What is inverse oflaplace e^{-5s} 1/(1+s) ?

The answer to inverse oflaplace e^{-5s} 1/(1+s) is \H(t-5)e^{-t+5}

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inverse oflaplace e^{-5s} 1/(1+s)

inverse laplace

e^{- 5 s} \frac{1}{1 + s}

H (t - 5) e^{- t + 5}

Solution steps

L^{- 1} {e^{- 5 s} \frac{1}{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 - 5) L^{- 1} {\frac{1}{1 + s}} (t - 5)

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

= H (t - 5) e^{- (t - 5)}

= H (t - 5) e^{- t + 5}

x y^{^{'}} - y = \frac{1}{2} x ln (x)

\int \frac{e ^{a x} + b}{e ^{a x} - b} d x

\int 10 x^{4} - 2 x - 6 d x

\int x x + 6 d x

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

What is inverse oflaplace e^{-5s} 1/(1+s) ?
The answer to inverse oflaplace e^{-5s} 1/(1+s) is \H(t-5)e^{-t+5}