What is inverse oflaplace (e^{-4s})/(s(s^2+16)) ?

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

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inverse oflaplace (e^{-4s})/(s(s^2+16))

inverse laplace

\frac{e ^{- 4 s}}{s ( s ^{2} + 16 )}

H (t - 4) (\frac{1}{16} - \frac{1}{16} cos (4 (t - 4)))

Solution steps

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

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

= H (t - 4) (\frac{1}{16} H (t - 4) - \frac{1}{16} cos (4 (t - 4)))

H (t - 4) = {01 t < 4 t \geq 4

= {0 \frac{1}{16} - \frac{1}{16} cos (4 (t - 4)) t < 4 t \geq 4

= H (t - 4) (\frac{1}{16} - \frac{1}{16} cos (4 (t - 4)))

\frac{d y}{d x} = - 3 y

(6 x + 7)^{x}

\frac{d}{d x} (x^{2} arccos (x))

\int \frac{11 x + 17}{2 x ^{2} + 7 x - 4} d x

\int \frac{1}{3 x + 7 x} d x

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