What is inverse oflaplace 3/(s^2-2s-15) ?

The answer to inverse oflaplace 3/(s^2-2s-15) is -3/8 e^{-3t}+3/8 e^{5t}

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inverse oflaplace 3/(s^2-2s-15)

inverse laplace

\frac{3}{s ^{2} - 2 s - 15}

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

Solution steps

L^{- 1} {\frac{3}{s ^{2} - 2 s - 15}}

Take the partial fraction of

\frac{3}{s ^{2} - 2 s - 15} : - \frac{3}{8 ( s + 3 )} + \frac{3}{8 ( s - 5 )}

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

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

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

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

\int (\frac{7}{x ^{2} + 1}) d x

\frac{d}{d x} (csc^{3} (x^{2}))

\frac{d}{d x} ((1 + x^{3})^{2020})

\frac{d}{d x} (4 e^{- x^{2}})

(8, 8), (4, 6)

What is inverse oflaplace 3/(s^2-2s-15) ?
The answer to inverse oflaplace 3/(s^2-2s-15) is -3/8 e^{-3t}+3/8 e^{5t}