Determine the inverse Laplace transform of the function below. \frac{3}{(2s+7)^4} L^{-1}\{\frac{3}{(2s+7)^4}\}=???

Jessie Lee 2021-12-19 Answered
Determine the inverse Laplace transform of the function below.
\(\displaystyle{\frac{{{3}}}{{{\left({2}{s}+{7}\right)}^{{4}}}}}\)
\(\displaystyle{L}^{{-{1}}}{\left\lbrace{\frac{{{3}}}{{{\left({2}{s}+{7}\right)}^{{4}}}}}\right\rbrace}=\)???

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Expert Answer

ramirezhereva
Answered 2021-12-20 Author has 1321 answers
\(\displaystyle{L}^{{-{1}}}{\left\lbrace{\frac{{{3}}}{{{\left({2}{s}+{7}\right)}^{{4}}}}}\right\rbrace}={L}^{{-{1}}}{\left\lbrace{\frac{{{3}}}{{{2}^{{4}}{\left({s}+{\frac{{{7}}}{{{2}}}}\right)}^{{4}}}}}\right\rbrace}\)
\(\displaystyle={\frac{{{3}}}{{{2}^{{4}}}}}{L}^{{-{1}}}{\left\lbrace{\frac{{{1}}}{{{\left({s}+{\frac{{{7}}}{{{2}}}}\right)}^{{4}}}}}\right\rbrace}\)
\(\displaystyle={\frac{{{3}}}{{{2}^{{4}}}}}\times{L}^{{-{1}}}{\left\lbrace{\frac{{{1}}}{{{3}!}}}\times{\frac{{{3}!}}{{{\left({s}+{\frac{{{7}}}{{{2}}}}\right)}^{{{3}+{1}}}}}}\right\rbrace}\)
\(\displaystyle={\frac{{{3}}}{{{2}^{{4}}}}}\times{\frac{{{1}}}{{{3}!}}}\times{L}^{{-{1}}}{\left\lbrace{\frac{{{3}!}}{{{\left({s}+{\frac{{{7}}}{{{2}}}}\right)}^{{{3}+{1}}}}}}\right\rbrace}\)
\(\displaystyle={\frac{{{3}}}{{{2}^{{4}}}}}\times{\frac{{{1}}}{{{3}!}}}\times{e}^{{{\frac{{{7}}}{{{2}}}}{t}}}\times{t}^{{3}}\)
\(\displaystyle={\frac{{{3}}}{{{16}\times{3}\times{2}\times{1}}}}\times{e}^{{{\frac{{{7}}}{{{2}}}}{t}}}\times{t}^{{3}}\)
\(\displaystyle{L}^{{-{1}}}{\left\lbrace{\frac{{{3}}}{{{\left({2}{s}+{7}\right)}^{{4}}}}}\right\rbrace}={\frac{{{1}}}{{{32}}}}\times{e}^{{{\frac{{{7}}}{{{2}}}}{t}}}\times{t}^{{3}}\)
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Bubich13
Answered 2021-12-21 Author has 4612 answers
How did you get to the 16?
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nick1337
Answered 2021-12-28 Author has 9672 answers

by taking common 2 from denominator we got denominator as \(2^4=16\)

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