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

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

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inverse oflaplace 3/(s-9)

inverse laplace

\frac{3}{s - 9}

3 e^{9 t}

Solution steps

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

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

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

Use Inverse Laplace Transform table:

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

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

= 3 e^{9 t}

\frac{d}{d x} (f (x) g (x))

\frac{y}{y + \frac{a}{y}}

n = 0 \sum \infty \frac{n}{e ^{2 n}}

\int \frac{e ^{4 x} - e ^{2 x} + 1}{e ^{x}} d x

y^{^{'}} = y

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