Logistic Models (College Board AP® Calculus BC): Revision Note

Logistic models

What is a logistic model?

• The standard logistic growth model is represented by the first-order logistic differential equation

  $\frac{\mathit{d}y}{\mathit{d}t}=ky(a-y)$

  • $y$ represents the size of a quantity at time $t$

    • Usually this will be the size of a population of some sort

    • This means $y\ge 0$ will always be true

  • ​$k$ is a constant determining the relative rate of change of $y$

    • In general $k>0$, with a larger value of $k$ representing a faster rate of change

  • $a$ is a constant known as the carrying capacity

    • This is also usually a positive constant

    • It places a limit on the size to which $y$ can grow (if $y$ is increasing) or shrink (if $y$ is decreasing)

    • $y$ will converge to the value $a$ as $t\to \infty$

What information can I derive from the logistic differential equation?

• Consider different values of $y$ in relation to the equation $\frac{\mathit{d}y}{\mathit{d}t}=ky(a-y)$ with $k,a>0$ and $y\ge 0$

  • If $y=0$ or $y=a$ then the rate of change $\frac{dy}{dt}=0$

    • i.e. $y$ is constant and will stay at that value forever

  • If $y<a$, then $\frac{dy}{dt}>0$

    • i.e. $y$ is increasing (and will continue to increase towards the value of $a$)

  • If $y>a$, then $\frac{dy}{dt}<0$

    • i.e. $y$ is decreasing (and will continue to decrease towards the value of $a$)

• Also note that $ky(a-y)=-k{y}^2+aky$ is a quadratic function in $y$ with a maximum value when $y=\frac{a}{2}$

  • That is when the maximum value of $\frac{dy}{dt}$ occurs

  • So for the case where $y<a$, and the initial value satisfies $y_0\le \frac{a}{2}$, the value of $y$ will be changing the fastest when $y$ is equal to half the carrying capacity

  • If you need to know the time $t$ when that occurs then you will need to solve the logistic equation

    • and then solve for the value of $t$ when $y=\frac{a}{2}$

How do I solve a logistic differential equation?

• Solving the differential equation uses separation of variables

  • It also requires partial fractions to evaluate one of the integrals

• Start by separating the variables

$\frac{\mathit{d}y}{\mathit{d}t}=ky(a-y)\Rightarrow \frac{1}{y\left(a-y\right)}\frac{dy}{dt}=k\Rightarrow \int \frac{1}{y\left(a-y\right)}dy=\int kdt$

• Use partial fractions to rewrite the integral on the left

  • See the 'Integration Using Partial Fractions' study guide

$\begin{array}{rcl}\frac{1}{a}\int \left(\frac{1}{y}+\frac{1}{a-y}\right)dy& =& \int kdt\\ \int \left(\frac{1}{y}+\frac{1}{a-y}\right)dy& =& \int akdt\end{array}$

• Integrate both sides of the equation

  • Don't forget a constant of integration

  • $\frac{1}{a-y}$ can be integrated using the substitution $u=a-y$

    • See the 'Integration Using Substitution' study guide

$\begin{array}{rcl}\ln \left|y\right|-\ln \left|a-y\right|& =& akt+C\\ \ln \left|\frac{y}{a-y}\right|& =& akt+C\\ \left|\frac{y}{a-y}\right|& =& e^{akt+C}\end{array}$

• As long as $y>0$ and $y<a$ initially, then $0<y<a$ will always remain true

  • In that case $\frac{y}{a-y}>0$, so that $\left|\frac{y}{a-y}\right|=\frac{y}{a-y}$

$$\begin{gathered} \frac{y}{a-y}=e^{akt+C}=e^C{e}^{akt}=B{e}^{akt} \\ \left(\mathrm{where}B=e^C\right) \end{gathered}$$

• That can be solved algebraically for $y$ by making $y$ the subject to get

$y=\frac{aB{e}^{akt}}{1+B{e}^{akt}}$

• You can multiply top and bottom by $\frac{1}{B}{e}^{-akt}$ and rewrite $\frac{1}{B}$ and a different constant $A$

$y=\frac{a}{1+A{e}^{-akt}}$

• If you are given the initial condition that $y=y_0$ when $t=0$, then you can also work out the value of $A$

  • Note that $A$ is just a rewritten form of the constant of integration

$y_0=\frac{a}{1+A{e}^{-ak\left(0\right)}}\Rightarrow y_0=\frac{a}{1+A}\Rightarrow A=\frac{a-y_0}{y_0}$

• Be careful if $y>a$ initially

  • In that case $0<a<y$ will always remain true

  • And $\frac{y}{a-y}<0$, so that $\left|\frac{y}{a-y}\right|=-\frac{y}{a-y}$

    • The minus sign means the last few steps, along with the final solution, will be slightly different

  • You get $y=\frac{a}{1-A{e}^{-akt}}$