Second-order initial and boundary value problems
Problem 6.616 · medium
Solve \( \displaystyle y - 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y(1) = 2 \).
- \[ m^{2} - 2 m + 1 = \left(m - 1\right)^{2} \]The characteristic equation has a repeated root 1.✓ Proved
- So y = C₁exp(x) + C₂x*exp(x); the two conditions fix C₁ and C₂.Reviewed
- \[ \left(- 4 e x + 2 x + 4 e\right) e^{x - 1} - 2 \frac{d}{d x} \left(- 4 e x + 2 x + 4 e\right) e^{x - 1} + \frac{d^{2}}{d x^{2}} \left(- 4 e x + 2 x + 4 e\right) e^{x - 1} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. 2 \left(- 2 e x + x + 2 e\right) e^{x - 1} \right|_{\substack{ x=0 }} = 4 \]... y(0) = 4 ...✓ Proved
- \[ \left. 2 \left(- 2 e x + x + 2 e\right) e^{x - 1} \right|_{\substack{ x=1 }} = 2 \]y(1) = 2.✓ Proved
Answer \( y = 2 \left(- 2 e x + x + 2 e\right) e^{x - 1} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly identifies the repeated root of the characteristic equation and the resulting general form. The specific constants are determined correctly to satisfy the boundary conditions, and the final expression is verified algebraically.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the repeated root of the characteristic equation and the resulting general form. The specific constants are determined correctly to satisfy the boundary conditions, and the final expression is verified algebraically.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the general form for a repeated root and verifies that the stated answer satisfies the differential equation and boundary conditions.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the repeated root of the characteristic equation, sets up the general solution form, and verifies that the stated answer satisfies both the differential equation and the boundary conditions.gpt-oss:20b: pass 2026-10-15
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.