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

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

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

inverse laplace

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

e^{t} \cdot 2 t

Solution steps

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

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

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

= e^{t} \cdot 2 t

2 cos (5 t)

(s + 2)^{2}

y = 5 + 5 x^{2} - 2 x^{3}

\frac{\partial}{\partial x} (x^{2} + y^{2} + 2 z^{2})

\frac{\partial}{\partial x} (2 x y + x)

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