Assuming the equation is $$ x \frac{d^2}{dx^2} y(x) + \frac{d}{dx} y(x) + x y(x) =0 $$ Dividing by $x$, checking if $x=0$ is regular singular point etc. I eventually, with $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+c}, \ c,a_n \ \ \text{t.b.d}, $$ arrive at $$ c^2a_0x^{c-1} + a_1 x^c (c+1)^2 + \sum_{m=1}^{\infty} x^{m+c} \left((m+c+1)^2 a_{m+1} + a_{m-1}\right) = 0, $$ I am pretty confident that this is correct so far, with the first independent solution being $$ y_1(x) = 1 + \sum_{m=1}^{\infty} \frac{(-1)^m}{m!^2} \left(\frac{x}{2}\right)^{2m}, $$ or in other words the Bessel function of order zero of first kind.
However, if I look at how one is supposed to get the second independent solution, that is $\frac{\partial y}{\partial c}$ (I know I could just use the bessel function of second kind or reduction of order, but I want to use that method), I see a problem.
The way we did it in class (here adapted to my example, hence the problem as you will see) was to expand the original ode like (with $\frac{d}{dx}\left(\frac{\partial}{\partial c}y\right) = \frac{\partial}{\partial c}\left(\frac{d}{dx}y\right)$, reccurance relation $= 0$,differentiating the right hand side with respect to $c$)
$$ x \frac{d^2}{dx^2}\left(\frac{\partial}{\partial c}y\right) + \frac{d}{dx}\left(\frac{\partial}{\partial c}y\right) + x \frac{\partial}{\partial c}y = a_0\left(2cx^{c-1}+c^2 x^{c-1}\ln x\right) + a_1\left(2(c+1)x^c + (c+1)^2x^c \ln x\right). $$
Now, for the example we did in class, the right hand side of the equation was only $$a_0\left(2cx^{c-1}+c^2 x^{c-1}\ln x\right)$$ which then with $c\rightarrow0$ obviously gives that $\frac{\partial}{\partial c}y|_{c=0}$ is a solution to the ode.
With the right hand side as is, however, after $c\rightarrow0$, there remains the $a_1$ term, which makes the following procedure useless (since that means that $\frac{\partial}{\partial c}y|_{c=0}$ is not a solution to the ode).
Any thoughts?