Question

求微分方程y′′+4y=xsin²x的通解

Answer 1

y′′+4y = xsin²x = x/2 - (1/2)xcos2x
特征方程 r^2 + 4 = 0, r = ±2i
特解应设为 y = ax+b + x(px+q)cos2x + x(ux+v)sin2x
= ax+b + (px^2+qx)cos2x + (ux^2+vx)sin2x
则 y' = a + [2ux^2+(2p+2v)x+q]cos2x + [-2px^2+(2u-2q)x+v]sin2x
y'' = [-4px^2+(8u-4q)x+2p+4v]cos2x + [-4ux^2-(8p+4v)x+2u-4q]sin2x
代入微分方程，得
[8ux+2p+4v]cos2x + [-8px+2u-4q]sin2x + 4ax+4b = x/2 - (1/2)xcos2x
得 a = 1/8， b = 0， u = -1/16, p = 0, v = 0, q = -1/32
特解为 y = x/8 - (x/32)cos2x - (x^2/16)sin2x
通解是 y = C1cos2x + C2sin2x + x/8 - (x/32)cos2x - (x^2/16)sin2x