The problem asks to find the Laplace transforms of two functions: (a) $f(t) = e^{-3t}$ (b) $f(t) = \sin(kt)$

1. Problem Description

The problem asks to find the Laplace transforms of two functions:

(a) $f(t) = e^{-3t}$

(b) $f(t) = \sin(kt)$

2. Solution Steps

(a) We need to find the Laplace transform of $f(t) = e^{-3t}$.

The Laplace transform is defined as:

$F(s) = \int_0^{\infty} e^{-st} f(t) dt$

So, for $f(t) = e^{-3t}$, we have:

$F(s) = \int_0^{\infty} e^{-st} e^{-3t} dt = \int_0^{\infty} e^{-(s+3)t} dt$

$F(s) = \left[ \frac{e^{-(s+3)t}}{-(s+3)} \right]_0^{\infty} = \frac{0 - 1}{-(s+3)} = \frac{1}{s+3}$

provided that $Re(s+3) > 0$ or $Re(s) > -3$.

(b) We need to find the Laplace transform of $f(t) = \sin(kt)$.

$F(s) = \int_0^{\infty} e^{-st} \sin(kt) dt$

We can use the following integral:

$\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) + C$

Therefore,

$\int_0^{\infty} e^{-st} \sin(kt) dt = \left[ \frac{e^{-st}}{s^2 + k^2} (-s \sin(kt) - k \cos(kt)) \right]_0^{\infty}$

$F(s) = \lim_{t \to \infty} \frac{e^{-st}}{s^2 + k^2} (-s \sin(kt) - k \cos(kt)) - \frac{1}{s^2 + k^2} (-s \sin(0) - k \cos(0))$

If $s > 0$, the first term goes to

0. So,

$F(s) = 0 - \frac{1}{s^2 + k^2} (0 - k) = \frac{k}{s^2 + k^2}$

3. Final Answer

(a) $\frac{1}{s+3}$

(b) $\frac{k}{s^2 + k^2}$