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Cubic Nonlinear ODE

$\displaystyle \frac{\textrm{d}u}{\textrm{d}t} = u(k-u^{2})$ (D.2.1)

Solved using the method of seperation of variables and some fancy algebra...

$\displaystyle \frac{\textrm{d}u}{\textrm{d}t}$ $\displaystyle = u(k-u^{2})$    
$\displaystyle \frac{\textrm{d}u}{u(k-u^{2})}$ $\displaystyle = \textrm{d}t$    
$\displaystyle \int \frac{\textrm{d}u}{u(k-u^{2})}$ $\displaystyle = \int \textrm{d}t$    

Now, we will break down $ \frac{1}{u(k-u^{2})}$

First note:

$\displaystyle (k - u^{2}) = (\sqrt{k} - u)(\sqrt{k} + u)$    

and thus,

$\displaystyle \frac{1}{k - u^{2}} = \frac{1}{2\sqrt{k}} \left(\frac{1}{\sqrt{k} - u} + \frac{1}{\sqrt{k} + u} \right)$ (D.2.2)

Also note,

$\displaystyle \frac{1}{u(\sqrt{k} - u)} = \frac{1}{\sqrt{k}} \left( \frac{1}{u} + \frac{1}{\sqrt{k} - u} \right)$ (D.2.3)

and

$\displaystyle \frac{1}{u(\sqrt{k} + u)} = \frac{1}{\sqrt{k}} \left( \frac{1}{u} - \frac{1}{\sqrt{k} + u} \right)$ (D.2.4)

Putting together the general concepts in Equations D.2.2, D.2.3, and D.2.4 we get...

$\displaystyle \frac{1}{u(k-u^{2})} = \frac{1}{2k}\left[ \left( \frac{1}{u} + \f...
...}{\sqrt{k}-u} \right) + \left(\frac{1}{u} - \frac{1}{\sqrt{k}+u}\right) \right]$    

So, to solve our non-linear differential equation, we make the substitution...

$\displaystyle \int \frac{1}{2k}\left[ \left( \frac{1}{u} + \frac{1}{\sqrt{k}-u} \right) + \left(\frac{1}{u} - \frac{1}{\sqrt{k}+u}\right) \right] \textrm{d}u$ $\displaystyle = \int \textrm{d}t$    
$\displaystyle \frac{1}{2k} \int \left[ \left( \frac{1}{u} + \frac{1}{\sqrt{k}-u} \right) + \left(\frac{1}{u} - \frac{1}{\sqrt{k}+u}\right) \right] \textrm{d}u$ $\displaystyle = t + C$    
$\displaystyle \frac{1}{2k}\left( \log \vert u\vert - \log \vert\sqrt{k}-u\vert + \log \vert u\vert - \log \vert\sqrt{k}+u\vert \right)$ $\displaystyle = t + C$    
$\displaystyle \log \left( \frac{\vert u\vert}{\vert\sqrt{k}-u\vert} \right) + \log \left( \frac{\vert u\vert}{\vert\sqrt{k}+u\vert} \right)$ $\displaystyle = 2tk + C$    
$\displaystyle \log \left( \frac{u^2}{\left(\vert\sqrt{k}-u\vert\right)\left(\vert\sqrt{k}+u\vert\right)} \right)$ $\displaystyle = 2tk + C$    
$\displaystyle \log \left( \frac{u^{2}}{\left(\vert k-u^{2}\vert\right)} \right)$ $\displaystyle = 2tk + C$    
$\displaystyle \frac{u^{2}}{\vert k-u^{2}\vert}$ $\displaystyle = e^{2tk + C} = e^{2tk}e^{C} = Ce^{2tk}$    
$\displaystyle \frac{\vert k-u^{2}\vert}{u^{2}}$ $\displaystyle = Ce^{-2tk}$    
$\displaystyle \frac{k}{u^{2}} - 1$ $\displaystyle = Ce^{-2tk}$    
$\displaystyle \frac{k}{u^{2}}$ $\displaystyle = 1 + Ce^{-2tk}$    
$\displaystyle u^{2}$ $\displaystyle = \frac{k}{1 + Ce^{-2tk}}$    
$\displaystyle u$ $\displaystyle = \sqrt{\frac{k}{1 + Ce^{-2tk}}}$    


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Next: Vector and Matrix Calculus Up: Derivations For the Curious Previous: Nernst-Planck Equation   Index

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Frank Starmer 2004-05-19
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