What is inverse oflaplace ((s+5))/((s+6)(s+10)) ?

The answer to inverse oflaplace ((s+5))/((s+6)(s+10)) is -1/4 e^{-6t}+5/4 e^{-10t}

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inverse oflaplace ((s+5))/((s+6)(s+10))

inverse laplace

\frac{( s + 5 )}{( s + 6 ) ( s + 10 )}

- \frac{1}{4} e^{- 6 t} + \frac{5}{4} e^{- 10 t}

Solution steps

L^{- 1} {\frac{( s + 5 )}{( s + 6 ) ( s + 10 )}}

Take the partial fraction of

\frac{s + 5}{( s + 6 ) ( s + 10 )} : - \frac{1}{4 ( s + 6 )} + \frac{5}{4 ( s + 10 )}

= L^{- 1} {- \frac{1}{4 ( s + 6 )} + \frac{5}{4 ( s + 10 )}}

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}{4 ( s + 6 )}} + L^{- 1} {\frac{5}{4 ( s + 10 )}}

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

L^{- 1} {\frac{5}{4 ( s + 10 )}} : \frac{5}{4} e^{- 10 t}

= - \frac{1}{4} e^{- 6 t} + \frac{5}{4} e^{- 10 t}

n = 0 \sum \infty \frac{n}{n ^{2} + 4}

\int_{0}^{1} 9

\int 4 x^{9} e^{- x^{5}} d x

y = 2 x, y = 5 x^{2}

\int (- 4 x) d x

What is inverse oflaplace ((s+5))/((s+6)(s+10)) ?
The answer to inverse oflaplace ((s+5))/((s+6)(s+10)) is -1/4 e^{-6t}+5/4 e^{-10t}