# Can this be integrated in any way?

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!

• do you need to solve it or is an approximation near a point satisfactory? – Saketh Malyala Jun 6 at 21:56
• I guess that could be helpful, do you mean expanding in a Taylor series? I have already numerically integrated it. Thanks! – Aye Prado Jun 7 at 12:19

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

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

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

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

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

• Thank you so so much! – Aye Prado Jun 10 at 12:48