Solution
= Solution
Since $a(\tau)=(\tau/\tau_0)^2$, the conformal Hubble parameter is $\mathcal H=2/\tau$. With $\Omega_M=1$, the mode equation becomes
$$
\delta''+\frac2\tau\delta'-\frac6{\tau^2}\delta=0.
$$
Substitution of $\delta=\tau^\beta$ gives
$$
\beta(\beta-1)+2\beta-6
=(\beta-2)(\beta+3)=0.
$$
The two independent solutions are therefore $\tau^2$ and $\tau^{-3}$, so
$$
\boxed{\delta(\mathbf k,\tau)
=A(\mathbf k)\tau^2+B(\mathbf k)\tau^{-3}}.
$$