The equation governing density perturbation modes $\delta_{\mathbf{k}}(t)$ in a matter-dominated universe (with $a(t)=\left(t / t_{0}\right)^{2 / 3}$ ) is

$$\ddot{\delta}_{\mathbf{k}}+2 \frac{\dot{a}}{a} \dot{\delta}_{\mathbf{k}}-\frac{3}{2}\left(\frac{\dot{a}}{a}\right)^{2} \delta_{\mathbf{k}}=0$$

where $\mathbf{k}$ is the comoving wavevector. Find the general solution for the perturbation, showing that there is a growing mode such that

$$\delta_{\mathbf{k}}(t) \approx \frac{a(t)}{a\left(t_{i}\right)} \delta_{\mathbf{k}}\left(t_{i}\right) \quad\left(t \gg t_{i}\right)$$

Show that the physical wavelength corresponding to the comoving wavenumber $k=|\mathbf{k}|$ crosses the Hubble radius $c H^{-1}$ at a time $t_{k}$ given by

$$\frac{t_{k}}{t_{0}}=\left(\frac{k_{0}}{k}\right)^{3} \quad, \quad \text { where } \quad k_{0}=\frac{2 \pi}{c H_{0}^{-1}}$$

According to inflationary theory, the amplitude of the variance at horizon-crossing is constant, that is, $\left\langle\left|\delta_{\mathbf{k}}\left(t_{k}\right)\right|^{2}\right\rangle=A V^{-1} / k^{3}$ where $A$ and $V$ (the volume) are constants. Given this amplitude and the results obtained above, deduce that the power spectrum today takes the form

$$P(k) \equiv V\left\langle\left|\delta_{\mathbf{k}}\left(t_{0}\right)\right|^{2}\right\rangle=\frac{A}{k_{0}^{4}} k$$