What is inverse oflaplace 4/((s-1)^3) ?

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

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inverse oflaplace 4/((s-1)^3)

inverse laplace

\frac{4}{( s - 1 ) ^{3}}

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

Solution steps

L^{- 1} {\frac{4}{( 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{4}{( s - 1 ) ^{3}} : a = 1, F (s) = \frac{4}{s ^{3}}

= e^{t} L^{- 1} {\frac{4}{s ^{3}}}

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

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

y^{^{'}} = a - b y^{2}

\frac{\partial}{\partial x} (cos (4 x))

\frac{\partial}{\partial y} (e^{x} + z^{2})

f (x) = \frac{1}{x + 4}

\int e^{x} + 2 cos (x) - \frac{3}{cos ^{2} ( x )} d x

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