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Initial-value problems needing partial fractions

Problem 6.765 · hard

Solve \( \displaystyle \left(t^{2} - 2 t\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(4) = 0 \) (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
\[ - \left. \frac{\ln{\left(\frac{t - 2}{t} \right)}}{2} \right|_{\substack{ t=4 }} + 0 = \frac{\ln{\left(2 \right)}}{2} \]
x(4) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(2 - \frac{4}{t} \right)}}{2} \right|_{\substack{ t=4 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \frac{\ln{\left(2 - \frac{4}{t} \right)}}{2} \)

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: pass
qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution ignores the domain constraint t > 2. For t > 2, the argument of the logarithm in the antiderivative, (t-2)/t, is positive, but the stat

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-30 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution ignores the domain constraint t > 2. For t > 2, the argument of the logarithm in the antiderivative, (t-2)/t, is positive, but the stat
gpt-oss:20b: pass 2026-10-30
qwen3.6:27b-mlx: fail (style) 2026-10-29 — [domain objection, downgraded to style] The solution ignores the domain constraint t > 2, where t - 2 > 0. The antiderivative log((t - 2)/t) is valid, but the final answer log(2 - 4/t) is undefined for t > 2 because 2 - 4/t < 0. The correct form should involve absolute values or be written as log((t-2)/t) to remain real-valued.
gpt-oss:20b: pass 2026-10-29
qwen3.6:27b-mlx: fail (style) 2026-10-29 — [domain objection, downgraded to style] The solution fails to account for the absolute value in the integral of 1/u, which is critical given the domain t > 2 where (t-2)/t is positive but the standard antiderivative form requires careful handling of signs or absolute values to be rigorous. More importantly, the stated answer x(t) = ln(2 - 4/t)/2 is undefined for t > 2 because 2 - 4/t is negative for 2 < t < 2 (impossible) wait, for t > 2, 4/t < 2, so 2 - 4/t > 0. Let's re-evaluate. For t > 2, t-2 > 0 and t > 0, so (t-2)/t > 0. The term 2 - 4/t = 2(t-2)/t. So ln(2 - 4/t) = ln(2) + ln((t-2)/t). The constant of integration absorbs the ln(2). The solution steps are algebraically consistent with the stated answer. However, step 1 says 'Separate: dx = dt / (t**2 - 2*t)'. This is correct. Step 2 is correct. Step 3 checks the derivative. Step 4 calculates the constant. Step 5 checks the IC. The main issue is that the problem asks to solve for t > 2. The expression 2 - 4/t is positive for t > 2. So the log is defined. Is there an error? Let's check the constant. Integral is 1/2 ln|(t-2)/t| + C. x(4) = 1/2 ln(2/4) + C = 1/2 ln(1/2) + C = -1/2 ln(2) + C = 0 => C = 1/2 ln(2). So x(t) = 1/2 ln((t-2)/t) + 1/2 ln(2) = 1/2 ln(2(t-2)/t) = 1/2 ln(2 - 4/t). The solution is correct. Why would it fail? Maybe the 'unchecked' equations are not the issue, but the sentences. Sentence 1 is fine. Sentence 2 is fine. Sentence 3 is fine. Sentence 4 is fine. Sentence 5 is fine. Wait, look at the domain. t > 2. The term (t-2)/t is positive. The absolute value is not strictly needed if we know the sign, but usually one writes ln|...|. The solution writes log((t-2)/t). Since t>2, (t-2)/t > 0, so this is valid. The final answer is valid. Is there a subtle error? The prompt says 'find what is wrong'. Let's look closer. Step 4: '-Subs(log((t - 2)/t)/2, t, 4) + 0 = log(2)/2'. This implies C = log(2)/2. Step 5: 'Subs(log(2 - 4/t)/2, t, 4) = 0'. This checks the final form. The logic holds. Is it possible the verdict should be pass? Let's re-read the instructions. 'Each line... is either an EQUATION... or a SENTENCE'. The review is for the sentences. The sentences are: 1. Separate... 2. Partial fractions. 3. An antiderivative... 4. x(4)=0 fixes... 5. The initial condition holds. These are all correct statements. However, often in these adversarial reviews, if the absolute value is omitted in a general context, it might be flagged. But here the domain is specified t > 2, making the argument positive. So it's technically correct. Is there a sign error? 1/(t^2-2t) = 1/(t(t-2)) = A/t + B/(t-2). A(t-2) + Bt = 1. t=0 -> -2A=1 -> A=-1/2. t=2 -> 2B=1 -> B=1/2. So 1/(2(t-2)) - 1/(2t). Correct. Integral is 1/2 ln|t-2| - 1/2 ln|t| = 1/2 ln|(t-2)/t|. Correct. The solution seems sound. Let me look for a 'style' or 'misleading' issue. The notation 'log' usually means natural log in calculus contexts, but sometimes base 10. If base 10, it's wrong. But standard calculus assumes ln. The problem uses 'ln' in the stated answer and 'log' in the steps. This inconsistency is a style issue. Also, the step 4 equation is written as '-Subs(...) + 0 = ...'. This is a bit weirdly phrased as a sentence justification for an equation, but the equation itself is marked unchecked. The sentence is 'x(4) = 0 fixes the constant.' This is a correct sentence. I will flag the log/ln inconsistency as a style issue, or perhaps pass if it's considered trivial. Let's look for a deeper error. What if t < 0? No, t > 2. What if the question implies a general solution? No, IVP. I will mark it as pass because the math is correct and the domain ensures the argument of the log is positive.
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.