Get the answer to "Find the Laplace transforms of the following L{e−6tsinh⁡(2t)e−6tcosh⁡(2t)}"

Tahmid Knox

Answered question

2020-10-27

Find the Laplace transforms of the following
$L {\frac{e^{- 6 t} \sinh (2 t)}{e^{- 6 t} \cosh (2 t)}}$

grbavit

Skilled2020-10-28Added 109 answers

Step 1
We have to evaluate the Laplace transformation

L {\frac{e^{- 6 t} \sinh (2 t)}{e^{- 6 t} \cosh (2 t)}}

Use the property of Laplace transformation

L {\frac{e^{- 6 t} \sinh (2 t)}{e^{- 6 t} \cosh (2 t)}} = \frac{L (e^{- 6 t} \sinh (2 t))}{L (e^{- 6 t} \cosh (2 t))}

Use the property

L (e^{a t} \sinh (b t)) = \frac{b}{{(s - a)}^{2} - b^{2}} and L (e^{a t} \cosh (b t)) = \frac{s - a}{{(s - a)}^{2} - b^{2}}

Step 2
Hence,

L (e^{- 6 t} \sinh (2 t)) = \frac{2}{{(s - (- 6))}^{2} - 2^{2}}

= \frac{2}{{(s + 6)}^{2} - 4}

Hence,

L (e^{- 6 t} \sinh (2 t)) = \frac{2}{{(s + 6)}^{2} - 4}

And,

L (e^{- 6 t} \cosh (2 t)) = \frac{s - (- 6)}{{(s - (- 6))}^{2} - 2^{2}}

= \frac{s + 6}{{(s + 6)}^{2} - 4}

Hence,

L (e 6 (- 6 t) \cosh (2 t)) = \frac{s + 6}{{(s + 6)}^{2} - 4}

Step 3
Hence,

L {\frac{e^{- 6 t} \sinh (2 t)}{e^{- 6 t} \cosh (2 t)}} = \frac{(\frac{2}{{(s + 6)}^{2}} - 4)}{\frac{s + 6}{{(s + 6)}^{2}} - 4}

= \frac{2}{s + 6}

Hence,

L {\frac{e^{- 6 t} \sinh (2 t)}{e^{- 6 t} \cosh (2 t)}} = \frac{2}{s + 6}

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