What is inverse oflaplace (s-1)/(s^2-2s+1) ?

The answer to inverse oflaplace (s-1)/(s^2-2s+1) is e^t

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inverse oflaplace (s-1)/(s^2-2s+1)

inverse laplace

\frac{s - 1}{s ^{2} - 2 s + 1}

e^{t}

Solution steps

L^{- 1} {\frac{s - 1}{s ^{2} - 2 s + 1}}

\frac{s - 1}{s ^{2} - 2 s + 1} = \frac{s - 1}{( s - 1 ) ^{2}}

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{s - 1}{( s - 1 ) ^{2}} : a = 1, F (s) = \frac{s}{s ^{2}}

= e^{t} L^{- 1} {\frac{s}{s ^{2}}}

L^{- 1} {\frac{s}{s ^{2}}} : 1

= e^{t}

\frac{\partial}{\partial y} (ln (x + 6 y + 5 z))

\int \frac{10}{( x + 1 ) ( x ^{2} + 9 ) ^{2}} d x

x \to 8 - lim (\frac{x - 8}{∣ x - 8 ∣})

\frac{d}{d x} (2 (4 - 9 x)^{4})

(x y + y + y^{2}) d x + (x + 2 y) d y = 0

What is inverse oflaplace (s-1)/(s^2-2s+1) ?
The answer to inverse oflaplace (s-1)/(s^2-2s+1) is e^t