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

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

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

inverse laplace

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

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

Solution steps

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

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

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

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

\frac{\partial}{\partial x} (3 cos (2 x))

h (x) = \frac{1}{x ^{2}} + \frac{1}{x} + x

\int \frac{cos ( a x )}{b - a sin ( a x )} d x

x \to 5 lim (3 x + 6)

\frac{d}{d x} (arcsin (\frac{x}{1 - x}))

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