I am trying to understand how to get "another solution" from an ODE problem. I was reading that for example in the Bessel function of order zero (i.e. when v = 0 in the following equation, so it says it has a repeated root):
$L[y] = x^2 y'' + x y' + (x^2 - v^2)y = 0$,
We are given that one solution is found to be: $y_1(x) =a_0[1+ \sum_{m=1}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m}(m!)^2}] $
And then, it says to find anther solution, we could differentiate the original ODE equation and set the r in the newly differentiated equation to be the value of the root that solves the indicial equation. Then we will get the second solution this way.
But I don't quite understand why this procedure would yield another solution in this case. The only explanation I got from the book is that since the derivative of the indicial equation at the first root is also zero, so we could differentiate the original ODE and set the r (the parameter in the indicial equation) to be the root of the indicial equation in this new equation, and this will give a second solution. But I don't quite understand why.
