Home›Calculus 3›Second-order initial and boundary value problems›Problem 6.807

Second-order initial and boundary value problems

Problem 6.807 · medium

Solve \( \displaystyle 6 y - 5 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y(2) = 1 \).

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

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

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: fail (error) — The solution never actually determines the constants C₁ and C₂ from the boundary conditions; it merely plugs in the claimed final formula without showing how it was obtained. The reasoning is incomplete and would mislead a student into thinking the answer was derived correctly.
qwen3.6:27b-mlx: fail (error) — The solution claims the general solution is y = C₁exp(2*x) + C₂exp(3*x), but the stated answer contains terms like exp(2x-4) and exp(x) (which is exp(1*x)). The term exp(x) is not a solution to the differential equation y'' - 5y' + 6y = 0, as its characteristic root is 1, not 2 or 3. The stated answer is mathematically incorrect for the given ODE.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-11-07 — The solution claims the general solution is y = C₁exp(2*x) + C₂exp(3*x), but the stated answer contains terms like exp(2x-4) and exp(x) (which is exp(1*x)). The term exp(x) is not a solution to the differential equation y'' - 5y' + 6y = 0, as its characteristic root is 1, not 2 or 3. The stated answer is mathematically incorrect for the given ODE.
gpt-oss:20b: fail (error) 2026-11-07 — The solution never actually determines the constants C₁ and C₂ from the boundary conditions; it merely plugs in the claimed final formula without showing how it was obtained. The reasoning is incomplete and would mislead a student into thinking the answer was derived correctly.
qwen3.6:27b-mlx: fail (error) 2026-11-06 — The solution claims the general solution is y = C₁exp(2*x) + C₂exp(3*x) based on roots 2 and 3, but the stated answer contains terms like exp(x) and exp(2x-4), which correspond to roots 1 and 2. The stated answer does not satisfy the differential equation derived from the roots 2 and 3.
gpt-oss:20b: pass 2026-11-06
qwen3.6:27b-mlx: fail (error) 2026-11-06 — The solution fails to actually solve for the constants C1 and C2 using the boundary conditions. It merely asserts the final answer and verifies it, skipping the derivation steps required to show how the answer was obtained from the general solution.
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.