Use $d/dt=aH,d/da$. Then
$$
\dot\delta_m=aH\frac{d\delta_m}{da},
$$
$$
\ddot\delta_m=a^2H^2\frac{d^2\delta_m}{da^2}
+aH(H+aH')\frac{d\delta_m}{da},
$$
where now $H'=dH/da$. Also
$$
4\pi G\bar\rho_m
=\frac32\Omega_{m,0}H_0^2a^{-3}.
$$
Substitution into part (a) and division by $a^2H^2$ gives the <matter growth equation as a function of scale factor>
$$
\boxed{
\frac{d^2\delta_m}{da^2}
+\left(\frac{d\log H}{da}+\frac3a\right)
\frac{d\delta_m}{da}
-\frac{3\Omega_{m,0}H_0^2}{2a^5H^2}\delta_m=0}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 310 / 3 / c / Solution

In a matter-plus-cosmological-constant universe, changing the independent variable from cosmic time to $a$ gives
$$
\frac{d^2\delta_m}{da^2}
+\left(\frac{d\log H}{da}+\frac3a\right)\frac{d\delta_m}{da}
-\frac{3\Omega_{m,0}H_0^2}{2a^5H^2}\delta_m=0.
$$


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