Click for printer friendely version of this HowTo Cubic Nonlinear ODE $$\frac{\textrm{d}u}{\textrm{d}t} = u(k-u^{2})$$ (D.2.1) Solved using the method of seperation of variables and some fancy algebra... $$\frac{\textrm{d}u}{\textrm{d}t} = u(k-u^{2})$$ $$\frac{\textrm{d}u}{u(k-u^{2})} = \textrm{d}t$$ $$\int \frac{\textrm{d}u}{u(k-u^{2})} = \int \textrm{d}t$$ Now, we will break down $\frac{1}{u(k-u^{2})}$ First note: $$(k - u^{2}) = (\sqrt{k} - u)(\sqrt{k} + u)$$ and thus, $$\frac{1}{k - u^{2}} = \frac{1}{2\sqrt{k}} \left(\frac{1}{\sqrt{k} - u} + \frac{1}{\sqrt{k} + u} \right)$$ (D.2.2) Also note, $$\frac{1}{u(\sqrt{k} - u)} = \frac{1}{\sqrt{k}} \left( \frac{1}{u} + \frac{1}{\sqrt{k} - u} \right)$$ (D.2.3) and $$\frac{1}{u(\sqrt{k} + u)} = \frac{1}{\sqrt{k}} \left( \frac{1}{u} - \frac{1}{\sqrt{k} + u} \right)$$ (D.2.4) Putting together the general concepts in Equations D.2.2, D.2.3, and D.2.4 we get... $$\frac{1}{u(k-u^{2})} = \frac{1}{2k}\left[ \left( \frac{1}{u} + \f... ...}{\sqrt{k}-u} \right) + \left(\frac{1}{u} - \frac{1}{\sqrt{k}+u}\right) \right]$$ So, to solve our non-linear differential equation, we make the substitution... $$\int \frac{1}{2k}\left[ \left( \frac{1}{u} + \frac{1}{\sqrt{k}-u} \right) + \left(\frac{1}{u} - \frac{1}{\sqrt{k}+u}\right) \right] \textrm{d}u = \int \textrm{d}t$$ $$\frac{1}{2k} \int \left[ \left( \frac{1}{u} + \frac{1}{\sqrt{k}-u} \right) + \left(\frac{1}{u} - \frac{1}{\sqrt{k}+u}\right) \right] \textrm{d}u = t + C$$ $$\frac{1}{2k}\left( \log \vert u\vert - \log \vert\sqrt{k}-u\vert + \log \vert u\vert - \log \vert\sqrt{k}+u\vert \right) = t + C$$ $$\log \left( \frac{\vert u\vert}{\vert\sqrt{k}-u\vert} \right) + \log \left( \frac{\vert u\vert}{\vert\sqrt{k}+u\vert} \right) = 2tk + C$$ $$\log \left( \frac{u^2}{\left(\vert\sqrt{k}-u\vert\right)\left(\vert\sqrt{k}+u\vert\right)} \right) = 2tk + C$$ $$\log \left( \frac{u^{2}}{\left(\vert k-u^{2}\vert\right)} \right) = 2tk + C$$ $$\frac{u^{2}}{\vert k-u^{2}\vert} = e^{2tk + C} = e^{2tk}e^{C} = Ce^{2tk}$$ $$\frac{\vert k-u^{2}\vert}{u^{2}} = Ce^{-2tk}$$ $$\frac{k}{u^{2}} - 1 = Ce^{-2tk}$$ $$\frac{k}{u^{2}} = 1 + Ce^{-2tk}$$ $$u^{2} = \frac{k}{1 + Ce^{-2tk}}$$ $$u = \sqrt{\frac{k}{1 + Ce^{-2tk}}}$$ Click for printer friendely version of this HowTo

>