Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.474

Second-order initial and boundary value problems

Problem 6.474 · medium

Solve \( \displaystyle 8 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 2 \).

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

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

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 verifies the final answer but fails to show the derivation of the constants C1 and C2 from the general solution. This skips the actual problem-solving process, teaching the student that verification is equivalent to solving.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (misleading) 2026-10-17 — The solution verifies the final answer but fails to show the derivation of the constants C1 and C2 from the general solution. This skips the actual problem-solving process, teaching the student that verification is equivalent to solving.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the general form of the solution and verifies that the stated answer satisfies both the differential equation and the initial conditions.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the characteristic roots and general form, and the provided specific solution satisfies both the differential equation and the initial conditions.
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.