<Neutrino free streaming> gives $\delta_\nu\simeq0$ on the stated scales, so $\delta_m=(1-f_\nu)\delta_{cb}$. With derivatives with respect to $N=\log a$, the growth equation becomes
$$
\delta_{cb,NN}+\frac12\delta_{cb,N}
-\frac32(1-f_\nu)\delta_{cb}=0.
$$
For $\delta_{cb}\propto a^p$,
$$
p=\frac{-1+\sqrt{25-24f_\nu}}4
=1-\frac35f_\nu+O(f_\nu^2).
$$
Hence
$$
\boxed{X(f_\nu)=1-\frac35f_\nu}
$$
to first order.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 310 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-310/3/c/solution