Frobenius solution of $xy"+y=0$

Question

$$xy"+y=0$$

Since $x=0$ is singular point: $y(x)=\sum_{n=0}^\infty a_nx^{n+r}$

$$\implies \sum_{n=0}^\infty a_n(n+r)(n+r-1)x^{n+r-1}+\sum_{n=0}^\infty a_nx^{n+r}=0$$

How to proceed further?

Answer 1

Rewrite the first series so that the powers match the other:

$$\sum_{n=0}^\infty a_n(n+r)(n+r-1)x^{n+r-1}=a_0r(r-1)x^{r-1}+\sum_{n=0}^\infty a_{n+1}(n+r+1)(n+r)x^{n+r}$$

so that we get

$$a_0r(r-1)x^{r-1}+\sum_{n=0}^\infty[a_{n+1}(n+r+1)(n+r)+a_n]x^{n+r}=0$$

implying that $r=0$ (appears twice modulo 1) so that we will need to use this solution to form a second solution, and that we have

$$a_{n+1}=-\frac{a_n}{(n+r+1)(n+r)}=-\frac{a_n}{n(n+1)}$$

$$a_n=\frac{(-1)^n}{n!(n+1)!}a_0$$

which gives a solution in terms of the Bessel function of the first kind. Since the root is repeated, we then consider

$$y_2=Cy_1\ln(x)+\sum_{n=0}^\infty b_nx^{n+r_2}$$