Question

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}

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

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, $${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}=\frac{2}{{{s}+{6}}}$$