Consider the following initial value problem: y" - 6y' - 27y = sin (8t) y (0) = 6, y' (0) = 7 Using Y for the Laplace transform of y (t), i. e., Y = L{y (t) }, find the equation you get by taking the Laplace transform of the differential equation and solve for Y (s) =

Y (s) = ((8) / (s^2-6s+27) (s^2+64)) + (6s-29) / (s^2-6s+27)

Step-by-step explanation:

Let's find the Laplace transform of each component of the differential equation.

Y'' = L{y'' (t) } = s^2Y (s) - sy (0) - y' (0) = s^2Y (s) - 6s - 7

Y' = L{-6y' (t) } = - 6L{y' (t) } = - 6 (sY (s) - y (0)) = - 6sY (s) + 36

Y = L{27y (t) } = 27L{y (t) } = 27Y (s)

L{sin (at) } = (a) / (s^2 + a^2), so

L{sin (8t) } = 8 / (s^2 + 64)

Now, putting all of this into the original differential equation, we have:

s^2Y (s) - 6s - 7 - 6sY (s) + 36 + 27Y (s) = 8 / (s^2+64)

Y (s) (s^2-6s+27) = (8 / (s^2+64)) + 6s - 29

Solving for Y (s), we end up with:

Y (s) = ((8) / (s^2-6s+27) (s^2+64)) + (6s-29) / (s^2-6s+27)