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?
Best Answer
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.