I have tried for the power series but when suppose the solution is $y(x)=\sum _{n=0}^\infty a_nx^n$ appeared sums dependents of $a_n$, $a_{n+2}$ and $a_{n-1}$ and this is not possible.


Taking $$ \begin{align} y&=\sum_{n=0}^\infty a_nx^n\\ y'&= \sum_{n=1}^\infty na_nx^{n-1}\\ y''&= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}\\ \end{align}$$

Then $$(1+ x^2)y''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=2}^\infty n(n-1)a_nx^n$$ and the equation is $$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=2}^\infty n(n-1)a_nx^n- \sum_{n=0}^\infty 6a_nx^{n+2}= 0$$

We can shift the dummy indices, taking $i= n- 2$ in the first sum, $i= n$ in the second sum and $i= n+ 2$ in the last sum: $$\sum_{i=0}^\infty (i+2)(i+1)a_{i+1}x^i+ \sum_{i= 2}^\infty i(i-1)a_{i}x^i- \sum_{i= 2}6a_{i-2}x^i= 0$$

Taking $i= 0$, we have $2a_1= 0$ so $a_1= 0$. Taking $i= 1$, $6a_2= 0$ so $a_2= 0$. For $i> 1$, we have $$(i+2)(i+1)a_{i+1}+ i(i-1)a_i- 6a_{i-2}= 0$$ So $$a_{i+1}= \frac{6a_{i-2}- i(i-1)a_i}{(i+2)(i+1)}$$ Since this is a second order linear differential equation, the general solution will have two undetermined constants. Since $a_1= a_2= 0$ we can take $a_3$ and $a_4$ as those constants and use that recursion to determine all $a_n$ for $n> 4$.

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    $\begingroup$ but in the last sum is $-\sum _{n=0}^\infty 6a_nx^{n+1}$ and you wrote $-\sum _{n=0}^\infty 6a_nx^{n+2}$ $\endgroup$ – Gabriel Chileider May 23 '18 at 17:16

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