Find the form of a second linear independent solution when the two roots of indicial equation are different by a integer Consider the differential equation
$$x^2y''+3(x-x^2)y'-3y=0$$
$(a)$ Find the recurrence equation and first three nonzero terms of the series solution in powers of $$ corresponding to the larger root of the indicial equation.
$(b)$ What would be the form of a second linearly independent solution of this differential equation?
I found the indicial equation to be $r(r-1)+3r-3 = 0$, so the two roots are $r_1=-3$ and $r_2=1$.
And the recurrence relation is $a_n = \dfrac{3(n-1+r)a_{n-1}}{(n+r)(n+r-1)+3(n+r)-3}$
set $r=1$, then $a_n = \dfrac{3na_{n-1}}{(n+3)(n+1)-3}$. And I can figure out first linearly independent solution.
But for $r=-3$,then $a_n = \dfrac{3(n-4)a_{n-1}}{(n-3)(n-1)-3}$, if let $a_0$ be an arbitrary constant, then $a_1 =\dfrac{3(-3)a_0}{(-2)(0)}$, which doesn't work.
Then how do I figure out the second linearly independent solution?

Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$ ,
Then $y'=\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}$
$y''=\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}$
$\therefore x^2\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r-2}+3(x-x^2)\sum\limits_{n=0}^\infty(n+r)a_nx^{n+r-1}-3\sum\limits_{n=0}^\infty a_nx^{n+r}=0$
$\sum\limits_{n=0}^\infty(n+r)(n+r-1)a_nx^{n+r}+\sum\limits_{n=0}^\infty3(n+r)a_nx^{n+r}-\sum\limits_{n=0}^\infty3(n+r)a_nx^{n+r+1}-\sum\limits_{n=0}^\infty 3a_nx^{n+r}=0$
$\sum\limits_{n=0}^\infty(n+r+3)(n+r-1)a_nx^{n+r}-\sum\limits_{n=1}^\infty3(n+r-1)a_{n-1}x^{n+r}=0$
$(r+3)(r-1)a_0x^r+\sum\limits_{n=1}^\infty((n+r+3)(n+r-1)a_n-3(n+r-1)a_{n-1})x^{n+r}=0$
The indicial equation is $(r+3)(r-1)=0$ , the two roots are $r_1=-3$ and $r_2=1$ .
The recurrence relation is $(n+r+3)(n+r-1)a_n-3(n+r-1)a_{n-1}=0$ , i.e. $(n+r+3)(n+r-1)a_n=3(n+r-1)a_{n-1}$
Each root from the indical equations may not always only find one group of the linearly independent solutions, sometimes we can find more than one group of the linearly independent solutions at the same time.
When we choose $r=-3$ , the recurrence relation becomes $n(n-4)a_n=3(n-4)a_{n-1}$ , which has a value of $n$ at $n=4$ so that it is independent to the recurrence relation. Besides $a_0$ can chooes arbitrary, $a_4$ can also chooes arbitrary. This makes the effect that we can find two groups of the linearly independent solutions that basing on $n=0$ to $n=3$ and $n=4$ to $n=+\infty$ respectively.
