What is inverse oflaplace (3se^{-3s})/(4s^2+64) ?

The answer to inverse oflaplace (3se^{-3s})/(4s^2+64) is \H(t-3) 3/4 cos(4(t-3))

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inverse oflaplace (3se^{-3s})/(4s^2+64)

inverse laplace

\frac{3 s e ^{- 3 s}}{4 s ^{2} + 64}

H (t - 3) \frac{3}{4} cos (4 (t - 3))

Solution steps

L^{- 1} {\frac{3 e ^{- 3 s} s}{4 s ^{2} + 64}}

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 - 3) L^{- 1} {\frac{3 s}{4 s ^{2} + 64}} (t - 3)

L^{- 1} {\frac{3 s}{4 s ^{2} + 64}} : \frac{3}{4} cos (4 t)

= H (t - 3) \frac{3}{4} cos (4 (t - 3))

\frac{d}{d x} (x^{\frac{4}{3}} - x^{\frac{1}{3}})

f (x) = (7 - x) (\frac{5}{x ^{2}} - 8)

\frac{d y}{d x}, y = x^{2} + 1

\int e^{6 θ} sin (7 θ) d θ

x \to \infty lim ((3)^{x})

What is inverse oflaplace (3se^{-3s})/(4s^2+64) ?
The answer to inverse oflaplace (3se^{-3s})/(4s^2+64) is \H(t-3) 3/4 cos(4(t-3))