Logistic function, from differential to solution

Jan 2019
Here is a explanation of differential to solution equation of generalized logistic function. I haven't reached from y to z with simple change of variables. I need a simple explanation in order to go first from second equation dy(t) to 5 and to 6. Thanks for helps.


MHF Helper
Apr 2005
Initially, the differential equation \(\displaystyle dx= bx\left(1- \left(\frac{x}{F}\right)^m\right)dt\) (I have dropped the subscript "i" since it is not relevant to this question). As suggested, dividing both sides by F gives \(\displaystyle \frac{dx}{F}= b \frac{x}{F} \left ( 1- \frac{x}{F} \right )^m dt\) and then, letting \(\displaystyle y= \frac{dx}{F}\), \(\displaystyle dy= by(1- y)dt\).

Now, let \(\displaystyle z= y^{-m}\). Then \(\displaystyle y= z^m\) so that \(\displaystyle dy= mz^{m-1}dz= bz^m(1- z)dt\) which reduces to \(\displaystyle mdz= bz(1- z)dt\) or \(\displaystyle \frac{mdz}{z(1- z)}= dt\). On the left use "partial fractions": Find numbers A and B such that \(\displaystyle \frac{1}{z(1- z)}= \frac{A}{z}+ \frac{B}{1- z}\) for all z. Multiply on both sides by z(1- z) to get \(\displaystyle 1= A(1- z)+ Bz\). When z= 1, 1= B. When z= 0, 1= A. So \(\displaystyle \frac{m}{z}dz + \frac{m}{1- z}dz= dt\). Integrating now, \(\displaystyle mln(z)- mlog(1- z)= m ln \left ( \frac{z}{1-z} \right )= t+ C\)
Last edited by a moderator: