What is inverse oflaplace (s+7)/(s^2-3s-10) ?

The answer to inverse oflaplace (s+7)/(s^2-3s-10) is -5/7 e^{-2t}+12/7 e^{5t}

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inverse oflaplace (s+7)/(s^2-3s-10)

inverse laplace

\frac{s + 7}{s ^{2} - 3 s - 10}

- \frac{5}{7} e^{- 2 t} + \frac{12}{7} e^{5 t}

Solution steps

L^{- 1} {\frac{s + 7}{s ^{2} - 3 s - 10}}

Take the partial fraction of

\frac{s + 7}{s ^{2} - 3 s - 10} : - \frac{5}{7 ( s + 2 )} + \frac{12}{7 ( s - 5 )}

= L^{- 1} {- \frac{5}{7 ( s + 2 )} + \frac{12}{7 ( 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{5}{7 ( s + 2 )}} + L^{- 1} {\frac{12}{7 ( s - 5 )}}

L^{- 1} {\frac{5}{7 ( s + 2 )}} : \frac{5}{7} e^{- 2 t}

L^{- 1} {\frac{12}{7 ( s - 5 )}} : \frac{12}{7} e^{5 t}

= - \frac{5}{7} e^{- 2 t} + \frac{12}{7} e^{5 t}

What is inverse oflaplace (s+7)/(s^2-3s-10) ?
The answer to inverse oflaplace (s+7)/(s^2-3s-10) is -5/7 e^{-2t}+12/7 e^{5t}