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

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

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

inverse laplace

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

\frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} cos (2 t) - \frac{1}{4} e^{- 2 t} sin (2 t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

L^{- 1} {\frac{1}{4 s}} : \frac{1}{4} H (t)

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)

= \frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} cos (2 t) - \frac{1}{2} e^{- 2 t} \frac{1}{2} sin (2 t)

Refine

\frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} cos (2 t) - \frac{1}{2} e^{- 2 t} \frac{1}{2} sin (2 t) : \frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} cos (2 t) - \frac{1}{4} e^{- 2 t} sin (2 t)

= \frac{1}{4} H (t) - \frac{1}{4} e^{- 2 t} cos (2 t) - \frac{1}{4} e^{- 2 t} sin (2 t)

\int_{2}^{3} π (4 - \frac{1}{2} x)^{2} d x

\int - 4 x^{3} e^{- 2 x^{2}} d x

\frac{( s + 2 )}{s + 1}

(2 y + 2 t - 3) d t + (4 y + 2 t - 1) d y = 0

\frac{\partial}{\partial x} (7 e^{x^{10} y})

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