What is inverse oflaplace (-2e^{-2s})/((s^2-2s+1)) ?

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

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inverse oflaplace (-2e^{-2s})/((s^2-2s+1))

inverse laplace

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

- 2 e^{t - 2} H (t - 2) (t - 2)

Solution steps

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

Factor

s^{2} - 2 s + 1 : (s - 1)^{2}

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

Apply inverse transform rule:

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

then

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

Where

H (t)

is Heaviside step function

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

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

= H (t - 2) (- 2 e^{t - 2} (t - 2))

= - 2 e^{t - 2} H (t - 2) (t - 2)

\frac{1}{s ^{3} ( s + 4 )}

\frac{d}{d x} (10 x^{10} e^{x})

\int \frac{1}{- t ^{2} + 8 t - 15} d t

x + e^{x} - \frac{1}{x}

\int \frac{( 4 x - 5 )}{( x - 2 ) ( x + 3 )} d x

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