I want to solve this equation
$$ y'' + (\frac{1}{x} + 4x)y' + (5+4x^2)y = 0 $$ Where $y''$ is second derivative and so on.
This equation has singuar point at $x=0$. And this is regular singular point. So, I used Frobenius method http://en.wikipedia.org/wiki/Frobenius_method. When I find indicial (or characteristic) equation of the solution, $$ p(0)=\lim_{x \to 0} x \bigg (\frac{1}{x} + 4x \bigg) = 1 \\ q(0)=\lim_{x \to 0} x^2 \bigg (5+4x^2\bigg) = 0 \\ $$ Indicial equation $r(r-1) + p(0)r + q(0)=0$ becomes $ r^2=0 \implies r=0,0$.
Now, how to proceed with this equation? Any ideas, much appreciated.
I don't know how much this could help you but if you start setting $$y=e^{-x^2}z$$ the differential equation $$y'' + (\frac{1}{x} + 4x)y' + (5+4x^2)y = 0$$ reduces to $$x z''+z'+x z=0$$ which probably much easier to handle (it effectively corresponds to Bessel differential equation).