Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.711

Second-order initial and boundary value problems

Problem 6.711 · medium

Solve \( \displaystyle 4 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 2 \).

\[ m^{2} + 4 m + 4 = \left(m + 2\right)^{2} \]
The characteristic equation has a repeated root -2.✓ Proved
So y = C₁exp(-2*x) + C₂x*exp(-2*x); the two conditions fix C₁ and C₂.
\[ 4 \left(- 4 x - 3\right) e^{- 2 x} + 4 \frac{d}{d x} \left(- 4 x - 3\right) e^{- 2 x} + \frac{d^{2}}{d x^{2}} \left(- 4 x - 3\right) e^{- 2 x} = 0 \]
It satisfies the equation ...✓ Proved
\[ \left. \left(- 4 x - 3\right) e^{- 2 x} \right|_{\substack{ x=0 }} = -3 \]
... y(0) = -3 ...✓ Proved
\[ \left. \frac{d}{d x} \left(- 4 x - 3\right) e^{- 2 x} \right|_{\substack{ x=0 }} = 2 \]
y′(0) = 2.✓ Proved

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

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 verifies the final answer but fails to actually solve for the constants C1 and C2 using the initial conditions. It asserts that the conditions fix the constants without showing the work, which is the core of the problem.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-20 — The solution verifies the final answer but fails to actually solve for the constants C1 and C2 using the initial conditions. It asserts that the conditions fix the constants without showing the work, which is the core of the problem.
gpt-oss:20b: pass 2026-10-20
qwen3.6:27b-mlx: fail (misleading) 2026-10-19 — The solution skips the derivation of the constants C₁ and C₂, presenting the final answer without showing how the initial conditions were used to determine them. This fails to demonstrate the method for solving initial value problems.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution skips the derivation of the constants C1 and C2, jumping directly to the final answer without showing how the initial conditions were used to determine them. This makes the solution incomplete and fails to demonstrate the method for solving the initial value problem.
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.