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

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

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

inverse laplace

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

H (t - 4) \cdot 3 e^{- 2 (t - 4)}

Solution steps

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

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{3}{s + 2}} (t - 4)

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

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

\int \frac{x ^{2} - 25}{x ^{2}} d x

\int 3 + (x^{2} + 2) e^{x^{3} + 6 x} d x

\int \frac{8}{6 + 2 x} d x

\frac{\partial}{\partial x} (x e^{- t} sin (θ))

x \to 1 lim (\frac{\frac{2}{x} - 2}{x - 1})

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