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Thread: Logistic function, from differential to solution

  1. #1
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    Logistic function, from differential to solution

    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.
    Logistic function, from differential to solution-capture.jpg
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  2. #2
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    Re: Logistic function, from differential to solution

    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 topsquark; Jan 23rd 2019 at 08:23 AM. Reason: Tweaked LaTeX
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