Chapter 6.5, Problem 4QQ

(a)

To determine

To find: The value of

Answer to Problem 4QQ

The answer:

Explanation of Solution

Given information:

The logistic differential equation is $\frac{dP}{dt}=\frac{P}{5}\left(1-\frac{P}{10}\right)$

And

  $P(0)=3$

Calculation:

Know that,

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

The above equation is known as the logistic DE and its solution is:

  $P(t)=\frac{M}{1+A{e}^{-kt}}\mathrm{.....}(1)$

Where,

  $A=\frac{M-P_0}{P_0}$

Given that,

  $\frac{dP}{dt}=\frac{P}{5}\left(1-\frac{P}{10}\right)$

To have,

  $\begin{array}{c}M=10,\\ k=1/5\\ P_0=3\end{array}$

Therefore,

  $\begin{array}{c}A=\frac{10-3}{3}\\ =\frac{7}{3}\end{array}$

Put this in equation $(1)$ .

And get,

  

Therefore, the required value of

(b)

To determine

To find: The value of

Answer to Problem 4QQ

The answer:

Explanation of Solution

Given information:

The logistic differential equation is $\frac{dP}{dt}=\frac{P}{5}\left(1-\frac{P}{10}\right)$

And

  $P(0)=20$

Calculation:

Know that,

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

The above equation is known as the logistic DE and its solution is:

  $P(t)=\frac{M}{1+A{e}^{-kt}}\mathrm{.....}(1)$

Where,

  $A=\frac{M-P_0}{P_0}$

Given that,

  $\frac{dP}{dt}=\frac{P}{5}\left(1-\frac{P}{10}\right)$

To have,

  $\begin{array}{c}M=20,\\ k=1/5\\ P_0=3\end{array}$

Therefore,

  $\begin{array}{c}A=\frac{20-3}{3}\\ =\frac{17}{3}\end{array}$

Put this in equation $(1)$ .

And get,

  

Therefore, the required value of

(c)

To determine

To find: The value of $Y(t)$ .

Answer to Problem 4QQ

The answer: $Y=3\exp \left[\frac{1}{5}\left(t-\frac{t^2}{20}\right)\right]$

Explanation of Solution

Given information:

The logistic differential equation is $\frac{dY}{dt}=\frac{Y}{5}\left(1-\frac{t}{10}\right)$

And

  $Y(0)=3$

Calculation:

Given that,

  $\frac{dY}{dt}=\frac{Y}{5}\left(1-\frac{t}{10}\right)$

Then,

  $\begin{array}{c}\frac{dY}{Y}=\frac{1}{5}\left(1-\frac{t}{10}\right)dt\\ {{\displaystyle \to \int }}^{\text{}}\frac{dY}{Y}=\frac{1}{5}{{\displaystyle \int }}^{\text{}}1-\frac{t}{10}dt\\ \to \ln \left|Y\right|=\frac{1}{5}\left(t-\frac{t^2}{20}\right)+C\\ \to \ln \left|3\right|=\frac{1}{5}\left(0-\frac{0^2}{20}\right)+C\\ \to \ln 3=0+C\end{array}$

Put the value of $C$ and get,

  $\begin{array}{c}\ln \left|Y\right|=\frac{1}{5}\left(t-\frac{t^2}{20}\right)+\ln 3\\ \to \ln Y=\frac{1}{5}\left(t-\frac{t^2}{20}\right)+\ln 3\\ \to Y=3\exp \left[\frac{1}{5}\left(t-\frac{t^2}{20}\right)\right]\end{array}$

Therefore, the required value of $Y=3\exp \left[\frac{1}{5}\left(t-\frac{t^2}{20}\right)\right]$ .

(d)

To determine

To find: The value of

Answer to Problem 4QQ

The answer:

Explanation of Solution

Given information:

The logistic differential equation is $\frac{dY}{dt}=\frac{Y}{5}\left(1-\frac{t}{10}\right)$

Calculation:

From part $(c)$ :

  $Y=3\exp \left[\frac{1}{5}\left(t-\frac{t^2}{20}\right)\right]$

Then,

  

The concerned polynomial's limit approaches negative infinity when moves closer to positive infinity because its leading coefficient is negative.

  

Therefore, the required value of