Question

use the annihilator method to determinethe form of a particular solution for the given equation.

$y\text{'}\text{'}\text{'}+2y\text{'}\text{'}-y\text{'}-2y=e^x-1$

Short answer

$y_p\left(x\right)=c_1+c_{5}x{e}^x$

Step-by-step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

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

The solution of the homogeneous is

$y_h\left(x\right)=c_1{e}^{-2x}+c_2{e}^{-x}+c_3{e}^x$ (1)

Now $e^x-1$is annihilated by$\left(D^2-D\right)$

Then, every solution to the given nonhomogeneous equation also satisfies

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

Then

$y\left(x\right)=c_1+c_2{e}^{-2x}+c_3{e}^{-x}+c_4{e}^x+c_{5}x{e}^x$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

$y_p\left(x\right)=c_1+c_{5}x{e}^x$