Frobenius Method general solution not providing an expected solution I'm trying to solve the following ODE:
$$3xy''+(3x+1)y'+y=0$$
So I started by assuming a solution(and its derivates) with the form:
$$
\tag{1}y = \sum_{n=0}^{\infty } a_{n} x^{n + r}$$
$$y' = \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}$$
$$y'' = \sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-2}$$
Replacing in original equation:
$$3x\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-2} + 3x\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}+\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + \sum_{n=0}^{\infty } a_{n} x^{n + r} = 0$$
$$ 3\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} + \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r} + \sum_{n=0}^{\infty } a_{n} x^{n + r} = 0   $$
To transform all $x^{n+r}$ into $x^{n+r-1}$ I've rewritten the equation as:
$$
3\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} +\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} +  3\sum_{n=1}^{\infty } {(n+r-1)}a_{n-1} x^{n + r -1} + \sum_{n=1}^{\infty } a_{n-1} x^{n + r - 1} = 0 $$
In order to make all sums start from the $n=1$:
$$ \tag{2}
 3r(r-1)a_{0}x^{r-1} + ra_{0}x^{r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} + \sum_{n=1}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)}a_{n-1} x^{n + r -1} + \sum_{n=1}^{\infty } a_{n-1} x^{n + r - 1} = 0   
$$
If I understood correctly, the indicial equation should then be:
$$ 3r(r-1)a_{0}x^{r-1} + ra_{0}x^{r-1} = 0 $$
and since $x^{r-1} \neq 0 $ and $a_{0} \neq 0 $:
$$ 3r(r-1) + r = 0 $$
$$ r_{1} = 0 $$
$$ r_{2} = \frac{2}{3} $$
From $(2)$:
$$ [3(n+r)(n+r-1)+(n+r)]a_{n} + [3(n+r-1)+1]a_{n-1} = 0
$$
which seems to give for both solutions of $r$:
$$ a_{n} = \frac{-a_{n-1}}{n}  $$
It seems to me that this is in terms of $a_{0}$:
$$ a_{n} = \frac{(-1)^{n}}{n!}a_{0} $$
So rewriting $(1)$ now would give me the solution and $y$ would be an infinite sum since all $a_{n} \neq 0$. And I know this can't be true because by deduction one can see that $y = e^{-x}$ is a solution to the ODE, which makes me think I made a mistake in some of the steps shown above, I'd be glad if anyone could show me where I made the mistake and give me some help in finding the correct general solution for this ODE.

For solution $r_1$: $$a_n=\frac{(-1)^n}{n!}a_0$$$$y_1=a_0(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots)\ =\ a_0e^{-x}$$
For solution $r_2$: $$b_n=\frac{(-3)^n}{(5)(8)(11)\cdots(3n+2)}b_0$$
$$y_2=b_0(1-\frac{3}{5}x+\frac{9}{40}x^2-\frac{27}{440}x^3+\cdots)$$
The general solution is  $\ y_1+y_2$.
