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

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

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inverse oflaplace 1/(s^3)

inverse laplace

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

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

Solution steps

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

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

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

x^{2} y^{^{'}} - 2 x y = 3 y^{4}, y (1) = \frac{1}{2}

\int \frac{1}{e ^{x} + e ^{2 x}} d x

\int \frac{tan ^{3} ( ln ( x ) )}{7 x} d x

8 e^{3 x} + e^{- x^{2}}

\int \frac{x ^{5}}{( 1 - x ^{6} ) ^{7}} d x

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