Home›Calculus 2›Initial-value problems needing partial fractions›Problem 6.603

Initial-value problems needing partial fractions

Problem 6.603 · hard

Solve \( \displaystyle \left(t^{2} - 5 t + 6\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(5) = 0 \) (for \( \displaystyle t > 3 \)).

Separate: dx = dt / (t**2 - 5*t + 6).
\[ \frac{1}{t^{2} - 5 t + 6} = - \frac{1}{t - 2} + \frac{1}{t - 3} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \ln{\left(\frac{t - 3}{t - 2} \right)} = \frac{1}{t^{2} - 5 t + 6} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \ln{\left(\frac{t - 3}{t - 2} \right)} \right|_{\substack{ t=5 }} + 0 = - \ln{\left(2 \right)} + \ln{\left(3 \right)} \]
x(5) = 0 fixes the constant.✓ Proved
\[ \left. \ln{\left(\frac{3 \left(t - 3\right)}{2 \left(t - 2\right)} \right)} \right|_{\substack{ t=5 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \ln{\left(\frac{3 \left(t - 3\right)}{2 \left(t - 2\right)} \right)} \)

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 > 3, where t - 2 > 0 and t - 3 > 0, so the absolute values in the logarithm are unnecessary but not wron

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 ignores the domain constraint t > 3, where t - 2 > 0 and t - 3 > 0, so the absolute values in the logarithm are unnecessary but not wron
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the antiderivative and applies the initial condition to determine the constant of integration. The domain constraint t > 3 ensures the argument of the logarithm is positive, making the absolute value unnecessary.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly separates variables, applies partial fraction decomposition, and determines the constant of integration using the initial condition. The final answer matches the stated answer.
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.