What is inverse oflaplace 4/((3s-1)^5) ?

The answer to inverse oflaplace 4/((3s-1)^5) is (e^{t/3}t^4)/(1458)

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inverse oflaplace 4/((3s-1)^5)

inverse laplace

\frac{4}{( 3 s - 1 ) ^{5}}

\frac{e ^{\frac{t}{3}} t ^{4}}{1458}

Solution steps

L^{- 1} {\frac{4}{( 3 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{4}{( 3 s - 1 ) ^{5}} : a = \frac{1}{3}, F (s) = \frac{4}{( 3 s ) ^{5}}

= e^{\frac{1}{3} t} L^{- 1} {\frac{4}{( 3 s ) ^{5}}}

L^{- 1} {\frac{4}{( 3 s ) ^{5}}} : \frac{t ^{4}}{1458}

= \frac{e ^{\frac{t}{3}} t ^{4}}{1458}

y^{^{'}} + \frac{y}{x} = 2 y^{2}

\frac{d}{d x} (\frac{4 x + 2}{( 3 - 2 x ) ^{3}})

\frac{\partial}{\partial x} (x e^{14 x y})

y^{^{'}} = 3 - \frac{3 x}{10000}

\int t^{2} e^{2 t} d t

What is inverse oflaplace 4/((3s-1)^5) ?
The answer to inverse oflaplace 4/((3s-1)^5) is (e^{t/3}t^4)/(1458)