Euler's method
Problem 6.378 · medium
Use Euler's method with step size \( \displaystyle h = \frac{1}{4} \) to approximate \( \displaystyle y(\frac{3}{2}) \) for \( \displaystyle y' = t - y \), \( \displaystyle y(1) = 1 \).
- Each step follows the tangent line: yₙ₊₁ = yₙ + h·f(tₙ, yₙ), tₙ₊₁ = tₙ + h.
- \[ 1 \cdot \frac{1}{4} \cdot 0 + 1 = 1 \]y1 = y0 + h·f(t0, y0) with t0 = 1, slope 0.✓ Proved
- \[ 1 \cdot \frac{1}{4} \cdot 1 \cdot \frac{1}{4} + 1 = \frac{17}{16} \]y2 = y1 + h·f(t1, y1) with t1 = 5/4, slope 1/4.✓ Proved
Answer \( y(\frac{3}{2}) \approx \frac{17}{16} = 1.0625 \)
Lines: 2 proved, 1 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.
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 | ✓ 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: fail (error) — The solution stops after two steps, reaching t = 1.5, but labels the result y2. While t2 = 1.5 is correct, the notation y2 implies the second approximation, which is correct for the value, but the solution fails to explicitly state that t2 = 3/2, leaving the connection to the target point y(3/2) implicit and potentially confusing regarding the step count (n=0 to n=2 is 2 steps, but usually indexed 1, 2). More critically, the problem asks for y(3/2). t0=1, t1=1.25, t2=1.5. So y2 is indeed the approximation for y(1.5). The calculation is correct. Wait, let me re-read carefully. t0=1, h=1/4. t1 = 1 + 1/4 = 5/4. t2 = 5/4 + 1/4 = 6/4 = 3/2. So y2 corresponds to t=3/2. The calculation is correct. Why would it fail? Let's check the slope calculation. f(t,y) = t-y. Step 1: t0=1, y0=1. f(1,1) = 0. y1 = 1 + 1/4(0) = 1. t1 = 1.25. Step 2: t1=1.25, y1=1. f(1.25, 1) = 1.25 - 1 = 0.25 = 1/4. y2 = 1 + 1/4(1/4) = 1 + 1/16 = 17/16. The math is correct. The setup is correct. The answer is correct. Is there a subtle error? "Each step follows the tangent line" is a bit loose but acceptable. The indices are consistent. The final value is correct. I will mark it as pass.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-11 — The solution stops after two steps, reaching t = 1.5, but labels the result y2. While t2 = 1.5 is correct, the notation y2 implies the second approximation, which is correct for the value, but the solution fails to explicitly state that t2 = 3/2, leaving the connection to the target point y(3/2) implicit and potentially confusing regarding the step count (n=0 to n=2 is 2 steps, but usually indexed 1, 2). More critically, the problem asks for y(3/2). t0=1, t1=1.25, t2=1.5. So y2 is indeed the approximation for y(1.5). The calculation is correct. Wait, let me re-read carefully. t0=1, h=1/4. t1 = 1 + 1/4 = 5/4. t2 = 5/4 + 1/4 = 6/4 = 3/2. So y2 corresponds to t=3/2. The calculation is correct. Why would it fail? Let's check the slope calculation. f(t,y) = t-y. Step 1: t0=1, y0=1. f(1,1) = 0. y1 = 1 + 1/4(0) = 1. t1 = 1.25. Step 2: t1=1.25, y1=1. f(1.25, 1) = 1.25 - 1 = 0.25 = 1/4. y2 = 1 + 1/4(1/4) = 1 + 1/16 = 17/16. The math is correct. The setup is correct. The answer is correct. Is there a subtle error? "Each step follows the tangent line" is a bit loose but acceptable. The indices are consistent. The final value is correct. I will mark it as pass.gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: fail (error) 2026-10-11 — The solution stops after two steps, reaching t = 1.5, but labels the result y2. While the final value is correct for t=1.5, the indexing is confusing (usually y0 is t=1, y1 is t=1.25, y2 is t=1.5). More critically, the problem asks for y(3/2). The solution computes y2 at t=1.5. However, the step size is 1/4. From t=1 to t=1.5 is 0.5, which is 2 steps. So y2 is indeed the value at t=1.5. The calculation is: y0=1, t0=1. f(1,1)=0. y1 = 1 + 0.25*0 = 1. t1=1.25. f(1.25, 1) = 1.25 - 1 = 0.25. y2 = 1 + 0.25*0.25 = 1.0625. t2=1.5. The result is correct. Why did I think it was wrong? Let me re-read. Ah, the prompt says 'Stated answer: y(3/2) approx 17/16'. The solution provides this. Is there an error? The solution has 3 lines. Line 1 is setup. Line 2 is step 1. Line 3 is step 2. It seems correct. Wait, let me check the algebra in line 2. (1/4)*0 + 1 = 1. Correct. Line 3: (1/4)*(1/4) + 1 = 17/16. Correct. The indices are y1 and y2. t0=1, t1=5/4, t2=6/4=3/2. The question asks for y(3/2). The solution provides y2 which corresponds to t=3/2. It seems correct. Is there a subtle error? 'Each step follows the tangent line...' is a bit loose but acceptable. The equations are marked 'unchecked' in the prompt description but the instructions say 'equation... checked... do not re-check'. The sentences are what I check. Sentence 1 is fine. The equations are just calculations. The solution is sound. Wait, looking at the provided solution text again. Line 2 says 'y1 = ... with t0=1'. Line 3 says 'y2 = ... with t1=5/4'. This is consistent. The final answer is y2. The question asks for y(3/2). t2 = 1 + 2*(1/4) = 1.5 = 3/2. So y2 is the answer. It seems correct. Why would it fail? Maybe the prompt implies I should find an error? Let me look really closely. y' = t - y. y(1)=1. h=1/4. Step 1: t0=1, y0=1. f(1,1) = 1-1=0. y1 = 1 + 0.25*0 = 1. t1 = 1.25. Step 2: t1=1.25, y1=1. f(1.25, 1) = 1.25 - 1 = 0.25. y2 = 1 + 0.25*0.25 = 1.0625. t2 = 1.5. The solution is correct. I will mark it as pass.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.