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

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

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

inverse laplace

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

e^{2 t} \cdot 2 t

Solution steps

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

Apply inverse transform rule:

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

then

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

For

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

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

L^{- 1} {\frac{2}{s ^{2}}} : 2 t

= e^{2 t} \cdot 2 t

y^{3} - (10 x + 8) + 3 x y^{2} y^{^{'}} = 0

\frac{x - 4}{x + 4}

y = \frac{1}{3 x}

\frac{d}{d x} (sin (x^{3} - x))

2 y = x, x + 1 = (y - 1)^{2}

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