Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.717

Second-order initial and boundary value problems

Problem 6.717 · medium

Solve \( \displaystyle 6 y - 5 y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = -1 \).

\[ m^{2} - 5 m + 6 = \left(m - 3\right) \left(m - 2\right) \]
The characteristic equation has distinct real roots 3 and 2.✓ Proved
So y = C₁exp(3*x) + C₂exp(2*x); the two conditions fix C₁ and C₂.
\[ 6 \left(3 e^{x} - 5\right) e^{2 x} - 5 \frac{d}{d x} \left(3 e^{x} - 5\right) e^{2 x} + \frac{d^{2}}{d x^{2}} \left(3 e^{x} - 5\right) e^{2 x} = 0 \]
It satisfies the equation ...✓ Proved
\[ \left. \left(3 e^{x} - 5\right) e^{2 x} \right|_{\substack{ x=0 }} = -2 \]
... y(0) = -2 ...✓ Proved
\[ \left. \frac{d}{d x} \left(3 e^{x} - 5\right) e^{2 x} \right|_{\substack{ x=0 }} = -1 \]
y′(0) = -1.✓ Proved

Answer \( y = \left(3 e^{x} - 5\right) e^{2 x} \)

Lines: 4 proved, 1 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	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Not checked	—	a sentence; read, not computed
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
4	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	finite differences of the answer satisfy the equation, and the conditions hold in floating point

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to actually solve for the constants C₁ and C₂. It states that the conditions fix them, but then immediately asserts the final answer without showing the system of equations or the algebraic steps to derive C₁=3 and C₂=-5.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-20 — The solution fails to actually solve for the constants C₁ and C₂. It states that the conditions fix them, but then immediately asserts the final answer without showing the system of equations or the algebraic steps to derive C₁=3 and C₂=-5.
gpt-oss:20b: pass 2026-10-20
qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly identifies the general solution from the characteristic equation and verifies that the stated answer satisfies both the differential equation and the initial conditions.
gpt-oss:20b: fail (misleading) 2026-10-19 — The sentence claiming that the two conditions fix C₁ and C₂ does not actually determine them; the solution skips the computation of C₁=3 and C₂=‑5, which could mislead a student into thinking the constants are automatically determined without work.
qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution skips the derivation of the constants C₁ and C₂, instead just verifying that the final answer satisfies the conditions. While the final answer is correct, the solution fails to show the work required to solve the initial value problem (setting up and solving the system of linear equations for C₁ and C₂).
gpt-oss:20b: pass 2026-10-19

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_ivp_bvp, checked 2026-10-20 with SymPy 1.14.0.