What is inverse oflaplace 1/(8s+1) ?

The answer to inverse oflaplace 1/(8s+1) is 1/8 e^{-t/8}

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inverse oflaplace 1/(8s+1)

inverse laplace

\frac{1}{8 s + 1}

\frac{1}{8} e^{- \frac{t}{8}}

Solution steps

L^{- 1} {\frac{1}{8 s + 1}}

= L^{- 1} {\frac{1}{8} \cdot \frac{1}{s + \frac{1}{8}}}

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}{8} L^{- 1} {\frac{1}{s + \frac{1}{8}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + \frac{1}{8}}} = e^{- \frac{t}{8}}

= \frac{1}{8} e^{- \frac{t}{8}}

\int 5 t e^{5 t} d t

\frac{d y}{d x} = 2 (y^{2} - y) x

y = e^{(3 - x)}

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

\int \frac{e ^{5 x}}{e ^{5 x} + 9} d x

What is inverse oflaplace 1/(8s+1) ?
The answer to inverse oflaplace 1/(8s+1) is 1/8 e^{-t/8}