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

The answer to inverse oflaplace (1+s^2)/(s^2(s-3)^2) is 2/27 \H(t)+t/9-2/27 e^{3t}+(10e^{3t}t)/9

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

inverse laplace

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

\frac{2}{27} H (t) + \frac{t}{9} - \frac{2}{27} e^{3 t} + \frac{10 e ^{3 t} t}{9}

Solution steps

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

Take the partial fraction of

\frac{1 + s ^{2}}{s ^{2} ( s - 3 ) ^{2}} : \frac{2}{27 s} + \frac{1}{9 s ^{2}} - \frac{2}{27 ( s - 3 )} + \frac{10}{9 ( s - 3 ) ^{2}}

= L^{- 1} {\frac{2}{27 s} + \frac{1}{9 s ^{2}} - \frac{2}{27 ( s - 3 )} + \frac{10}{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{10}{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{10}{9 ( s - 3 ) ^{2}}} : \frac{10 e ^{3 t} t}{9}

= \frac{2}{27} H (t) + \frac{t}{9} - \frac{2}{27} e^{3 t} + \frac{10 e ^{3 t} t}{9}

3 x + 5

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

\frac{1}{u}

f (x) = - \frac{2 e ^{2 x}}{( e ^{2 x} - 1 ) ^{2}}

What is inverse oflaplace (1+s^2)/(s^2(s-3)^2) ?
The answer to inverse oflaplace (1+s^2)/(s^2(s-3)^2) is 2/27 \H(t)+t/9-2/27 e^{3t}+(10e^{3t}t)/9