What is inverse oflaplace 4/((s-2)^2) ?

The answer to inverse oflaplace 4/((s-2)^2) is e^{2t}*4t

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inverse oflaplace 4/((s-2)^2)

inverse laplace

\frac{4}{( s - 2 ) ^{2}}

e^{2 t} \cdot 4 t

Solution steps

L^{- 1} {\frac{4}{( 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{4}{( s - 2 ) ^{2}} : a = 2, F (s) = \frac{4}{s ^{2}}

= e^{2 t} L^{- 1} {\frac{4}{s ^{2}}}

L^{- 1} {\frac{4}{s ^{2}}} : 4 t

= e^{2 t} \cdot 4 t

\frac{d}{d x} (\frac{e ^{2 x}}{x + 1})

\int 2 (x + 1) d x

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = - 1, y^{^{'}} (0) = - 3

\int sin (x) cos (n x) d x

\frac{d y}{d x} y = 8 x

What is inverse oflaplace 4/((s-2)^2) ?
The answer to inverse oflaplace 4/((s-2)^2) is e^{2t}*4t