Laplace Transforms & Inverse Step by step Solution with Explanation

Your question:

In Probs. 33-36 find the transform. In Probs. 37-45 find the inverse transform. Show the details of your work.

33. t2.e-3t

Laplace Transforms & Inverse Answers and Explanation

Finding Laplace Transforms:

t^2 * e^(-3t):

0.5e^(-4.5t) * sin(2πt):

Since the Laplace transform of sin(ωt) is ω / (s^2 + ω^2), we can't directly transform the product. However, a common trick is to decompose the product using trigonometric identities. In this case, we can use the product-to-sum identity: sin(a)sin(b) = (1/2) * [cos(a - b) + cos(a + b)] Apply the identity: 0.5e^(-4.5t) * sin(2πt) = (1/4) * [e^(-4.5t)cos(-πt) + e^(-4.5t)cos(3πt)] Take the Laplace transforms of each term: L {(1/4) * e^(-4.5t)cos(-πt)} = (s + 4.5) / [(s + 4.5)^2 + π^2] (using the time-shifting property) L {(1/4) * e^(-4.5t)cos(3πt)} = (s + 4.5) / [(s + 4.5)^2 + 9π^2] Combine the results: L {0.5e^(-4.5t) * sin(2πt)} = (s + 4.5) / [(s + 4.5)^2 + π^2] + (s + 4.5) / [(s + 4.5)^2 + 9π^2]

Apply the inverse Laplace transform property: L^-1 {1 / (s + a)} = e^(-at) Take the inverse transform: L^-1 {π / (s + π)^2} = π * e^(-πt) - πt * e^(-πt) (use the derivative property for the second term) = πe^(-πt) (1 - t)

Thus,