Take the Laplace transform of the following initial value problem and solve for $Y(s) = \mathcal{L}\lbrace y(t) \rbrace$: $$y'' -{6} y' - {27} y=\begin{cases} 1, & 0 \leq t < 1 \cr 0, & 1 \leq t \end{cases} \hspace{0.5in} y(0)=0, \; y'(0)=0$$

$Y(s)=$ .
Now find the inverse transform to find $y(t) =$ . (Use step(t-c) for $u_c(t)$ .) Note: $$\frac{1}{s(s-9)(s+3)} = \frac{-\frac{1}{27}}{s} + \frac{\frac{1}{36}}{s+3} + \frac{\frac{1}{108}}{s - 9}$$