y'-y/x=2 y∧2.lnx的通解怎么求,不好分啊

解微分方程首先要分清微分方程的类型,本题为伯努利方程.
y'-y/x = 2y^2lnx,令 z=y^(1-2)= 1/y,则 y=1/z,
-z'/z^2 -1/(xz) = 2lnx/z^2,即 z'+z/x = -2lnx 为一阶线性微分方程,
z = e^(-∫dx/x)[∫-2lnxe^(∫dx/x)dx+C]
= (1/x)[∫-2xlnxdx+C] = (1/x)[∫-lnxdx^2+C]
= (1/x)[-x^2lnx+∫xdx+C] = (1/x)(-x^2lnx+x^2/2+C)
= x/2-xlnx+C/x.
于是通解为 y(x/2-xlnx+C/x)=1