∫Calc Practice

Second-order initial and boundary value problems

Problem 6.481 · medium

Solve \( \displaystyle 13 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = 2 \).
  1. \[ m^{2} - 4 m + 13 = \left(m - 2 - 3 i\right) \left(m - 2 + 3 i\right) \]
    The characteristic equation has complex roots 2 ± 3i.✓ Proved
  2. So y = C₁exp(2*x)*cos(3*x) + C₂exp(2*x)*sin(3*x); the two conditions fix C₁ and C₂.
  3. \[ \frac{13 \left(- 4 \sin{\left(3 x \right)} + 9 \cos{\left(3 x \right)}\right) e^{2 x}}{3} - 4 \frac{d}{d x} \frac{\left(- 4 \sin{\left(3 x \right)} + 9 \cos{\left(3 x \right)}\right) e^{2 x}}{3} + \frac{d^{2}}{d x^{2}} \frac{\left(- 4 \sin{\left(3 x \right)} + 9 \cos{\left(3 x \right)}\right) e^{2 x}}{3} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{\left(- 4 \sin{\left(3 x \right)} + 9 \cos{\left(3 x \right)}\right) e^{2 x}}{3} \right|_{\substack{ x=0 }} = 3 \]
    ... y(0) = 3 ...✓ Proved
  5. \[ \left. \frac{d}{d x} \frac{\left(- 4 \sin{\left(3 x \right)} + 9 \cos{\left(3 x \right)}\right) e^{2 x}}{3} \right|_{\substack{ x=0 }} = 2 \]
    y′(0) = 2.✓ Proved
Answer \( y = \frac{\left(- 4 \sin{\left(3 x \right)} + 9 \cos{\left(3 x \right)}\right) e^{2 x}}{3} \)

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 skips the step of solving for the constants C₁ and C₂ using the initial conditions. It jumps from the general form to the specific answer and then verifies it, rather than deriving it. While the verification proves the answer is correct, the solution fails to demonstrate the method for finding the constants, which is the core of the problem.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution skips the step of solving for the constants C₁ and C₂ using the initial conditions. It jumps from the general form to the specific answer and then verifies it, rather than deriving it. While the verification proves the answer is correct, the solution fails to demonstrate the method for finding the constants, which is the core of the problem.
  • gpt-oss:20b: pass 2026-10-17
  • qwen3.6:27b-mlx: fail (error) 2026-10-15 — The solution fails to actually solve for the constants C1 and C2. It states the general form and then immediately jumps to verifying the final answer without showing the derivation of the constants from the initial conditions.
  • gpt-oss:20b: pass 2026-10-15
  • qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the characteristic roots, sets up the general solution, and verifies that the specific answer satisfies both the differential equation and the initial conditions.
  • gpt-oss:20b: pass 2026-10-14

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 2026-10-17 with SymPy 1.14.0.