Respuesta :
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)=1
Substitute 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.
