Home›Calculus 2›Initial-value problems needing partial fractions›Problem 6.761

Initial-value problems needing partial fractions

Problem 6.761 · hard

Solve \( \displaystyle \left(t^{2} - 9 t + 20\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(7) = 0 \) (for \( \displaystyle t > 5 \)).

Separate: dx = dt / (t**2 - 9*t + 20).
Reviewed
\[ \frac{1}{t^{2} - 9 t + 20} = - \frac{1}{t - 4} + \frac{1}{t - 5} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \ln{\left(\frac{t - 5}{t - 4} \right)} = \frac{1}{t^{2} - 9 t + 20} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \ln{\left(\frac{t - 5}{t - 4} \right)} \right|_{\substack{ t=7 }} + 0 = - \ln{\left(2 \right)} + \ln{\left(3 \right)} \]
x(7) = 0 fixes the constant.✓ Proved
\[ \left. \ln{\left(\frac{3 \left(t - 5\right)}{2 \left(t - 4\right)} \right)} \right|_{\substack{ t=7 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \ln{\left(\frac{3 \left(t - 5\right)}{2 \left(t - 4\right)} \right)} \)

Lines: 4 proved, 1 reviewed. 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	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: pass
qwen3.6:27b-mlx: pass — The solution correctly separates variables, applies partial fraction decomposition, and determines the constant of integration using the initial condition. The final expression satisfies both the differential equation and the initial condition for t > 5.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-30 — The solution correctly separates variables, applies partial fraction decomposition, and determines the constant of integration using the initial condition. The final expression satisfies both the differential equation and the initial condition for t > 5.
gpt-oss:20b: pass 2026-10-30
qwen3.6:27b-mlx: inconclusive 2026-10-29 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution fails to account for the absolute value in the logarithm of the integral of 1/u. Since the domain is t > 5, t-5 is positive but t-4 is
gpt-oss:20b: pass 2026-10-29
qwen3.6:27b-mlx: inconclusive 2026-10-29 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The partial fraction decomposition in step 2 is incorrect; the correct decomposition is 1/(t-5) - 1/(t-4), not -1/(t-4) + 1/(t-5) (which is actually
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.