求(1+x)y''+y'=(1+x)^2满足初始条件y(0)=1,y'(0)=-1的特解

x+2y=2x+y+1=7x-y 求：2x-y? (x-y)/(x+y)=3求( 3x-2y-1)/(x+y-5) 若x*x+y*y=x+y-1/2,求x,y的值如题 已知x+y=a,2x-y=-2a,求[（x/y-y/x)/(x+y)-x(1/x-1/y)]/[(x+1)/y]的值 若2/x-1/y=3,求[y/x-y/x-y(x-y/x-x+y)]/x-2y/x的值 {(x+y)/2+(x-y)/3=1,(x+y)-5(x-y)=2.求x=?,y=? x+y=1,xy=-1/2,求x(x+y)(x-y)-x(x+y)2 已知:【(x²+y²)-(x-y)²+2y(x-y)】/4y=1,求4x/4x²-y²-1/2x+y 已知：2x+y=6,x-3y=1 求7y(x-3y)(x-3y)-2(3y-x)(3y-x) 已知x-y=1,求[(x+2y)^2+(2x+y)(x-4y)-3(x+y)(x-y)]除以y的值大神们帮帮忙 已知x-y=1,求[(x+2y)²+(2x+y)(x-4y)-3(x+y)(x-y)]÷y的值 y=arcsin(x/√1+x^2),求y' y=(1+x^2)的x次方,求y ' 求y‘-(1/x)y=x^2 的通解 y=ln(x+√x^2+1),求y y'-(1/x)y=x^2求通解 y=1/(x^2-x),求y^(n) 对于(y一撇）y`=y/x(y-1)求导,我求的结果是y``=-y^2-y`x+y/x^2(y-1)^2 怎么不对啊,我的求导过程是：y`x(y-1) - [x(y-1)]` y y`xy-y`x-(y-1+xy`)y -y^2-y`x+yy``=-------------------------=----------------------=-----------------x^2 (y-1