Home›Calculus 3›Second-order linear equations›Problem 6.158

Second-order linear equations

Problem 6.158 · hard

Solve \( \displaystyle y'' - 4y' + 8y = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = 1 \).

Try y = e^(mx): the characteristic equation is m² + bm + c = 0.
\[ \left(m - 2 - 2 i\right) \left(m - 2 + 2 i\right) = m^{2} - 4 m + 8 \]
Roots: 2 - 2*I, 2 + 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.
\[ \left(- 12 \sin{\left(2 x \right)} + 16 \cos{\left(2 x \right)}\right) e^{2 x} - 4 \frac{d}{d x} \frac{\left(- 3 \sin{\left(2 x \right)} + 4 \cos{\left(2 x \right)}\right) e^{2 x}}{2} + \frac{d^{2}}{d x^{2}} \frac{\left(- 3 \sin{\left(2 x \right)} + 4 \cos{\left(2 x \right)}\right) e^{2 x}}{2} = 0 \]
The solution satisfies the equation.✓ Proved
\[ 2 \]
y(0).
\[ 1 \]
y'(0).

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

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; it merely states that the final expression satisfies the differential equation and the initial values without showing the derivation, which could mislead a student into thinking the constants were found correctly.
qwen3.6:27b-mlx: pass — The solution correctly identifies the characteristic equation, roots, and general form. The verification steps confirm the solution satisfies the differential equation and initial conditions.

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 verification steps confirm the solution satisfies 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; it merely states that the final expression satisfies the differential equation and the initial values without showing the derivation, which could mislead a student into thinking the constants were found correctly.
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the characteristic equation, roots, and general form. The final answer satisfies the differential equation and initial conditions as verified by the algebraic checks.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: pass 2026-09-28 — 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 steps are logically sound and mathematically correct.
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.