Since log(1-F) and log(F-1) differ by a constant, there is
no need to special case 0 1.
If you want to solve for F, you'll need to use some tricks:
(%i1) de : 'diff(f,t) = f*(1-f)$
(%i2) ode2(de,f,t);
(%o2) log(f)-log(f-1)=t+%C
(%i3) logcontract(%);
(%o3) log(f/(f-1))=t+%C
(%i4) map('exp,%);
(%o4) f/(f-1)=%e^(t+%C)
(%i5) solve(%,f);
(%o5) [f=%e^(t+%C)/(%e^(t+%C)-1)]
The replacement exp(%c) --> %c makes the solution nicer:
(%i6) ratsubst(%c, exp(%c),%);
(%o6) [f=(%c*%e^t)/(%c*%e^t-1)]
Maxima's solution does miss the two singular solutions (F = 0 and F = 1).
Yes, (%o6) does include F = 0, but (%o6) should carry the proviso
that %c is not equal to zero (zero is the only complex number not in the
range of exp.)
Barton
-----maxima-admin@math.utexas.edu wrote: -----
>When Maxima solves a simple ode (logistic
>equation):
>
>dF/dt = F*(1-F)
>
>it only comes up with a single answer:
>
>log(F)-log(F-1)=t+%C
>
>which is correct but does not include the region
>of practical intrest, i.e. 0<1.
>
>Can I ask or force Maxima to produce the other
>solution: log(F)-log(1-F)=t+%C ?