\[\frac{dP}{dt}=kP(P_0-P)\]
where
\[\frac{dP}{dt}\]
is the rte of population increase\[k \ge 0\]
is constant of proportionality\[P \ge 0\]
is the population\[P_0 \ge 0\]
is the maximum stable populationWe can solve this equation by separation of variables.
\[\frac{dP}{(P(P_0-P)}=dt\]
Write nbsp;
\[\frac{dP}{(P(P_0-P)}\]
as partial fractions\[\frac{dP}{(P(P_0-P)}=\frac{1/P_0}{P}+ \frac{1/P_0}{P_0-P}\]
Now integrate.
\[\int \frac{1/P_0}{P}+\frac{1/P_0}{P_0-P} dP = \int dt\]
\[\frac{1}{P_0}ln(P)- \frac{1}{P_0} ln(P_0-P)=kt+C\]
\[ln(P)- ln(P_0-P)=P_0(kt+C)\]
\[ln(\frac{P}{P_0-P})=P_0(kt+C)\]
\[\frac{P}{P_0-P}=e^{P_0(kt+C)}=Ae^{kt}\]
\[P=(P_0-P)Ae^{kt}=P_0Ae^{kt}-PAe^{kt}\]
\[P+PAe^{kt}=P_0Ae^{kt}\]
\[P(1+Ae^{kt})=P_0Ae^{kt}\]
\[P=\frac{P_0Ae^{kt}}{1+Ae^{kt}}=\frac{AP_0}{e^{-kt}+A}\]
\[A\]
is an arbitrary constant.As
\[t \rightarrow m\infty , \: P \rightarrow P_0\]
.