What is inverse oflaplace 4/((s^3+4s^2+8s)) ?

The answer to inverse oflaplace 4/((s^3+4s^2+8s)) is 1/2 \H(t)-1/2 e^{-2t}cos(2t)-1/2 e^{-2t}sin(2t)

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inverse oflaplace 4/((s^3+4s^2+8s))

inverse laplace

\frac{4}{( s ^{3} + 4 s ^{2} + 8 s )}

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

Solution steps

L^{- 1} {\frac{4}{( s ^{3} + 4 s ^{2} + 8 s )}}

Take the partial fraction of

\frac{4}{s ^{3} + 4 s ^{2} + 8 s} : \frac{1}{2 s} + \frac{- s - 4}{2 ( s ^{2} + 4 s + 8 )}

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

Expand

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

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

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

Refine

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

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

\int csc^{4} (x) cos (x) d x

\int x^{4} 3 x^{5} - 5 d x

\frac{d}{d x} (\frac{5}{x x})

(z, 2), (- 2, - 6)

\int 6 x sec (3 x^{2}) tan (3 x^{2}) d x

What is inverse oflaplace 4/((s^3+4s^2+8s)) ?
The answer to inverse oflaplace 4/((s^3+4s^2+8s)) is 1/2 \H(t)-1/2 e^{-2t}cos(2t)-1/2 e^{-2t}sin(2t)