Second-order initial and boundary value problems
Problem 6.471 · medium
Solve \( \displaystyle 8 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y(1) = 1 \).
- \[ m^{2} + 6 m + 8 = \left(m + 2\right) \left(m + 4\right) \]The characteristic equation has distinct real roots -2 and -4.✓ Proved
- So y = C₁exp(-2*x) + C₂exp(-4*x); the two conditions fix C₁ and C₂.
- \[ \frac{8 \left(1 - e^{2 x}\right) e^{4 - 4 x}}{1 - e^{2}} + 6 \frac{d}{d x} \frac{\left(1 - e^{2 x}\right) e^{4 - 4 x}}{1 - e^{2}} + \frac{d^{2}}{d x^{2}} \frac{\left(1 - e^{2 x}\right) e^{4 - 4 x}}{1 - e^{2}} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. \frac{\left(1 - e^{2 x}\right) e^{4 - 4 x}}{1 - e^{2}} \right|_{\substack{ x=0 }} = 0 \]... y(0) = 0 ...✓ Proved
- \[ \left. \frac{\left(1 - e^{2 x}\right) e^{4 - 4 x}}{1 - e^{2}} \right|_{\substack{ x=1 }} = 1 \]y(1) = 1.✓ Proved
Answer \( y = \frac{\left(1 - e^{2 x}\right) e^{4 - 4 x}}{1 - e^{2}} \)
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: fail (error) — The solution skips the crucial step of determining the constants C₁ and C₂ from the boundary conditions. The given formula is asserted to satisfy the ODE and the conditions without showing how it was derived from the general solution, leaving the reasoning incomplete and potentially misleading.qwen3.6:27b-mlx: pass — The solution correctly identifies the general form of the solution for the constant coefficient linear ODE and verifies that the stated specific solution satisfies both the differential equation and the boundary conditions.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly identifies the general form of the solution for the constant coefficient linear ODE and verifies that the stated specific solution satisfies both the differential equation and the boundary conditions.gpt-oss:20b: fail (error) 2026-10-17 — The solution skips the crucial step of determining the constants C₁ and C₂ from the boundary conditions. The given formula is asserted to satisfy the ODE and the conditions without showing how it was derived from the general solution, leaving the reasoning incomplete and potentially misleading.qwen3.6:27b-mlx: pass 2026-10-15gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the characteristic roots, sets up the general solution, and verifies that the specific answer satisfies both the differential equation and the boundary 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.