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

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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Elberte

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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