$$k(x)y'' + p(x)y' + q(x)y = 0$$
its a simple form of singularity
$p(x) = \lim\limits_{x\to 0}\left(x\times\dfrac{p(x)}{k(x)}\right)=p_0 $
$q(x) = \lim\limits_{x\to 0}\left(x^2\times\dfrac{q(x)}{k(x)}\right)=q_0 $
as known we find roots from these limits equations and make index equation.
And these roots must be real number for regular singularity. My question is what if roots wouldn't be real number ? What would happen if we found one of these limits goes to infinity ?
Its like;
$$(x^2)y'' + (x^3 + 2)y' - y = 0$$
We find here $ p_0 = x\times\dfrac{x^3 + 2}{x^2} = x^2 + \dfrac{2}{x} = \infty $ at $ x_0=0 $
and we stuck here.. Is there someone have idea for this solution?
If you help me I'd be appreciate.. Best regards!