when $\tanh\kappa t_{\max}=\lambda$, giving
$$
\boxed{t_{\max}=\kappa^{-1}\operatorname{artanh}\lambda,\quad
a_{\max}^2=\frac3{2\Lambda}(1-\sqrt{1-\lambda^2}),\quad
t_{\rm crunch}=2t_{\max}}.
$$
The physical solution expands from zero and then symmetrically recollapses to zero; negative $a^2$ beyond this interval is not a continuation of a real scale factor.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-1-radiation-universes.png]
{title=Scale-factor histories for closed radiation universes with positive cosmological constant: eternal expansion for lambda above one and finite-time recollapse for lambda below one}
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