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

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

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

inverse laplace

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

H (t - 2) sinh (2 (t - 2))

Solution steps

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

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

= H (t - 2) sinh (2 (t - 2))

e^{t} - t

\int \frac{2}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

x \to 4 lim (0.1 x^{2} + 3 x + 2)

\frac{\partial}{\partial x} (cos (x) e^{2 x y} \cdot 2 x)

\int \frac{x ^{5}}{x} d x

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