What is inverse oflaplace 1/(s^2-4s+4) ?

The answer to inverse oflaplace 1/(s^2-4s+4) is e^{2t}t

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inverse oflaplace 1/(s^2-4s+4)

inverse laplace

\frac{1}{s ^{2} - 4 s + 4}

e^{2 t} t

Solution steps

L^{- 1} {\frac{1}{s ^{2} - 4 s + 4}}

\frac{1}{s ^{2} - 4 s + 4} = \frac{1}{( s - 2 ) ^{2}}

= 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} (\frac{2}{x y})

\int_{0}^{\frac{π}{4}} 2 sec (x) d x

f^{^{'}} (x) = x^{2} e^{x^{2}}

\frac{3 x}{x + 1}

2 x^{2} - \frac{1}{16} ln (x)

What is inverse oflaplace 1/(s^2-4s+4) ?
The answer to inverse oflaplace 1/(s^2-4s+4) is e^{2t}t