Second-order linear equations
Problem 6.133 · medium
Solve \( \displaystyle y'' + 0y' + 1y = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 0 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m - i\right) \left(m + i\right) = m^{2} + 1 \]Roots: -I, I.✓ Proved
- Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ - 3 \cos{\left(x \right)} + \frac{d^{2}}{d x^{2}} \left(- 3 \cos{\left(x \right)}\right) = 0 \]The solution satisfies the equation.✓ Proved
- \[ -3 \]y(0).
- \[ 0 \]y'(0).
Answer \( y = - 3 \cos{\left(x \right)} \)
Lines: 2 proved, 5 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. How we verify
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 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 7 | Not checked | — | 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: fail (error) — Sentences 6 and 7 incorrectly state the initial values: y(0) is -3 and y'(0) is 0, but the solution lists -3, -3 and 0, 0, respectively.qwen3.6:27b-mlx: pass — The solution correctly identifies the characteristic equation, roots, and general form. The constants are implicitly determined by the final answer, and the verification steps confirm the solution satisfies both 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, roots, and general form. The constants are implicitly determined by the final answer, and the verification steps confirm the solution satisfies both the differential equation and initial conditions.gpt-oss:20b: fail (error) 2026-09-28 — Sentences 6 and 7 incorrectly state the initial values: y(0) is -3 and y'(0) is 0, but the solution lists -3, -3 and 0, 0, respectively.qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic equation, roots, and general form. The constants are determined correctly via initial conditions, and the final answer is verified.gpt-oss:20b: fail (error) 2026-09-27 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions. It also presents the verification step in a confusing way and lists duplicate values for y(0) and y'(0) instead of showing the actual calculations. This omission makes the reasoning incomplete and potentially misleading.qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic equation, finds the complex roots, applies the general solution formula for complex conjugate roots, and verifies the initial conditions.gpt-oss:20b: pass 2026-09-27
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.