Question

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) =

Answer 1

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

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)