What is inverse oflaplace ((e^{-2s}))/(s-9) ?

The answer to inverse oflaplace ((e^{-2s}))/(s-9) is \H(t-2)e^{9(t-2)}

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inverse oflaplace ((e^{-2s}))/(s-9)

inverse laplace

\frac{( e ^{- 2 s} )}{s - 9}

H (t - 2) e^{9 (t - 2)}

Solution steps

L^{- 1} {\frac{( e ^{- 2 s} )}{s - 9}}

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 - 9}} (t - 2)

L^{- 1} {\frac{1}{s - 9}} : e^{9 t}

= H (t - 2) e^{9 (t - 2)}

f (x) = cos (a^{4} + x^{4})

\int cos (x) (sin^{2} (x)) d x

\int (sin (t))^{2} d t

\frac{\partial}{\partial x} (e^{2 x} + e^{- 2 x})

x \to 0 lim (\frac{3 ^{x} - 1}{x})

What is inverse oflaplace ((e^{-2s}))/(s-9) ?
The answer to inverse oflaplace ((e^{-2s}))/(s-9) is \H(t-2)e^{9(t-2)}