The logistic equation
Problem 6.226 · hard
Solve \( \displaystyle P' = \frac{1}{10}P\left(1 - \frac{P}{100}\right) \), \( \displaystyle P(0) = 20 \). Find the equilibrium solutions, \( \displaystyle P(7) \), and the time when the population reaches half its carrying capacity.
- Equilibria: P = 0 and P = K = 100. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 4 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{100}{1 + 4 e^{- \frac{t}{10}}} = \frac{10 - \frac{10}{1 + 4 e^{- \frac{t}{10}}}}{1 + 4 e^{- \frac{t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 20 \]and the initial condition.✓ Proved
- \[ \frac{100}{1 + \frac{4}{e^{\frac{7}{10}}}} = \frac{100 e^{\frac{7}{10}}}{e^{\frac{7}{10}} + 4} \]P(7).✓ Proved
- \[ 10 \ln{\left(4 \right)} = \ln{\left(1048576 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{100}{1 + 4 e^{- \frac{t}{10}}},\ P(7) \approx 33.5,\ t = \ln{\left(1048576 \right)} \approx 13.863 \)
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 calculation for the time to reach half carrying capacity is incorrect. The condition A e^{-rt} = 1 yields t = (1/r) ln(A). With r = 1/10 and A =
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 calculation for the time to reach half carrying capacity is incorrect. The condition A e^{-rt} = 1 yields t = (1/r) ln(A). With r = 1/10 and A =gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The calculation for the time to reach half carrying capacity is incorrect. The condition is A e^{-rt} = 1, which implies t = (1/r) ln(A). With r = 1/gpt-oss:20b: pass 2026-10-04
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.