What is inverse oflaplace s*+5 ?

The answer to inverse oflaplace s*+5 is \delta'(t)+5\delta(t)

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inverse oflaplace s*+5

inverse laplace

s \cdot + 5

δ' (t) + 5 δ (t)

Solution steps

L^{- 1} {s + 5}

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} {5}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {5} : 5 δ (t)

= δ^{^{'}} (t) + 5 δ (t)

x \to 2 lim (\frac{x - 1}{x ^{2} - 1})

\int x^{4} (ln (x))^{2} d x

\int \frac{9}{x} + 9 x d x

\int \frac{1}{( x - 3 ) ^{2}} d x

y = 2 sin (π x - y), (1, 0)

What is inverse oflaplace s*+5 ?
The answer to inverse oflaplace s*+5 is \delta'(t)+5\delta(t)