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Second-order linear equations

Problem 6.39 · hard

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

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m + 1 - i\right) \left(m + 1 + i\right) = m^{2} + 2 m + 2 \]
Roots: -1 - I, -1 + I.✓ Proved
Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).
Use y(0) and y'(0) to find C₁ and C₂.
\[ 2 \frac{d}{d x} \left(- 3 e^{- x} \cos{\left(x \right)}\right) + \frac{d^{2}}{d x^{2}} \left(- 3 e^{- x} \cos{\left(x \right)}\right) - 6 e^{- x} \cos{\left(x \right)} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -3 \]
y(0).
\[ 3 \]
y'(0).

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

Lines: 2 proved, 5 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	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	Not checked	—	a sentence; read, not computed
4	Not checked	—	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
7	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	sympy.checkodesol substitutes the solution back; both initial conditions match

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly determine the constants C1 and C2 from the initial conditions, jumping directly to the final answer without showing the necessary algebraic steps. While the final answer is correct, the logical gap between setting up the general solution and stating the specific solution makes the derivation incomplete and potentially misleading for a student learning the method.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to explicitly determine the constants C1 and C2 from the initial conditions, jumping directly to the final answer without showing the necessary algebraic steps. While the final answer is correct, the logical gap between setting up the general solution and stating the specific solution makes the derivation incomplete and potentially misleading for a student learning the method.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to explicitly derive the constants C1 and C2 from the initial conditions, jumping directly to the final answer. While the final answer is correct, the logical step of solving the system for C1 and C2 is missing, making the derivation incomplete.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly determine the constants C1 and C2 from the initial conditions, jumping directly to the final answer without showing the necessary algebraic steps. Additionally, line 1 is misleading as it implies a generic form m^2 + bm + c = 0 without specifying b=2 and c=2 for this specific problem.
gpt-oss:20b: pass 2026-09-26

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_linear, checked 2026-10-13 with SymPy 1.14.0.