What is inverse oflaplace 1/(4s^2) ?

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

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inverse oflaplace 1/(4s^2)

inverse laplace

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

\frac{t}{4}

Solution steps

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

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

= \frac{1}{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

= \frac{1}{4} t

= \frac{t}{4}

y^{^{'}} - y = 3 t e^{2 t}, y (0) = 1

2 sin (t)

\frac{\partial}{\partial x} (y e^{(\frac{x}{y})})

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 4 x}{x + 1})

\frac{d}{d x} (3 x^{2} 5 x + 3)

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