= Solution
{c}
Let $t_0$ be the initial epoch, let $r(t)$ be the physical radius of the outer matter shell, and put $r(t_0)=r_0$. The background <Einstein-de Sitter universe> has <critical density> $\rho_b=3H_i^2/(8\pi G)$ at that epoch. Its mass inside $r_0$ is $M_b=M-\delta M$, where $M$ is the actual conserved mass. Hence the background <Friedmann equation> implies
$$
H_i^2=\frac{2G(M-\delta M)}{r_0^3},\qquad
\dot r_0^2=H_i^2r_0^2=\frac{2G(M-\delta M)}{r_0}.
$$
For <spherical collapse> of pressureless matter, with no <shell crossing>, the <shell theorem> gives $\ddot r=-GM/r^2$. Its <conservation of energy> first integral is
$$
\frac12\dot r^2-\frac{GM}{r}
=\frac12\dot r_0^2-\frac{GM}{r_0}
=-\frac{G\delta M}{r_0}.
$$
Consequently
$$
\dot r^2=\frac Ar-B,\qquad
\boxed{A=2GM,\qquad B=\frac{2G\delta M}{r_0}.}
$$
For $\delta M>0$, the initially expanding branch reaches <cosmological turnaround> at $r_{\rm ta}=A/B=Mr_0/\delta M$. Since $dt=dr/\dot r$ on that branch, the elapsed <cosmic time> is
$$
\boxed{\tau=t_{\rm ta}-t_0
=\int_{r_0}^{A/B}\sqrt{\frac r{A-Br}}\,dr.}
$$
This is <spherical turnaround from an unperturbed Hubble velocity>. The integral measures the comoving matter observers' time interval in the pressureless model. The integration variable is the physical radius, not the radius divided by the background <scale factor>. If an absolute <cosmic time> is wanted, add $t_0$ to the displayed interval.
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