The logistic equation
Problem 6.267 · hard
Solve \( \displaystyle P' = \frac{3}{10}P\left(1 - \frac{P}{100}\right) \), \( \displaystyle P(0) = 2 \). Find the equilibrium solutions, \( \displaystyle P(2) \), 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.
- \[ 49 \]A.✓ Proved
- \[ \frac{d}{d t} \frac{100}{1 + 49 e^{- \frac{3 t}{10}}} = \frac{30 - \frac{30}{1 + 49 e^{- \frac{3 t}{10}}}}{1 + 49 e^{- \frac{3 t}{10}}} \]This P(t) satisfies the differential equation.✓ Proved
- \[ 2 \]and the initial condition.✓ Proved
- \[ \frac{100}{1 + \frac{49}{e^{\frac{3}{5}}}} = \frac{100 e^{\frac{3}{5}}}{e^{\frac{3}{5}} + 49} \]P(2).✓ Proved
- \[ \frac{10 \ln{\left(49 \right)}}{3} = \frac{20 \ln{\left(7 \right)}}{3} \]Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.✓ Proved
Answer \( P(t) = \frac{100}{1 + 49 e^{- \frac{3 t}{10}}},\ P(2) \approx 3.6,\ t = \frac{20 \ln{\left(7 \right)}}{3} \approx 12.973 \)
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(2) ≈ 3.6, but the correct value is approximately 2.59. Additionally, the time to reach half the carrying capacity is calculated as t = (20 ln(7))/3 ≈ 15.97, not 12.97.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution claims P(2) ≈ 3.6, but the correct value is approximately 2.59. Additionally, the time to reach half the carrying capacity is calculated as t = (20 ln(7))/3 ≈ 15.97, not 12.97.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The calculation for P(2) is incorrect because the exponent in the solution is -3t/10, so at t=2 it should be -3/5, but the numerical approximation 3.6 corresponds to t=2 only if the exponent was -3t/5 (or similar error). Actually, let's re-evaluate: P(2) = 100 / (1 + 49*e^(-0.6)). e^(-0.6) ≈ 0.5488. 49*0.5488 ≈ 26.89. 100/27.89 ≈ 3.58. So 3.6 is correct. Wait, let me check the time for half capacity. Half capacity is P=50. 50 = 100/(1+Ae^(-rt)) => 1+Ae^(-rt)=2 => Ae^(-rt)=1 => e^(rt)=A => rt=ln(A) => t=ln(A)/r. A=49, r=3/10. t = ln(49)/(3/10) = 10*ln(49)/3 = 20*ln(7)/3. This is correct. Let me re-read the prompt's stated answer vs solution. Stated answer: P(2) approx 3.6. Solution line 5 computes P(2) symbolically. Line 6 computes t. The solution text says 'Equilibria: P = 0 and P = K = 100'. This is correct. 'The solution has the form... with A = (K-P0)/P0'. This is correct. Line 2 checks A=49. Correct. Line 3 checks DE. Correct. Line 4 checks IC. Correct. Line 5 checks P(2) expression. Correct. Line 6 checks t expression. Correct. Why did I think it was wrong? Let me check the numerical value of t. 20*ln(7)/3 ≈ 20*1.9459/3 ≈ 12.97. Correct. Is there a subtle error? The problem asks for 'the time when the population reaches half its carrying capacity'. The carrying capacity is 100. Half is 50. The logic holds. Wait, look at line 1. 'Equilibria: P = 0 and P = K = 100'. The DE is P' = 0.3 P (1 - P/100). Equilibria are P=0 and P=100. Correct. Is the form P = K/(1+Ae^(-rt)) correct for logistic growth P' = rP(1-P/K)? Yes, if r is the intrinsic growth rate. Here r=3/10. The solution uses r=3/10 in the exponent. Correct. A = (100-2)/2 = 49. Correct. Everything seems correct. Why would it fail? Maybe the 'unchecked' equations are not the issue, but the sentences. Sentence 1: 'Equilibria: P = 0 and P = K = 100. The solution has the form P = K/(1 + Ae^(−rt)) with A = (K − P₀)/P₀.' This is a standard result. It is correct. Sentence 6: 'Half the carrying capacity when Ae^(−rt) = 1, i.e. t = ln(A)/r.' This is correct. Is there a missing unit? No. Is the approximation 3.6 misleading? 3.58 rounds to 3.6. It's fine. Let me look closer at the prompt's 'Stated answer'. P(2) approx 3.6. t approx 12.973. The solution derives the exact forms. The sentences justify the steps. I don't see an error. Wait, is the direction of rotation mentioned? No, it's a population model. Is the region specified? No. Is the theorem application valid? Yes. Is it possible the user considers the lack of explicit integration steps as a 'style' issue? Or is there a hidden error? Let's check the derivative check in line 3. P = 100(1+49e^-0.3t)^-1. P' = -100(1+49e^-0.3t)^-2 * 49(-0.3)e^-0.3t = 1470 e^-0.3t / (1+49e^-0.3t)^2. RHS: 0.3 P (1 - P/100) = 0.3 * 100/(1+49e^-0.3t) * (1 - 1/(1+49e^-0.3t)) = 30/(1+49e^-0.3t) * (49e^-0.3t)/(1+49e^-0.3t) = 1470 e^-0.3t / (1+49e^-0.3t)^2. They match. The solution is sound. I will pass it.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.