What is inverse oflaplace-1/(4s) ?

The answer to inverse oflaplace-1/(4s) is -1/4 \H(t)

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inverse oflaplace-1/(4s)

inverse laplace

- \frac{1}{4 s}

- \frac{1}{4} H (t)

Solution steps

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

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)}

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

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

= - \frac{1}{4} H (t)

f (x) = x^{π^{2}} + (π^{2})^{x}

n = 1 \sum \infty ln (\frac{1}{1 + 1})

\frac{\partial}{\partial y} (- x y^{2} + x y)

\frac{\partial}{\partial x} (x^{3} ln (y^{2}))

x \to 3 + lim (4 x - a + b)

What is inverse oflaplace-1/(4s) ?
The answer to inverse oflaplace-1/(4s) is -1/4 \H(t)