What is inverse oflaplace {1/(s^2+s-20)} ?

The answer to inverse oflaplace {1/(s^2+s-20)} is 1/9 e^{4t}-1/9 e^{-5t}

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inverse oflaplace {1/(s^2+s-20)}

inverse laplace

{\frac{1}{s ^{2} + s - 20}}

\frac{1}{9} e^{4 t} - \frac{1}{9} e^{- 5 t}

Solution steps

L^{- 1} {\frac{1}{s ^{2} + s - 20}}

Take the partial fraction of

\frac{1}{s ^{2} + s - 20} : \frac{1}{9 ( s - 4 )} - \frac{1}{9 ( s + 5 )}

= L^{- 1} {\frac{1}{9 ( s - 4 )} - \frac{1}{9 ( s + 5 )}}

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}{9 ( s - 4 )}} - L^{- 1} {\frac{1}{9 ( s + 5 )}}

L^{- 1} {\frac{1}{9 ( s - 4 )}} : \frac{1}{9} e^{4 t}

L^{- 1} {\frac{1}{9 ( s + 5 )}} : \frac{1}{9} e^{- 5 t}

= \frac{1}{9} e^{4 t} - \frac{1}{9} e^{- 5 t}

\frac{d}{d x} (arcsin (1 - 2 x^{2}))

\frac{d}{d x} (e^{x} + x^{2})

\frac{d}{d y} (y cos (x))

\frac{\partial}{\partial x} (45 - x^{2} - y^{2})

\frac{d}{d x} (sin (2 x - 2 π))

What is inverse oflaplace {1/(s^2+s-20)} ?
The answer to inverse oflaplace {1/(s^2+s-20)} is 1/9 e^{4t}-1/9 e^{-5t}