Euler's method
Problem 6.372 · medium
Use Euler's method with step size \( \displaystyle h = \frac{1}{2} \) to approximate \( \displaystyle y(2) \) for \( \displaystyle y' = t^{2} + y \), \( \displaystyle y(1) = -1 \).
- Each step follows the tangent line: yₙ₊₁ = yₙ + h·f(tₙ, yₙ), tₙ₊₁ = tₙ + h.Reviewed
- \[ -1 + 1 \cdot \frac{1}{2} \cdot 0 = -1 \]y1 = y0 + h·f(t0, y0) with t0 = 1, slope 0.✓ Proved
- \[ -1 + 1 \cdot \frac{1}{2} \cdot 5 \cdot \frac{1}{4} = - \frac{3}{8} \]y2 = y1 + h·f(t1, y1) with t1 = 3/2, slope 5/4.✓ Proved
Answer \( y(2) \approx - \frac{3}{8} = -0.375 \)
Lines: 2 proved, 1 reviewed. 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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ 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 | the steps rerun in floating point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies Euler's method with the given step size and initial conditions. The intermediate calculations for the slopes and updates are accurate.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly applies Euler's method with the given step size and initial conditions. The intermediate calculations for the slopes and updates are accurate.gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: fail (error) 2026-10-11 — The solution incorrectly calculates the slope for the first step. f(t0, y0) = 1^2 + (-1) = 0, so y1 = -1. However, for the second step, t1 = 1.5 and y1 = -1, so f(t1, y1) = (1.5)^2 + (-1) = 2.25 - 1 = 1.25 = 5/4. The calculation in line 3 uses the correct slope 5/4, but the previous line implies y1 was calculated using slope 0, which is correct. Wait, let me re-read carefully. Step 1: t0=1, y0=-1. f(1, -1) = 1^2 + (-1) = 0. y1 = -1 + 0.5*0 = -1. t1 = 1.5. Step 2: t1=1.5, y1=-1. f(1.5, -1) = 1.5^2 + (-1) = 2.25 - 1 = 1.25 = 5/4. y2 = -1 + 0.5*(5/4) = -1 + 5/8 = -3/8. The solution states: "-1 + (1/2)*0 = -1" for y1. This is correct. The solution states: "-1 + (1/2)*(5/4) = -3/8" for y2. This is correct. Why did I think it was wrong? Let me check the prompt's stated answer. y(2) approx -3/8. My calculation yields -3/8. The steps are correct. The sentences are correct. The equations are correct. Wait, look at line 2: "-1 + (1/2)*0 = -1 -- y1 = y0 + h·f(t0, y0) with t0 = 1, slope 0." This is correct. Look at line 3: "-1 + (1/2)*(5/4) = -3/8 -- y2 = y1 + h·f(t1, y1) with t1 = 3/2, slope 5/4." This is correct. Is there a subtle error? t0 = 1, y0 = -1. t1 = 1 + 0.5 = 1.5. y1 = -1 + 0.5 * (1^2 + (-1)) = -1 + 0 = -1. t2 = 1.5 + 0.5 = 2. y2 = -1 + 0.5 * (1.5^2 + (-1)) = -1 + 0.5 * (2.25 - 1) = -1 + 0.5 * 1.25 = -1 + 0.625 = -0.375 = -3/8. The solution is correct. Why would I fail it? Maybe the format? "Each step follows the tangent line" is a bit loose but acceptable for Euler's method explanation. Let me double check the problem statement. y' = t^2 + y, y(1) = -1. h=1/2. Approximate y(2). Steps: 1 -> 1.5 -> 2. Two steps. Step 1: y(1.5) approx -1. Step 2: y(2) approx -3/8. The solution is sound. I will pass it.gpt-oss:20b: pass 2026-10-11
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/eulers_method, checked 2026-10-11 with SymPy 1.14.0.