Home›Calculus 2›Initial-value problems needing partial fractions›Problem 6.695

Initial-value problems needing partial fractions

Problem 6.695 · hard

Solve \( \displaystyle \left(t^{2} - 2 t - 8\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(6) = 0 \) (for \( \displaystyle t > 4 \)).

Separate: dx = dt / (t**2 - 2*t - 8).
\[ \frac{1}{t^{2} - 2 t - 8} = - \frac{1}{6 t + 12} + \frac{1}{6 t - 24} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 4}{t + 2} \right)}}{6} = \frac{1}{t^{2} - 2 t - 8} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 4}{t + 2} \right)}}{6} \right|_{\substack{ t=6 }} + 0 = \frac{\ln{\left(2 \right)}}{3} \]
x(6) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(\frac{4 \left(t - 4\right)}{t + 2} \right)}}{6} \right|_{\substack{ t=6 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \frac{\ln{\left(\frac{4 \left(t - 4\right)}{t + 2} \right)}}{6} \)

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: fail (error) — The solution incorrectly determines the constant of integration. Step 4 calculates the value of the particular antiderivative at t=6 as log(2)/3, but then Step 5 presents a final solution with a factor of 4 inside the logarithm that was not derived from the integration constant. The correct general solution is x(t) = (1/6)ln((t-4)/(t+2)) + C. Using x(6)=0 gives C = -(1/6)ln(1/4) = (1/3)ln(2). Thus x(t) = (1/6)ln((t-4)/(t+2)) + (1/3)ln(2) = (1/6)ln(4(t-4)/(t+2)). While the final formula in Step 5 is actually correct, Step 4 is mathematically nonsensical as written ('-Subs(...) + 0 = log(2)/3' does not explain how the constant is added or why the argument changes to 4*(t-4)/(t+2)). The jump from Step 4 to Step 5 is unjustified and confusing, implying the constant is handled by magic rather than algebra. Specifically, Step 4 claims to fix the constant but the equation shown is just an evaluation, not the determination of C for the general form. A student would be confused about how to combine the particular integral and the constant.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution incorrectly determines the constant of integration. Step 4 calculates the value of the particular antiderivative at t=6 as log(2)/3, but then Step 5 presents a final solution with a factor of 4 inside the logarithm that was not derived from the integration constant. The correct general solution is x(t) = (1/6)ln((t-4)/(t+2)) + C. Using x(6)=0 gives C = -(1/6)ln(1/4) = (1/3)ln(2). Thus x(t) = (1/6)ln((t-4)/(t+2)) + (1/3)ln(2) = (1/6)ln(4(t-4)/(t+2)). While the final formula in Step 5 is actually correct, Step 4 is mathematically nonsensical as written ('-Subs(...) + 0 = log(2)/3' does not explain how the constant is added or why the argument changes to 4*(t-4)/(t+2)). The jump from Step 4 to Step 5 is unjustified and confusing, implying the constant is handled by magic rather than algebra. Specifically, Step 4 claims to fix the constant but the equation shown is just an evaluation, not the determination of C for the general form. A student would be confused about how to combine the particular integral and the constant.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution fails to account for the absolute value in the integral of 1/u, leading to an incorrect constant of integration. For t > 4, (t-4)/(t+2)
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (style) 2026-10-17 — [domain objection, downgraded to style] The solution fails to account for the absolute value in the logarithm arising from the integration of 1/(t-4). Since the problem specifies t > 4, the term ln|t-4| is ln(t-4), but the term ln|t+2| is ln(t+2). The partial fraction decomposition leads to (1/6)ln|(t-4)/(t+2)|. The proposed answer uses ln(4(t-4)/(t+2))/6. While the constant factor 4 inside the log is absorbed into the integration constant, the sign of the argument inside the log must be positive for the log to be real. For t > 4, (t-4)/(t+2) is positive, so the absolute value is not strictly necessary for the final form if the constant is chosen correctly, BUT the derivation in step 4 is flawed. Step 4 calculates the constant C such that x(6)=0. If x(t) = (1/6)ln((t-4)/(t+2)) + C, then 0 = (1/6)ln(2/8) + C => C = -(1/6)ln(1/4) = (1/6)ln(4). So x(t) = (1/6)ln((t-4)/(t+2)) + (1/6)ln(4) = (1/6)ln(4(t-4)/(t+2)). This matches the stated answer. However, the sentence in step 1 says 'Separate: dx = dt / ...' which is fine. The issue is that step 3 claims 'Derivative(log((t - 4)/(t + 2))/6, t) = ...' which is true for t>4. Step 4 says '-Subs(log((t - 4)/(t + 2))/6, t, 6) + 0 = log(2)/3'. Let's check: log((6-4)/(6+2))/6 = log(2/8)/6 = log(1/4)/6 = -log(4)/6. So -(-log(4)/6) = log(4)/6. The equation says this equals log(2)/3. log(4)/6 = 2log(2)/6 = log(2)/3. This is correct. Step 5 checks the final answer. The logic seems sound. Wait, is there a domain issue? t>4. The argument 4(t-4)/(t+2) is positive. The log is defined. The derivative is correct. The initial condition is satisfied. Why would it fail? Let me re-read carefully. 'Separate: dx = dt / (t**2 - 2*t - 8).' This is correct. 'Partial fractions.' Correct. 'An antiderivative...' Correct. 'x(6)=0 fixes the constant.' The calculation in step 4 is: - (value at 6) + 0 = constant? No, usually x(t) = F(t) + C. x(6) = F(6) + C = 0 => C = -F(6). The equation in step 4 is '-Subs(...) + 0 = log(2)/3'. This implies C = log(2)/3. Let's check C = -F(6). F(6) = log(2/8)/6 = -log(4)/6. So C = log(4)/6 = log(2)/3. This is correct. The final answer x(t) = F(t) + C = log((t-4)/(t+2))/6 + log(2)/3 = log((t-4)/(t+2))/6 + log(4)/6 = log(4(t-4)/(t+2))/6. This matches. Is there any subtle error? The problem asks to solve for t > 4. The solution is valid. I see no error. Wait, look at step 4 again. '-Subs(log((t - 4)/(t + 2))/6, t, 6) + 0 = log(2)/3'. This line is calculating the constant C. It says C = log(2)/3. Then step 5 substitutes this C into the expression. But step 5 writes 'Subs(log(4*(t - 4)/(t + 2))/6, t, 6) = 0'. This checks the final form. The final form combines the log terms. log(A) + log(B) = log(AB). log((t-4)/(t+2))/6 + log(4)/6 = log(4(t-4)/(t+2))/6. This is correct. I will mark it as pass.
gpt-oss:20b: pass 2026-10-17

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 2026-10-19 with SymPy 1.14.0.