Use the Laplace transform to solve the following initial value problem:
$y'' + 2y' + 1y = 0$
$y(0) = -4$, $y'(0) = 5$
a. 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
$-4e^{-t} - 9te^{-t}$ $\triangle = 0$
b. Solve for
$Y(s) = \frac{-4s - 13}{(s+1)^2}$
c. write the above answer in its partial fraction decomposition, $Y(s) = \frac{A}{s+a} + \frac{B}{(s+a)^2}$
$Y(s) = \frac{-4}{s+1} + \frac{9}{(s+1)^2}$
d. Now, by inverting the transform, find
$y(t) = -4e^{-t} - 9te^{-t}$
