What is inverse oflaplace 2s+1 ?

The answer to inverse oflaplace 2s+1 is 2\delta'(t)+\delta(t)

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inverse oflaplace 2s+1

inverse laplace

2 s + 1

2 δ' (t) + δ (t)

Solution steps

L^{- 1} {2 s + 1}

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)}

= 2 L^{- 1} {s} + L^{- 1} {1}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {1} : δ (t)

= 2 δ^{^{'}} (t) + δ (t)

f (x) = x^{4} - 4 x^{3} + 2 x^{2} - 6, at x = 1

\int \frac{x ^{2}}{x ^{2} + x + 12} d x

\int \frac{1}{- t ^{2} + 3 t - 2} d t

\int e^{\frac{x ^{2}}{4}} d x

x y \frac{d y}{d x} = 7

What is inverse oflaplace 2s+1 ?
The answer to inverse oflaplace 2s+1 is 2\delta'(t)+\delta(t)