Home›Calculus 3›Second-order linear equations›Problem 6.134

Second-order linear equations

Problem 6.134 · hard

Solve \( \displaystyle y'' + 2y' + 10y = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 1 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m + 1 - 3 i\right) \left(m + 1 + 3 i\right) = m^{2} + 2 m + 10 \]
Roots: -1 - 3*I, -1 + 3*I.✓ Proved
Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \frac{\left(- 20 \sin{\left(3 x \right)} - 90 \cos{\left(3 x \right)}\right) e^{- x}}{3} + 2 \frac{d}{d x} \frac{\left(- 2 \sin{\left(3 x \right)} - 9 \cos{\left(3 x \right)}\right) e^{- x}}{3} + \frac{d^{2}}{d x^{2}} \frac{\left(- 2 \sin{\left(3 x \right)} - 9 \cos{\left(3 x \right)}\right) e^{- x}}{3} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -3 \]
y(0).
\[ 1 \]
y'(0).

Answer \( y = - \frac{\left(2 \sin{\left(3 x \right)} + 9 \cos{\left(3 x \right)}\right) e^{- x}}{3} \)

Lines: 2 proved, 5 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. 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 (style) — The solution skips the explicit determination of the constants C₁ and C₂ from the initial conditions and does not show how the final expression is obtained. While the final formula is correct, the omitted steps could mislead a student into thinking the result follows automatically from the earlier statements.
qwen3.6:27b-mlx: pass — The solution correctly identifies the characteristic roots, applies the standard form for complex conjugate roots, and verifies that the final answer satisfies both the differential equation and the initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the characteristic roots, applies the standard form for complex conjugate roots, and verifies that the final answer satisfies both the differential equation and the initial conditions.
gpt-oss:20b: fail (style) 2026-09-28 — The solution skips the explicit determination of the constants C₁ and C₂ from the initial conditions and does not show how the final expression is obtained. While the final formula is correct, the omitted steps could mislead a student into thinking the result follows automatically from the earlier statements.
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic roots, applies the general form for complex conjugate roots, and verifies that the specific solution satisfies both the differential equation and the initial conditions.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic equation, finds the complex roots, applies the standard form for complex conjugate roots, and verifies the initial conditions and differential equation. The logic is sound and complete.
gpt-oss:20b: pass 2026-09-27

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.