= Solution
Multiply the growth equation by $2\eta^2(1+\eta)$:
$$
2\eta^2(1+\eta)D_{\eta\eta}+\eta(3+4\eta)D_\eta-3D=0.
$$
Use the <Frobenius method>, writing $D=\sum_{n\geq0}b_n\eta^{\alpha+n}$ with $b_0\ne0$. The lowest power gives the <indicial equation>
$$
2\alpha(\alpha-1)+3\alpha-3=(\alpha-1)(2\alpha+3)=0.
$$
Thus the leading powers are $\alpha=1$ and $\alpha=-3/2$. For $n\geq1$, equating the coefficient of $\eta^{\alpha+n}$ gives
$$
(\alpha+n-1)[2(\alpha+n)+3]b_n
+2(\alpha+n-1)(\alpha+n)b_{n-1}=0.
$$
For the growing branch $\alpha=1$, $b_n=-2(n+1)b_{n-1}/(2n+5)$, so $b_1=-4b_0/7$ and $b_2=8b_0/21$. Taking $b_0=A_k$,
$$
\boxed{\delta_C=A_k\eta\left[1-\frac47\eta+\frac8{21}\eta^2+O(\eta^3)\right].}
$$
Since $\eta\propto a$, the leading behavior is precisely the usual growing <density contrast> $\delta_C\propto a\propto t^{2/3}$ during <matter domination>. The negative first correction shows the beginning of suppressed growth as the string fraction increases. The other branch starts as $a^{-3/2}$, the familiar decaying matter mode; its exact form is proportional to $\sqrt{1+\eta}/\eta^{3/2}$.
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