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

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

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

inverse laplace

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

H (t - 2) cos (4 (t - 2))

Solution steps

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

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

= H (t - 2) cos (4 (t - 2))

\int_{0}^{1} cos^{2} (π x) d x

\int \frac{2 - x}{x ^{2}} d x

\frac{d}{d y} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{2}})

\int 54 - 18 z - 6 z^{2} + 2 z^{3} d z

\int_{0}^{1} 2 x 5^{x} d x

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