Short Answer

Use the annihilator method to show that if $f (x)$ in (4) has the form $f (x) = a c o s β x + b s i n β x,$
then equation (4) has a particular solution of the form
(18) $y_{p} (x) = x^{s} {A c o s β x + B s i n β x}$ ,where s is chosen to be the smallest nonnegative integer such that $x^{3} c o s β x$ and $x^{3} \sin β x$ are not solutions to the corresponding homogeneous equation

$y_{p} = x^{s} (A \cos β x + B \sin β x)$ is the form of particular solution.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition

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

Step 2: For particular solution

Equation (4) is given by $a_{n} y^{(n)} + a_{n - 1} y^{n - 1} + \dots \dots . . + a_{0} y = f$

Also given $f (x) = a \cos β x + b \sin β x$

Then, $(a_{n} D^{n} + a_{n - 1} D^{n - 1} + \dots \dots . . + a_{0}) y = a \cos β x + b \sin β x$

So, $\sin β x \cos β x$ is annihilated by $(D^{2} + β^{2})$

So, $(D^{2} + β^{2}) (a_{n} D^{n} + a_{n - 1} D^{n - 1} + \dots \dots . . + a_{0}) y = 0$

For particular solution i.e; $y_{p}$ check if $\cos β x, \sin β x$ are solutions to homogeneous, particular solution is different than homogeneous solution choose

$y_{p} = x^{s} (A \cos β x + B \sin β x)$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Fundamentals Of Differential Equations And Boundary Value Problems

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