The logistic equation
Problem 6.224 · hard
Solve \( \displaystyle P' = \frac{1}{5}P\left(1 - \frac{P}{5000}\right) \), \( \displaystyle P(0) = 1250 \). Find the equilibrium solutions, \( \displaystyle P(1) \), and the time when the population reaches half its carrying capacity.
- Equilibria: P = 0 and P = K = 5000. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.
- \[ 3 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{5000}{1 + 3 e^{- \frac{t}{5}}} = \frac{1000 - \frac{1000}{1 + 3 e^{- \frac{t}{5}}}}{1 + 3 e^{- \frac{t}{5}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 1250 \]and the initial condition.✓ Proved
- \[ \frac{5000}{1 + \frac{3}{e^{\frac{1}{5}}}} = \frac{5000 e^{\frac{1}{5}}}{e^{\frac{1}{5}} + 3} \]P(1).✓ Proved
- \[ 5 \ln{\left(3 \right)} = \ln{\left(243 \right)} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{5000}{1 + 3 e^{- \frac{t}{5}}},\ P(1) \approx 1446.7,\ t = \ln{\left(243 \right)} \approx 5.493 \)
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 the time to reach half the carrying capacity is t = ln(243), but this is incorrect. Half the carrying capacity is P = 2500. Since the initial population P(0) = 1250 is already half the carrying capacity, the time required is t = 0. The formula t = ln(A)/r calculates the time to reach the carrying capacity K (where the denominator term Ae^(-rt) becomes 1 relative to the '1' in the denominator? No, Ae^(-rt)=1 implies P = K/2? Let's re-evaluate. P(t) = K / (1 + A e^{-rt}). If P = K/2, then 1 + A e^{-rt} = 2 => A e^{-rt} = 1 => e^{-rt} = 1/A => -rt = -ln(A) => t = ln(A)/r. Here A=3, r=1/5. So t = ln(3)/(1/5) = 5 ln(3) = ln(243). Wait. P(0) = 1250. K = 5000. K/2 = 2500. P(0) is NOT K/2. P(0) = 1250 = K/4. So the population starts at 1/4 of carrying capacity and grows to 1/2. The calculation t = ln(243) is the time to reach P=2500. The question asks for the time when the population reaches half its carrying capacity. So the calculation is actually correct for the question asked. Let me re-read carefully. P(0)=1250. K=5000. Half K is 2500. We need t such that P(t)=2500. 2500 = 5000 / (1 + 3e^{-t/5}) => 1 + 3e^{-t/5} = 2 => 3e^{-t/5} = 1 => e^{-t/5} = 1/3 => -t/5 = -ln(3) => t = 5ln(3) = ln(3^5) = ln(243). The math is correct. Why did I think it was wrong? Ah, I misread P(0) as 2500 in my head initially. P(0)=1250 is correct. The solution is sound.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims the time to reach half the carrying capacity is t = ln(243), but this is incorrect. Half the carrying capacity is P = 2500. Since the initial population P(0) = 1250 is already half the carrying capacity, the time required is t = 0. The formula t = ln(A)/r calculates the time to reach the carrying capacity K (where the denominator term Ae^(-rt) becomes 1 relative to the '1' in the denominator? No, Ae^(-rt)=1 implies P = K/2? Let's re-evaluate. P(t) = K / (1 + A e^{-rt}). If P = K/2, then 1 + A e^{-rt} = 2 => A e^{-rt} = 1 => e^{-rt} = 1/A => -rt = -ln(A) => t = ln(A)/r. Here A=3, r=1/5. So t = ln(3)/(1/5) = 5 ln(3) = ln(243). Wait. P(0) = 1250. K = 5000. K/2 = 2500. P(0) is NOT K/2. P(0) = 1250 = K/4. So the population starts at 1/4 of carrying capacity and grows to 1/2. The calculation t = ln(243) is the time to reach P=2500. The question asks for the time when the population reaches half its carrying capacity. So the calculation is actually correct for the question asked. Let me re-read carefully. P(0)=1250. K=5000. Half K is 2500. We need t such that P(t)=2500. 2500 = 5000 / (1 + 3e^{-t/5}) => 1 + 3e^{-t/5} = 2 => 3e^{-t/5} = 1 => e^{-t/5} = 1/3 => -t/5 = -ln(3) => t = 5ln(3) = ln(3^5) = ln(243). The math is correct. Why did I think it was wrong? Ah, I misread P(0) as 2500 in my head initially. P(0)=1250 is correct. The solution is sound.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 solution claims the time to reach half the carrying capacity is t = ln(243), but the correct time is t = 5 ln(3) = ln(243) only if r=1. Here 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.