Euler's method
Problem 6.215 · medium
Use Euler's method with step size \( \displaystyle h = \frac{1}{2} \) to approximate \( \displaystyle y(\frac{5}{2}) \) for \( \displaystyle y' = 2 t + y \), \( \displaystyle y(1) = 2 \).
- Each step follows the tangent line: yₙ₊₁ = yₙ + h·f(tₙ, yₙ), tₙ₊₁ = tₙ + h.Reviewed
- \[ 2 + 1 \cdot \frac{1}{2} \cdot 4 = 4 \]y1 = y0 + h·f(t0, y0) with t0 = 1, slope 4.✓ Proved
- \[ 1 \cdot \frac{1}{2} \cdot 7 + 4 = \frac{15}{2} \]y2 = y1 + h·f(t1, y1) with t1 = 3/2, slope 7.✓ Proved
- \[ 1 \cdot \frac{1}{2} \cdot 23 \cdot \frac{1}{2} + \frac{15}{2} = \frac{53}{4} \]y3 = y2 + h·f(t2, y2) with t2 = 2, slope 23/2.✓ Proved
Answer \( y(\frac{5}{2}) \approx \frac{53}{4} = 13.25 \)
Lines: 3 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 |
| 4 | ✓ 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 specified step size and initial conditions. The arithmetic and logic are sound.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Euler's method with the specified step size and initial conditions. The arithmetic and logic are sound.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution performs 3 steps, reaching t = 2.5, but the intermediate calculations are incorrect. Specifically, step 2 calculates y1 = 4, but f(1, 2) = 2(1)+2 = 4, so y1 = 2 + 0.5*4 = 4 is correct. However, step 3 uses t1 = 1.5 and y1 = 4. f(1.5, 4) = 2(1.5) + 4 = 7. y2 = 4 + 0.5*7 = 7.5 = 15/2. This is correct. Step 4 uses t2 = 2 and y2 = 7.5. f(2, 7.5) = 2(2) + 7.5 = 11.5 = 23/2. y3 = 7.5 + 0.5*11.5 = 7.5 + 5.75 = 13.25 = 53/4. Wait, let me re-read the provided solution text carefully. Line 2: `2 + (1/2)*4 = 4`. Correct. t0=1, y0=2. f(1,2)=4. y1=4. t1=1.5. Line 3: `(1/2)*7 + 4 = 15/2`. Correct. t1=1.5, y1=4. f(1.5,4)=3+4=7. y2=4+3.5=7.5. t2=2. Line 4: `(1/2)*(23/2) + 15/2 = 53/4`. Correct. t2=2, y2=7.5. f(2,7.5)=4+7.5=11.5=23/2. y3=7.5 + 5.75 = 13.25. t3=2.5. The target is y(5/2) = y(2.5). The steps are h=0.5. t0=1, t1=1.5, t2=2, t3=2.5. So 3 steps are needed. The solution provides 3 steps. The arithmetic is correct. The final answer is correct. Why did I think it was wrong? Let me double check the problem statement. y' = 2t + y, y(1)=2. Step 1: t=1, y=2. f=4. y_new = 2 + 0.5*4 = 4. t_new = 1.5. Step 2: t=1.5, y=4. f=3+4=7. y_new = 4 + 0.5*7 = 7.5. t_new = 2. Step 3: t=2, y=7.5. f=4+7.5=11.5. y_new = 7.5 + 0.5*11.5 = 7.5 + 5.75 = 13.25. t_new = 2.5. The solution is actually correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.