What is inverse oflaplace s-1 ?

The answer to inverse oflaplace s-1 is \delta'(t)-\delta(t)

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inverse oflaplace s-1

inverse laplace

s - 1

δ' (t) - δ (t)

Solution steps

L^{- 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} {s} - L^{- 1} {1}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {1} : δ (t)

= δ^{^{'}} (t) - δ (t)

\int e^{- x} cos (x) d x

\frac{d}{d x} (a x^{2} + b)

\frac{d}{d x} (4 x ln (7 x) - 4 x)

\int - \frac{sec ^{2} ( x ) - 1}{tan ( x )} d x

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

What is inverse oflaplace s-1 ?
The answer to inverse oflaplace s-1 is \delta'(t)-\delta(t)