Home›Calculus 3›Second-order linear equations›Problem 6.162

Second-order linear equations

Problem 6.162 · hard

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

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
Reviewed
\[ \left(m + 1\right)^{2} = m^{2} + 2 m + 1 \]
Roots: -1.✓ Proved
A repeated root r gives y = (C₁ + C₂x)e^(rx).
Reviewed
Use y(0) and y'(0) to find C₁ and C₂.
Reviewed
\[ \left(- 4 x - 2\right) e^{- x} + 2 \frac{d}{d x} \left(- 4 x - 2\right) e^{- x} + \frac{d^{2}}{d x^{2}} \left(- 4 x - 2\right) e^{- x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -2 \]
y(0).Reviewed
\[ -2 \]
y'(0).Reviewed

Answer \( y = 2 \left(- 2 x - 1\right) e^{- x} \)

Lines: 2 proved, 5 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
4	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
7	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	sympy.checkodesol substitutes the solution back; both initial conditions match

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the repeated root, states the general form for repeated roots, and verifies that the specific solution satisfies both the differential equation and the initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly identifies the repeated root, states the general form for repeated roots, and verifies that the specific solution satisfies both the differential equation and the initial conditions.
gpt-oss:20b: pass 2026-10-03
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the final answer without showing the system of equations derived from the initial conditions, making the derivation incomplete.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the repeated root, applies the standard form for repeated roots, and verifies the final answer against the differential equation and initial conditions.
gpt-oss:20b: pass 2026-09-28

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_linear, checked 2026-10-13 with SymPy 1.14.0.