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Unit 7: Differential Equations

The Logistic Model

Modeling bounded growth with the logistic differential equation, unique to BC.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

  • Separable Differential Equations

Solving the Differential Equation Behind Bounded Growth

dP/dt = kP describes unbounded exponential growth — the larger P gets, the faster it grows, forever. A population that levels off near a carrying capacity L needs an equation whose growth rate shrinks as P approaches L. Before reading on: what would multiplying kP by a factor that goes to 0 as P → L actually look like?

Definition — The Logistic Differential Equation

dP/dt = kP(1 − P/L), where L is the carrying capacity. Near P = 0, the factor (1 − P/L) is close to 1 and growth looks exponential; as P approaches L, that factor shrinks toward 0 and growth slows to a stop.

Worked Example — Solving the Logistic Equation by Separation and Partial Fractions

Separate dP/dt = kP(1−P/L): since P(1−P/L) = P(L−P)/L, the equation becomes dP/[P(L−P)/L] = k dt, or L·dP/[P(L−P)] = k dt. Decompose L/[P(L−P)] into partial fractions: L/[P(L−P)] = A/P + B/(L−P). Multiplying through by P(L−P): L = A(L−P) + BP. Setting P = 0 gives A = 1; setting P = L gives B = 1. So the equation becomes [1/P + 1/(L−P)]dP = k dt. Integrating: ln|P| − ln|L−P| = kt + C₁, so ln|P/(L−P)| = kt + C₁, giving P/(L−P) = Ae^(kt). Solving for P: P = Ae^(kt)(L−P), so P(1+Ae^(kt)) = ALe^(kt), and dividing through by Ae^(kt) gives P(t) = L/(1 + Be^(−kt)), where B = 1/A — the logistic function, now derived from its differential equation rather than assumed.
dPdt=kP(1−PL)  ⇒  P(t)=L1+Be−kt\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right) \;\Rightarrow\; P(t) = \frac{L}{1 + Be^{-kt}}

Worked Example — Applying the Logistic Model

A population follows dP/dt = 0.3P(1 − P/1000), with P(0) = 100. Since P(0) = L/(1+B) = 100, B = (L−P₀)/P₀ = (1000−100)/100 = 9, so P(t) = 1000/(1 + 9e^(−0.3t)). Find P(5): e^(−1.5) ≈ 0.22313, so 9(0.22313) ≈ 2.0082, and P(5) = 1000/3.0082 ≈ 332.4. Find when growth is fastest — when P = L/2 = 500: solve 9e^(−0.3t) = 1, giving t = ln(9)/0.3 ≈ 7.32. Check: at that t, e^(−0.3t) = 1/9 exactly, so P = 1000/(1+1) = 500 — matching.

Function Explorer

Transform: g(x) = a·f(b(x − h)) + k
1
1
0
0
Composition
Analysis (of the transformed function, in view)
y-intercept
(0, 0)
x-intercepts
(-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Extrema
local min at (-7.85, -1); local max at (-4.71, 1); local min at (-1.57, -1); local max at (1.57, 1); local min at (4.71, -1); local max at (7.85, 1)
Inflection points
(-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Vertical asymptotes
none found in view
Horizontal asymptotes
none found
Domain
all real numbers in view
Range (estimated)
approximately [-1, 1]

Tip

A logistic curve is steepest exactly at P = L/2 — before that point, growth accelerates like an exponential; after it, growth decelerates toward the carrying capacity.

Common Mistakes

  • Applying the sign rule ∫1/(L−P)dP = ln|L−P| without the chain-rule-driven minus sign.

    Since d/dP[L−P] = −1, the correct antiderivative is −ln|L−P|, not +ln|L−P| — dropping that sign flips the whole derivation.

  • Treating the logistic differential equation and the plain exponential equation dP/dt = kP as interchangeable.

    The logistic equation's extra factor (1 − P/L) is exactly what makes growth slow down as P approaches L — omitting it collapses the model back to unbounded exponential growth.

Key Takeaways

  • The logistic differential equation dP/dt = kP(1 − P/L) builds in its own slowdown as P approaches the carrying capacity L.
  • Separating variables and applying partial fractions solves it exactly, producing the same logistic function form seen algebraically in earlier courses — now derived from its governing equation.
  • Growth is fastest exactly at P = L/2, the logistic curve's point of symmetry.

Summary

This closes Unit 7. Unit 8 returns to definite integrals as tools for measuring area and volume, closing with a technique unique to BC — arc length.

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