What is inverse oflaplace 1+s ?

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

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inverse oflaplace 1+s

inverse laplace

1 + s

δ (t) + δ' (t)

Solution steps

L^{- 1} {1 + s}

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} {s}

L^{- 1} {1} : δ (t)

L^{- 1} {s} : δ^{'} (t)

= δ (t) + δ^{^{'}} (t)

f (x) = 23 sin (x) + 2 cos (x)

\frac{d}{d x} (cosh (\frac{4 x ^{3}}{x - 2}))

y = ln (ln (x^{32}))

\frac{d}{d t} (cos (t) (1 - sin (t)))

\int_{0}^{2} \frac{1}{x ^{4} + 1} d x

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