Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.803

Second-order initial and boundary value problems

Problem 6.803 · medium

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

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

Answer \( y = 2 e^{- 2 x} \)

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

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	Not checked	—	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: fail (error) — The solution claims the general solution is y = C₁exp(x) + C₂exp(-2*x) but then only verifies the specific case y = 2exp(-2*x) without solving for the constants. It fails to show that C₁ must be 0 and C₂ must be 2, skipping the actual determination of the constants from the initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-11-07 — The solution claims the general solution is y = C₁exp(x) + C₂exp(-2*x) but then only verifies the specific case y = 2exp(-2*x) without solving for the constants. It fails to show that C₁ must be 0 and C₂ must be 2, skipping the actual determination of the constants from the initial conditions.
gpt-oss:20b: pass 2026-11-07
qwen3.6:27b-mlx: fail (error) 2026-11-06 — The solution claims the general solution is y = C₁exp(x) + C₂exp(-2*x) but then only verifies the specific case y = 2exp(-2*x). It fails to solve for C₁ and C₂ using the initial conditions, effectively skipping the determination that C₁ must be 0.
gpt-oss:20b: pass 2026-11-06
qwen3.6:27b-mlx: fail (error) 2026-11-06 — The solution fails to determine the constants C₁ and C₂. It claims the conditions fix them but then only verifies that the specific function 2e^{-2x} satisfies the conditions, ignoring the general solution form derived in step 2. The correct solution requires solving the system for C₁ and C₂, which yields C₁=0 and C₂=2, but the logic presented skips this derivation entirely.
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.