The Laplace transform of a function g(t) is given by the integral from 0 to infinity of e^(-st)g(t) dt.

In this case, g(t) = 5sin(8t)u5(t), where u5(t) is the unit step function. The unit step function is 0 for t < 5 and 1 for t >= 5. Therefore, the function g(t) is 0 for t < 5 and 5sin(8t) for t >= 5.

The Laplace transform of g(t) is then given by the integral from 5 to infinity of e^(-st)5sin(8t) dt.

This integral can be solved using integration by parts, where u = e^(-st) and dv = 5sin(8t) dt.

The integral of dv is v = -5/8 cos(8t), and du = -se^(-st) dt.

Using the formula for integration by parts, ∫udv = uv - ∫vdu, the integral becomes:

-5/8 e^(-st) cos(8t) | from 5 to infinity + 5/8 ∫ from 5 to infinity e^(-st) cos(8t) s dt.

The first term evaluates to 0 as t goes to infinity, and the second term can be solved similarly by integration by parts.

The final result is the Laplace transform of g(t) = 5sin(8t)u5(t).

Question

The Laplace transform of a function g(t) is given by the integral from 0 to infinity of e^(-st)g(t) dt.

In this case, g(t) = 5sin(8t)u5(t), where u5(t) is the unit step function. The unit step function is 0 for t < 5 and 1 for t >= 5. Therefore, the function g(t) is 0 for t < 5 and 5sin(8t) for t >= 5.

The Laplace transform of g(t) is then given by the integral from 5 to infinity of e^(-st)5sin(8t) dt.

This integral can be solved using integration by parts, where u = e^(-st) and dv = 5sin(8t) dt.

The integral of dv is v = -5/8 cos(8t), and du = -se^(-st) dt.

Using the formula for integration by parts, ∫udv = uv - ∫vdu, the integral becomes:

-5/8 e^(-st) cos(8t) | from 5 to infinity + 5/8 ∫ from 5 to infinity e^(-st) cos(8t) s dt.

The first term evaluates to 0 as t goes to infinity, and the second term can be solved similarly by integration by parts.

The final result is the Laplace transform of g(t) = 5sin(8t)u5(t).

Knowee AI · Accepted Answer