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

The answer to inverse oflaplace 1/((s-2)(s+1)(s-3)) is -1/3 e^{2t}+1/12 e^{-t}+1/4 e^{3t}

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

inverse laplace

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

- \frac{1}{3} e^{2 t} + \frac{1}{12} e^{- t} + \frac{1}{4} e^{3 t}

Solution steps

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

Take the partial fraction of

\frac{1}{( s - 2 ) ( s + 1 ) ( s - 3 )} : - \frac{1}{3 ( s - 2 )} + \frac{1}{12 ( s + 1 )} + \frac{1}{4 ( s - 3 )}

= L^{- 1} {- \frac{1}{3 ( s - 2 )} + \frac{1}{12 ( s + 1 )} + \frac{1}{4 ( 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 - 2 )}} + L^{- 1} {\frac{1}{12 ( s + 1 )}} + L^{- 1} {\frac{1}{4 ( s - 3 )}}

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

L^{- 1} {\frac{1}{12 ( s + 1 )}} : \frac{1}{12} e^{- t}

L^{- 1} {\frac{1}{4 ( s - 3 )}} : \frac{1}{4} e^{3 t}

= - \frac{1}{3} e^{2 t} + \frac{1}{12} e^{- t} + \frac{1}{4} e^{3 t}

What is inverse oflaplace 1/((s-2)(s+1)(s-3)) ?
The answer to inverse oflaplace 1/((s-2)(s+1)(s-3)) is -1/3 e^{2t}+1/12 e^{-t}+1/4 e^{3t}