Second-order initial and boundary value problems
Problem 6.801 · medium
Solve \( \displaystyle 5 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = 4 \).
- \[ m^{2} - 4 m + 5 = \left(m - 2 - i\right) \left(m - 2 + i\right) \]The characteristic equation has complex roots 2 ± 1i.✓ Proved
- So y = C₁exp(2*x)*cos(x) + C₂exp(2*x)*sin(x); the two conditions fix C₁ and C₂.Reviewed
- \[ 5 \left(2 \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{2 x} - 4 \frac{d}{d x} \left(2 \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{2 x} + \frac{d^{2}}{d x^{2}} \left(2 \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{2 x} = 0 \]It satisfies the equation ...✓ Proved
- \[ \left. \left(2 \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{2 x} \right|_{\substack{ x=0 }} = 1 \]... y(0) = 1 ...✓ Proved
- \[ \left. \frac{d}{d x} \left(2 \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{2 x} \right|_{\substack{ x=0 }} = 4 \]y′(0) = 4.✓ Proved
Answer \( y = \left(2 \sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{2 x} \)
✓ Nihil obstat Lines: 4 proved, 1 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
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | finite differences of the answer satisfy the equation, and the conditions hold in floating point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-11-07gpt-oss:20b: pass 2026-11-07qwen3.6:27b-mlx: pass 2026-11-06gpt-oss:20b: pass 2026-11-06qwen3.6:27b-mlx: pass 2026-11-06gpt-oss:20b: pass 2026-11-06
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 2026-11-07 with SymPy 1.14.0.