Find a particular solution to the given higher-order equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{t}}^2\mathbf{+}\mathbf{4}\mathbf{t}\mathbf{-}\mathbf{9}$

The given differential equation is,

$\mathrm{y}\text{'}\text{'}\text{'}-2\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+2\mathrm{y}=2{\mathrm{t}}^2+4\mathrm{t}-9......\left(1\right)$

Consider the particular solution is,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}....\left(2\right)$

Take the first, second, and third derivatives of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=2\mathrm{At}+\mathrm{B}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=2\mathrm{A}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=0\end{array}$

Substitute value of ${\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)$in the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}\text{'}-2\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}+2\mathrm{y}=2{\mathrm{t}}^2+4\mathrm{t}-9\\ 0-2\left[2\mathrm{A}\right]-[2\mathrm{At}+\mathrm{B}]+2[{\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}]=2{\mathrm{t}}^2+4\mathrm{t}-9\\ 2{\mathrm{At}}^2+(-2\mathrm{A}+2\mathrm{B})\mathrm{t}+(-4\mathrm{A}-\mathrm{B}+2\mathrm{C})=2{\mathrm{t}}^2+4\mathrm{t}-9\end{array}$

Comparing all coefficients of the above equation;

$\begin{array}{c}2\mathrm{A}=2\Rightarrow \mathrm{A}=1\\ -2\mathrm{A}+2\mathrm{B}=4......\left(3\right)\\ -4\mathrm{A}-\mathrm{B}+2\mathrm{C}=-9......\left(4\right)\end{array}$

Substitute the value A in the equation (3),

$\begin{array}{c}-2\left(1\right)+2\mathrm{B}=4\\ 2\mathrm{B}=6\\ \mathrm{B}=3\end{array}$

Substitute the value A and B in the equation (4),

$\begin{array}{c}-4\left(1\right)-3+2\mathrm{C}=-9\\ 2\mathrm{C}=-2\\ \mathrm{C}=-1\end{array}$

Therefore, the particular solution of the equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{At}}^2+\mathrm{Bt}+\mathrm{C}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\left(1\right){\mathrm{t}}^2+\left(3\right)\mathrm{t}+(-1)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^2+3\mathrm{t}-1\end{array}$