What is inverse oflaplace 3s+2 ?

The answer to inverse oflaplace 3s+2 is 3\delta'(t)+2\delta(t)

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inverse oflaplace 3s+2

inverse laplace

3 s + 2

3 δ' (t) + 2 δ (t)

Solution steps

L^{- 1} {3 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)}

= 3 L^{- 1} {s} + L^{- 1} {2}

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {2} : 2 δ (t)

= 3 δ^{^{'}} (t) + 2 δ (t)

x \to \infty lim (e^{x} - 3)

\frac{d}{d x} (\frac{z ( x )}{x})

\int_{0}^{2} (y - 2)^{2} d y

\frac{d}{d x} ((2 x + 1)^{2} (x^{2} + x + 1))

\int \frac{1}{( x + 6 ) ^{2}} d x

What is inverse oflaplace 3s+2 ?
The answer to inverse oflaplace 3s+2 is 3\delta'(t)+2\delta(t)