The logistic equation
Problem 6.306 · hard
Solve \( \displaystyle P' = \frac{1}{5}P\left(1 - \frac{P}{10000}\right) \), \( \displaystyle P(0) = 2000 \). Find the equilibrium solutions, \( \displaystyle P(9) \), and the time when the population reaches half its carrying capacity.
- Equilibria: P = 0 and P = K = 10000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 4 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{10000}{1 + 4 e^{- \frac{t}{5}}} = \frac{2000 - \frac{2000}{1 + 4 e^{- \frac{t}{5}}}}{1 + 4 e^{- \frac{t}{5}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 2000 \]and the initial condition.✓ Proved
- \[ \frac{10000}{\frac{4}{e^{\frac{9}{5}}} + 1} = \frac{10000 e^{\frac{9}{5}}}{4 + e^{\frac{9}{5}}} \]P(9).✓ Proved
- \[ 5 \ln{\left(4 \right)} = \ln{\left(1024 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{10000}{1 + 4 e^{- \frac{t}{5}}},\ P(9) \approx 6019.8,\ t = \ln{\left(1024 \right)} \approx 6.931 \)
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 t = ln(1024) ≈ 6.931, but ln(1024) ≈ 6.931 is incorrect (ln(1024) ≈ 6.93147 is close, but the exact value derived in line 6 is 5*
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution claims t = ln(1024) ≈ 6.931, but ln(1024) ≈ 6.931 is incorrect (ln(1024) ≈ 6.93147 is close, but the exact value derived in line 6 is 5*gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution incorrectly identifies the time to reach half the carrying capacity. For the logistic equation, half the carrying capacity is reached atgpt-oss:20b: pass 2026-10-08
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 2026-10-08 with SymPy 1.14.0.