$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 …