What is inverse oflaplace 1/(9s) ?

The answer to inverse oflaplace 1/(9s) is 1/9 \H(t)

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inverse oflaplace 1/(9s)

inverse laplace

\frac{1}{9 s}

\frac{1}{9} H (t)

Solution steps

L^{- 1} {\frac{1}{9 s}}

= L^{- 1} {\frac{1}{9} \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}{9} 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}{9} H (t)

\frac{d v}{d t} = \frac{457.8 - 7 v}{2180}

y^{^{'}} = 2 y + x^{2} e^{2 x}

e^{- 3 x} (y + 1) d y = x e^{3 x} (1 + y^{3}) d x

\frac{4 x}{x + 3}

y = \frac{2}{x}

What is inverse oflaplace 1/(9s) ?
The answer to inverse oflaplace 1/(9s) is 1/9 \H(t)