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Initial-value problems needing partial fractions

Problem 6.468 · hard

Solve \( \displaystyle \left(t^{2} - 4 t - 5\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(6) = 0 \) (for \( \displaystyle t > 5 \)).

Separate: dx = dt / (t**2 - 4*t - 5).
\[ \frac{1}{t^{2} - 4 t - 5} = - \frac{1}{6 t + 6} + \frac{1}{6 t - 30} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 5}{t + 1} \right)}}{6} = \frac{1}{t^{2} - 4 t - 5} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 5}{t + 1} \right)}}{6} \right|_{\substack{ t=6 }} + 0 = \frac{\ln{\left(7 \right)}}{6} \]
x(6) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{7 \left(t - 5\right)}{t + 1} \right)}}{6} \right|_{\substack{ t=6 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \frac{\ln{\left(\frac{7 \left(t - 5\right)}{t + 1} \right)}}{6} \)

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: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution fails to account for the absolute value in the logarithm arising from the integration of 1/(t-5). Since t > 5, ln|t-5| = ln(t-5), but th

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution fails to account for the absolute value in the logarithm arising from the integration of 1/(t-5). Since t > 5, ln|t-5| = ln(t-5), but th
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies separation of variables and partial fraction decomposition. The handling of the absolute value in the logarithm is consistent with the domain t > 5, and the constant of integration is correctly determined to satisfy the initial condition.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly separates variables, applies partial fraction decomposition, integrates, and determines the constant of integration to satisfy the initial condition. The final expression is algebraically equivalent to the stated answer.
gpt-oss:20b: pass 2026-10-14

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 2026-10-17 with SymPy 1.14.0.