∫Calc Practice

Second-order linear equations

Problem 6.134 · hard

Solve \( \displaystyle y'' + 2y' + 10y = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 1 \).
  1. Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
    Reviewed
  2. \[ \left(m + 1 - 3 i\right) \left(m + 1 + 3 i\right) = m^{2} + 2 m + 10 \]
    Roots: -1 - 3*I, -1 + 3*I.✓ Proved
  3. Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).
    Reviewed
  4. Use y(0) and y'(0) to find C₁ and C₂.
    Reviewed
  5. \[ \frac{\left(- 20 \sin{\left(3 x \right)} - 90 \cos{\left(3 x \right)}\right) e^{- x}}{3} + 2 \frac{d}{d x} \frac{\left(- 2 \sin{\left(3 x \right)} - 9 \cos{\left(3 x \right)}\right) e^{- x}}{3} + \frac{d^{2}}{d x^{2}} \frac{\left(- 2 \sin{\left(3 x \right)} - 9 \cos{\left(3 x \right)}\right) e^{- x}}{3} = 0 \]
    The solution satisfies the equation.✓ Proved
  6. \[ -3 \]
    y(0).✓ Proved
  7. \[ 1 \]
    y'(0).✓ Proved
Answer \( y = - \frac{\left(2 \sin{\left(3 x \right)} + 9 \cos{\left(3 x \right)}\right) e^{- x}}{3} \)

✓ Nihil obstat Lines: 4 proved, 3 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

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
6✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
7✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
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 characteristic roots, applies the general form for complex conjugate roots, and verifies that the specific solution satisfies both the differential equation and the initial conditions.
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic roots, applies the general form for complex conjugate roots, 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 characteristic equation, finds the complex roots, applies the standard form for complex conjugate roots, and verifies the initial conditions and differential equation. The logic is sound and complete.
  • 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 2026-09-27 with SymPy 1.14.0.