$Now,\ {(D}^2-2D+1)x=e^t \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}{(D}^2x-2Dx+x)=\ e^t \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}L\left[D^2x\right]-2L\left[Dx\right]+L(x)=L[e^t] \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}s^2X-sx\left(0\right)-x^{'}\left(0\right)-2[sX-x\left(0\right)]+X=L\left[e^t\right] \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}s^2X-s\left(2\right)-\left(-1\right)-2sX+4+X=\frac{1}{s-1} \\[2ex] \textit{ } \displaystyle \{since\ at\ t=0,\ x=2\ gives\ x(0)=2,\ and\ Dx=-1,\ gives\ x’(0)=-1\} \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}{X(s}^2-2s+1)=\frac{1}{s-1}-1-4+2s \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}X{\left(s-1\right)}^2\ =\frac{1+(-5+2s)(s-1)}{s-1} \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}X=\frac{1-5s+5+2s^2-2s}{{\left(s-1\right)}^3} \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}x=L^{-1}\left[\frac{2s^2-7s+6}{{\left(s-1\right)}^3}\right] \\[2ex] \displaystyle \frac{2s^2-7s+6}{{\left(s-1\right)}^3}=\frac{A}{s-1}+\frac{B}{{\left(s-1\right)}^2}+\frac{C}{{\left(s-1\right)}^3\ } \\[2ex] \displaystyle 2s^2-7s+6=A{\left(s-1\right)}^2+B(s-1)+C \\[2ex] \displaystyle Put\ s=1, \\[2ex] \displaystyle 2-7+6=C \\[2ex] \displaystyle C=1 \\[2ex] \displaystyle 2s^2-7s+6\ =As^2-2As+A+Bs-B+1 \\[2ex] \displaystyle Comparing\ the\ co-efficients\ of\ s^2,A=2\ \\[2ex] \displaystyle Comparing\ the\ co-efficients\ of\ s,\ \\[2ex] \displaystyle -7=\ -2A+B,\ -1=\ -4+B \\[2ex] \displaystyle B=-3 \\[2ex]$

$\begin{align*} \displaystyle \frac{2s^2-s}{\left(s-1\right)\left(s^2+1\ \right)}&=\frac{A}{s-1}+\frac{B}{{\left(s-1\right)}^2}+\frac{C}{{\left(s-1\right)}^3}\\[2ex] \displaystyle &=\frac{2}{(s-1)}-\ \frac{3}{{\left(s-1\right)}^2}+\frac{1}{{\left(s-1\right)}^3\ }) \\[2ex] \end{align*}\\ \displaystyle L^{-1}\left[\frac{2}{(s-1)}-\ \frac{3}{{\left(s-1\right)}^2}+\frac{1}{{\left(s-1\right)}^3\ }\right]=2\frac{t^0}{0!}\ e^t-3\frac{t}{1!}\ e^t+\frac{t^2}{2!}e^t\ \\[2ex] \displaystyle \{As,\ L^{-1}\left[X\left(s-a\right)\right]=e^{at}x\left(t\right)\ \ and\ \ L^{-1}\left[\frac{1}{s^n}\right]=\frac{t^{n-1}}{(n-1)!}\} \\[2ex] x=\ 2\ e^t-3t\ e^t+\frac{e^tt^2}{2}=\frac{1}{2}\ e^t\ \left(t^2-6t+4\right) $