Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.629

Second-order initial and boundary value problems

Problem 6.629 · medium

Solve \( \displaystyle y + 2 y' + y'' = 0 \) with \( \displaystyle y(0) = -1,\ y(1) = -3 \).

\[ m^{2} + 2 m + 1 = \left(m + 1\right)^{2} \]
The characteristic equation has a repeated root -1.✓ Proved
So y = C₁exp(-x) + C₂x*exp(-x); the two conditions fix C₁ and C₂.
Reviewed
\[ \left(x \left(1 - 3 e\right) - 1\right) e^{- x} + 2 \frac{d}{d x} \left(x \left(1 - 3 e\right) - 1\right) e^{- x} + \frac{d^{2}}{d x^{2}} \left(x \left(1 - 3 e\right) - 1\right) e^{- x} = 0 \]
It satisfies the equation ...✓ Proved
\[ \left. \left(x \left(1 - 3 e\right) - 1\right) e^{- x} \right|_{\substack{ x=0 }} = -1 \]
... y(0) = -1 ...✓ Proved
\[ \left. \left(x \left(1 - 3 e\right) - 1\right) e^{- x} \right|_{\substack{ x=1 }} = -3 \]
y(1) = -3.✓ Proved

Answer \( y = \left(x \left(1 - 3 e\right) - 1\right) e^{- x} \)

Lines: 4 proved, 1 reviewed. 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	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: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the general solution for the repeated root and verifies that the stated answer satisfies both the differential equation and the boundary conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the general solution for the repeated root and verifies that the stated answer satisfies both the differential equation and the boundary conditions.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the repeated root, constructs the general solution, and verifies that the specific answer satisfies both the differential equation and the boundary conditions.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the repeated root of the characteristic equation, sets up the general solution, and verifies that the specific answer satisfies both the differential equation and the boundary conditions.
gpt-oss:20b: fail (misleading) 2026-10-15 — The solution claims the constants are fixed by the boundary conditions but never actually solves for them; the subsequent verification uses a specific form without justification, misleading a student about how to determine C₁ and C₂.

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.