Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.475

Second-order initial and boundary value problems

Problem 6.475 · medium

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

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

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

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 skips the step of determining the constants C₁ and C₂ from the initial conditions, instead jumping directly to the final answer and verifying it. While the verification is correct, the solution fails to demonstrate the actual solving process requested by 'Solve'. A student would not learn how to find the constants.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution skips the step of determining the constants C₁ and C₂ from the initial conditions, instead jumping directly to the final answer and verifying it. While the verification is correct, the solution fails to demonstrate the actual solving process requested by 'Solve'. A student would not learn how to find the constants.
gpt-oss:20b: pass 2026-10-17
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. It merely states that the conditions fix them and then verifies the final answer, rather than deriving it. A student would not learn how to actually solve for the constants.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution skips the derivation of the constants C₁ and C₂. It states the general solution and then immediately verifies the final answer without showing how the initial conditions determine the specific coefficients, which is the core of the problem.
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.