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?

Here is the procedure.

Denote the two roots by $r_1$ and $r_2$, with $r_1 \gt r_2$.

The Method of Frobenius will always generate a solution corresponding to $r_1$, but may generate a solution for the smaller second root $r_2$ of the indicial equation.

If the method fails for $r_2$, then an approach is to keep the recursion solution in terms of $r$ and use it to find the coefficients $a_n$ (for $n \ge 1$), in terms of both $r$ and $a_0$, where $a_0 \ne 0$. For ease, $a_0$ is typically chose to be one.

Using this more general form and the coefficients, the two independent solutions can be written as:

$$y_1(r, x) = x^r \sum_{n=0}^\infty ~ a_n(r)x^n = \sum_{n=0}^\infty ~ a_n(r)x^{n+r} \\y_2(r,x) = \dfrac{\partial}{\partial r}[(r - r_2)y_1(r, x)]~\Bigr|_{r=r_2}$$

You should be able to use this approach and show:

$$y(x) = y_1(x) + y_2(x) = \dfrac{c_1(3x(3x^2 + 3x+ 2) + 2)}{x^3} + \dfrac{c_2e^{3x}}{x^3}$$