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

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

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inverse oflaplace e^{-4(t-1)}

inverse laplace

e^{- 4 (t - 1)}

e^{t} δ (t - 4)

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

e^{- 4 (t - 1)} : a = 1, F (t) = e^{- 4 t}

= e^{t} L^{- 1} {e^{- 4 t}}

L^{- 1} {e^{- 4 t}} : δ (t - 4)

= e^{t} δ (t - 4)

y^{^{'''}} + 4 y^{^{'}} = cot (2 x)

f (x) = x^{2}, (0, - 4)

3 x^{3} + \frac{4}{x}

\frac{\partial}{\partial x} (x ln (y) - y^{2} 2^{x y})

y^{^{''}} + 9 y = - sin (3 x)

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