For , the slow-roll approximation gives
Taking the expanding branch and using gives the first-order linear differential equation
Thus the quartic-potential slow-roll solution with the normalization used in this question is
If , the field remains negative and , so it rolls toward the potential minimum at zero.
For and positive , the slow-roll Friedmann equation gives
The scalar equation then reduces to
With ,
Furthermore,
Thus the quartic-potential slow-roll solution is