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

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

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

inverse laplace

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

H (t - 2) \cdot 2 e^{t - 2} cos (t - 2)

Solution steps

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

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

= H (t - 2) \cdot 2 e^{t - 2} cos (t - 2)

y = x^{2} + 2, (- 4, 18)

y = 7 x, y = \frac{7}{4} x, y = 78 - x^{2}

\frac{\partial}{\partial t} (r \cdot cos (t))

ln (, at^{c})

\int - \frac{1}{50} d x

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