Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.472

Second-order initial and boundary value problems

Problem 6.472 · medium

Solve \( \displaystyle 8 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = 0 \).

\[ m^{2} + 6 m + 8 = \left(m + 2\right) \left(m + 4\right) \]
The characteristic equation has distinct real roots -2 and -4.✓ Proved
So y = C₁exp(-2*x) + C₂exp(-4*x); the two conditions fix C₁ and C₂.
\[ 8 \left(6 e^{2 x} - 3\right) e^{- 4 x} + 6 \frac{d}{d x} \left(6 e^{2 x} - 3\right) e^{- 4 x} + \frac{d^{2}}{d x^{2}} \left(6 e^{2 x} - 3\right) e^{- 4 x} = 0 \]
It satisfies the equation ...✓ Proved
\[ \left. 3 \left(2 e^{2 x} - 1\right) e^{- 4 x} \right|_{\substack{ x=0 }} = 3 \]
... y(0) = 3 ...✓ Proved
\[ \left. \frac{d}{d x} 3 \left(2 e^{2 x} - 1\right) e^{- 4 x} \right|_{\substack{ x=0 }} = 0 \]
y′(0) = 0.✓ Proved

Answer \( y = 3 \left(2 e^{2 x} - 1\right) e^{- 4 x} \)

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 skips the actual derivation of the constants C1 and C2, jumping from the general form to the specific answer. A student would not learn how to apply the initial conditions to solve for the coefficients.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (misleading) 2026-10-17 — The solution skips the actual derivation of the constants C1 and C2, jumping from the general form to the specific answer. A student would not learn how to apply the initial conditions to solve for the coefficients.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution fails to actually solve for the constants C1 and C2 using the initial conditions. Instead, it merely verifies that the stated answer satisfies the differential equation and initial conditions, which does not constitute a derivation or solution process.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to actually solve for the constants C1 and C2 using the initial conditions. Instead, it merely verifies that the stated answer satisfies the differential equation and initial conditions, which does not constitute a derivation or solution of the problem as requested.
gpt-oss:20b: pass 2026-10-14

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 2026-10-17 with SymPy 1.14.0.