∫Calc Practice

Second-order linear equations

Problem 6.182 · hard

Solve \( \displaystyle y'' - 4y' + 8y = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = -2 \).
  1. Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
    Reviewed
  2. \[ \left(m - 2 - 2 i\right) \left(m - 2 + 2 i\right) = m^{2} - 4 m + 8 \]
    Roots: 2 - 2*I, 2 + 2*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. \[ - 8 e^{2 x} \cos{\left(2 x \right)} - 4 \frac{d}{d x} \left(- e^{2 x} \cos{\left(2 x \right)}\right) + \frac{d^{2}}{d x^{2}} \left(- e^{2 x} \cos{\left(2 x \right)}\right) = 0 \]
    The solution satisfies the equation.✓ Proved
  6. \[ -1 \]
    y(0).Reviewed
  7. \[ -2 \]
    y'(0).Reviewed
Answer \( y = - e^{2 x} \cos{\left(2 x \right)} \)

Lines: 2 proved, 5 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
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 characteristic roots, applies the standard form for complex conjugate roots, and verifies that the resulting function satisfies both the differential equation and the initial conditions.
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the characteristic roots, applies the standard form for complex conjugate roots, and verifies that the resulting function satisfies both the differential equation and the initial conditions.
  • gpt-oss:20b: pass 2026-10-04
  • qwen3.6:27b-mlx: fail (error) 2026-09-29 — The solution fails to explicitly derive the constants C1 and C2 from the initial conditions, jumping directly to the final answer without showing the system of equations. While the final answer is correct, the logical gap between step 4 and step 5 makes it unclear how the specific coefficients were determined, which is the core of the initial value problem.
  • gpt-oss:20b: pass 2026-09-29
  • qwen3.6:27b-mlx: pass 2026-09-29 — The solution correctly identifies the characteristic equation, roots, and general form. The verification steps confirm that the stated answer satisfies both the differential equation and the initial conditions.
  • gpt-oss:20b: pass 2026-09-29

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.