>>12858029If you want the next term, include all terms involving 1/e^2t in the taylor series.
In general write y(t) = e^f(t), y' = f'e^f, y'' = (f'^2 +f'')e^f, etc.
This lets e^f factor out.
0th order gives y' = y so f' ~ 1 so f ~ t and f = t + g where g<<t and g'<<1.
1st order gives 1 + g' ~ 1 + (1 + g')t/e^t
cancelling and taking the most significant terms gives g' ~ t/e^t so g ~ -(t+1)/e^t and g = -(t+1)/e^t + h where h,h',h''<<t/e^t.
2nd order gives 1 + t/e^t + h' ~ 1 + (1 + t/e^t + h')(t/e^t - (t^2)/(2e^2t)) + (t^2)/(2e^2t)((1 + t/e^t + h')^2 + (1-t)/e^t + h'')
cancelling and taking significant terms gives h' ~ (t^2)/e^2t so h ~ -(2t^2 + 2t +1)/(4e^(2t))
the 3rd order term is -(18t^3 +9t^2 + 6t +2)/(54e^(3t))
y~ e^[t - (t+1)/e^t - (2t^2 + 2t +1)/(4e^(2t)) - (18t^3 +9t^2 + 6t +2)/(54e^(3t))]
You can probably Pade the series to do even better
