Question

If \( \mathcal{L}\{f(t)\} = \frac{1 - e^{-1/s}}{s} \), then \( \mathcal{L}\{e^{-t}f(3t)\} \) is

• A. \( \frac{e^{3/s + 1}}{s + 1} \)
• B. \( \frac{e^{-3/s}}{s + 1} \)
• C. \( \frac{e^{-1/s + 1}}{s + 1} \)
• D. \( \frac{e^{3/s - 1}}{s + 1} \)

Hint:

Use Laplace transform properties: time scaling \( f(at) \), and shifting \( e^{-bt}f(t) \), systematically.

Answer 1

The Correct Option is A

Given that the Laplace Transform \( \mathcal{L}\{f(t)\} = \frac{1 - e^{-1/s}}{s} \), we want to find the Laplace Transform of \( e^{-t}f(3t) \), denoted as \( \mathcal{L}\{e^{-t}f(3t)\} \).

To solve this, we use the following properties of the Laplace Transform:

1. For a function scaled in time, \( \mathcal{L}\{f(at)\} = \frac{1}{a}F\left(\frac{s}{a}\right) \), where \( F(s) \) is the Laplace Transform of \( f(t) \).
2. The exponential shift property: \( \mathcal{L}\{e^{-bt}f(t)\} = F(s + b) \).

Combining these steps, first find the Laplace Transform of \( f(3t) \):

\(\mathcal{L}\{f(3t)\} = \frac{1}{3}F\left(\frac{s}{3}\right) = \frac{1}{3}\left(\frac{1 - e^{-3/s}}{\frac{s}{3}}\right) = \frac{1 - e^{-3/s}}{s}\).

Next, consider the exponential term in \( e^{-t}f(3t) \):

\(\mathcal{L}\{e^{-t}f(t)\} = F(s + 1)\). Applying this to \( f(3t) \), we have:

\(\mathcal{L}\{e^{-t}f(3t)\} = \left.\left(\frac{1 - e^{-3/s}}{s}\right)\right\vert_{s = s + 1} = \frac{1 - e^{-3/(s+1)}}{s+1}\).

Therefore, combining everything together:

\(\frac{1 - e^{-3/(s+1)}}{s+1} \),

which simplifies to the option \( \frac{e^{3/s + 1}}{s + 1} \).