Difficulty: Intermediate | Prerequisites: Separable differential equations, basic integration, properties of exponential functions.
Exponential decay and logistic growth are two of the most common models built from differential equations. Exponential decay describes populations or quantities that shrink at a rate proportional to their current size, while logistic growth describes populations that grow toward a ceiling (carrying capacity) imposed by limited resources.
Both models start from a separable DE that you solve with the same technique: separate, integrate, apply initial conditions. The difference is in the DE itself. Exponential decay is the simpler case, dP/dt = −kP, and logistic growth adds a cap, dP/dt = kP(M − P) or equivalently dP/dt = k(M − P). These problems appear regularly on the AP exam, often as a multi-part question that walks from one model to the other.
You should already be comfortable separating variables and integrating. If not, review those notes before continuing.
Exponential decay (dP/dt = −kP) gives solutions of the form P = P₀e^(−kt), where the quantity shrinks toward zero. Logistic growth (dP/dt = k(M − P)) gives solutions that rise toward a carrying capacity M. Both are solved by separation of variables, and the long-run behaviour comes from taking the limit as t → ∞.
Exponential decay
A model in which a quantity decreases at a rate proportional to its current value: dP/dt = −kP, where k > 0. The solution is P(t) = P₀e^(−kt). Think of it as a population that loses a fixed percentage of itself each year, so the actual number lost gets smaller as the population shrinks.
Carrying capacity (M)
The maximum population (or quantity) that an environment can sustain indefinitely. In the logistic model, the population grows toward M but never exceeds it. In simple terms, it is the ceiling the population levels off at.
Logistic growth model
A model where the growth rate is proportional to both the current population and the remaining room to grow: dP/dt = k(M − P). As P approaches M, the factor (M − P) shrinks and growth slows to zero. The solution has the form P(t) = M + Ce^(−kt), where C is determined by initial conditions.
Proportionality constant (k)
The positive constant that scales the rate of change. A larger k means faster decay or faster approach to the carrying capacity. You typically find k by using a second data point (a known population at a known later time).
Verifying a solution
Showing that a proposed function satisfies a DE by differentiating it and checking that both sides of the equation match. This is a common exam task: differentiate the candidate, substitute, and confirm equality.
Given: dP/dt = −0.1P and the proposed solution P = 495e^(−0.1t).
Differentiate: dP/dt = 495 · (−0.1)e^(−0.1t) = −0.1 · 495e^(−0.1t) = −0.1P.
The derivative equals the right side of the DE, confirming that P = 495e^(−0.1t) is indeed a solution.
To find the population after 5 years: P(5) = 495e^(−0.5) ≈ 495(0.6065) ≈ 300 deer.
After the initial 5-year decay period, the deer population grows again according to dP/dt = k(650 − P), where 650 is the carrying capacity and the initial population for this phase is P(0) = 300.
Separate variables: dP/(650 − P) = k dt.
Integrate both sides: −ln|650 − P| = kt + C.
Exponentiate: 650 − P = Ae^(−kt), so P = 650 − Ae^(−kt).
Apply P(0) = 300: 300 = 650 − A, giving A = 350.
Particular solution: P(t) = 650 − 350e^(−kt).
Given that P(1) = 350 (one year later, the population is 350):
350 = 650 − 350e^(−k)
350e^(−k) = 300
e^(−k) = 300/350 = 6/7
−k = ln(6/7)
k = −ln(6/7) = ln(7/6)
So k = ln(7/6) ≈ 0.1542.
With P(t) = 650 − 350e^(−kt) and k > 0:
As t → ∞, e^(−kt) → 0.
Therefore lim(t→∞) P(t) = 650 − 0 = 650.
The population approaches the carrying capacity of 650 deer in the long run.
\text{Exponential decay: } \frac{dP}{dt} = -kP \quad \Rightarrow \quad P(t) = P_0 e^{-kt}\text{Logistic growth: } \frac{dP}{dt} = k(M - P) \quad \Rightarrow \quad P(t) = M - Ae^{-kt}\text{Finding } k \text{ from a data point: } k = -\ln\left(\frac{M - P(t_1)}{M - P(0)}\right) \cdot \frac{1}{t_1}\text{Long-run behaviour: } \lim_{t \to \infty} P(t) = M \quad \text{(carrying capacity)}Exponential decay models radioactive half-life, drug metabolism in the body, and depreciation of assets. If a doctor prescribes a drug whose plasma concentration drops by a fixed fraction per hour, the concentration follows dC/dt = −kC.
