Question

use the annihilator method to determinethe form of a particular solution for the given equation. $z\text{'}\text{'}\text{'}-2z\text{'}\text{'}+z\text{'}=x-e^x$

Short answer

$z_p\left(x\right)=c_3{x}^2{e}^x+c_{5}x+c_6{x}^2$

Step-by-step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

$(D^3-2D^2+D)\left[z\right]=0$

The solution of the homogeneous is

$z_h\left(x\right)=c_1{e}^x+c_{2}x{e}^x+c_3$ (1)

Now$x-e^x$ is annihilated by$\left(D^3-D^2\right)$

Then, every solution to the given nonhomogeneous equation also satisfies

.$\left(D^3-D^2\right)$$(D^3-2D^2+D)\left[z\right]=0$

Then

$z\left(x\right)=c_1{e}^x+c_{2}x{e}^x+c_3{x}^2{e}^x+c_4+c_{5}x+c_6{x}^2$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

$z_p\left(x\right)=c_3{x}^2{e}^x+c_{5}x+c_6{x}^2$