This will likely be an exceedingly simple question, but the text's use of the operator is confusing me.
To preface, the operator $D$ is defined as: $$ \frac{d^{n} y}{d x^{n}}=D^{n} y $$ and can be used as a polynomial operator similarly as, $$ L=a_{n}(x) D^{n}+a_{n-1}(x) D^{n-1}+\cdots+a_{1}(x) D+a_{0}(x) $$ Problem: $$ \begin{array}{l}\text { Let } n=1,2,3, \ldots \text { Discuss how the observations } D^{n} x^{n-1}=0 \\ \text { and } D^{n} x^{n}=n ! \text { can be used to find the general solutions of the } \\ \text { given differential equations. }\end{array} $$ $$ y^{\prime \prime}=0 $$
Answer:$$ \begin{array}{l}\text { since } D^{2} x=0, x \text { and } 1 \text { are solutions of } y^{\prime \prime}=0 . \text { since they are linearly independent, the } \\ \text { general solution is } y=c_{1} x+c_{2} \text { . }\end{array} $$
The operator makes sense applied to some $f(x)$. But $D^2x$ doesn't make sense to me. As I understand it any polynomial expression of $D$ is still just an operator. In other words $$ D^n = \frac{d^n}{dx^n} $$
If that's correct, how do I get to the provided answer?
You say that the operator makes sense applied to some $f(x)$, but $D^2x$ does not make sense.
So I infer that you don't realize that $x$ plays the role of $f(x)$, and
$$D^2x=0$$ because $$x''=0.$$
Beware that the operator applies to the function on its right.