What is inverse oflaplace (3s+4)/(s^2-1) ?

The answer to inverse oflaplace (3s+4)/(s^2-1) is -1/2 e^{-t}+7/2 e^t

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inverse oflaplace (3s+4)/(s^2-1)

inverse laplace

\frac{3 s + 4}{s ^{2} - 1}

- \frac{1}{2} e^{- t} + \frac{7}{2} e^{t}

Solution steps

L^{- 1} {\frac{3 s + 4}{s ^{2} - 1}}

Take the partial fraction of

\frac{3 s + 4}{s ^{2} - 1} : - \frac{1}{2 ( s + 1 )} + \frac{7}{2 ( s - 1 )}

= L^{- 1} {- \frac{1}{2 ( s + 1 )} + \frac{7}{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}{2 ( s + 1 )}} + L^{- 1} {\frac{7}{2 ( s - 1 )}}

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

L^{- 1} {\frac{7}{2 ( s - 1 )}} : \frac{7}{2} e^{t}

= - \frac{1}{2} e^{- t} + \frac{7}{2} e^{t}

\int 3 e^{- x} - 2 sin (\frac{π}{2} - x) d x

\frac{d}{d x} (3 x^{2} - 6 x - 9)

y = 2 sec (x) - 4 cos (x)

t \to 0 lim (\frac{e ^{4 t} - 1}{t})

5 x^{9} - 2 x^{6}

What is inverse oflaplace (3s+4)/(s^2-1) ?
The answer to inverse oflaplace (3s+4)/(s^2-1) is -1/2 e^{-t}+7/2 e^t