Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.622

Second-order initial and boundary value problems

Problem 6.622 · medium

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

\[ m^{2} - 6 m + 9 = \left(m - 3\right)^{2} \]
The characteristic equation has a repeated root 3.✓ Proved
So y = C₁exp(3*x) + C₂x*exp(3*x); the two conditions fix C₁ and C₂.
\[ 9 \left(12 x - 3\right) e^{3 x} - 6 \frac{d}{d x} \left(12 x - 3\right) e^{3 x} + \frac{d^{2}}{d x^{2}} \left(12 x - 3\right) e^{3 x} = 0 \]
It satisfies the equation ...✓ Proved
\[ \left. \left(12 x - 3\right) e^{3 x} \right|_{\substack{ x=0 }} = -3 \]
... y(0) = -3 ...✓ Proved
\[ \left. \frac{d}{d x} \left(12 x - 3\right) e^{3 x} \right|_{\substack{ x=0 }} = 3 \]
y′(0) = 3.✓ Proved

Answer \( y = \left(12 x - 3\right) e^{3 x} \)

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 (misleading) — The solution verifies that the stated answer is correct but fails to demonstrate the derivation of the constants C₁ and C₂ from the initial conditions. A student would not learn how to solve for the constants, only how to check a given answer.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (misleading) 2026-10-18 — The solution verifies that the stated answer is correct but fails to demonstrate the derivation of the constants C₁ and C₂ from the initial conditions. A student would not learn how to solve for the constants, only how to check a given answer.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (misleading) 2026-10-16 — The solution skips the derivation of the constants C1 and C2, jumping directly to the final answer. While the final answer is verified, the solution fails to demonstrate how the initial conditions determine the specific coefficients, which is the core of solving an IVP.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution skips the step of determining the constants C₁ and C₂ from the initial conditions, instead jumping directly to the final answer. While the final answer is correct, the logical flow is broken because the derivation of the specific constants is missing.
gpt-oss:20b: pass 2026-10-15

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 2026-10-18 with SymPy 1.14.0.