求微分方程dy=ky(N-y)dx (N,k>0 为常数）的解?

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

dy = ky(n-y) dx => ( 1/ y(n-y))dy = k dx => 1/n( 1/(n-y) + 1/y )dy = k dx
=> ln( y/(n-y) ) + C1 = nk dx
=> y/(n-y) = C2 EXP( nk x )
解完！：）