What is inverse oflaplace s/((s^2-1^2)(s^2+17)) ?

The answer to inverse oflaplace s/((s^2-1^2)(s^2+17)) is 1/36 e^{-t}+1/36 e^t-1/18 cos(sqrt(17)t)

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inverse oflaplace s/((s^2-1^2)(s^2+17))

inverse laplace

\frac{s}{( s ^{2} - 1 ^{2} ) ( s ^{2} + 17 )}

\frac{1}{36} e^{- t} + \frac{1}{36} e^{t} - \frac{1}{18} cos (17 t)

Solution steps

L^{- 1} {\frac{s}{( s ^{2} - 1 ^{2} ) ( s ^{2} + 17 )}}

Take the partial fraction of

\frac{s}{( s ^{2} - 1 ^{2} ) ( s ^{2} + 17 )} : \frac{1}{36 ( s + 1 )} + \frac{1}{36 ( s - 1 )} - \frac{s}{18 ( s ^{2} + 17 )}

= L^{- 1} {\frac{1}{36 ( s + 1 )} + \frac{1}{36 ( s - 1 )} - \frac{s}{18 ( s ^{2} + 17 )}}

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}{36 ( s + 1 )}} + L^{- 1} {\frac{1}{36 ( s - 1 )}} - L^{- 1} {\frac{s}{18 ( s ^{2} + 17 )}}

L^{- 1} {\frac{1}{36 ( s + 1 )}} : \frac{1}{36} e^{- t}

L^{- 1} {\frac{1}{36 ( s - 1 )}} : \frac{1}{36} e^{t}

L^{- 1} {\frac{s}{18 ( s ^{2} + 17 )}} : \frac{1}{18} cos (17 t)

= \frac{1}{36} e^{- t} + \frac{1}{36} e^{t} - \frac{1}{18} cos (17 t)

x \to - \infty lim (e^{x} \cdot x)

\int \frac{x + 10}{x ^{2} + 6 x + 13} d x

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

\int x^{2} tan^{3} (x^{3}) d x

What is inverse oflaplace s/((s^2-1^2)(s^2+17)) ?
The answer to inverse oflaplace s/((s^2-1^2)(s^2+17)) is 1/36 e^{-t}+1/36 e^t-1/18 cos(sqrt(17)t)