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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

x \to - 2 + lim (\frac{2 x - 1}{x + 2})

y^{^{''}} + 4 y = sin^{2} (x)

\int 3 tan (4 x) sec^{2} (4 x) d x

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

x \to 0 lim (\frac{sin ( x ^{2} )}{x})

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