Question

What is the inverse laplace transform of the following function, F(s) = α/(s-1)² e⁻³ˢ where alpha is a real constant.

Answer 1

The inverse Laplace transform of the function F(s) =
`(\alpha)/((s-1)^2) e^(-3s)\)` is f(t) =
`\alpha t e^t u(t-3)\)`.

Step-by-step explanation:

The inverse Laplace transform involves converting a function from the Laplace domain back to the time domain. For F(s) =
`(\alpha)/((s-1)^2) e^(-3s)\)`, we recognize that the function resembles a shifted exponentia
`l \(e^(at)\)` in the time domain, implying
`\(u(t-a)\)` where u(t) is the unit step function.

Breaking down the function, the term {1}/{(s-1
`)^2`} corresponds to the Laplace transform of t multiplied by a constant
`\(\alpha\)`. This suggests the original function involves t in the time domain. The exponential term
`e^(-3s)\)` signifies a time delay, shifting the function to the right by 3 units.

The inverse Laplace transform of {1}/{(s-1
`)^2`} is t
`e^t` (without the constant
`(\alpha\))`, which is adjusted by the unit step function u(t-3) due to the shift. The function is zero for t < 3 and takes the value of t
`e^t` for (t
`\geq` 3). Finally, multiplying this with the constant
`\(\alpha\)` provides the complete inverse Laplace transform f(t) =
`\alpha t e^t u(t-3)\)`.

This transformation involves recognizing the shifting and scaling properties in both the exponential and function terms, resulting in a time-domain function that incorporates the initial function's components and their adjustments due to the Laplace transform.