7.9 Logistic Models with Differential Equations

Learning objectives

—FUN-7.H—Interpret the meaning of the logistic growth model in context• FUN-7.H Interpret the meaning of the logistic growth model in context.• FUN-7.H.1 The model for logistic growth that arises from the statement “The rate of change of a quantity is jointly proportional to the size of the quantity and the difference between the quantity and the carrying capacity” is dy/dt = ky(a − y).• FUN-7.H.2 The logistic differential equation and initial conditions can be interpreted without solving the differential equation.• FUN-7.H.3 The limiting value (carrying capacity) of a logistic differential equation as the independent variable approaches infinity can be determined using the logistic growth model and initial conditions.• FUN-7.H.4 The value of the dependent variable in a logistic differential equation at the point when it is changing fastest can be determined using the logistic growth model and initial conditions.• Enduring understanding FUN-7: Solving differential equations allows us to determine functions and develop models.

For $k>0$ , the logistic model $dy/dt=ky(a-y)$ says the rate is jointly proportional to the current amount $y$ and the unused capacity $a-y$ . Here $a$ is the carrying capacity; in this form, $k$ has units $1/$ (quantity $\cdot$ time).

Current value	Sign of $dy/dt$	Model behavior
$y=0$	$0$	equilibrium at zero
$0<y<a$	positive	$y$ increases toward $a$
$y=a$	$0$	equilibrium at carrying capacity
$y>a$	negative	$y$ decreases toward $a$

The initial condition tells which row applies. If $0<y(0)<a$ , the solution increases but the factor $a-y$ shrinks, so growth slows and $y$ approaches $a$ as $t\to\infty$ . If $y(0)=0$ or $y(0)=a$ , the rate remains zero and the solution stays at that equilibrium.

For values between $0$ and $a$ , the growth rate as a function of $y$ is $k(ay-y^2)$ , a downward-opening quadratic. Its vertex occurs at $y=a/2$ , so the quantity is increasing fastest when it reaches half the carrying capacity.

In the illustrative model $dy/dt=ky(1000-y)$ with $k>0$ and $y(0)=100$ , the quantity initially grows, approaches the carrying capacity $1000$ , and has its greatest growth rate when $y=500$ . None of these conclusions requires solving for $y(t)$ .

The carrying capacity is the value that makes the capacity factor zero, not the coefficient $k$ . Also distinguish “largest value of $y$ ” from “largest rate”: for a growth path below capacity, $y$ approaches $a$ , while $dy/dt$ is largest earlier at $y=a/2$ .

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—AP Calculus BCRead