(y'-y/x)/(1+y'y/x)=√3/3 x=0时,y=0 这个微分方程怎么解请问……

y'/y=1/x x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2 先化简 再求值 (2x-y)(y+x)-(x-2y)(2y+x)-(-3y+x)^2其(√x+1)+y^2+4=-4y 化简 x / y(x+y) - y / x(x+y) = (x+y)(x-y)+4(y-1) x+x+y+y+y=21 ,x+x+y+y+y+y+y=27 ,x ,y=? x+x+y+y+y=21 ,x+x+y+y+y+y+y=27 ,x ,y=? x(x+y)(x-y)-y(y+x)(y-x)=(x-y)( )填空 y'-(1/x)y=xe^-x {(x,y) |x|+|y| x>y?x:y [(y*y*y-y)/(xy+1)*(xy+1)-(x+y)*(x+y)]/[(x+y)*y/(x+1)]其中x=-11,y=1/12,化简再求值. y(x+y)+(x-y)²-（x+y）（x-y）,其中x=-1/3,y=3 （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 解常微分方程：y/x=y'+√(1+y'^2),y=f(x) 常微分方程y'=(x+y)ln(x+y)-1 y'=(x-y+1)/(x+y-3)通解