I am given $3x^2(x+2)y''+7xy'-2y=0, x \geq 0$. I am asked to solve this differential equation with a series solution around $x=0$. Note, however that $x=0$ is a regular singular point since:
$$ P(x) = 3x^2(x+2) \implies 3(0)^2(0+2)= 0, $$ $$ \lim_{x \to 0}x\frac{Q(x)}{P(x)} = \lim_{x \to 0}x\frac{7x}{3x^2(x+2)} = \frac{7}{6} \text{ and} $$ $$ \lim_{x \to 0}x^2\frac{R(x)}{P(x)} = \lim_{x \to 0}x^2\frac{-2}{3x^2(x+2)} = -\frac{1}{3} $$ are both finite. Thus we assume a solution of the form
$$ y = \sum_{n=0}^\infty a_n x^{n+r}, y' = \sum_{n=0}^\infty a_n(n+r) x^{n+r-1}, \\y''= \sum_{n=0}^\infty a_n(n+r)(n+r-1) x^{n+r-2} $$
Which, after some algebra, yields the indicial equation $$ (2r+1)(3r+5)=0 \implies r_1 = -\frac{1}{2}, r_2 = -\frac{3}{5} $$ And the recurrence relation $$ a_n=-\frac{-3a_{n-1}(n-3/2)(n-5/2)}{(6n+1)(n-1)}, \text{for} \ n \geq 1 $$ and $a_0$ is arbitrary. However, I soon realized that $a_1$ is undetermined as the denominator becomes 0. I was wondering if I got my algebra wrong or the entire approach is not correct. Thanks
The collected coefficients for the power $n+r$ are $$ 0=6(n+r)(n+r-1)a_n+3(n+r-1)(n+r-2)a_{n-1}+7(n+r)a_n-2a_n\\ =[6(n+r)^2+(n+r)-2]a_n+3(n+r-1)(n+r-2)a_{n-1} $$ with the convention that $a_{-1}=0$. Then the indicial equation for $n=0$ is $$ 0=6r^2+r-2=(2r-1)(3r+2) $$ with recursion $$ a_n=-\frac{3(n+r-1)(n+r-2)}{(2n+2r-1)(3n+3r+2)}. $$ The recursion for $r=\frac12$ reduces to $$ a_n=-\frac{3(2n-1)(2n-3)}{4n(6n+7)}a_{n-1}, $$ and for $r=-\frac23$ $$ a_n=-\frac{(3n-5)(3n-8)}{3n(6n-7)}a_{n-1}. $$ This gives totally problem free recursions for the coefficient sequences of the two basis solutions.