Home›Calculus 3›Second-order linear equations›Problem 6.35

Second-order linear equations

Problem 6.35 · hard

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

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m + 2\right) \left(m + 3\right) = m^{2} + 5 m + 6 \]
Roots: -2, -3.✓ Proved
Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \left(24 - 36 e^{x}\right) e^{- 3 x} + 5 \frac{d}{d x} \left(4 - 6 e^{x}\right) e^{- 3 x} + \frac{d^{2}}{d x^{2}} \left(4 - 6 e^{x}\right) e^{- 3 x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -2 \]
y(0).
\[ 0 \]
y'(0).

Answer \( y = 2 \left(2 - 3 e^{x}\right) e^{- 3 x} \)

Lines: 2 proved, 5 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	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	Not checked	—	a sentence; read, not computed
4	Not checked	—	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
7	Not checked	—	both sides are the same expression: the line states a value, and the working that produced it is not on the line
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	sympy.checkodesol substitutes the solution back; both initial conditions match

Reviewers

gpt-oss:20b: fail (error) — The solution never actually solves for the constants C₁ and C₂; instead it tests a function that is 12 times the correct solution, which does not satisfy the differential equation. The boundary conditions are also mis‑reported (two values for y(0) and y′(0)).
qwen3.6:27b-mlx: fail (error) — The solution skips the crucial step of solving for the constants C1 and C2 using the initial conditions. It jumps from the general form to the specific answer without showing the system of equations or the derivation, making it impossible to verify the logic.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution skips the crucial step of solving for the constants C1 and C2 using the initial conditions. It jumps from the general form to the specific answer without showing the system of equations or the derivation, making it impossible to verify the logic.
gpt-oss:20b: fail (error) 2026-09-28 — The solution never actually solves for the constants C₁ and C₂; instead it tests a function that is 12 times the correct solution, which does not satisfy the differential equation. The boundary conditions are also mis‑reported (two values for y(0) and y′(0)).
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to explicitly solve for the constants C1 and C2 using the initial conditions, jumping directly to the final answer. While the final answer is verified to be correct, the logical step of deriving the specific coefficients from the general solution is missing.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the specific answer without showing the system of equations derived from the initial conditions, making the derivation incomplete.
gpt-oss:20b: pass 2026-09-26

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_linear, checked 2026-10-13 with SymPy 1.14.0.