What is inverse oflaplace 1/((s-2)(s-2)) ?

The answer to inverse oflaplace 1/((s-2)(s-2)) is e^{2t}t

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inverse oflaplace 1/((s-2)(s-2))

inverse laplace

\frac{1}{( s - 2 ) ( s - 2 )}

e^{2 t} t

Solution steps

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

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s - 2 ) ( s - 2 )} : a = 2, F (s) = \frac{1}{ss}

= e^{2 t} L^{- 1} {\frac{1}{ss}}

L^{- 1} {\frac{1}{ss}} : t

= e^{2 t} t

(x^{2} + x) e^{x}

x \to 1 + lim (\frac{1}{1 - x})

8 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

\int \frac{e ^{2 y} - y}{e ^{y}} d y

\int 9.8 \cdot e^{- 0.4 x} d x

What is inverse oflaplace 1/((s-2)(s-2)) ?
The answer to inverse oflaplace 1/((s-2)(s-2)) is e^{2t}t