What is inverse oflaplace-2/(s^3(s-3)) ?

The answer to inverse oflaplace-2/(s^3(s-3)) is 2/27 \H(t)+(2t)/9+(t^2)/3-2/27 e^{3t}

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inverse oflaplace-2/(s^3(s-3))

inverse laplace

- \frac{2}{s ^{3} ( s - 3 )}

\frac{2}{27} H (t) + \frac{2 t}{9} + \frac{t ^{2}}{3} - \frac{2}{27} e^{3 t}

Solution steps

L^{- 1} {- \frac{2}{s ^{3} ( s - 3 )}}

Take the partial fraction of

\frac{2}{s ^{3} ( s - 3 )} : - \frac{2}{27 s} - \frac{2}{9 s ^{2}} - \frac{2}{3 s ^{3}} + \frac{2}{27 ( s - 3 )}

= L^{- 1} {\frac{2}{27 s} + \frac{2}{9 s ^{2}} + \frac{2}{3 s ^{3}} - \frac{2}{27 ( s - 3 )}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{2}{27 s}} + L^{- 1} {\frac{2}{9 s ^{2}}} + L^{- 1} {\frac{2}{3 s ^{3}}} - L^{- 1} {\frac{2}{27 ( s - 3 )}}

L^{- 1} {\frac{2}{27 s}} : \frac{2}{27} H (t)

L^{- 1} {\frac{2}{9 s ^{2}}} : \frac{2 t}{9}

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

L^{- 1} {\frac{2}{27 ( s - 3 )}} : \frac{2}{27} e^{3 t}

= \frac{2}{27} H (t) + \frac{2 t}{9} + \frac{t ^{2}}{3} - \frac{2}{27} e^{3 t}

\int_{0}^{N} e^{- s t} e^{- 2 t} d t

f (x) = (x^{2} + 3 x - 5) (1 - x), at x = 2

\frac{3 x}{x + 1}

\int \frac{1}{2 x ^{2} - 3} d x

\int_{0}^{1} (x^{2} + 2 x + 3) e^{2 x} d x

What is inverse oflaplace-2/(s^3(s-3)) ?
The answer to inverse oflaplace-2/(s^3(s-3)) is 2/27 \H(t)+(2t)/9+(t^2)/3-2/27 e^{3t}