What is inverse oflaplace (e^{-2s})/((s^2+16)) ?

The answer to inverse oflaplace (e^{-2s})/((s^2+16)) is \H(t-2) 1/4 sin(4(t-2))

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inverse oflaplace (e^{-2s})/((s^2+16))

inverse laplace

\frac{e ^{- 2 s}}{( s ^{2} + 16 )}

H (t - 2) \frac{1}{4} sin (4 (t - 2))

Solution steps

L^{- 1} {\frac{e ^{- 2 s}}{( s ^{2} + 16 )}}

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{1}{s ^{2} + 16}} (t - 2)

L^{- 1} {\frac{1}{s ^{2} + 16}} : \frac{1}{4} sin (4 t)

= H (t - 2) \frac{1}{4} sin (4 (t - 2))

y = 4 x, x = 4, y = \frac{4}{x}

\int_{0}^{1} (x^{3} - 1) d x

x \to - 3 lim (\frac{x ^{2}}{x + 3})

\int \frac{3 y}{2 y ^{2} + 5} d y

arctan (8 x)

What is inverse oflaplace (e^{-2s})/((s^2+16)) ?
The answer to inverse oflaplace (e^{-2s})/((s^2+16)) is \H(t-2) 1/4 sin(4(t-2))