The logistic equation
Problem 6.228 · hard
Solve \( \displaystyle P' = \frac{1}{2}P\left(1 - \frac{P}{1000}\right) \), \( \displaystyle P(0) = 100 \). Find the equilibrium solutions, \( \displaystyle P(4) \), and the time when the population reaches half its carrying capacity.
- Equilibria: P = 0 and P = K = 1000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 9 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{1000}{1 + 9 e^{- \frac{t}{2}}} = \frac{500 - \frac{500}{1 + 9 e^{- \frac{t}{2}}}}{1 + 9 e^{- \frac{t}{2}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 100 \]and the initial condition.✓ Proved
- \[ \frac{1000}{1 + \frac{9}{e^{2}}} = \frac{1000 e^{2}}{e^{2} + 9} \]P(4).✓ Proved
- \[ 2 \ln{\left(9 \right)} = \ln{\left(81 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{1000}{1 + 9 e^{- \frac{t}{2}}},\ P(4) \approx 450.9,\ t = \ln{\left(81 \right)} \approx 4.394 \)
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: fail (error) — The solution claims P(4) ≈ 450.9, but the proven equation in step 5 is for P(2) (since exp(-4/2) = exp(-2)). The value 450.9 corresponds to P(2), not P(4). Additionally, the time to reach half carrying capacity is calculated as ln(81) ≈ 4.394, but the formula t = ln(A)/r with A=9 and r=1/2 yields t = 2*ln(9) = ln(81), which is correct, but the label in step 6 implies t=4 is the target for the previous step, creating confusion. The primary error is labeling the evaluation at t=2 as P(4).
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims P(4) ≈ 450.9, but the proven equation in step 5 is for P(2) (since exp(-4/2) = exp(-2)). The value 450.9 corresponds to P(2), not P(4). Additionally, the time to reach half carrying capacity is calculated as ln(81) ≈ 4.394, but the formula t = ln(A)/r with A=9 and r=1/2 yields t = 2*ln(9) = ln(81), which is correct, but the label in step 6 implies t=4 is the target for the previous step, creating confusion. The primary error is labeling the evaluation at t=2 as P(4).gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims P(4) ≈ 450.9, but the correct value is P(4) = 1000/(1 + 9e^(-2)) ≈ 233.4. The error likely stems from using t=2 instead of t=4 in the exponent or a calculation mistake, as 1000/(1+9e^(-2)) is not 450.9.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.