What is inverse oflaplace (s+10)/(s^2-s-2) ?

The answer to inverse oflaplace (s+10)/(s^2-s-2) is -3e^{-t}+4e^{2t}

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inverse oflaplace (s+10)/(s^2-s-2)

inverse laplace

\frac{s + 10}{s ^{2} - s - 2}

- 3 e^{- t} + 4 e^{2 t}

Solution steps

L^{- 1} {\frac{s + 10}{s ^{2} - s - 2}}

Take the partial fraction of

\frac{s + 10}{s ^{2} - s - 2} : - \frac{3}{s + 1} + \frac{4}{s - 2}

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

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

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

L^{- 1} {\frac{4}{s - 2}} : 4 e^{2 t}

= - 3 e^{- t} + 4 e^{2 t}

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

\int e^{x} - e^{2 x} d x

\frac{1}{s - 2}

sin (3 x^{5} + 5 x^{4})

x \frac{d y}{d x} + \frac{y}{x + 1} = x (x + 1)

What is inverse oflaplace (s+10)/(s^2-s-2) ?
The answer to inverse oflaplace (s+10)/(s^2-s-2) is -3e^{-t}+4e^{2t}