What is inverse oflaplace 1/((s-1)^5) ?

The answer to inverse oflaplace 1/((s-1)^5) is (e^tt^4)/(24)

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inverse oflaplace 1/((s-1)^5)

inverse laplace

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

\frac{e ^{t} t ^{4}}{24}

Solution steps

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

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 ) ^{5}} : a = 1, F (s) = \frac{1}{s ^{5}}

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

L^{- 1} {\frac{1}{s ^{5}}} : \frac{t ^{4}}{24}

= \frac{e ^{t} t ^{4}}{24}

\frac{d}{d x} (2 x^{3} - 5 x)

\int \frac{2}{( x - 2 ) ^{2}} d x

x = y^{2} - 8, x = 8

\int (\frac{x + 4}{x}) d x

\frac{\partial}{\partial x} (4 x^{\frac{1}{2}} + y)

What is inverse oflaplace 1/((s-1)^5) ?
The answer to inverse oflaplace 1/((s-1)^5) is (e^tt^4)/(24)