What is inverse oflaplace 1/(s^2+4s+8) ?

The answer to inverse oflaplace 1/(s^2+4s+8) is e^{-2t} 1/2 sin(2t)

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inverse oflaplace 1/(s^2+4s+8)

inverse laplace

\frac{1}{s ^{2} + 4 s + 8}

e^{- 2 t} \frac{1}{2} sin (2 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 4 s + 8}}

\frac{1}{s ^{2} + 4 s + 8} = \frac{1}{( s + 2 ) ^{2} + 4}

= L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 4}}

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s + 2 ) ^{2} + 4} : a = - 2, F (s) = \frac{1}{s ^{2} + 4}

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

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

= e^{- 2 t} \frac{1}{2} sin (2 t)

What is inverse oflaplace 1/(s^2+4s+8) ?
The answer to inverse oflaplace 1/(s^2+4s+8) is e^{-2t} 1/2 sin(2t)