= Solution
For the <open matter-dominated Friedmann solution>, write $\Omega=\Omega_{m,0}$, so $0<\Omega<1$. The present <Friedmann equation> gives $-K=H_0^2(1-\Omega)$. With $A=H_0^2\Omega$ and $B=H_0^2(1-\Omega)>0$, the expanding equation becomes
$$
\dot a^2=\frac Aa+B,\qquad
dt=\frac1{\sqrt B}\sqrt{\frac a{a+A/B}}\,da.
$$
Choose the <hyperbolic substitution>
$$
a=\frac AB\sinh^2\frac\eta2
=\frac A{2B}(\cosh\eta-1),\qquad \eta\geq0.
$$
Then $da=(A/(2B))\sinh\eta\,d\eta$ and $\sqrt{a/(a+A/B)}=\tanh(\eta/2)$. Their product gives
$$
\frac{dt}{d\eta}=\frac A{2B^{3/2}}(\cosh\eta-1)
=\frac a{\sqrt B}.
$$
Take the <Big Bang> to be $t=0$, $\eta=0$, and integrate. The \b[parametric solution] is
$$
\boxed{a(\eta)=\frac{\Omega}{2(1-\Omega)}(\cosh\eta-1),
\qquad
t(\eta)=\frac{\Omega}{2H_0(1-\Omega)^{3/2}}(\sinh\eta-\eta).}
$$
The parameter is determined explicitly by
$$
\boxed{\eta=2\operatorname{arsinh}\sqrt{\frac{(1-\Omega)a}{\Omega}},
\qquad \eta=\sqrt{-K}\int_0^t\frac{dt'}{a(t')}.}
$$
Thus $\eta$ is <conformal time> multiplied by $\sqrt{-K}$, with its origin at the <Big Bang>; it is not equal to unscaled <cosmic time>. In particular today $\eta_0=\operatorname{arcosh}(2/\Omega-1)$. Small $\eta$ gives $a\propto\eta^2$ and $t\propto\eta^3$, hence $a\propto t^{2/3}$. Large $\eta$ gives $a/t\to H_0\sqrt{1-\Omega}=\sqrt{-K}$, in agreement with the <curvature-dominated universe> limit.
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