What is inverse oflaplace 1/(1-LS) ?

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

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inverse oflaplace 1/(1-LS)

inverse laplace

\frac{1}{1 - L S}

- \frac{e ^{\frac{t}{L}}}{L}

Solution steps

L^{- 1} {\frac{1}{1 - L S}}

= L^{- 1} {- \frac{1}{L} \cdot \frac{1}{S - \frac{1}{L}}}

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}{L} L^{- 1} {\frac{1}{S - \frac{1}{L}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{S - \frac{1}{L}}} = e^{\frac{t}{L}}

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

= - \frac{e ^{\frac{t}{L}}}{L}

\frac{d}{d x} (\frac{- 9}{x ^{2}})

\int e^{- 2 t} d t

y^{^{'}} - \frac{8}{x} y = \frac{y ^{4}}{x ^{17}}, y (1) = 1

x \to 3 + lim (- \frac{2}{x ^{2} - 9})

\int 5 e^{- 0.4 x} d x

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