The logistic equation
Problem 6.270 · hard
A population follows the logistic model with growth rate \( \displaystyle r = \frac{1}{2} \) per year, carrying capacity 200, and initial population 10. Find \( \displaystyle P(t) \), \( \displaystyle P(2) \), and when the population reaches 100.
- Equilibria: P = 0 and P = K = 200. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 19 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{200}{1 + 19 e^{- \frac{t}{2}}} = \frac{100 - \frac{100}{1 + 19 e^{- \frac{t}{2}}}}{1 + 19 e^{- \frac{t}{2}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 10 \]and the initial condition.✓ Proved
- \[ \frac{200}{1 + \frac{19}{e}} = \frac{200 e}{e + 19} \]P(2).✓ Proved
- \[ 2 \ln{\left(19 \right)} = \ln{\left(361 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{200}{1 + 19 e^{- \frac{t}{2}}},\ P(2) \approx 25.0,\ t = \ln{\left(361 \right)} \approx 5.889 \)
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 calculates t = ln(361) ≈ 5.889, but the correct time is t = 2*ln(19) ≈ 11.778 because r = 1/2. The equation in line 6 is algebraically correct, but the final numerical answer in the stated answer is wrong (it missed the division by r).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution calculates t = ln(361) ≈ 5.889, but the correct time is t = 2*ln(19) ≈ 11.778 because r = 1/2. The equation in line 6 is algebraically correct, but the final numerical answer in the stated answer is wrong (it missed the division by r).qwen3.6:27b-mlx: fail (error) 2026-10-06 — The calculation for t when P(t)=100 is incorrect. The solution claims t = ln(361), but substituting P=100 into the model yields 1 + 19e^(-t/2) = 2, so 19e^(-t/2) = 1, which gives t = 2 ln(19). The solution incorrectly states t = ln(361) = 2 ln(19) is the answer, but the numerical approximation 5.889 corresponds to ln(361), whereas 2 ln(19) is approximately 5.889. Wait, 2*ln(19) is indeed ln(361). Let me re-read carefully. The text says 't = ln(361) approx 5.889'. 2*ln(19) = ln(19^2) = ln(361). So the symbolic answer is correct. However, line 6 says 't = ln(A)/r'. A=19, r=1/2. ln(19)/(1/2) = 2 ln(19). This is correct. Why did I think it was wrong? Let me check the numerical value. ln(361) is approx 5.88887. The solution says 5.889. This is correct. Is there an error? Let's check P(2). P(2) = 200 / (1 + 19 e^{-1}). 19/e approx 6.99. 1+6.99 = 7.99. 200/7.99 approx 25.03. The solution says 25.0. This is a reasonable approximation. Let's look closer at the sentence in line 1. 'The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.' This is the standard form for logistic growth dP/dt = rP(1-P/K). The problem states 'growth rate r=1/2'. Usually, the logistic equation is dP/dt = rP(1-P/K). The solution uses this form. Is there any ambiguity? Sometimes 'growth rate' refers to the intrinsic rate r, sometimes to the initial per capita growth rate. Given the standard phrasing, r is the parameter in the ODE. The solution seems correct. Let me re-read the prompt's instruction: 'find what is wrong'. If it's correct, I should pass. But wait, look at line 6 again. 'Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' This sentence is technically a statement about the time to reach half capacity. It is correct. Is there a subtle error? Let's check the initial condition check in line 4. P(0) = 200/(1+19) = 10. Correct. Let's check the derivative check in line 3. It says 'This P(t) satisfies the differential equation.' It doesn't specify which DE, but implies the logistic one. This is fine. Let's look at the numerical approximation for P(2). 200/(1 + 19/e) = 200e / (e + 19). e approx 2.718. 19+2.718 = 21.718. 200*2.718 = 543.6. 543.6 / 21.718 approx 25.029. Rounding to 25.0 is acceptable. Let's look at t. t = 2 ln(19). ln(19) approx 2.944. 2*2.944 = 5.888. Rounding to 5.889 is acceptable. Is there a 'style' issue? Line 1 says 'Equilibria: P=0 and P=K=200'. This is true for the ODE, but not directly used to find the specific solution, though it justifies the form. Line 6 says 'Half the carrying capacity...'. This is a correct observation. I don't see an error. Wait, let me look at the JSON output format again. If it is sound, pass. But usually these adversarial reviews have a trap. Let's look at the formula for A. A = (K-P0)/P0 = (200-10)/10 = 19. Correct. Let's look at the exponent. e^(-rt). r=1/2. e^(-t/2). Correct. Let's look at the time to reach 100. P(t)=100. 100 = 200/(1+19e^(-t/2)). 1+19e^(-t/2) = 2. 19e^(-t/2) = 1. e^(-t/2) = 1/19. -t/2 = ln(1/19) = -ln(19). t = 2 ln(19). The solution says t = ln(361). Since 361 = 19^2, ln(361) = 2 ln(19). This is correct. Is it possible the problem implies a different logistic model? e.g. dP/dt = rP(K-P)? No, standard is rP(1-P/K). If it were dP/dt = rP(K-P), the solution would be different. But 'growth rate r' usually implies the per capita rate at low density, which is r in the standard form. If the model was dP/dt = k P (K-P), then the solution is P = K / (1 + (K/P0 - 1) e^{-kKt}). Here r would be kK? Or is r the k? Standard calculus texts define logistic as dP/dt = rP(1-P/K). I will assume this standard. The solution is correct. Why would it fail? Maybe the approximation of P(2) is considered too rough? 25.0 vs 25.03. Usually 3 sig figs are expected. 25.0 is 3 sig figs. 25.03 is 4. 25.0 is acceptable. Maybe the time? 5.889 is 4 sig figs. 2 ln(19) is exact. The solution provides both. I see no error. I will pass.gpt-oss:20b: pass 2026-10-06
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 2026-10-06 with SymPy 1.14.0.