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I am reading my differential equations book and it is going over the differential inverse operator and more specifically this case where $y_p=\frac{1}{D-a_0}(bx^k)$. So then they do these two steps and I do not know how they did the second one. $$=\frac{1}{-a_0(1-\frac{D}{a_0})}(bx^k)$$ $$=-\frac{1}{a_0}\left[1+\frac{D}{a_0}+\frac{D^2}{a_0^2}+\frac{D^3}{a_0^3}+...+\frac{D^k}{a_0^k}\right](bx^k)$$

I don't know how the two are equal. I feel like it might just be something I forgot from my last math classes. The book says it is the series expansion of the inverse operator by ordinary division. I do not know what it means that the series expansion is by "ordinary division"

Can someone please explain to me what they did there in that step and what is the "ordinary division" part or maybe point me to some place I can learn about it?

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  • $\begingroup$ What is $x$ here, exactly? $\endgroup$
    – Bernard
    Mar 29, 2020 at 23:20
  • $\begingroup$ The independent variable of our function. You know like y(x) and x is our input. Sorry if I am not clear haha $\endgroup$
    – Angel Borge
    Mar 29, 2020 at 23:22
  • $\begingroup$ And what does $x^k$ denote? $\endgroup$
    – Bernard
    Mar 29, 2020 at 23:26
  • $\begingroup$ x to any power of k im pretty sure $\endgroup$
    – Angel Borge
    Mar 29, 2020 at 23:34

1 Answer 1

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You don't prove this expansion rigorously at all. Just compute $y_p$ using this heuristic argument and then verify that the answer you got really satisfies the equation $(D-a_0)y_p=bx^{k}$.

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  • $\begingroup$ Sorry my I edited my question now to make what I was asking clearer. Im mostly just asking about what exactly they are doing there! And if perhaps you could remind what it is so I can go get refreshed about it. $\endgroup$
    – Angel Borge
    Mar 29, 2020 at 23:20
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    $\begingroup$ They are just using the formula $\frac 1 {1-t} =1+t+t^{2}+...$ with $t$ replaced by $\frac D {a_0}$. @AngelBorge $\endgroup$
    – Kavi Rama Murthy
    Mar 29, 2020 at 23:23
  • $\begingroup$ Oh my gosh thank you so much! I knew it was something simple like that which I had forgotten from Calc 2. Thank you! $\endgroup$
    – Angel Borge
    Mar 29, 2020 at 23:24
  • $\begingroup$ And what does the "ordinary division" part mean? $\endgroup$
    – Angel Borge
    Mar 29, 2020 at 23:24
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    $\begingroup$ @AngelBorge Ordinary divison simply means you are treating $\frac 1 {1-D/a_0}$ as if $D/a_0$ was just an ordinary number $t$ . It is not, but the method does yield a solution to the DE. $\endgroup$
    – Kavi Rama Murthy
    Mar 29, 2020 at 23:27

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