Laplace Transform Calculation for $e^{-2t} \sin 3t$

This solution explains how to find the Laplace transform of the function $f(t) = e^{-2t} \sin 3t$. We will use a key property of Laplace transforms known as the first shifting theorem (or frequency shifting theorem).

Understanding the Problem

We need to compute $L\{f(t)\} = L\{e^{-2t} \sin 3t\}$.

Step 1: Laplace Transform of the Sine Function

First, let's find the Laplace transform of the basic sine function, $\sin(\omega t)$. The standard formula is:

$L\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}$

In our case, $\omega = 3$. So, the Laplace transform of $\sin 3t$ is:

$L\{\sin 3t\} = \frac{3}{s^2 + 3^2} = \frac{3}{s^2 + 9}$

Let's denote this transform as $F(s)$, so $F(s) = \frac{3}{s^2 + 9}$.

Step 2: Applying the First Shifting Theorem

The first shifting theorem states that if $L\{f(t)\} = F(s)$, then the Laplace transform of $e^{at}f(t)$ is given by $F(s-a)$.

$L\{e^{at}f(t)\} = F(s-a)$

In our function $e^{-2t} \sin 3t$, we have $f(t) = \sin 3t$ and $a = -2$. We need to compute $L\{e^{-2t} \sin 3t\}$.

Using the theorem, we substitute $(s - a)$ for $s$ in $F(s)$. Since $a = -2$, we replace $s$ with $s - (-2) = s+2$.

Step 3: Calculation

We apply the shift to our $F(s) = \frac{3}{s^2 + 9}$:

$L\{e^{-2t} \sin 3t\} = F(s+2) = \frac{3}{(s+2)^2 + 9}$

Step 4: Simplifying the Denominator

Now, we expand the denominator $(s+2)^2 + 9$:

$(s+2)^2 + 9 = (s^2 + 2(s)(2) + 2^2) + 9$

$= (s^2 + 4s + 4) + 9$

$= s^2 + 4s + 13$

So, the final Laplace transform is:

$L\{e^{-2t} \sin 3t\} = \frac{3}{s^2 + 4s + 13}$

Conclusion

The Laplace transform of the function $e^{-2t} \sin 3t$ is $\frac{3}{s^2 + 4s + 13}$. This result is obtained by first finding the transform of $\sin 3t$ and then applying the frequency shifting theorem due to the presence of the $e^{-2t}$ term.