Second-order linear equations
Problem 6.82 · hard
Solve \( \displaystyle y'' + 2y' + 2y = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = -1 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m + 1 - i\right) \left(m + 1 + i\right) = m^{2} + 2 m + 2 \]Roots: -1 - I, -1 + I.✓ Proved
- Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ \left(- 6 \sin{\left(x \right)} - 4 \cos{\left(x \right)}\right) e^{- x} + 2 \frac{d}{d x} \left(- 3 \sin{\left(x \right)} - 2 \cos{\left(x \right)}\right) e^{- x} + \frac{d^{2}}{d x^{2}} \left(- 3 \sin{\left(x \right)} - 2 \cos{\left(x \right)}\right) e^{- x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ -2 \]y(0).
- \[ -1 \]y'(0).
Answer \( y = - \left(3 \sin{\left(x \right)} + 2 \cos{\left(x \right)}\right) e^{- x} \)
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: passqwen3.6:27b-mlx: fail (error) — The solution fails to explicitly derive the constants C1 and C2 from the initial conditions, jumping directly to the final answer. While the final answer is verified to satisfy the ODE and ICs, the logical step of solving for the constants is missing, making the solution incomplete as a demonstration of the method.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to explicitly derive the constants C1 and C2 from the initial conditions, jumping directly to the final answer. While the final answer is verified to satisfy the ODE and ICs, the logical step of solving for the constants is missing, making the solution incomplete as a demonstration of the method.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to actually solve for the constants C1 and C2 using the initial conditions. It jumps from the general form to the final answer without showing the calculation (C1 = -2, C2 = -3), making the derivation incomplete and logically disconnected.gpt-oss:20b: fail (error) 2026-09-27 — The proposed solution is scaled by an extra factor of –2, which makes it inconsistent with the initial conditions. The correct solution is –(3 sin x + 2 cos x) e^(–x).qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly derive the constants C1 and C2 from the initial conditions, jumping directly to the final answer. While the final answer is correct, the logical step of solving the system for C1 and C2 is missing, making the derivation incomplete and potentially misleading about how the constants were determined.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.