What is inverse oflaplace 7/(2s^{2+8)} ?

The answer to inverse oflaplace 7/(2s^{2+8)} is (t^9)/(103680)

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inverse oflaplace 7/(2s^{2+8)}

inverse laplace

\frac{7}{2 s ^{2 + 8}}

\frac{t ^{9}}{103680}

Solution steps

L^{- 1} {\frac{7}{2 s ^{2 + 8}}}

= L^{- 1} {\frac{1}{103680} \cdot \frac{362880}{s ^{10}}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{1}{103680} L^{- 1} {\frac{362880}{s ^{10}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{( n ) !}{s ^{n + 1}}} = t^{n}

L^{- 1} {\frac{362880}{s ^{10}}} = t^{9}

= \frac{1}{103680} t^{9}

= \frac{t ^{9}}{103680}

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\int \frac{- sin ( x ) tan ( x )}{1} d x

\frac{d}{d x} (\frac{14}{x})

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What is inverse oflaplace 7/(2s^{2+8)} ?
The answer to inverse oflaplace 7/(2s^{2+8)} is (t^9)/(103680)