Second solution of ODE $xy''+y'-y=0$? Suppose we have the following equation $$xy''+y'-y=0$$ where it has a regular singular point at $x=0$ and we want to derive the series solution near $x=0$. We write the ODE in canonical form: $y''+\frac{1}{x}y'-\frac{1}{x}y=0$ and then we set $$y=\sum_{n=0}^{\infty}a_nx^{n+r}.$$ This will gives us $$y=\sum_{n=0}^{\infty}a_n(n+r)(n+r-1)x^{n+r-1}+a_n(n+r)x^{n+r-1}-a_nx^{n+r}.$$ We continue deriving the indicial equation from the coefficients of the lowest power of $x$ and thus $$a_0r(r-1)+a_0r=0\implies r^2=0$$ and so $r=0$ (repeated). Also we have for $r=0,~a_n=\frac{a_{n-1}}{n^2}=\ldots=\frac{a_0}{(n!)^2}.$ Hence $$y_1=\sum_{n=0}^{\infty}\frac{x^n}{(n!)^2}.$$ For the second solution $y_2$ we may proceed either with the Wronskian technique (in this case it's rather difficult to do that partly because the calculations are quite hard to do) or by differentiating with respect to $r$. But I struggle to understand the steps and exactly what to do. So we have $$L[y]=\sum_{n=0}^{\infty}a_n(n+r)^2x^{n+r-1}-a_nx^{n+r}.$$ But how do we continue from here to get an equation of $L[y]$ and differentiate it w.r.t $r$.
Thank you in advance for your help.

$$xy'' +y' -y=0\qquad ......(1)$$
$~x=0~$ is a regular singular point of equation $(1)$.
So the equation admits of a Frobenius series of the form $$y=\sum_{n=0}^{\infty}C_n~x^{n+r},\qquad C_0\neq 0 \qquad ..........(2)$$ 
which converges for all $~x~$.
From $(2)$,
$$y'(x)=\sum_{n=0}^{\infty}(n+r)C_n~x^{n+r-1};\qquad \qquad y''(x)=\sum_{n=0}^{\infty}(n+r-1)(n+r)C_n~x^{n+r-2}\qquad .....(3)$$
Substituting $(2)$ and $(3)$ in $(1)$ we get,
$$x~\sum_{n=0}^{\infty}(n+r-1)(n+r)C_n~x^{n+r-2}+\sum_{n=0}^{\infty}(n+r)C_n~x^{n+r-1}-\sum_{n=0}^{\infty}C_n~x^{n+r}=0$$
$$\implies \sum_{n=0}^{\infty}(n+r)^2~C_n~x^{n+r-1}~-~\sum_{n=0}^{\infty}C_n~x^{n+r}=0\qquad .....(4)$$
Lowest power of $~x~$ in equation $(4)$ is $~{r-1}~$, so coefficient of $~x^{r-1}~=0$ gives the  indicial equation $~r^2~=0\implies r=0,~0$
From equation $(4)$ we have the following recursive formula,
$$(n+r+1)^2~C_{n+1}~-~C_{n}=0$$
$$\implies C_{n+1}=\frac{1}{(n+r+1)^2}~C_{n}\qquad ........(5)$$
From $(5)$ we have 
$C_1=\frac{1}{(r+1)^2}~C_{0}$
$C_2=\frac{1}{(r+2)^2}~C_{1}=\frac{1}{(r+1)^2~(r+2)^2}~C_{0}$
$C_3=\frac{1}{(r+3)^2}~C_{2}=\frac{1}{(r+1)^2~(r+2)^2~(r+3)^2}~C_{0}$
$\cdots$
Therefore 
$$y(x)=C_0~x^r \left[1+\frac{1}{(r+1)^2}~x+\frac{1}{(r+1)^2~(r+2)^2}~x^2+\frac{1}{(r+1)^2~(r+2)^2~(r+3)^2}~x^3~+\cdots\right]$$
For $~r=0~$, $$y_1(x)= \left[1+~x~+\frac{x^2}{4}+\frac{x^3}{36}+\cdots\right]$$
$$\implies y_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{(n!)^2}=J_0 (2\sqrt{x})$$
$J_0(X)~$ is the modified Bessel function of first kind and order $~0~$.
The other independent solution of equation $(1)$ is  $$y_2(x)=\left[\frac{\partial y}{\partial r}\right]_{r=0}$$
$$\implies y_2(x)=y_1(x)~\log x~-~\left[2~x+\frac{3}{4}~x^2~+\cdots\right]$$
$$\implies y_2(x)=Y_0 (2\sqrt{x})$$
$Y_0(X)~$ is the modified Bessel function of second kind and order $~0~$.
General solution is $$y(x)=A~y_1(x)~+~B~y_2(x)\qquad \text{where $~A,~B~$are constants.}$$
