find inverse Laplace transform of the given Function. L^(-1) frac(s-5)((s-5)^2-9)

Tabansi 2021-09-23 Answered
find inverse Laplace transform of the given Function
\(\displaystyle{L}^{{-{1}}}{\frac{{{s}-{5}}}{{{\left({s}-{5}\right)}^{{2}}-{9}}}}\)

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Expert Answer

AGRFTr
Answered 2021-09-24 Author has 2995 answers
Step 1
We have to find inverse Laplace transform of the given Function
Step 2
We have to find
\(\displaystyle{L}^{{-{1}}}{\left[{\frac{{{s}-{5}}}{{{\left({s}-{5}\right)}^{{2}}-{9}}}}\right]}\)
Solution: \(\displaystyle{L}^{{-{1}}}{\left[{\frac{{{s}-{5}}}{{{\left({s}-{5}\right)}^{{2}}-{9}}}}\right]}\)
By shifting theorem inverse Laplace Transform.
\(\displaystyle{L}^{{-{1}}}{\left[{F}{\left({s}-{a}\right)}\right]}={e}^{{{a}{t}}}{L}^{{-{1}}}{\left[{F}{\left({s}\right)}\right]}={e}^{{{a}{t}}}{f{{\left({t}\right)}}}\)
\(\displaystyle\therefore{L}^{{-{1}}}{\left[{\frac{{{s}-{5}}}{{{\left({s}-{5}\right)}^{{2}}-{9}}}}\right]}={e}^{{{s}{t}}}{L}^{{-{1}}}{\left[{\frac{{{s}}}{{{s}^{{2}}-{9}}}}\right]}{\left({1}\right)}\)
\(\displaystyle\therefore{L}^{{-{1}}}{\left[{\frac{{{s}}}{{{s}^{{2}}-{9}}}}\right]}={L}^{{-{1}}}{\left[{\frac{{{s}}}{{{s}^{{2}}-{3}^{{2}}}}}\right]}\)
\(\displaystyle={\cos{{h}}}{\left({3}{t}\right)}{\left[\because{L}^{{-{1}}}{\left[{\frac{{{s}}}{{{s}^{{2}}-{a}^{{2}}}}}\right]}={\cos{{h}}}{\left({a}{t}\right)}\right]}\)
use this value in (1)
\(\displaystyle\therefore{L}^{{-{1}}}{\left[{\frac{{{s}-{5}}}{{{\left({s}-{5}\right)}^{{2}}-{9}}}}\right]}={e}^{{{5}{t}}}{\cos{{h}}}{\left({3}{t}\right)}\)
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