In the book of Differential Equations: A Dynamical Systems Approach: Ordinary Differential Equations by West&Hubbard, at page 73, it is states for the Method of Undetermined coefficients that
Consider the differential equation $x' = p(t)x + q(t)$. Suppose that $q(t)$ is an element of a finite-dimensional vector space $V$ of junctions $f$, closed under the operation
$$f \to f' - p(t) f$$ [...]
I'm really confused about the definiton of $V$ because, for example, the operation that is stated is a unary operation (I see only one input), so how can we define a vector space with a unary operation ? Moreover, I coulnd't understand how exactly show we determined when a given function $g$ is in $V$ or not. I mean as far as I understood, to check whether $g\in V$ or not, I need to check that $g' - p(t)g$ (where $p(t)$ is given in the ODE) is in $V$, but by authors' definition, to check whether $g' - p(t)g \in V$, I need to check whether $[g' - p(t)g]' - p(t)[g' - p(t)g] \in V$ or not..., so I really found this confusing and I'm not even sure that I understood the author correctly.
Therefore, I'm looking for some clarifications for this proposition.
Plus, what is the motivation behind this vector space ? I mean I couldn't see why do need "the guesses solutions" to a first order linear DEs to be in this space ?
Edit:
I have put a link of the book.
Edit 2:
Note that my main question is about the structure of $V$, and not about it's existance or something, so you can assume $V$ exists for the purpuse of the question.
The "unary operation", as you called it, is not going to be one of the operations of the vector space. More precisely, you have to find a vector space $(V, +, \cdot)$ such that $$ f\in V \quad \Rightarrow\quad Df \in V, $$ where $Df:=f' -p(x)f.$ (Verifying that this is indeed the case amounts exactly to the task that you "find confusing", unfortunately).
Let me make a toy model. Redefine $D$ as $Df:=f'$. The vector space $V$ of real polynomials is closed$^{[1]}$ under the operation $D$, because if $f$ is a polynomial, its derivative is again a polynomial. Notice that this space is infinite dimensional. If $n\in\mathbb N$, the vector space of polynomials of degree up to $n$ is finite dimensional and closed under $D$.
In general, constructing such vector spaces is difficult. The framework is called spectral theory.
[1] This choice of terminology is unfortunate, according to me. I would have said that $V$ is invariant under $D$.