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

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

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

inverse laplace

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

e^{t} t

Solution steps

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

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

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

= e^{t} t

\int_{0}^{1} k x d x

y = cos (x^{2})

\frac{d}{d x} (\frac{- 5 x - 9}{2 x + 8})

2.5 x^{3} - 3 x {2}

y = x^{3} - 9 x

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