Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.480

Second-order initial and boundary value problems

Problem 6.480 · medium

Solve \( \displaystyle 20 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = 2 \).

\[ m^{2} + 4 m + 20 = \left(m + 2 - 4 i\right) \left(m + 2 + 4 i\right) \]
The characteristic equation has complex roots -2 ± 4i.✓ Proved
So y = C₁exp(-2*x)*cos(4*x) + C₂exp(-2*x)*sin(4*x); the two conditions fix C₁ and C₂.
Reviewed
\[ 20 \left(\sin{\left(4 x \right)} + \cos{\left(4 x \right)}\right) e^{- 2 x} + 4 \frac{d}{d x} \left(\sin{\left(4 x \right)} + \cos{\left(4 x \right)}\right) e^{- 2 x} + \frac{d^{2}}{d x^{2}} \left(\sin{\left(4 x \right)} + \cos{\left(4 x \right)}\right) e^{- 2 x} = 0 \]
It satisfies the equation ...✓ Proved
\[ \left. \left(\sin{\left(4 x \right)} + \cos{\left(4 x \right)}\right) e^{- 2 x} \right|_{\substack{ x=0 }} = 1 \]
... y(0) = 1 ...✓ Proved
\[ \left. \frac{d}{d x} \left(\sin{\left(4 x \right)} + \cos{\left(4 x \right)}\right) e^{- 2 x} \right|_{\substack{ x=0 }} = 2 \]
y′(0) = 2.✓ Proved

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

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

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the characteristic roots, states the general form, and verifies that the specific answer satisfies both the differential equation and the initial conditions.
gpt-oss:20b: fail (misleading) 2026-10-15 — The solution claims that y=(sin4x+cos4x)e^{-2x} satisfies the differential equation and the initial conditions, but it never determines the constants C₁ and C₂ from the initial data. The verification steps (3–5) simply check that this particular combination works, which is not justified by the earlier general solution. A student would incorrectly conclude that the constants are automatically 1 without solving for them.
qwen3.6:27b-mlx: fail (misleading) 2026-10-14 — The solution skips the derivation of the constants C₁ and C₂, jumping directly to the final answer. This fails to demonstrate how the initial conditions determine the specific solution, which is the core of the problem.
gpt-oss:20b: fail (error) 2026-10-14 — The constants C₁ and C₂ are not correctly determined. Solving the initial conditions gives C₁=1/3 and C₂=2/3, so the correct solution is (1/3 cos 4x + 2/3 sin 4x)e^{-2x}, not (sin 4x + cos 4x)e^{-2x}.

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.