Differential Equations · Study Guide

The Logistic
Equation

From first principles to real-world word problems. Master every question type that appears on exams.

20
Questions
5
Units
3
Difficulty Levels
0 / 20
Score: 0
Unit 1
The Logistic Equation — Structure & Basics
⚡ Instant Memory Point

LOGISTIC = LIMITED GROWTH
The equation is dP/dt = kP(1 − P/M).
Think: k = speed, M = ceiling (carrying capacity), P = population now.
When P = M/2 → growth is fastest. When P → M → growth → 0.

Q 01 Medium Concept
Which of the following is the standard logistic differential equation, where \(P\) is population, \(t\) is time, \(k\) is the growth rate constant, and \(M\) is the carrying capacity?
Q 02 Medium Equilibrium
For the logistic equation \(\dfrac{dP}{dt} = 0.04P\!\left(1 - \dfrac{P}{500}\right)\), what are the two equilibrium (constant) solutions?
Q 03 Hard Max Growth Rate
At what population value \(P\) does the logistic equation \(\dfrac{dP}{dt} = 0.06P\!\left(1 - \dfrac{P}{800}\right)\) achieve its maximum rate of growth?
\(\text{Hint: treat } \frac{dP}{dt} \text{ as a function of } P \text{ and differentiate w.r.t. } P.\)
Q 04 Hard Inflection Point
The solution \(P(t)\) to a logistic equation has an inflection point. Which statement correctly describes this point?
Unit 2
Separation of Variables & The General Solution
⚡ Instant Memory Point

SEPARATE → PARTIAL FRACTIONS → INTEGRATE → EXPONENTIATE
General solution: \(\displaystyle P(t) = \frac{M}{1 + Ae^{-kt}}\) where \(A = \dfrac{M - P_0}{P_0}\).
A = "how far from M at start". If \(P_0 = M/2\), then \(A = 1\).

Q 05 Medium General Solution Form
Which expression gives the general solution to the logistic equation \(\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)\)?
Q 06 Hard Initial Value Problem
Given \(\dfrac{dP}{dt} = 0.05P\!\left(1 - \dfrac{P}{1000}\right)\) with \(P(0) = 200\), what is the value of the constant \(A\) in the solution \(P(t) = \dfrac{1000}{1 + Ae^{-0.05t}}\)?
Q 07 Expert Tricky Algebra
A logistic equation has solution \(P(t) = \dfrac{600}{1 + 5e^{-0.3t}}\). What is the initial population \(P(0)\), and what is the carrying capacity \(M\)?
Unit 3
Slope Fields, Phase Line & Long-Run Behavior
⚡ Instant Memory Point

ABOVE M → decreasing. BELOW M → increasing. AT 0 or M → flat.
Phase line arrows: \(0 \xrightarrow{+} M/2 \xrightarrow{+} M\). Concave up below \(M/2\), concave down above \(M/2\).
Long-run: P(t) → M as t → ∞ (always, if P₀ > 0)

Q 08 Medium Long-Run Behavior
A population follows a logistic model with carrying capacity \(M = 1200\) and \(P(0) = 50\). Which statement about \(\lim_{t \to \infty} P(t)\) is correct?
Q 09 Hard Concavity Trap
For a logistic model with \(M = 500\), a solution curve passes through \(P(0) = 400\). Which describes the concavity of \(P(t)\) for \(t > 0\) (before reaching equilibrium)?
Q 10 Expert Reading dP/dt Graph
A graph of \(\dfrac{dP}{dt}\) vs \(P\) is an upside-down parabola with roots at \(P = 0\) and \(P = 400\), and maximum at \(P = 200\). Which logistic equation matches this graph?
Unit 4
Word Problems — Biology, Ecology & Medicine
⚡ Instant Memory Point — Word Problem Attack Plan

STEP 1: ID the M (carrying capacity) — look for "max," "limit," "can support."
STEP 2: ID the k (rate) — look for "grows at rate __ % per year."
STEP 3: Plug into P(t) = M / (1 + Ae^{-kt}) using P(0) = P₀ to find A.
STEP 4: Solve for t or P as asked.

