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

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

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

inverse laplace

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

H (t - 1) \cdot 5 e^{- t + 1}

Solution steps

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

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

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

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

f (x) = 3 x + 7, at a = 6

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

\frac{d}{d x} (\frac{5 - x}{x})

\int x^{3} - \frac{1}{x ^{2}} - x^{\frac{2}{3}} d x

x \to \infty lim (5 x^{2})

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