What is inverse oflaplace 1/(4s^3) ?

The answer to inverse oflaplace 1/(4s^3) is (t^2)/8

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inverse oflaplace 1/(4s^3)

inverse laplace

\frac{1}{4 s ^{3}}

\frac{t ^{2}}{8}

Solution steps

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

= L^{- 1} {\frac{1}{8} \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)}

= \frac{1}{8} 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}

= \frac{1}{8} t^{2}

= \frac{t ^{2}}{8}

x \to 1 lim (\frac{x - 1}{1 - x})

\int \frac{1}{( x ^{2} + 2 x + 3 ) ^{2}} d x

\frac{d}{d x} (\frac{4}{1 - x})

\frac{d}{d x} (sin (x) + x cos (x))

7 t^{2}

What is inverse oflaplace 1/(4s^3) ?
The answer to inverse oflaplace 1/(4s^3) is (t^2)/8