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

The answer to inverse oflaplace 5/((s-3)^3) is (5e^{3t}t^2)/2

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

inverse laplace

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

\frac{5 e ^{3 t} t ^{2}}{2}

Solution steps

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

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

L^{- 1} {\frac{5}{s ^{3}}} : \frac{5 t ^{2}}{2}

= \frac{5 e ^{3 t} t ^{2}}{2}

\int x (x + 5)^{6} d x

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

\frac{\partial}{\partial y} (\frac{t ^{2}}{4 y ^{4}})

\int \frac{3 x + 1}{( x ^{2} + 2 ) ( x + 3 )} d x

\int 2 x x - 3 d x

What is inverse oflaplace 5/((s-3)^3) ?
The answer to inverse oflaplace 5/((s-3)^3) is (5e^{3t}t^2)/2