Finding a particular solution using an inverse operator

**mezy** · June 5th 2011, 10:05 AM

I am trying to figure out the concept behind inverse operators, but I am getting stuck. My book gives the definition as:

$y_{p} = \frac{b}{a_{0}}(1 + b_{1}D + b_{2}D^{2} + \ldots + b_{k}D^{k})x^{k}), where a_{0} \neq 0$

where $(1 + b_{1}D + b_{2}D^{2} + \ldots + b_{k}D^{k})x^{k})$ is the series expansion of the inverse operator 1/P(D) obtained by ordinary division.

Okay - that sounds great, but what are they dividing by?

Here is the example in the book:

$4y'' - 3y' + 9y = 5x^{2}, P(D) = 4D^{2} - 3D + 9$

$Step 1: y_{p} = \frac{1}{9(1-\frac{D}{3}+\frac{4D^{2}}{9})}(5x^{2})$

$Step 2: = \frac{5}{9}(1+\frac{D}{3}-\frac{D^2}{3})x^{2}$

This is where I get lost. How did they go from step 1 to step 2?

Any help is greatly appreciated!

**General** · June 5th 2011, 04:33 PM

First, for constants a&b : $\displaystyle \frac{1}{a \cdot f(D)} \, b \cdot g(x) = \frac{b}{a} \, \frac{1}{f(D)} \, g(x)$

so this will explain from where did "5/9" come in step two.

To get $\displaystyle \frac{1}{1-\frac{D}{3}+\frac{4D^2}{9}} = 1 + \frac{D}{3} - \frac{D^2}{3}$ , you will use long division.

You will devide 1 by 1 - (D/3) + (4D^2/9) , note that it should be arranged in increasing order. you will the constant term then the term contain D then term contain D^2 and so on.
Surely, this arrangement will make division result has infinite terms.
But you will step when the division result produces the term D^2 since your R(x) is x^2.
In general, if your R(x)=x^n then you will stop when you reach D^n.
I hope I've explained this in good way. Sorry for bad E, it is my 3rd language.
See the following picture to know how to do the long division :

http://www.m5zn.com/uploads/2011/6/5...morwehouzq.jpg

**TheEmptySet** · June 5th 2011, 05:09 PM

Originally Posted by mezy

I am trying to figure out the concept behind inverse operators, but I am getting stuck. My book gives the definition as:

$y_{p} = \frac{b}{a_{0}}(1 + b_{1}D + b_{2}D^{2} + \ldots + b_{k}D^{k})x^{k}), where a_{0} \neq 0$

where $(1 + b_{1}D + b_{2}D^{2} + \ldots + b_{k}D^{k})x^{k})$ is the series expansion of the inverse operator 1/P(D) obtained by ordinary division.

Okay - that sounds great, but what are they dividing by?

Here is the example in the book:

$4y'' - 3y' + 9y = 5x^{2}, P(D) = 4D^{2} - 3D + 9$

$Step 1: y_{p} = \frac{1}{9(1-\frac{D}{3}+\frac{4D^{2}}{9})}(5x^{2})$

$Step 2: = \frac{5}{9}(1+\frac{D}{3}-\frac{D^2}{3})x^{2}$

This is where I get lost. How did they go from step 1 to step 2?

Any help is greatly appreciated!

Another way to think of this is to formally expand

$f(D)=\frac{1}{4D^2-3D+9}$ as its Taylor series about zero We will only need the first three terms

$f(0)=\frac{1}{9}$

$f'(0)=\frac{1}{27}$

$f''(0)=\frac{-2}{27}$

This gives

$f(D)=\frac{1}{4D^2-3D+9}=\frac{1}{9}+\frac{1}{27}D-\frac{2}{27(2!)}D^2+\mathcal{O}(D^3)$

This will give

$\left(\frac{1}{9}+\frac{1}{27}D-\frac{2}{27(2!)}D^2+\mathcal{O}(D^3)\right)(5x^3)= \frac{5}{9}(1+\frac{D}{3}-\frac{D^2}{3}+\mathcal{O}(D^3))x^2$

**mezy** · June 5th 2011, 06:35 PM

Thank you! This is exactly the piece I was missing!

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