∫Calc Practice

Second-order initial and boundary value problems

Problem 6.718 · medium

Solve \( \displaystyle 6 y + 5 y' + y'' = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = -4 \).
  1. \[ m^{2} + 5 m + 6 = \left(m + 2\right) \left(m + 3\right) \]
    The characteristic equation has distinct real roots -3 and -2.✓ Proved
  2. So y = C₁exp(-3*x) + C₂exp(-2*x); the two conditions fix C₁ and C₂.
  3. \[ 6 \left(6 - 7 e^{x}\right) e^{- 3 x} + 5 \frac{d}{d x} \left(6 - 7 e^{x}\right) e^{- 3 x} + \frac{d^{2}}{d x^{2}} \left(6 - 7 e^{x}\right) e^{- 3 x} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \left(6 - 7 e^{x}\right) e^{- 3 x} \right|_{\substack{ x=0 }} = -1 \]
    ... y(0) = -1 ...✓ Proved
  5. \[ \left. \frac{d}{d x} \left(6 - 7 e^{x}\right) e^{- 3 x} \right|_{\substack{ x=0 }} = -4 \]
    y′(0) = -4.✓ Proved
Answer \( y = \left(6 - 7 e^{x}\right) e^{- 3 x} \)

Lines: 4 proved, 1 not checked. 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
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0finite differences of the answer satisfy the equation, and the conditions hold in floating point

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution fails to actually solve for the constants C₁ and C₂. It states that the conditions fix them, but then jumps to verifying a specific answer without showing the derivation. This skips the core of the problem-solving process.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (error) 2026-10-20 — The solution fails to actually solve for the constants C₁ and C₂. It states that the conditions fix them, but then jumps to verifying a specific answer without showing the derivation. This skips the core of the problem-solving process.
  • gpt-oss:20b: pass 2026-10-20
  • qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly identifies the characteristic roots and general form, and the subsequent steps verify that the stated answer satisfies both the differential equation and the initial conditions.
  • gpt-oss:20b: pass 2026-10-19
  • qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution fails to actually solve for the constants C₁ and C₂. It asserts the general form and then verifies the final answer without showing the derivation (solving the system of equations from the initial conditions), which is the core task of the problem.
  • gpt-oss:20b: pass 2026-10-19

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_ivp_bvp, checked 2026-10-20 with SymPy 1.14.0.