What is inverse oflaplace (s^2-1)/((s-2)^6) ?

The answer to inverse oflaplace (s^2-1)/((s-2)^6) is (e^{2t}t^3)/6+(e^{2t}t^4)/6+(e^{2t}t^5)/(40)

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inverse oflaplace (s^2-1)/((s-2)^6)

inverse laplace

\frac{s ^{2} - 1}{( s - 2 ) ^{6}}

\frac{e ^{2 t} t ^{3}}{6} + \frac{e ^{2 t} t ^{4}}{6} + \frac{e ^{2 t} t ^{5}}{40}

Solution steps

L^{- 1} {\frac{s ^{2} - 1}{( s - 2 ) ^{6}}}

Take the partial fraction of

\frac{s ^{2} - 1}{( s - 2 ) ^{6}} : \frac{1}{( s - 2 ) ^{4}} + \frac{4}{( s - 2 ) ^{5}} + \frac{3}{( s - 2 ) ^{6}}

= L^{- 1} {\frac{1}{( s - 2 ) ^{4}} + \frac{4}{( s - 2 ) ^{5}} + \frac{3}{( s - 2 ) ^{6}}}

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 - 2 ) ^{4}}} + L^{- 1} {\frac{4}{( s - 2 ) ^{5}}} + L^{- 1} {\frac{3}{( s - 2 ) ^{6}}}

L^{- 1} {\frac{1}{( s - 2 ) ^{4}}} : \frac{e ^{2 t} t ^{3}}{6}

L^{- 1} {\frac{4}{( s - 2 ) ^{5}}} : \frac{e ^{2 t} t ^{4}}{6}

L^{- 1} {\frac{3}{( s - 2 ) ^{6}}} : \frac{e ^{2 t} t ^{5}}{40}

= \frac{e ^{2 t} t ^{3}}{6} + \frac{e ^{2 t} t ^{4}}{6} + \frac{e ^{2 t} t ^{5}}{40}

What is inverse oflaplace (s^2-1)/((s-2)^6) ?
The answer to inverse oflaplace (s^2-1)/((s-2)^6) is (e^{2t}t^3)/6+(e^{2t}t^4)/6+(e^{2t}t^5)/(40)