What is inverse oflaplace 4/(s^2) ?

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

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inverse oflaplace 4/(s^2)

inverse laplace

\frac{4}{s ^{2}}

4 t

Solution steps

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

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

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{1}{s ^{2}}}

Use Inverse Laplace Transform table:

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

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

= 4 t

\frac{d}{d x} (ln (e^{a x + b}))

\int \frac{8}{x ^{7}} - \frac{6}{x ^{4}} + 3 x d x

y = 0.4 x^{1.2}

y = x (x - 8), (16, 32)

\int_{0}^{1} x^{2} - 2 x + 1 d x

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