Home›Calculus 3›Second-order linear equations›Problem 6.45

Second-order linear equations

Problem 6.45 · hard

Solve \( \displaystyle y'' + 2y' + 5y = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = 3 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m + 1 - 2 i\right) \left(m + 1 + 2 i\right) = m^{2} + 2 m + 5 \]
Roots: -1 - 2*I, -1 + 2*I.✓ Proved
Roots α ± βi give y = e^(αx)(C₁cos βx + C₂sin βx).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \left(10 \sin{\left(2 x \right)} + 5 \cos{\left(2 x \right)}\right) e^{- x} + 2 \frac{d}{d x} \left(2 \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{- x} + \frac{d^{2}}{d x^{2}} \left(2 \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{- x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ 1 \]
y(0).
\[ 3 \]
y'(0).

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

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: pass
qwen3.6:27b-mlx: fail (error) — The solution skips the crucial step of determining the constants C1 and C2 using the initial conditions. It jumps from the general form to the final answer without showing the calculation that yields C1=1 and C2=2, making the derivation incomplete and logically disconnected.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution skips the crucial step of determining the constants C1 and C2 using the initial conditions. It jumps from the general form to the final answer without showing the calculation that yields C1=1 and C2=2, making the derivation incomplete and logically disconnected.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the characteristic roots, applies the standard form for complex conjugate roots, and verifies that the final expression satisfies both the differential equation and the initial conditions.
gpt-oss:20b: fail (error) 2026-09-27 — The proposed solution y=(2sin(2x)+cos(2x))e^{-x} satisfies the differential equation but does not satisfy the initial condition y'(0)=3 (it gives y'(0)=1). The constants C₁ and C₂ were never determined, so the conclusion is incorrect.
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the characteristic equation, finds the complex roots, applies the standard form for the general solution, and verifies that the specific constants satisfy both the differential equation and the initial conditions.
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.