由已知有
dy/(ky(N-y)) = dx
两边积分有
∫dy/(ky(N-y)) = x
左边那个积分= (1/k)∫dy/(y(N-y)) = (1/(kN))(∫dy/y + ∫dy/(N-y))
=ln|y/(N-y)|/(kN)+C
所以有 （y/(N-y)）^(1/(kN)) =Cx