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

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

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

inverse laplace

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

e^{2 t} t

Solution steps

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

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

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

= e^{2 t} t

\frac{d}{d x} (1 - ∣ x - 1 ∣)

\int_{- 1}^{1} (4 - 4 x^{2})^{2} d x

\frac{\partial}{\partial x} (tan (\frac{u}{v}))

\frac{d}{d x} ((x + y)^{2})

\int_{0}^{1} 1 + 9 x^{2} d x

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