What is inverse oflaplace 1/(s+1)-s-1 ?

The answer to inverse oflaplace 1/(s+1)-s-1 is e^{-t}-\delta'(t)-\delta(t)

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inverse oflaplace 1/(s+1)-s-1

inverse laplace

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

e^{- t} - δ' (t) - δ (t)

Solution steps

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

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

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

L^{- 1} {s} : δ^{'} (t)

L^{- 1} {1} : δ (t)

= e^{- t} - δ^{^{'}} (t) - δ (t)

y = x^{5}, (- 1, - 1)

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{4 x ^{\frac{3}{4}}})

\frac{\partial}{\partial y} (\frac{3 x - y}{5 x y + 9})

x y^{^{'}} - y = x^{2} sin (x)

\frac{d x}{d y} = 5 x (6 - x)

What is inverse oflaplace 1/(s+1)-s-1 ?
The answer to inverse oflaplace 1/(s+1)-s-1 is e^{-t}-\delta'(t)-\delta(t)