Equations of the form dy/dx = g(y) and logistic equations

Solving dy/dx = g(y) by taking reciprocals, exponential growth, the logistic equation dP/dt = kP(1 − P/K), its solution, carrying capacity, constant solutions, and where the growth is fastest.

Solving dy/dx = g(y)

Take reciprocals: $\dfrac{dx}{dy} = \dfrac{1}{g(y)}$, then integrate with respect to $y$. This is the same as separating the variables. Check for constant solutions first.

Example 1●●

Solve $\dfrac{dy}{dx} = 1 + y^2$, given $y = 0$ when $x = 0$.

$\dfrac{dx}{dy} = \dfrac{1}{1 + y^2}$, so $x = \tan^{-1} y + C$.

At $x = 0$, $C = 0$. So $y = \tan x$, for $-\tfrac{\pi}{2} \lt x \lt \tfrac{\pi}{2}$.

The logistic equation

Exponential growth $\dfrac{dP}{dt} = kP$ never stops. In reality, food and space run out. The logistic equation slows the growth as $P$ approaches a carrying capacity $K$:

Logistic equation

$$\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right) \qquad P = \frac{K}{1 + Ae^{-kt}}$$

The constant solutions are $P = 0$ and $P = K$.
If $0 \lt P(0) \lt K$, then $P$ increases towards $K$. If $P(0) \gt K$, then $P$ decreases towards $K$.
$\dfrac{dP}{dt}$ is a quadratic in $P$, greatest at $P = \tfrac{K}{2}$. That is where the graph has its point of inflection.

Logistic growth

The curve rises steeply, slows down, and levels off at K. The steepest point is at P = K/2. Try a starting value above K.

Try this

How does the starting value affect the long-run value?

Example 2●●●

Show that $P = \dfrac{K}{1 + Ae^{-kt}}$ satisfies the logistic equation.

$P = K(1 + Ae^{-kt})^{-1}$, so $\dfrac{dP}{dt} = KkAe^{-kt}(1 + Ae^{-kt})^{-2}$.

Also $1 - \dfrac{P}{K} = \dfrac{Ae^{-kt}}{1 + Ae^{-kt}}$, so $kP\left(1 - \dfrac{P}{K}\right) = \dfrac{kKAe^{-kt}}{(1 + Ae^{-kt})^2}$, which is the same.

Example 3●●

$\dfrac{dP}{dt} = 0.3P\left(1 - \dfrac{P}{600}\right)$ with $P(0) = 100$. Find $A$, the carrying capacity, and the greatest growth rate.

$100 = \dfrac{600}{1 + A}$, so $A = 5$. The carrying capacity is $600$.

The fastest growth is at $P = 300$: $\dfrac{dP}{dt} = 0.3 \times 300 \times \tfrac12 = 45$.

Flashcards

Solving $dfrac{dy}{dx} = g(y)$

$\dfrac{dx}{dy} = \dfrac{1}{g(y)}$, then integrate.

Logistic equation

$\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{K}\right)$

Logistic solution

$P = \dfrac{K}{1 + Ae^{-kt}}$

Fastest logistic growth

At $P = \tfrac{K}{2}$, the point of inflection.

Practice

Solve $\dfrac{dy}{dx} = 1 + y^2$ with $y = 0$ when $x = 0$. Find $y$ when $x = \tfrac{\pi}{4}$.
Solve $\dfrac{dy}{dx} = 2y$ with $y = 5$ when $x = 0$. For what $x$ is $y = 5e^4$?
In Example 3, what value does $P$ approach?
In Example 3, at what population is the growth fastest?
In Example 3, find $A$.
In Example 3, find the greatest rate of growth.
The constant solutions of $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{K}\right)$ are:
A logistic population starts above its carrying capacity $K$. Then $P$: