find the inverse Laplace transform of the given function. {F}{left({s}right)}=frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}

find the inverse Laplace transform of the given function. {F}{left({s}right)}=frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}

Question
Laplace transform
asked 2021-02-16
find the inverse Laplace transform of the given function.
\({F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}\)

Answers (1)

2021-02-17
Step 1
Given function is:
\({F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}\)
Since the Laplace transform of:
\({L}{\left( \sinh{{a}}{t}\right)}=\frac{a}{{{s}^{2}-{a}^{2}}}\)
And
\({L}{\left( \cosh{{a}}{t}\right)}=\frac{s}{{{s}^{2}-{a}^{2}}}\)
Step 2
Now,
Using the above transform:
\({F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}\)
\(=\frac{{{2}{s}}}{{{s}^{2}-{4}}}-\frac{3}{{{s}^{2}-{4}}}\)
\(={2}\cdot\frac{s}{{{s}^{2}-{2}^{2}}}-\frac{3}{{2}}\cdot\frac{2}{{{s}^{2}-{2}^{2}}}\)
\(={2}{L}{\left( \cosh{{2}}{t}\right)}-\frac{3}{{2}}{L}{\left( \sinh{{2}}{t}\right)}\)
\(={L}{\left({2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}\right)}\)
inverse Laplace transform of \({L}{\left({2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}\right)}\):
\({L}^{ -{{1}}}{\left({L}{\left({2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}\right)}\right)}={2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}\)
THEREFORE Inverse Laplace transform of given function is \({2} \cosh{{2}}{t}-\frac{3}{{2}} \sinh{{2}}{t}\)
0

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