Assume a solution of the form
[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
with derivatives
[tex]y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n[/tex]
[tex]y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n[/tex]
Substituting into the ODE, which appears to be
[tex](x^2-4)y''+3xy'+y=0,[/tex]
gives
[tex]\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}x^{n+2}-4(n+2)(n+1)a_{n+2}x^n+3(n+1)a_{n+1}x^{n+1}+a_nx^n\bigg)=0[/tex]
[tex]\displaystyle\sum_{n\ge2}n(n-1)a_nx^n-4\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n+3\sum_{n\ge1}na_nx^n+\sum_{n\g0}a_nx^n=0[/tex]
[tex](a_0-8a_2)+(4a_1-24a_3)x+\displaystyle\sum_{n\ge2}\bigg[(n+1)^2a_n-4(n+2)(n+1)a_{n+2}\bigg]x^n=0[/tex]
which gives the recurrence for the coefficients [tex]a_n[/tex],
[tex]\begin{cases}a_0=a_0\\a_1=a_1\\4(n+2)a_{n+2}=(n+1)a_n&\text{for }n\ge0\end{cases}[/tex]
There's dependency between coefficients that are 2 indices apart, so we consider 2 cases.
[tex]k=0\implies n=0\implies a_0=a_0[/tex]
[tex]k=1\implies n=2\implies a_2=\dfrac1{4\cdot2}a_0=\dfrac2{4\cdot2^2}a_0=\dfrac{2!}{2^4}a_0[/tex]
[tex]k=2\implies n=4\implies a_4=\dfrac3{4\cdot4}a_2=\dfrac3{4^2\cdot4\cdot2}a_0=\dfrac{4!}{2^8(2!)^2}a_0[/tex]
[tex]k=3\implies n=6\implies a_6=\dfrac5{4\cdot6}a_4=\dfrac{5\cdot3}{4^3\cdot6\cdot4\cdot2}a_0=\dfrac{6!}{2^{12}(3!)^2}a_0[/tex]
and so on, with the general pattern
[tex]a_{2k}=\dfrac{(2k)!}{2^{4k}(k!)^2}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{4\cdot3}a_1=\dfrac{2^2}{2^2\cdot3\cdot2}a_1=\dfrac1{(3!)^2}a_1[/tex]
[tex]k=2\implies n=5\implies a_5=\dfrac4{4\cdot5}a_3=\dfrac{4\cdot2}{4^2\cdot5\cdot3}a_1=\dfrac{(2!)^2}{5!}a_1[/tex]
[tex]k=3\implies n=7\implies a_7=\dfrac6{4\cdot7}a_5=\dfrac{6\cdot4\cdot2}{4^3\cdot7\cdot5\cdot3}a_1=\dfrac{(3!)^2}{7!}a_1[/tex]
and so on, with
[tex]a_{2k+1}=\dfrac{(k!)^2}{(2k+1)!}a_1[/tex]
Then the two independent solutions to the ODE are
[tex]\boxed{y_1(x)=\displaystyle a_0\sum_{k\ge0}\frac{(2k)!}{2^{4k}(k!)^2}x^{2k}}[/tex]
and
[tex]\boxed{y_2(x)=\displaystyle a_1\sum_{k\ge0}\frac{(k!)^2}{(2k+1)!}x^{2k+1}}[/tex]
By the ratio test, both series converge for [tex]|x|<2[/tex], which also can be deduced from the fact that [tex]x=\pm2[/tex] are singular points for this ODE.
