Question

use the annihilator method to determinethe form of a particular solution for the given equation. $\mathit{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{\theta }\mathbf{=}\mathit{x}{\mathit{e}}^{\mathbf{x}}$

Short answer

${\theta }_p\left(x\right)=c_{3}x{e}^x+c_4{x}^2{e}^x$

Step-by-step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

$\left(D^2-1\right)\left[\theta \right]=(D-1)(D+1)\left[\theta \right]=0$

The solution of the homogeneous is

${\theta }_h\left(x\right)=c_1{e}^{-x}+c_2{e}^x$ (1)

Now $x{e}^x$is annihilated by$(D-1{)}^2$

Then, every solution to the given nonhomogeneous equation also satisfies

.$(D-1{)}^2(D-1)(D+1)\left[\theta \right]=(D-1{)}^3(D+1)\left[\theta \right]=0$

Then

$\theta \left(x\right)=c_1{e}^{-x}+c_2{e}^x+c_{3}x{e}^x+c_4{x}^2{e}^x$ (2)

is the general solution to this homogeneous equation

We know$u\left(x\right)=u_h+u_p$

Comparing (1) & (2)

${\theta }_p\left(x\right)=c_{3}x{e}^x+c_4{x}^2{e}^x$