Let's say there is a stable population size Q that p(t) approaches as time passes. Thus the speed at which the population is growing will approach zero as the population size approaches Q. One way to model this is via the differential equation
p' = kp(Q-p) p(0) = A.
The solution of this initial value problem is p(t) =

Question

Let's say there is a stable population size Q that p(t) approaches as time passes. Thus the speed at which the population is growing will approach zero as the population size approaches Q. One way to model this is via the differential equation
p' = kp(Q-p)         p(0) = A.
The solution of this initial value problem is p(t) =

anonymous · Answer

If $$\frac{dp}{dt}=kp(Q-p)$$then$$\frac{dt}{dp}=\frac{1}{k}\frac{1}{p(Q-p)}=\frac{1}{k}[\frac{1}{Q(Q-p)}+\frac{1}{Qp}]$$thus integrate both sides with respect to p$$t=\frac{1}{kQ}[ln(p)-ln(Q-P)]=\frac{1}{kQ}ln(\frac{p}{Q-p})$$so$$kQt = ln(\frac{p}{Q-p})$$$$e^{kQt}=\frac{p}{Q-p}$$or$$p=\frac{Qe^{kQt}}{1+e^{kQt}}$$at t = 0$$p(0) = A = \frac{Q 1}{1+1}=Q/2$$or Q = 2A, so$$p(t)= \frac{2A e^{kQt}}{1+e^{kQt}}$$

anonymous · Answer

ohhh i see where i went wrong, thanks for the help!