What is inverse oflaplace 1/(9s-1) ?

The answer to inverse oflaplace 1/(9s-1) is 1/9 e^{t/9}

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inverse oflaplace 1/(9s-1)

inverse laplace

\frac{1}{9 s - 1}

\frac{1}{9} e^{\frac{t}{9}}

Solution steps

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

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

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 - \frac{1}{9}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{s - \frac{1}{9}}} = e^{\frac{t}{9}}

= \frac{1}{9} e^{\frac{t}{9}}

x \to 0 lim (\frac{cos ( 2 x )}{x})

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

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

\int 2 x^{4} x^{5} + 6 d x

\frac{\partial}{\partial x} (sin (7 x - 4 y))

What is inverse oflaplace 1/(9s-1) ?
The answer to inverse oflaplace 1/(9s-1) is 1/9 e^{t/9}