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I've seen people refer to $n$th order linear differential equations as an "$n$th order linear polynomial of the function $y$ and its derivatives."

I can see how it definitely share the same form as a polynomial:

$a_0(x)y+a_1(x)y'+a_2(x)y''+\cdots+a_n(x)y^{(n)}=b(x)$

I also see this in how people characterize homogeneous LDE's as such because they are homogeneous polynomials (i.e all the terms equal $0$). I get the analog being drawn but these aren't truly polynomials right? How can we say these are polynomials if they are not being exponentiated?

Is there a more general notion of polynomial or '$n$th order' that would encompass $n$th order differential equations and $n$th order polynomials?

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  • $\begingroup$ I see, so the differential operator is being multiplied by itself. So I guess a polynomial of an operator is still a polynomial? $\endgroup$ – Ozaner Hansha May 28 '18 at 0:28
  • $\begingroup$ It definitely depends on your point of view. The Wikipedia article is viewing this as a linear polynomial (degree one) with $y$ and its derivatives being your "variables", and arbitrary functions of $x$ as your "coefficients". So they are thinking in a different way than I was talking about. $\endgroup$ – Morgan Rodgers May 28 '18 at 0:37
  • $\begingroup$ But I would keep in mind that this is Wikipedia, which is not always a reliable source of mathematical information (in fact is often quite bad). Whether or not this is a common interpretation of linear DEqs, I can't say. $\endgroup$ – Morgan Rodgers May 28 '18 at 0:39
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A polynomial of degree one in the variables $z_0,z_1,\dots,z_n$ has the form $$a_0 z_0 + a_1 z_1 + \dots + a_n z_n + b.$$ Now substitute $z_0=y$, $z_1=y'$ etc. (and allow all coefficients to depend on $x$).

(By the way, in the Wikipedia page that you link to there is no mention of an “$n$th order linear polynomial”, whatever that would be; it just says “linear polynomial”.)

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