What is inverse oflaplace (s^4+1)/(s^3-s) ?

The answer to inverse oflaplace (s^4+1)/(s^3-s) is \delta'(t)-\H(t)+e^{-t}+e^t

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inverse oflaplace (s^4+1)/(s^3-s)

inverse laplace

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

δ' (t) - H (t) + e^{- t} + e^{t}

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {s - \frac{1}{s} + \frac{1}{s + 1} + \frac{1}{s - 1}}

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} {s} - L^{- 1} {\frac{1}{s}} + L^{- 1} {\frac{1}{s + 1}} + L^{- 1} {\frac{1}{s - 1}}

L^{- 1} {s} : δ^{'} (t)

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

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

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

= δ^{^{'}} (t) - H (t) + e^{- t} + e^{t}

\int sin (8 x) cos (4 x) d x

f (x) = \frac{4 x}{x - 3}, at x = 5

(\frac{10}{9}) t^{2}

5 y^{^{''}} + y^{^{'}} = - 6 x

\frac{\partial}{\partial x} (y - 3 x)

What is inverse oflaplace (s^4+1)/(s^3-s) ?
The answer to inverse oflaplace (s^4+1)/(s^3-s) is \delta'(t)-\H(t)+e^{-t}+e^t