Home›Calculus 3›Second-order linear equations›Problem 6.160

Second-order linear equations

Problem 6.160 · hard

Solve \( \displaystyle y'' + 6y' + 9y = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = -1 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
Reviewed
\[ \left(m + 3\right)^{2} = m^{2} + 6 m + 9 \]
Roots: -3.✓ Proved
A repeated root r gives y = (C₁ + C₂x)e^(rx).
Reviewed
Use y(0) and y'(0) to find C₁ and C₂.
Reviewed
\[ \left(18 x + 9\right) e^{- 3 x} + 6 \frac{d}{d x} \left(2 x + 1\right) e^{- 3 x} + \frac{d^{2}}{d x^{2}} \left(2 x + 1\right) e^{- 3 x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ 1 \]
y(0).Reviewed
\[ -1 \]
y'(0).Reviewed

Answer \( y = \left(2 x + 1\right) e^{- 3 x} \)

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 the specific solution against the differential equation and 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 the specific solution against the differential equation and initial conditions.
gpt-oss:20b: pass 2026-10-03
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the repeated root, applies the standard form for the general solution, and verifies the initial conditions and differential equation.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the repeated root of the characteristic equation, applies the standard form for repeated roots, and verifies that the final answer satisfies both the differential equation and the initial conditions.
gpt-oss:20b: fail (misleading) 2026-09-28 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions and merely states that the proposed function satisfies the differential equation and the initial values without showing the calculations. This could mislead a student into thinking the verification is automatic rather than derived.

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.