Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.799

Second-order initial and boundary value problems

Problem 6.799 · medium

Solve \( \displaystyle 4 y + 5 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y(2) = -4 \).

\[ m^{2} + 5 m + 4 = \left(m + 1\right) \left(m + 4\right) \]
The characteristic equation has distinct real roots -4 and -1.✓ Proved
So y = C₁exp(-4*x) + C₂exp(-x); the two conditions fix C₁ and C₂.
Reviewed
\[ \frac{4 \left(4 \left(1 + e^{8}\right) e^{3 x} - 4 \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} + 5 \frac{d}{d x} \frac{\left(4 \left(1 + e^{8}\right) e^{3 x} - 4 \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} + \frac{d^{2}}{d x^{2}} \frac{\left(4 \left(1 + e^{8}\right) e^{3 x} - 4 \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} = 0 \]
It satisfies the equation ...✓ Proved
\[ \left. \frac{4 \left(\left(1 + e^{8}\right) e^{3 x} - \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} \right|_{\substack{ x=0 }} = 4 \]
... y(0) = 4 ...✓ Proved
\[ \left. \frac{4 \left(\left(1 + e^{8}\right) e^{3 x} - \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} \right|_{\substack{ x=2 }} = -4 \]
y(2) = -4.✓ Proved

Answer \( y = \frac{4 \left(\left(1 + e^{8}\right) e^{3 x} - \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} \)

✓ 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: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the general form of the solution and verifies that the stated answer satisfies both the differential equation and the boundary conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-11-07 — The solution correctly identifies the general form of the solution and verifies that the stated answer satisfies both the differential equation and the boundary conditions.
gpt-oss:20b: pass 2026-11-07
qwen3.6:27b-mlx: pass 2026-11-06
gpt-oss:20b: pass 2026-11-06
qwen3.6:27b-mlx: pass 2026-11-06 — The solution correctly identifies the general form of the solution to the homogeneous linear ODE and verifies that the stated answer satisfies both the differential equation and the boundary conditions.
gpt-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.