What is inverse oflaplace 1/(2s) ?

The answer to inverse oflaplace 1/(2s) is 1/2 \H(t)

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inverse oflaplace 1/(2s)

inverse laplace

\frac{1}{2 s}

\frac{1}{2} H (t)

Solution steps

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

= L^{- 1} {\frac{1}{2} \cdot \frac{1}{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)}

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

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s}} = a H (t)

Where

H (t)

is Heaviside step function

= \frac{1}{2} H (t)

f (x) = cos (\frac{1}{x})

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

(3^{2} + 5 x)^{4} (2 x - 2)^{3}

\frac{d y}{d x} + y^{3} x + y = 0

- 2 cos (2 x)

What is inverse oflaplace 1/(2s) ?
The answer to inverse oflaplace 1/(2s) is 1/2 \H(t)