What is inverse oflaplace-(12)/(s^2-5) ?

The answer to inverse oflaplace-(12)/(s^2-5) is 6/(sqrt(5))e^{-sqrt(5)t}-6/(sqrt(5))e^{sqrt(5)t}

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inverse oflaplace-(12)/(s^2-5)

inverse laplace

- \frac{12}{s ^{2} - 5}

\frac{6}{5} e^{- 5 t} - \frac{6}{5} e^{5 t}

Solution steps

L^{- 1} {- \frac{12}{s ^{2} - 5}}

Take the partial fraction of

\frac{12}{s ^{2} - 5} : - \frac{6}{5 ( s + 5 )} + \frac{6}{5 ( s - 5 )}

= L^{- 1} {\frac{6}{5 ( s + 5 )} - \frac{6}{5 ( 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{6}{5 ( s + 5 )}} - L^{- 1} {\frac{6}{5 ( s - 5 )}}

L^{- 1} {\frac{6}{5 ( s + 5 )}} : \frac{6}{5} e^{- 5 t}

L^{- 1} {\frac{6}{5 ( s - 5 )}} : \frac{6}{5} e^{5 t}

= \frac{6}{5} e^{- 5 t} - \frac{6}{5} e^{5 t}

\int 17 θ sec^{2} (θ) d θ

y = e^{c x}

\int_{0}^{4} (5 - x) x d x

x e^{y}

\int \frac{1}{x ^{2} + 4 x - 5} d x

What is inverse oflaplace-(12)/(s^2-5) ?
The answer to inverse oflaplace-(12)/(s^2-5) is 6/(sqrt(5))e^{-sqrt(5)t}-6/(sqrt(5))e^{sqrt(5)t}