What is inverse oflaplace (4se^{-4s})/(3s^2+27) ?

The answer to inverse oflaplace (4se^{-4s})/(3s^2+27) is \H(t-4) 4/3 cos(3(t-4))

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inverse oflaplace (4se^{-4s})/(3s^2+27)

inverse laplace

\frac{4 s e ^{- 4 s}}{3 s ^{2} + 27}

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

Solution steps

L^{- 1} {\frac{4 e ^{- 4 s} s}{3 s ^{2} + 27}}

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{4 s}{3 s ^{2} + 27}} (t - 4)

L^{- 1} {\frac{4 s}{3 s ^{2} + 27}} : \frac{4}{3} cos (3 t)

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

\int [x ln (x^{2} + 1)] d x

y^{^{''}} + y = sin (t), y (0) = 0, y^{^{'}} (0) = 0

x y y^{^{'}} = x^{2} e^{\frac{y}{x}} + y^{2}

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

(x + 3) (\frac{d y}{d x}) + y = ln (x)

What is inverse oflaplace (4se^{-4s})/(3s^2+27) ?
The answer to inverse oflaplace (4se^{-4s})/(3s^2+27) is \H(t-4) 4/3 cos(3(t-4))