Solution
= Solution
Substitution gives $\dot\phi=2M_{\rm Pl}/t$, $\ddot\phi=-2M_{\rm Pl}/t^2$ and $V=V_*e^{-\phi_*/M_{\rm Pl}}(t_*/t)^2=10M_{\rm Pl}^2/t^2$, exactly satisfying both equations when $\phi_*=M_{\rm Pl}\log(V_*t_*^2/(10M_{\rm Pl}^2))$. Since $a\propto t^2$, $\ddot a>0$: it inflates. $N=2\log(t_*/t)$ and $\Delta\phi=2M_{\rm Pl}\log(t_*/t)=NM_{\rm Pl}$, so 60 e-folds require a field excursion about $60M_{\rm Pl}$.