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

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

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

inverse laplace

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

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

Solution steps

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

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

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

\int x - csc (x) \cdot co tan (x)

\int arcsin (1) d x

f (x) = - x^{3}

\frac{d}{d x} (\frac{- 1}{2 x ^{\frac{3}{2}}})

\frac{d}{d x} (sec (x^{3}))

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