Use the Laplace transform to solve the following initial value problem: $$y'' -{4} y' + {13} y= 0 \hspace{0.5in} y(0)=0, \; y'(0)=3$$

First, using $Y$ for the Laplace transform of $y(t)$, i.e., $Y = \mathcal{L} \lbrace y(t) \rbrace$,
find the equation you get by taking the Laplace transform of the differential equation
$=0$

Now solve for $Y(s) =$

By completing the square in the denominator and inverting the transform, find
$y(t)=$ .