What is inverse oflaplace 9/(3s-1) ?

The answer to inverse oflaplace 9/(3s-1) is 3e^{t/3}

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inverse oflaplace 9/(3s-1)

inverse laplace

\frac{9}{3 s - 1}

3 e^{\frac{t}{3}}

Solution steps

L^{- 1} {\frac{9}{3 s - 1}}

= L^{- 1} {3 \cdot \frac{1}{s - \frac{1}{3}}}

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)}

= 3 L^{- 1} {\frac{1}{s - \frac{1}{3}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s - \frac{1}{3}}} = e^{\frac{t}{3}}

= 3 e^{\frac{t}{3}}

\int x^{1} d x

\frac{d}{d x} (ln (\frac{e ^{x} arcsin ( 2 x )}{x + 2}))

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

What is inverse oflaplace 9/(3s-1) ?
The answer to inverse oflaplace 9/(3s-1) is 3e^{t/3}