What is inverse oflaplace {s/((s-2)(s-3)(s-6))} ?

The answer to inverse oflaplace {s/((s-2)(s-3)(s-6))} is 1/2 e^{2t}-e^{3t}+1/2 e^{6t}

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inverse oflaplace {s/((s-2)(s-3)(s-6))}

inverse laplace

{\frac{s}{( s - 2 ) ( s - 3 ) ( s - 6 )}}

\frac{1}{2} e^{2 t} - e^{3 t} + \frac{1}{2} e^{6 t}

Solution steps

L^{- 1} {\frac{s}{( s - 2 ) ( s - 3 ) ( s - 6 )}}

Take the partial fraction of

\frac{s}{( s - 2 ) ( s - 3 ) ( s - 6 )} : \frac{1}{2 ( s - 2 )} - \frac{1}{s - 3} + \frac{1}{2 ( s - 6 )}

= L^{- 1} {\frac{1}{2 ( s - 2 )} - \frac{1}{s - 3} + \frac{1}{2 ( s - 6 )}}

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 - 2 )}} - L^{- 1} {\frac{1}{s - 3}} + L^{- 1} {\frac{1}{2 ( s - 6 )}}

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

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

L^{- 1} {\frac{1}{2 ( s - 6 )}} : \frac{1}{2} e^{6 t}

= \frac{1}{2} e^{2 t} - e^{3 t} + \frac{1}{2} e^{6 t}

\frac{d}{d x} ((x + a) e^{3 x} + b)

4 x^{- 3} + 6 x^{- \frac{8}{3}}

\frac{\partial}{\partial x} (arcsin (3 x + y))

f (x) = 6 e^{x} cos (x)

y^{2} = x + 2, y^{2} = 2 - x

What is inverse oflaplace {s/((s-2)(s-3)(s-6))} ?
The answer to inverse oflaplace {s/((s-2)(s-3)(s-6))} is 1/2 e^{2t}-e^{3t}+1/2 e^{6t}