Q 11 Medium Word Problem
A national park can support a maximum of 2,000 deer. Initially there are 500 deer, and the population grows logistically at rate \(k = 0.08\) per year. Which expression gives the deer population after \(t\) years?
Q 12 Hard Word Problem — Solve for t
A bacterial colony grows logistically with \(M = 10{,}000\) cells, \(k = 0.5\) per hour, and \(P(0) = 1{,}000\). At approximately what time \(t\) will the population reach 5,000 cells (the inflection point)?
\(\text{Use: } P(t) = \frac{10000}{1 + 9e^{-0.5t}} = 5000\)
Q 13 Hard Word Problem — Find k
A fish population in a lake follows logistic growth with carrying capacity \(M = 5{,}000\). At time \(t = 0\), the population is \(P(0) = 500\). At \(t = 2\) years, the population is \(P(2) = 1{,}000\). Which equation would you solve to find \(k\)?
Q 14 Expert Word Problem — Epidemiology
In an epidemic model, the number of infected people \(I(t)\) in a town of 8,000 satisfies:
\(\frac{dI}{dt} = 0.3\,I\!\left(1 - \frac{I}{8000}\right), \quad I(0) = 80\)
What is the rate of spread \(\dfrac{dI}{dt}\) when exactly half the town is infected (\(I = 4{,}000\))?
Unit 5
Advanced — Harvesting, Modification & Traps
⚡ Instant Memory Point — Common Traps

TRAP 1: Confusing k with the max growth rate. Max rate = kM/4, not k.
TRAP 2: Forgetting A can be negative if \(P_0 > M\) (population overshoots, then decreases).
TRAP 3: "Grows 4% per year" → k = 0.04, but that's the intrinsic rate. Real rate at P=P₀ is \(0.04\cdot P_0(1-P_0/M)\).
HARVESTING: dP/dt = kP(1−P/M) − H shifts equilibria. New M is lower.

Q 15 Hard Tricky — Max Growth Rate Value
For the logistic equation \(\dfrac{dP}{dt} = 0.1P\!\left(1 - \dfrac{P}{600}\right)\), what is the maximum value of \(\dfrac{dP}{dt}\)? (Not where — but the actual value.)
Q 16 Expert Overshoot Trap
A population \(P\) satisfies the logistic equation with \(M = 300\). If \(P(0) = 450\) (population above carrying capacity), what happens as \(t \to \infty\)?
Q 17 Expert Word Problem — Harvesting
A fish population follows \(\dfrac{dP}{dt} = 0.2P\!\left(1 - \dfrac{P}{4000}\right)\). Fishermen harvest fish at a constant rate of 80 fish per year. The new equation is:
\(\frac{dP}{dt} = 0.2P\!\left(1 - \frac{P}{4000}\right) - 80\)
Which best describes the effect of harvesting on the carrying capacity?
Q 18 Expert Word Problem — Social Media
A viral social media post spreads through a college of 12,000 students. Let \(S(t)\) be the number who have seen it. Initially 120 students saw it. The spread rate is proportional to both those who've seen it and those who haven't. After 1 day, 600 students have seen it.

Which differential equation models this situation?
Q 19 Expert Word Problem — Medicine (Drug Diffusion)
A drug diffuses into a tissue. Let \(C(t)\) be the drug concentration (in mg/L). The concentration grows logistically toward a saturation level of 50 mg/L. The growth constant is \(k = 0.4\,\text{hr}^{-1}\) and \(C(0) = 2\) mg/L.

At what concentration \(C\) is the drug absorption rate \(\dfrac{dC}{dt}\) at its maximum?
Q 20 Expert Grand Synthesis
Final Boss. A population of endangered condors on an island is modeled by:
\(\frac{dP}{dt} = 0.12P\!\left(1 - \frac{P}{250}\right), \quad P(0) = 25\)
Conservation biologists observe that population growth is fastest when \(P = 125\). They want to know: at \(P = 25\) (the start), is the solution curve concave up or concave down, and what does this mean biologically?
0/20
Questions Correct

Review the explanations above for any questions you missed — mastery comes from understanding the why, not just the answer.