What is inverse oflaplace 1/(25s^2-1) ?

The answer to inverse oflaplace 1/(25s^2-1) is -1/10 e^{-t/5}+1/10 e^{t/5}

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inverse oflaplace 1/(25s^2-1)

inverse laplace

\frac{1}{25 s ^{2} - 1}

- \frac{1}{10} e^{- \frac{t}{5}} + \frac{1}{10} e^{\frac{t}{5}}

Solution steps

L^{- 1} {\frac{1}{25 s ^{2} - 1}}

Take the partial fraction of

\frac{1}{25 s ^{2} - 1} : - \frac{1}{2 ( 5 s + 1 )} + \frac{1}{2 ( 5 s - 1 )}

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

L^{- 1} {\frac{1}{2 ( 5 s + 1 )}} : \frac{1}{10} e^{- \frac{t}{5}}

L^{- 1} {\frac{1}{2 ( 5 s - 1 )}} : \frac{1}{10} e^{\frac{t}{5}}

= - \frac{1}{10} e^{- \frac{t}{5}} + \frac{1}{10} e^{\frac{t}{5}}

\int t^{2} e^{t^{3}} d t

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

t \cdot \frac{d y}{d t} + 14 y = \frac{1}{t}

\frac{d}{d x} (a^{x^{2}})

\int \frac{( ln ( x ) ) ^{2}}{x} d x

What is inverse oflaplace 1/(25s^2-1) ?
The answer to inverse oflaplace 1/(25s^2-1) is -1/10 e^{-t/5}+1/10 e^{t/5}