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Suppose we have the following differential equation using operator notation: $$(D-x)(D+x)y=0\tag{1}$$ where $$D=\frac{d}{dx}$$

Now I could rewrite $(1)$ as $$\begin{align}\require{enclose}(D-x)(D+x)y&=\left(\frac{d}{dx}-x\right)\left(y^{\prime}+xy\right)\\&=y^{\prime\prime}+\bbox[#AFA]{\left(xy\right)^{\prime}}-xy^{\prime}-x^2y\\&=y^{\prime\prime}+\bbox[#AFA]{y+\enclose{downdiagonalstrike}{xy^{\prime}}}-\enclose{downdiagonalstrike}{xy^{\prime}}-x^2y\\&\implies \fbox{$y^{\prime\prime}-x^2y + y = 0$}\tag{a}\end{align}$$

Or, by switching the order of the brackets; I could rewrite $(1)$ as $$\begin{align}\require{enclose}(D-x)(D\color{blue}{\textbf{ + }}x)y&=(D+x)(D\color{red}{\textbf{ - }}x)y\\&=\left(\frac{d}{dx}+x\right)\left(y^{\prime}-xy\right)\\&=y^{\prime\prime}\bbox[#FAA]{-\left(xy\right)^{\prime}}+xy^{\prime}-x^2y\\&=y^{\prime\prime}\bbox[#FAA]{-y-\enclose{downdiagonalstrike}{xy^{\prime}}}+\enclose{downdiagonalstrike}{xy^{\prime}}-x^2y\\&\implies \fbox{$y^{\prime\prime}-x^2y - y = 0$}\tag{b}\end{align}$$

Lastly, I could rewrite $(1)$ as $$\begin{align}\require{enclose}(D-x)(D+x)y&=(D^2-x^2)y\\&=\left(\frac{d^2}{dx^2}-x^2\right)y\\&=y^{\prime\prime}-x^2y\\&\implies\ \fbox{$y^{\prime\prime}-x^2y = 0$}\tag{c}\end{align}$$

There's no doubt there's most probably a simple explanation for it; but how can the same differential equation $(1)$ be written in three different ways: $(\mathrm{a})$, $(\mathrm{b})$, $(\mathrm{c})$?

Many thanks.

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    $\begingroup$ Now explain the downvote $\endgroup$ – BLAZE Jun 24 '16 at 7:30
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    $\begingroup$ Did not down vote, but the answer is obvious. Operators $D-x$ and $D+x$ do not commute. (You cannot switch the orders) $\endgroup$ – mastrok Jun 24 '16 at 7:32
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    $\begingroup$ @mastrok Thanks for that; it was not obvious to me. Do you have any sources/links that explain/prove why operators are not commutative? $\endgroup$ – BLAZE Jun 24 '16 at 7:34
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    $\begingroup$ @BLAZE: look e.g. at the quantum harmonic oscillator. Your observation derives from the elementary fact that $[D,x] =1$ with $[D,x]= Dx -xD$ the commutator. This statement is a statement about operators, thus it means that $[D,x]y =y$ or $(Dx-xD)y=y$. $\endgroup$ – Fabian Jun 24 '16 at 7:38
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    $\begingroup$ Since you already figured it out, why don't you answer the question yourself? :) $\endgroup$ – mastrok Jun 24 '16 at 7:43
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$D+x$ and $D-x$ do not commute. Your (b) is a demonstration of this.

In (c) you are equating $(D+x)(D-x)$ and $D^2-x^2$, but that is wrong for a similar reason - in $D^2-x^2$ the first $D$ is no longer acting on the second $x$.

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  • $\begingroup$ Many thanks for your answer; best regards. $\endgroup$ – BLAZE Jun 24 '16 at 10:31
  • $\begingroup$ Apart from the not getting any points; What is the difference between Community Wiki answers and those that are not? $\endgroup$ – BLAZE Jun 24 '16 at 10:46
  • $\begingroup$ You would have to check the help pages. I think you need less reputation to edit them. $\endgroup$ – almagest Jun 24 '16 at 10:47
  • $\begingroup$ Okay, thanks; I'll take a look. $\endgroup$ – BLAZE Jun 24 '16 at 10:51

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