Second-order initial and boundary value problems
Problem 6.802 · medium
Solve \( \displaystyle 9 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 1 \).
- \[ m^{2} + 6 m + 9 = \left(m + 3\right)^{2} \]The characteristic equation has a repeated root -3.✓ Proved
- So y = C₁exp(-3*x) + C₂x*exp(-3*x); the two conditions fix C₁ and C₂.
- \[ 9 x e^{- 3 x} + 6 \frac{d}{d x} x e^{- 3 x} + \frac{d^{2}}{d x^{2}} x e^{- 3 x} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. x e^{- 3 x} \right|_{\substack{ x=0 }} = 0 \]... y(0) = 0 ...✓ Proved
- \[ \left. \frac{d}{d x} x e^{- 3 x} \right|_{\substack{ x=0 }} = 1 \]y′(0) = 1.✓ Proved
Answer \( y = x e^{- 3 x} \)
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 fails to actually solve for the constants C₁ and C₂. It assumes the form of the solution is just the second basis function (implicitly setting C₁=0 and C₂=1) and verifies that this specific function satisfies the conditions, but it does not demonstrate that this is the unique solution derived from the general form.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-11-07 — The solution fails to actually solve for the constants C₁ and C₂. It assumes the form of the solution is just the second basis function (implicitly setting C₁=0 and C₂=1) and verifies that this specific function satisfies the conditions, but it does not demonstrate that this is the unique solution derived from the general form.gpt-oss:20b: pass 2026-11-07qwen3.6:27b-mlx: fail (misleading) 2026-11-06 — The solution skips the derivation of the constants C1 and C2, jumping directly to the final answer. While the final answer is correct, the text claims 'the two conditions fix C1 and C2' without showing that C1=0 and C2=1, which is the core of the initial value problem solution.gpt-oss:20b: pass 2026-11-06qwen3.6:27b-mlx: fail (error) 2026-11-06 — The solution fails to explicitly solve for the constants C₁ and C₂. It asserts that the conditions fix them and then verifies the final answer, but skips the actual derivation step (showing C₁=0 and C₂=1), which is the core of the problem.gpt-oss:20b: fail (error) 2026-11-06 — The solution omits determining the constants C₁ and C₂. It checks the differential equation and initial conditions only for the particular function x e^{−3x}, but does not show that the general solution y=C₁e^{−3x}+C₂x e^{−3x} reduces to y=x e^{−3x}. Consequently the reasoning is incomplete and could mislead a student into thinking the verification is sufficient without solving for the constants.
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 2026-11-07 with SymPy 1.14.0.