= Solution
Put $a=\alpha\widetilde a$ with
$$
\boxed{\alpha=\frac cR},
\qquad
\boxed{\Gamma=\frac{8\pi G\rho_0R^4}{3c^6}}.
$$
Then
$$
\left(\frac{\dot{\widetilde a}}{\widetilde a}\right)^2
=\frac\Gamma{\widetilde a^4}-\frac1{\widetilde a^2}+\frac\Lambda3.
$$
For $y=\widetilde a^2$ this becomes
$$
\dot y^2=4\Gamma-4y+\frac{4\Lambda}{3}y^2.
$$
The expanding big-bang branch has
$$
y(0)=0,\qquad \dot y(0)=2\sqrt\Gamma.
$$
Differentiation gives
$$
\ddot y=\frac{4\Lambda}{3}y-2.
$$
Thus, with $\omega=\sqrt{4\Lambda/3}$,
$$
\boxed{y(t)=\frac3{2\Lambda}\left[1-\cosh(\omega t)+\lambda\sinh(\omega t)\right]},
$$
where the initial derivative gives
$$
\boxed{\lambda=2\sqrt{\frac{\Lambda\Gamma}{3}}}.
$$
For $\lambda>1$, $\widetilde a$ expands forever and becomes asymptotically exponential. For $\lambda<1$, it reaches a maximum and recollapses to zero. At $\lambda=1$, $y$ increases monotonically to $3/(2\Lambda)$, the limiting static radius.
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