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

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

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

inverse laplace

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

H (t - 1) \frac{1}{2} sinh (2 (t - 1))

Solution steps

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

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

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

= H (t - 1) \frac{1}{2} sinh (2 (t - 1))

\frac{d}{d x} (e^{2 x} cos (2 x))

h \to 0 lim ((1 + \frac{h}{x})^{\frac{1}{h}})

\frac{d}{d x} (6 x^{- \frac{5}{6}})

\frac{\partial}{\partial x} (5 x + 2 y)

x \to 0 lim (x^{- 5})

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