Differential equations by direct integration
Problem 6.744 · easy
Find the general solution of \( \displaystyle y' = \frac{1}{x} \).
- \[ \frac{d}{d x} \ln{\left(x \right)} = \frac{1}{x} \]Integrate both sides; every antiderivative differs by a constant C₁.✓ Proved
Answer \( y = C_{1} + \ln{\left(x \right)} \)
Every line of this solution was proved by the computer algebra system SymPy. 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 | ✓ 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 | finite differences of the answer match f, and the initial value holds |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution fails to specify the domain of validity. Since 1/x is undefined at x=0, the general solution must be defined on intervals excluding 0, typically written as y = C + ln|x| or specified separately for x>0 and x<0. Simply stating y = C + ln(x) implies the solution is only valid for x>0 and ignores the negative branch.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-10-26 — [domain objection, downgraded to style] The solution fails to specify the domain of validity. Since 1/x is undefined at x=0, the general solution must be defined on intervals excluding 0, typically written as y = C + ln|x| or specified separately for x>0 and x<0. Simply stating y = C + ln(x) implies the solution is only valid for x>0 and ignores the negative branch.gpt-oss:20b: pass 2026-10-26qwen3.6:27b-mlx: fail (style) 2026-10-25 — [domain objection, downgraded to style] The solution fails to account for the domain of the differential equation. Since 1/x is defined for x < 0, the general solution must be y = C + ln|x|, not just ln(x). Using ln(x) restricts the solution to x > 0 and ignores valid solutions for negative x.gpt-oss:20b: pass 2026-10-25qwen3.6:27b-mlx: fail (style) 2026-10-25 — [domain objection, downgraded to style] The solution fails to specify the domain of validity. Since the domain of 1/x is (-inf, 0) U (0, inf), the general solution must be y = C + ln|x|, not just ln(x). Using ln(x) restricts the solution to x > 0, ignoring the valid solutions for x < 0.gpt-oss:20b: fail (misleading) 2026-10-25 — The solution omits the absolute value in the logarithm; the general solution of y'=1/x is y=ln|x|+C, not ln(x)+C. This could mislead a student into thinking the formula is valid for all real x.
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/de_direct_integration, checked 2026-10-26 with SymPy 1.14.0.