= Solution
The <flatness problem> is that the observed <cosmological density parameter> is close to $\Omega=1$, although in an ordinary decelerating universe any departure from one grows with time. Extrapolation backward therefore requires implausibly precise early cancellation of the curvature term.
The <cosmological perfect-fluid continuity equation> is
$$
\dot\rho+3H(\rho+P)=0.
$$
Consequently
$$
\frac d{dt}(\rho a^2)
=a^2\dot\rho+2a\dot a\rho
=-Ha^2(\rho+3P).
$$
For an expanding universe $H>0$. If
$$
\rho+3P<0,
$$
then $\rho a^2$ increases. The <Friedmann equation> implies
$$
\left|\Omega^{-1}-1\right|
=\frac{\text{constant}\times |k|}{\rho a^2}.
$$
It therefore decreases during this period, driving $\Omega$ toward one. By the <Friedmann acceleration equation>, the same condition gives $\ddot a>0$ when the cosmological constant is included in the effective fluid. This is the <inflationary solution of the flatness problem>.
For the scalar field, the <slow-roll approximation> means
$$
\dot\phi^2\ll V(\phi),
\qquad
|\ddot\phi|\ll3H|\dot\phi|.
$$
The equations reduce to
$$
3H^2\simeq m^2\phi^2,
\qquad
3H\dot\phi\simeq-2m^2\phi.
$$
Taking the positive-field branch gives $H\simeq m\phi/\sqrt3$, and therefore
$$
\dot\phi\simeq-\frac{2m}{\sqrt3}.
$$
With $\phi(0)=\phi_i$,
$$
\boxed{\phi(t)=\phi_i-\frac{2m}{\sqrt3}t.}
$$
Now integrate $H=\dot a/a\simeq m\phi/\sqrt3$:
$$
\log\frac{a(t)}{a_i}
=\int_0^t\frac m{\sqrt3}
\left(\phi_i-\frac{2m}{\sqrt3}s\right)ds
=\frac{m\phi_i}{\sqrt3}t-\frac{m^2t^2}{3}.
$$
Thus the <quadratic-potential slow-roll solution> is
$$
\boxed{
a(t)=a_i\exp\left[
\frac{m\phi_i}{\sqrt3}t-\frac{m^2t^2}{3}
\right].}
$$
Finally,
$$
\phi_i^2-\phi(t)^2
=\frac{4m\phi_i}{\sqrt3}t-\frac{4m^2}{3}t^2,
$$
so equivalently
$$
\boxed{
a(t)=a_i\exp\left[
\frac{\phi_i^2-\phi(t)^2}{4}
\right].}
$$
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