Question

use the annihilator method to determinethe form of a particular solution for the given equation$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{x}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}$

Short answer

$y\text{'}\text{'}-5y\text{'}+6y=e^{3x}-x^2$

Step-by-step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

$\left(D^2-5D+6\right)\left[y\right]=(D-2)(D-3)\left[y\right]=0$

The solution of the homogeneous is

$y_h\left(x\right)=c_1{e}^{2x}+c_2{e}^{3x}$ (1)

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

Then

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

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

Then

$D^3\left[h\right]=0$

Hence

$D^3(D-3)[g-h]=0$

Then, every solution to the given nonhomogeneous equation also satisfies

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

Then

$y\left(x\right)=c_1{e}^{2x}+c_2{e}^{3x}+c_{3}x{e}^{3x}+c_4+c_{5}x+c_6{x}^2$ (2)

is the general solution to this homogeneous equation

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

Comparing (1) & (2)

$y_p\left(x\right)=c_{3}x{e}^{3x}+c_4+c_{5}x+c_6{x}^2$