Chapter 7.5, Problem 17E

i.

To determine

To Prove: That the solution is increasing when $m<P<M$ and decreasing when $0<P<m$ .

Explanation of Solution

Given: The differential equation,

  $\frac{dP}{dt}=kP(1-\frac{P}{M})(1-\frac{m}{P})$

Calculation:

From the given information,

If $m<P<M$ , then $(1-\frac{P}{M})>0$ and $(1-\frac{m}{P})>0$ , thus the product of the two will also be positive.

So $\frac{dP}{dt}>0$ this means that the solution is increasing.

If $0<P<m$ , then $(1-\frac{P}{M})>0$ assuming ( $P>M$ ) and $(1-\frac{m}{P})<0$ , thus the product of the two will also be negative.

So $\frac{dP}{dt}<0$ this means that the solution is decreasing.

ii.

To determine

To solve: The differential equation.

Answer to Problem 17E

The solution of the differential equation is,

  $\ln \frac{P-1000}{P-200}=-\frac{8t}{125}+\ln \frac{P_0-1000}{P_0-200}$ .

Explanation of Solution

Given:

The differential equation is

  $\frac{dP}{dt}=kP(1-\frac{P}{M})(1-\frac{m}{P})$ .

Calculation:

From the given information,

  $\frac{dP}{dt}=kP(1-\frac{P}{M})(1-\frac{m}{P})$

  $\frac{dP}{dt}=\frac{-1}{12500}(P-1000)(P-200)$

  $\frac{dP}{(P-1000)(P-200)}=\frac{-1}{12500}dt$

Integrating both sides,

  ${\displaystyle \int \frac{dP}{(P-1000)(P-200)}}={\displaystyle \int \frac{-1}{12500}dt}$

  $\ln \frac{P-1000}{P-200}=-\frac{8t}{125}+c$

Put $t=0$

  $\ln \frac{P_0-1000}{P_0-200}=c$

Therefore,

  $\ln \frac{P-1000}{P-200}=-\frac{8t}{125}+\ln \frac{P_0-1000}{P_0-200}$ .

So the solution of the differential equation is,

  $\ln \frac{P-1000}{P-200}=-\frac{8t}{125}+\ln \frac{P_0-1000}{P_0-200}$ .