 Solution

Set $y=a^2$. Multiplication of the Friedmann equation by $4a^4$ givesFor the expanding branch emerging from $y(0)=0$, $\dot{y}(0)=2\sqrt{\Gamma }$. Differentiating where $\dot{y}\ne 0$ 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 givesInitially $a(t)\sim (2\sqrt{\Gamma }t{)}^{1/2}$. 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 $\tanh \kappa t_{\max }=\lambda$, givingThe 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.