We want to find a solution such that
[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]
[tex]y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n[/tex]
[tex]y''=\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n[/tex]
With these conditions, [tex]y(0)=a_0[/tex] and [tex]y'(0)=a_1[/tex].
Substituting the series into the ODE gives
[tex]\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n+3\sum_{n\ge0}(n+1)a_{n+1}x^n=0[/tex]
[tex]\displaystyle\sum_{n\ge0}\bigg((n+1)(n+2)a_{n+2}+3(n+1)a_{n+1}\bigg)x^n=0[/tex]
so that the coefficients of [tex]y[/tex] satisfy
[tex](n+1)(n+2)a_{n+2}+3(n+1)a_{n+1}=0[/tex]
for [tex]n\ge0[/tex], or
[tex](n-1)na_n+3(n-1)a_{n-1}=0\implies a_n=-\dfrac3na_{n-1}[/tex]
for [tex]n\ge2[/tex]. Notice that this implies a dependency of all [tex]a_n[/tex] beyond [tex]n=1[/tex] on [tex]a_1[/tex], while [tex]a_0[/tex] takes on whatever initial value [tex]y(0)[/tex] is given. In particular,
[tex]a_2=-\dfrac32a_1[/tex]
[tex]a_3=-\dfrac33a_2=\dfrac{3^2}{3\cdot2}a_1=\dfrac{3^2}{3!}a_1[/tex]
[tex]a_4=-\dfrac34a_3=-\dfrac{3^3}{4\cdot3\cdot2}a_1=-\dfrac{3^3}{4!}a_1[/tex]
and so on up to
[tex]a_n=\dfrac{(-3)^{n-1}}{n!}a_1[/tex]
So we can extract two fundamental solutions [tex]y_1,y_2[/tex] such that [tex]y=y_1+y_2[/tex], where
[tex]y_1=a_0[/tex]
[tex]y_2=\displaystyle-3a_1\sum_{n\ge1}\frac{(-3x)^n}{n!}[/tex]
Recall that
[tex]e^x=\displaystyle\sum_{n\ge0}\frac{x^n}{n!}[/tex]
which tells us
[tex]y_2=-3a_1(e^{-3x}-1)=-3a_1e^{-3x}+3a_1[/tex]
but [tex]y_1[/tex] is a constant solution and already accounts for the constant term in [tex]y_2[/tex], and [tex]-3a_1[/tex] can be reduced to a simpler constant [tex]a_1[/tex], leaving us with
[tex]y_2=a_1e^{-3x}[/tex]
The Wronskian is
[tex]W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=\begin{vmatrix}a_0&a_1e^{-3x}\\0&-3a_1e^{-3x}\end{vmatrix}=-3a_0a_1e^{-3x}[/tex]
[tex]\implies W(y_1,y_2)(0)=-3a_0a_1[/tex]
so the two solutions are indeed independent as long as neither initial value is 0.
