What is inverse oflaplace (s+5)/(s^2+4s+3) ?

The answer to inverse oflaplace (s+5)/(s^2+4s+3) is 2e^{-t}-e^{-3t}

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inverse oflaplace (s+5)/(s^2+4s+3)

inverse laplace

\frac{s + 5}{s ^{2} + 4 s + 3}

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

Solution steps

L^{- 1} {\frac{s + 5}{s ^{2} + 4 s + 3}}

Take the partial fraction of

\frac{s + 5}{s ^{2} + 4 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} (e^{x} + c)

f (x) = x^{2} + x + 1

y^{^{'''}} - 6 y^{^{''}} + 8 y^{^{'}} = 0

\int \frac{1}{p - p ^{2}} d p

n = 0 \sum \infty 2^{- n} + 3^{- n}

What is inverse oflaplace (s+5)/(s^2+4s+3) ?
The answer to inverse oflaplace (s+5)/(s^2+4s+3) is 2e^{-t}-e^{-3t}