Find the Laplace transforms of the following {L}{leftlbracefrac{{{e}^{{-{6}{t}}} sinh{{left({2}{t}right)}}}}{{{e}^{{-{6}{t}}} cosh{{left({2}{t}right)}}}}rightrbrace}

Question
Laplace transform
asked 2020-10-27
Find the Laplace transforms of the following
\({L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}\)

Answers (1)

2020-10-28
Step 1
We have to evaluate the Laplace transformation \({L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}\)
Use the property of Laplace transformation
\({L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}=\frac{{{L}{\left({e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}\right)}}}{{{L}{\left({e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}\right)}}}\)
Use the property
\({L}{\left({e}^{{{a}{t}}} \sinh{{\left({b}{t}\right)}}\right)}=\frac{b}{{{\left({s}-{a}\right)}^{2}-{b}^{2}}}\ \text{ and }\ {L}{\left({e}^{{{a}{t}}} \cosh{{\left({b}{t}\right)}}\right)}=\frac{{{s}-{a}}}{{{\left({s}-{a}\right)}^{2}-{b}^{2}}}\)
Step 2
Hence,
\({L}{\left({e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}\right)}=\frac{2}{{{\left({s}-{\left(-{6}\right)}\right)}^{2}-{2}^{2}}}\)
\(=\frac{2}{{{\left({s}+{6}\right)}^{2}-{4}}}\)
Hence, \({L}{\left({e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}\right)}=\frac{2}{{{\left({s}+{6}\right)}^{2}-{4}}}\)
And,
\({L}{\left({e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}\right)}=\frac{{{s}-{\left(-{6}\right)}}}{{{\left({s}-{\left(-{6}\right)}\right)}^{2}-{2}^{2}}}\)
\(=\frac{{{s}+{6}}}{{{\left({s}+{6}\right)}^{2}-{4}}}\)
Hence, \({L}{\left({e}{6}{\left(-{6}{t}\right)} \cosh{{\left({2}{t}\right)}}\right)}=\frac{{{s}+{6}}}{{{\left({s}+{6}\right)}^{2}-{4}}}\)
Step 3
Hence,\({L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}=\frac{{{\left(\frac{2}{{\left({s}+{6}\right)}^{2}}-{4}\right)}}}{{\frac{{{s}+{6}}}{{\left({s}+{6}\right)}^{2}}-{4}}}\)
\(=\frac{2}{{{s}+{6}}}\)
Hence, \({L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}=\frac{2}{{{s}+{6}}}\)
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