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

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

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inverse oflaplace ((-2s-1))/(s+1)

inverse laplace

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

- 2 δ (t) + e^{- t}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {2} : 2 δ (t)

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

= - 2 δ (t) + e^{- t}

\int \frac{1}{y ^{2} + x ^{2}} d x

2 x - 1, at x = 5

\int ln (22 x) d x

\int \frac{3 x ^{3} - 3 x + 4}{4 x ^{2} - 4} d x

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

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