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

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

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

inverse laplace

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

H (t - 2) e^{- 3 (t - 2)}

Solution steps

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

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

= H (t - 2) e^{- 3 (t - 2)}

\frac{d}{d x} (x^{2} \cdot tan (x))

x = - 2, x = 2, y = 4 x, y = x^{2} - 5

t^{2}

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

\frac{s + 3}{s ^{2} + 6 s + 13}

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