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

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

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

inverse laplace

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

\frac{e ^{3 t} t ^{2}}{2}

Solution steps

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

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

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

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

= \frac{e ^{3 t} t ^{2}}{2}

\frac{d v}{d t} = - k t v^{2}

f (x) = 15 x ln (8 x)

\frac{\partial}{\partial x} (- 6 + 5 - 6 x)

\frac{d}{d x} (\frac{2}{x + 3})

n = 0 \sum \infty r^{n}

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