$$y'' +f(r)y' + g(r)y = 0$$
Is the form of the equation I'm given. I'm using $r$ instead of a more conventional $t$ or $x$ because it pertains to fluid dynamics with $r$ being some radius. I'm just wondering if I can simply use an auxiliary equation? Or is there more to it, I'm a little rusty at solving these
I don't know if it matters, but
$$f(r)= \frac {1-Re} {r}$$ and $$g(r) = -\frac {1+Re} {r^2}$$
($Re$ is the Reynolds number).
Any help is appreciated, thank you.
The form of the coefficient functions does matter: in the forms you've given, you can multiply by $r^2$ to turn the equation into one of the form $$ r^2y'' + ar y' + by = 0, $$ where $a,b$ are constants. This is called an Euler–Cauchy equation, which has a well-known form of general solution: try $y=r^m$, and you can derive a quadratic equation for $m$. With two distinct roots, this gives the general solution, $y=Ar^{m_1}+Br^{m_2}$. If the roots are the same, try $r^m \log{r}$ to find the other one. (The Wikipedia article linked explains how to arrive at these trial solutions.)