Consider the initial value problem for function y given by,
$y'' - y - 6y = 3t^2 \delta(t - 2)$,
$y(0) = 0$,
$y'(0) = 0$.
(a) Find the Laplace Transform of the source function, $F(s) = \mathcal{L} [3t^2 \delta(t - 2)]$.
F(s) =
?
(b) Find the Laplace Transform of the solution, $Y(s) = \mathcal{L}[y(t)]$.
Y(s) =
?
(c) Find the solution y(t) of the initial value problem above.
y(t) =
?
Recall: If needed, the step function at c is denoted as u(t - c).
