= Solution
For $r=A(1-\cos\theta)$ and $t+T=B(\theta-\sin\theta)$, differentiation gives
$$
\dot r=\frac AB\frac{\sin\theta}{1-\cos\theta},\qquad
\ddot r=-\frac A{B^2(1-\cos\theta)^2}.
$$
Comparison with $-GM/[A^2(1-\cos\theta)^2]$ proves $A^3=GMB^2$. Substitution into the conserved <specific orbital energy> gives $\epsilon=-GM/(2A)$. Therefore
$$
\boxed{A=\frac{r_i(1+\delta_i)}{2\delta_i},\qquad
B=\frac{1+\delta_i}{2H_i\delta_i^{3/2}}.}
$$
These constants exactly reproduce both the initial radius and expanding <velocity>. The initial parameter obeys
$$
\cos\theta_i=\frac{1-\delta_i}{1+\delta_i},\qquad
\theta_i=2\arctan\sqrt{\delta_i}.
$$
In an <Einstein-de Sitter universe> whose background big bang is $t=0$, $t_i=2/(3H_i)$. Thus the exact required offset is
$$
\boxed{T=B(\theta_i-\sin\theta_i)-t_i.}
$$
Expanding $\theta_i-\sin\theta_i=\frac43\delta_i^{3/2}-\frac85\delta_i^{5/2}+O(\delta_i^{7/2})$ gives
$$
T=-\frac15t_i\delta_i+O(t_i\delta_i^2).
$$
\b[One cannot set $T=0$ exactly at finite $\delta_i$ while retaining the stated initial Hubble velocity and background time origin.] For example $\delta_i=0.01$, $t_i=1$ gives $T=-0.00199148$. Shifting to $t'=t+T$ removes the offset from the local orbit, but the background then scales as $a\propto(t'-T)^{2/3}$, not $t'^{2/3}$. Setting $T\simeq0$ is justified in the early-initial-time limit for subsequent evolution; the following standard growing-mode formulae use that approximation.
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