What is inverse oflaplace s/(s+1000) ?

The answer to inverse oflaplace s/(s+1000) is \delta(t)-1000e^{-1000t}

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inverse oflaplace s/(s+1000)

inverse laplace

\frac{s}{s + 1000}

δ (t) - 1000 e^{- 1000 t}

Solution steps

L^{- 1} {\frac{s}{s + 1000}}

Take the partial fraction of

\frac{s}{s + 1000} : 1 - \frac{1000}{s + 1000}

= L^{- 1} {1 - \frac{1000}{s + 1000}}

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} {1} - L^{- 1} {\frac{1000}{s + 1000}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1000}{s + 1000}} : 1000 e^{- 1000 t}

= δ (t) - 1000 e^{- 1000 t}

\frac{d}{d y} (\frac{e ^{y} + e ^{- y}}{2})

y^{^{''}} - 2 y^{^{'}} + 3 y = 0

\frac{d}{d x} (\frac{1}{3} sin^{3} (x) - sin (x) + 1)

x \to 4 lim (\frac{x + 4}{x + 7})

\frac{\partial}{\partial x} (e^{- 2 x} cos (6 y))

What is inverse oflaplace s/(s+1000) ?
The answer to inverse oflaplace s/(s+1000) is \delta(t)-1000e^{-1000t}