Solution
= Solution
The slow-roll equations imply
$$
\frac1a\frac{da}{d\phi}
=\frac{H}{\dot\phi}
=-\frac{3H^2}{c^2V'(\phi)}
=-\frac{8\pi G}{c^4}\frac{V}{V'}.
$$
For $V=\lambda\phi^n/n$, $V/V'=\phi/n$. Integrating from $\phi_i$ and using $a(\phi_i)=1$ gives
$$
\boxed{
a(\phi)=\exp\left[
\frac{4\pi G}{c^4n}(\phi_i^2-\phi^2)
\right]}.
$$