The Logistic Equation
A petri dish starts with 100 bacteria. They double every hour — but the dish can only hold 10 000. Does growth just stop at capacity? What does the population curve look like? The logistic equation $\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$ captures this S-shaped reality. In this lesson you'll solve it using partial fractions, sketch the solution curve, and answer HSC-style questions on carrying capacity and inflection.
A population starts at $P_0 = 200$ with carrying capacity $M = 1000$ and growth rate $k = 0.4$. Without using a formula — sketch or describe what you expect the population-vs-time curve to look like. Does growth speed up, slow down, or both?
Exponential growth ($\tfrac{dP}{dt} = kP$) predicts unlimited growth — unrealistic for any real population. The logistic model adds a braking term $\bigl(1 - \tfrac{P}{M}\bigr)$ that shrinks to zero as $P \to M$:
When $P \ll M$ the factor $\approx 1$ and growth is nearly exponential. When $P \approx M$ the factor $\approx 0$ and growth stalls. The inflection point — fastest growth — occurs at exactly $P = M/2$.
$\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$
Key facts
- Logistic DE: $\tfrac{dP}{dt} = kP(1 - P/M)$; $M$ is carrying capacity
- Inflection point at $P = M/2$; solution is an S-curve
- General solution: $P(t) = \dfrac{M}{1 + Ae^{-kt}}$ where $A = \dfrac{M - P_0}{P_0}$
Concepts
- Why the braking factor $(1 - P/M)$ produces bounded growth
- How partial fractions decompose $\tfrac{1}{P(M-P)}$
- The biological meaning of $k$, $M$, and the inflection point
Skills
- Solve the logistic DE by separation of variables and partial fractions
- Apply initial conditions to find $A$ and interpret the solution
- Identify $k$, $M$, $P_0$ from a context and find $P(t)$
Start with $\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$. Rewrite as $\dfrac{dP}{dt} = \dfrac{k}{M} P(M - P)$ and separate variables:
Partial fractions: Write $\dfrac{1}{P(M-P)} = \dfrac{A}{P} + \dfrac{B}{M-P}$. Multiplying both sides by $P(M-P)$ and comparing coefficients gives $A = \tfrac{1}{M}$, $B = \tfrac{1}{M}$. So:
Multiplying both sides by $M$ and exponentiating:
where $A = \dfrac{M - P_0}{P_0}$ is determined by the initial condition $P(0) = P_0$.
Start with $\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$. Rewrite as $\dfrac{dP}{dt} = \dfrac{k}{M} P(M - P)$ and separate variables:
Pause — copy the logistic solution $P(t)=\frac{M}{1+Ae^{-kt}}$ with $A=(M-P_0)/P_0$, and the partial-fraction step that generates it into your book.
Quick check: A population satisfies $\dfrac{dP}{dt} = 0.3P\!\left(1 - \dfrac{P}{500}\right)$ with $P(0) = 50$. What is the value of $A$ in the general solution $P(t) = \dfrac{M}{1+Ae^{-kt}}$?
We just saw that separating $\frac{dP}{dt}=kP(1-P/M)$ and using partial fractions gives $P(t)=\frac{M}{1+Ae^{-kt}}$ where $A=(M-P_0)/P_0$. That raises a question: the growth rate $\frac{dP}{dt}$ starts small, rises, then falls back to zero — at what exact value of $P$ is it maximised, and what is that maximum rate? This card answers it → the inflection occurs at $P=M/2$, and the maximum growth rate is $\frac{dP}{dt}=kM/4$.
The inflection point of $P(t)$ is where $\dfrac{d^2P}{dt^2} = 0$, i.e.\ where the growth rate $\dfrac{dP}{dt}$ is a maximum.
Define $f(P) = kP(1-P/M)$. Differentiate with respect to $P$:
$f'(P) = k - \dfrac{2kP}{M} = 0 \quad \Rightarrow \quad P = \dfrac{M}{2}$
At this population the growth rate equals $\dfrac{dP}{dt}\Big|_{P=M/2} = k \cdot \dfrac{M}{2} \cdot \dfrac{1}{2} = \dfrac{kM}{4}$.
The inflection point of $P(t)$ is where $\dfrac{d^2P}{dt^2} = 0$, i.e.\ where the growth rate $\dfrac{dP}{dt}$ is a maximum.
Pause — copy the inflection-point result: $P=M/2$ gives maximum growth rate $kM/4$; above $M/2$ the growth decelerates; $P=M$ is the asymptote into your book.
Did you get this? True or false: for the logistic equation with $M = 800$, the population growth rate is greatest when $P = 400$.
