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

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

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

inverse laplace

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

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

Solution steps

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

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y^{^{'}} = \frac{- 1 + x + y}{1 + x}

\frac{9 - ln ( x )}{ln ( x ) - 1}

\int_{0}^{1} \frac{1}{x ^{\frac{1}{3}}} d x

2 x^{2}, 8 x - 2 x^{2}

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