y"-y=-x ,y'(2)=-2 ,y(2)=0,试求微分方程的幂级数解,我都想2礼拜了,

x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 y+y'-2y=2x 微分方程y - 2y' + y = x 填空：x-y/x+y=( )/x^2-y^2 （x-y)2•(y-x)³(x-y)=? (x-y) (x-2y)+(x+2y) (x-3y)-2(x-3y) (x-4y)= (x+y)(x-y)+(x-y)平方-(x-y)(3x-y) 其中x=-2 y=3分之一 已知x-y/x+y=3,求代数式5(x-y)/x+y-x+y/2(x-y) (x-4y)(2x+y)= (x+y)(2x-3y)=? x+y>=2,x-y x+y>=2,x-y y'-2y=(e^x)-x y'+2y/x=x通解 dy/dx=(2x*x*x*y-y*y*y*y)/(x*x*x*x-2x*y*y*y) 【2(x-y)^3-8(x-y)^2(x+y)+6y(x-y)^2】/2(x-y)^2= （x-y)(x+y)-(x+y)^2+2y(y-x),其中x=1,y=3. 2y(x+二分之一y)-[(x+y)(x-y)+2y(y+x)],其中|x-1|=2