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I am doing a model of something, and this differential equation popped up. It is not separable, not exact. I can't see any way I could integrate this. I am not asking someone to integrate it for me but just to guide me in the right direction. I am not even sure it can be integrated explicitly.This is the equation:

$$ \frac{dx}{dt}=\frac{1}{(1-x)x}\left(\frac{a-x}{1-a}+b e^{-t}\right) $$

Thanks!

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  • $\begingroup$ do you need to solve it or is an approximation near a point satisfactory? $\endgroup$ – Saketh Malyala Jun 6 at 21:56
  • $\begingroup$ I guess that could be helpful, do you mean expanding in a Taylor series? I have already numerically integrated it. Thanks! $\endgroup$ – Aye Prado Jun 7 at 12:19
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Assume $b\neq0$ and $a\neq1$ for the key case:

Hint:

$\dfrac{dx}{dt}=\dfrac{1}{(1-x)x}\left(\dfrac{a-x}{1-a}+be^{-t}\right)$

$\left(be^{-t}+\dfrac{x-a}{a-1}\right)\dfrac{dt}{dx}=x(1-x)$

Let $u=e^{-t}$ ,

Then $t=-\ln u$

$\dfrac{dt}{dx}=-\dfrac{1}{u}\dfrac{du}{dx}$

$\therefore\left(bu+\dfrac{x-a}{a-1}\right)\left(-\dfrac{1}{u}\dfrac{du}{dx}\right)=x(1-x)$

$\left(u+\dfrac{x-a}{(a-1)b}\right)\dfrac{du}{dx}=\dfrac{x(x-1)u}{b}$

This belongs to an Abel equation of the second kind.

Let $v=u+\dfrac{x-a}{(a-1)b}$ ,

Then $u=v-\dfrac{x-a}{(a-1)b}$

$\dfrac{du}{dx}=\dfrac{dv}{dx}-\dfrac{1}{(a-1)b}$

$\therefore v\left(\dfrac{dv}{dx}-\dfrac{1}{(a-1)b}\right)=\dfrac{x(x-1)}{b}\left(v-\dfrac{x-a}{(a-1)b}\right)$

$v\dfrac{dv}{dx}-\dfrac{v}{(a-1)b}=\dfrac{x(x-1)v}{b}-\dfrac{x(x-1)(x-a)}{(a-1)b^2}$

$v\dfrac{dv}{dx}=\dfrac{((a-1)x^2-(a-1)x+1)v}{(a-1)b}-\dfrac{x(x-1)(x-a)}{(a-1)b^2}$

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  • $\begingroup$ Thank you so so much! $\endgroup$ – Aye Prado Jun 10 at 12:48

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