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Population Dynamics

Differential Equations · Section 2.5 · generated practice set

Logistic growth, threshold models and constant harvesting, all read off the shape of one parabola.

Practice this set → Fresh numbers on every attempt. No account needed.

Method

The logistic equation is 1.1's phase-line analysis applied to one specific $f$.

$$\frac{dy}{dt} = ry\left(1 - \frac{y}{K}\right)$$

Equilibria $y = 0$ (unstable) and $y = K$ (asymptotically stable). Any positive initial value converges to the carrying capacity $K$; the curve has an inflection at $y = K/2$ where growth is fastest.

The expanded form $y' = ay - by^2$ is the same equation with $r = a$ and $K = a/b$. Exams use both.

Explicit solution

$$y(t) = \frac{Ky_0}{y_0 + (K - y_0)e^{-rt}}$$

Threshold model

$y' = -ry\left(1 - \frac{y}{T}\right)$ flips the stability: $0$ is stable, $T$ is an unstable threshold. Starting below $T$ means extinction.

Harvesting

$y' = ry(1 - y/K) - h$ is a downward parabola in $y$. Factor it: the smaller root is an unstable threshold, the larger is stable. When the roots merge the population collapses — that is the maximum sustainable yield.

Definitions and theorems

Logistic solution
$y' = ry(1 - y/K)$, $y(0) = y_0 > 0$ gives $y(t) = \dfrac{Ky_0}{y_0 + (K-y_0)e^{-rt}} \to K$.
Parabola rule
If $f(y)$ is a downward parabola with roots $p < q$, then $p$ is unstable and $q$ is asymptotically stable.

Worked example

Analyze $y' = 5y - y^2 - 6$.

  1. Factor: $f(y) = -(y^2 - 5y + 6) = -(y-2)(y-3)$.
  2. Equilibria $y = 2$ and $y = 3$.
  3. Between them, $f > 0$: take $y = 2.5$, $f = -(0.5)(-0.5) = 0.25 > 0$. Solutions rise.
  4. Outside them $f < 0$. So arrows point away from 2 and toward 3.
  5. $y = 2$ is unstable (a threshold); $y = 3$ is asymptotically stable. Any start above 2 converges to 3; below 2 the population dies out.

Common mistakes

  1. Reading $K$ off the wrong form. In $y' = ay - by^2$ the carrying capacity is $a/b$, not $b$ or $a$.
  2. Sign of the exponent in the logistic formula. It is $e^{-rt}$ in the denominator; a dropped minus sends the answer to 0 instead of $K$.
  3. Assuming the larger equilibrium is always stable. True for a downward parabola; the threshold model flips it.
  4. Ignoring $y_0 > K$. The solution still tends to $K$, just decreasing — the formula handles it, your sketch should too.

Practice it

Reading the method is not the same as being able to run it under time pressure. This set generates a new problem with new coefficients every attempt, marks each part separately, and shows the full worked solution.

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