Step 1

Consider the provided question,

We have to find \({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}\) Step 2

Now, the given Laplace transform is find as,

\({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}={L}{\left\lbrace{e}^{{-{t}}}\right\rbrace}+{L}{\left\lbrace{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}\)

\(={\left(\frac{1}{{{s}+{1}}}\right)}\cdot\frac{{{s}-{1}}}{{{\left({s}-{1}\right)}^{2}+{1}^{2}}}{\left[{U}{s}{e},{L}{\left({e}^{{{a}{t}}}\right)}=\frac{1}{{{s}-{a}}}{\quad\text{and}\quad}{L}{\left({e}^{{{a}{t}}} \cos{{b}}{t}\right)}=\frac{{{s}-{a}}}{{{\left({s}-{a}\right)}^{2}+{b}^{2}}}\right]}\)

\(=\frac{{{s}-{1}}}{{{\left({s}+{1}\right)}{\left[{\left({s}-{1}\right)}^{2}+{1}\right]}}}\)

Thus , \({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}=\frac{{{s}-{1}}}{{{\left({s}+{1}\right)}{\left[{\left({s}-{1}\right)}^{2}+{1}\right]}}}\)

Consider the provided question,

We have to find \({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}\) Step 2

Now, the given Laplace transform is find as,

\({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}={L}{\left\lbrace{e}^{{-{t}}}\right\rbrace}+{L}{\left\lbrace{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}\)

\(={\left(\frac{1}{{{s}+{1}}}\right)}\cdot\frac{{{s}-{1}}}{{{\left({s}-{1}\right)}^{2}+{1}^{2}}}{\left[{U}{s}{e},{L}{\left({e}^{{{a}{t}}}\right)}=\frac{1}{{{s}-{a}}}{\quad\text{and}\quad}{L}{\left({e}^{{{a}{t}}} \cos{{b}}{t}\right)}=\frac{{{s}-{a}}}{{{\left({s}-{a}\right)}^{2}+{b}^{2}}}\right]}\)

\(=\frac{{{s}-{1}}}{{{\left({s}+{1}\right)}{\left[{\left({s}-{1}\right)}^{2}+{1}\right]}}}\)

Thus , \({L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}=\frac{{{s}-{1}}}{{{\left({s}+{1}\right)}{\left[{\left({s}-{1}\right)}^{2}+{1}\right]}}}\)