What is inverse oflaplace s+2 ?

The answer to inverse oflaplace s+2 is \delta'(t)+2\delta(t)

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inverse oflaplace s+2

inverse laplace

s + 2

δ' (t) + 2 δ (t)

Solution steps

L^{- 1} {s + 2}

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} {2}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {2} : 2 δ (t)

= δ^{^{'}} (t) + 2 δ (t)

y^{^{''}} - 6 y^{^{'}} + 9 y = 30 t^{2} e^{3 t}

\frac{d}{d x} (\frac{27 x ^{3}}{8})

\frac{1}{1 + a s}

\int 8 x^{5} d x

\frac{d}{d x} (ln (x^{12}))

What is inverse oflaplace s+2 ?
The answer to inverse oflaplace s+2 is \delta'(t)+2\delta(t)