What is inverse oflaplace t-1 ?

The answer to inverse oflaplace t-1 is \delta'(t)-\delta(t)

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inverse oflaplace t-1

inverse laplace

t - 1

δ' (t) - δ (t)

Solution steps

L^{- 1} {t - 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)}

= L^{- 1} {t} - L^{- 1} {1}

L^{- 1} {t} : δ^{'} (t)

L^{- 1} {1} : δ (t)

= δ^{^{'}} (t) - δ (t)

(12 x^{2} - 24 x + 24) e^{x}

\frac{\partial}{\partial y} (sin (\frac{2 y}{x}))

\int \frac{x}{x ^{2} - 2 x + 1} d x

\int_{0}^{3} 2^{x} d x

n = 0 \sum \infty \frac{n ^{3}}{2 ^{n}}

What is inverse oflaplace t-1 ?
The answer to inverse oflaplace t-1 is \delta'(t)-\delta(t)