Question

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

$y\text{'}\text{'}+6y\text{'}+8y=e^{3x}-sinx$

Short answer

$y_p\left(x\right)=c_3{e}^{3x}+c_4\sin x+c_5\cos x$

Step-by-step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

$$\begin{gathered} \left(D^2+6D+8\right)\left[y\right]=0 \\ \Rightarrow (D+2)(D+4)\left[y\right]=0 \end{gathered}$$

The solution of the homogeneous is

$y_h\left(x\right)=c_1{e}^{-2x}+c_2{e}^{-4x}$ (1)

Let$g\left(x\right)=e^{3x}$

Then

$(D-3)\left[g\right]=0$

Let$h\left(x\right)=\sin 2x$

Then

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

Hence

$(D-3)\left(D^2+1\right)[g-h]=0$

Then, every solution to the given nonhomogeneous equation also satisfies

.$(D-3)\left(D^2+1\right)(D+2)(D+4)$$\left[y\right]=0$

Then

$y\left(x\right)=c_1{e}^{-2x}+c_2{e}^{-4x}+c_3{e}^{3x}+c_4\sin x+c_5\cos x$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

$y_p\left(x\right)=c_3{e}^{3x}+c_4\sin x+c_5\cos x$