Second-order linear equations
Problem 6.79 · medium
Solve \( \displaystyle y'' + 4y' + 4y = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = -3 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m + 2\right)^{2} = m^{2} + 4 m + 4 \]Roots: -2.✓ Proved
- A repeated root r gives y = (C₁ + C₂x)e^(rx).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ \left(4 - 4 x\right) e^{- 2 x} + 4 \frac{d}{d x} \left(1 - x\right) e^{- 2 x} + \frac{d^{2}}{d x^{2}} \left(1 - x\right) e^{- 2 x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ 1 \]y(0).
- \[ -3 \]y'(0).
Answer \( y = \left(1 - x\right) e^{- 2 x} \)
Lines: 2 proved, 5 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 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | Not checked | — | a sentence; read, not computed |
| 4 | Not checked | — | a sentence; read, not computed |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| 7 | Not checked | — | both sides are the same expression: the line states a value, and the working that produced it is not on the line |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | sympy.checkodesol substitutes the solution back; both initial conditions match |
Reviewers
gpt-oss:20b: fail (misleading) — The solution never actually solves for the constants C₁ and C₂; it simply plugs in the final answer and checks it, which does not demonstrate how the constants were determined from the initial conditions.qwen3.6:27b-mlx: pass — The solution correctly identifies the repeated root, states the general form for repeated roots, and verifies the specific solution against the differential equation and initial conditions.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the repeated root, states the general form for repeated roots, and verifies the specific solution against the differential equation and initial conditions.gpt-oss:20b: fail (misleading) 2026-09-28 — The solution never actually solves for the constants C₁ and C₂; it simply plugs in the final answer and checks it, which does not demonstrate how the constants were determined from the initial conditions.qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the final answer against the differential equation and initial conditions.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the repeated root, applies the standard form for repeated roots, and verifies the initial conditions and differential equation.gpt-oss:20b: fail (error) 2026-09-26 — The proposed solution multiplies the differential equation by 4 in step 5, leading to an incorrect verification. The correct solution is (1‑x)e^(‑2x); the verification should use the original equation without the extraneous factor.
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_linear, checked 2026-10-13 with SymPy 1.14.0.