What is inverse oflaplace 1/(5s) ?

The answer to inverse oflaplace 1/(5s) is 1/5 \H(t)

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inverse oflaplace 1/(5s)

inverse laplace

\frac{1}{5 s}

\frac{1}{5} H (t)

Solution steps

L^{- 1} {\frac{1}{5 s}}

= L^{- 1} {\frac{1}{5} \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}{5} 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}{5} H (t)

\frac{\partial}{\partial x} (x yz e^{x^{2} + y^{2} + z^{2}})

\int x 4 a^{2} - x^{2} d x

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

x \to \infty lim (- 1 x)

y = x, y = 0.2 x^{2} - 1.2 x + 3

What is inverse oflaplace 1/(5s) ?
The answer to inverse oflaplace 1/(5s) is 1/5 \H(t)