Understanding the Constant Field of a Differential Field in $k((x))$

[mathmari]

Hey!

I am looking at the following part:

View attachment 5078 The constant field of the differential field $k((x))$ is $k((x^p))$.
Hence if $(1)_p$ has a solution in $k((x))$, multiplication by a suitable constant yields a solution in $k[[x]]$. What exactly is a constant field of a field? (Wondering)

 

[Turgul]

Here, they almost surely mean the "functions" in $k((x))$ which are zero under differentiation. In characteristic zero, this would just be $k$, but in characteristic $p$, the monomial $x^p$ differentiates as
\[\frac{d}{dx}x^p = px^{p-1} = 0x = 0.\]
One consequence of this observation is that for any $f(x) \in k((x))$, then
\[\frac{d}{dx}x^pf = px^{p-1}f + x^p \frac{d}{dx}f = x^p\frac{df}{dx}\]
which means that if $f$ is a solution to a linear differential equation, then $x^pf$ will also be a solution to the same equation.