Home›Calculus 3›Second-order linear equations›Problem 6.161

Second-order linear equations

Problem 6.161 · hard

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

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
Reviewed
\[ \left(m + 1 - i\right) \left(m + 1 + i\right) = m^{2} + 2 m + 2 \]
Roots: -1 - I, -1 + I.✓ Proved
Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).
Reviewed
Use y(0) and y'(0) to find C₁ and C₂.
Reviewed
\[ \left(- 6 \sin{\left(x \right)} - 2 \cos{\left(x \right)}\right) e^{- x} + 2 \frac{d}{d x} \left(- 3 \sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{- x} + \frac{d^{2}}{d x^{2}} \left(- 3 \sin{\left(x \right)} - \cos{\left(x \right)}\right) e^{- x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -1 \]
y(0).Reviewed
\[ -2 \]
y'(0).Reviewed

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

Lines: 2 proved, 5 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
4	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
5	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
6	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	both sides are the same expression: the line states a value, and the working that produced it is not on the line
7	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass — The solution correctly identifies the characteristic equation, roots, and general form. The constants are implicitly determined to satisfy the initial conditions, and the final verification steps confirm the result is correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly identifies the characteristic equation, roots, and general form. The constants are implicitly determined to satisfy the initial conditions, and the final verification steps confirm the result is correct.
gpt-oss:20b: pass 2026-10-03
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution claims the general form is y = e^(αx)(C₁cos βx + C₂sin βx) but fails to explicitly determine the constants C₁ and C₂. It jumps directly to verifying a specific function without showing the derivation of the coefficients from the initial conditions, leaving the logical gap of how the specific solution was obtained.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to show the derivation of the constants C1 and C2 from the initial conditions, jumping directly to the final answer without demonstrating how the specific values were obtained.
gpt-oss:20b: fail (misleading) 2026-09-28 — The solution skips the key step of determining the constants C₁ and C₂ from the initial conditions and presents an incorrect intermediate expression (the extra factor of –2). A student could be misled into thinking the verification step is complete without actually showing the substitution.

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.