Where do the constants in this formula come from? I do not exactly know what they are. I am reading about inverse operators and the book is going over this one.
After proving some stuff about this operator it finally says if $$P(D)y=(a_nD^n+...+a_1D+a_0)y=bx^k$$
then
$$y_p=\frac{1}{P(D)}(bx^k)$$
$$=\frac{1}{a_0(1+\frac{a_1}{a_0}D+\frac{a_2}{a_0}D^2+...+\frac{a_n}{a_0}D^n)}bx^k$$
$$=\frac{b}{a_0}(1+b_1D+b_2D^2+...+b_kD^k)x^k$$
In the last step, I do not understand what the constants $b_1,b_2,...,b_k$ are. What are these values supposed to be?
To be clear in what I mean, here is an example where we are the finding the particular solution of $4y^{''}-3y^{'}+9y=5x^2$
$$y_p=\frac{1}{9(1-\frac{D}{3}+\frac{4}{9}D^2)}5x^2$$
$$=\frac{5}{9}\left(1+\frac{D}{3}-\frac{D^2}{3}\right)x^2$$
You see there that by using the previously established property in that last step they got some constants in front of all the D operators. 
How did they arrive to those constants and why did the signs change?

There are two key points here. 


*

*First, the computation is expanding the formal function (in $D$) 
$$ \frac{1}{1 + c_1 D + c_2 D^2 + \cdots + c_n D^n} $$
using its Maclaurin series, which would look like
$$ 1 + b_1 D + b_2 D^2 + b_3 D^3 + \ldots $$

*When looking at the action on $x^k$, the Maclaurin series can be truncated to the $k$th Maclaurin polynomial because $D^{k+m} x^k = 0$ for all $m \geq 1$.  

To do the example you included, suppose we are looking at the function 
$$ \frac{1}{1 - \frac13 D + \frac49 D^2} $$
We want to find its Maclaurin series, which is the series $\sum b_j D^j$ such that 
$$ (1 - \frac13 D + \frac49 D^2) \sum_{j = 0}^\infty b_j D^j = 1 $$
This requires then, by matching coefficients
$$ b_0 = 1 $$
$$ - \frac13 b_0 + b_1 = 0 $$
$$  \frac49 b_j - \frac13 b_{j+1} + b_{j+2} = 0  \qquad \forall j \geq 0 $$
which we can solve to get $b_1 = \frac13$ and 
$$ \frac49 + \frac19 + b_{2} = 0 \implies b_2 = - \frac13 $$
as was claimed. 
