- #1

reed2100

- 49

- 1

## Homework Statement

Give the inverse Laplace transform of F(s) = (-3/s) + (e^-4s)/(s^2) + (3e^-4s)/s

## Homework Equations

Inverse Laplace [e^(-cs) F(s)] = f(x-c)u(x-c)

## The Attempt at a Solution

I'll break this into 3 parts.

Part 1 - (-3/s)

-3/s = -3(1/s) -> inverse laplace of -3(1/s) = -3

Part 2 - (e^(-4s))/(s^2)

By looking at the e component, I see that in this case "c" = 4.

Rearrange to (e^(-4s)) (1/s^2)

So the inverse laplace = u(x-4)*x [ u(x-4) being for the heaviside component and x being the inverse laplace of 1/s^2]

Part 3 - (3e^-4s / s)

e component still says that c = 4.

Rearrange into 3(e^-4s) (1/s)

Inverse laplace becomes 3 * u(x-4) * 1 = 3u(x-4)

So MY answer would be that the inverse laplace of the complete F(s) is

-3 + u(x-4)x + 3u(x-4)

The CORRECT answer that wolfram spat out and that is an answer choice on the online practice test is

f(x) = u(x-4)x - 3 - u(x-4)

I see the similarities but that makes it even more confusing to me how they got

-u(x-4) instead of +3u(x-4). I don't see how that works out.

Any and all help is greatly appreciated.