Fit a logistic model using both printed integration methods

For $0<x<1000$, $1/[x(1000-x)]=\tfrac1{1000}[1/x+1/(1000-x)]$. Integration gives $\ln|x|-\ln|1000-x|=1000kt+C_0$, where the combined constant can also be written as 1000C, as in the source.

Since $0<x<1000$, remove the modulus signs. “Initially” means t=0: x=1 gives $C_0=-\ln999$. At t=5, x=120, so $5000k=\ln120-\ln880+\ln999=\ln(2997/22)$. Setting x=500 gives $1000kt=\ln999$, hence $t=5\ln999/\ln(2997/22)\approx7.03$ days.

$\ln|x/(1000-x)|=1000kt+C_0$ gives $x/(1000-x)=Ae^{1000kt}$ after absorbing sign and constant magnitude into A. At t=0, A=1/999. At t=5, $e^{5000k}=120(999)/880=2997/22$, so $k=\ln(2997/22)/5000$. At x=500 the ratio is 1, yielding the same 7.03 days. One source substitution line drops the last 2 in the ratio denominator and shows 2997/2; the earlier fitted k and later calculation both use 2997/22. The correct fitted value is retained here.

The population remains constant: nobody dies, leaves or enters the community. This is the printed closed-community assumption. The model also represents infected population continuously and has no recovery term; its solution from x=1 stays between 0 and 1000. The mathematical equilibria x=0 and x=1000 were excluded during separation but do not match the initial condition.