The logistic equation
Problem 6.345 · hard
Solve \( \displaystyle P' = \frac{1}{5}P\left(1 - \frac{P}{10000}\right) \), \( \displaystyle P(0) = 2000 \). Find the equilibrium solutions, \( \displaystyle P(4) \), 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
- \[ \frac{10000}{1 + \frac{4}{e^{\frac{4}{5}}}} = \frac{10000 e^{\frac{4}{5}}}{e^{\frac{4}{5}} + 4} \]P(4).✓ 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 \( P(t) = \frac{10000}{1 + 4 e^{- \frac{t}{5}}},\ P(4) \approx 3574.9,\ t = \ln{\left(1024 \right)} \approx 6.931 \)
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 formula for the time to reach half carrying capacity is incorrect. The solution claims t = ln(A)/r, which yields ln(4)/0.2 = 5 ln(4). However, th
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-10-10 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The formula for the time to reach half carrying capacity is incorrect. The solution claims t = ln(A)/r, which yields ln(4)/0.2 = 5 ln(4). However, thgpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: inconclusive 2026-10-10 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution contains a critical algebraic error in step 6. The condition for half carrying capacity is $Ae^{-rt} = 1$, which implies $e^{-rt} = 1/A$gpt-oss:20b: pass 2026-10-10
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 2026-10-10 with SymPy 1.14.0.