Home›Calculus 2›Initial-value problems needing partial fractions›Problem 6.458

Initial-value problems needing partial fractions

Problem 6.458 · hard

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

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

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

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 matches the stated answer.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly separates variables, applies partial fraction decomposition, and determines the constant of integration using the initial condition. The final expression matches the stated answer.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution incorrectly determines the constant of integration. Step 4 calculates the constant as log(2)/2, but Step 5 claims that substituting this constant into the expression log(2*(t-1)/(t+1))/2 yields 0 at t=3. However, log(2*(3-1)/(3+1))/2 = log(1)/2 = 0, which implies the constant added was 0, not log(2)/2. The stated answer x(t) = ln(2(t-1)/(t+1))/2 does not satisfy x(3)=0 if derived from the standard antiderivative ln((t-1)/(t+1))/2 + C with C=log(2)/2, because ln((3-1)/(3+1))/2 + log(2)/2 = -log(2)/2 + log(2)/2 = 0. Wait, let's re-evaluate. Antiderivative F(t) = 1/2 ln((t-1)/(t+1)). F(3) = 1/2 ln(2/4) = 1/2 ln(1/2) = -1/2 ln(2). We need x(3) = 0, so C = -F(3) = 1/2 ln(2). So x(t) = 1/2 ln((t-1)/(t+1)) + 1/2 ln(2) = 1/2 ln(2(t-1)/(t+1)). Step 4 says: -Subs(log((t - 1)/(t + 1))/2, t, 3) + 0 = log(2)/2. Subs(...) is -1/2 ln(2). So -(-1/2 ln(2)) = 1/2 ln(2). This is correct. Step 5 says: Subs(log(2*(t - 1)/(t + 1))/2, t, 3) = 0. log(2*(2)/4)/2 = log(1)/2 = 0. This is correct. The logic holds. The constant C is determined correctly in step 4, and the final form in step 5 is consistent with that constant. The solution is actually correct. Why did I think it was wrong? Let's re-read carefully. Step 4: `-Subs(log((t - 1)/(t + 1))/2, t, 3) + 0 = log(2)/2`. This equation is asserting that the constant C is equal to `-F(3)`. Since `x(t) = F(t) + C` and `x(3)=0`, `C = -F(3)`. `F(3) = 1/2 ln(2/4) = -1/2 ln(2)`. `-F(3) = 1/2 ln(2)`. So `C = log(2)/2`. This is correct. Step 5: `Subs(log(2*(t - 1)/(t + 1))/2, t, 3) = 0`. This checks the final answer. `log(2*(3-1)/(3+1))/2 = log(4/4)/2 = log(1)/2 = 0`. This is correct. The solution is sound. The steps are logically consistent and algebraically correct.
gpt-oss:20b: pass 2026-10-14
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly separates variables, applies partial fraction decomposition, and determines the constant of integration to satisfy the initial condition. The final expression matches the stated answer.
gpt-oss:20b: pass 2026-10-14

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 2026-10-17 with SymPy 1.14.0.