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Second-order linear equations

Problem 6.83 · hard

Solve \( \displaystyle y'' + 1y' - 6y = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 3 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m - 2\right) \left(m + 3\right) = m^{2} + m - 6 \]
Roots: 2, -3.✓ Proved
Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \frac{\left(18 - 18 e^{5 x}\right) e^{- 3 x}}{5} + \frac{d}{d x} \frac{\left(3 e^{5 x} - 3\right) e^{- 3 x}}{5} + \frac{d^{2}}{d x^{2}} \frac{\left(3 e^{5 x} - 3\right) e^{- 3 x}}{5} = 0 \]
The solution satisfies the equation.✓ Proved
\[ 0 \]
y(0).
\[ 3 \]
y'(0).

Answer \( y = \frac{\left(3 e^{5 x} - 3\right) e^{- 3 x}}{5} \)

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: pass
qwen3.6:27b-mlx: fail (misleading) — The solution skips the crucial step of solving for the constants C1 and C2 using the initial conditions. It jumps from the general form to the final answer without showing the system of equations, which is the core of the problem.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (misleading) 2026-09-28 — The solution skips the crucial step of solving for the constants C1 and C2 using the initial conditions. It jumps from the general form to the final answer without showing the system of equations, which is the core of the problem.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic equation, roots, and general form. The final answer satisfies the differential equation and initial conditions as verified by the algebraic checks.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the characteristic roots, sets up the general solution, and verifies that the specific solution satisfies both the differential equation and the initial conditions.
gpt-oss:20b: pass 2026-09-26

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.