What is inverse oflaplace (s+2)/(s^2-s-12) ?

The answer to inverse oflaplace (s+2)/(s^2-s-12) is 1/7 e^{-3t}+6/7 e^{4t}

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inverse oflaplace (s+2)/(s^2-s-12)

inverse laplace

\frac{s + 2}{s ^{2} - s - 12}

\frac{1}{7} e^{- 3 t} + \frac{6}{7} e^{4 t}

Solution steps

L^{- 1} {\frac{s + 2}{s ^{2} - s - 12}}

Take the partial fraction of

\frac{s + 2}{s ^{2} - s - 12} : \frac{1}{7 ( s + 3 )} + \frac{6}{7 ( s - 4 )}

= L^{- 1} {\frac{1}{7 ( s + 3 )} + \frac{6}{7 ( 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}{7 ( s + 3 )}} + L^{- 1} {\frac{6}{7 ( s - 4 )}}

L^{- 1} {\frac{1}{7 ( s + 3 )}} : \frac{1}{7} e^{- 3 t}

L^{- 1} {\frac{6}{7 ( s - 4 )}} : \frac{6}{7} e^{4 t}

= \frac{1}{7} e^{- 3 t} + \frac{6}{7} e^{4 t}

x \to 0 lim (3 x^{7 x})

\int \frac{x ^{2} + 24 x - 16}{x ^{3} - 16 x} d x

\frac{6}{7 t ^{4}}

x \to - 2 lim (\frac{3}{( x + 2 )})

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

What is inverse oflaplace (s+2)/(s^2-s-12) ?
The answer to inverse oflaplace (s+2)/(s^2-s-12) is 1/7 e^{-3t}+6/7 e^{4t}