What is inverse oflaplace 1/(s^2(s-3)^2) ?

The answer to inverse oflaplace 1/(s^2(s-3)^2) is 2/27 \H(t)+t/9-2/27 e^{3t}+(e^{3t}t)/9

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inverse oflaplace 1/(s^2(s-3)^2)

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

L^{- 1} {\frac{1}{9 ( s - 3 ) ^{2}}} : \frac{e ^{3 t} t}{9}

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

\int \frac{1}{t - 2} d t

\int \frac{3 x ^{3}}{4 - x ^{2}} d x

\int \frac{1}{x ^{n}} d x

\int t^{3} e^{- 2 t} d t

\frac{d}{d x} (π ln (e))

What is inverse oflaplace 1/(s^2(s-3)^2) ?
The answer to inverse oflaplace 1/(s^2(s-3)^2) is 2/27 \H(t)+t/9-2/27 e^{3t}+(e^{3t}t)/9