Question

Use convolution theorem to find the inverse Laplace Transform of the function F(s) = 1 / (s^2 * (s^2 + k^2)).

(a) 1 - k^2 * t * e^(-k * t) / 2
(b) 1 - e^(-k * t) / (k * t)
(c) 1 - cos(k * t)
(d) 1 / (k * t) - k * t * e^(-k * t)

Answer 1

To find the inverse Laplace Transform of the function F(s) = 1 / (s^2 * (s^2 + k^2)), use the convolution theorem. The inverse Laplace transform of F(s) is 1 - k^2 * t * e^(-k * t) / 2.

Step-by-step explanation:

To find the inverse Laplace Transform of the function F(s) = 1 / (s^2 * (s^2 + k^2)), we can use the convolution theorem. The convolution theorem states that the inverse Laplace transform of the product of two functions is equal to the convolution of their inverse Laplace transforms. In this case, we can find the inverse Laplace transform of 1 / (s^2) and 1 / (s^2 + k^2), and then convolve them to find the inverse Laplace transform of F(s).

1. Let's find the inverse Laplace transform of 1 / (s^2). The inverse Laplace transform of 1 / (s^2) is t. So, the inverse Laplace transform of F(s) = 1 / (s^2 * (s^2 + k^2)) is equal to the convolution of t and the inverse Laplace transform of 1 / (s^2 + k^2).
2. Next, let's find the inverse Laplace transform of 1 / (s^2 + k^2). The inverse Laplace transform of 1 / (s^2 + k^2) is (1 - e^(-k * t)) / (k * t).
3. Now, we can convolve t and (1 - e^(-k * t)) / (k * t) to find the inverse Laplace transform of F(s). The convolution of t and (1 - e^(-k * t)) / (k * t) is equal to 1 - k^2 * t * e^(-k * t) / 2.

Therefore, the correct answer is (a) 1 - k^2 * t * e^(-k * t) / 2.