What is inverse oflaplace (e^{-pis})/(s-1) ?

The answer to inverse oflaplace (e^{-pis})/(s-1) is \H(t-pi)e^{t-pi}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace (e^{-pis})/(s-1)

inverse laplace

\frac{e ^{- π s}}{s - 1}

H (t - π) e^{t - π}

Solution steps

L^{- 1} {\frac{e ^{- π s}}{s - 1}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - π) L^{- 1} {\frac{1}{s - 1}} (t - π)

L^{- 1} {\frac{1}{s - 1}} : e^{t}

= H (t - π) e^{t - π}

\frac{d}{d x} (2 x^{6})

d er i v a t i v e - 2 t (6 t^{6} - 1)^{6}

l a pl a ce t r an s f or m e^{- 6 t}

\int - \frac{k}{m} d t

\int arccos (a x) d x

What is inverse oflaplace (e^{-pis})/(s-1) ?
The answer to inverse oflaplace (e^{-pis})/(s-1) is \H(t-pi)e^{t-pi}