Home›Calculus 3›Second-order linear equations›Problem 6.185

Second-order linear equations

Problem 6.185 · medium

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

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

Answer \( y = \left(1 - 2 x\right) e^{x} \)

Lines: 2 proved, 5 reviewed. 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	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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
4	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
7	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
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: pass — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the initial conditions.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the final result against the differential equation and initial conditions.
gpt-oss:20b: fail (misleading) 2026-09-29 — The solution skips the explicit determination of the constants C₁ and C₂ from the initial conditions, merely stating that they are found. A student would be left unsure how the final expression (1‑2x)eˣ was obtained. Additionally, the first sentence gives a generic form of the characteristic equation without specifying the actual coefficients, which could confuse readers.
qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly identifies the repeated root of the characteristic equation, applies the standard form for repeated roots, and verifies the initial conditions and differential equation.
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 2026-10-13 with SymPy 1.14.0.