Home›Calculus 2›Initial-value problems needing partial fractions›Problem 6.539

Initial-value problems needing partial fractions

Problem 6.539 · hard

Solve \( \displaystyle \left(t^{2} - 5 t\right)\dfrac{dx}{dt} = 1 \) with \( \displaystyle x(6) = 0 \) (for \( \displaystyle t > 5 \)).

Separate: dx = dt / (t**2 - 5*t).
\[ \frac{1}{t^{2} - 5 t} = \frac{1}{5 t - 25} - \frac{1}{5 t} \]
Partial fractions.✓ Proved
\[ \frac{d}{d t} \frac{\ln{\left(\frac{t - 5}{t} \right)}}{5} = \frac{1}{t^{2} - 5 t} \]
An antiderivative, checked by differentiating.✓ Proved
\[ - \left. \frac{\ln{\left(\frac{t - 5}{t} \right)}}{5} \right|_{\substack{ t=6 }} + 0 = \frac{\ln{\left(6 \right)}}{5} \]
x(6) = 0 fixes the constant.✓ Proved
\[ \left. \frac{\ln{\left(6 - \frac{30}{t} \right)}}{5} \right|_{\substack{ t=6 }} = 0 \]
The initial condition holds.✓ Proved

Answer \( x(t) = \frac{\ln{\left(6 - \frac{30}{t} \right)}}{5} \)

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 (style) — [domain objection, downgraded to style] The solution fails to account for the absolute value in the logarithmic integral, which is critical for the domain t > 5. Specifically, for 5 < t < 6, the term (t-5)/t is positive but less than 1, while the argument of the log in the final answer (6 - 30/t) becomes negative, making the solution undefined in part of the requested domain. The correct general solution involves ln|(t-5)/t|, and the constant must be chosen carefully to handle the sign change or split domains.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-18 — [domain objection, downgraded to style] The solution fails to account for the absolute value in the logarithmic integral, which is critical for the domain t > 5. Specifically, for 5 < t < 6, the term (t-5)/t is positive but less than 1, while the argument of the log in the final answer (6 - 30/t) becomes negative, making the solution undefined in part of the requested domain. The correct general solution involves ln|(t-5)/t|, and the constant must be chosen carefully to handle the sign change or split domains.
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 of the antiderivative. Since the domain is t > 5, the term (t-5)/t is positive, so ln((t-5)/t) is valid. However, the final answer is given as ln(6 - 30/t)/5. Note that 6 - 30/t = (6t - 30)/t = 6(t-5)/t. The constant factor 6 inside the log corresponds to the integration constant, but the derivation in step 4 calculates the constant based on ln((t-5)/t), not ln(6-30/t). Specifically, step 4 computes the constant C such that x(t) = (1/5)ln((t-5)/t) + C. With x(6)=0, C = -(1/5)ln(1/6) = (1/5)ln(6). Thus x(t) = (1/5)ln((t-5)/t) + (1/5)ln(6) = (1/5)ln(6(t-5)/t) = (1/5)ln(6 - 30/t). The algebra is actually correct, but the presentation is disjointed. Step 3 identifies an antiderivative of ln((t-5)/t)/5. Step 4 calculates the constant for *that specific* antiderivative form. Step 5 checks a *different* form (which is equivalent). The error is not in the math, but the logic flow is confusing. Wait, let me re-read the prompt constraints. 'Check that... each conclusion follows from the lines before it'. Step 4 concludes 'x(6)=0 fixes the constant' resulting in 'log(6)/5'. This implies the general solution is x(t) = log((t-5)/t)/5 + log(6)/5. Step 5 checks that 'Subs(log(6 - 30/t)/5, t, 6) = 0'. It does not explicitly state that log(6 - 30/t)/5 is the same as the previous expression. It just checks the IC for the final answer. Is this an error? It's a gap in logic. The student might think the antiderivative in step 3 transforms directly into step 5 without showing the constant absorption. However, looking closer at Step 4: '-Subs(log((t - 5)/t)/5, t, 6) + 0 = log(6)/5'. This calculates C = log(6)/5. So x(t) = log((t-5)/t)/5 + log(6)/5 = log(6(t-5)/t)/5 = log(6 - 30/t)/5. The steps are mathematically sound, but Step 5 is just a verification, not a derivation of the final form from the previous steps. The final answer is not explicitly derived in a step, only verified. This is a 'style' or 'misleading' issue if it implies the form in step 5 is the direct result of step 4 without algebraic manipulation. But is it an error? No. Is it misleading? Yes, a student might not see how to get from step 4 to the stated answer. Let's look for a harder error. Domain: t > 5. (t-5)/t > 0. 6 - 30/t > 0 for t > 5. Everything is defined. The partial fraction is correct. The integration is correct. The constant is correct. The final form is correct. The only issue is the logical gap between step 4 and the final answer. Step 5 verifies the final answer satisfies the IC, but doesn't show it equals the expression from step 4. This is a 'style' issue or 'misleading' because it skips the algebraic combination of logs. I will mark it as 'misleading' because it doesn't show the equivalence.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to include the constant of integration in the general solution before applying the initial condition. Step 4 incorrectly assumes the constant is zero by equating the negative of the particular antiderivative at t=6 to 0, rather than solving C - F(6) = 0. This leads to an incorrect final answer (missing the constant term ln(5)/5).
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 2026-10-18 with SymPy 1.14.0.