What is inverse oflaplace 5/(2(s+2)^2+8) ?

The answer to inverse oflaplace 5/(2(s+2)^2+8) is e^{-2t} 5/4 sin(2t)

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inverse oflaplace 5/(2(s+2)^2+8)

inverse laplace

\frac{5}{2 ( s + 2 ) ^{2} + 8}

e^{- 2 t} \frac{5}{4} sin (2 t)

Solution steps

L^{- 1} {\frac{5}{2 ( s + 2 ) ^{2} + 8}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{5}{2 ( s + 2 ) ^{2} + 8} : a = - 2, F (s) = \frac{5}{2 s ^{2} + 8}

= e^{- 2 t} L^{- 1} {\frac{5}{2 s ^{2} + 8}}

L^{- 1} {\frac{5}{2 s ^{2} + 8}} : \frac{5}{4} sin (2 t)

= e^{- 2 t} \frac{5}{4} sin (2 t)

f (x) = 2 x^{3} + 4 x^{2} - 5 x - 3, at x = - 1

\frac{\partial}{\partial y} (\frac{3 x}{2 y})

\int_{- \infty}^{\infty} 1 d x

\int 5 t \cdot sin^{2} (t) d t

\frac{d}{d x} (arctan (e^{3 x}))

What is inverse oflaplace 5/(2(s+2)^2+8) ?
The answer to inverse oflaplace 5/(2(s+2)^2+8) is e^{-2t} 5/4 sin(2t)