Second-order linear equations
Problem 6.77 · hard
Solve \( \displaystyle y'' + 1y' - 2y = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 3 \).
- Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
- \[ \left(m - 1\right) \left(m + 2\right) = m^{2} + m - 2 \]Roots: 1, -2.✓ Proved
- Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
- Use y(0) and y'(0) to find C₁ and C₂.
- \[ - 2 e^{x} + \frac{d}{d x} \left(e^{x} - e^{- 2 x}\right) + \frac{d^{2}}{d x^{2}} \left(e^{x} - e^{- 2 x}\right) + 2 e^{- 2 x} = 0 \]The solution satisfies the equation.✓ Proved
- \[ 0 \]y(0).
- \[ 3 \]y'(0).
Answer \( y = e^{x} - e^{- 2 x} \)
Lines: 2 proved, 5 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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 claims to determine constants C₁ and C₂ from the initial conditions, but it never actually performs that calculation; it simply states the final function and verifies it. This omission could mislead a student into thinking the constants were derived when they were not shown.qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly solve for the constants C1 and C2. While the final answer is correct, the steps jump from setting up the system of equations to verifying the result, omitting the crucial algebraic step of determining that C1=1 and C2=-1.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to explicitly solve for the constants C1 and C2. While the final answer is correct, the steps jump from setting up the system of equations to verifying the result, omitting the crucial algebraic step of determining that C1=1 and C2=-1.gpt-oss:20b: fail (misleading) 2026-09-28 — The solution claims to determine constants C₁ and C₂ from the initial conditions, but it never actually performs that calculation; it simply states the final function and verifies it. This omission could mislead a student into thinking the constants were derived when they were not shown.qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the specific answer without showing the system of equations derived from the initial conditions, making the derivation incomplete and logically disconnected.gpt-oss:20b: fail (misleading) 2026-09-27 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions. It merely states that the proposed function satisfies the differential equation and the initial conditions without showing the algebra that leads to C₁=1 and C₂=−1, which could mislead a student into thinking the verification is sufficient.qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the specific answer without showing the system of equations derived from the initial conditions, which is the core of the problem.gpt-oss:20b: fail (misleading) 2026-09-26 — The solution never shows how the constants C₁ and C₂ are determined from the initial conditions; it jumps straight to a proposed solution without demonstrating that y(0)=0 and y'(0)=3 actually lead to C₁=1, C₂=−1. This omission could mislead a student into thinking the constants were found automatically.
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.