Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace tranform of the function below.

Cabiolab 2021-09-23 Answered
Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace tranform of the function below.
\(\displaystyle{e}^{{-{2}{t}}}{\cos{{6}}}{t}+{e}^{{{5}{t}}}-{1}\)

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Expert Answer

Elberte
Answered 2021-09-24 Author has 17801 answers

Step 1
\(L\left\{e^{-2t}\cos 6t + e^{5t}-1\right\}\)
\(L\left\{e^{at}\right\}=\frac{1}{s-a}\)
\(L\left\{\cos bt \right\}= \frac{s}{s^2+b^2}\)
\(L\left\{e^{at} \cos bt\right\} =\frac{s-a}{(s-a)^2+b^2}\)
\(\displaystyle={\frac{{{s}+{2}}}{{{\left({s}+{2}\right)}^{{2}}+{6}^{{2}}}}}+{\frac{{{1}}}{{{s}-{5}}}}-{\frac{{{1}}}{{{s}}}}\)
\(L\left\{e^{-2t}\cos 6t+e^{5t}-1\right\}=\frac{s+2}{(s+2)^2+36}+\frac{1}{s-5}-\frac{1}{s}\)

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