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chambek
Group Title
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(Qp) p(0) = A.
The solution of this initial value problem is p(t) =
 one year ago
 one year ago
chambek Group Title
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(Qp) p(0) = A. The solution of this initial value problem is p(t) =
 one year ago
 one year ago

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hewsmike Group TitleBest ResponseYou've already chosen the best response.1
If \[\frac{dp}{dt}=kp(Qp)\]then\[\frac{dt}{dp}=\frac{1}{k}\frac{1}{p(Qp)}=\frac{1}{k}[\frac{1}{Q(Qp)}+\frac{1}{Qp}]\]thus integrate both sides with respect to p\[t=\frac{1}{kQ}[ln(p)ln(QP)]=\frac{1}{kQ}ln(\frac{p}{Qp})\]so\[kQt = ln(\frac{p}{Qp})\]\[e^{kQt}=\frac{p}{Qp}\]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\[p(t)= \frac{2A e^{kQt}}{1+e^{kQt}}\]
 one year ago

chambek Group TitleBest ResponseYou've already chosen the best response.0
ohhh i see where i went wrong, thanks for the help!
 one year ago
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