Differential-Equations · Unit 5 · Video 4 · Interactive Practice

A Downward Parabola Beats Partial Fractions: The Logistic Equation's Carrying Capacity

IKey Formulas

Formula	Name	What it gives
$\dfrac{dy}{dt} = (a - by)\,y = ay - by^2$	Logistic equation, $a, b > 0$	The growth rate $k = a - by$ starts at $a$ and falls by $b$ per unit of population
$y\,(a - by) = 0 \iff y = 0 \ \text{or}\ y = \dfrac{a}{b}$	Critical points (Step 1)	Two constant solutions, the barriers; $a/b$ is the carrying capacity
$y' < 0,\ \ y' > 0,\ \ y' < 0$ on $y < 0,\ \ 0 < y < \tfrac{a}{b},\ \ y > \tfrac{a}{b}$	Sign of the parabola (Step 2)	Arrows along the $y$ -axis: left, right, left
$y'' = (a - 2by)(ay - by^2)$	Concavity along a solution	Zero at $y = \tfrac{a}{2b}$ : the S-curve is steepest where the parabola peaks

Key Insight: No integral is needed. The roots of $ay - by^2$ are the barriers, its sign sets the arrows, and every population that starts above $0$ tends to the carrying capacity $a/b$ — the growth rate of a small population divided by the crowding coefficient.

IIThe Growth Rate's Root Is the Carrying Capacity

Where the net birth rate $k = a - by$ hits zero, the parabola $y' = y\,(a - by)$ has its second root.

Growth rate a 2 Crowding b 0.25

IIIThree Bands, Three Fates

The parabola's sign at the starting value $y_0$ decides the solution: rise toward $a/b$ , fall, or stay constant.

Start y₀ 1

IVSteepest at Half the Capacity

The S-curve's slope at height $y$ is the parabola's height at $y$ , so the climb is fastest at $y = a/(2b)$ .

Time t 0.80

VQuiz Questions

Problem 1 · Find the Critical Points

Given: the logistic equation $\dfrac{dy}{dt} = 3y - 0.5y^2$ — find all of its critical points.

✅ Correct! Factoring gives

y\,(3 - 0.5y) = 0

, so

y = 0

y = a/b = 3/0.5 = 6

: two constant solutions.

❌ A root was lost. Dividing by

y

throws away the solution

y = 0

. Factor instead:

y\,(3 - 0.5y) = 0

holds when either factor is zero, and

y \equiv 0

is a constant solution too.

❌ That is $b/a$ . From

3 - 0.5y = 0

you get

0.5y = 3

, so

y = 3/0.5 = 6

. The capacity is

a/b

, the growth rate over the crowding coefficient.

❌ $y = 3$ is the peak, not a root. Halfway between the roots, at

a/(2b)

, the parabola is at its highest:

3(3) - 0.5(3)^2 = 9 - 4.5 = 4.5 \neq 0

, so

y \equiv 3

is not a solution.

Show solution

Step 1 of the method: set the right-hand side to zero and pull out the common factor $y$ .

3y - 0.5y^2 = y\,(3 - 0.5y) = 0

A product is zero only when one of its factors is, so either $y = 0$ or $3 - 0.5y = 0$ , which gives $y = \dfrac{3}{0.5} = 6$ .

Check the constant $y \equiv 6$ : the left side $dy/dt$ is $0$ , and the right side is $3(6) - 0.5(6)^2 = 18 - 18 = 0$ ✓.

The critical points are $y = 0$ and $y = a/b = 6$ .

Problem 2 · Read the Right Axis

Given: $\dfrac{dy}{dt} = 2y - 0.25y^2$ with $y(0) = 2$ . Plotted against $y$ , the right-hand side rises to a peak at $y = 4$ and falls back to $0$ at $y = 8$ . What does the population do for $t > 0$ ?

✅ Correct! On

0 < y < 8

the parabola is above the axis, so

y

increases, and the barrier

y \equiv 8

keeps it from crossing: it levels off beneath

8

❌ That reads the parabola as a graph against time. Its horizontal axis is

y

, not

t

, and its height is the rate

y'

. That height is positive all the way from

0

8

, so the population keeps increasing and never turns around.

❌ Integral curves cannot meet.

y \equiv 8

is itself a solution. A curve that reached

8

at a finite time would touch that constant solution, so it can only approach

8

❌ The peak is where $y'$ is largest, not zero. At

y = 4

y' = 2(4) - 0.25(4)^2 = 8 - 4 = 4 > 0

: the population passes

4

at its fastest and keeps climbing toward

8

Show solution

Step 2 of the method reads the sign of $y' = 2y - 0.25y^2 = y\,(2 - 0.25y)$ band by band; its roots are $0$ and $a/b = 2/0.25 = 8$ .

