What is inverse oflaplace (2s^2)/(s^4-1) ?

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

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inverse oflaplace (2s^2)/(s^4-1)

inverse laplace

\frac{2 s ^{2}}{s ^{4} - 1}

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

\int_{0}^{1} 2 x (x^{2} + 1)^{2} d x

f (x) = \frac{6}{x ^{2}}

x, 64

\int + \frac{x}{3 - x ^{4}} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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