Use the Laplace transform to solve the following initial value problem:
$y'' - 7y' - 18y = 0$ $y(0) = -3$, $y'(0) = 4$
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
$= 0$
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 $y(t) =$ .
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