= Solution
Differentiate the <Friedmann equation> and use <energy> conservation to obtain $\dot H=-4\pi G(\rho+P/c^2)$ on the expanding branch, where $H\ne0$. Since $\ddot a/a=\dot H+H^2$,
$$
\boxed{\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3P/c^2)+\frac{\Lambda c^2}{3}}.
$$
For <pressureless matter>, conservation gives $\rho_M=\rho_{M0}a^{-3}$. In terms of the specified <cosmological density parameters>, flatness implies $\Omega_M+\Omega_\Lambda=1$, and
$$
H=H_0\sqrt{\Omega_Ma^{-3}+\Omega_\Lambda},\qquad
\frac{\ddot a}{a}=-\frac{H_0^2}2(\Omega_Ma^{-3}-2\Omega_\Lambda).
$$
Set $b=a^{3/2}$ and take $\Omega_M>0$. Then $\dot b=\tfrac32H_0\sqrt{\Omega_M+\Omega_\Lambda b^2}$. Integrating with the <Big Bang> at $t=0$ gives
$$
\operatorname{arsinh}\left(\sqrt{\frac{\Omega_\Lambda}{\Omega_M}}b\right)
=\frac32H_0\sqrt{\Omega_\Lambda}\,t,
\qquad
\boxed{a(t)=\left(\frac{\Omega_M}{\Omega_\Lambda}\right)^{1/3}
\sinh^{2/3}\left(\frac32H_0\sqrt{\Omega_\Lambda}\,t\right)}.
$$
For another time origin replace $t$ by $t-t_B$. The <cosmological constant> has effective density $\rho_\Lambda=3H_0^2\Omega_\Lambda/(8\pi G)$. Equality with the matter density means $a^3=\Omega_M/\Omega_\Lambda$, so
$$
\boxed{\bar t=\frac{2}{3H_0\sqrt{\Omega_\Lambda}}\operatorname{arsinh}1}.
$$
If $\Omega_M=0$, the expanding solution is pure exponential <de Sitter scale factor in flat slicing> and there is no finite matter-density equality time or dust Big Bang; the stated hyperbolic-sine formula presumes a nonzero dust component.
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