Q36E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the annihilator method to show that if $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ in (4) has the form $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{B}{\mathit{e}}^{\mathbf{\alpha }\mathbf{x}}$, then equation (4) has a particular solution of the form ${\mathit{y}}_{\mathbf{p}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{x}}^{\mathbf{s}}\mathit{B}{\mathit{e}}^{\mathbf{\alpha }\mathbf{x}}$, where $\mathit{s}$is chosen to be the smallest nonnegative integer such that ${\mathit{x}}^{\mathbf{s}}{\mathit{e}}^{\mathbf{\alpha }\mathbf{x}}$ is not a solution to the corresponding homogeneous equation

Step-by-Step Solution

Answer

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

Step 1: Definition

A linear differential operator $\mathit{A}$is said to annihilate a function$\mathit{f}$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 $\mathit{f}$if$\mathit{f}$is a solution to the homogeneous linear differential equation (2) on.

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

Step 2: Find particular solution

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

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

$(D-\alpha )$$\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 .

$y_p=x^s\lambda e^{\alpha x}$