From Poisson's first Memoire sur la distribution d'electricite (1812), $\S$39. Suppose $b$ is very small relative to $c-a$ where $a,b,c$ are positive real scalars. Consider the equations \begin{align*} af(x)+\frac{b^2}{c-ax} F\left(\frac{b}{c-ax}\right)&=h,\\ bF(x)+\frac{a^2}{c-bx}f\left(\frac{a}{c-bx}\right)&=g; \end{align*} for arbitrary functions $f(x)$ and $F(x)$ (representing the "number of electric molecules" on a surface in the original memoir), for $x\in[-1,1]$. The variables $g$ and $h$ are arbitrary constants. With the assumption that $b$ is very smaller relative to $c-a$, we may neglect the cubes of $\frac{b}{c}$.
Given the above, Poisson says that the expansion of $bF(x)$ (presumably about $x=\frac{a}{c}$) takes the form: \begin{align*} bF&(x)=g-\frac{a^2}{c}f\left(\frac{a}{c}\right)-\left[\frac{a^2}{c}f\left(\frac{a}{c}\right)+\frac{a^3}{c^2}f'\left(\frac{a}{c}\right)\right]\frac{bx}{c}\\ &-\left[\frac{a^2}{c}f\left(\frac{a}{c}\right)+\frac{2a^3}{c^2} f'\left(\frac{a}{c}\right)+\frac{a^4}{2c^3}f''\left(\frac{a}{c}\right)\right]\frac{b^2x}{c^2}; \end{align*}
Some notes before I go on: Poisson used the notation $fx$ for function $f$, which I have assumed is intended to represent the form $f(x)$. There may be other results related to the above series expansion earlier (some time after $\S$15) but I haven't found anything.
Now, the presence of derivatives brings Taylor series to mind, but expanding about $x=\frac{c}{a}$ gives $$ bF(x) \approx g-\frac{a^2}{c-bx}\left[ f\left(\frac{c}{a}\right)+f'\left(\frac{c}{a}\right) \left(x-\frac{c}{a}\right)+\frac{1}{2} f''\left(\frac{c}{a}\right) \left(x-\frac{c}{a}\right)^2 \right] $$ Which is far different.
Question: How do you go from the above equations to the expansion for $bF(x)$ given?
For context, I'm writing a translation of the memoir (which is available here). The closest to useful result I've found is this equation: $$ af\left(\frac{x}{a}\right)-\frac{a^2b}{c^2-b^2-cx}f\left(\frac{ac-ax}{c^2-b^2-cx}\right)=h-\frac{gb}{c-x}\quad (3) $$ An equation which Poisson says must be integrated, in general, to solve for $f(x)$, which can then be used to solve $F(x)$. But it's not clear still what is happening above.
The expansion is around $x=0$. Then $$bF(x)=g-\frac{a^2}{c-bx}f\left(\frac a{c-bx}\right)$$ Expanding in Taylor series at $x=0$, the constant term is $$g-\frac{a^2}cf\left(\frac ac\right)$$ The linear term is given by the first derivative: $$\left(g-\frac{a^2}{c-bx}f\left(\frac a{c-bx}\right)\right)'=\frac{a^2}{(c-bx)^2}(-b)f\left(\frac a{c-bx}\right)-\frac{a^2}{c-bx}f'\left(\frac a{c-bx}\right)\left(\frac a{c-bx}\right)'\\=-\frac{a^2b}{(c-bx)^2}f\left(\frac a{c-bx}\right)-\frac{a^3b}{(c-bx)^3}f'\left(\frac a{c-bx}\right)$$ Taking the value at $x=0$ you get $$\left(g-\frac{a^2}{c-bx}f\left(\frac a{c-bx}\right)\right)'_{x=0}=-\frac{a^2b}{c^2}f\left(\frac a{c}\right)-\frac{a^3b}{c^3}f'\left(\frac a{c}\right)$$ You can do the same for the next derivative. Up to the first order, $$bF(x)=g-\frac{a^2}cf\left(\frac ac\right)-\left[\frac{a^2b}{c^2}f\left(\frac a{c}\right)+\frac{a^3b}{c^3}f'\left(\frac a{c}\right)\right]x\\=g-\frac{a^2}cf\left(\frac ac\right)-\left[\frac{a^2}{c}f\left(\frac a{c}\right)+\frac{a^3}{c^2}f'\left(\frac a{c}\right)\right]\frac{bx}c$$