Saturday, 13 May 2017

calculus - Why can this differential equation be written in 3 different ways?



Suppose we have the following differential equation using operator notation: (Dx)(D+x)y=0 where D=ddx




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.


Answer




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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