= Solution
Separate background <stress-energy conservation> gives $\rho_C'=-3\mathcal H\rho_C$ and $\rho_S'=-2\mathcal H\rho_S$. Therefore
$$
\rho_C\propto a^{-3},\quad \rho_S\propto a^{-2},\quad
\eta=\frac{\rho_S}{\rho_C}\propto a,\quad
\eta'=\mathcal H\eta,\quad \Omega_C=\frac1{1+\eta}.
$$
For consistency, compute the expansion derivative from the <Friedmann equation>, rather than assume the printed second hint. Differentiating $\mathcal H^2=(8\pi G/3)a^2\rho_{\rm tot}$ and using total <stress-energy conservation> yields
$$
\mathcal H'=-\frac{4\pi G}{3}a^2(\rho_{\rm tot}+3P_{\rm tot})
=-\frac{4\pi G}{3}a^2\rho_C
=-\frac{\mathcal H^2}{2(1+\eta)}.
$$
The PDF's hint has $\rho_{\rm tot}+P_{\rm tot}$ where $\rho_{\rm tot}+3P_{\rm tot}$ is required. It is incompatible with the first hint and conservation, and would not give the requested equation. The expression above is the corrected conformal-time acceleration identity.
For a function $D(\eta)=\delta_C$, the chain rule gives
$$
\delta_C'=\mathcal H\eta D_\eta,\qquad
\delta_C''=\mathcal H^2\eta^2D_{\eta\eta}
+\eta(\mathcal H'+\mathcal H^2)D_\eta.
$$
Substituting in the cold-matter equation and dividing by $\mathcal H^2\eta^2$ produces
$$
D_{\eta\eta}+\frac{2+\mathcal H'/\mathcal H^2}{\eta}D_\eta
-\frac{3}{2\eta^2(1+\eta)}D=0,
$$
or
$$
\boxed{D_{\eta\eta}+\frac{3+4\eta}{2\eta(1+\eta)}D_\eta
-\frac{3D}{2\eta^2(1+\eta)}=0.}
$$
This is <cold-matter growth in a matter-coasting-fluid universe>; its coefficients depend on the background density ratio, not on $k$.
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