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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

y^{^{'}} = \frac{e ^{x - y}}{1 + e ^{x}}, y (1) = 0

y = \frac{6 x ^{2} + 2 x + 8}{x}

\int \frac{1 + sin ( x )}{cos ( x )} d x

\int_{0}^{9} x^{2} + 81 d x

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