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

The answer to inverse oflaplace (2s+5)/(s^2-s-6) is -1/5 e^{-2t}+11/5 e^{3t}

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

inverse laplace

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

- \frac{1}{5} e^{- 2 t} + \frac{11}{5} e^{3 t}

Solution steps

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

Take the partial fraction of

\frac{2 s + 5}{s ^{2} - s - 6} : - \frac{1}{5 ( s + 2 )} + \frac{11}{5 ( s - 3 )}

= L^{- 1} {- \frac{1}{5 ( s + 2 )} + \frac{11}{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{1}{5 ( s + 2 )}} + L^{- 1} {\frac{11}{5 ( s - 3 )}}

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

L^{- 1} {\frac{11}{5 ( s - 3 )}} : \frac{11}{5} e^{3 t}

= - \frac{1}{5} e^{- 2 t} + \frac{11}{5} e^{3 t}

x \to 0 lim ((1 + \frac{3}{x ^{5}})^{x})

\frac{d}{d x} (6 e^{- x})

y = 5 x, y = \frac{1}{7} x, y = 50 - x^{2}

\frac{d}{d θ} (sin (θ) + tan (θ))

\int \frac{1}{x ^{2} - 2 x - 8} d x

What is inverse oflaplace (2s+5)/(s^2-s-6) ?
The answer to inverse oflaplace (2s+5)/(s^2-s-6) is -1/5 e^{-2t}+11/5 e^{3t}