First order dif. eq with second order polynomial

Quote:

Originally Posted by given

Hello,
I’m trying to solve for the integral of B as a function of time.

Please forgive the simple-text formatting of the Eqs, for some reason they are losing formatting when I paste them from word.

dB/dt=c-eB-aB^2

I have reworked it to:

dB/(c-eB-aB^2 )=dt

But am unable to simplify it to a point where I can integrate it.

I appreciate any guidance on how to approach this. I have looked through engineering solutions on ordinary Dif eqs. But couldn’t find an example that was helpful.
Thanks

$\displaystyle \begin{align*} \int{\frac{dB}{c - e\,B - a\,B^2}} &= \int{dt} \\ \int{-\frac{1}{a}\left(\frac{dB}{B^2 + \frac{e}{a}\,B - \frac{c}{a}}\right)} &= \int{dt} \\ \int{\frac{dB}{B^2 + \frac{e}{a}\,B - \frac{c}{a}}} &= \int{-a\,dt} \\ \int{\frac{dB}{B^2 + \frac{e}{a}\,B + \left(\frac{e}{2a}\right)^2 - \left(\frac{e}{2a}\right)^2 - \frac{c}{a}}} &= \int{-a\,dt} \\ \int{\frac{dB}{\left(B + \frac{e}{2a}\right)^2 - \frac{e^2}{4a^2} - \frac{4ac}{4a^2}}} &= \int{-a\,dt} \\ \int{\frac{dB}{\left(B + \frac{e}{2a}\right)^2 - \left(\frac{e^2 - 4ac}{4a^2}\right)}} &= \int{-a\,dt} \\ \int{\frac{dB}{\left(B + \frac{e}{2a}\right)^2 - \left(\frac{\sqrt{e^2 - 4ac}}{2a}\right)^2}} &= \int{-a\,dt} \\ \int{\frac{dB}{ \left(\frac{\sqrt{e^2 - 4ac}}{2a}\right)^2 - \left(B + \frac{e}{2a}\right)^2}} &= \int{a\,dt} \\ \int{\frac{dB}{\left(\frac{\sqrt{e^2 - 4ac}}{2a}\right)^2 - \left(B + \frac{e}{2a}\right)^2}} &= a\,t + C_1 \end{align*}$

Now make the substitution $\displaystyle \begin{align*} B + \frac{e}{2a} = \frac{\sqrt{e^2 - 4ac}}{2a}\sin{\theta} \implies dB = \frac{\sqrt{e^2 - 4ac}}{2a}\cos{\theta}\,d\theta \end{align*}$ and the DE becomes

$\displaystyle \begin{align*} \int{\frac{\frac{\sqrt{e^2 - 4ac}}{2a}\cos{\theta}\,d\theta}{\left(\frac{\sqrt{ e^2 - 4ac}}{2a}\right)^2 - \left(\frac{\sqrt{e^2 - 4ac}}{2a}\sin{\theta}\right)^2}} &= a\,t + C_1 \\ \frac{\sqrt{e^2 - 4ac}}{2a}\int{\frac{\cos{\theta}\,d\theta}{\frac{e ^2 - 4ac}{4a^2} - \frac{e^2 - 4ac}{4a^2}\sin^2{\theta}}} &= a\,t + C_1 \\ \frac{\sqrt{e^2 - 4ac}}{2a}\int{\frac{\cos{\theta}\,d\theta}{\frac{e ^2 - 4ac}{4a^2}\left(1 - \sin^2{\theta}\right)}} &= a\,t + C_1 \\ \frac{2a}{\sqrt{e^2 - 4ac}}\int{\frac{\cos{\theta}\,d\theta}{1 - \sin^2{\theta}}} &= a\,t + C_1 \end{align*}$

Now make the substitution $\displaystyle \begin{align*} u = \sin{\theta} \implies du = \cos{\theta}\,d\theta \end{align*}$ and the DE becomes

$\displaystyle \begin{align*} \frac{2a}{\sqrt{e^2 - 4ac}}\int{\frac{du}{1 - u^2}} &= a\,t + C_1 \\ \int{\frac{du}{1 - u^2}} &= \frac{\sqrt{e^2 - 4ac}}{2a}\left(a\,t + C_1\right) \\ \int{\frac{du}{(1 - u)(1 + u)}} &= \frac{\sqrt{e^2 - 4ac}}{2a}\left(a\,t + C_1\right) \\ \frac{1}{2}\int{\frac{1}{1 - u} + \frac{1}{1 + u}\,du} &= \frac{\sqrt{e^2 - 4ac}}{2a}\left(a\,t + C_1\right) \\ \int{\frac{1}{1 - u} + \frac{1}{1 + u}\,du} &= \frac{\sqrt{e^2 - 4ac}}{a}\left(a\,t + C_1\right) \\ -\ln{|1 - u|} + \ln{|1 + u|} + C_2 &= \frac{\sqrt{e^2 - 4ac}}{a}\left(a\,t + C_1\right) \\ \ln{\left|\frac{1 + u}{1 - u}\right|} &= \sqrt{e^2 - 4ac}\,t + \frac{C_1\sqrt{e^2 - 4ac}}{a} - C_2 \\ \ln{\left|-\left(\frac{u + 1}{u - 1}\right)\right|} &= \sqrt{e^2 - 4ac}\,t + \frac{C_1\sqrt{e^2 - 4ac}}{a} - C_2 \\ \ln{\left|-\left(1 + \frac{2}{u - 1}\right)\right|} &= \sqrt{e^2 - 4ac}\,t + \frac{C_1\sqrt{e^2 - 4ac}}{a} - C_2 \end{align*}$

$\begin{align*} \left|-\left(1 + \frac{2}{u - 1}\right)\right| &= \exp{\left(\sqrt{e^2 - 4ac}\,t + \frac{C_1\sqrt{e^2 - 4ac}}{a} - C_2\right)} \\ \left|-\left(1 + \frac{2}{u - 1}\right)\right| &= \exp{\left(\frac{C_1\sqrt{e^2 - 4ac}}{a} - C_2\right)} \exp{\left(\sqrt{e^2 - 4ac}\,t\right)} \\ 1 + \frac{2}{u - 1} &= A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} \textrm{ where } A = \pm \exp{\left(\frac{C_1\sqrt{e^2 - 4ac}}{a} - C_2\right)} \\ \frac{2}{u - 1} &= A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} - 1 \\ u - 1 &= \frac{2}{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} - 1} \\ u &= \frac{2}{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} - 1} + 1 \\ u &= \frac{2 + A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} - 1}{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} - 1} \\ u &= \frac{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} + 1}{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} - 1} \\ \sin{\theta} &= \frac{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} + 1}{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} - 1} \end{align*}$

$\begin{align*} \frac{2a}{\sqrt{e^2 - 4ac}}\left(B + \frac{e}{2a}\right) &= \frac{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} + 1}{A\exp{\left(\sqrt{e^2 - 4ac}\,t\right)} - 1} \end{align*}$

You could now write B in terms of t if you wish.