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Differential equations by direct integration

Problem 6.528 · easy

Solve \( \displaystyle y' = \frac{1}{x} \) with \( \displaystyle y(1) = -3 \).

\[ \frac{d}{d x} \ln{\left(x \right)} = \frac{1}{x} \]
Integrate: y = this + C.✓ Proved
\[ - \left. \ln{\left(x \right)} \right|_{\substack{ x=1 }} - 3 = -3 \]
y(1) = -3 fixes C.✓ Proved
\[ \left. \ln{\left(x \right)} - 3 \right|_{\substack{ x=1 }} = -3 \]
The initial condition holds.✓ Proved

Answer \( y = \ln{\left(x \right)} - 3 \)

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
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ 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: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The general solution to y' = 1/x is y = ln|x| + C. The provided solution y = ln(x) - 3 is only valid for x > 0. While the initial condition x=1 is in this domain, the solution fails to specify the domain of validity or use the absolute value, which is mathematically imprecise for the general antiderivative. More critically, step 1 implies y = ln(x) + C without absolute values, which is technically incorrect for x < 0, though often accepted in introductory contexts if domain is restricted. However, the bigger issue is that the 'Stated answer' is y = ln(x) - 3, but the derivation in step 2 calculates C by substituting into ln(x) + C. Let's check the math: y(1) = ln(1) + C = 0 + C = C. Given y(1) = -3, C = -3. So y = ln(x) - 3. The algebra is correct for x>0. Is it an error? Strictly speaking, the antiderivative of 1/x is ln|x|. If the problem implies a real-valued function on an interval containing 1, ln(x) is fine. But usually, 'error' is reserved for wrong math. Is there a subtle error? Step 1 says 'Derivative(log(x), x) = 1/x'. This is true for x>0. Step 2 checks the IC. Step 3 checks the IC. The solution seems correct for the domain x>0. Why would it fail? Perhaps the lack of absolute value is considered an error in a rigorous context? Or is there a sign error? No. Let's look closer. 'Stated answer: y = ln(x) - 3'. If x can be negative, this is undefined. The differential equation y'=1/x is defined for x!=0. The solution should be y = ln|x| - 3. By omitting the absolute value, the solution is incomplete/wrong for x<0. In many calculus courses, this is a 'style' or 'misleading' issue, but often marked as an error if rigor is expected. Let's look for a harder error. Did it solve for C correctly? y = ln(x) + C. y(1) = -3 => 0 + C = -3 => C = -3. Correct. Is the final answer y = ln(x) - 3? Yes. Is it wrong? It's missing the absolute value. I will mark it as error because the general antiderivative requires absolute value, and stating y=ln(x) restricts the domain unnecessarily without justification.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-18 — [domain objection, downgraded to style] The general solution to y' = 1/x is y = ln|x| + C. The provided solution y = ln(x) - 3 is only valid for x > 0. While the initial condition x=1 is in this domain, the solution fails to specify the domain of validity or use the absolute value, which is mathematically imprecise for the general antiderivative. More critically, step 1 implies y = ln(x) + C without absolute values, which is technically incorrect for x < 0, though often accepted in introductory contexts if domain is restricted. However, the bigger issue is that the 'Stated answer' is y = ln(x) - 3, but the derivation in step 2 calculates C by substituting into ln(x) + C. Let's check the math: y(1) = ln(1) + C = 0 + C = C. Given y(1) = -3, C = -3. So y = ln(x) - 3. The algebra is correct for x>0. Is it an error? Strictly speaking, the antiderivative of 1/x is ln|x|. If the problem implies a real-valued function on an interval containing 1, ln(x) is fine. But usually, 'error' is reserved for wrong math. Is there a subtle error? Step 1 says 'Derivative(log(x), x) = 1/x'. This is true for x>0. Step 2 checks the IC. Step 3 checks the IC. The solution seems correct for the domain x>0. Why would it fail? Perhaps the lack of absolute value is considered an error in a rigorous context? Or is there a sign error? No. Let's look closer. 'Stated answer: y = ln(x) - 3'. If x can be negative, this is undefined. The differential equation y'=1/x is defined for x!=0. The solution should be y = ln|x| - 3. By omitting the absolute value, the solution is incomplete/wrong for x<0. In many calculus courses, this is a 'style' or 'misleading' issue, but often marked as an error if rigor is expected. Let's look for a harder error. Did it solve for C correctly? y = ln(x) + C. y(1) = -3 => 0 + C = -3 => C = -3. Correct. Is the final answer y = ln(x) - 3? Yes. Is it wrong? It's missing the absolute value. I will mark it as error because the general antiderivative requires absolute value, and stating y=ln(x) restricts the domain unnecessarily without justification.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution fails to determine the constant of integration C. It assumes the particular solution is ln(x) - 3 without showing that C = -3 derived from the general solution y = ln(x) + C. Step 2 checks a value but does not logically derive C.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to explicitly determine the constant of integration C. It jumps from the general form y = ln(x) + C to the specific solution y = ln(x) - 3 without showing the step where C is calculated as -3. While the final answer is correct, the logical derivation is incomplete and skips the crucial step of solving for C.
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/de_direct_integration, checked 2026-10-18 with SymPy 1.14.0.