What is inverse oflaplace (e^{-4s})/(s^2+4s+3) ?

The answer to inverse oflaplace (e^{-4s})/(s^2+4s+3) is \H(t-4)(1/2 e^{-t+4}-1/2 e^{-3(t-4)})

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inverse oflaplace (e^{-4s})/(s^2+4s+3)

inverse laplace

\frac{e ^{- 4 s}}{s ^{2} + 4 s + 3}

H (t - 4) (\frac{1}{2} e^{- t + 4} - \frac{1}{2} e^{- 3 (t - 4)})

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

= H (t - 4) L^{- 1} {\frac{1}{s ^{2} + 4 s + 3}} (t - 4)

L^{- 1} {\frac{1}{s ^{2} + 4 s + 3}} : \frac{1}{2} e^{- t} - \frac{1}{2} e^{- 3 t}

= H (t - 4) (\frac{1}{2} e^{- (t - 4)} - \frac{1}{2} e^{- 3 (t - 4)})

= H (t - 4) (\frac{1}{2} e^{- t + 4} - \frac{1}{2} e^{- 3 (t - 4)})

- 2 x + 3

y^{^{'}} = \frac{x - y}{2}

y = 17 - x^{2}, y = - 2 x + 14

\int \frac{1}{x ^{2} + 2 x - 15} d x

4 y^{^{''}} - 4 y^{^{'}} - 3 y = cos (2 t)

What is inverse oflaplace (e^{-4s})/(s^2+4s+3) ?
The answer to inverse oflaplace (e^{-4s})/(s^2+4s+3) is \H(t-4)(1/2 e^{-t+4}-1/2 e^{-3(t-4)})