Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.618

Second-order initial and boundary value problems

Problem 6.618 · medium

Solve \( \displaystyle y + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y(\frac{\pi}{4}) = -1 \).

\[ m^{2} + 1 = \left(m - i\right) \left(m + i\right) \]
The characteristic equation has complex roots 0 ± 1i.✓ Proved
So y = C₁cos(x) + C₂sin(x); the two conditions fix C₁ and C₂.
Reviewed
\[ \left(-4 - \sqrt{2}\right) \sin{\left(x \right)} + 4 \cos{\left(x \right)} + \frac{d^{2}}{d x^{2}} \left(\left(-4 - \sqrt{2}\right) \sin{\left(x \right)} + 4 \cos{\left(x \right)}\right) = 0 \]
It satisfies the equation ...✓ Proved
\[ \left. \left(-4 - \sqrt{2}\right) \sin{\left(x \right)} + 4 \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 4 \]
... y(0) = 4 ...✓ Proved
\[ \left. \left(-4 - \sqrt{2}\right) \sin{\left(x \right)} + 4 \cos{\left(x \right)} \right|_{\substack{ x=\frac{\pi}{4} }} = -1 \]
y(pi/4) = -1.✓ Proved

Answer \( y = - \left(\sqrt{2} + 4\right) \sin{\left(x \right)} + 4 \cos{\left(x \right)} \)

Lines: 4 proved, 1 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass — The solution correctly identifies the general form, verifies the differential equation and boundary conditions, and presents the correct final answer.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the general form, verifies the differential equation and boundary conditions, and presents the correct final answer.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the general form of the solution for the given differential equation and verifies that the stated answer satisfies both the differential equation and the boundary conditions.
gpt-oss:20b: fail (misleading) 2026-10-16 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions; it merely states that they are fixed and then presents a particular function without showing the derivation, which could mislead a student into thinking the solution is complete.
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution fails to demonstrate how the constants C1 and C2 were determined. It merely states that the conditions fix them and then verifies the final answer, omitting the actual derivation steps required to solve the problem.
gpt-oss:20b: pass 2026-10-15

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 2026-10-18 with SymPy 1.14.0.