The logistic equation
Problem 6.223 · hard
A population follows the logistic model with growth rate \( \displaystyle r = \frac{1}{2} \) per year, carrying capacity 5000, and initial population 1250. Find \( \displaystyle P(t) \), \( \displaystyle P(2) \), and when the population reaches 2500.
- Equilibria: P = 0 and P = K = 5000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 3 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{5000}{1 + 3 e^{- \frac{t}{2}}} = \frac{2500 - \frac{2500}{1 + 3 e^{- \frac{t}{2}}}}{1 + 3 e^{- \frac{t}{2}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 1250 \]and the initial condition.✓ Proved
- \[ \frac{5000}{1 + \frac{3}{e}} = \frac{5000 e}{e + 3} \]P(2).✓ Proved
- \[ 2 \ln{\left(3 \right)} = \ln{\left(9 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{5000}{1 + 3 e^{- \frac{t}{2}}},\ P(2) \approx 2376.8,\ t = \ln{\left(9 \right)} \approx 2.197 \)
Lines: 5 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the differential equation solved numerically (Taylor-series ODE solver) from P(0) |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution claims P(2) is approximately 2376.8, but the equation in step 5 evaluates P(1) (since exp(-1) corresponds to t=1 given the exponent -t/
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution claims P(2) is approximately 2376.8, but the equation in step 5 evaluates P(1) (since exp(-1) corresponds to t=1 given the exponent -t/gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution calculates P(2) symbolically but fails to provide the numerical approximation requested in the problem statement (P(2) ≈ 2376.8). Additionally, the final time t is given as ln(9) ≈ 2.197, but the derivation in line 6 implies t = ln(A)/r = ln(3)/(1/2) = 2ln(3) = ln(9), which is correct, yet the text 'Half the carrying capacity when Ae^(−rt) = 1' is slightly imprecise as it doesn't explicitly show the substitution P=2500 leading to that condition, though the result is correct. The primary failure is the missing numerical value for P(2).gpt-oss:20b: fail (misleading) 2026-10-04 — Sentence 5 incorrectly rewrites P(2) as 5000*E/(E+3), which is not equivalent to the correct expression 5000/(1+3e^{-1}).
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/logistic_equation, checked 2026-10-04 with SymPy 1.14.0.