Second-order initial and boundary value problems
Problem 6.484 · medium
Solve \( \displaystyle - 2 y + y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = 1 \).
- \[ m^{2} + m - 2 = \left(m - 1\right) \left(m + 2\right) \]The characteristic equation has distinct real roots 1 and -2.✓ Proved
- So y = C₁exp(x) + C₂exp(-2*x); the two conditions fix C₁ and C₂.
- \[ 2 e^{x} + \frac{d}{d x} \left(- e^{x} - e^{- 2 x}\right) + \frac{d^{2}}{d x^{2}} \left(- e^{x} - e^{- 2 x}\right) + 2 e^{- 2 x} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. - e^{x} - e^{- 2 x} \right|_{\substack{ x=0 }} = -2 \]... y(0) = -2 ...✓ Proved
- \[ \left. \frac{d}{d x} \left(- e^{x} - e^{- 2 x}\right) \right|_{\substack{ x=0 }} = 1 \]y′(0) = 1.✓ Proved
Answer \( y = - e^{x} - e^{- 2 x} \)
Lines: 4 proved, 1 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ 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 | finite differences of the answer satisfy the equation, and the conditions hold in floating point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (misleading) — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions, instead jumping to the final answer and verifying it. This fails to demonstrate the method for solving the initial value problem.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (misleading) 2026-10-17 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions, instead jumping to the final answer and verifying it. This fails to demonstrate the method for solving the initial value problem.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (misleading) 2026-10-15 — The solution verifies the final answer but fails to show the derivation of the constants C1 and C2 from the initial conditions. A student would not learn how to solve for the coefficients, only how to check a given answer.gpt-oss:20b: fail (misleading) 2026-10-15 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions. Without showing that C₁ = C₂ = –1, the subsequent verification of the differential equation and initial conditions is incomplete and could mislead a student into thinking the solution was derived correctly.qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the characteristic equation and its roots, constructs the general solution, and verifies that the specific answer satisfies both the differential equation and the initial conditions.gpt-oss:20b: pass 2026-10-14
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_ivp_bvp, checked 2026-10-17 with SymPy 1.14.0.