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

Short answer

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

Step-by-step solution

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.

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

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 .

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