Using the result from part (a) and the general properties of Laplace transforms, find the Laplace transforms of the following functions: f1(t) = e3t cosh(t) f2(t) = sinh(t) f3(t) = t cosh(t)

To find the Laplace transforms of the given functions using the result from part (a) and the general properties of Laplace transforms, we'll use several properties of Laplace transforms:

1. **Shifting Theorem**: \( \mathcal{L}\{e^{at}f(t)\} = F(s-a) \)
2. **Linearity**: \( \mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s) \)
3. **Differentiation in the Time Domain**: \( \mathcal{L}\{t f(t)\} = -\frac{d}{ds}F(s) \)

From part (a), we know that:
\[
\mathcal{L}\{\cosh(t)\} = \frac{s}{s^2 - 1}
\]

### Finding the Laplace Transforms:

#### 1. \( f_1(t) = e^{3t} \cosh(t) \)

Using the shifting theorem:
\[
\mathcal{L}\{e^{3t} \cosh(t)\} = \mathcal{L}\{\cosh(t)\} \Big|_{s \to s - 3}
\]
Substituting \( s - 3 \) for \( s \) in \( \frac{s}{s^2 - 1} \):
\[
\mathcal{L}\{e^{3t} \cosh(t)\} = \frac{s-3}{(s-3)^2 - 1} = \frac{s-3}{s^2 - 6s + 9 - 1} = \frac{s-3}{s^2 - 6s + 8}
\]

#### 2. \( f_2(t) = \sinh(t) \)

We can use the definition of \( \sinh(t) \) and the linearity property:
\[
\sinh(t) = \frac{e^t - e^{-t}}{2}
\]
Thus:
\[
\mathcal{L}\{\sinh(t)\} = \frac{1}{2} \mathcal{L}\{e^t\} - \frac{1}{2} \mathcal{L}\{e^{-t}\}
\]
Using the Laplace transform of \( e^{at} \):
\[
\mathcal{L}\{e^{at}\} = \frac{1}{s-a}
\]
For \( e^t \) and \( e^{-t} \):
\[
\mathcal{L}\{e^t\} = \frac{1}{s-1} \quad \text{and} \quad \mathcal{L}\{e^{-t}\} = \frac{1}{s+1}
\]
So:
\[
\mathcal{L}\{\sinh(t)\} = \frac{1}{2} \left( \frac{1}{s-1} - \frac{1}{s+1} \right)
\]
Combining the fractions:
\[
\mathcal{L}\{\sinh(t)\} = \frac{1}{2} \left( \frac{(s+1) - (s-1)}{(s-1)(s+1)} \right) = \frac{1}{2} \left( \frac{s+1-s+1}{s^2 - 1} \right) = \frac{1}{2} \left( \frac{2}{s^2 - 1} \right) = \frac{1}{s^2 - 1}
\]

#### 3. \( f_3(t) = t \cosh(t) \)

Using the differentiation property:
\[
\mathcal{L}\{t \cosh(t)\} = -\frac{d}{ds} \mathcal{L}\{\cosh(t)\}
\]
We already have:
\[
\mathcal{L}\{\cosh(t)\} = \frac{s}{s^2 - 1}
\]
Differentiate this with respect to \( s \):
\[


Explore how to find the Laplace transforms of exponential and hyperbolic functions using the Shifting Theorem, Linearity, and Differentiation property of Laplace transforms. Detailed solutions for f1(t) = e^3t cosh(t), f2(t) = sinh(t), and f3(t) = t cosh(t) provided.

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