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} ((1 + \frac{1}{x})^{3})

\int \frac{- 30}{( 2 x + 1 ) ^{2}} + 30 d x

\frac{\partial}{\partial x} ((ln (x))^{x})

\int - 8 sin (x) d x

\int - x^{4} + \frac{( 9 x ^{2} )}{4} - 2 x + 3 d x

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))