求方程x(x2-1)y'-(2*x2-1)y+x3=0求方程

如果是
x(x2-1)y'-(3*x2-1)y+x3=0，
解法如下

x(x^2 - 1)y' - (3x^2 - 1)y = - x^3,

[(x^3 - x)y' - (3x^2 - 1)y]/(x^3 - x) = -x^3/(x^3 - x),

d[y/(x^3 - x)]/dx = -x^3/(x^3 - x)

-x^3/(x^3 - x) 的不定积分

= S (x-x^3)dx/(x^3 - x) - S xdx/(x^3 - x)

= -S dx - S xdx/[(x^2-1)x]

= -x - S dx/(x^2-1)

= -x + [S dx/(x+1) - S dx/(x-1)]/2

= -x + [ln|x+1| - ln|x-1|]/2 + C

所以，
y/(x^3 - x) = -x + [ln|x+1| - ln|x-1|]/2 + C

y = {-x + [ln|x+1| - ln|x-1|]/2 + C}(x^3 - x), C = const.

你这星号是啥？ y丿是求导符号 写清楚点啊