Singularities classification in DE's

[fluidistic]

According to Mathworld, if in $y''+P(x)y'+Q(x)y=0$, P diverges at $x=x_0$ quicker than $\frac{1}{(x-x_0)}$ or Q diverges at $x=x_0$ quicker than $\frac{1}{(x-x_0)^2}$ then $x_0$ is called an essential singularity.
What I don't understand is that let's suppose Q diverges like $\frac{1}{(x-x_0)^5}$. In that case x_0 would be called an essential singularity. But what I don't understand is that to me it looks like a pole of order 5, not an essential singularity (pole of order infinity).
Am I missing something?

 

[Dickfore]

If $x_0$ is a pole of order k for the function, then it should behave as:
$$y(x) \sim \frac{1}{(x - x_0)^k}$$
then:
$$y'(x) \sim \frac{-k}{(x - x_0)^{k + 1}}$$
and
$$y''(x) \sim \frac{k (k + 1)}{(x - x_0)^{k + 2}}$$
How do you propose to make an equality using:
$$Q(x) \, y(x) \sim \frac{1}{(x - x_0)^{k + 5}}$$
?

 

[fluidistic]

If $x_0$ is a pole of order k for the function, then it should behave as:
$$y(x) \sim \frac{1}{(x - x_0)^k}$$
then:
$$y'(x) \sim \frac{-k}{(x - x_0)^{k + 1}}$$
and
$$y''(x) \sim \frac{k (k + 1)}{(x - x_0)^{k + 2}}$$
How do you propose to make an equality using:
$$Q(x) \, y(x) \sim \frac{1}{(x - x_0)^{k + 5}}$$
?

Hmm I don't really understand your question. We're talking about a pole/singularity for Q or P right? Not y(x)... or I'm wrong on this?
So it would be "let's say Q(x) behaves like $\frac{1}{(x-x_0)^5}$. It's an essential singularity because it diverges quicker than $\frac{1}{(x-x_0)^2}$ when $x$ tends to $x_0$."
But if I use the definition of a pole of order n for a function, namely that $\lim _{x\to x_0} (x-x_0)^nf(x)$ is differentiable at $x=x_0$, where n is the smallest integer and where $f(x)=Q(x)$, I get that $\lim _{x\to x_0} (x-x_0)^5Q(x)=1$ which is clearly differentiable at $x=x_0$. For n=4, it isn't differentiable in $x=x_0$.
I know I'm missing something but I still don't see it. Could you be more specific please?
Thank you so far for your answer!