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

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

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inverse oflaplace s/(s^2-pi)

inverse laplace

\frac{s}{s ^{2} - π}

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

Solution steps

L^{- 1} {\frac{s}{s ^{2} - π}}

Take the partial fraction of

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

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

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

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

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

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

y = x^{12} e^{x}

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

\int (2 - sin (θ))^{2} d θ

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

\int 6 cos^{2} (x) d x

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