Find the Inverse Laplace transform of function by using Convolution Theorem F(s) = 1/(S+2)(S-1)

Sure, here are the steps to find the inverse Laplace transform of the function F(s) = 1/(s+2)(s-1) using the Convolution Theorem:

1. First, we need to express the function F(s) as a product of two simpler functions that we can easily find the inverse Laplace transform of. In this case, we can write F(s) = 1/(s+2)(s-1) as F(s) = A/(s+2) + B/(s-1).

2. To find the constants A and B, we can use the method of partial fractions. We equate the function to its partial fraction decomposition and solve for A and B. This gives us A = -1/3 and B = 1/3.

3. Now, we have F(s) = -1/3 * 1/(s+2) + 1/3 * 1/(s-1). The inverse Laplace transform of 1/(s+a) is e^(-at), so the inverse Laplace transforms of the two terms are -1/3 * e^(-2t) and 1/3 * e^t.

4. According to the Convolution Theorem, the inverse Laplace transform of a product of two functions is the convolution of the inverse Laplace transforms of the two functions. The convolution of two functions f(t) and g(t) is defined as ∫_0^t f(τ)g(t-τ) dτ.

5. Therefore, the inverse Laplace transform of F(s) is the convolution of -1/3 * e^(-2t) and 1/3 * e^t, which is ∫_0^t [-1/3 * e^(-2τ)] * [1/3 * e^(τ-t)] dτ.

6. Evaluating this integral gives the inverse Laplace transform of F(s).