Second-order linear equations
Problem 6.161 · hard
Solve \( \displaystyle y'' + 2y' + 2y = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = -2 \).
- 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.
- \[ \left(- 6 \sin{\left(x \right)} - 2 \cos{\left(x \right)}\right) e^{- x} + 2 \frac{d}{d x} \left(- 3 \sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{- x} + \frac{d^{2}}{d x^{2}} \left(- 3 \sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{- x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ -1 \]y(0).✓ Proved
- \[ -2 \]y'(0).✓ Proved
Answer \( y = - \left(3 \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x} \)
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 claims the general form is y = e^(αx)(C₁cos βx + C₂sin βx) but fails to explicitly determine the constants C₁ and C₂. It jumps directly to verifying a specific function without showing the derivation of the coefficients from the initial conditions, leaving the logical gap of how the specific solution was obtained.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution claims the general form is y = e^(αx)(C₁cos βx + C₂sin βx) but fails to explicitly determine the constants C₁ and C₂. It jumps directly to verifying a specific function without showing the derivation of the coefficients from the initial conditions, leaving the logical gap of how the specific solution was obtained.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to show the derivation of the constants C1 and C2 from the initial conditions, jumping directly to the final answer without demonstrating how the specific values were obtained.gpt-oss:20b: fail (misleading) 2026-09-28 — The solution skips the key step of determining the constants C₁ and C₂ from the initial conditions and presents an incorrect intermediate expression (the extra factor of –2). A student could be misled into thinking the verification step is complete without actually showing the substitution.
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.