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

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

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

inverse laplace

\frac{( e ^{- 4 s} )}{s + 2}

H (t - 4) e^{- 2 (t - 4)}

Solution steps

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

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}{s + 2}} (t - 4)

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

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

\frac{d}{d x} ((4^{x})^{9})

\frac{\partial}{\partial y} (5 x^{\frac{4}{5}} y^{\frac{1}{5}})

\int cos^{2} (t) + sin^{2} (t) d t

\frac{\partial}{\partial z} (8 y arctan (\frac{x}{z}))

x \to 8 lim (5 x - 2)

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