The small-$\theta$ expansions are
$$
a=\frac A2\theta^2\left(1-\frac{\theta^2}{12}+\cdots\right),
\qquad
t=\frac B6\theta^3\left(1-\frac{\theta^2}{20}+\cdots\right).
$$
Writing $x=(6t/B)^{1/3}$ and reverting the second series gives $\theta=x(1+x^2/60+\cdots)$. Therefore
$$
\frac a{t^{2/3}}
=\boxed{\frac A2\left(\frac6B\right)^{2/3}}
\left[1-\frac1{20}\left(\frac{6t}{B}\right)^{2/3}+\cdots\right],
$$
so $C=(A/2)(6/B)^{2/3}$. The leading $a\propto t^{2/3}$ is the flat matter-dominated limit.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 310 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-310/1/d/solution