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 sin(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} sin (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} sin (2 t)

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

\frac{d}{d x} (6 sin^{3} (x))

\frac{d y}{d x} = 4 y e^{x}

\frac{\partial}{\partial x} (ln (9 x^{2} + 2 y^{2} + 4))

\frac{\partial}{\partial x} (x + \frac{y}{z})

y^{^{''}} - 8 y^{^{'}} + 17 y = 0

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 sin(2(t-1))