What is inverse oflaplace (s+5)/((s+1)(s+3)) ?

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

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inverse oflaplace (s+5)/((s+1)(s+3))

inverse laplace

\frac{s + 5}{( s + 1 ) ( s + 3 )}

2 e^{- t} - e^{- 3 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= 2 e^{- t} - e^{- 3 t}

\frac{d}{d x} (\frac{12}{x})

ln (3) - ln (x^{3} - 4 x + 2)

\int_{0}^{4} \frac{1}{x ^{2} - 16} d x

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

6 x, 8 x^{2}, (0, 0.75)

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