What is inverse oflaplace x^{-1}+2x ?

The answer to inverse oflaplace x^{-1}+2x is \H(t)+2\delta'(t)

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inverse oflaplace x^{-1}+2x

inverse laplace

x^{- 1} + 2 x

H (t) + 2 δ' (t)

Solution steps

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

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

L^{- 1} {x^{- 1}} : H (t)

L^{- 1} {x} : δ^{'} (t)

= H (t) + 2 δ^{^{'}} (t)

\int e^{x} e^{- x} d x

x \to - 299999 lim (x^{2})

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 1

\frac{\partial}{\partial x} (4 x (x^{2} + y^{2}) - 4 x)

t \to 1 lim (sin (3 π t))

What is inverse oflaplace x^{-1}+2x ?
The answer to inverse oflaplace x^{-1}+2x is \H(t)+2\delta'(t)