What is inverse oflaplace 1/((s-3)(s-9)) ?

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

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inverse oflaplace 1/((s-3)(s-9))

inverse laplace

\frac{1}{( s - 3 ) ( s - 9 )}

- \frac{1}{6} e^{3 t} + \frac{1}{6} e^{9 t}

Solution steps

L^{- 1} {\frac{1}{( s - 3 ) ( s - 9 )}}

Take the partial fraction of

\frac{1}{( s - 3 ) ( s - 9 )} : - \frac{1}{6 ( s - 3 )} + \frac{1}{6 ( s - 9 )}

= L^{- 1} {- \frac{1}{6 ( s - 3 )} + \frac{1}{6 ( s - 9 )}}

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 ( s - 3 )}} + L^{- 1} {\frac{1}{6 ( s - 9 )}}

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

L^{- 1} {\frac{1}{6 ( s - 9 )}} : \frac{1}{6} e^{9 t}

= - \frac{1}{6} e^{3 t} + \frac{1}{6} e^{9 t}

(x, y) \to (0, 0) lim (\frac{( sin ( 2 x ) - 2 x + y )}{x ^{3} + y})

\int ln (x) \cdot ln (x) d x

y^{^{''}} x = - 5 sin (8 x)

\frac{\partial}{\partial x} (x ln (z^{7} + x^{5}))

f (x) = (x^{2} + 4 x + 3)^{2}

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