Second-order initial and boundary value problems
Problem 6.623 · medium
Solve \( \displaystyle 3 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = -2 \).
- \[ m^{2} + 4 m + 3 = \left(m + 1\right) \left(m + 3\right) \]The characteristic equation has distinct real roots -3 and -1.✓ Proved
- So y = C₁exp(-3*x) + C₂exp(-x); the two conditions fix C₁ and C₂.
- \[ \frac{3 \left(5 - 11 e^{2 x}\right) e^{- 3 x}}{2} + 4 \frac{d}{d x} \frac{\left(5 - 11 e^{2 x}\right) e^{- 3 x}}{2} + \frac{d^{2}}{d x^{2}} \frac{\left(5 - 11 e^{2 x}\right) e^{- 3 x}}{2} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. \frac{\left(5 - 11 e^{2 x}\right) e^{- 3 x}}{2} \right|_{\substack{ x=0 }} = -3 \]... y(0) = -3 ...✓ Proved
- \[ \left. \frac{d}{d x} \frac{\left(5 - 11 e^{2 x}\right) e^{- 3 x}}{2} \right|_{\substack{ x=0 }} = -2 \]y′(0) = -2.✓ Proved
Answer \( y = \frac{\left(5 - 11 e^{2 x}\right) e^{- 3 x}}{2} \)
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: passqwen3.6:27b-mlx: fail (misleading) — The solution verifies the final answer but fails to show the derivation of the constants C₁ and C₂ from the initial conditions. A student would not learn how to solve for the constants, which is the core of the initial value problem.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (misleading) 2026-10-18 — The solution verifies the final answer but fails to show the derivation of the constants C₁ and C₂ from the initial conditions. A student would not learn how to solve for the constants, which is the core of the initial value problem.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the characteristic equation and general form, and the provided specific solution is verified to satisfy both the differential equation and the initial conditions.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution skips the derivation of the constants C₁ and C₂, merely asserting that the conditions fix them and then verifying the final answer. This fails to demonstrate the method for solving the initial value problem.gpt-oss:20b: fail (error) 2026-10-15 — The constants C₁ and C₂ are not correctly determined; the resulting solution does not satisfy the differential equation and initial conditions.
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.