Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.482

Second-order initial and boundary value problems

Problem 6.482 · medium

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

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

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

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: pass
qwen3.6:27b-mlx: fail (misleading) — The solution verifies the final answer but fails to show the derivation of the constants C₁ and C₂ from the initial conditions. A student would not learn how to solve for the constants, only how to check a given answer.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (misleading) 2026-10-17 — The solution verifies the final answer but fails to show the derivation of the constants C₁ and C₂ from the initial conditions. A student would not learn how to solve for the constants, only how to check a given answer.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (misleading) 2026-10-15 — The solution verifies the final answer but fails to demonstrate the derivation of the constants C₁ and C₂ from the initial conditions, which is the core of the problem. A student would not learn how to solve for the constants.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to actually solve for the constants C₁ and C₂. It states that the conditions fix them, but then jumps to verifying the final answer without showing the derivation (C₁=1, C₂=-3). This skips the core logical step of the problem.
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.