Let F(s) be the Laplace tranform of f(x)=t cos (2t) find the value of F(1)

he298c 2021-09-27 Answered
Let F(s) be the Laplace tranform of \(\displaystyle{f{{\left({x}\right)}}}={t}{\cos{{\left({2}{t}\right)}}}\)
find the value of F(1)

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Expert Answer

Elberte
Answered 2021-09-28 Author has 17801 answers

Step 1
Determine the Laplace transform of \(\displaystyle{f{{\left({t}\right)}}}={t}{\cos{{\left({2}{t}\right)}}}\)
\(L\left\{f(t)\right\}=L\left\{t \cos (2t)\right\}\)
\(\displaystyle{F}{\left({s}\right)}=-{\frac{{{d}}}{{{d}{s}}}}{\left({\frac{{{s}}}{{{s}^{{2}}+{2}^{{2}}}}}\right)}\)
\(\displaystyle=-{\frac{{{\left({s}^{{2}}+{4}\right)}-{2}{s}^{{2}}}}{{{\left({s}^{{2}}+{4}\right)}^{{2}}}}}\)
\(\displaystyle={\frac{{{s}^{{2}}-{4}}}{{{\left({s}^{{2}}+{4}\right)}^{{2}}}}}\)
step 2
Substitute 1 for s in the equation \(\displaystyle{F}{\left({s}\right)}={\frac{{{s}^{{2}}-{4}}}{{{\left({s}^{{2}}+{4}\right)}^{{2}}}}}\)
\(\displaystyle{F}{\left({1}\right)}={\frac{{{1}^{{2}}-{4}}}{{{\left({1}^{{2}}+{4}\right)}^{{2}}}}}\)
\(\displaystyle={\frac{{{1}-{4}}}{{{5}^{{2}}}}}\)
\(\displaystyle={\frac{{-{3}}}{{{25}}}}\)
Therefore \(\displaystyle{F}{\left({1}\right)}={\frac{{-{3}}}{{{25}}}}\)

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