I guess the ODE is supposed to be
[tex]y''+2y=0[/tex]
So if
[tex]y=\displaystyle\sum_{n\ge0}a_nz^n\implies y''=\sum_{n\ge0}(n+2)(n+1)a_{n+2}z^n[/tex]
then
[tex]\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}z^n+2\sum_{n\ge0}a_nz^n=0[/tex]
[tex]\displaystyle\sum_{n\ge0}\bigg[(n+2)(n+1)a_{n+2}+2a_n\bigg]z^n=0[/tex]
so that
[tex](n+2)(n+1)a_{n+2}+2a_n=0[/tex]
for [tex]n\ge0[/tex], or equivalently,
[tex]n(n-1)a_n+2a_{n-2}=0\implies a_n=-\dfrac2{n(n-1)}a_{n-2}[/tex]
for [tex]n\ge2[/tex]. Note the dependency between every other coefficient - this means we can consider two cases:
[tex]k=0\implies n=0\implies a_0=a_0[/tex]
[tex]k=1\implies n=2\implies a_2=-\dfrac2{2\cdot1}a_0[/tex]
[tex]k=2\implies n=4\implies a_4=-\dfrac2{4\cdot3}a_2=\dfrac{(-2)^2}{4!}a_0[/tex]
[tex]k=3\implies n=6\implies a_6=-\dfrac2{6\cdot5}a_4=\dfrac{(-2)^3}{6!}a_1[/tex]
and so on, with
[tex]a_{2k}=\dfrac{(-2)^k}{(2k)!}a_0[/tex]
[tex]k=0\implies n=1\implies a_1=a_1[/tex]
[tex]k=1\implies n=3\implies a_3=-\dfrac2{3\cdot2}a_1[/tex]
[tex]k=2\implies n=5\implies a_5=-\dfrac2{5\cdot4}a_3=\dfrac{(-2)^2}{5!}a_1[/tex]
and so on, with
[tex]a_{2k+1}=\dfrac{(-2)^k}{(2k+1)!}a_1[/tex]
Then the ODE has solution
[tex]\displaystyle y(x)=\sum_{k\ge0}(a_{2k}z^{2k}+a_{2k+1}z^{2k+1})[/tex]
and the first six non-zero terms occur for [tex]0\le n\le5[/tex], for which we get
[tex]\boxed{a_0\left(1-x^2+\dfrac{x^4}6\right)+a_1\left(x-\dfrac{x^3}3\right)}[/tex]
