Second-order linear equations
Problem 6.137 · hard
Solve \( \displaystyle y'' + 2y' + 10y = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = 1 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.Reviewed
- \[ \left(m + 1 - 3 i\right) \left(m + 1 + 3 i\right) = m^{2} + 2 m + 10 \]Roots: -1 - 3*I, -1 + 3*I.✓ Proved
- Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).Reviewed
- Use y(0) and y'(0) to find C₁ and C₂.Reviewed
- \[ \left(\frac{20 \sin{\left(3 x \right)}}{3} + 10 \cos{\left(3 x \right)}\right) e^{- x} + 2 \frac{d}{d x} \left(\frac{2 \sin{\left(3 x \right)}}{3} + \cos{\left(3 x \right)}\right) e^{- x} + \frac{d^{2}}{d x^{2}} \left(\frac{2 \sin{\left(3 x \right)}}{3} + \cos{\left(3 x \right)}\right) e^{- x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ 1 \]y(0).✓ Proved
- \[ 1 \]y'(0).✓ Proved
Answer \( y = \left(\frac{2 \sin{\left(3 x \right)}}{3} + \cos{\left(3 x \right)}\right) e^{- x} \)
Lines: 4 proved, 3 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.
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 | ✓ 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: passqwen3.6:27b-mlx: pass — The solution correctly identifies the characteristic roots, applies the standard form for complex conjugate roots, and verifies that the final expression satisfies both the differential equation and the initial conditions.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic roots, applies the standard form for complex conjugate roots, and verifies that the final expression satisfies both the differential equation and the initial conditions.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic equation, roots, and general form. The constants are determined correctly to satisfy the initial conditions, and the final answer matches the stated answer.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 merely states that the proposed function satisfies the differential equation and lists the initial values without showing how those values were obtained or verifying that the constants indeed yield y(0)=1 and y'(0)=1. This omission means the reasoning is incomplete and could mislead a student into thinking the constants were derived automatically.
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 2026-09-27 with SymPy 1.14.0.