What is inverse oflaplace 1/(s(s^2+16))e^{-21s} ?

The answer to inverse oflaplace 1/(s(s^2+16))e^{-21s} is \H(t-21)(1/16-1/16 cos(4(t-21)))

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inverse oflaplace 1/(s(s^2+16))e^{-21s}

inverse laplace

\frac{1}{s ( s ^{2} + 16 )} e^{- 21 s}

H (t - 21) (\frac{1}{16} - \frac{1}{16} cos (4 (t - 21)))

Solution steps

L^{- 1} {e^{- 21 s} \frac{1}{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 - 21) L^{- 1} {\frac{1}{s ( s ^{2} + 16 )}} (t - 21)

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

= H (t - 21) (\frac{1}{16} H (t - 21) - \frac{1}{16} cos (4 (t - 21)))

H (t - 21) = {01 t < 21 t \geq 21

= {0 \frac{1}{16} - \frac{1}{16} cos (4 (t - 21)) t < 21 t \geq 21

= H (t - 21) (\frac{1}{16} - \frac{1}{16} cos (4 (t - 21)))

n = 0 \sum \infty (\frac{- 3}{5})^{n}

x^{\frac{1}{3}} + y^{\frac{1}{3}} = 13

\int_{1}^{5} 3 2 x - 1 d x

\int (1 + e^{x}) x d x

\frac{d y}{d x} = 6 x^{7} y^{2} - \frac{y}{x}

What is inverse oflaplace 1/(s(s^2+16))e^{-21s} ?
The answer to inverse oflaplace 1/(s(s^2+16))e^{-21s} is \H(t-21)(1/16-1/16 cos(4(t-21)))