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

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

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

inverse laplace

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

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

Solution steps

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

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

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

\frac{\partial}{\partial y} (x^{\frac{1}{2}} + y^{\frac{1}{2}})

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

\frac{u - 2}{u ^{2} - 2 u + 1} d u = - \frac{1}{x} d x

f (x) = (9 x - 1)^{- \frac{1}{3}}, at x = 1

t, \frac{d y}{d t} = y^{3} - y^{2} - 12 y

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