At the start, $y' = 2(2) - 0.25(2)^2 = 4 - 1 = 3 > 0$ , and $y' > 0$ everywhere on $0 < y < 8$ , so $y$ increases. The constant solution $y \equiv 8$ is a barrier the curve can neither cross nor touch, so $y$ rises toward $8$ and flattens beneath it.

The peak of the parabola at $y = 4$ marks the fastest growth, not a turning point. The partial-fractions formula agrees: $y = \dfrac{8}{1 + 3e^{-2t}}$ stays below $8$ because $3e^{-2t} > 0$ , and tends to $8$ because $3e^{-2t} \to 0$ .

Problem 3 · Capacity and the Bend

Given: a fish population obeys $\dfrac{dy}{dt} = 0.6y - 0.002y^2$ ( $t$ in years) with $y(0) = 50$ . Find the carrying capacity and the population at which the graph of $y(t)$ switches from concave up to concave down.

What is the carrying capacity?

Where does the curve switch from concave up to concave down?

✅ Correct! The capacity is

a/b = 0.6/0.002 = 300

fish, and the curve switches from bending up to bending down at

a/(2b) = 150

, where the parabola peaks and

y''

changes sign.

❌ $150$ is only halfway.

a/(2b) = 150

is where the parabola peaks. The capacity is its second root, where

0.6 - 0.002y = 0

❌ That is $b/a$ . Solve

0.6 - 0.002y = 0

for

y

y = 0.6/0.002

, growth rate over crowding coefficient.

❌ Check the division.

\dfrac{0.6}{0.002} = \dfrac{600}{2} = 300

, not

3000

❌ $50$ is where the population starts. There

y'' = (0.6 - 0.004 \cdot 50)(0.6 \cdot 50 - 0.002 \cdot 50^2) = (0.4)(25) = 10 > 0

, so the curve is still bending upward.

❌ The curve never gets to $300$ . It flattens toward

300

without reaching it, so nothing can switch at

300

. The bend flips where the slope

y'

is largest: the vertex of the parabola.

❌ That holds only for a start between $150$ and $300$ . From

y(0) = 50

the slope

y' = 0.6y - 0.002y^2

keeps growing while

y < 150

, because the parabola is still climbing there, so the curve starts out concave up.

❌ Check the capacity. Set

0.6y - 0.002y^2 = y\,(0.6 - 0.002y) = 0

and take the nonzero root.

❌ Check the bend. Differentiate

y' = 0.6y - 0.002y^2

along the solution and find where

y''

changes sign.

Show solution

Capacity. $0.6y - 0.002y^2 = y\,(0.6 - 0.002y) = 0$ gives $y = 0$ or $y = \dfrac{0.6}{0.002} = 300$ .

Concavity. Differentiate $y' = f(y)$ along the solution with the chain rule:

y'' = f'(y)\,y' = (0.6 - 0.004y)(0.6y - 0.002y^2)

On $0 < y < 300$ the second factor is positive, so $y''$ has the sign of $0.6 - 0.004y$ : positive for $y < 150$ (concave up) and negative for $y > 150$ (concave down).

Starting from $50 < 150$ , the curve bends upward until $y = 150 = a/(2b)$ , the vertex of the parabola, and then bends downward as it flattens toward $300$ .

Problem 4 · Build the Model

Given: a bacterial culture's net birth rate is $0.5$ per hour while the culture is small, and every additional $1000$ cells lower it by $0.05$ per hour, a linear decline $k = a - by$ . Find the carrying capacity.

✅ Correct!

b = 0.05/1000 = 0.00005

, so

a/b = 0.5/0.00005 = 10\,000

cells: the population at which births exactly balance deaths.

❌ $0.05$ is the drop per $1000$ cells, not per cell. Per cell,

b = 0.05/1000 = 0.00005

, and then

a/b = 0.5/0.00005 = 10\,000

❌ That is $a/(2b)$ , half the capacity.

5000

cells is where the culture grows fastest. The capacity is where

k = 0.5 - 0.00005y

reaches zero.

❌ That uses $b = 0.05 \times 1000 = 50$ . A drop of

0.05

spread over

1000

cells means dividing:

b = 0.00005

per cell.

Show solution

The small-population rate is $a = 0.5$ per hour. The rate falls $0.05$ per hour for every $1000$ cells, so per cell it falls by

b = \frac{0.05}{1000} = 0.00005 \text{ per cell per hour}.

So $k = 0.5 - 0.00005y$ , and the model is $\dfrac{dy}{dt} = (0.5 - 0.00005y)\,y = 0.5y - 0.00005y^2$ .

The carrying capacity is the nonzero critical point, where the growth rate reaches zero:

\frac{a}{b} = \frac{0.5}{0.00005} = 10\,000 \text{ cells}.

Check: at $y = 10\,000$ , $k = 0.5 - 0.00005 \cdot 10\,000 = 0.5 - 0.5 = 0$ ✓.

Solved: 0 / 4