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

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

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

inverse laplace

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

H (t - 1) e^{5 (t - 1)}

Solution steps

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

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

L^{- 1} {\frac{1}{s - 5}} : e^{5 t}

= H (t - 1) e^{5 (t - 1)}

\frac{d}{d x} (\frac{( x - 1 ) ^{2}}{x ^{2}})

x \to 0 lim (e^{2 x} + 10 x)

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

\frac{\partial}{\partial x} (arctan (x^{4} y^{3}))

f (x) = e^{- x^{2}} - 2 e^{- x^{2}} x^{2}

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