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

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

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

inverse laplace

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

\frac{1}{2} e^{- t} - \frac{1}{2} e^{t} + e^{t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

\frac{\partial}{\partial y} (x^{2} + 5 y^{2})

x \to 3 - lim (- \frac{3}{7} x + \frac{4}{11})

\frac{\partial}{\partial z} (\frac{x}{y + z})

x \to 0 - lim (\frac{( x + 4 )}{( 3 )})

9 y^{^{''}} - 30 y^{^{'}} + 25 y = 0

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