The Logistic Equation
\(dP/dt\)\( = kP(1 - P/M)\) models growth that starts out exponential but levels off at a carrying capacity M, producing the famous S-shaped curve.
By the end you'll be able to read the logistic equation qualitatively — where growth speeds up, where it peaks (at \(P = M/2\)), and why every solution levels off at M — and use its closed-form solution to predict a population.
Predict: at what fraction of the carrying capacity M is \(dP/dt\) largest — near the bottom of the S, at the top, or somewhere in between? Slide M and k and watch the marked dot to check.
The solid curve solves \(dP/dt = kP(1 - P/M)\) starting from \(P_0 = 50\); the dashed exponential solves \(dP/dt = kP\) from the same start. The dot marks the inflection point, where growth is fastest. Hover or tab to the curve and dot for exact values.
Fastest growth at P = M/2 = 250 (peak rate kM/4 = 62.50 per unit time), reached at t ≈ 4.39
The logistic equation \(dP/dt = kP(1 - P/M)\) models bounded growth toward a carrying capacity M, producing an S-shaped curve that grows fastest at \(P = M/2\).
Why not plain exponential growth?
You've already met \(dP/dt = kP\): growth proportional to size, which explodes without bound. Real populations run into food, space, and resource limits. The logistic equation patches the exponential model with a braking factor \((1 - P/M)\) that switches on as the population approaches its carrying capacity:
$$\frac{dP}{dt} = kP\left(1 - \frac{P}{M}\right).$$
Read the equation off the picture above. While \(P \ll M\), the braking factor \((1 - P/M) \approx 1\), so \(dP/dt \approx kP\) — the solid curve hugs the dashed exponential. As \(P \to M\) the factor shrinks toward 0 and the curve flattens into the top of the S; at \(P = M\) exactly, \(dP/dt = 0\), so M is an equilibrium. And if a population ever overshoots (\(P > M\)), the factor turns negative and the population declines back toward M. Steepening, then flattening, with a single moment of steepest growth in between — that's the sigmoid shape.
Where the curve grows fastest
You don't need to solve the equation to find the point of steepest growth. Treat the rate as a function of P alone: \(r(P) = kP(1 - P/M) = kP - \frac{k}{M}P^2\) is a downward-opening parabola in P. It peaks where \(r'(P) = k - \frac{2k}{M}P = 0\), i.e. at \(P = M/2\) — exactly half the carrying capacity, no matter what k is. That's why the marked dot in the demo always sits at half of M, and why the S-curve's inflection point is the midpoint of the climb.
The closed-form solution
The equation is separable, and solving with \(P(0) = P_0\) gives
$$P(t) = \frac{P_0 M e^{kt}}{(M - P_0) + P_0 e^{kt}}.$$
As \(t \to \infty\), \(e^{kt} \to \infty\) and the expression approaches M — the carrying capacity is a horizontal asymptote for every solution starting with \(0 < P_0 < M\) (or above it).
A study of Kentucky's deer population used \(P_0 = 900{,}000\), \(M = 1{,}072{,}764\), and \(k = 0.2311\) per year. The logistic model predicts \(P(3) \approx 978{,}830\) deer, while a pure exponential with the same k would predict \(900{,}000\,e^{0.6933} \approx 1{,}800{,}000\) — nearly double. The braking term matters enormously once the population is already near M.
Kentucky deer: \(P_0 = 900{,}000\), \(M = 1{,}072{,}764\), \(k = 0.2311\)/yr. Plug \(t = 3\) into the closed form: \(P(3) = \dfrac{(900{,}000)(1{,}072{,}764)\,e^{0.2311(3)}}{(1{,}072{,}764 - 900{,}000) + 900{,}000\,e^{0.2311(3)}} \approx\) 978,830 deer. The population creeps toward the carrying capacity instead of doubling exponentially.
A fish population follows \(dP/dt = 0.4\,P(1 - P/800)\). At what population does it grow fastest, and what is that peak growth rate? Fastest growth is at \(P = M/2 = \) ____, and plugging that back in gives \(dP/dt = 0.4 \cdot 400 \cdot (1 - 400/800) = \) ____ fish per unit time.
Reveal the answer
Fastest growth at \(P = 800/2 = \) 400, and the peak rate is \(0.4 \cdot 400 \cdot \tfrac{1}{2} = \) 80 fish per unit time — matching the shortcut \(kM/4 = 0.4 \cdot 800 / 4 = 80\). Set M = 800 and k = 0.4 in the demo above and check the dot's tooltip.
More info — the logistic curve as a "corrected" exponential
Rewrite the closed form as \(P(t) = \dfrac{P_0\,e^{kt}}{1 + \frac{P_0}{M}(e^{kt} - 1)}\): the numerator is exactly the exponential solution \(P_0 e^{kt}\) you saw in the previous lesson, and the denominator is the correction that grows from 1 toward \(\frac{P_0}{M}e^{kt}\), capping the whole thing at M. When \(P_0 \ll M\) and t is small the denominator is ≈ 1 and the two models agree — which is why the two curves in the demo start out glued together. The OpenStax section linked in Dive deeper derives this solution step by step via separation of variables.
Check your understanding
For \(dP/dt = kP(1 - P/500)\), at what population size is growth fastest?
A population follows \(dP/dt = kP(1 - P/M)\) with \(k > 0\), but it starts ABOVE carrying capacity: \(P_0 > M\). What happens?
Early on, while \(P \ll M\), how does a logistic population compare to a pure exponential one with the same k, \(dP/dt = kP\)?
A logistic model has carrying capacity \(M = 1200\) and \(0 < P_0 < M\). What is \(\lim_{t\to\infty} P(t)\), and at what population is growth fastest along the way?
Recap
- \(dP/dt = kP(1 - P/M)\): exponential-style growth \(kP\) times a braking factor \((1 - P/M)\) that shuts growth off as \(P \to M\).
- Qualitative behavior: \(P \ll M\) → near-exponential; \(P = M\) → equilibrium (\(dP/dt = 0\)); \(P > M\) → decline back toward M.
- Growth is fastest at \(P = M/2\) — the inflection point of the S-curve — with peak rate \(kM/4\).
- Closed form: \(P(t) = \dfrac{P_0 M e^{kt}}{(M - P_0) + P_0 e^{kt}}\), which approaches M as \(t \to \infty\).
Dive deeper
- OpenStax Calculus Volume 2, §4.4 The Logistic Equation Study the closed-form logistic solution and the Kentucky deer-population example.
Sources
- The logistic differential equation