Find the recurrence relationship and the general solution at the point $xy^{''}+y^{'}+xy=0$ at $x_{0}=0$
I have solved it to a certain point, but I want to know if I am good to continue.
$y(x)=\sum_{n=0}^∞a_{n}x^{n}$
$y′(x)=\sum_{n=0}^∞na_{n}x^{n−1}$
$y″(x)=\sum_{n=0}^∞n(n−1)a_{n}x^{n−2}$
$$ a_1 + \sum_{n=2}^{\infty} (n^2a_n + a_{n-2})x^{n-1}=0$$ then $a_1=0$. Recursive relation is $$n^2a_n = -a_{n-2}$$ Take $a_0 = a_0$ and $a_1 = a_1$
$a_2 = \displaystyle\frac{-a_0}{2^2}$
$a_3 = \displaystyle\frac{-a_1}{3^2}$
$a_4 = \displaystyle\frac{a_0}{2^2.4^2}$
$a_5 = \displaystyle\frac{a_1}{3^2.5^2}$
$a_6 = \displaystyle\frac{-a_0}{2^2.4^2.6^2}$
$a_7 = \displaystyle\frac{-a_1}{3^2.5^2.7^2}$
Hence solution will be
$$y(x) = a_0 + a_0\sum_{k=1}^{\infty} \frac{(-1)^kx^{2k}}{\Pi_{n=1}^{k} (2n)^2}$$
Excuse me if I have mistakes I'm starting on the subject. Thank you in advance for your help.
Hint $$xy^{''}+y^{'}+xy=0$$ multiply by x $$x^2y^{''}+xy^{'}+x^2y=0$$ Thats Bessel's equation of order zero
The solution is $$y_1=J_0=\sum_{k=0}^{\infty} \frac{(-1)^k(x/2)^{2k}}{(k!)^2}$$ The second solution of your equation should have the form since $r=0$ is a double root $$y_2=y_1\ln(x)+\sum_{n=1}^\infty b_nx^n$$ And then $$y(x)=C_1y_1(x)+C_2y_2(x)$$