What is inverse oflaplace s+6 ?

The answer to inverse oflaplace s+6 is \delta'(t)+6\delta(t)

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inverse oflaplace s+6

inverse laplace

s + 6

δ' (t) + 6 δ (t)

Solution steps

L^{- 1} {s + 6}

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} {6}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {6} : 6 δ (t)

= δ^{^{'}} (t) + 6 δ (t)

\frac{d}{d x} (\frac{1}{4 x})

\int_{1}^{2} e^{x^{2}} d x

\int_{1}^{3} x arctan (x) d x

x \to - 2 lim (- x^{2} + 7 x - 5)

\int e^{- 2 x} cos (2 x) d x

What is inverse oflaplace s+6 ?
The answer to inverse oflaplace s+6 is \delta'(t)+6\delta(t)