I am dealing with Bessel's equation of order $0$.
$$x\frac{d^2y}{dx^2}+\frac{dy}{dx}+xy=0$$
We assume that this is a first solution to the Bessel's equation,
$$J_0(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(n!)^2}({\frac{x}{2}})^{2n}$$
Writing out a few terms
$$J_0(x)=1-\frac{x^2}{4}+\frac{x^4}{64}-\frac{x^6}{2304}+...$$
We want to seek another linearly independent solution to the differential equation,
Now using method of reduction of order we found that,
$$y_2(x)=J_0(x)\int \frac{e^{-\int\frac{dx}{x}}}{[J_0(x)]^2}dx$$
$$y_2(x)=J_0(x)\int \frac{dx}{x[J_0(x)]^2}$$
$$[J_0(x)]^2=1-\frac{x^2}{2}+\frac{3x^4}{32}-\frac{5x^6}{576}+...$$
$$\frac{1}{[J_0(x)]^2}=1+\frac{x^2}{2}+\frac{5x^4}{32}+\frac{23x^6}{576}+...$$
I want to know how to get the reciprocal?
Other than long division, I prefer undermined coefficient:
Let $a_0, a_1, a_2, ....$ be given, we wish to find $b_0, b_1, b_2, ...$ such that $$(a_0+a_1x+a_2x^2+a_3x^3+...)(b_0+b_1x+b_2x^2+b_3x^3+...) = 1$$
Such $b_i$'s can always be sought given that $a_0 \neq 0$.
Comparing the constant term: $a_0b_0 = 1$, since $a_0 \neq 0$, we can find $b_0$
Comparing the coefficient of $x$: $a_0b_1+b_0a_1 = 0$, since we already found $b_0$ and $a_0, a_1$ is known from the start, we can find $b_1$ from this equation.
Comparing the coefficient of $x^2$: $a_0b_2+b_1a_1+b_0a_2= 0$, since we already found $b_0, b_1$ and $a_0, a_1, a_2$ is known from the start, we can find $b_2$ from this equation.
Continue this process, you can find all $b_i$'s, they might not have a nice general formula, but they can always be found.