What is inverse oflaplace ((e^{-s}))/((s-1)^2) ?

The answer to inverse oflaplace ((e^{-s}))/((s-1)^2) is \H(t-1)e^{t-1}(t-1)

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inverse oflaplace ((e^{-s}))/((s-1)^2)

inverse laplace

\frac{( e ^{- s} )}{( s - 1 ) ^{2}}

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

Solution steps

L^{- 1} {\frac{( e ^{- s} )}{( s - 1 ) ^{2}}}

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 - 1 ) ^{2}}} (t - 1)

L^{- 1} {\frac{1}{( s - 1 ) ^{2}}} : e^{t} t

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

\frac{d y}{d x} = \frac{( x - 2 )}{y - 4}

\frac{\partial}{\partial x} (y \cdot e^{- x y})

\frac{d}{d x} (3 + sin (x))

\frac{d}{d x} (2 cos (π x))

x \to 0 lim (\frac{1}{( x - 1 ) ^{2}})

What is inverse oflaplace ((e^{-s}))/((s-1)^2) ?
The answer to inverse oflaplace ((e^{-s}))/((s-1)^2) is \H(t-1)e^{t-1}(t-1)