I have to solve the following ODE using the table of Laplace transforms
$y''-y'+y= δ(t − 2) , y(0)=1$ and $y'(0)= 0$
This is my work, yet I am not even close to the answer. If someone could help me understand what I'm doing wrong this would be very helpful!!
$y'' - y' + y = \delta (t-2)\\ s^2 Y - sy(0) - y'(0) - sY + y(0) + Y = e^{-2s}\\ (s^2 - s +1) Y = e^{-2s}+ s - 1\\ Y = \frac {e^{-2s} - s+ 1}{(s^2 - s +1)}\\ (s^2 - s +1) = (s-\frac 12)^2 + \frac 34\\ Y = \frac {e^{-2s} - \frac 12}{(s-\frac12)^2 +\frac 34} + \frac {s-\frac 12}{(s-\frac12)^2 +\frac 34}\\ \mathcal L^{-1} \{\frac {s-b}{(s-b)^2 + a^2}\} = e^{bt} \cos at\\ \mathcal L^{-1} \{\frac {a}{(s-b)^2 + a^2}\} = e^{bt} \sin at\\ y = u(t-2) \frac 2{\sqrt 3} e^{\frac 12 t}\sin \frac {\sqrt {3}}{2} t - \frac 1{\sqrt 3} e^{\frac 12 t}\sin \frac {\sqrt {3}}{2} t + e^{\frac 12 t}\cos \frac {\sqrt {3}}{2} t$