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In this site, Bessel differential equation is presented as:

$$x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + (x^2 - n^2)y = 0 \label{a} \tag{1}$$

and "equivalently, dividing through by $x^2$",

$$\frac{d^2y}{dx^2} + \frac{1}{x} \frac{dy}{dx} + \left(1 - \frac{n^2}{x^2} \right)y = 0 \label{b} \tag{2}$$

The unknown function is $y(x)$. For $x = 0$, the equation in the first form trivially becomes $y = 0$ (considering the parameter $n \neq 0$).

Why is $x = 0$ exluded and so why are \ref{a} and \ref{b} said to be equivalent?

BowPark
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Don't forget that $y = y(x)$ is dependent on $x$. So, when $x=0$, the equation does not trivially become $y = 0$. It becomes $\lim\limits_{x\to 0} x^2y''(x)+xy'(x)+(x^2-n^2)y(x) = 0$. This is because $y(x)$ is not necessarily defined at $x=0$. In fact, solving the first equation results in a solution which is not defined at $x=0$, so dividing by $x$ doesn't change the equation at all. Hence, they are equivalent.

Alex Jones
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  • First, thank you. I have still some doubts. First, ok, you can't state that $y$ is defined in $x = 0$, but can't even be sure that it is not defined. So we can't make assumptions on $y$ I guess. Moreover: "solving the first equation results in a solution which is not defined at $x=0$". How did you get to this conclusion? – BowPark Sep 21 '18 at 11:50
  • We can employ power series techniques to solve the first equation, and show that the series solution does not converge for $x=0$. We can also use the fact that the coefficient functions $x^2, x,$ and $x^2-n^2$ don't have any common factors between them, so zeros of the leading coefficient $x^2$ are called singular points; in this case, $x=0$. This fact is equivalent to the fact that, when dividing the equation by $x^2$, the coefficients are not all defined at $x=0$. If a point is singular, then the solution will not be defined at that point. Thus, the division is justified. – Alex Jones Sep 21 '18 at 17:57
  • This is very helpful to better understand your answer and thank you. I have a small knowledge about ODEs. If you are interested in these topics, check out this new question. – BowPark Sep 26 '18 at 10:12