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

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

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

inverse laplace

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

- \frac{3}{2} H (t) - t + \frac{3}{2} e^{2 t} - e^{2 t} \cdot 2 t + e^{2 t} t^{2}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

= - \frac{3}{2} H (t) - t + \frac{3}{2} e^{2 t} - e^{2 t} \cdot 2 t + e^{2 t} t^{2}

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