Worked examples · 3 in a row, reveal as you go
A fish population satisfies $\dfrac{dP}{dt} = 0.2P\!\left(1 - \dfrac{P}{1000}\right)$ with $P(0) = 100$. Find $P(t)$.
Using the solution from Problem 1, at what time $t$ does the population reach the inflection point?
Show that $P(t) = \dfrac{M}{1 + Ae^{-kt}}$ satisfies $\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$.
Fill the gap: For the logistic equation with $M = 600$ and $k = 0.5$, the maximum growth rate equals $\dfrac{kM}{4} = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for a logistic population with $P_0 = M/2$, the constant $A$ in $P(t) = M/(1+Ae^{-kt})$ equals 1.
Activities · practice with the ideas
A population satisfies $\tfrac{dP}{dt} = 0.1P(1 - P/2000)$ with $P(0) = 200$. Write down the values of $k$, $M$, and $A$, then state $P(t)$.
For the DE in Activity 1, at what population value is the growth rate a maximum? What is that maximum growth rate?
Use partial fractions to show that $\displaystyle\int \frac{dP}{P(1000-P)} = \frac{1}{1000}\ln\left|\frac{P}{1000-P}\right| + C$.
A population with $M = 500$, $k = 0.6$, $P(0) = 50$ satisfies the logistic model. Find the time when $P = 400$. Give your answer in exact form.
Explain why a logistic population with $P_0 > M$ (population starts above carrying capacity) would decrease towards $M$. What is the sign of $\tfrac{dP}{dt}$ in this case?
Odd one out: Three of these statements about the logistic equation are correct. Which one is NOT?
Earlier you sketched or described the population curve for $P_0 = 200$, $M = 1000$, $k = 0.4$.
The exact solution is $P(t) = \dfrac{1000}{1 + 4e^{-0.4t}}$ (here $A = \tfrac{1000-200}{200} = 4$). Growth accelerates up to $P = 500$ (inflection at $t = \tfrac{\ln 4}{0.4} \approx 3.47$), then decelerates, asymptoting to 1000. Did your sketch capture the S-shape with deceleration near $M$?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Write down the general solution of $\dfrac{dP}{dt} = 0.4P\!\left(1 - \dfrac{P}{800}\right)$ with $P(0) = 160$. (2 marks)
Q2. For the logistic equation in Q1, find the time at which the growth rate is a maximum. Give your answer in exact form. (3 marks)
Q3. Show that if $P(t) = \dfrac{M}{1+Ae^{-kt}}$, then $\dfrac{dP}{dt} = \dfrac{k}{M}P(M-P)$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $k=0.1$, $M=2000$, $A=(2000-200)/200=9$, $P(t)=2000/(1+9e^{-0.1t})$ · 2. Inflection at $P=1000$; max rate $=0.1\times2000/4=50$ per unit time · 3. Partial fractions: $A=B=1/1000$; integrate to get $\tfrac{1}{1000}\ln|P/(1000-P)|+C$ · 4. $A=(500-50)/50=9$; set $400=500/(1+9e^{-0.6t})$, solve $e^{-0.6t}=1/36$, $t=\tfrac{\ln 36}{0.6}=\tfrac{5\ln 36}{3}$ · 5. When $P>M$, $(1-P/M)<0$ so $dP/dt<0$ — population decreases towards $M$.
Q1 (2 marks): $A=(800-160)/160=4$ [1]. $P(t)=\dfrac{800}{1+4e^{-0.4t}}$ [1].
Q2 (3 marks): Inflection at $P=M/2=400$ [1]. Set $\dfrac{800}{1+4e^{-0.4t}}=400$: $1+4e^{-0.4t}=2$, $e^{-0.4t}=\tfrac{1}{4}$ [1]. $t=\dfrac{\ln 4}{0.4}=\dfrac{5\ln 4}{2}$ [1].
Q3 (3 marks): Differentiate $P=M/(1+Ae^{-kt})$: $dP/dt = MkAe^{-kt}/(1+Ae^{-kt})^2$ [1]. Note $P = M/(1+Ae^{-kt})$, so $P/M = 1/(1+Ae^{-kt})$, hence $1-P/M = Ae^{-kt}/(1+Ae^{-kt})$ [1]. Multiply: $\tfrac{k}{M}P(M-P) = k \cdot \tfrac{M}{1+Ae^{-kt}} \cdot \tfrac{MAe^{-kt}}{1+Ae^{-kt}} \cdot \tfrac{1}{M} = \tfrac{MkAe^{-kt}}{(1+Ae^{-kt})^2} = dP/dt$ ✓ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering logistic equation questions. Lighter alternative to the boss.
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