Second-order linear equations
Problem 6.76 · hard
Solve \( \displaystyle y'' - 2y' + 5y = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = -3 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.Reviewed
- \[ \left(m - 1 - 2 i\right) \left(m - 1 + 2 i\right) = m^{2} - 2 m + 5 \]Roots: 1 - 2*I, 1 + 2*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
- \[ \frac{\left(- 25 \sin{\left(2 x \right)} + 20 \cos{\left(2 x \right)}\right) e^{x}}{2} - 2 \frac{d}{d x} \frac{\left(- 5 \sin{\left(2 x \right)} + 4 \cos{\left(2 x \right)}\right) e^{x}}{2} + \frac{d^{2}}{d x^{2}} \frac{\left(- 5 \sin{\left(2 x \right)} + 4 \cos{\left(2 x \right)}\right) e^{x}}{2} = 0 \]The solution satisfies the equation.✓ Proved
- \[ 2 \]y(0).Reviewed
- \[ -3 \]y'(0).Reviewed
Answer \( y = \frac{\left(- 5 \sin{\left(2 x \right)} + 4 \cos{\left(2 x \right)}\right) e^{x}}{2} \)
Lines: 2 proved, 5 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. How we verify
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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 7 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly identifies the characteristic equation, roots, and general form. The final answer satisfies both the differential equation and the initial conditions, as verified by the provided checks.
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 final answer satisfies both the differential equation and the initial conditions, as verified by the provided checks.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic equation, finds the complex roots, applies the general form for complex conjugate roots, and verifies the final solution against the differential equation and initial conditions.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the characteristic equation, finds the complex roots, applies the standard form for complex conjugate roots, and verifies the initial conditions and differential equation.gpt-oss:20b: fail (error) 2026-09-26 — The solution never actually determines the constants C₁ and C₂ from the initial conditions; step 4 is merely a statement, not a calculation. Moreover, the verification in step 5 uses an incorrect second derivative notation and does not show that the proposed function satisfies the given initial values. Consequently the solution is incomplete and potentially misleading.
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.