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Initial-value problems needing partial fractions

Problem 6.607 · hard

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

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

Answer \( x(t) = \frac{\ln{\left(\frac{2 \left(t - 1\right)}{t + 2} \right)}}{3} \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. 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 the differential equation and the initial condition for t > 1.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly separates variables, applies partial fraction decomposition, and determines the constant of integration using the initial condition. The final expression satisfies the differential equation and the initial condition for t > 1.
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. Since the domain is t > 1, the term ln(t-1) is valid, but the term ln(t+2) is also positive, so the argument (t-1)/(t+2) is positive. However, the constant of integration step (Line 4) is algebraically confused and does not clearly derive the final form. More critically, Line 4 calculates the constant C such that x(t) = F(t) + C. If F(t) = (1/3)ln((t-1)/(t+2)), then F(4) = (1/3)ln(1/2) = -ln(2)/3. To satisfy x(4)=0, we need C = ln(2)/3. Thus x(t) = (1/3)ln((t-1)/(t+2)) + ln(2)/3 = (1/3)ln(2(t-1)/(t+2)). While the final answer is correct, Line 4's notation '-Subs(...)' is ambiguous and potentially misleading regarding how the constant is added vs subtracted, but the main issue is that the derivation of the constant is presented as a black-box check rather than a clear derivation, and Line 3's antiderivative misses the absolute value which is technically required for general validity, though harmless here. The primary error is the lack of clarity in deriving the constant C in Line 4, which reads like a verification rather than a derivation step, and the notation is confusing.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies separation of variables and partial fraction decomposition. The integration constant is determined correctly using the initial condition, and the final expression satisfies both the differential equation and the initial value.
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.