Introduction: In this article , we present the solution of the question how to find general solution of logistic differential equation dP/dt =(1/4)(P(t)-2)(4-P(t)) and specific solutions with initial conditions P(0)=2, P(0)=3
Question: Find general solution of logistic differential equation
\[\frac{dP}{dt}=\frac{1}{4}\left(P-2\right)\left(4-P\right)\]
and specific solutions satisfying initial conditions \(P(0)=2, P(0)=3\)
Solution:
First let’s find general solution of logistic differential equation
\[\frac{dP}{dt}=\frac{1}{4}\left(P-2\right)\left(4-P\right)\]
\[\frac{4}{\left(P-2\right)\left(4-P\right)}dP=dt\]
\[\frac{1}{2}\left(\frac{1}{P-2}+\frac{1}{4-P}\right)dP=dt\]
integrating both sides
\[\log\left(P-2\right)-\log\left(4-P\right)=2t+2c\]
where c is integrating constant.
\[\log\left(\frac{P-2}{4-P}\right)=2t+2c\]
\[\frac{P-2}{4-P}=e^{2t+2c}\]
\[P(t)=\frac{4e^{2t+2c}+2}{1+e^{2t+2c}}\]
This is the required general solution of the logistic differential equation.
Case 1 :
Now to find specific solution corresponding to initial condition \(P(0)=2\)
substituting t =0 into P(t) and using initial condition
\[2=\frac{4e^{2c}+2}{1+e^{2c}}\]
\[2+2e^{2c}=4e^{2c}+2\]
\[2e^{2c}=0\]
\[e^{2c}=0\]
Substituting this into expression of P(t) we get specific solution as
\[P(t)=\frac{4e^{2t+2c}+2}{1+e^{2t+2c}}\]
\[P(t)=\frac{4e^{2t}*0+2}{1+e^{2t}*0}\]
\[P(t) =2\]
That is a constant solution for all t .
Case 2:
Now to find specific solution corresponding to initial condition \(P(0)=3\)
\[3=\frac{4e^{2c}+2}{1+e^{2c}}\]
\[3+3e^{2c}=4e^{2c}+2\]
\[e^{2c}=1\]
Substituting this value in to the expression of P(t)
\[P(t)=\frac{4e^{2t}+2}{1+e^{2t}}\]
is the required specific solution of the logistic differential equation.