What is inverse oflaplace (-2s+1)/(s^2+2s-8) ?

The answer to inverse oflaplace (-2s+1)/(s^2+2s-8) is -1/2 e^{2t}-3/2 e^{-4t}

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inverse oflaplace (-2s+1)/(s^2+2s-8)

inverse laplace

\frac{- 2 s + 1}{s ^{2} + 2 s - 8}

- \frac{1}{2} e^{2 t} - \frac{3}{2} e^{- 4 t}

Solution steps

L^{- 1} {\frac{- 2 s + 1}{s ^{2} + 2 s - 8}}

Take the partial fraction of

\frac{- 2 s + 1}{s ^{2} + 2 s - 8} : - \frac{1}{2 ( s - 2 )} - \frac{3}{2 ( s + 4 )}

= L^{- 1} {- \frac{1}{2 ( s - 2 )} - \frac{3}{2 ( s + 4 )}}

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}{2 ( s - 2 )}} - L^{- 1} {\frac{3}{2 ( s + 4 )}}

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

L^{- 1} {\frac{3}{2 ( s + 4 )}} : \frac{3}{2} e^{- 4 t}

= - \frac{1}{2} e^{2 t} - \frac{3}{2} e^{- 4 t}

x \to 0 lim (\frac{x + 4}{x})

\int \frac{3}{( x + 5 ) ^{2} + 4} d x

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

(\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) (y + 9 y^{3})

h \to 0 lim (\frac{( x + h ) ^{2} - 9}{h})

What is inverse oflaplace (-2s+1)/(s^2+2s-8) ?
The answer to inverse oflaplace (-2s+1)/(s^2+2s-8) is -1/2 e^{2t}-3/2 e^{-4t}