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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

x \to 3 - lim (x 9 - x^{2})

\int tan (x) - sin (x) sin (y) d x

\int_{0}^{\frac{1}{2}} 7 arccos (x) d x

y^{^{'}} = 14 x y - 2 x

\frac{\partial}{\partial y} (\frac{1 + y}{1 + x})

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