What is inverse oflaplace (e^{-4x})/((x-2)^2) ?

The answer to inverse oflaplace (e^{-4x})/((x-2)^2) is \H(t-4)e^{2(t-4)}(t-4)

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inverse oflaplace (e^{-4x})/((x-2)^2)

inverse laplace

\frac{e ^{- 4 x}}{( x - 2 ) ^{2}}

H (t - 4) e^{2 (t - 4)} (t - 4)

Solution steps

L^{- 1} {\frac{e ^{- 4 x}}{( x - 2 ) ^{2}}}

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 - 4) L^{- 1} {\frac{1}{( x - 2 ) ^{2}}} (t - 4)

L^{- 1} {\frac{1}{( x - 2 ) ^{2}}} : e^{2 t} t

= H (t - 4) e^{2 (t - 4)} (t - 4)

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

\int (10 x^{9} - \frac{9}{x ^{9}}) d x

x \to 1 lim (x^{4} e^{- x^{2}})

\frac{d}{d x} (- 2 x e^{- x})

\int 5 sin (\frac{t}{2}) - 4 cos (2 t) d t

What is inverse oflaplace (e^{-4x})/((x-2)^2) ?
The answer to inverse oflaplace (e^{-4x})/((x-2)^2) is \H(t-4)e^{2(t-4)}(t-4)