Question

Suppose a certain population satisfies the logistic equation given by dP

Answer 1

Answer:

The population when t = 3 is 10.

Step-by-step explanation:

Suppose a certain population satisfies the logistic equation given by

$\frac{dP}{dt}=10P-P^2$

with P(0)=1. We need to find the population when t=3.

Using variable separable method we get

$\frac{dP}{10P-P^2}=dt$

Integrate both sides.

$\int \frac{dP}{10P-P^2}=\int dt$             .... (1)

Using partial fraction

$\frac{1}{P(10-P)}=\frac{A}{P}+\frac{B}{(10-P)}$

$A=\frac{1}{10},B=\frac{1}{10}$

Using these values the equation (1) can be written as

$\int (\frac{1}{10P}+\frac{1}{10(10-P)})dP=\int dt$

$\int \frac{dP}{10P}+\int \frac{dP}{10(10-P)}=\int dt$

On simplification we get

$\frac{1}{10}\ln P-\frac{1}{10}\ln (10-P)=t+C$

$\frac{1}{10}(\ln \frac{P}{10-P})=t+C$

We have P(0)=1

Substitute t=0 and P=1 in above equation.

$\frac{1}{10}(\ln \frac{1}{10-1})=0+C$

$\frac{1}{10}(\ln \frac{1}{9})=C$

The required equation is

$\frac{1}{10}(\ln \frac{P}{10-P})=t+\frac{1}{10}(\ln \frac{1}{9})$

Multiply both sides by 10.

$\ln \frac{P}{10-P}=10t+\ln \frac{1}{9}$

$e^{\ln \frac{P}{10-P}}=e^{10t+\ln \frac{1}{9}}$

$\frac{P}{10-P}=\frac{1}{9}e^{10t}$

Reciprocal it

$\dfrac{10-P}{P}=9e^{-10t}$

$P(t)=\dfrac{10}{1+9e^{-10t}}$

The population when t = 3 is

$P(3)=\dfrac{10}{1+9e^{-10\cdot 3}}$

Using calculator,

$P=9.999\approx 10$

Therefore, the population when t = 3 is 10.