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 \frac{4 x ^{2}}{( 1 - 8 x ^{3} ) ^{4}} d x

y^{^{'}} - \frac{y}{x} = 1

\int \frac{x}{a - x} d x

(x - 2 y)^{3} = 2 y^{2} - 3

\frac{d y}{d x}, x + y = 9

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)}