由y’+xy²-y²=1-x得dy/(1+y^2)=(1-x)dx,积分得arctany=x-x^2/2+c,所以y=tan(x-x^2/2+c).
dy/dx = y' = 1-x + (1-x)y^2 = (1 - x) (1 + y^2)1/(1 + y^2) * dy = (1 - x) * dxarctan(y) = x -0.5 x ^2 + C
