∫Calc Practice

Second-order linear equations

Problem 6.138 · hard

Solve \( \displaystyle y'' + 6y' + 9y = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = -2 \).
  1. Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
    Reviewed
  2. \[ \left(m + 3\right)^{2} = m^{2} + 6 m + 9 \]
    Roots: -3.✓ Proved
  3. A repeated root r gives y = (C₁ + C₂x)e^(rx).
    Reviewed
  4. Use y(0) and y'(0) to find C₁ and C₂.
    Reviewed
  5. \[ \left(- 45 x - 9\right) e^{- 3 x} + 6 \frac{d}{d x} \left(- 5 x - 1\right) e^{- 3 x} + \frac{d^{2}}{d x^{2}} \left(- 5 x - 1\right) e^{- 3 x} = 0 \]
    The solution satisfies the equation.✓ Proved
  6. \[ -1 \]
    y(0).Reviewed
  7. \[ -2 \]
    y'(0).Reviewed
Answer \( y = \left(- 5 x - 1\right) e^{- 3 x} \)

✓ Nihil obstat Lines: 2 proved, 5 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. 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
3Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
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
7Reviewedreviewers: 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; both initial conditions match

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly identifies the repeated root, applies the standard form for the general solution, and verifies that the stated answer satisfies both the differential equation and the initial conditions.
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the repeated root, applies the standard form for the general solution, and verifies that the stated answer satisfies both the differential equation and the initial conditions.
  • gpt-oss:20b: pass 2026-09-28
  • qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the repeated root, states the general form, and verifies that the specific solution satisfies both the differential equation and the initial conditions.
  • gpt-oss:20b: pass 2026-09-27
  • qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the repeated root of the characteristic equation, applies the standard form for repeated roots, and verifies that the final answer satisfies both the differential equation and the initial conditions.
  • gpt-oss:20b: pass 2026-09-27

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/second_order_linear, checked 2026-10-13 with SymPy 1.14.0.