Question

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

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

Short answer

$y_p\left(x\right)=c_1+c_{2}x+c_3{x}^2+c_{6}x{e}^{-x}\cos x+c_{7}x{e}^{-x}\sin x$

Step-by-step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

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

The solution of the homogeneous is

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

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

Then, every solution to the given nonhomogeneous equation also satisfies

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

Then

$y\left(x\right)=c_1+c_{2}x+c_3{x}^2+c_4{e}^{-x}\cos x+c_5{e}^{-x}\sin x+$ $c_3{x}^2+c_{6}x{e}^{-x}\cos x+c_{7}x{e}^{-x}\sin x$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

$y_p\left(x\right)=c_1+c_{2}x+c_3{x}^2+c_{6}x{e}^{-x}\cos x+c_{7}x{e}^{-x}\sin x$