What is inverse oflaplace 8/((s-1)^3) ?

The answer to inverse oflaplace 8/((s-1)^3) is e^t*4t^2

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inverse oflaplace 8/((s-1)^3)

inverse laplace

\frac{8}{( s - 1 ) ^{3}}

e^{t} \cdot 4 t^{2}

Solution steps

L^{- 1} {\frac{8}{( 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{8}{( s - 1 ) ^{3}} : a = 1, F (s) = \frac{8}{s ^{3}}

= e^{t} L^{- 1} {\frac{8}{s ^{3}}}

L^{- 1} {\frac{8}{s ^{3}}} : 4 t^{2}

= e^{t} \cdot 4 t^{2}

y^{^{'}} = \frac{1}{1 + x}

\int x^{3} x^{2} - 16 d x

\frac{d}{d x} (5 - ∣ x ∣)

\int tan (3 x se) c^{2} 3 x d x

x \to - 3 lim (x^{2} + x - x)

What is inverse oflaplace 8/((s-1)^3) ?
The answer to inverse oflaplace 8/((s-1)^3) is e^t*4t^2