Question: "Find the Laplace Transform of cos h 3t..."

Name: Find the Laplace Transform of cos h 3t cos 2t
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the Laplace Transform of cos h 3t cos 2t

Answered question

2022-02-08

Find the Laplace Transform of cos h 3t cos 2t

Answer & Explanation

alenahelenash

Expert2023-04-23Added 556 answers

To find the Laplace Transform of $\cosh (3 t) \cos (2 t)$ , we first need to use the product-to-sum identity:

$\cos (A) \cos (B) = \frac{1}{2} [\cos (A + B) + \cos (A - B)]$

Using this identity, we can write:

$\cosh (3 t) \cos (2 t) = \frac{1}{2} [\cos ((3 t + 2 t)) + \cos ((3 t - 2 t))]$
$= \frac{1}{2} [\cos (5 t) + \cos (t)]$

Now, we can use the definition of the Laplace Transform to find its expression:

$L [\cosh (3 t) \cos (2 t)] (s) = \int [0, \infty) e^{- s t} \cdot \frac{1}{2} [\cos (5 t) + \cos (t)] d t$

We can split this integral into two parts:

$L [\cosh (3 t) \cos (2 t)] (s) = \frac{1}{2} \int [0, \infty) e^{- s t} \cos (5 t) d t + \frac{1}{2} \int [0, \infty) e^{- s t} \cos (t) d t$

To evaluate these integrals, we can use the Laplace Transform of cosine:

$L [\cos (a t)] (s) = \frac{s}{s^{2} + a^{2}}$

Using this formula, we get:

$L [\cosh (3 t) \cos (2 t)] (s) = \frac{1}{2} [\frac{s}{s^{2} + 5^{2}} + \frac{s}{s^{2} + 1^{2}}]$

Simplifying this expression, we can write:

$L [\cosh (3 t) \cos (2 t)] (s) = \frac{s}{2} [\frac{s^{2} + 26}{(s^{2} + 5^{2}) (s^{2} + 1^{2})}]$

Therefore, the Laplace Transform of $\cosh (3 t) \cos (2 t)$ is $\frac{s}{2} [\frac{s^{2} + 26}{(s^{2} + 5^{2}) (s^{2} + 1^{2})}]$ , which can also be written in partial fraction form if needed.

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