What is inverse oflaplace s+3 ?

The answer to inverse oflaplace s+3 is \delta'(t)+3\delta(t)

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inverse oflaplace s+3

inverse laplace

s + 3

δ' (t) + 3 δ (t)

Solution steps

L^{- 1} {s + 3}

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} {3}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {3} : 3 δ (t)

= δ^{^{'}} (t) + 3 δ (t)

f (x) = \frac{x \frac{1}{9}}{2}

\int (2 cos (x) + 4 sin (x) + 6) d x

\int_{1}^{2} \frac{x ^{3} + 3 x ^{6}}{x ^{4}} d x

\int_{0}^{2} 5 x d x

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

What is inverse oflaplace s+3 ?
The answer to inverse oflaplace s+3 is \delta'(t)+3\delta(t)