= Solution
For conserved nonrelativistic <dark matter> and <radiation>, $\rho_{\rm DM}\propto a^{-3}$ and $\rho_\gamma\propto a^{-4}$, so $y=\rho_{\rm DM}/\rho_\gamma=a/a_{\rm eq}$ obeys $\dot y=yH$. Denote differentiation with respect to $y$ by a prime. Then
$$
\dot\delta=yH\delta',\qquad
\ddot\delta=y^2H^2\delta''+y(\dot H+H^2)\delta'.
$$
The supplied <Friedmann equations> imply
$$
\frac{\dot H}{H^2}=-\frac{4+3y}{2(1+y)},\qquad
\frac{4\pi G\rho_{\rm DM}}{H^2}=\frac{3y}{2(1+y)}.
$$
Substitution in the linear growth equation and division by $y^2H^2$ give the <Mészáros equation>
$$
\boxed{\delta''+\frac{2+3y}{2y(1+y)}\delta'-\frac{3}{2y(1+y)}\delta=0.}
$$
For $D_1=1+3y/2$, $D_1'=3/2$ and $D_1''=0$, making the two remaining terms cancel exactly. For <reduction of order>, put $\delta=fD_1$ and $P=(2+3y)/[2y(1+y)]=1/y+1/[2(1+y)]$. The equation reduces to $f''+(2D_1'/D_1+P)f'=0$, whence
$$
f'=\frac{C}{y\sqrt{1+y}(1+3y/2)^2}.
$$
Using the supplied antiderivative and reversing an irrelevant overall sign gives the independent solution
$$
\boxed{D_2(y)=\left(1+\frac32y\right)\ln\frac{\sqrt{1+y}+1}{\sqrt{1+y}-1}-3\sqrt{1+y},\qquad
\delta=A_1D_1+A_2D_2.}
$$
Its <Wronskian> is $D_1D_2'-D_1'D_2=-1/[y\sqrt{1+y}]\ne0$. The asymptotic expansions are
$$
\begin{array}{c|cc}
&D_1&D_2\\ \hline
y\ll1&1+3y/2&\ln(4/y)-3+O(y\ln(1/y))\\
y\gg1&3y/2+1&\frac4{15}y^{-3/2}+O(y^{-5/2})
\end{array}
$$
Thus subhorizon <dark matter> during <radiation domination> has constant and logarithmic modes, rather than the superhorizon $a^2$ growth of <6/b>. A combination with negative $A_2$ supplies positive logarithmic growth. During <matter domination>, the growing mode is proportional to $a$, agreeing with <6/b>, while the second mode decays as $a^{-3/2}$. These results neglect subhorizon radiation clustering; the <Mészáros equation> is the approximation stated in the question, not a full photon–matter perturbation system through horizon entry.
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