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Second-order linear equations

Problem 6.184 · medium

Solve \( \displaystyle y'' - 2y' + 1y = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = 3 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m - 1\right)^{2} = m^{2} - 2 m + 1 \]
Roots: 1.✓ Proved
A repeated root r gives y = (C₁ + C₂x)e^(rx).
Use y(0) and y'(0) to find C₁ and C₂.
\[ 3 e^{x} - 2 \frac{d}{d x} 3 e^{x} + \frac{d^{2}}{d x^{2}} 3 e^{x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ 3 \]
y(0).✓ Proved
\[ 3 \]
y'(0).✓ Proved

Answer \( y = 3 e^{x} \)

Lines: 4 proved, 3 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.

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	Not checked	—	a sentence; read, not computed
4	Not checked	—	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
7	✓ 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	sympy.checkodesol substitutes the solution back; both initial conditions match

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution claims the general solution for a repeated root is y = (C₁ + C₂x)e^(rx), but then provides a final answer of 3*exp(x) which corresponds

Every verdict on record (4)

qwen3.6:27b-mlx: inconclusive 2026-09-29 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution claims the general solution for a repeated root is y = (C₁ + C₂x)e^(rx), but then provides a final answer of 3*exp(x) which corresponds
gpt-oss:20b: pass 2026-09-29
qwen3.6:27b-mlx: fail (error) 2026-09-29 — The solution claims the general solution for a repeated root is y = (C₁ + C₂x)e^(rx), but then presents y = 3e^x as the final answer without determining C₂. Since y'(0)=3 and y(0)=3 for y=3e^x, C₂ must be 0, but the solution fails to show this calculation or justify why the x term is absent, effectively skipping the step of solving for the constants.
gpt-oss:20b: pass 2026-09-29

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/second_order_linear, checked 2026-09-29 with SymPy 1.14.0.