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

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

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

inverse laplace

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

H (t) - 2 t + 2 t^{2} - e^{- 2 t}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {\frac{1}{s} - \frac{2}{s ^{2}} + \frac{4}{s ^{3}} - \frac{1}{s + 2}}

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

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

L^{- 1} {\frac{2}{s ^{2}}} : 2 t

L^{- 1} {\frac{4}{s ^{3}}} : 2 t^{2}

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

= H (t) - 2 t + 2 t^{2} - e^{- 2 t}

4 - x^{2}, (- 2, 2)

\frac{7}{2 7 x + 2}

f (x) = 3 x^{- 4}

\int \frac{3 t}{3 t ^{2} + 3} d t

f (x) = e^{- 4 x} cos (- 4 x)

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