What is inverse oflaplace x/((x^2-9)*x) ?

The answer to inverse oflaplace x/((x^2-9)*x) is -1/6 e^{-3t}+1/6 e^{3t}

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inverse oflaplace x/((x^2-9)*x)

inverse laplace

\frac{x}{( x ^{2} - 9 ) \cdot x}

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

Solution steps

L^{- 1} {\frac{x}{x ( x ^{2} - 9 )}}

Take the partial fraction of

\frac{x}{( x ^{2} - 9 ) x} : - \frac{1}{6 ( x + 3 )} + \frac{1}{6 ( x - 3 )}

= L^{- 1} {- \frac{1}{6 ( x + 3 )} + \frac{1}{6 ( x - 3 )}}

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}{6 ( x + 3 )}} + L^{- 1} {\frac{1}{6 ( x - 3 )}}

L^{- 1} {\frac{1}{6 ( x + 3 )}} : \frac{1}{6} e^{- 3 t}

L^{- 1} {\frac{1}{6 ( x - 3 )}} : \frac{1}{6} e^{3 t}

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

\frac{d}{d x} (\frac{sin ( x )}{2})

\frac{d ^{2} x}{d t ^{2}} + k^{2} x = 0

x \to 0 lim (\frac{1}{cos ( x )})

x \to 4 lim (\frac{5}{( x - 4 ) ^{2}})

\frac{d}{d x} (2 cos (x) - 3 ln (x) - 1)

What is inverse oflaplace x/((x^2-9)*x) ?
The answer to inverse oflaplace x/((x^2-9)*x) is -1/6 e^{-3t}+1/6 e^{3t}