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Using laplace trasnform solve the following differential equation with given condition. (D2-3D+2) y=e2t, y(0)=-3, y'(0)=5
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$\displaystyle Now,\ {(D}^2-3D+2)y={4e}^{2t} \\[2ex] \displaystyle \therefore{}{(D}^2y-3Dy+2y)=\ {4e}^{2t} \\[2ex] \displaystyle \therefore{}L\left[D^2y\right]-3L\left[Dy\right]+2L(y)=L[{4e}^{2t}] \\[2ex] \displaystyle \therefore{}s^2Y-sy\left(0\right)-y^{'}\left(0\right)-3[sY-y\left(0\right)]+2Y=4L\left[e^{2t}\right] \\[2ex] \displaystyle \therefore{}s^2Y-s\left(-3\right)-\left(5\right)-3sY+3(-3)+2Y=4\frac{1}{s-2} \\[2ex] \textit{ } \displaystyle \{since\ y\left(0\right)=-3\ ,y^{'}\left(0\right)=5\} \\[2ex] $

$\displaystyle \therefore{}{Y(s}^2-3s+2)=\frac{4}{s-2}-3s+5+9\ \\[2ex] \displaystyle \therefore{}Y\left(s-2\right)\left(s-1\right)=\frac{4+\left(14-3s\right)\left(s-2\right)}{s-2}=\frac{-3s^2+20s-24}{s-2} \\[2ex] \displaystyle \therefore{}Y=\frac{-3s^2+20s-24}{{\left(s-2\right)}^2(s-1)} \\[2ex] \displaystyle \therefore{}y=L^{-1}\left[\frac{-3s^2+20s-24}{{\left(s-2\right)}^2(s-1)}\right] \\[2ex] $

$\displaystyle \frac{-3s^2+20s-24}{{\left(s-2\right)}^2(s-1)}=\frac{A}{s-1}+\frac{B}{(s-2)}+\frac{C}{{\left(s-2\right)}^2\ } \\[4ex] \displaystyle -3s^2+20s-24=A{\left(s-2\right)}^2+B(s-1)(s-2)+C(s-1) \\[3ex] \displaystyle Put\ s=1, \\[2ex] \displaystyle -3+20-24=A\ {\left(-1\right)}^2 \\[2ex] \displaystyle A=\ -7 \\[2ex] \displaystyle Put\ s=2, \\[2ex] \displaystyle -3\left(4\right)+40-24=C\left(1\right) \\[2ex] \displaystyle C=4\ \\[2ex] \displaystyle Put\ s=0 \\[2ex] \displaystyle -24=4A+2B-C\ \\[2ex] \displaystyle -24=\ -28+2B-4 \\[2ex] \displaystyle 2B=8 \\[2ex] \displaystyle B=4\ \\[2ex] $

$\displaystyle {y=\ L}^{-1}\left[\frac{-3s^2+20s-24}{{\left(s-2\right)}^2\left(s-1\right)}\right]\\[2ex] \displaystyle y=L^{-1}\left[\frac{-7}{s-1}+\frac{4}{\left(s-2\right)}+\frac{4}{{\left(s-2\right)}^2\ }\right] \\[2ex] \displaystyle y\ =L^{-1}\left[\frac{-7}{s-1}\right]\ +L^{-1}\left[\frac{4}{\left(s-2\right)}\right]+L^{-1}\left[\frac{4}{{\left(s-2\right)}^2\ }\right] \\[2ex] \displaystyle y=-7\ e^tL^{-1}\left[\frac{1}{s}\right]+4\ e^{2t}L^{-1}\left[\frac{1}{s}\right]+4e^{2t}L^{-1}\left[\frac{1}{s^2}\right] \\[2ex] \displaystyle y=\ -7\ e^t+4e^{2t}+4te^{2t}\ \\[2ex] $

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