After the Big Bang

We will use $H_0=1$ as its value does not influence the answer. That is equivalent of measuring the time in units of $1/H_0$ .

The equation can be solved by separation of variables (or using one of the online services) and the solution is

$a=\frac{1}{(2\Omega)^{2/3}}e^{-\sqrt{\Omega}(t+c)}\left(e^{3\sqrt{\Omega}(t+c)}-(1-\Omega)\Omega\right)^{2/3}$

To make sure that the Universe starts at $t=0$ (i.e. $a(t=0)=0$ ) we have to solve

$e^{3\sqrt{\Omega}c}-(1-\Omega)\Omega=0$

that yields

$c=\frac{1}{3\sqrt{\Omega}} ln (1-\Omega)\Omega$

With this value of $e^{3\sqrt{\Omega}c}=(1-\Omega)\Omega$ , and $e^{-\sqrt{\Omega}c}=[(1-\Omega)\Omega]^{-1/3}$ . The function $a$ can be re-expressed $^*$ as

$a= \left(\frac{1-\Omega}{\Omega}\right)^{1/3} \sinh \frac{3}{2} \left(\sqrt{\Omega} t\right)^{2/3}$ .

The age of the current Universe is obtained by finding the time when $a=1$ . This can be done by solving

$1= \left(\frac{1-\Omega}{\Omega}\right)^{1/3} \sinh \frac{3}{2} \left(\sqrt{\Omega} t\right)^{2/3}$

or by integrating the original differential equation

$t_0= \int_0^1{ \sqrt{\frac{a}{1-\Omega +\Omega a^3} }}da$

Either way, the result is

$t_0=\frac{1}{3\sqrt{\Omega}} ln \frac {1+\sqrt{\Omega}}{1-\sqrt{\Omega}}=0.9468$

We evaluate $a(t)$ at $t=t_0/2$ and we get

$a=0.5653$

The graphs below show the evolution of the scale factor $a$ , the density of dark energy, the density of matter and the second time derivative of the scale factor. The red arrow points to the present time. Notice that in this model the dark energy takes over the matter at around $t=0.78$ . The Universe's expansion turns from slowing down (negative $a''$ ) to accelerating (positive $a''$ ) around $t=0.54$ .

$*$ Thanks to Mark Hennings for pointing this out.