Second-order linear equations
Problem 6.84 · hard
Solve \( \displaystyle y'' - 2y' + 5y = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = 1 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m - 1 - 2 i\right) \left(m - 1 + 2 i\right) = m^{2} - 2 m + 5 \]Roots: 1 - 2*I, 1 + 2*I.✓ Proved
- Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ \frac{\left(15 \sin{\left(2 x \right)} - 20 \cos{\left(2 x \right)}\right) e^{x}}{2} - 2 \frac{d}{d x} \frac{\left(3 \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right) e^{x}}{2} + \frac{d^{2}}{d x^{2}} \frac{\left(3 \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right) e^{x}}{2} = 0 \]The solution satisfies the equation.✓ Proved
- \[ -2 \]y(0).
- \[ 1 \]y'(0).
Answer \( y = \frac{\left(3 \sin{\left(2 x \right)} - 4 \cos{\left(2 x \right)}\right) e^{x}}{2} \)
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: fail (error) — Step 6 incorrectly lists two values for y(0); it should only give y(0) = -2. The rest of the reasoning is sound.qwen3.6:27b-mlx: pass — 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. The steps are logically sound and algebraically correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28 — 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. The steps are logically sound and algebraically correct.gpt-oss:20b: fail (error) 2026-09-28 — Step 6 incorrectly lists two values for y(0); it should only give y(0) = -2. The rest of the reasoning is sound.qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic equation, finds the complex roots, applies the standard form for complex conjugate roots, and verifies the initial conditions. The steps are logically sound and mathematically correct.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. The steps are logically sound and algebraically correct.gpt-oss:20b: fail (misleading) 2026-09-26 — The solution incorrectly reports the initial values: it lists "-2, -2" for y(0) and "1, 1" for y'(0), whereas the problem specifies y(0) = -2 and y'(0) = 1. This misstatement could mislead a student about the correct initial conditions.
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.