y' = y/x - (y/x)^2
设y/x = u
y = xu
y' = u + xu'
所以u + xu' = u - u^2
xu' = -u^2
- du / u^2 = dx/x
积分得
1/u = lnx + C
所以x/y = lnx + C
通解为
y = x / (lnx +C)