Home›Calculus 3›Second-order linear equations›Problem 6.88

Second-order linear equations

Problem 6.88 · hard

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

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
Reviewed
\[ \left(m - 3\right) \left(m + 2\right) = m^{2} - m - 6 \]
Roots: 3, -2.✓ Proved
Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
Reviewed
Use y(0) and y'(0) to find C₁ and C₂.
Reviewed
\[ \frac{\left(6 e^{5 x} - 36\right) e^{- 2 x}}{5} - \frac{d}{d x} \frac{\left(6 - e^{5 x}\right) e^{- 2 x}}{5} + \frac{d^{2}}{d x^{2}} \frac{\left(6 - e^{5 x}\right) e^{- 2 x}}{5} = 0 \]
The solution satisfies the equation.✓ Proved
\[ 1 \]
y(0).Reviewed
\[ -3 \]
y'(0).Reviewed

Answer \( y = \frac{\left(6 - e^{5 x}\right) e^{- 2 x}}{5} \)

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 characteristic equation, finds the roots, sets up the general solution, 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, finds the roots, sets up the general solution, and verifies the final answer against the differential equation and initial conditions.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The characteristic equation in step 1 is stated as m² + bm + c = 0, but the roots found in step 2 correspond to m² - m - 6 = 0. The sign of the linear term is incorrect in the setup description, implying b=1 instead of b=-1.
gpt-oss:20b: fail (misleading) 2026-09-27 — The solution verifies the differential equation and the initial conditions, but it never actually shows how the constants C₁ and C₂ are determined from y(0)=1 and y'(0)=-3. A student would be misled into thinking the constants were found, when in fact the step is omitted.
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the characteristic equation, finds the 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.