What is inverse oflaplace (s+1)e^{-spi} ?

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

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inverse oflaplace (s+1)e^{-spi}

inverse laplace

(s + 1) e^{- s π}

H (t - π) (δ' (t - π) + δ (t - π))

Solution steps

L^{- 1} {e^{- π s} (s + 1)}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - π) L^{- 1} {(s + 1)} (t - π)

L^{- 1} {(s + 1)} : δ^{'} (t) + δ (t)

= H (t - π) (δ^{^{'}} (t - π) + δ (t - π))

\frac{d y}{d x} = (x^{2})

\int (0) d x

3 y^{^{'}} + 7 y = 4 e^{- x}

y = ln (6 (ln (x))^{8})

\frac{d y}{d x} = - 2 x tan (y)

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