 Solution

Write $\delta ={\delta }_{C,\mathbf{k}}$ and $\mathcal{H}=a^{\prime }/a$, the conformal Hubble parameter. The flat Friedmann equation givesOutside the horizon, the stated adiabatic cosmological perturbation relation is ${\delta }_R=(4/3)\delta$. It expresses equality of ${\delta }_i/(1+w_i)$ for dust and radiation in the density convention of the supplied equations. During radiation domination, $a\propto \tau$, so $\mathcal{H}=1/\tau$ and ${\rho }_C$ is negligible. Consequently $8\pi G{a}^2{\rho }_R=3/{\tau }^2$, and the linear cosmological density perturbation equation becomesThe trial solution $\delta \propto {\tau }^q$ gives $q(q-1)+q=4$, hence $q=2,-2$.

During matter domination, $a\propto {\tau }^2$, so $\mathcal{H}=2/\tau$. Now the radiation source is negligible and $4\pi G{a}^2{\rho }_C=(3/2){\mathcal{H}}^2=6/{\tau }^2$. The equation becomeswhose exponents satisfy $q(q-1)+2q=6$, or $(q-2)(q+3)=0$. Thus the growing solutions in both eras areThe other modes are ${\tau }^{-2}$ in the radiation limit and ${\tau }^{-3}$ in the matter limit. These are statements in the density and gauge convention of the supplied evolution equation; a density contrast outside the horizon is not a gauge-independent observable by itself.