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

The answer to inverse oflaplace 1/((s-3)^4) is (e^{3t}t^3)/6

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

inverse laplace

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

\frac{e ^{3 t} t ^{3}}{6}

Solution steps

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

Apply inverse transform rule:

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

then

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

For

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

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

L^{- 1} {\frac{1}{s ^{4}}} : \frac{t ^{3}}{6}

= \frac{e ^{3 t} t ^{3}}{6}

3 x (x^{3} - 3 x + 2)

\frac{d y}{d x} = (4 x^{4} - 7 x^{2} + 6)

\int e^{2 x} + e^{- 2 x} d x

h \to 0 lim (2.5 h^{2} + 6 h)

3 t + 2 e^{3 t}

What is inverse oflaplace 1/((s-3)^4) ?
The answer to inverse oflaplace 1/((s-3)^4) is (e^{3t}t^3)/6