求方程y''+y=xcos2x的通解在设y*=(Ax+B)cos2x+(Cx+D)sin2x之后怎么代入原方程求ABCD

Question

求方程y''+y=xcos2x的通解在设y*=(Ax+B)cos2x+(Cx+D)sin2x之后怎么代入原方程求ABCD

Answer 1

y*'=acos2x-(2ax+2b)sin2x+csin2x+(2cx+2d)cos2x=(-2ax-2b+c)sin2x+(2cx+2d+a)cos2x
y*"=-4asin2x-(4ax+4b)cos2x+4ccos2x-(4cx-4d)sin2x=(-4a-4cx+4d)sin2x+(-4ax-4b+4c)cos2x
代入原方程：
(cx+d-4a-4cx+4d)sin2x+(ax+b-4ax-4b+4c)cos2x=xcos2x
即(-3cx-4a+5d)sin2x+(-3ax-3b+4c)cos2x=xcos2x
对比系数得：
-3c=0,-4a+5d=0,-3a=1,-3b+4c=0
解得：a=-1/3,b=0,c=0,d=-4/15