# I do not know what this series expansion is.

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?

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

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 $$(D-a_0)y_p=bx^{k}$$.
• They are just using the formula $\frac 1 {1-t} =1+t+t^{2}+...$ with $t$ replaced by $\frac D {a_0}$. @AngelBorge Mar 29, 2020 at 23:23
• @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. Mar 29, 2020 at 23:27