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

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

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

inverse laplace

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

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

Solution steps

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

Expand

\frac{s}{s ^{2} + 4 s + 8} : \frac{s + 2}{( s + 2 ) ^{2} + 4} - 2 \cdot \frac{1}{( s + 2 ) ^{2} + 4}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

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

Refine

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

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

g (x) = - x^{2} + 7 x + 7

y^{^{''}} + 16 y^{^{'}} + 65 y = 0, y (0) = 1, y^{^{'}} (0) = 0

f (x) = 8 x^{\frac{1}{3}} - 5, at x = 8

3 y y^{^{'}} = x

\frac{d}{d t} (2^{t} - 2 t)

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