What is inverse oflaplace (10)/(s^2) ?

The answer to inverse oflaplace (10)/(s^2) is 10t

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inverse oflaplace (10)/(s^2)

inverse laplace

\frac{10}{s ^{2}}

10 t

Solution steps

L^{- 1} {\frac{10}{s ^{2}}}

= L^{- 1} {10 \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)}

= 10 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

= 10 t

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\int \frac{1}{2 x ln ( x )} d x

\frac{d}{d x} (\frac{7 x}{x ^{2} + 49})

\frac{d ^{4}}{d x ^{4}} (4 e^{x^{2}})

What is inverse oflaplace (10)/(s^2) ?
The answer to inverse oflaplace (10)/(s^2) is 10t