What is inverse oflaplace ((s+5))/((s+3)(s+7)) ?

The answer to inverse oflaplace ((s+5))/((s+3)(s+7)) is 1/2 e^{-3t}+1/2 e^{-7t}

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inverse oflaplace ((s+5))/((s+3)(s+7))

inverse laplace

\frac{( s + 5 )}{( s + 3 ) ( s + 7 )}

\frac{1}{2} e^{- 3 t} + \frac{1}{2} e^{- 7 t}

Solution steps

L^{- 1} {\frac{( s + 5 )}{( s + 3 ) ( s + 7 )}}

Take the partial fraction of

\frac{s + 5}{( s + 3 ) ( s + 7 )} : \frac{1}{2 ( s + 3 )} + \frac{1}{2 ( s + 7 )}

= L^{- 1} {\frac{1}{2 ( s + 3 )} + \frac{1}{2 ( s + 7 )}}

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}{2 ( s + 3 )}} + L^{- 1} {\frac{1}{2 ( s + 7 )}}

L^{- 1} {\frac{1}{2 ( s + 3 )}} : \frac{1}{2} e^{- 3 t}

L^{- 1} {\frac{1}{2 ( s + 7 )}} : \frac{1}{2} e^{- 7 t}

= \frac{1}{2} e^{- 3 t} + \frac{1}{2} e^{- 7 t}

f (x) = 2 x - \frac{16}{x}

6 cos (x) + 3 sin^{2} (x)

\frac{d}{d x} (\frac{1}{2 x})

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

\frac{d}{d x} (x^{4} - 4 x^{2})

What is inverse oflaplace ((s+5))/((s+3)(s+7)) ?
The answer to inverse oflaplace ((s+5))/((s+3)(s+7)) is 1/2 e^{-3t}+1/2 e^{-7t}