What is inverse oflaplace (e^{-4s})/(2s-1) ?

The answer to inverse oflaplace (e^{-4s})/(2s-1) is \H(t-4) 1/2 e^{(t-4)/2}

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inverse oflaplace (e^{-4s})/(2s-1)

inverse laplace

\frac{e ^{- 4 s}}{2 s - 1}

H (t - 4) \frac{1}{2} e^{\frac{t - 4}{2}}

Solution steps

L^{- 1} {\frac{e ^{- 4 s}}{2 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 - 4) L^{- 1} {\frac{1}{2 s - 1}} (t - 4)

L^{- 1} {\frac{1}{2 s - 1}} : \frac{1}{2} e^{\frac{t}{2}}

= H (t - 4) \frac{1}{2} e^{\frac{t - 4}{2}}

x^{2} sin (x)

x \to \frac{1}{5} + lim (\frac{x}{10 x - 2})

\int \frac{2 x}{3 x + 1} d x

\int x (cos^{3} (x^{2}) - sin^{3} (x^{2})) d x

\frac{d}{d x} (x^{3} - 9 x^{2} + 27 x)

What is inverse oflaplace (e^{-4s})/(2s-1) ?
The answer to inverse oflaplace (e^{-4s})/(2s-1) is \H(t-4) 1/2 e^{(t-4)/2}