What is inverse oflaplace 8/(s^3) ?

The answer to inverse oflaplace 8/(s^3) is 4t^2

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inverse oflaplace 8/(s^3)

inverse laplace

\frac{8}{s ^{3}}

4 t^{2}

Solution steps

L^{- 1} {\frac{8}{s ^{3}}}

= L^{- 1} {4 \cdot \frac{2}{s ^{3}}}

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)}

= 4 L^{- 1} {\frac{2}{s ^{3}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{2}{s ^{3}}} = t^{2}

= 4 t^{2}

\frac{d}{d x} (- \frac{4}{x})

x \to \frac{π}{2} lim (2 tan (x))

\int \frac{8 - 128 x}{25 - 64 x ^{2}} d x

f (x) = x^{- 6}

\int \frac{1}{x ^{2} + 8 x + 17} d x

What is inverse oflaplace 8/(s^3) ?
The answer to inverse oflaplace 8/(s^3) is 4t^2