What is inverse oflaplace (5se^{-s})/(2s^2+32) ?

The answer to inverse oflaplace (5se^{-s})/(2s^2+32) is \H(t-1) 5/2 cos(4(t-1))

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inverse oflaplace (5se^{-s})/(2s^2+32)

inverse laplace

\frac{5 s e ^{- s}}{2 s ^{2} + 32}

H (t - 1) \frac{5}{2} cos (4 (t - 1))

Solution steps

L^{- 1} {\frac{5 e ^{- s} s}{2 s ^{2} + 32}}

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 - 1) L^{- 1} {\frac{5 s}{2 s ^{2} + 32}} (t - 1)

L^{- 1} {\frac{5 s}{2 s ^{2} + 32}} : \frac{5}{2} cos (4 t)

= H (t - 1) \frac{5}{2} cos (4 (t - 1))

\int \frac{x + 1}{( x + 2 ) ^{3}} d x

9 y^{^{''}} - 12 y^{^{'}} + 4 y = e^{- 3 x}

\int \frac{e ^{x}}{e ^{x} + e ^{2 x} - 2} d x

\frac{d}{d x} (x^{2} + 8)

(x + y) d x + x ln (x) d y = 0

What is inverse oflaplace (5se^{-s})/(2s^2+32) ?
The answer to inverse oflaplace (5se^{-s})/(2s^2+32) is \H(t-1) 5/2 cos(4(t-1))