What is inverse oflaplace 5/((s-1)^3) ?

The answer to inverse oflaplace 5/((s-1)^3) is (5e^tt^2)/2

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inverse oflaplace 5/((s-1)^3)

inverse laplace

\frac{5}{( s - 1 ) ^{3}}

\frac{5 e ^{t} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{5}{( s - 1 ) ^{3}}}

Apply inverse transform rule:

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

then

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

For

\frac{5}{( s - 1 ) ^{3}} : a = 1, F (s) = \frac{5}{s ^{3}}

= e^{t} L^{- 1} {\frac{5}{s ^{3}}}

L^{- 1} {\frac{5}{s ^{3}}} : \frac{5 t ^{2}}{2}

= \frac{5 e ^{t} t ^{2}}{2}

\int x \frac{2}{3} d x

\frac{\partial}{\partial x} (x^{2} + x ln (yz))

x \to 0 lim (ln (\frac{x}{x + 1}))

\int 2^{1 - x} d x

n = 1 \sum \infty \frac{n}{4 ^{n}}

What is inverse oflaplace 5/((s-1)^3) ?
The answer to inverse oflaplace 5/((s-1)^3) is (5e^tt^2)/2