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Expert-verified Found in: Page 338 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Use the annihilator method to show that if ${\mathbit{f}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}$ in (4) has the form ${\mathbit{f}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{=}}{\mathbit{B}}{{\mathbit{e}}}^{\mathbf{\alpha }\mathbf{x}}$, then equation (4) has a particular solution of the form ${{\mathbit{y}}}_{{\mathbf{p}}}{\mathbf{\left(}}{\mathbit{x}}{\mathbf{\right)}}{\mathbf{=}}{{\mathbit{x}}}^{{\mathbf{s}}}{\mathbit{B}}{{\mathbit{e}}}^{\mathbf{\alpha }\mathbf{x}}$, where ${\mathbit{s}}$is chosen to be the smallest nonnegative integer such that ${{\mathbit{x}}}^{{\mathbf{s}}}{{\mathbit{e}}}^{\mathbf{\alpha }\mathbf{x}}$ is not a solution to the corresponding homogeneous equation

${y}_{p}={x}^{s}\lambda {e}^{\alpha x}$is the form of particular solution.

See the step by step solution

## Step 1: Definition

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

${\mathbf{\left(}}{\mathbf{-}}{\mathbf{\infty }}{\mathbf{,}}{\mathbf{\infty }}{\mathbf{\right)}}$

## Step 2: Find particular solution

Given that $f\left(x\right)=B{e}^{\alpha x}$

Since ${e}^{\alpha x}$ is annihilated by $\left(D-\alpha \right)$so we have:

$\left(D-\alpha \right)$$\left({a}_{n}{y}^{\left(n\right)}\left(x\right)+\dots ..+{a}_{0}y\right)=B{e}^{\alpha x}$

To check if${e}^{\alpha x}$is solution of homogeneous equation then ${y}_{p}\ne {y}_{\text{homogeneous}}$ So take .

${y}_{p}={x}^{s}\lambda {e}^{\alpha x}$ ### Want to see more solutions like these? 