What is inverse oflaplace (s+2)/(s^2+4s+6) ?

The answer to inverse oflaplace (s+2)/(s^2+4s+6) is e^{-2t}cos(sqrt(2)t)

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inverse oflaplace (s+2)/(s^2+4s+6)

inverse laplace

\frac{s + 2}{s ^{2} + 4 s + 6}

e^{- 2 t} cos (2 t)

Solution steps

L^{- 1} {\frac{s + 2}{s ^{2} + 4 s + 6}}

\frac{s + 2}{s ^{2} + 4 s + 6} = \frac{s + 2}{( s + 2 ) ^{2} + 2}

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

Apply inverse transform rule:

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

then

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

For

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

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

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

= e^{- 2 t} cos (2 t)

What is inverse oflaplace (s+2)/(s^2+4s+6) ?
The answer to inverse oflaplace (s+2)/(s^2+4s+6) is e^{-2t}cos(sqrt(2)t)