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.
- \[ \left(m - 1\right)^{2} = m^{2} - 2 m + 1 \]Roots: 1.✓ Proved
- A repeated root r gives y = (C₁ + C₂x)e^(rx).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ \left(1 - 2 x\right) e^{x} - 2 \frac{d}{d x} \left(1 - 2 x\right) e^{x} + \frac{d^{2}}{d x^{2}} \left(1 - 2 x\right) e^{x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ 1 \]y(0).✓ Proved
- \[ -1 \]y'(0).✓ Proved
Answer \( y = \left(1 - 2 x\right) e^{x} \)
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: fail (misleading) — 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 — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the final result against the differential equation and initial conditions.
Every verdict on record (4)
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 2026-09-29 with SymPy 1.14.0.