∫Calc Practice

First-order linear equations

Problem 6.67 · hard Video

Solve \( \displaystyle y' + 2y = 3 e^{x} \) with \( \displaystyle y(0) = -2 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫2 dx) = e^(2x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{2 x} = 2 Y{\left(x \right)} e^{2 x} + e^{2 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 3 e^{3 x}\, dx = e^{3 x} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = -2 fixes the constant of integration: C = -3.
    Reviewed
  5. \[ 2 e^{x} + \frac{d}{d x} \left(e^{x} - 3 e^{- 2 x}\right) - 6 e^{- 2 x} = 3 e^{x} \]
    The solution satisfies the equation.✓ Proved
  6. \[ -2 \]
    And the initial condition.Reviewed
Answer \( y = e^{x} - 3 e^{- 2 x} \)
The slope field of y' = -2y + 3e^(x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, -2) is highlighted, y = e^(x) - 3e^(-2x).

Lines: 3 proved, 3 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. How we verify

The full receipt
LineStatusChecked byDetail
1Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
6Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxboth 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 independentlysympy 1.14.0 + mpmath 1.3.0sympy.checkodesol substitutes the solution back; y(0) matches

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly identifies the integrating factor, performs the integration, determines the constant using the initial condition, and verifies the final result.
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the integrating factor, performs the integration, determines the constant using the initial condition, and verifies the final result.
  • gpt-oss:20b: pass 2026-09-28
  • qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution skips the crucial step of dividing by the integrating factor to isolate y(x). It jumps from the integrated form to the final answer without showing the algebraic manipulation, and implicitly assumes the constant C is determined correctly without showing the intermediate equation y*e^(2x) = e^(3x) + C.
  • gpt-oss:20b: pass 2026-09-27
  • qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the integrating factor, performs the integration, determines the constant of integration, and verifies the final result against both the differential equation and the initial condition.
  • gpt-oss:20b: pass 2026-09-26

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 2026-10-13 with SymPy 1.14.0.