Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 55 / 1 / iii / Solution

Multiply the growth equation by $2{\eta }^2(1+\eta )$:Use the Frobenius method, writing $D={\sum }_{n\ge 0}{b}_n{\eta }^{\alpha +n}$ with $b_0\ne 0$. The lowest power gives the indicial equationThus the leading powers are $\alpha =1$ and $\alpha =-3/2$. For $n\ge 1$, equating the coefficient of ${\eta }^{\alpha +n}$ givesFor the growing branch $\alpha =1$, $b_n=-2(n+1)b_{n-1}/(2n+5)$, so $b_1=-4b_0/7$ and $b_2=8b_0/21$. Taking $b_0=A_k$,Since $\eta \propto a$, the leading behavior is precisely the usual growing density contrast ${\delta }_C\propto a\propto t^{2/3}$ during matter domination. The negative first correction shows the beginning of suppressed growth as the string fraction increases. The other branch starts as $a^{-3/2}$, the familiar decaying matter mode; its exact form is proportional to $\sqrt{1+\eta }/{\eta }^{3/2}$.