Home›Calculus 3›Second-order linear equations›Problem 6.186

Second-order linear equations

Problem 6.186 · medium

Solve \( \displaystyle y'' - 1y' - 6y = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = 2 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m - 3\right) \left(m + 2\right) = m^{2} - m - 6 \]
Roots: 3, -2.✓ Proved
Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
Use y(0) and y'(0) to find C₁ and C₂.
\[ - \frac{d}{d x} \left(- e^{- 2 x}\right) + \frac{d^{2}}{d x^{2}} \left(- e^{- 2 x}\right) + 6 e^{- 2 x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -1 \]
y(0).
\[ 2 \]
y'(0).

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

Lines: 2 proved, 5 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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 (misleading) — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions, leaving the reader without a clear justification for why y = –e^(–2x) is the unique solution. This omission could mislead a student into thinking the answer follows automatically without solving for the constants.
qwen3.6:27b-mlx: fail (error) — The solution fails to actually solve for the constants C1 and C2 using the initial conditions. It merely verifies that the stated answer satisfies the conditions, but does not derive it, leaving the student without the method to find the solution.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to actually solve for the constants C1 and C2 using the initial conditions. It merely verifies that the stated answer satisfies the conditions, but does not derive it, leaving the student without the method to find the solution.
gpt-oss:20b: fail (misleading) 2026-10-04 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions, leaving the reader without a clear justification for why y = –e^(–2x) is the unique solution. This omission could mislead a student into thinking the answer follows automatically without solving for the constants.
qwen3.6:27b-mlx: fail (error) 2026-09-29 — The solution claims the answer is -exp(-2*x) but never solves for the constants C1 and C2 using the initial conditions. It simply verifies that the proposed answer satisfies the ODE and initial conditions, skipping the actual derivation step implied by lines 3 and 4.
gpt-oss:20b: pass 2026-09-29
qwen3.6:27b-mlx: fail (error) 2026-09-29 — The solution claims the answer is -exp(-2*x) without solving for the constants C1 and C2. The actual solution satisfying y(0)=-1 and y'(0)=2 is y = (1/5)e^(3x) - (6/5)e^(-2x). The provided function fails to satisfy the initial condition y'(0)=2 (it gives y'(0)=4).
gpt-oss:20b: pass 2026-09-29

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.