Chapter 7, Problem 15RE

a.

To determine

To calculate : the population when $t=20$ and also write the solution of the initial value problem

Answer to Problem 15RE

The population when $t=20$ is $560$

Explanation of Solution

Given information : The question is

  $\frac{dP}{dt}=0.1P\left(1-\frac{P}{2000}\right);P\left(0\right)=100$

Calculation :

  $\frac{dP}{dt}=0.1P\left(1-\frac{P}{2000}\right)$

The above equation is logistic differential equation

So, $\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)$

  $\begin{array}{l}k=0.1;M=2000\&P_0=P\left(0\right)=100\\ P\left(20\right)=\frac{2000}{1+19e^{-0.1\times 20}}\\ P\left(20\right)=\frac{2000}{3.571}\approx 560\\ P\left(t\right)=\frac{2000}{1+19e^{-0.1\times t}}\end{array}$

b.

To determine

When does the population reach $1200$

Answer to Problem 15RE

The population reaches $1200$ in $33.5$

Explanation of Solution

Given information : The question is

  $\frac{dP}{dt}=0.1P\left(1-\frac{P}{2000}\right);P\left(0\right)=100$

Calculation :

From part (a)

  $\begin{array}{l}P\left(t\right)=\frac{2000}{1+19e^{-0.1\times t}}\\ 1200=\frac{2000}{1+19e^{-0.1\times t}}\\ \frac{5}{3}=1+19e^{-0.1\times t}\\ \frac{2}{57}=e^{-0.1t}\\ \ln \frac{2}{57}=\ln e^{-0.1t}\\ t=-\frac{\ln \frac{2}{57}}{\ln e^{-0.1t}}\\ t\approx 33.5\end{array}$