Second-order initial and boundary value problems
Problem 6.473 · medium
Solve \( \displaystyle 5 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = -4 \).
- \[ m^{2} + 4 m + 5 = \left(m + 2 - i\right) \left(m + 2 + i\right) \]The characteristic equation has complex roots -2 ± 1i.✓ Proved
- So y = C₁exp(-2*x)*cos(x) + C₂exp(-2*x)*sin(x); the two conditions fix C₁ and C₂.
- \[ 4 \frac{d}{d x} 2 e^{- 2 x} \cos{\left(x \right)} + \frac{d^{2}}{d x^{2}} 2 e^{- 2 x} \cos{\left(x \right)} + 10 e^{- 2 x} \cos{\left(x \right)} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. 2 e^{- 2 x} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 2 \]... y(0) = 2 ...✓ Proved
- \[ \left. \frac{d}{d x} 2 e^{- 2 x} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = -4 \]y′(0) = -4.✓ Proved
Answer \( y = 2 e^{- 2 x} \cos{\left(x \right)} \)
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: fail (misleading) — The solution skips the step of determining the constants C₁ and C₂ from the initial conditions; it merely states that the two conditions fix them and then presents a particular solution without showing the calculation, which could mislead a student into thinking the constants are arbitrary or that the reasoning is complete.qwen3.6:27b-mlx: fail (error) — The solution fails to determine the constants C₁ and C₂. It asserts the general form and then verifies that a specific function (with C₁=2, C₂=0) satisfies the conditions, but it does not show the derivation or justification for why C₂ must be 0. A student would not learn how to solve for the constants from this.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution fails to determine the constants C₁ and C₂. It asserts the general form and then verifies that a specific function (with C₁=2, C₂=0) satisfies the conditions, but it does not show the derivation or justification for why C₂ must be 0. A student would not learn how to solve for the constants from this.gpt-oss:20b: fail (misleading) 2026-10-17 — The solution skips the step of determining the constants C₁ and C₂ from the initial conditions; it merely states that the two conditions fix them and then presents a particular solution without showing the calculation, which could mislead a student into thinking the constants are arbitrary or that the reasoning is complete.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution assumes the specific form of the answer (with C1=2, C2=0) without deriving the constants from the general solution. It verifies that this specific function is a solution, but fails to demonstrate that it is the unique solution satisfying the initial conditions, omitting the necessary step of solving for C1 and C2.gpt-oss:20b: fail (error) 2026-10-15 — The solution never determines the constants C₁ and C₂ from the initial conditions; it simply asserts that the conditions fix them and then jumps to the particular solution 2e^{-2x}cos x without justification. This omission means the reasoning is incomplete and potentially misleading.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution fails to actually solve for the constants C1 and C2. It asserts the general form and then verifies that a specific function (with C1=2, C2=0) satisfies the conditions, but it does not demonstrate the derivation of these constants from the initial value 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.