Question

微分方程y''+2y=sinx的通解

Answer 1

特征方程：

r² + 2 = 0

r = ±√2i

y = C₁sin(√2x) + C₂cos(√2x)

令特解p = Asinx + Bcosx

p'' = - Asinx - Bcosx,代入方程得

(- Asinx - Bcosx) + 2(Asinx + Bcosx) = sinx

{ - A + 2A = 1 => A = 1

{ - B + 2B = 0 => B = 0

特p = sinx

∴方程通y = C₁sin(√2x) + C₂cos(√2x) + sinx