Initial-value problems needing partial fractions
Problem 6.611 · hard
Solve \( \displaystyle \left(t^{2} - 2 t\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(3) = 0 \) (for \( \displaystyle t > 2 \)).
- Separate: dx = dt / (t**2 - 2*t).
- \[ \frac{1}{t^{2} - 2 t} = \frac{1}{2 t - 4} - \frac{1}{2 t} \]Partial fractions.✓ Proved
- \[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 2}{t} \right)}}{2} = \frac{1}{t^{2} - 2 t} \]An antiderivative, checked by differentiating.✓ Proved
- \[ - \left. \frac{\ln{\left(\frac{t - 2}{t} \right)}}{2} \right|_{\substack{ t=3 }} + 0 = \frac{\ln{\left(3 \right)}}{2} \]x(3) = 0 fixes the constant.✓ Proved
- \[ \left. \frac{\ln{\left(3 - \frac{6}{t} \right)}}{2} \right|_{\substack{ t=3 }} = 0 \]The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(3 - \frac{6}{t} \right)}}{2} \)
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: passqwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution ignores the absolute value required for the integral of 1/u, leading to a complex-valued function for t in (2, 3) where (t-2)/t is negative. The correct real-valued solution for t > 2 is x(t) = (1/2)ln(|(t-2)/t|), which simplifies to (1/2)ln((2-t)/t) for 2 < t < 3 and (1/2)ln((t-2)/t) for t > 3. The stated answer is undefined for 2 < t < 3.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-10-18 — [domain objection, downgraded to style] The solution ignores the absolute value required for the integral of 1/u, leading to a complex-valued function for t in (2, 3) where (t-2)/t is negative. The correct real-valued solution for t > 2 is x(t) = (1/2)ln(|(t-2)/t|), which simplifies to (1/2)ln((2-t)/t) for 2 < t < 3 and (1/2)ln((t-2)/t) for t > 3. The stated answer is undefined for 2 < t < 3.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (style) 2026-10-16 — [domain objection, downgraded to style] The solution fails to account for the absolute value in the integral of 1/u, which is critical for t in (2, 3) where (t-2)/t is positive but the log argument in the stated answer 3 - 6/t is negative. The stated answer x(t) = ln(3 - 6/t)/2 is undefined for 2 < t < 3, whereas the correct solution involves ln((t-2)/t) or ln(|3 - 6/t|) to satisfy the domain t > 2.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution fails to account for the absolute value in the integral of 1/u, which is critical because (t-2)/t is negative for t in (2, 3). The stated answer yields complex values for t < 3, whereas the real-valued solution requires ln(|(t-2)/t|).gpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.