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Initial-value problems needing partial fractions

Problem 6.540 · hard

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

Separate: dx = dt / (t**2 - 4*t).
\[ \frac{1}{t^{2} - 4 t} = \frac{1}{4 t - 16} - \frac{1}{4 t} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 4}{t} \right)}}{4} = \frac{1}{t^{2} - 4 t} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 4}{t} \right)}}{4} \right|_{\substack{ t=6 }} + 0 = \frac{\ln{\left(3 \right)}}{4} \]
x(6) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(3 - \frac{12}{t} \right)}}{4} \right|_{\substack{ t=6 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \frac{\ln{\left(3 - \frac{12}{t} \right)}}{4} \)

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 integral of 1/u, leading to an incorrect domain of validity. For t > 4, (t-4)/t is posit

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-18 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution fails to account for the absolute value in the integral of 1/u, leading to an incorrect domain of validity. For t > 4, (t-4)/t is posit
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (style) 2026-10-16 — [domain objection, downgraded to style] The solution fails to account for the absolute value in the logarithm arising from the integration of 1/t terms. For t > 4, (t-4)/t is positive, but the general antiderivative requires ln|(t-4)/t|. More critically, the final answer form ln(3 - 12/t)/4 is algebraically equivalent to ln((3t-12)/t)/4 = ln(3(t-4)/t)/4 = (ln 3 + ln((t-4)/t))/4. The constant of integration was handled incorrectly in the transition from step 4 to 5; step 4 implies C = -ln(1/3)/4 = ln(3)/4, leading to x(t) = (1/4)ln((t-4)/t) + (1/4)ln(3) = (1/4)ln(3(t-4)/t). While the final expression is correct, the intermediate step 4 is confusingly presented as fixing the constant by substitution without showing the additive constant structure clearly, and step 5 just checks the result rather than deriving it. However, the main issue is that step 3 presents a specific antiderivative without absolute values, which is technically valid for t>4, but the derivation of the constant in step 4 is opaque. Let's look closer. Step 4 says `-Subs(..., t, 6) + 0 = log(3)/4`. This calculates the constant C such that x(t) = F(t) + C. F(6) = ln(2/6)/4 = ln(1/3)/4 = -ln(3)/4. So -F(6) = ln(3)/4. Thus C = ln(3)/4. Then x(t) = F(t) + C = (1/4)ln((t-4)/t) + (1/4)ln(3) = (1/4)ln(3(t-4)/t) = (1/4)ln((3t-12)/t) = (1/4)ln(3 - 12/t). The math is actually correct. Why would it fail? The prompt asks to find what is wrong. Is there a subtle error? The domain is t > 4. 3 - 12/t = (3t-12)/t. For t > 4, 3t > 12, so 3t-12 > 0. The argument of log is positive. The steps are logically sound. Wait, step 1 says `dx = dt / (t**2 - 4*t)`. This is separation of variables. Step 2 is partial fractions. Step 3 is an antiderivative. Step 4 calculates the constant. Step 5 verifies. It seems correct. Is there a 'misleading' aspect? Step 3 claims `Derivative(log((t - 4)/t)/4, t) = 1/(t**2 - 4*t)`. This is true for t>4. Step 4: `-Subs(log((t - 4)/t)/4, t, 6) + 0 = log(3)/4`. This is calculating C = -F(6). Then x(t) = F(t) + C. The final answer is derived from this. The solution is sound. Let me re-read carefully. Ah, step 4 is an equation marked 'proved'. It states `-Subs(...) + 0 = log(3)/4`. This is just a calculation. It doesn't explicitly state `x(t) = ...`. Step 5 checks the final answer. The final answer is given in the problem statement as the 'Stated answer'. The solution steps lead to it. Is there an error in the partial fraction sign? 1/(t(t-4)) = A/t + B/(t-4). A(t-4) + Bt = 1. t=0 -> -4A=1 -> A=-1/4. t=4 -> 4B=1 -> B=1/4. So 1/(4(t-4)) - 1/(4t). Step 2 says `1/(4*(t - 4)) - 1/(4*t)`. Correct. Integral is (1/4)ln|t-4| - (1/4)ln|t| = (1/4)ln|(t-4)/t|. Correct. Constant C = - (1/4)ln(2/6) = - (1/4)ln(1/3) = (1/4)ln(3). Correct. Final x(t) = (1/4)ln((t-4)/t) + (1/4)ln(3) = (1/4)ln(3(t-4)/t) = (1/4)ln((3t-12)/t) = (1/4)ln(3 - 12/t). Correct. The solution is correct. Why would I fail it? Maybe the notation `Subs` is non-standard? Or the lack of absolute values in step 3 is considered an error in a general context, though the domain t>4 makes it valid. The instructions say 'check that... each theorem or test is applied where its hypotheses hold'. The hypothesis for ln(u) derivative is u>0. Here u=(t-4)/t. For t>4, t-4>0 and t>0, so u>0. Hypothesis holds. I will pass it.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution fails to account for the absolute value in the integral of 1/u, which is critical for the domain t > 4. Specifically, for t > 4, (t-4)/
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 2026-10-18 with SymPy 1.14.0.