Home›Calculus 2›Initial-value problems needing partial fractions›Problem 6.609

Initial-value problems needing partial fractions

Problem 6.609 · hard

Solve \( \displaystyle \left(t^{2} - 3 t\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(4) = 0 \) (for \( \displaystyle t > 3 \)).

Separate: dx = dt / (t**2 - 3*t).
\[ \frac{1}{t^{2} - 3 t} = \frac{1}{3 t - 9} - \frac{1}{3 t} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 3}{t} \right)}}{3} = \frac{1}{t^{2} - 3 t} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 3}{t} \right)}}{3} \right|_{\substack{ t=4 }} + 0 = \frac{2 \ln{\left(2 \right)}}{3} \]
x(4) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(4 - \frac{12}{t} \right)}}{3} \right|_{\substack{ t=4 }} = 0 \]
The initial condition holds.✓ Proved

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

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: fail (error) — The stated answer x(t) = ln(4 - 12/t)/3 is incorrect. The correct solution derived from the partial fractions and initial condition is x(t) = (1/3)ln((t-3)/t) + (2/3)ln(2), which simplifies to (1/3)ln(2(t-3)/t). The stated answer fails to satisfy the differential equation or the initial condition properly (e.g., at t=4, ln(4-3)/3 = 0, which works for IC, but derivative check fails: d/dt ln(4-12/t) = (12/t^2)/(4-12/t) = 12/(4t-12) = 3/(t-3), so dx/dt = 1/(t-3), but equation requires dx/dt = 1/(t(t-3)). The term 1/t is missing.)

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-18 — The stated answer x(t) = ln(4 - 12/t)/3 is incorrect. The correct solution derived from the partial fractions and initial condition is x(t) = (1/3)ln((t-3)/t) + (2/3)ln(2), which simplifies to (1/3)ln(2(t-3)/t). The stated answer fails to satisfy the differential equation or the initial condition properly (e.g., at t=4, ln(4-3)/3 = 0, which works for IC, but derivative check fails: d/dt ln(4-12/t) = (12/t^2)/(4-12/t) = 12/(4t-12) = 3/(t-3), so dx/dt = 1/(t-3), but equation requires dx/dt = 1/(t(t-3)). The term 1/t is missing.)
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution claims x(t) = ln(4 - 12/t)/3 satisfies the initial condition x(4)=0, but ln(4 - 12/4)/3 = ln(1)/3 = 0 is correct, however, the general
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The solution fails to account for the absolute value in the integral of 1/u, which is necessary because (t-3)/t is negative for 3 < t < 4. The stated answer ln(4 - 12/t) is undefined for t in (3, 4), whereas the correct solution involves ln|1 - 3/t| or ln((3-t)/t) in that interval.
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.