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}

1 - 4 x

y^{^{'}} = \frac{y}{x} + \frac{y ^{2}}{x ^{2}}

- \frac{1}{2} y^{^{'}} + y = 4

28 x (4 x^{2} + 9)^{\frac{5}{2}}

x lo g_{3} (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}