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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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}{3 ( s + 1 )}}

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

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

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

(6, - 2), (12, 1)

x \to 2 lim (x^{2} + 3 x)

\frac{d}{d x} (2 + 3 cos (2 x))

f (x) = x + 3, at x = 13

\frac{d}{d x} (x^{2} + π^{2} + x^{π})

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