求方程(y^2+xy^2)dx+(x^2-yx^2)dy=0的通解

∵(y^2+xy^2)dx+(x^2-yx^2)dy=0==>y²(1+x)dx+x²(1-y)dy=0==>[(y-1)/y²]dy=[(1+x)/x²]dx==>(1/y-1/y²)dy=(1/x+1/x²)dx==>ln|y|+i/y=ln|x|-1/x+C1 (C1是积分常数)==>ln|y|-ln|x|=-1/x-1...