Initial-value problems needing partial fractions
Problem 6.768 · hard
Solve \( \displaystyle \left(t^{2} - 9\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(4) = 0 \) (for \( \displaystyle t > 3 \)).
- Separate: dx = dt / (t**2 - 9).Reviewed
- \[ \frac{1}{t^{2} - 9} = - \frac{1}{6 t + 18} + \frac{1}{6 t - 18} \]Partial fractions.✓ Proved
- \[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 3}{t + 3} \right)}}{6} = \frac{1}{t^{2} - 9} \]An antiderivative, checked by differentiating.✓ Proved
- \[ - \left. \frac{\ln{\left(\frac{t - 3}{t + 3} \right)}}{6} \right|_{\substack{ t=4 }} + 0 = \frac{\ln{\left(7 \right)}}{6} \]x(4) = 0 fixes the constant.✓ Proved
- \[ \left. \frac{\ln{\left(\frac{7 \left(t - 3\right)}{t + 3} \right)}}{6} \right|_{\substack{ t=4 }} = 0 \]The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{7 \left(t - 3\right)}{t + 3} \right)}}{6} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly applies separation of variables and partial fraction decomposition. The integration constant is determined correctly using the initial condition, and the final form satisfies both the differential equation and the initial condition for t > 3.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-30 — The solution correctly applies separation of variables and partial fraction decomposition. The integration constant is determined correctly using the initial condition, and the final form satisfies both the differential equation and the initial condition for t > 3.gpt-oss:20b: pass 2026-10-30qwen3.6:27b-mlx: pass 2026-10-29 — The solution correctly separates variables, applies partial fraction decomposition, and determines the constant of integration using the initial condition. The final expression satisfies both the differential equation and the initial condition for t > 3.gpt-oss:20b: pass 2026-10-29qwen3.6:27b-mlx: pass 2026-10-29 — The solution correctly applies separation of variables, partial fraction decomposition, and the initial condition. The algebraic manipulation to combine the constant into the logarithm argument is valid for t > 3.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.