Use the Laplace transform to solve the following initial value problem: $$y'' +{3} y' = 0 \hspace{0.5in} y(0)=3, \; y'(0)=-2$$

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 $Y(s) =$
and write the above answer in its partial fraction decomposition, $Y(s) = \frac{A}{s+a}+\frac{B}{s+b}$ where $a < b$
$Y(s) =$ +

Now by inverting the transform, find $y(t)=$ .