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?