this is my first question on math.stackexchange, I hope to have phrased it correctly!
I have a differential equation
$\text{$\frac{\text{d}x}{\text{d}t} = \alpha t^{-3}\frac{f'(x)}{f(x)}$ with $\alpha > 0\,\,,$}$
where $f(x)$ has an expansion around a maximum $x_\text{max}$ (that I'll put to zero for simplicity) given by
$f(x) = \sum_{i = 0}^{+\infty}\frac{F^{(i)}}{i!}x^i\,\,.$
Around the maximum the RHS of the differential equation can be expanded as
$\frac{f'(x)}{f(x)} = \frac{F^{(2)}}{F^{(0)}}x + \frac{F^{(3)}}{2F^{(0)}}x^2 + \mathcal{O}(x^3)\,\,,$
and if I stay around the maximum with my initial condition and neglect orders greater than $1$ I get the solution
$x(t) = x_0\exp\Big(-\frac{\alpha}{2}\frac{F^{(2)}}{F^{(0)}}t^{-2}\Big)\,\,,$
which says that $x(t)\to 0 = x_\text{max}$ for $t\to 0^+$.
Now: the question is the following:
Original question
Is it possible to find a generic power series expansion for $x(t)$ around $t = 0$ when I include also the $\mathcal{O}(x^n),\,n > 1$ terms?
Answer: there is no power series expansion, since $t^{-3}$ is a too strong divergence for $t\to 0^+$ (leads to essential singularity).
Second question
What I just need, however, is to evaluate the limit:
$\lim_{t\to 0^+}\frac{\dfrac{\text{d}}{\text{d}t}\bigg(\dfrac{1}{t^2}\dfrac{f''(x)}{f(x)}\bigg)}{\dfrac{\text{d}}{\text{d}t}\bigg(\dfrac{1}{t^2}\bigg(\dfrac{f'(x)}{f(x)}\bigg)^2\bigg)}\,\,.$
This is the reason why I looked for some power series expansion: in that way I would have been able to control what was happening in that limit.
Attempt at answering - 1: I was wondering if, since it seems that the $t^{-2}$ factors at the numerator and denominator could lead to a indeterminate form $\frac{\infty}{\infty}$, one could use the Hopital rule to compute, instead, the limit
$\lim_{t\to 0^+}\frac{\dfrac{1}{t^2}\dfrac{f''(x)}{f(x)}}{\dfrac{1}{t^2}\bigg(\dfrac{f'(x)}{f(x)}\bigg)^2} = \dots\,\,,$
but almost certainly I am using the Hopital rule in the wrong way.
Attempt at answering - 2: can I just plug in the form $x(t) = x_0\exp\Big(-\frac{\alpha}{2}\frac{F^{(2)}}{F^{(0)}}t^{-2}\Big)$ in the limit and calculate it in this way?
EDIT
I solved the problem: if I choose as solution $x(t) = F(c\times e^{-\alpha t^{-2}})$, after some manipulation I can obtain a power series for $F$. I also solved the limit in this way.