What is inverse oflaplace 1/((s+6)(s-4)) ?

The answer to inverse oflaplace 1/((s+6)(s-4)) is -1/10 e^{-6t}+1/10 e^{4t}

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inverse oflaplace 1/((s+6)(s-4))

inverse laplace

\frac{1}{( s + 6 ) ( s - 4 )}

- \frac{1}{10} e^{- 6 t} + \frac{1}{10} e^{4 t}

Solution steps

L^{- 1} {\frac{1}{( s + 6 ) ( s - 4 )}}

Take the partial fraction of

\frac{1}{( s + 6 ) ( s - 4 )} : - \frac{1}{10 ( s + 6 )} + \frac{1}{10 ( s - 4 )}

= L^{- 1} {- \frac{1}{10 ( s + 6 )} + \frac{1}{10 ( s - 4 )}}

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}{10 ( s + 6 )}} + L^{- 1} {\frac{1}{10 ( s - 4 )}}

L^{- 1} {\frac{1}{10 ( s + 6 )}} : \frac{1}{10} e^{- 6 t}

L^{- 1} {\frac{1}{10 ( s - 4 )}} : \frac{1}{10} e^{4 t}

= - \frac{1}{10} e^{- 6 t} + \frac{1}{10} e^{4 t}

\frac{d}{d x} (4^{(8 x^{4} + c o s (x))^{3}})

\int 4 x^{3} - 3 x^{2} + 4 x - 3 d x

t^{2} - t \cdot y \cdot y^{^{'}} + y^{2} = 0

\frac{t ^{6}}{( t ^{3} + 1 ) ^{2}}

\int 6 tan^{5} (x) sec^{2} (x) d x

What is inverse oflaplace 1/((s+6)(s-4)) ?
The answer to inverse oflaplace 1/((s+6)(s-4)) is -1/10 e^{-6t}+1/10 e^{4t}