What is inverse oflaplace e^{-2s}*1/(s-3) ?

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

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inverse oflaplace e^{-2s}*1/(s-3)

inverse laplace

e^{- 2 s} \cdot \frac{1}{s - 3}

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

Solution steps

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

\int \frac{1}{( x ^{2} - 25 ) ^{\frac{1}{2}}} d x

f (x) = - x^{3}, at x = 2

x \to 0 lim (2 x e^{2 x} + k x)

y^{^{'}} = 3 x + y, y (0) = 6

y = ln (e^{- x} + x e^{- x})

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