Initial-value problems needing partial fractions
Problem 6.457 · hard
Solve \( \displaystyle \left(t^{2} + 2 t - 3\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(2) = 0 \) (for \( \displaystyle t > 1 \)).
- Separate: dx = dt / (t**2 + 2*t - 3).Reviewed
- \[ \frac{1}{t^{2} + 2 t - 3} = - \frac{1}{4 t + 12} + \frac{1}{4 t - 4} \]Partial fractions.✓ Proved
- \[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 1}{t + 3} \right)}}{4} = \frac{1}{t^{2} + 2 t - 3} \]An antiderivative, checked by differentiating.✓ Proved
- \[ - \left. \frac{\ln{\left(\frac{t - 1}{t + 3} \right)}}{4} \right|_{\substack{ t=2 }} + 0 = \frac{\ln{\left(5 \right)}}{4} \]x(2) = 0 fixes the constant.✓ Proved
- \[ \left. \frac{\ln{\left(\frac{5 \left(t - 1\right)}{t + 3} \right)}}{4} \right|_{\substack{ t=2 }} = 0 \]The initial condition holds.✓ Proved
Answer \( x(t) = \frac{\ln{\left(\frac{5 \left(t - 1\right)}{t + 3} \right)}}{4} \)
✓ 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 separates variables, applies partial fraction decomposition, and determines the constant of integration to satisfy the initial condition. The final expression matches the stated answer.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly separates variables, applies partial fraction decomposition, and determines the constant of integration to satisfy the initial condition. The final expression matches the stated answer.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly separates variables, applies partial fraction decomposition, integrates, and determines the constant of integration using the initial condition. The final form matches the stated answer.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution fails to account for the absolute value in the integral of 1/u, which is necessary because the term (t-1) is negative for t in (1, 2). The derived formula yields a complex logarithm for t < 1 (or rather, for t where the argument is negative, though here t>1 so t-1>0, wait. Let's re-evaluate. t>1 implies t-1 > 0 and t+3 > 0. So (t-1)/(t+3) is positive. The absolute value is not strictly needed for the domain t>1. However, the constant calculation in step 4 is weirdly phrased but algebraically leads to the correct result. Let's look closer. Step 4 says '-Subs(..., t, 2) + 0 = log(5)/4'. This implies C = log(5)/4. Then x(t) = F(t) + C. Step 5 checks Subs(log(5*(t-1)/(t+3))/4, t, 2) = 0. log(5*1/5)/4 = log(1)/4 = 0. This is correct. Is there an error? The problem states t > 1. In this domain, t-1 > 0. So ln((t-1)/(t+3)) is real. The solution seems correct. Why would it fail? Let's check the partial fraction signs. 1/((t-1)(t+3)) = A/(t-1) + B/(t+3). A(t+3) + B(t-1) = 1. t=1: 4A=1 => A=1/4. t=-3: -4B=1 => B=-1/4. So 1/(4(t-1)) - 1/(4(t+3)). Integral is 1/4 ln|t-1| - 1/4 ln|t+3| = 1/4 ln|(t-1)/(t+3)|. For t>1, this is 1/4 ln((t-1)/(t+3)). x(t) = 1/4 ln((t-1)/(t+3)) + C. x(2) = 1/4 ln(1/5) + C = 0 => C = -1/4 ln(1/5) = 1/4 ln(5). x(t) = 1/4 ln((t-1)/(t+3)) + 1/4 ln(5) = 1/4 ln(5(t-1)/(t+3)). The solution is correct. Wait, did I miss something? 'unchecked' equations are not to be checked. Sentences are to be checked. Sentence 1: 'Separate: dx = dt / (t**2 + 2*t - 3)'. This is correct. Sentence 2 is an equation. Sentence 3 is an equation. Sentence 4 is an equation. Sentence 5 is an equation. There are no other sentences. The setup is correct. The logic is correct. The answer is correct. Verdict: pass.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.