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

$Y(s)=$ . Hint: write the right hand side in terms of the Heaviside function.
Now find the inverse transform to find $y(t) =$ . (Use step(t-c) for $u_c(t)$ .) Note: $$\frac{\pi}{(s^2+\pi^2)(s^2+1)} = \frac{\pi}{\pi^2-1}\left( \frac{1}{s^2+1} - \frac{1}{s^2+\pi^2}\right)$$