Second-order linear equations
Problem 6.81 · hard
Solve \( \displaystyle y'' + 4y' + 4y = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = 2 \).
- 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.
- \[ \left(24 x + 8\right) e^{- 2 x} + 4 \frac{d}{d x} \left(6 x + 2\right) e^{- 2 x} + \frac{d^{2}}{d x^{2}} \left(6 x + 2\right) e^{- 2 x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ 2 \]y(0).
- \[ 2 \]y'(0).
Answer \( y = 2 \left(3 x + 1\right) e^{- 2 x} \)
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 (error) — The proposed solution y=2(3x+1)e^{-2x} does not satisfy the initial conditions y(0)=2 and y'(0)=2; the correct solution is y=(2+2x)e^{-2x}.qwen3.6:27b-mlx: pass — The solution correctly identifies the repeated root, states the correct general form for the homogeneous solution, and verifies that the specific answer satisfies both the differential equation and the initial conditions.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the repeated root, states the correct general form for the homogeneous solution, and verifies that the specific answer satisfies both the differential equation and the initial conditions.gpt-oss:20b: fail (error) 2026-09-28 — The proposed solution y=2(3x+1)e^{-2x} does not satisfy the initial conditions y(0)=2 and y'(0)=2; the correct solution is y=(2+2x)e^{-2x}.qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to actually solve for the constants C1 and C2. It jumps from the general form to the final answer without showing the system of equations derived from the initial conditions, making the derivation incomplete and logically disconnected.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the repeated root, states the general form, and verifies the specific solution against the differential equation and initial conditions.gpt-oss:20b: fail (error) 2026-09-26 — The proposed solution y=2(3x+1)e^{-2x} does not satisfy the differential equation; the algebra in step 5 is incorrect and the constants C₁,C₂ are not correctly determined from the 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_linear, checked 2026-10-13 with SymPy 1.14.0.