What is inverse oflaplace (4se^{-4s})/(2s^2+50) ?

The answer to inverse oflaplace (4se^{-4s})/(2s^2+50) is \H(t-4)*2cos(5(t-4))

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inverse oflaplace (4se^{-4s})/(2s^2+50)

inverse laplace

\frac{4 s e ^{- 4 s}}{2 s ^{2} + 50}

H (t - 4) \cdot 2 cos (5 (t - 4))

Solution steps

L^{- 1} {\frac{4 e ^{- 4 s} s}{2 s ^{2} + 50}}

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}{2 s ^{2} + 50}} (t - 4)

L^{- 1} {\frac{4 s}{2 s ^{2} + 50}} : 2 cos (5 t)

= H (t - 4) \cdot 2 cos (5 (t - 4))

\frac{s + 1}{s ^{2} + s + 1}

\frac{d}{d t} (ln (sec (t) + tan (t)))

y^{^{'}} = y (2 x - 5)

\int 24 (1 - x)^{5} d x

\frac{d}{d x} (A (2 x e^{2 x} + e^{2 x} \cdot 2 x^{2}))

What is inverse oflaplace (4se^{-4s})/(2s^2+50) ?
The answer to inverse oflaplace (4se^{-4s})/(2s^2+50) is \H(t-4)*2cos(5(t-4))