Question

Use the annihilator method to show that if$a_0\ne 0$in equation (4) and $f\left(x\right)$has the form (17) $f\left(x\right)=b_m{x}^m+b_{m-1}{x}^{m-1}+\cdots +b_{1}x+b_0$, then$y_p\left(x\right)=B_{mr}{x}^m+B_{m-1}{x}^{m-1}+\cdots +B_{1}x+B_0$is the form of a particular solution to equation (4).

Short answer

$y_p=B_m{x}^m+\dots \dots \dots +B_{1}x+B_0$is the form of particular solution.

Step-by-step solution

Definition

A linear differential operator $\mathit{A}$is said to annihilate a function$\mathbf{}\mathit{f}$if $\mathit{A}\mathbf{\left[}\mathit{f}\mathbf{\right]}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$for all x. That is,$\mathit{A}$ annihilates fif fis a solution to the homogeneous linear differential equation (2) on$\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$.

Check for particular solution

It is given that$f\left(x\right)=b_m{x}^m+\dots \dots \dots +b_{1}x+b_0$ and $a_0\ne 0$.

Then the $y_g$is given by:

$\left(a_n{y}^{\left(n\right)}+\dots ..+a_1{y}^{\text{'}}+a_{0}y\right)=f$

So$y_p=B_m{x}^m+\dots \dots \dots +B_{1}x+B_0$

(Then $y_p\ne y_g$)

Therefore Homogeneous auxiliary equation is not particular solution for $f$'s, annihilator.