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}

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