What is inverse oflaplace f(s)= 1/(s(s-3)) ?

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

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inverse oflaplace f(s)= 1/(s(s-3))

inverselaplace

- \frac{1}{3} H (t) + \frac{1}{3} e^{3 t}

Solution steps

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

Take the partial fraction of

\frac{1}{s ( s - 3 )} : - \frac{1}{3 s} + \frac{1}{3 ( s - 3 )}

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

L^{- 1} {\frac{1}{3 s}} : \frac{1}{3} H (t)

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

= - \frac{1}{3} H (t) + \frac{1}{3} e^{3 t}

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