Suppose a certain population satisfies the logistic equation given by dPdt = 10P − P2 with P(0) = 1.Determine the population when t = 3

Answer:The population when t = 3 is 10.Step-by-step explanation:Suppose a certain population satisfies the logistic equation given by[tex]\frac{dP}{dt}=10P-P^2[/tex]with P(0)=1. We need to find the population when t=3.Using variable separable method we get [tex]\frac{dP}{10P-P^2}=dt[/tex]Integrate both sides. [tex]\int \frac{dP}{10P-P^2}=\int dt[/tex]             .... (1)Using partial fraction[tex]\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}[/tex][tex]A=\frac{1}{10},B=\frac{1}{10}[/tex]Using these values the equation (1) can be written as [tex]\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt[/tex] [tex]\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt[/tex]On simplification we get [tex]\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C[/tex] [tex]\frac{1}{10}(\ln \frac{P}{10-P})=t+C[/tex]We have P(0)=1Substitute t=0 and P=1 in above equation. [tex]\frac{1}{10}(\ln \frac{1}{10-1})=0+C[/tex] [tex]\frac{1}{10}(\ln \frac{1}{9})=C[/tex]The required equation is [tex]\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})[/tex]Multiply both sides by 10.[tex]\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}[/tex][tex]e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}[/tex][tex]\frac{P}{10-P}=\frac{1}{9}e^{10t}[/tex]Reciprocal it[tex]\dfrac{10-P}{P}=9e^{-10t}[/tex][tex]P(t)=\dfrac{10}{1+9e^{-10t}}[/tex]The population when t = 3 is [tex]P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}[/tex]Using calculator,[tex]P=9.999\approx 10[/tex]Therefore, the population when t = 3 is 10.