Where do the constants in this formula come from? I do not exactly know what they are.

Question

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?

Answer 1

There are two key points here.

1. 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 $$ 1 + b_1 D + b_2 D^2 + b_3 D^3 + \ldots $$

2. 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 $m \geq 1$.

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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.