[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\displaystyle\sum_{n\ge1}na_nx^{n-1}\implies4x\dfrac{\mathrm dy}{\mathrm dx}=4\sum_{n\ge1}na_nx^n=4\sum_{n\ge0}na_nx^n[/tex]
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n[/tex]
Substituting into the ODE
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}-4x\dfrac{\mathrm dy}{\mathrm dx}+12y=0[/tex]
gives
[tex]\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}-4na_n+12a_n\bigg)x^n=0[/tex]
so that the coefficients of the series are given according to
[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\dfrac{4(n-3)a_n}{(n+2)(n+1)}&\text{for }n\ge0\end{cases}[/tex]
We can shift the index in the recursive part of this definition to get
[tex]a_n=\dfrac{4(n-5)a_{n-2}}{n(n-1)}[/tex]
for [tex]n\ge2[/tex]. There's dependency between coefficients that are 2 indices apart, so we can consider 2 cases:
[tex]k=0\implies n=0\implies a_0=a_0[/tex]
but since [tex]y(0)=0[/tex], we have [tex]a_0=0[/tex] and [tex]a_{2k}=0[/tex] for all [tex]k\ge0[/tex].
[tex]k=0\implies n=1\implies a_1=a_1[/tex]
[tex]k=1\implies n=3\implies a_3=\dfrac{4(-2)a_1}{3\cdot2}=-\dfrac43a_1[/tex]
[tex]k=2\implies n=5\implies a_5=0[/tex]
and so [tex]a_{2k+1}=0[/tex] for all [tex]k\ge2[/tex]. If [tex]y'(0)=1[/tex], we then have [tex]a_1=1[/tex] and [tex]a_3=-\dfrac43[/tex].
So the ODE has solution
[tex]y(x)=\displaystyle\sum_{k\ge0}(a_{2k}x^{2k}+a_{2k+1}x^{2k+1})\implies\boxed{y(x)=x-\dfrac43x^3}[/tex]
