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}

y = x x, (16, 64)

\int 36 - x^{2} d x

\frac{d ^{2}}{d x ^{2}} (\frac{6}{x})

\int \frac{1}{( 1 + x ^{3} )} d x

x \to 2 + lim (ln (x^{2} - 4))

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}