Home›Calculus 3›Second-order linear equations›Problem 6.131

Second-order linear equations

Problem 6.131 · hard

Solve \( \displaystyle y'' + 6y' + 9y = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 0 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m + 3\right)^{2} = m^{2} + 6 m + 9 \]
Roots: -3.✓ Proved
A repeated root r gives y = (C₁ + C₂x)e^(rx).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \left(- 81 x - 27\right) e^{- 3 x} + 6 \frac{d}{d x} \left(- 9 x - 3\right) e^{- 3 x} + \frac{d^{2}}{d x^{2}} \left(- 9 x - 3\right) e^{- 3 x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -3 \]
y(0).
\[ 0 \]
y'(0).

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

Lines: 2 proved, 5 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	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	Not checked	—	a sentence; read, not computed
4	Not checked	—	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
7	Not checked	—	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: fail (error) — The verification step multiplies the correct solution by 27, and the initial condition checks list two values instead of one, indicating a mistake in the solution’s verification and reporting.
qwen3.6:27b-mlx: pass — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the final answer against the differential equation and initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the final answer against the differential equation and initial conditions.
gpt-oss:20b: fail (error) 2026-09-28 — The verification step multiplies the correct solution by 27, and the initial condition checks list two values instead of one, indicating a mistake in the solution’s verification and reporting.
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the repeated root, applies the standard form for repeated roots, and verifies the initial conditions and differential equation. The steps are logically sound and mathematically correct.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-27 — 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 2026-09-27

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.