What is inverse oflaplace (s-5)/(s^2+s-6) ?

The answer to inverse oflaplace (s-5)/(s^2+s-6) is -3/5 e^{2t}+8/5 e^{-3t}

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inverse oflaplace (s-5)/(s^2+s-6)

inverse laplace

\frac{s - 5}{s ^{2} + s - 6}

- \frac{3}{5} e^{2 t} + \frac{8}{5} e^{- 3 t}

Solution steps

L^{- 1} {\frac{s - 5}{s ^{2} + s - 6}}

Take the partial fraction of

\frac{s - 5}{s ^{2} + s - 6} : - \frac{3}{5 ( s - 2 )} + \frac{8}{5 ( s + 3 )}

= L^{- 1} {- \frac{3}{5 ( s - 2 )} + \frac{8}{5 ( 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{3}{5 ( s - 2 )}} + L^{- 1} {\frac{8}{5 ( s + 3 )}}

L^{- 1} {\frac{3}{5 ( s - 2 )}} : \frac{3}{5} e^{2 t}

L^{- 1} {\frac{8}{5 ( s + 3 )}} : \frac{8}{5} e^{- 3 t}

= - \frac{3}{5} e^{2 t} + \frac{8}{5} e^{- 3 t}

\frac{d}{d x} (sin (x) d x)

\int x 3 x - 2 d x

\frac{d}{d x} ((3 x)^{l n (3 x)})

\frac{d r}{d t} = - 2 t r, r (0) = 2

\int sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

What is inverse oflaplace (s-5)/(s^2+s-6) ?
The answer to inverse oflaplace (s-5)/(s^2+s-6) is -3/5 e^{2t}+8/5 e^{-3t}