Use the Laplace transform to solve the following initial value problem:
$y'' + 5y' = 0$ $y(0) = -4$, $y'(0) = 1$
First, using $Y$ for the Laplace transform of $y(t)$, i.e., $Y = \mathcal{L}\{y(t)\}$.
Find the equation you get by taking the Laplace transform of the differential equation
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) = $
