Second-order initial and boundary value problems
Problem 6.655 · medium
Solve \( \displaystyle 4 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y(2) = 0 \).
- \[ m^{2} + 4 m + 4 = \left(m + 2\right)^{2} \]The characteristic equation has a repeated root -2.✓ Proved
- So y = C₁exp(-2*x) + C₂x*exp(-2*x); the two conditions fix C₁ and C₂.
- \[ 4 \frac{d}{d x} 0 + \frac{d^{2}}{d x^{2}} 0 = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. 0 \right|_{\substack{ x=0 }} = 0 \]... y(0) = 0 ...✓ Proved
- \[ \left. 0 \right|_{\substack{ x=2 }} = 0 \]y(2) = 0.✓ Proved
Answer \( y = 0 \)
Lines: 4 proved, 1 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. 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: fail (misleading) — The solution claims the two boundary conditions determine the constants, but never actually solves for C₁ and C₂. Without showing that C₁=C₂=0, it is unclear why y=0 is the unique solution; a student could incorrectly assume any linear combination of the basis functions satisfies the conditions.qwen3.6:27b-mlx: fail (error) — The solution fails to actually solve for the constants C1 and C2. It asserts that the conditions fix them, but then checks the trivial solution y=0 without demonstrating that it is the unique solution (i.e., showing the system for C1 and C2 has only the trivial solution).
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution fails to actually solve for the constants C1 and C2. It asserts that the conditions fix them, but then checks the trivial solution y=0 without demonstrating that it is the unique solution (i.e., showing the system for C1 and C2 has only the trivial solution).gpt-oss:20b: fail (misleading) 2026-10-19 — The solution claims the two boundary conditions determine the constants, but never actually solves for C₁ and C₂. Without showing that C₁=C₂=0, it is unclear why y=0 is the unique solution; a student could incorrectly assume any linear combination of the basis functions satisfies the conditions.qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution fails to actually solve for the constants C1 and C2. It asserts that the conditions fix them but then verifies the trivial solution y=0 without showing that it is the unique solution. A student would not learn how to determine the constants from the boundary conditions.gpt-oss:20b: fail (misleading) 2026-10-16 — The solution claims that the two boundary conditions determine the constants, yet it never actually solves for C₁ and C₂. Instead, it substitutes y=0 into the differential equation and the boundary conditions, which is not justified. A student would incorrectly conclude that the zero function is the only solution without seeing the proper determination of the constants.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution fails to actually solve for the constants C1 and C2 using the boundary conditions. It merely states that the conditions fix them and then verifies that the zero function is a solution, without proving it is the unique solution.gpt-oss:20b: fail (misleading) 2026-10-15 — The solution claims that the constants are fixed by the two boundary conditions but never actually solves for them. Instead it substitutes the zero function into the differential equation and boundary conditions, which is not a valid verification of the general solution. This could mislead a student into thinking the steps were justified when they were not.
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 2026-10-19 with SymPy 1.14.0.