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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

L^{- 1} {\frac{2}{( s + 2 ) ^{3}}} : e^{- 2 t} t^{2}

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

\int e^{l n (t)} d t

\int - cot (x) d x

\int \frac{35}{x ^{3} - 27} d x

y = x^{3} - 10 x^{2} + 16 x, y = - x^{3} + 10 x^{2} - 16 x

y^{^{'}} = \frac{( x y ^{3} )}{1 + x ^{2}}

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