Question

advanced mathmatics~~~高等数学】特解系列-2，非齐次微分方程的特解

Answer 1

非齐次项若是 (e^2)xcos2a ，
设特解 y = Ax+B，代入微分方程得
A = (e^2)cos2a, B = 0
特解是 y = (e^2)xcos2a。

非齐次项若是 (e^2)xcos2x ，
设特解 y = (ax+b)cos2x+(cx+d)sin2x，
y' = acos2x-2(ax+b)sin2x+csin2x+2(cx+d)cos2x
= (2cx+2d+a)cos2x + (-2ax-2b+c)sin2x
y'' = 2ccos2x-2(2cx+2d+a)sin2x-2asin2x+2(-2ax-2b+c)cos2x
= (-4ax-4b+4c)cosx + (-4cx-4d-4a)sin2x
代入微分方程得
-3ax-3b+4c = (e^2)x
-3cx-3d-4a = 0
得 a = e^2/3, c = 0， b = 0, d = -4e^2/9
特解是 y = (e^2/3)xcos2x - (4e^2/9)sin2x。

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