Home›Calculus 3›Second-order linear equations›Problem 6.34

Second-order linear equations

Problem 6.34 · hard

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

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m - 2\right) \left(m + 1\right) = m^{2} - m - 2 \]
Roots: 2, -1.✓ Proved
Two real roots give y = C₁e^(r₁x) + C₂e^(r₂x).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \frac{\left(- 2 e^{3 x} - 10\right) e^{- x}}{3} - \frac{d}{d x} \frac{\left(e^{3 x} + 5\right) e^{- x}}{3} + \frac{d^{2}}{d x^{2}} \frac{\left(e^{3 x} + 5\right) e^{- x}}{3} = 0 \]
The solution satisfies the equation.✓ Proved
\[ 2 \]
y(0).
\[ -1 \]
y'(0).

Answer \( y = \frac{\left(e^{3 x} + 5\right) e^{- x}}{3} \)

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: pass
qwen3.6:27b-mlx: fail (error) — The characteristic equation in line 1 is stated as m² + bm + c = 0, but the factorization in line 2 corresponds to m² - m - 2 = 0. The solution uses the wrong sign for the linear term in the characteristic equation description, although the roots found are correct for the problem. This is a contradiction in the setup description.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The characteristic equation in line 1 is stated as m² + bm + c = 0, but the factorization in line 2 corresponds to m² - m - 2 = 0. The solution uses the wrong sign for the linear term in the characteristic equation description, although the roots found are correct for the problem. This is a contradiction in the setup description.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The characteristic equation in line 1 is stated as m² + bm + c = 0, but line 2 shows m² - m - 2, implying b = -1. However, the problem states y'' - 1y' - 2y = 0, so the characteristic equation should be m² - m - 2 = 0. While the roots found (2, -1) are correct for m² - m - 2, the generic formula in line 1 is misleadingly vague and doesn't explicitly connect b=-1 and c=-2 to the problem coefficients. More critically, the solution jumps from finding roots to the final answer without showing the determination of constants C1 and C2, making the logical flow incomplete. The CAS checks confirm the final answer is correct, but the pedagogical steps are missing or poorly explained.
gpt-oss:20b: inconclusive 2026-09-27 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The proposed solution does not satisfy the differential equation or the initial conditions. The correct solution is \(y=\tfrac12e^{2x}+\tfrac32e^{-x}\), n
qwen3.6:27b-mlx: fail (error) 2026-09-26 — The characteristic equation in line 1 is stated as m² + bm + c = 0, but line 2 solves m² - m - 2 = 0. The coefficient of y' is -1, so b should be -1, not left generic or implied positive. More critically, the roots found are 2 and -1, but the general solution form in line 3 uses r₁ and r₂ without specifying which is which, leading to potential confusion, though the final answer is correct. However, the primary error is that line 1 does not explicitly substitute the coefficients from the specific ODE (b=-1, c=-2), making the transition to line 2 unjustified as written. A student might think b and c are generic placeholders that magically become -1 and -2. It is misleading to state the generic form without immediately instantiating it.
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.