Find the Laplace transforms of the given functions. f{{left({t}right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}

Find the Laplace transforms of the given functions. f{{left({t}right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}

Question
Laplace transform
asked 2021-03-11
Find the Laplace transforms of the given functions.
\(f{{\left({t}\right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}\)

Answers (1)

2021-03-12

Step 1
It can be solve using Laplace transformation table
We know that
\({L}{\left\lbrace{e}^{{{a}{t}}}\right\rbrace}=\frac{1}{{{s}-{a}}},{L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}\)
\(f{{\left({t}\right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}\)
\({L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={6}{L}{\left\lbrace{e}^{{-{5}{t}}}\right\rbrace}+{L}{\left\lbrace{e}^{{{3}{t}}}+{5}{L}{\left\lbrace{t}^{3}\right\rbrace}-{9}{L}{\left\lbrace{1}\right\rbrace}\right.}\)
\(=\frac{6}{{{s}+{5}}}+\frac{1}{{{s}-{3}}}+{5}\cdot\frac{6}{{s}^{3}}-{9}{H}{\left({t}\right)}\)
\(=\frac{{{7}{s}-{13}}}{{{\left({s}+{5}\right)}{\left({s}-{3}\right)}}}+\frac{30}{{{s}^{3}}}-{9}{H}{\left({t}\right)}\) where
\({H}{\left({t}\right)}={\left\lbrace\begin{matrix}{1}&{t}\ge{0}\\{0}&{t}<{0}\end{matrix}\right.}\)

0

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