Question

ANSWER ALL AS SOON AS POSSIBLE. WILL UPVOTE

Find the Laplace transform of the following: 1. f(t) = (14t) - 3e-7t cos(8t) - (2t - 6)² 2. 2y+3y"-23y' - 12y = 5t-7, 3. g(t)=(3+2te - 4t² - 2t³)u(t) (8t-7t³, (t² + 3t-4, 0≤t

Answer 1

1. Applying Laplace transform to each term:

L{(14t)} = 14/s^2

L{(3e^(-7t)cos(8t))} = 3/(s+7)^2 + 64/(s+7)^2

L{((2t-6)^2)} = 4(2/s^3 - 6/s^2 + 6/s)

Therefore,

L{f(t)} = 14/s^2 + 3/(s+7)^2 + 64/(s+7)^2 + 4(2/s^3 - 6/s^2 + 6/s)

Simplifying, we get

L{f(t)} = (28s - 24)/(s^3) + (3s^2 + 98s + 198)/(s^2 + 14s + 49) + (2/s^3)

2. Applying Laplace transform to the given differential equation, we get:

2L{y} + 3(s^2)L{y} - 23sL{y} - 12L{y} = 5/s^2 - 7/s

Simplifying and solving for L{y}, we get:

L{y} = (5/(2s^3)) - (7/(2s^2)) + (23/(4(s-1))) + (3/(4(s+1)))

3. Applying Laplace transform to each term:

L{(3+2te^(-4t))u(t)} = 3/s + 2/(s+4)^2

L{(8t-7t^3)} = 8/s^2 - 42/(s^4)

L{((t^2+3t-4))u(t)} = (1/s^3) + (3/s^2) - (4/s)

Therefore,

L{g(t)} = 3/s + 2/(s+4)^2 + 8/s^2 - 42/(s^4) + (1/s^3) + (3/s^2) - (4/s)

Simplifying, we get

L{g(t)} = (7s^4 - 76s^3 + 20s^2 + 96s - 24)/(s^4(s+4)^2)