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

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

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inverse oflaplace (s-1)/(s+1)

inverse laplace

\frac{s - 1}{s + 1}

δ (t) - 2 e^{- t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {1} : δ (t)

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

= δ (t) - 2 e^{- t}

\frac{\partial}{\partial y} (x^{2} - x y + 4 y^{2})

x \to 2 lim (2 x - 7)

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

\frac{d}{d x} (4^{c o s (x)})

\int \frac{x ^{3}}{( x ^{2} + 1 ) ^{2}} d x

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