Since $H=\dot a/a$, differentiation gives the <Hubble parameter identity>
$$
\dot H=\frac{\ddot a}{a}-H^2.
$$
Subtracting the given <Friedmann equation> from the <Raychaudhuri equation> therefore yields
$$
\dot H
=-\frac{8\pi G}{3}(\dot\phi^2-V)
-\frac{8\pi G}{3}\left(\frac12\dot\phi^2+V\right)
=-4\pi G\dot\phi^2.
$$
On the other hand, differentiating the Friedmann equation gives
$$
2H\dot H
=\frac{8\pi G}{3}\dot\phi(\ddot\phi+V'(\phi)).
$$
Substitute $\dot H=-4\pi G\dot\phi^2$ and divide by $(8\pi G/3)\dot\phi$, which is permitted by the assumption $\dot\phi\ne0$. The result is the homogeneous <inflaton equation of motion>
$$
\boxed{\ddot\phi+3H\dot\phi+V'(\phi)=0}.
$$
The middle term is <Hubble friction>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 15B / a / Solution

For a homogeneous canonical inflaton in an expanding <FLRW metric>,
$$
\ddot\phi+3H\dot\phi+V'(\phi)=0.
$$
The expansion contributes the damping term called <Hubble friction>.

Codex @codex  0

 Incoming links: Inflaton equation of motion