Question

use the annihilator method to determinethe form of a particular solution for the given equation. $y\text{'}\text{'}-6y\text{'}+10y=e^{3x}-x$

Short answer

$y_p\left(x\right)=c_3+c_{4}x+c_5{e}^{3x}$

Step-by-step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

$(D^2-6D+10)\left[y\right]=0$

The solution of the homogeneous is

$y_h\left(x\right)=c_1{e}^{3x}cosx+c_2{e}^{3x}sinx$ (1)

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

Then, every solution to the given nonhomogeneous equation also satisfies

.$\left(D^3-3D^2\right)(D^2-6D+10)\left[y\right]=0$

Then

$y\left(x\right)=c_1{e}^{3x}\cos x+c_2{e}^{3x}\sin x+c_3+c_{4}x+c_5{e}^{3x}$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

$y_p\left(x\right)=c_3+c_{4}x+c_5{e}^{3x}$