Chapter 16: Chapter 16

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 $$ f(t)=\cosh k t, \sinh k t, \cos \mathrm{kt}, \text { and } \sin k t $$

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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. $L\\{\sin k t}\\} = \frac{2k}{s^2 + k^2}$

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TABLE OF CONTENTS :

TABLE OF CONTENTS

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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Chapter 16: Chapter 16

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Step 1: Understand the basic Laplace Transform for a function

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

Step 3: Find the Laplace transforms for the hyperbolic functions

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

Step 5: Find the Laplace transforms for the circular functions

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