Second-order linear equations
Problem 6.162 · hard
Solve \( \displaystyle y'' + 2y' + 1y = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = -2 \).
- 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.
- \[ \left(- 4 x - 2\right) e^{- x} + 2 \frac{d}{d x} \left(- 4 x - 2\right) e^{- x} + \frac{d^{2}}{d x^{2}} \left(- 4 x - 2\right) e^{- x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ -2 \]y(0).✓ Proved
- \[ -2 \]y'(0).✓ Proved
Answer \( y = 2 \left(- 2 x - 1\right) e^{- x} \)
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: passqwen3.6:27b-mlx: fail (error) — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the final answer without showing the system of equations derived from the initial conditions, making the derivation incomplete.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the final answer without showing the system of equations derived from the initial conditions, making the derivation incomplete.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the repeated root, applies the standard form for repeated roots, and verifies the final answer against the differential equation and initial conditions.gpt-oss:20b: pass 2026-09-28
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 2026-09-28 with SymPy 1.14.0.