Question

Use appropriate algebra and to find the given inverse Laplace transform. (Write your answer as a function of t.) Ꮭ⁻¹={4s/4s²+1}

Answer 1

The inverse Laplace transform of {4s/4s^2+1} is L^-1={4/√2}e^(-t/2)sin(t/√2).

Step-by-step explanation:

To find the inverse Laplace transform of L^-1={4s/4s^2+1}, we can use the table of Laplace transforms. We can see that the numerator in the Laplace transform is 4s and the denominator is 4s^2 + 1. Comparing this to the standard form in the table, we can identify that the inverse Laplace transform is L^-1={4/√2}e^(-t/2)sin(t/√2).