Second-order linear equations
Problem 6.44 · hard
Solve \( \displaystyle y'' - 6y' + 9y = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = -3 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.Reviewed
- \[ \left(m - 3\right)^{2} = m^{2} - 6 m + 9 \]Roots: 3.✓ Proved
- A repeated root r gives y = (C₁ + C₂x)e^(rx).Reviewed
- Use y(0) and y'(0) to find C₁ and C₂.Reviewed
- \[ \left(54 x - 27\right) e^{3 x} - 6 \frac{d}{d x} \left(6 x - 3\right) e^{3 x} + \frac{d^{2}}{d x^{2}} \left(6 x - 3\right) e^{3 x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ -3 \]y(0).Reviewed
- \[ -3 \]y'(0).Reviewed
Answer \( y = \left(6 x - 3\right) e^{3 x} \)
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: passqwen3.6:27b-mlx: pass — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the final answer against the differential equation and initial conditions.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the final answer against the differential equation and initial conditions.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the repeated root, applies the standard form for the general solution, and verifies that the specific constants satisfy both the differential equation and the initial conditions.gpt-oss:20b: fail (misleading) 2026-09-27 — The solution skips the key step of determining the constants C₁ and C₂ from the initial conditions, and the initial sentence gives a generic characteristic equation that could confuse students. The verification steps are also presented without showing the actual calculations, which may mislead readers about how the constants were found.qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the repeated root, applies the standard form for the general solution, and verifies that the stated answer satisfies both the differential equation and the initial conditions.gpt-oss:20b: pass 2026-09-26
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.