= Solution
For $V=\lambda\phi^4/4$ and positive $\phi$, the slow-roll Friedmann equation gives
$$
H\simeq\phi^2\sqrt{\frac{2\pi G\lambda}{3c^2}}.
$$
The scalar equation then reduces to
$$
\dot\phi\simeq
-c\sqrt{\frac{\lambda}{6\pi G}}\,\phi.
$$
With $\phi(t_0)=\phi_0$,
$$
\boxed{
\phi(t)=\phi_0
\exp\left[-c\sqrt{\frac{\lambda}{6\pi G}}(t-t_0)\right]
}.
$$
Furthermore,
$$
\frac{d\log a}{d\phi}
=\frac H{\dot\phi}
\simeq-\frac{3H^2}{V'}
=-\frac{2\pi G}{c^2}\phi.
$$
Thus the <quartic-potential slow-roll solution> is
$$
\boxed{
a(\phi)=a_0
\exp\left[
\frac{\pi G}{c^2}(\phi_0^2-\phi^2)
\right]
}.
$$
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