For the given differential equation Y'' - Y' - 30 = sin(9t), with initial conditions Y(0) = 6 and Y'(0) = 4, what is the Laplace transform equation for Y(s)? a) Y(s) = (s^2 + 4s + 6)/(s^2 + s - 30) - (9s)/(s^2 + 81) b) Y(s) = (s^2 - 4s - 6)/(s^2 - s + 30) + (9s)/(s^2 - 81) c) Y(s) = (s^2 + 4s - 6)/(s^2 - s - 30) - (9s)/(s^2 + 81) d) Y(s) = (s^2 - 4s + 6)/(s^2 + s + 30) + (9s)/(s^2 + 81)

Answer:

None of the answer choices (a, b, c, d) match this result exactly, but the correct choice closest to this result would be:

a) Y(s) = (s^2 + 4s + 6)/(s^2 + s - 30) - (9s)/(s^2 + 81)

Step-by-step explanation:

To find the Laplace transform of the given differential equation, we'll first take the Laplace transform of each term in the equation. Let's denote Y(s) as the Laplace transform of Y(t).

The differential equation is:

Y''(t) - Y'(t) - 30Y(t) = sin(9t)

Taking the Laplace transform of each term, we get:

L{Y''(t)} - L{Y'(t)} - 30L{Y(t)} = L{sin(9t)}

Using the properties of the Laplace transform, we can find the Laplace transforms of the derivatives as follows:

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

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

So the equation becomes:

s^2Y(s) - sy(0) - y'(0) - sY(s) + y(0) - 30Y(s) = 9/(s^2 + 81)

Now, substitute the initial conditions Y(0) = 6 and Y'(0) = 4:

s^2Y(s) - 6s - 4 - sY(s) + 6 - 30Y(s) = 9/(s^2 + 81)

Now, group like terms:

(s^2 - s - 30)Y(s) - 6s - 4 + 6 = 9/(s^2 + 81)

(s^2 - s - 30)Y(s) - 6s + 2 = 9/(s^2 + 81)

Now, solve for Y(s):

Y(s) = [9/(s^2 + 81) + 6s - 2] / (s^2 - s - 30)

Factoring the denominators:

Y(s) = [9/(s^2 + 81) + 6s - 2] / [(s - 6)(s + 5)]

Now, we have the Laplace transform equation for Y(s):

Y(s) = [9/(s^2 + 81) + 6s - 2] / [(s - 6)(s + 5)]

None of the answer choices (a, b, c, d) match this result exactly, but the correct choice closest to this result would be:

a) Y(s) = (s^2 + 4s + 6)/(s^2 + s - 30) - (9s)/(s^2 + 81)