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Initial-value problems needing partial fractions

Problem 6.760 · hard

Solve \( \displaystyle \left(t^{2} - 3 t\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(6) = 0 \) (for \( \displaystyle t > 3 \)).

Separate: dx = dt / (t**2 - 3*t).
\[ \frac{1}{t^{2} - 3 t} = \frac{1}{3 t - 9} - \frac{1}{3 t} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 3}{t} \right)}}{3} = \frac{1}{t^{2} - 3 t} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 3}{t} \right)}}{3} \right|_{\substack{ t=6 }} + 0 = \frac{\ln{\left(2 \right)}}{3} \]
x(6) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(2 - \frac{6}{t} \right)}}{3} \right|_{\substack{ t=6 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \frac{\ln{\left(2 - \frac{6}{t} \right)}}{3} \)

Lines: 4 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. How we verify

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
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	mpmath's ODE solver from t₀ to t₀ + 1

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution ignores the absolute value required in the logarithmic integral of 1/t, leading to an incorrect constant of integration. For t > 3, the term (t-3)/t is positive, but the integration of 1/t yields ln|t|. The correct antiderivative is (1/3)ln|(t-3)/t|. Evaluating at t=6 gives (1/3)ln(1/2) = -ln(2)/3. The solution incorrectly calculates the constant as ln(2)/3, resulting in a sign error in the final answer.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-30 — The solution ignores the absolute value required in the logarithmic integral of 1/t, leading to an incorrect constant of integration. For t > 3, the term (t-3)/t is positive, but the integration of 1/t yields ln|t|. The correct antiderivative is (1/3)ln|(t-3)/t|. Evaluating at t=6 gives (1/3)ln(1/2) = -ln(2)/3. The solution incorrectly calculates the constant as ln(2)/3, resulting in a sign error in the final answer.
gpt-oss:20b: pass 2026-10-30
qwen3.6:27b-mlx: fail (error) 2026-10-29 — The solution fails to account for the absolute value in the integration of 1/t, resulting in an incorrect sign for the constant of integration. For t > 3, (t-3)/t is positive, so the antiderivative is (1/3)ln((t-3)/t). Applying x(6)=0 yields C = -ln(2)/3, but the solution incorrectly derives C = +ln(2)/3 (or equivalent error in step 4) leading to the wrong final expression.
gpt-oss:20b: pass 2026-10-29
qwen3.6:27b-mlx: inconclusive 2026-10-29 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution fails to account for the absolute value in the integral of 1/u, leading to an incorrect domain of validity. For t > 3, (t-3)/t is posit
gpt-oss:20b: pass 2026-10-29

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/ivp_partial_fractions, checked 2026-10-30 with SymPy 1.14.0.