Home›Calculus 2›Initial-value problems needing partial fractions›Problem 6.763

Initial-value problems needing partial fractions

Problem 6.763 · hard

Solve \( \displaystyle \left(t^{2} - 4\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(5) = 0 \) (for \( \displaystyle t > 2 \)).

Separate: dx = dt / (t**2 - 4).
Reviewed
\[ \frac{1}{t^{2} - 4} = - \frac{1}{4 t + 8} + \frac{1}{4 t - 8} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 2}{t + 2} \right)}}{4} = \frac{1}{t^{2} - 4} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 2}{t + 2} \right)}}{4} \right|_{\substack{ t=5 }} + 0 = - \frac{\ln{\left(3 \right)}}{4} + \frac{\ln{\left(7 \right)}}{4} \]
x(5) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{7 \left(t - 2\right)}{3 \left(t + 2\right)} \right)}}{4} \right|_{\substack{ t=5 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \frac{\ln{\left(\frac{7 \left(t - 2\right)}{3 \left(t + 2\right)} \right)}}{4} \)

✓ 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 applies partial fraction decomposition and integration, properly handling the domain t > 2. The constant of integration is determined correctly using the initial condition, and the final form matches the stated answer.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-30 — The solution correctly applies partial fraction decomposition and integration, properly handling the domain t > 2. The constant of integration is determined correctly using the initial condition, and the final form matches the stated answer.
gpt-oss:20b: pass 2026-10-30
qwen3.6:27b-mlx: pass 2026-10-29 — The solution correctly applies separation of variables, partial fraction decomposition, and the initial condition. The final answer matches the stated answer and satisfies the differential equation and initial condition for t > 2.
gpt-oss:20b: pass 2026-10-29
qwen3.6:27b-mlx: pass 2026-10-29 — 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 the differential equation and the initial value.
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.