# Logistic function, from differential to solution

#### iskenderc

Hi,
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.

#### HallsofIvy

MHF Helper
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, $$\displaystyle mln(z)- mlog(1- z)= m ln \left ( \frac{z}{1-z} \right )= t+ C$$

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