What is inverse oflaplace (4se^{-2s})/(3s^2+12) ?

The answer to inverse oflaplace (4se^{-2s})/(3s^2+12) is \H(t-2) 4/3 cos(2(t-2))

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inverse oflaplace (4se^{-2s})/(3s^2+12)

inverse laplace

\frac{4 s e ^{- 2 s}}{3 s ^{2} + 12}

H (t - 2) \frac{4}{3} cos (2 (t - 2))

Solution steps

L^{- 1} {\frac{4 e ^{- 2 s} s}{3 s ^{2} + 12}}

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{4 s}{3 s ^{2} + 12}} (t - 2)

L^{- 1} {\frac{4 s}{3 s ^{2} + 12}} : \frac{4}{3} cos (2 t)

= H (t - 2) \frac{4}{3} cos (2 (t - 2))

\frac{d y}{d t} = y + 2 sin (t) + 1

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

\frac{\partial}{\partial x} (x^{2} - 4 x y)

\int (- 0.98 wl x - 0.98 w l^{2}) d x

3 sin (t) cos (t)

What is inverse oflaplace (4se^{-2s})/(3s^2+12) ?
The answer to inverse oflaplace (4se^{-2s})/(3s^2+12) is \H(t-2) 4/3 cos(2(t-2))