Solution:

$(D^2 -3D +2)y=4e^2t$

Taking Laplace Transform,

$L(y'')=3(y')+2L(y)=4L(e^2t) \\[2ex] Let \ L(y)=\bar{y} \\[2ex] \therefore L(y')=s\bar{y}-y(0) =s\bar{y}+3 \\[2ex] And \ L(y'')=s^2 \bar{y}-sy(0)-y'(0)\\ =s^2\bar{y}+3s-5 \\[2ex] L(e^{at})=\dfrac{1}{s-2}$

Now equation is,

$(s^2 \bar{y}+3s-5)-3(s\bar{y}+3)+2\bar{y}=4\dfrac{1}{s-2}$

$\left(s^2-3s+2\right)\bar{y}=\ \dfrac{4}{s-2}+14-3s=\ \dfrac{4+14s-28-3s^2+6s}{s-2}=\dfrac{-3s^2+20s-24}{s-2}$

$\bar{y}=\ \dfrac{-3s^2+20s-24}{\left(s^2-3s+2\right)\left(s-2\right)}=\dfrac{-3s^2+20s-24}{\left(s-1\right){\left(s-2\right)}^2}$

By partial fraction,

$\bar{y}=\dfrac{-3s^2+20s-24}{\left(s-1\right){\left(s-2\right)}^2} = \dfrac{A}{s-1}+\ \dfrac{B}{s-2}+\dfrac{C}{{\left(s-2\right)}^2} \\ \therefore -3s^2+20s-24=A{\left(s-2\right)}^2+\ B\left(s-1\right)\left(s-2\right)+C(s-1)$

Put s = 1, we get A = -7

Put s = 2, we get C = 4

Put A = -7, C = 4 and s = 0 we get B = 4

$\bar{y}=\dfrac{-7}{s-1}+\dfrac{4}{s-2}+\dfrac{4}{(s-2)^2}$

Taking inverse Laplace transform,

$y = -7L^{-1}\dfrac{1}{s-1}+4L^{-1}\ \dfrac{1}{s-2}+4L^{-1}\dfrac{1}{{\left(s-2\right)}^2} \\ =-7{{\ e}^tL}^{-1}\dfrac{1}{s}+4{{\ e}^{2t}L}^{-1}\ \dfrac{1}{s}+4{{\ e}^{2t}L}^{-1}\dfrac{1}{{\left(s\right)}^2}$

$=-7{\ e}^t+4{\ e}^{2t}\ +4{\ te}^{2t} \\ \therefore \ the \ solution \ is \ 7{\ e}^t+4{\ e}^{2t}\ +4{\ te}^{2t}$