Home›Calculus 3›Second-order linear equations›Problem 6.159

Second-order linear equations

Problem 6.159 · hard

Solve \( \displaystyle y'' + 4y' + 4y = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 2 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m + 2\right)^{2} = m^{2} + 4 m + 4 \]
Roots: -2.✓ Proved
A repeated root r gives y = (C₁ + C₂x)e^(rx).
Use y(0) and y'(0) to find C₁ and C₂.
\[ \left(- 16 x - 12\right) e^{- 2 x} + 4 \frac{d}{d x} \left(- 4 x - 3\right) e^{- 2 x} + \frac{d^{2}}{d x^{2}} \left(- 4 x - 3\right) e^{- 2 x} = 0 \]
The solution satisfies the equation.✓ Proved
\[ -3 \]
y(0).
\[ 2 \]
y'(0).

Answer \( y = \left(- 4 x - 3\right) e^{- 2 x} \)

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 (misleading) — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions. Without showing how the constants are found, the conclusion that y = (−4x−3)e^(−2x) is the unique solution is unsupported, potentially misleading a student who may think the result follows automatically.
qwen3.6:27b-mlx: pass — The solution correctly identifies the repeated root, states the general form, and verifies the specific solution against the differential equation and initial conditions.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-03 — The solution correctly identifies the repeated root, states the general form, and verifies the specific solution against the differential equation and initial conditions.
gpt-oss:20b: fail (misleading) 2026-10-03 — The solution skips the crucial step of determining the constants C₁ and C₂ from the initial conditions. Without showing how the constants are found, the conclusion that y = (−4x−3)e^(−2x) is the unique solution is unsupported, potentially misleading a student who may think the result follows automatically.
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the characteristic equation, handles the repeated root case, and verifies the final answer against the differential equation and initial conditions.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution fails to explicitly solve for the constants C1 and C2. It jumps from the general form to the final answer without showing the system of equations derived from the initial conditions, making it impossible to verify the logic or learn the method.
gpt-oss:20b: pass 2026-09-28

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.