= Solution
Take $c=1$ and normalize the <Big Bang> to $t=0$. The <cosmological perfect-fluid continuity equation> for <pressureless matter> makes $C=8\pi G\rho_m a^3/3$ a positive constant. The flat <Friedmann equation> consequently reads
$$
\dot a^2=\frac Ca+\frac{\Lambda a^2}{3}.
$$
Putting $u=a^{3/2}$ gives $\dot u^2=\tfrac94(C+\Lambda u^2/3)$, which is directly integrable. The expanding <flat dust universe with a cosmological constant> has
$$
\boxed{
a(t)=\begin{cases}
(3C/\Lambda)^{1/3}\sinh^{2/3}(\sqrt{3\Lambda}\,t/2),&\Lambda>0,\\
(9C/4)^{1/3}t^{2/3},&\Lambda=0,\\
(3C/|\Lambda|)^{1/3}\sin^{2/3}(\sqrt{3|\Lambda|}\,t/2),&\Lambda<0.
\end{cases}}
$$
For negative <cosmological constant>, the last expression describes expansion and contraction on $0<t<2\pi/\sqrt{3|\Lambda|}$. Its maximum is $a_m=(3C/|\Lambda|)^{1/3}$ at $t_m=\pi/\sqrt{3|\Lambda|}$; it reaches zero again at $2t_m$. Positive or zero <cosmological constant> gives unbounded expansion and no finite maximum. All three have <infimum> zero at their initial singularity; negative <cosmological constant> also gives zero at the final singularity, with no positive minimum.
For positive <cosmological constant>, expanding the hyperbolic sine shows $a\sim(9C/4)^{1/3}t^{2/3}$ at early times: <pressureless matter> dominates. At late times,
$$
a\sim\left(\frac{3C}{4\Lambda}\right)^{1/3}e^{\sqrt{\Lambda/3}\,t},\qquad H\longrightarrow\sqrt{\Lambda/3}.
$$
Thus the expansion changes from matter-dominated deceleration to exponential acceleration. Indeed $\ddot a/a=-C/(2a^3)+\Lambda/3$ changes sign at $a^3=3C/(2\Lambda)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-67-scale-factor.png]
{title=Flat dust expansion for positive, zero and negative cosmological constant, with a common early-time normalization}
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