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

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

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

inverse laplace

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

2 e^{- t} + e^{t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= 2 e^{- t} + e^{t}

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

\frac{d y}{d x} = \frac{y x}{x - 1}

\int \frac{( cos ( x ) )}{sin ^{2} ( x )} d x

\int 2 sin^{2} (x co) s^{3} x d x

y^{^{'}} + 5 y = 2 x y^{5}

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