 Solution

Here $H=\dot{a}/a=2/(3t)$, so the perturbation equation becomes $\ddot{\delta }+(4/3t)\dot{\delta }-(2/3t^2)\delta =0$. A trial $t^s$ gives $s(s-1)+(4/3)s-2/3=0$, with roots $2/3,-1$. ThusThe pure growing mode obeys ${\delta }_k(t)/{\delta }_k(t_i)=a(t)/a(t_i)$; for generic data the decaying component eventually becomes negligible, though the growing-mode normalization must be determined from both initial data.

With $a(t_0)=1$, the physical wavelength is $2\pi a(t)/k$. Equating it to the Hubble radius $c/H=3ct/2$, and using $k_0=2\pi H_0/c$, givesGrowing-mode evolution from crossing to today multiplies the perturbation amplitude by $(t_0/t_k{)}^{2/3}=(k/k_0{)}^2$. Hence the supplied horizon-crossing variance givesThis conclusion uses the stated matter-dominated growing-mode approximation over the relevant evolution, rather than an additional radiation-era transfer function.