What is inverse oflaplace e^{-s} 1/(s-4) ?

The answer to inverse oflaplace e^{-s} 1/(s-4) is \H(t-1)e^{4(t-1)}

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inverse oflaplace e^{-s} 1/(s-4)

inverse laplace

e^{- s} \frac{1}{s - 4}

H (t - 1) e^{4 (t - 1)}

Solution steps

L^{- 1} {e^{- s} \frac{1}{s - 4}}

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 - 1) L^{- 1} {\frac{1}{s - 4}} (t - 1)

L^{- 1} {\frac{1}{s - 4}} : e^{4 t}

= H (t - 1) e^{4 (t - 1)}

x \to \infty + lim (\frac{2 x + 1}{4 x ^{2} + x})

\frac{y}{x} \cdot \frac{d y}{d x} = - 4 - \frac{y ^{2}}{x ^{2}}

x \to 3 lim (\frac{1}{x})

x y^{^{'}} - y = \frac{x ^{2}}{y} e^{\frac{y}{x}}

y = x^{3} - 2 x + 2, (3, 23)

What is inverse oflaplace e^{-s} 1/(s-4) ?
The answer to inverse oflaplace e^{-s} 1/(s-4) is \H(t-1)e^{4(t-1)}