Second-order initial and boundary value problems
Problem 6.617 · medium
Solve \( \displaystyle 4 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y(1) = -2 \).
- \[ m^{2} + 4 m + 4 = \left(m + 2\right)^{2} \]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.
- \[ 4 \left(- 2 x \left(1 + e^{2}\right) + 2\right) e^{- 2 x} + 4 \frac{d}{d x} \left(- 2 x \left(1 + e^{2}\right) + 2\right) e^{- 2 x} + \frac{d^{2}}{d x^{2}} \left(- 2 x \left(1 + e^{2}\right) + 2\right) e^{- 2 x} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. 2 \left(- x \left(1 + e^{2}\right) + 1\right) e^{- 2 x} \right|_{\substack{ x=0 }} = 2 \]... y(0) = 2 ...✓ Proved
- \[ \left. 2 \left(- x \left(1 + e^{2}\right) + 1\right) e^{- 2 x} \right|_{\substack{ x=1 }} = -2 \]y(1) = -2.✓ Proved
Answer \( y = 2 \left(- x \left(1 + e^{2}\right) + 1\right) e^{- 2 x} \)
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 (error) — The solution skips the derivation of the constants C1 and C2, jumping directly to the final answer. While the final answer is correct, the solution fails to show the work required to determine the constants from the boundary conditions, which is the core of the problem.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The solution skips the derivation of the constants C1 and C2, jumping directly to the final answer. While the final answer is correct, the solution fails to show the work required to determine the constants from the boundary conditions, which is the core of the problem.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution skips the derivation of the constants C1 and C2, jumping directly to the final answer. While the final answer is verified to be correct, the solution fails to show the work required to solve the boundary value problem, making it an incomplete solution rather than a valid derivation.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution skips the derivation of the constants C1 and C2, jumping directly to the final answer. While the final answer is correct, the solution fails to demonstrate how the boundary conditions were applied to determine the constants, which is the core of the problem.gpt-oss:20b: fail (misleading) 2026-10-15 — The solution skips the crucial step of determining the constants C₁ and C₂ from the boundary conditions before presenting the final expression. Without showing that C₁=2 and C₂=−2(1+e²), a student might incorrectly assume the given form is already correct.
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 2026-10-18 with SymPy 1.14.0.