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Initial-value problems needing partial fractions

Problem 6.536 · hard

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

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

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

✓ 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

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — 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 2026-10-16
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution fails to account for the domain restriction t > -1. The term ln(t+1) is undefined for t <= -1, and the partial fraction decomposition leads to ln|t+1| - ln|t+2|. While the final answer is correct for t > -1, the intermediate steps (specifically line 3 and 4) implicitly assume t+1 > 0 without stating it, and line 4's calculation of the constant is ambiguous regarding absolute values. More critically, line 4 calculates the constant C such that x(t) = ln((t+1)/(t+2)) + C. At t=0, ln(1/2) + C = 0 => C = ln(2). So x(t) = ln((t+1)/(t+2)) + ln(2) = ln(2(t+1)/(t+2)). This logic is sound, but the presentation in line 4 is confusing: '-Subs(log((t + 1)/(t + 2)), t, 0) + 0 = log(2)' implies C = -x_particular(0). This is correct. However, the main issue is that the problem specifies t > -1. The antiderivative of 1/(t+1) is ln|t+1|. For t > -1, this is ln(t+1). The solution uses log((t+1)/(t+2)). For t > -1, t+1 > 0 and t+2 > 1 > 0, so the argument of the log is positive. The solution is technically correct for the given domain. Let's look closer. Is there an error? Line 2: 1/((t+1)(t+2)) = 1/(t+1) - 1/(t+2). Line 3 says derivative of log((t+1)/(t+2)) is the integrand. This is correct. Line 4: x(t) = log((t+1)/(t+2)) + C. x(0) = log(1/2) + C = 0 => C = log(2). Line 5 checks x(0) for the final answer. The steps are logically sound. Why would it fail? Perhaps the 'unchecked' equations are not the issue, but the sentences. There are no sentences except line 1. Line 1 is 'Separate: dx = dt / (t**2 + 3*t + 2).' This is a valid separation. The rest are equations. Wait, the prompt says 'Each line... is either an EQUATION... or a SENTENCE'. Lines 2-5 are marked [equation, unchecked]. The prompt says 'do not re-check the algebra of a line marked proved'. But these are marked 'unchecked'. Does 'unchecked' mean I should check them? The prompt says 'a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. It implies lines marked 'proved' are trusted. Lines marked 'unchecked' might need checking? Or perhaps the prompt implies that ONLY sentences are my job, and equations are trusted? 'The sentences are what you are here for.' This suggests I should ignore algebraic errors in equations if they are marked as such? But they are marked 'unchecked'. This is ambiguous. However, usually in these tasks, if a line is an equation, I check if the sentence *leading* to it or the *interpretation* is correct. Here, there are no sentences between the equations. Line 1 is the only sentence. It is correct. The equations follow logically. Is there a subtle error? The domain t > -1. The function 1/(t^2+3t+2) has singularities at t=-1 and t=-2. The interval t > -1 avoids both. The solution is valid. Let's look at the 'severity' options. If it's correct, it's pass. Is it misleading? No. Is it style? No. I will mark it pass.
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.