Logistic growth models populations constrained by food, space, or predation. It also describes the spread of diseases or rumours through a finite group, and the adoption curve of new technologies.
Students often confuse the sign of k. In dP/dt = −kP, k is positive and the minus sign is written explicitly. Writing dP/dt = kP with k negative works algebraically but loses marks because the exam expects k > 0 with the sign shown.
When integrating dP/(650 − P), the result is −ln|650 − P|, not +ln|650 − P|. The negative comes from the chain rule (the derivative of 650 − P is −1). Missing this sign flips the entire solution.
Students sometimes write the limit of P(t) = 650 − 350e^(−kt) as t → ∞ by letting only part of the expression go to zero. The entire exponential term 350e^(−kt) approaches zero, not just the exponent.
A frequent error when finding k is dividing by the wrong quantity after isolating the exponential. Be methodical: isolate the exponential term, then take the natural log of both sides.
⚠️ "Show that P = ... is a solution" is a verification problem, not a solving problem. Differentiate the given function, substitute into the DE, and confirm both sides match. Do not solve the DE from scratch unless asked.
⚠️ Multi-part problems often shift the model partway through (decay first, then logistic growth). Read carefully which DE applies to which time period, and reset your initial condition at the transition.
⚠️ Finding k requires a second data point. The exam will always give you one. Set up the equation, isolate the exponential, take ln, and solve. Show every algebraic step for full marks.
⚠️ The limit as t → ∞ question is nearly free points. State that the exponential term decays to zero and conclude that P approaches the carrying capacity M.
True or False: The solution to dP/dt = −0.1P with P(0) = 495 is P = 495e^(0.1t).
Fill in the blank: In the logistic model dP/dt = k(650 − P), the carrying capacity is ____.
True or False: As t → ∞, the logistic solution P(t) = 650 − 350e^(−kt) approaches 300.
Fill in the blank: To verify a proposed solution, you ____ it and substitute into the DE.
True or False: If P(0) = 300 and P(1) = 350 in the model P(t) = 650 − 350e^(−kt), then k = ln(7/6).
Answers: 1. False (the exponent should be −0.1t, not +0.1t). 2. 650. 3. False (it approaches 650). 4. Differentiate. 5. True.
Q: Show that P = 495e^(−0.1t) is a solution to dP/dt = −0.1P.
A: Differentiate: dP/dt = 495(−0.1)e^(−0.1t) = −0.1(495e^(−0.1t)) = −0.1P. Both sides of the DE match, confirming the solution.
Q: Using P = 495e^(−0.1t), find the deer population after 5 years to the nearest whole number.
A: P(5) = 495e^(−0.5) = 495(0.6065…) ≈ 300 deer.
Q: Solve dP/dt = k(650 − P) with P(0) = 300 to find P(t) in terms of k.
A: Separate: dP/(650 − P) = k dt. Integrate: −ln|650 − P| = kt + C. At t = 0: −ln(350) = C. Exponentiate: 650 − P = 350e^(−kt). So P(t) = 650 − 350e^(−kt).
Q: Given P(t) = 650 − 350e^(−kt) and P(1) = 350, find k.
A: 350 = 650 − 350e^(−k), so 350e^(−k) = 300, e^(−k) = 6/7, and k = ln(7/6).
Q: Using the logistic model from above, find the limit of P(t) as t → ∞ and explain what it means.
A: As t → ∞, e^(−kt) → 0, so P(t) → 650 − 0 = 650. The deer population approaches the carrying capacity of 650, meaning the environment can sustain at most 650 deer in the long run.
Exponential decay is the flip side of exponential growth (dP/dt = kP with k > 0). The same separation technique applies; only the sign changes. Compound interest and continuously compounding models use the same DE.
Logistic growth connects to the idea of equilibrium solutions. Setting dP/dt = 0 gives P = M: this is a stable equilibrium because nearby solutions are attracted to it.
The technique of verifying a solution by substitution connects to the broader concept of checking work in differential equations, which reappears in BC Calculus with more complex DEs such as second-order linear equations.
Exponential decay, exponential growth, logistic growth, logistic differential equation, carrying capacity, proportionality constant, separation of variables, initial value problem, IVP, verify a solution, deer population problem, dP/dt = −kP, dP/dt = k(M − P), limit as t approaches infinity, long-run behaviour, equilibrium solution, stable equilibrium, Newton's Law of Cooling, radioactive decay, half-life, AB Calculus free response, AP Calc population model.