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

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

Popular >

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} (e^{x y} + x^{2} z)

\frac{x ^{4}}{1 - x ^{3}}

x^{2}, (- 2, 1)

\int \frac{x - 1}{x ( x + 1 )} d x

\frac{d}{d x} ((4 x^{3} + 1)^{2})

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