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

The answer to inverse oflaplace (s^2-1)/(s(s+2)(s-3)) is 1/6 \H(t)+3/10 e^{-2t}+8/15 e^{3t}

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

inverse laplace

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

\frac{1}{6} H (t) + \frac{3}{10} e^{- 2 t} + \frac{8}{15} e^{3 t}

Solution steps

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

Take the partial fraction of

\frac{s ^{2} - 1}{s ( s + 2 ) ( s - 3 )} : \frac{1}{6 s} + \frac{3}{10 ( s + 2 )} + \frac{8}{15 ( s - 3 )}

= L^{- 1} {\frac{1}{6 s} + \frac{3}{10 ( s + 2 )} + \frac{8}{15 ( 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{3}{10 ( s + 2 )}} + L^{- 1} {\frac{8}{15 ( s - 3 )}}

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

L^{- 1} {\frac{3}{10 ( s + 2 )}} : \frac{3}{10} e^{- 2 t}

L^{- 1} {\frac{8}{15 ( s - 3 )}} : \frac{8}{15} e^{3 t}

= \frac{1}{6} H (t) + \frac{3}{10} e^{- 2 t} + \frac{8}{15} e^{3 t}

What is inverse oflaplace (s^2-1)/(s(s+2)(s-3)) ?
The answer to inverse oflaplace (s^2-1)/(s(s+2)(s-3)) is 1/6 \H(t)+3/10 e^{-2t}+8/15 e^{3t}