ODE simplifying.

O

Overdriven33

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May 4, 2021
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I'm currently doing the question about ODEs in the picture below:
1620128555267.png
The solution to the problem is posted below, but I'm having trouble with the last part in (b) where they simplify it, I can't seem to figure out how they did it.
1620128921783.png
Thank you :)
 
[MATH]x=\frac{KCe^{Krt}}{1+Ce^{Krt}}[/MATH]
Divide top and bottom by [MATH]Ce^{Krt}[/MATH] (which is non-zero)

[MATH]x=\frac{K}{\tfrac{1}{Ce^{Krt}}+1}[/MATH]
[MATH]x=\frac{K}{\tfrac{1}{C}e^{-Krt}+1}[/MATH]
[MATH]x=\frac{K}{1+\hat{C}e^{-Krt}}[/MATH] where [MATH]\hat{C}[/MATH] is [MATH]\tfrac{1}{C}[/MATH]
 
[MATH]\dfrac{1}{K} * \{ ln(x) - ln(K - x) \} = rt + c, \text { provided } 0 < x < K.[/MATH]
[MATH]\therefore ln \left ( \dfrac{x}{K - x} \right ) = Krt + Kc \implies[/MATH]
[MATH]\dfrac{x}{K - x} = e \text {^} (Krt + Kc) = e \text {^}(Krt) * e \text{^}(Kc).[/MATH]
[MATH]\text {Let } u = e \text {^} (Krt) \text { and } C = e \text {^} (Kc) \implies \dfrac{x}{K - x} = Cu \implies[/MATH]
[MATH]x = CKu - Cxu \implies x + Cxu = CKu \implies x = \dfrac{CKu}{1 + Cu} \implies[/MATH]
[MATH]x = \dfrac{\dfrac{CKu}{Cu}}{\dfrac{1 + Cu}{Cu}} = \dfrac{K}{\dfrac{1}{Cu}+ \dfrac{Cu}{Cu}} = \dfrac{K}{(Cu)^{-1} + 1}.[/MATH]
[MATH]\text {Let } \hat C = C^{-1} \implies x = \dfrac{K}{1 + \hat C u^{-1}}.[/MATH]
 
* Note: [MATH]Ce^{Krt}[/MATH] is non-zero as clearly [MATH]e^{Krt}[/MATH] is, but also C is non-zero, as it is [MATH]e^{Kc}[/MATH] coming from the equation: [MATH]\frac{1}{K}[\ln{|x|}-\ln{|k-x|}]=rt+c[/MATH]
 
Yes, and given that they've swapped the order of the two terms on the bottom, while it appears they haven't - it's easy to miss what has happened!
 
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