First-order linear equations
Problem 6.200 · medium
Solve \( \displaystyle y' + 1y = 2 e^{x} \) with \( \displaystyle y(0) = 3 \).
- The equation is linear in standard form; the integrating factor is e^(∫1 dx) = e^(1x).Reviewed
- \[ \frac{d}{d x} Y{\left(x \right)} e^{x} = Y{\left(x \right)} e^{x} + e^{x} \frac{d}{d x} Y{\left(x \right)} \]Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
- \[ \int 2 e^{2 x}\, dx = e^{2 x} \]Integrate the right side.✓ Proved
- Setting x = 0 and y = 3 fixes the constant of integration: C = 2.Reviewed
- \[ e^{x} + \frac{d}{d x} \left(e^{x} + 2 e^{- x}\right) + 2 e^{- x} = 2 e^{x} \]The solution satisfies the equation.✓ Proved
- \[ 3 \]And the initial condition.✓ Proved
Answer \( y = e^{x} + 2 e^{- x} \)
Lines: 4 proved, 2 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | ✓ 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 | sympy.checkodesol substitutes the solution back; y(0) matches |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the integrating factor, performs the integration, applies the initial condition, and verifies the result. The logic is sound and the steps are valid.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly identifies the integrating factor, performs the integration, applies the initial condition, and verifies the result. The logic is sound and the steps are valid.gpt-oss:20b: pass 2026-10-03qwen3.6:27b-mlx: fail (error) 2026-10-03 — The integration step is incorrect: the integral of 2e^(2x) is e^(2x) + C, but the solution implies the particular solution is e^x (from e^(2x)/e^x), yet line 3 claims the integral is just exp(2x) without the constant, and line 4 jumps to C=2 without showing the intermediate step of dividing by the integrating factor or solving for C explicitly from y(x) = e^x + Ce^{-x}. More critically, line 3 is mathematically false as an indefinite integral statement (missing +C) and line 4's logic is opaque and skips the crucial algebraic step of isolating y(x) before applying the initial condition.gpt-oss:20b: pass 2026-10-03
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/linear_first_order, checked 2026-10-03 with SymPy 1.14.0.