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Problem 474

# Use the Laplace transform of $$f(t)=e^{k t}$$ where $$\mathrm{k}$$ is a complex constant of the form $\mathrm{k}=\operatorname{Re}\\{\mathrm{k}\\}+\mathrm{i} \operatorname{lm}\\{\mathrm{k}\\}$ with $$\operatorname{Re}\\{k\\}$$ the real part of $$k, \operatorname{lm}\\{k\\}$$ the imaginary part of $$k$$, and $$\mathrm{i} \equiv \sqrt{(-1)}$$ to find the Laplace transforms of $$f(t)=\cosh k t, \sinh k t, \cos \mathrm{kt}, \text { and } \sin k t$$

Expert verified
In summary, the Laplace transforms for the given functions are: 1. $$L\\{\cosh k t}\\} = \frac{2s}{s^2 - k^2}$$ 2. $$L\\{\sinh k t}\\} = \frac{2k}{s^2 - k^2}$$ 3. $$L\\{\cos k t}\\} = \frac{2s}{s^2 + k^2}$$ 4. $$L\\{\sin k t}\\} = \frac{2k}{s^2 + k^2}$$
See the step by step solution

## Step 1: Understand the basic Laplace Transform for a function

The Laplace transform of $$e^{k t}$$ is given by $L\\{e^{k t}\\}= \int_{0}^{\infty} e^{-s t}e^{k t} dt$. This is simplified to $$L\\{e^{k t}\\}= \int_{0}^{\infty} e^{(k-s) t} dt$$, and further simplified to $$L\\{e^{k t}\\}= \frac{1}{s-k}$$ for $$s > \mathrm{Re}\\{k\\}$$.

## Step 2: Break down the hyperbolic function terms using exponential function

Hyperbolic functions could be written in terms of exponential functions, i.e., $$\cosh k t = \frac{e^{k t} + e^{-k t}}{2}$$ and $$\sinh k t = \frac{e^{k t} - e^{-k t}}{2}$$. By knowing the Laplace Transform of $$e^{k t}$$, it is possible to find the Laplace transforms for these two functions.

## Step 3: Find the Laplace transforms for the hyperbolic functions

Using the similarity principle of the Laplace Transformation, we find: $L\\{e^{k t}+ e^{-k t}\\} = L\\{e^{k t}\\} + L\\{e^{-k t}\\}$ So, the Laplace Transform for $$\cosh k t$$ becomes: $L\\{\cosh k t}\\} = \frac{1}{s-k} + \frac{1}{s+k} = \frac{2s}{s^2 - k^2}$ Similarly, the Laplace Transform for $$\sinh k t$$ becomes: $L\\{\sinh k t}\\} = \frac{1}{s-k} - \frac{1}{s+k} = \frac{2k}{s^2 - k^2}$

## Step 4: Break down the circular function terms using Euler's formula

The circular functions can be written in terms of exponential functions using Euler's formula, i.e., $$\cos k t = \frac{e^{ikt} + e^{-ikt}}{2}$$ and $$\sin k t = \frac{e^{ikt} - e^{-ikt}}{2i}$$. By knowing the Laplace Transform of $$e^{ikt}$$, it is possible to find the Laplace transforms for these two functions.

## Step 5: Find the Laplace transforms for the circular functions

Using the similarity principle of the Laplace Transform, we can find the Laplace transforms for cosine and sine. The Laplace transform for $$\cos k t$$ becomes: $L\\{\cos k t}\\} = \frac{1}{s - ik} + \frac{1}{s + ik} = \frac{2s}{s^2 + k^2}$ Similarly, the Laplace transform for $$\sin k t$$ becomes: $L\\{\sin k t}\\} = \frac{1}{s - ik} - \frac{1}{s + ik} = \frac{-2k}{s^2 + k^2}$ Remember to divide by $$i$$ as $$\sin kt = \frac{e^{i k t} - e^{-i k t}}{2i}$$, so this complex number $$i$$ needs to be accounted for in the transform. This results in \(L\\{\sin k t}\\} = \frac{2k}{s^2 + k^2}\] End of solution.

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