Second-order linear equations
Problem 6.210 · hard
Solve \( \displaystyle y'' + 1y' - 2y = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 0 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m - 1\right) \left(m + 2\right) = m^{2} + m - 2 \]Roots: 1, -2.✓ Proved
- Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ 4 e^{x} + \frac{d}{d x} \left(- 2 e^{x} - e^{- 2 x}\right) + \frac{d^{2}}{d x^{2}} \left(- 2 e^{x} - e^{- 2 x}\right) + 2 e^{- 2 x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ -3 \]y(0).✓ Proved
- \[ 0 \]y'(0).✓ Proved
Answer \( y = - 2 e^{x} - e^{- 2 x} \)
Lines: 4 proved, 3 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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: fail (misleading) — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions; it simply states the final form without showing how it was obtained, which could mislead a student into thinking the answer was derived correctly.qwen3.6:27b-mlx: fail (error) — The solution fails to actually solve for the constants C1 and C2 using the initial conditions; it merely verifies that the stated answer satisfies the conditions without showing the derivation.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-03 — The solution fails to actually solve for the constants C1 and C2 using the initial conditions; it merely verifies that the stated answer satisfies the conditions without showing the derivation.gpt-oss:20b: fail (misleading) 2026-10-03 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions; it simply states the final form without showing how it was obtained, which could mislead a student into thinking the answer was derived correctly.qwen3.6:27b-mlx: fail (error) 2026-10-03 — The solution fails to show the derivation of the constants C1 and C2 from the initial conditions, jumping directly to the final answer without solving the system of equations. Additionally, the verification step (line 5) is algebraically nonsensical and does not properly verify the solution.gpt-oss:20b: fail (error) 2026-10-03 — The solution never actually determines the constants C₁ and C₂; step 5 contains a nonsensical equation instead of the correct substitution and solving for the constants. The subsequent checks merely verify the claimed answer without derivation.
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 2026-10-03 with SymPy 1.14.0.