What is inverse oflaplace (100)/(s^3(s-2)) ?

The answer to inverse oflaplace (100)/(s^3(s-2)) is -25/2 \H(t)-25t-25t^2+25/2 e^{2t}

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inverse oflaplace (100)/(s^3(s-2))

inverse laplace

\frac{100}{s ^{3} ( s - 2 )}

- \frac{25}{2} H (t) - 25 t - 25 t^{2} + \frac{25}{2} e^{2 t}

Solution steps

L^{- 1} {\frac{100}{s ^{3} ( s - 2 )}}

Take the partial fraction of

\frac{100}{s ^{3} ( s - 2 )} : - \frac{25}{2 s} - \frac{25}{s ^{2}} - \frac{50}{s ^{3}} + \frac{25}{2 ( s - 2 )}

= L^{- 1} {- \frac{25}{2 s} - \frac{25}{s ^{2}} - \frac{50}{s ^{3}} + \frac{25}{2 ( s - 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{25}{2 s}} - L^{- 1} {\frac{25}{s ^{2}}} - L^{- 1} {\frac{50}{s ^{3}}} + L^{- 1} {\frac{25}{2 ( s - 2 )}}

L^{- 1} {\frac{25}{2 s}} : \frac{25}{2} H (t)

L^{- 1} {\frac{25}{s ^{2}}} : 25 t

L^{- 1} {\frac{50}{s ^{3}}} : 25 t^{2}

L^{- 1} {\frac{25}{2 ( s - 2 )}} : \frac{25}{2} e^{2 t}

= - \frac{25}{2} H (t) - 25 t - 25 t^{2} + \frac{25}{2} e^{2 t}

x \to - 5 lim (217)

n = 1 \sum \infty 3^{n} \frac{1}{n !}

\frac{d}{d x} (e^{- 4 x} cos (4 x))

\frac{\partial}{\partial x} (3 x^{4})

y = 6 x^{2} - x^{3}

What is inverse oflaplace (100)/(s^3(s-2)) ?
The answer to inverse oflaplace (100)/(s^3(s-2)) is -25/2 \H(t)-25t-25t^2+25/2 e^{2t}