Short Answer

Use the annihilator method to show that if $a_{0} \neq 0$ in equation (4) and $f (x)$ has the form (17) $f (x) = b_{m} x^{m} + b_{m - 1} x^{m - 1} + \dots + b_{1} x + b_{0}$ , then $y_{p} (x) = B_{m r} x^{m} + B_{m - 1} x^{m - 1} + \dots + B_{1} x + B_{0}$ is the form of a particular solution to equation (4).

$y_{p} = B_{m} x^{m} + \dots \dots \dots + B_{1} x + B_{0}$ 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 x. That is, $A$ annihilates f if fis a solution to the homogeneous linear differential equation (2) on $(- \infty, \infty)$ .

Step 2: Check for particular solution

It is given that $f (x) = b_{m} x^{m} + \dots \dots \dots + b_{1} x + b_{0}$ and $a_{0} \neq 0$ .

Then the $y_{g}$ is given by:

$(a_{n} y^{(n)} + \dots . . + a_{1} y^{'} + a_{0} y) = f$

So $y_{p} = B_{m} x^{m} + \dots \dots \dots + B_{1} x + B_{0}$

(Then $y_{p} \neq y_{g}$ )

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Step 1: Definition

Step 2: Check for particular solution

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Use the annihilator method to show that ifa0≠0in equation (4) and fx has the form (17) f(x)=bmxm+bm-1xm-1+⋯+b1x+b0, then yp(x)=Bmrxm+Bm-1xm-1+⋯+B1x+B0 is the form of a particular solution to equation (4).

Step by Step Solution

Step 1: Definition

Step 2: Check for particular solution

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