Second-order initial and boundary value problems
Problem 6.714 · medium
Solve \( \displaystyle 3 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = -4 \).
- \[ m^{2} + 4 m + 3 = \left(m + 1\right) \left(m + 3\right) \]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.
- \[ \frac{3 \left(5 e^{2 x} + 1\right) e^{- 3 x}}{2} + 4 \frac{d}{d x} \frac{\left(5 e^{2 x} + 1\right) e^{- 3 x}}{2} + \frac{d^{2}}{d x^{2}} \frac{\left(5 e^{2 x} + 1\right) e^{- 3 x}}{2} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. \frac{\left(5 e^{2 x} + 1\right) e^{- 3 x}}{2} \right|_{\substack{ x=0 }} = 3 \]... y(0) = 3 ...✓ Proved
- \[ \left. \frac{d}{d x} \frac{\left(5 e^{2 x} + 1\right) e^{- 3 x}}{2} \right|_{\substack{ x=0 }} = -4 \]y′(0) = -4.✓ Proved
Answer \( y = \frac{\left(5 e^{2 x} + 1\right) e^{- 3 x}}{2} \)
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: fail (misleading) — The solution skips the step of determining the constants C₁ and C₂ from the initial conditions. It simply presents the final expression without showing how it was obtained, which could mislead a student into thinking the answer was guessed rather than derived.qwen3.6:27b-mlx: fail (error) — The solution fails to actually solve for the constants C1 and C2 using the initial conditions. Instead, it merely verifies that the stated answer satisfies the differential equation and initial conditions, skipping the derivation step required by the prompt 'Solve'. A student would not learn how to determine the constants from the general solution.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-20 — The solution fails to actually solve for the constants C1 and C2 using the initial conditions. Instead, it merely verifies that the stated answer satisfies the differential equation and initial conditions, skipping the derivation step required by the prompt 'Solve'. A student would not learn how to determine the constants from the general solution.gpt-oss:20b: fail (misleading) 2026-10-20 — The solution skips the step of determining the constants C₁ and C₂ from the initial conditions. It simply presents the final expression without showing how it was obtained, which could mislead a student into thinking the answer was guessed rather than derived.qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution fails to actually solve for the constants C₁ and C₂ using the initial conditions. Instead, it merely verifies that the stated answer satisfies the differential equation and initial conditions, skipping the required derivation steps.gpt-oss:20b: fail (error) 2026-10-19 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions. It simply presents the final formula and then verifies it, without showing how the constants were derived from y(0)=3 and y′(0)=−4. This omission means the reasoning is incomplete and could mislead a student into thinking the answer was obtained directly from the differential equation alone.qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution skips the derivation of the constants C1 and C2, jumping directly to the final answer. While the final answer is correct, the solution fails to show the system of equations used to determine the constants, making it an incomplete proof rather than a solution.gpt-oss:20b: pass 2026-10-19
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 2026-10-20 with SymPy 1.14.0.