What is inverse oflaplace (e^{-4s})/((5s^2+20)) ?

The answer to inverse oflaplace (e^{-4s})/((5s^2+20)) is \H(t-4) 1/10 sin(2(t-4))

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inverse oflaplace (e^{-4s})/((5s^2+20))

inverse laplace

\frac{e ^{- 4 s}}{( 5 s ^{2} + 20 )}

H (t - 4) \frac{1}{10} sin (2 (t - 4))

Solution steps

L^{- 1} {\frac{e ^{- 4 s}}{( 5 s ^{2} + 20 )}}

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 - 4) L^{- 1} {\frac{1}{5 s ^{2} + 20}} (t - 4)

L^{- 1} {\frac{1}{5 s ^{2} + 20}} : \frac{1}{10} sin (2 t)

= H (t - 4) \frac{1}{10} sin (2 (t - 4))

\int (\frac{7}{t ^{2}} - 5 t^{5}) d t

\frac{\partial}{\partial x} (e^{x} + e^{y})

\int (\frac{3}{x ^{4}}) d x

\frac{\partial}{\partial y} (e^{y x})

\frac{\partial}{\partial x} (4 x + 9 y - 360)

What is inverse oflaplace (e^{-4s})/((5s^2+20)) ?
The answer to inverse oflaplace (e^{-4s})/((5s^2+20)) is \H(t-4) 1/10 sin(2(t-4))